Solving a crazy iterated floor equation.

[@blackpenredpen]

Man! This is super crazy!! @20:47, I will try to see if I can solve any of that. If I can, I will be super hyped!

[@bhavya4917]

Finally got to see that bprp also watches this channel😝😝

[@friedrichotto5675]

The second: Since you know that floor(a)≤a, 2020=x*floor(x*floor(x*floor(x)))≤x*floor(x*floor(x^2))≤x*floor(x^3)≤x^4 2020^(1/4)≤x_1 and -2020^(1/4)≥x_2

[@blackpenredpen]

Hydra17 I do. And I do have to say that I can’t solve all the problems that professor Penn puts up.

[@emersonchaves567]

What is this, a crossover episode? Nice

[@cristofer2794]

Use the taylor series formula aparecida in the wiki with the summatory. Then where says x put the same formula inside three times multiply for x. Then put=2020 and solve the equation. Lie its a joke, is impossible.

[@DerBossLUEKU]

Nobody: Michael Penn: okay great

[@gabriel_talih]

He reminds me of chael sonnen from the ufc

[@lynxfl]

First Time a 'nobody' meme was funny

F A N T A S T I C

[@lgooch]

@@lynxfl false

[@bsharpmajorscale]

"Comment what the future is like" It sucks, don't come here. :P

[@MichaelPennMath]

I was hoping to get a comment from 2022, it hasn't happened yet....

[@snowfloofcathug]

Michael Penn well, twenty twenty-two is really just twenty twenty, too

[@Juhamakiviita2.0]

@@snowfloofcathug i C what u did there

[@websnarf]

@@MichaelPennMath -2022/305 --> 2022 if that's what you will be looking for. :)

[@osere6432]

2022 electric boogaloo

[@Davidamp]

Bold to assume there will be a future

[@joryjones6808]

Hi from the Near Future.

[@DelandaBaudLacanian]

Future reporting in, we all gone

[@sawyerwest3990]

I am from the future Michael from the past but now this comment is entering the past too.

[@jonastechmanski1592]

Doing strong

[@mjcat5000]

Still doing strong even with Heatwave Zoe in July 2022. 😎

[@cycklist]

What a great puzzle! It would be great to see more of these.

[@jroemling]

Wow, this was a random recommendation that was super-interesting with a good production quality and pleasant to listen to. And you have been churning those out many times a week for 9 months and gotten like 70 views per video, but you kept going and now lately some of them blew up a bit. Thanks for sticking to it, I will definitely subscribe to this one!

[@aditya95sriram]

I used to think that these high-level math videos rarely get enough appreciation from the general public and thus fly under the radar when it comes to being recommended. But I'm glad to have been proven wrong to some extent. I also discovered this channel because of a random recommendation.

[@friedrichotto5675]

At the example: You know 3

[@MichaelPennMath]

The warm up is easy to guess, but sometimes it is helpful to overwork an easier problem to provide insight into the harder problem..

[@Danicker]

Yes I thought that too but then I realised Michael's method was more useful for problems involving multiple floor operators

[@BlueRaja]

@@Danicker Another commentor extended this method to work with multiple floor operators, without requiring a computer: kzbin.info/www/bejne/hamnmYFojbeGsNU&lc=Ugz1zCW_bv9sX9G4Fit4AaABAg

[@the_lava_wielder6996]

@@Danicker not if you recalculate the range and guess and check

[@josir1994]

This "solving" just doesn't feel good. I worked on it by substituting each floor function by a subtraction of a small number between 0 and 1, so the thing becomes (n+a){(n+a)[(n+a)n-b]-c}=2020, where 0

[@RobsMiscellania]

This addresses every one of the questions at the end of the video. Very well done. This method also illustrates the floor function (as well as the ceiling function) in a different way than presented here. Again, well done.

