Wringing out one more result.

[@MathElite]

I think for the last part it should have been floor(sqrt(n)) instead of floor(n) when doing the sum of the squares formula

[@petersievert6830]

It was actually quite fascinating to see how many times he said it the right way while hovering above the mistake and even when moving it to the RHS doing it correctly then as well without realising the mistake. But that is maybe just how the human brain works running fully on auto-correct mode ;-)

[@richardfredlund3802]

@@petersievert6830 yeah it was like a real time hallucination. I think because he said it so clearly, the belief was there already.

[@MathElite]

@@aashsyed1277 hii

[@MathElite]

@@aashsyed1277 ofc

[@MathElite]

im subscribed to syber, bprp, flammable maths, and michael penn

[@aa-lr1jk]

Plot twist: when you see a normal video about a integral identity and think that Michael is tired of floor functions, but then you find out that the video is actually just a preparation for another video about the floor function.

[@goodplacetostop2973]

14:18 This homework is from the 6th APMO (1994). Prove that for any n > 1 there is either a power of 10 with n digits in base 2 or a power of 10 with n digits in base 5, but not both.

[@johannesh7610]

40_5, 110_2, 1300_5,... are the first few

[@MrRyanroberson1]

1300_5 is two hundred

[@johannesh7610]

@@MrRyanroberson1 oops, I missed a zero. Well, then the example for 4 digits is missing.

[@MrRyanroberson1]

@@johannesh7610 that lines up perfectly with 10 in binary beint 4 digits (1010, 8+2)

[@ikocheratcr]

I love the way you approached the problem, the reasoning, the problem solving. When I saw the initial problem, I was in "how on earth ones solves this", and you provided a very easy path to understand the solution. Great video.

[@anonymous_4276]

I'm amazed. Very interesting theorem!

[@SaveSoilSaveSoil]

I haven't seen this one before!

[@kevinmartin7760]

The "correction" should be even more complex. It is there because lattice points on the upper and right edges of the small 'ac' rectangle are double-counted. So the correction should subtract floor(a) if c is a non-zero integer and floor(c) if a is a non-zero integer and add 1 if both a and c are non-zero integers. The final "add 1" is there because if both a and c are non-zero integers the top-right corner of the rectangle is a lattice point and it would have been subtracted twice. Since we aren't counting lattice points on the axes we need the special "non-zero" conditions here. In this particular application of the formula, a and c are both zero (and thus integral) and all the corrections are zero.

[@anonymous_4276]

If a and c are both non-zero positive integers, shouldn't we subtract both floor(a) and floor(c) and then add 1?

[@tomatrix7525]

@Kevin Martin Shouldn’t all hiss latace point countings stacked up for each x value be floor(f(n)) +1 because we must count the x axis lattace? E.g. if y = 6.34, there are 7 latice points for the given x value corresponding to it. At y=0 to y=6 which is 7 points, so it is floor(f(n)) +1???? I am not talking about the corrections you made btw, they too seem correct, but I am asking is this another error he also included, which would have effected the sum result, whereas your correction wouldnt luckily

[@SSGranor]

Pretty sure all of this can be handled by having the subtracted term be ceiling(a-1)*ceiling(c-1) rather than trying to put everything in terms of floors. (This is, of course, because ceiling(a-1)=floor(a) except when a is an integer.)

[@anonymous_4276]

@@tomatrix7525 on both LHS and RHS lattice points lying on either axes are not counted so it shouldn't make a difference. The RHS is floor(b)floor(d)-floor(a)floor(c) (without any corrections). Here too we aren't counting the lattice points on the axes.

[@tomatrix7525]

@@anonymous_4276 you’re right. It makes sense now. In his video he said he was counting the axes points, but it was an error of speach, since his formula doesn’t count them and It doesn’t make sense to count them of course. Thanks again

[@tcoren1]

Shouldn't it be floor(f(n))+1 originally as the number of integers points for a specific x value in A? Why don't you count (n,0) as a point?

[@shikharmukherji1236]

a and c are strictly greater than 0

[@tcoren1]

@@shikharmukherji1236 I'm talking about the number of integer points (n,k) for a fixed n such that k is in the range [0,f(n)]. He said that the number is floor(f(n)) but it should be floor(f(n))+1.

[@kilimanjarocruz660]

I think there are some mistakes in your reasoning. The way I did it also counts the lattice points lying on the axis (i.e. (0,n) or (n,0)). Also, the correction factor should consider when a and c are simultaneously integers. My answer was: floor(b+1)*floor(d+1) - floor(a+1)*floor(c+1) = sum_{n=ceil(a)}^{floor(b)} [floor(f(n) + 1)] + sum_{m=ceil(c)}^{floor(d)} [floor(f^{-1}(m) + 1)] - L_f - L_a - L_c + L_ac Where L_f is the number of lattices points of the graph of f, L_a is the number of points on the segment from (0,a) to (c,a), L_c is the number of points on the segment from (a,0) to (a,c) and L_ac is 1 if both a and c are integers and 0 otherwise (this is to avoid double count of this point). This way, if we have f(x) = sqrt(x) - 1, a = 1 and b = N and after some simplifications we get: sum_{n=1}^{N} [floor(sqrt(n))] = (N + 1)*floor(sqrt(n)) - floor(sqrt(n))*(floor(sqrt(n)) + 1)*(2*floor(sqrt(n)) + 1)/6

[@xCorvus7x]

Yes, b and d should be b+1 and d+1 respectively but shouldn't a and c rather be a-1 and c-1? We are counting all integers on the closed intervals [a, b] and [c, d], so if a and c are integers, they as well as floor(f(a)) and floor(f(c)) will be counted for the "areas" A and B.