[@josir1994]

@Harsh Goel if you work out the whole equation of 1029a+...=381, the ... contains only terms with smaller coefficients. Sorry not have time to rework the thing but the second largest coefficient should be somewhat smaller than 100. Since 0

[@josir1994]

@Harsh Goel or in other words, we are ignoring those with smaller coefficients, not necessary because of a, b or c

[@irrelevant_noob]

@Harsh Goel the thing is, expanding those terms gets you to 381 = 1029a +49b -7c - 49a^2 -7ab - 49a^2 - 7ab - 49a^2 + 7a^3 + a^2b + ac, so it still leaves you with 1029a between 323 and 549, so a has to be between 0.313 and 0.534, so you only need to check 302 to 312.

[@saivivekpeta1696]

You don't even need to do that, just divide ewuate 1029a=381 and we'll get 0.37 and then it becomes -6.63, after that divide 2020÷-6.63, we'll roughly get 304, test one or two numbers ahead and before that and we'll get the answer. It's surprsing that this trivial method is more accurate than that.

[@TheOneThreeSeven]

I am totally floored with how cool this is!

[@tracyh5751]

nice

[@jishnubala7296]

FLOOR GANG!

[@roberttelarket4934]

TheOneThreeSeven:Very very very clever wording!!!

[@laszloliptak611]

A small correction at 18:44: Strict inequality means that you can't have equality, and you want to say the opposite: taking the ceiling (or floor) will now make the inequality not strict.

[@mushroomsteve]

Also at 22:25, he must have meant -2020/289.

[@mushroomsteve]

I also think the next step should say -337

[@Nukestarmaster]

He just kept doing this, it is fairly aggravating.

[@happyhappyguy5034]

@@Nukestarmaster I think it is super clear.

[@happyhappyguy5034]

@@mushroomsteve True, he is correct actually

[@samuelmeyer1884]

Question 2: For the negative values: floor(x) = -7 => x*floor(x*floor(-7x)) = 2020 floor(x) = x-1 => floor(-7x) x*floor(-7x) >= -7x^2 =>floor(x*floor(-7x))>=x*floor(-7x)-1 >=-7x^2-1 =>x*floor(x*floor(-7x))) =2020 x = -7x-1 x*floor(-7x)

[@niekteeuwen13]

I think I figured a way to tighten the bounds on x. When we fill in x=∜(2020) into the equation we will get an answer smaller than 2020, since floor(∜(2020)) < ∜(2020). This means ∜(2020) is a lower bound for the positive solution, so ∜(2020) < x < 7. In a similar way we can argue that the bounds for the negative solutions can be tightened to -∜(2020) < x < -6.

[@AlephThree]

This is brilliant. I started off by observing that 6

[@fplancke3336]

I did almost the same, except my enlightenment came at the beginning of the video, from reading the question as written on the board: "solve [this equation] for x" which seemed to imply that there would actually be a solution. (As opposed to the "can we solve" question on the thumbnail).

[@williamsean3469]

The left alternating floor ceiling has a solution 7 - 17/291, the right alternating floor ceiling has a solution 7 - 23/61. And wrapping another gives a range of solutions from -2021/305 to -2020/305 including -2020/305

[@bhanusri3732]

Range of solutions are -2020/m where 304.85

[@dzspdref]

5:19 : reciprocating the inequality should have been written 1/4 < m/10 < 1/3 ; oopsed on the 1/3 by writing 3.

[@irrelevant_noob]

For bonus points, find the other 5 oopses. ;-)

[@44hwxyz90]

Nice video! Just a little remark: at 12:05 m should technically be greater than or equal to floor(2020/7)+1 and less than or equal to ceil(2020/6)-1. In this case it doesn't matter but if, for example, instead of 2020 it were 2022, which is a multible of 6, then m being greater than 2022/6=337 doesn't imply that m is greater than or equal to ceiling(2022/6)=337.

[@newkid9807]

Hwxyz Wrong

[@luisbenites4825]

The search is reduced to 6 cases if you express x = -6.7 + d, where 0 < d < 0.1. Plus you only need to check 3 of those cases this way. Like you did, work your way out, the third floor operation gives you that range of 6 integers to explore (remember d0.1 so no checks needed. Of the other three, two of them don’t work: There’s no need to check the whole thing, just that the inner [-308.2 + 46d] doesn’t match the integer that produced that “d” Quite happy about this!

[@Flammewar]

Guys, I‘m from 2025. It’s crazy to see an old video of Michael before he proved the Riemann Hypothesis.

[@lostwizard]

Probably be a good idea not to say "two twenty" for "2020" since "two twenty" would be 220 instead. Just saying.

[@JonathonV]

William Astle Yup, that began to really bug me after a while too. Ah well, interesting problem, anyway.

[@NStripleseven]

Yeah, that was a little weird. We all knew what he meant, though, so I guess it worked.

[@JM-us3fr]

Or say "6" when pointing to 1/6. Or write "3" when it should have been a 1/3. Or say "strict-inequality" when pointing to a weak inequality (twice!).

[@d_mcg]

Reminds me of JibJab.com: "2-0-5" 2005 Year In Review kzbin.info/www/bejne/aYXLeaWnoayXic0

[@G.Aaron.Fisher]

His abbreviated form kind of grew on me. I assume 220 would be "two hundred twenty" and 1,032,800 would be "one thirty-two eight hundred."

[@hymanimy]

You present such interesting questions and have great production value. Keep it up 👍

[@bartoszpanek8606]

In 12:25 you probably want a write ceil(2020/7) = 289, not 2020/6.

[@spaanse]

For the fourth extra question: 2020/291 for the first equation and 2020/305 for the second equation. No other solutions exist. I will highlight my approach below: assume x in (1,inf) then floor(x) in (x-1,x] isct [1,inf) then x floor(x) in (x^2-x,x^2] isct [x,inf) = (x^2-x,x^2] isct (1,inf) then ceil(x floor(x)) in (x^2-1,x^2+1) isct [2,inf) ... We use these ranges to put bounds on x for which x does x^4-x^3-x < 2020 < x^4+x^2 hold? As I don't know how to solve 4-th order polynomials I use Wolfram Alpha to find that approximately 6.66687 < x < 6.97453 we convert these bounds on x to bounds on m and check the functions on all possible 2020/m A similar argument for x < 1. For x in [-1,1] such floor/ceiling functions remain in [-1,1] and therefore can't be a solution.

[@ivanklimov7078]

Great video, but why the long explanation for finding m at the 4:58 minute mark? If m is floor(x), and x is between 3 and 4 (not including 4), isn't that enough to confidently state that m=3?

[@TheDGomezzi]

On the simple problem, it’s easy to just know it. On the full problem, understanding his longer method is very helpful.

[@goseigentwitch3105]

The trick for me was realizing that the answer must be some integer of the form x = 2020/k for some integer k also for 6 < x < 7, floor(x floor( x)) is at most 41, so floor(x floor(x floor(x))) is at most 286, so our original equation is at most 2001 therefore -7 < x < -6 some small calculator works shows -6.7 < x < -6.6, so m can only be (negative) 304, 305, 306, or 307 (this just shrinks your exhaustive search, so i didn't need anything beyond a calculator to solve it in a few minutes) plug into calculator looking for your starting number to show up as your answer to floor(x floor(x floor(x))) voila, m is -305, so x = 2020/-305 check the answer, and it checks out :)

[@mskiptr]

Instead of plugging these guesses into the calculator one after another you can perform binary search and find the solution exponentially faster (since λx. x⌊x⌊x⌊x⌋⌋⌋ is monotonic for x≥0 (or x≤0)) - if you check that x from the middle of this sequence is too big|small you can discard a half of them at once.