[@Horinius]

We count also the lattice point at (n, 0), isn't it floor(f(n)) + 1 total pts instead of floor(f(n)) total pts? Suppose f(n) = 2.5. So we have lattice points at (n, 0), (n, 1) and (n, 2). There are three points.

[@yoav613]

great video.you can also find this sum by spliting it to intervals from m^2 to(m+1)^2-1 in which the floor of sqrtN equals to m so you get (2m+1)m. m starts in 1 and the last interval is between the (floor of sqrtN)^2 until N in which it equals to floor of sqrt N. adding all of this intervals you get the same formula

[@ruathak1106]

Needs more finger curls and crunch hangs.

[@alirezaghadami2929]

you're great! keep it up 👍I liked this video. and also please make more videos on topology.

[@guilhermefranco2949]

Michael "the floor function enthusiast" Penn

[@Huxya]

Nice theorem, although if you call [sql(N)] = m and if you note that [sq(n)] increases by 1 every perfect square. you can write the sum this way: 1 + (N-1) + (N-4) + N(N-9)+ ... + (N-m^2) = 1+ m*N - (1+4+9+...+m^2) or something like that. you got the idea

[@a.osethkin55]

Thank you!!

[@paulkohl9267]

Interesting floor identity. 3 points: as noted by @TheObserver in the first post, a = c = 0, so there is a correction term missing; the number of graph points is N + 1, not N (as we go from n = 0 to N); the final formula includes a dummy variable outside of the sum, which, can not be right... Can you do another video on this with corrections?

[@MrRemi1802]

It's funny, you post this video at the exact same time I'm reading Concrete Mathematics (Knuth, Graham, Patashnik) for the first time.

[@SaveSoilSaveSoil]

How are you liking the book?

[@MrRemi1802]

@@SaveSoilSaveSoil I like it! It forces me to think differently.

[@sophiophile]

Can someone explain why the formula he's written by 13:23 is equal to n*floor(sqrt(n))? I think I'm missing how you get that from floor(b)*floor(d)-floor(a)...etc. Thank you in advance.

[@hotlinkster123]

He moved the floor(sqrt(N)) from the LHS to the RHS and factored out the floor(sqrt(N))

[@Milan_Openfeint]

He wrote floor N but meant floor of sqrt N at 12:41. Then continued as if it was sqrt N.

[@MrRyanroberson1]

4:28 i think you made a +1 error. Consider f(n)=pi. There are four lattice points below f, at y values 3,2,1,0, since you included 0. That wod be floor(f)+1, or if you exclude f (n) itself then ceil(f)

[@tomatrix7525]

As somebody else mentioned, I think your ‘correction’ lacks some stuff but it luckily didn’t effect the sum at the end

[@lukevideckis2260]

Complexity is also correct: answer = O(n^(3/2)) as expected

[@bernieg5874]

If double-counting points on the graph can happen, doesn't that mean you should be using ceil(f(x)) not floor(f(x))?

[@merta7154]

sooo... is the main idea, magnitude of l(Gf) tells us how well, a function acts "nice" on integers? ("nice" as in integer)

[@teeweezeven]

I'm a little confused. Is it supposed to be the floor? Are you then not counting the points on the axes? Because e.g. (4,0) and (4,1) are below (4,1.4) so that'd be two points and not floor(1.4)

[@debjitmullick7004]

Sir ... Are the greatest integer fn. And floor fn. the same thing ???.... any one plz help

[@alexpavlov6754]

There was little mistake. There not may be n in the result formula.

[@MichaelGrantPhD]

For the general/initial statement, Is it necessary for f and its inverse to be differentiable as stated, or is continuity sufficient?

[@ThAlEdison]

He has a video on it, I think differentiability is used in the proof

[@Tuuuuusssjjjjjjnrnfnnfnfn]

I just love ur content...wholesome❤ U make maths magical💫 Greetings from india

[@danielmilyutin9914]

Woops. You've lost square root. with sN = sqrt(N), fsN = floor(sN) we will get (N+1)fsN - fsN(fsN+1)(2*fsN+1)/6

[@lisandro73]

I’m not sure how the 13:26 is true, what I missed?

[@non-inertialobserver946]

Because he wrote floor(N) instead of floor(√N)

[@lisandro73]

@@non-inertialobserver946 so it wasn't a simplification error, just a typo in the previous line, was it?

[@Milan_Openfeint]

@@lisandro73 He even says "square root of N" at 12:41 but writes just N.

[@lisandro73]

@@Milan_Openfeint great, thanks 🙂

[@non-inertialobserver946]

@@lisandro73 yep

[@twrk139]

You didn't finish saying "stop" at the end before ending the video. Shameless

[@indarajgochermaths5176]

Nice

[@romajimamulo]

Man, you both said the square root when you didn't write one, and didn't say it when you did write one

[@Iamnotyou29]

Nice interesting video👌👌🤘🤟👏👏

[@ritam8767]

Here a=c=0, which is an integer. So where's the correction factor?

[@luciuskhor554]

0

[@romajimamulo]

Notice that if a was zero, the correction factor isn't there

[@farhanofficial2307]

What we have put at "n"..in the formula