[@Oliver-nb2iu]

same. i also realized that the solution must be of form 2020/k. I did not expect though that there had to be trial and error search

[@nikolamarijanovic6261]

How did you get 6.7 and 6.6 in "some small calculator works shows -6.7 < x < -6.6"

[@goseigentwitch3105]

@@nikolamarijanovic6261 Well, I knew it had to be between 6 and 7 so I just plugged in 6.5 to the equation. I t was too small. I plugged in 6.6. It was too small. I plugged in 6.7. It was too big. The other response to my method has a more mathematical method. Essentially, take your whole list of possible numbers {289, ... , 336} or whatever it was. Take the number in the very center (308) try it (in the form 2020/308). If it is too large or too small, you can discard all numbers larger or smaller than it respectively. This way you basically try half of the remaining numbers each test.

[@yurenchu]

@@mskiptr , That binary search approach actually leads to a never-ending iteration; so no, it's not faster.

[@disgruntledtoons]

From the fact that x is multiplied by an integer to yield 2020, x must be a rational number of the form 2020/y. The absolute value of x falls between the floor and ceiling of the positive real fourth root of 2020, which restricts x to two ranges, either -7 to -6 or +6 to +7, which means that y is either from -336 to -289 or from 289 to 336. Using divide-and-conquer reveals that y = -305, giving a reduced value of -404/61 for x.

[@konstantinosalexiou6797]

This is amazing!! I hope for more videos like this.

[@Czeckie]

if this is from a competition, there should be a way how to solve the equation without using the computer or checking the possible solutions (even though you don't need to check them all, since the function is monotone, so you can find it by binary search and that might be manageable)

[@luisbenites4825]

Check my comment above, a simpler way where only 3 solutions need to be checked

[@ordanguric461]

Dude, your videos are just amazing, I learn from them a lot, thank you!!!

[@pianoclassico718]

i bet someone already solved this but i found that n needs to be less or equal to 10 so that the equation has positive solutions ( if x is less than 1 then it's trivial and there are no solutions , if x is 1 then there are no solutions either , if x is bigger than 1 the only solution starts at x =2 because if x=1.9 then x[x] =x as the definition of the floor function , we find that the first possible solution is for x=2 , and we notice that for n>10.98 the equation has no solutions because then root of 2020 becomes smaller than 2 , thus since n is an integer n=10 is the maximal n for which there are positive solutions , I will try to do the negative ones because they don't work by simple analogy and need some writing haha , great video , loving them !

[@ElchiKing]

Yes, I just wrote the same thing and then found your comment :D

[@SwapnilAnand1998]

When this guy starts to present solutions, there are moments when I actually pause the video and clap for the ingenious mathematical tools this guy provides us with. 🙌🙌🙌👏👏👏

[@paler1231]

As far as tightening bonds goes, you've used floor( n ^ (1 / no_of_iterations)) < x < ceil(n ^ (1 / no_of_iterations)), but clearly if |x| < n ^ ( 1 / no_of_iterations), then this floor multiplication will yield results less than n.

[@the_lava_wielder6996]

If -7

[@daniloguimaraes4619]

This is exactly what I needed for quarantine

[@cubicinfinity]

When I solved this I worked from the inside out from the beginning. You get the same kind of thing. I concluded there was no solution, but I forgot to consider negative values.

[@saiforos7928]

With everything floored you get at least extra solutions that slightly increase your value of x, since the multiplication of all of it will increase it by less than one. The goal is to be so small no intermediate floor increases (in absolute values due to sign). You can tell that gives you a half open/closed line with as least value your solution (below your value the original product is less than 2020 so the floor is 2019) Also happy to have this popped up in my recommendations, subscribed.

[@Avighna]

@3:51 but the floor function is increasing for negative values too. As you said, it behaves like the square function.

[@schrodingerbracat2927]

there's probably a more efficient way to search through all the possibilieis e.g. guess halfway, depending on whether it's and over- or under-estimate, you guess halfway of the appropriate half. something like a binary search algorithm.

[@ElchiKing]

Ok, some boundary on the 3rd question at 20:50 As soon as the number of floors exceeds log_2(2020)=10.9... (i.e. the number of floors is at least 11), we can't have any positive solution. Proof: If x is a solution, then x=2^11=2048>2020. But then floor(x)=1, so x floor(x)=x, hence we get floor(x floor(...)))=1. Negative values for x seem to be more subtle, since the rounding of the absolute value changes between floor and ceiling.

[@ElchiKing]

If the number of xs appearing on the lhs is greater than 10 and odd, there is no solution (since there can't be a positive solution, and each negative number would result in a negative number) Edit: Using your method, we can prove that there is also an upper bound on the number of floors for which the equation can be solved: When there are at least 11 floors, the absolute value of x must be smaller than 2. But then from x=2020/m, we get -1/2>m/2020>-1, so -1010>m>-2020. Hence, we only get finitely many values for m which can be a solution for any problem involving at least 11 floors. However, if x is between -1 and -2 floor(x*n)n. Hence, any x can only be the solution for at most one problem with rhs 2020. So, there are at most 1010+11=1021 possible numbers of floors, for which the problem x*floor....=2020 is solvable.

[@bollyfan1330]

For the positive solution there is a very easy way to prove it is impossible. You know that 6 < x < 7. We also know that LHS is a non decreasing function of x. Just assume x = 6.999999....., which is the largest possible value still less than 7 LHS = x * floor(x * floor(x * floor(x))) LHS = x * floor(x * floor(x * 6)) LHS = x * floor(x * (7 * 6 - 1)) LHS = x * (7 * (7 * 6 - 1) - 1) LHS = x * (7 * 41 - 1) LHS = x * 286 LHS < 7 * 286 LHS < 2002 Therefore LHS cannot ever equal 2020 and hence there are no positive solutions. For the tighter bound, you had asked in followup questions: Solving for x^4 = 2020 gives x = 2020 ^ 0.25 = 6.704059..... Hence if you use floor, the x value will be 6.704059...

[@easymathematik]

Hello, your second question and the end is in my opinion a very interesting and exciting question. "Can we tighten the bounds of the intervalls?" I haven´t prepared anything, but I can imagine, that following steps could lead to a solution. First we saw: The given equation has no solution for positive x because k = 2020 is bigger than the maximum possible value as shown, let´s say M. One could think about a solveble range of values M for this intervall. The goal is: Find a maximum epsilon e > 0, s. t. the existent solution x is in the open intervall (a + e ; b - e). And my feeling tells me, that possibly the relation between the range of possible M and maximum epsilon is something involving squareroot and the difference of M and K.

[@xCorvus7x]

At 19:11, shouldn't the bounds have the same absolute value as for the case that the solution is positive, 336 and 289 respectively? Taking the elevator down in the positive case should mirror taking the elevator up in the negative one, shouldn't it? Edit: Also, around 19:43 you incidentally go further negative from -2020/288, the next element in the set should be -2020/289 . But whatever, just a slip-up.

[@abushahid1150]

2009 is also far fetched. Extending you logic for finding (x(x)) I found that maximum it can take is 2002. Still. Loved it.

[@yurenchu]

Actually, it cannot reach 2002; 2002 is the _non-inclusive_ upper boundary of what the expression can reach when 6 < x < 7 .

[@TheDGomezzi]

He was just using a simple bound, it wasn’t a tight bound, it just had to be tight enough to show that 2020 was out of reach.

[@GauravKumar-bf2rq]

I love this channel. Thanks for the content!

[@kenhaley4]

At 5:00, note: We already know that m is floor(x) where x is between 3 and 4. The floor of any number between 3 and 4 (not inclusive) is 3. Done! We don't need all that other stuff up to 5:45 to get to m = 3.

[@jackhanke343]

This is a great channel, wish I found it sooner!

[@cubechessmanasmr3331]

What chalk and board are you using man, i want that clear look

[@LogosNigrum]

You are one of the world's greatest gifts.

[@AcaciaAvenue]

I tried to replace x with 6+a for positive x and with -7+a for negative x, with 0

when you know the interval of integers, I would do a binary search if I had to do it on paper.

[@ZedaZ80]

But only if the output is monotonic! It is in this case

[@markryback3797]

Well, it seams, that binary search require less math. 289 and 336 average is 312, so check if 2020/312 is too small or too big.

[@gilber78]

I missed the reasoning behind the transformation at about 12:50. What theorem or part of the vid says we can do that…?

[@gilber78]

Nvm I figured it out since m has to be an integer my b

[@xCorvus7x]

The floor function over the negative integers mirrors the ceiling function over the positive ones. So the equation x*ceiling(x*ceiling(x*ceiling(x))) = 2020 has a positive but no negative solution and it has the same absolute value as the solution here.

[@bekhaddaderrar2111]

Thank you professor for this video 👌👌👌👌👍👍👍

[@andmo90]

2020 is almost over, that's roughly all there is left to be grateful for.

[@KarampaEnglishSchool]

Setting constants as something relatable such as calendar dates makes math more relatable somehow.

[@mariobrito427]

Awesome problem! I got super hyped with this.

[@rishiraje]

In the first equation, how did you select 10 as the value. If you selected a diff value, say 100 what would happen?

[@mirkoschultz9347]

Regarding the third question: I quickly wrote a little Python script that gives me five solutions for n. I only checked for n

[@MichaelPennMath]

I don't know the answer to the questions I posed! Part of me wants to play around with it, but I also want to be surprised when someone comes up solutions.

[@mirkoschultz9347]

@@MichaelPennMath For you and for those who try this question here are my five results for (n,m): (1,-45) (3,-305) (8,815) (9,-979) (49,-1750) m

[@MrKytametam]

@@mirkoschultz9347 There is one more solution: (795, -2009). I checked all cases and there are no more.

[@MrKytametam]

​@@mirkoschultz9347 My mistake is one more :(2689, -2018)

[@050138]

@@MrKytametam Holy smookes! A chain of 2689 floor iterations to reach 2020 and the solution is -2020/2018 😲😯😮

[@050138]

Woohoo! What a beautiful problem! ❤️ I'm so excited I got this answer -404/61, took me more than half an hour, using the same approach and didn't use a computer (though used calculator for few checks).... I was almost about to give up at the end of 30 mins as I found no solutions, but then suddenly it struck me that there could be negative solutions.... I saw at the start that x had to lie between 6 and 7 and tried to reduce the bounds of the interval as you suggested at the end.... I quickly saw that the expression in question f(x) kept increasing with x.... But due to the discontinuous nature of function, there was a jump near 7.... If we approach 7 from the right we have 7*7*7*7, but if we approach from left, x = 6.999999........ x times fl.x = 41.999999....... x times fl of fl of x = 286.999999..... So final expression is 2001.999999..... = 2002 < 2020, so no positive solutions But for negative solutions, I started with x as fourth root of 2020 = -6.7041 and kept increasing x by 0.01 till x reached -6.2, then all the inside terms come to 305, making x = -2020/305 The thing to note as that for the second floor operation inside, the number is positive so smaller number should be taken for floor value.... ☺️

[@JamesLewis2]

I tried a condition that was a bit too strong (x itself is an integer multiple of a certain integer n, as in kn, plus an integer multiple of 1/n, as in j/n, where 0

[@moshadj]

The future was pretty brutal actually

[@iamadooddood4331]

For x > 0, 6 < x < 7 is too wide a range. We know that for any non-integer x, x^4 > ⌊x ⌊x ⌊x⌋⌋⌋, so 2020^.25 < x < 7 is a better range, then we have ⌈2020/7⌉ ≤ m ≤ ⌊2020^.75⌋, or 289 ≤ m ≤ 301. Similarly, for x < 0, we have -(2020^.25) < x < -6 as a better range, and ⌈-2020^.75⌉ ≤ m ≤ ⌊-2020/7⌋, or -301 ≤ m ≤ -289.

[@nosarcasm1]

Interesting problem. In the simpler solution, after finding the boundaries for x isn't it directly clear that integer of x has to be three without the way around m? I know, you need it for the other problem but you should mention more?

[@G.Aaron.Fisher]

Subscribed because I can tell good things are going to come from this channel. :D

[@phasm42]

The graph of x*floor(x) in Desmos is pretty interesting.

[@OMGclueless]

Wow, it's even cooler than I expected. Both halves (x < 0, x > 1) are discontinuous, strictly increasing, and piecewise linear which is obvious, but I didn't realize before looking that it looks like the domain of the function is the whole of y ≥ 0. And it looks like the function is even invertible except for at y = 0.

[@ExtraterrestrialIntelligence]

Acually doing numbers with theory is kinda stupid but a nice way to test someone anyways thanks for makeing things like this available for the public.

[@SlamminGraham]

If the Floor[] function were defined to return values which were the nearest integer to the operand always in the direction of zero, then the negative solutions would work.

[@ZedaZ80]

(positive no-solutions) Take x=7-a for a on (0,1) b = x*floor(x) = x*6 = 42-6a, so floor(b) is at most 41 c = x*floor(b) = x*41 = 287-41a, so floor(c) is at most 286. d = x*floor(c) = x*286 = 2002-286a. So for tight bounds, 1296

[@matron9936]

Great video, but i‘m watching it in the past so I can’t comment how the future is.

[@ДаниилРабинович-б9п]

12:29 that should be 2020/7, not 2020/6 19:00 should be -336≤m≤-228, not -336

[@lforlight]

5:00 that's weird. If you already said 3

[@Sachin27071998]

this was so that for the cases ahead where m = FL(FL(FL(x))) you could use the same logic to get m.

[@yujie.ho123]

9:04 Is -10/4 not a solution?

[@alvoi4379]

The floor of -10/4 is -3 :)

[@irrelevant_noob]

You know, for a teacher you make quite a LOT of slip-ups... :-s 4:44 you've just assigned the SAME mark to two different statements. -.- 5:22 that should be 1/3. 12:26 that should be ceiling of 2020/7. 18:41 those are NON-strict inequalities... 18:51 and in fact, the floors and ceilings of those negative numbers should be -336 and -289. Also the inequalities shouldn't be strict here. 19:23 the set should go through the denominators in INCREASING order, so (if using your -337 / -288 interval) 288, 289, 290, ... up to 337. [Technically, should be 289, 290, 291, ... up to 336.]

[@TaladrisKpop]

The fact that you found no negative in the first case ( x floor(x) = 10) does not mean no solution exist, but that the approximation x floor(x) = x^2 may be bad (it is a simplification of the true inequality x-1 < floor(x) 10, it is possible that there is a solution in (-3,-2)

[@angelogandolfo4174]

The future’s bright!

[@koenth2359]

If x[x[x[x]]]=2020 and x positive, we know that 6

[@dnre123]

Thank you for the video!

[@damonpalovaara4211]

You can actually get away with only checking 6 values instead of an exhaustive search by doing a binary search which only needs to do Ceil(Log_2(n)) searches at worst case where n is the number of possible solutions. Just start at the middle of the possible solutions and if it's less than 2020 every possible solution less than the one you just checked won't work either, if it's larger than 2020 then every possible solution larger than the one just checked can be eliminated.

[@DagothXil]

hmm. there are probably a couple complex solutions to this too but finding them wouldn't be trivial I think....in fact I think it'd end up looking a lot like the alternating floor/ceiling problem you proposed at the end

[@DagothXil]

I actually tried this and at least purely imaginary values miss the mark. x={-(2020/308)i, -(2020/307)i} are adjacent negative complex possible solutions that yield ~1974 and ~2033, respectively (with other possible integers moving even further away from 2020) and x={(2020/292)i, (2020/291)i} are adjacent positive complex possible solutions that yield ~2013 and ~2027 in the same regard. I might think on this some more later but trying this with polar angles that aren't multiples of pi/2 does not sound particularly.......possible

[@rafiihsanalfathin9479]

How do you floor a complex number?

[@DagothXil]

@@rafiihsanalfathin9479 I don't really remember this problem but I'm fairly sure you just floor the real and imaginary parts separately. floor(z) = floor(Re[z]) + i*floor(Im[z])

[@rafiihsanalfathin9479]

@@DagothXil hmm interesting

[@nandeesh2ninad]

When you say 3

[@jha180508]

Future is good as I found this channel.

[@juliocesarmarentesmosqueda1511]

What about the other two Fourth roots of 2020.?

[@bryanbischof4351]

Great suggested problems at the end.

[@danielharris1101]

what about for complex values of x? with floor(a+bi) = floor(a) + floor(b)i

[@ezequielangelucci1263]

that doesnt makes sense, ¿the v¡floor of a complex number?

[@fremsleysballoon]

18:52 Where did that arrow come from?

[@tdark987]

0:18 _Ooooooh boy......_ honestly, I doubt that even in your wildest dreams you could’ve imagined quite how f-ing crazy things have turned for us over the past 2.5yrs

[@archimendis8616]

Jeees, You lost me on "solve x" xD

[@diabl2master]

For positive x, it is easy to see that the function f(x)=x*floor(x*floor(x*floor(x))) is non-descreasing in x. Now since f(6.9999....)

[@samb5017]

Why not just note if 3 < x < 4 then m = floor(x) = 3?

[@kevinarturourrutiaalvarez2613]

I don't really understand why we can say that the floor funtion behaves like x^n. I mean is obvious for the upper value, but not so much for the lower one.

[@BlueRaja]

If x is an integer, x⌊x⌊x⌊x⌋⌋⌋ = x^4. 6^4 < 2020, so any value of x ≤ 6 will necessarily be too small. Similarly, 7^4 > 2020 so x ≥ 7 will be too big. Thus we must have 6 < x < 7

[@mrluchtverfrisser]

Noo, here I was working out the problem and watching the video to check whether there was a nicer/cleaner way to do this only to see you did the same thing. Oh well. One thing to note is I went a bit further to conclude that x=-6.6... so to make the gap and amount of things to check only about 5 values.

[@jamesyoung9311]

“The future ain’t what it used to be to be.”

[@SlidellRobotics]

Why all the rigamarole? If 3

[@Bemajster]

Future is better

[@SleepycoasterOrigamiUnicorn]

This sounds like a 2020 national math exam question

[@subpopulations]

I think all ceils would be more interesting than you thought as interlacing the ceils on the odd xs has the same effect as changing the sign as even products will always be positive and thus the floor will decrease the magnitude. x roof(x roof(x roof(x)))= x roof(x roof(7x)) from trying numbers: 47/7(returns 2121 from roofs)=6+5/7>x>6+4/7= 46/7(returns 1992 from roofs) so roof( 7x) = 47 but 46/7< any solving x so x roof(x roof(x roof(x)))> x roof(47 x) > 309x> 2030 so no positive x solves x roof(x roof(x roof(x)))=x roof(x roof(-6x)) by check x

[@CDChester]

This is pretty dope tbh. Never trained too much in number or graph theory but this was easy to follow.

[@scepticusverisimillimenonm8450]

Yeah I was also glad that my Graph Theory knowledge was sufficient for following this video! ;)

[@erickherrerapena8981]

Gran canal, buen video.

[@patrickk1855]

Future gang 🎉

[@MaorWertheim]

you have made a mistake at 13:04. you wrote that m is bigger than the ceiling of (2020/6) and smaller then the floor of (2020/6). i thing that you meant :"m is bigger than the ceiling of (2020/**7**) and smaller then the floor of (2020/6)"