The average distance between points on a square

[@nschloe]

For anyone wondering, the numerical value is about 0.5214

[@hrayz]

Thank you. I was just pulling out the calculator to find that out myself.

[@olumuyiwaasaolu7674]

Makes sense when the diagonals are considered

[@steveodonnell3030]

That is a good place to stop.

[@hrayz]

@@Eianex engineering math... Like when π=e=3

[@falconhawker]

@@Eianex I am an engineer . I guessed at ther answer...my guessd was 0.5

[@Bodyknock]

It is pretty funny how much more complicated the closed form of the answer is compared to what you'd expect from the layout of the problem. You would think the average distance between two points on a square would simplify down to something pretty rudimentary but instead it's got this weird sum of fractions involving the number 15 and a natural log out of seemingly nowhere.

[@luckyw4ss4bi]

I agree! It seems like a classic problem where a nice constant or solution would pop out but it's almost more interesting because of how nasty the closed form solution appears!

[@TomJones-tx7pb]

Love how he is 20 minutes in and then just jumps to the solution. Reminds me of my undergrad math professor for fluid mechanics.

[@TheBrutalDoomer]

It is interesting how adding more dimensions increases the average distance between points in a unit cube (it's a small exercise to check that for 1-dimensional cube the answer indeed is 1/3). Here is a listing of a small simulation in JavaScript (1 to 10 dimensions, 10^7 samples): dim=1: 0.3333 dim=2: 0.5215 dim=3: 0.6616 dim=4: 0.7777 dim=5: 0.8786 dim=6: 0.969 dim=7: 1.0515 dim=8: 1.1282 dim=9: 1.1999 dim=10: 1.2674 It feels like this can be attributed to the fact that with increasing dimensions, the outermost part of the unit cube takes more and more relative volume. However, the maximum possible distance √dim (diagonal) is also increasing, so these averages are likely to approach infinity, but there is one interesting thing: There seems to exist a limit of AvgDist(dim) / √dim, and it is around 0.4082 I have no idea what it means, maybe that is some known constant :D Upd: 0.4082 is pretty close to 1/√6, if that helps.

[@landsgevaer]

The fact that more dimensions increase the distance is logical, I think (for a cube), because the extra dimension can only increase the distance between points: it would only remain the same if the points have the same coordinate on that extra dimension, and can never decrease. Evaluating that constant scale factor relative to sqrt(dim) is a nice exercise though. You're essentially determining sqrt(sum(x_i^2)/N), where x_i is the difference between two uniformly distributed variates. This is the square root of the mean of that squared difference. That mean is 1/6, it turns out, so that constant is 1/sqrt(6) indeed.

[@popularmisconception1]

It means that the average segment length is for all practical purposes directly linearly proportional to the length of the longest diagonal - AvgDist(dim) = 0.4082 * √dim, especially for high dimensions . To be more precise, the average segment distance is somewhere between cca 33.33% and 40.82% of the length of the longest diagonal, smaller for low dimensions, higher for high dimensions, but that's still pretty well bounded for a quick estimate. If you don't know the dimension, you can just use the constant and you won't be much off. It has implication for high-dimensional stochastic processes, where increasing the dimensionality of representation increases the average length between randomly generated samples or noise or average length of vectors generated using such stochastic data, which means the errors/noise is growing with dimension. If that is a problem, you can maybe stabilize it by dividing or multiplying proper stuff in the equations by square root of dimension or it's powers (e.g. second power is just dim), by which you pretty much stabilize the average length regardless of the dimensionality of representation. Your constant is not the important thing here, the practically linear proportionality of average segment and longest diagonal is.

[@landsgevaer]

@@popularmisconception1 en.m.wikipedia.org/wiki/Curse_of_dimensionality

[@ekxo1126]

prove that the answer is exatly sqrt((n-1/3)/6) where n is the number of dimensions

[@ekxo1126]

ok that's probably wrong since it isn't what he gets in the video

[@Pharoom]

One interesting detail here is that if you add a third random point and examine the average area of the triangle with vertices at the points, the problem becomes much simpler. It's even possible to deduce the answer (11/144) without using integration.

[@philg4116]

But didn’t you need to crosscheck with the general solution?

[@nanamacapagal8342]

without integration? how exactly is that possible?

[@gabrielevezzosi4546]

How is possible without integration?

[@user-wx8bm1pg1d]

How?

[@samueldeandrade8535]

Hahahaha. What a little troll.

[@davidgillon6187]

The ‘editor’ still didn’t fix the limits properly!

[@benheideveld4617]

Look at a general (orthogonal) rectangle with horizontal side x and vertical side y inside the unit square. Without loss of generality assume P to be the bottom left corner and Q the top right corner. The required distance is √(x²+ y²). The weight of this value is the wiggle room of the rectangle within the square, which is (1-x) horizontally and (1-y) vertically. The average distance between two arbitrary points becomes (with integration boundaries x,y both in [0,1]) ∫∫(1-x)(1-y)√(x²+ y²)dxdy/∫∫(1-x)(1-y)dxdy because we have to divide by the average of the weight factors.

[@benheideveld4617]

Which on WolframAlpha gives the same answer!!

[@BachelorChowFlavour]

Excellent work! This is a better answer

[@abbeleon]

Elegant!

[@leif1075]

Doesn't anyone agree that sub he does of y equals x tan theta shouldn't even be allowed..you don't know y will ever equal x times tangent theta..y will only.ewual x sometimes for some points..serjosuly I dknt see why anyone would.ever think of thst..why nkt just y equals tantheta and x is sec theta or cosine and sine..something sensible and more logical!!

[@xulerpi]

A similar approach. If you consider the positive lengths in the whole unit square then you get 4∫∫(1-x)(1-y)√(x²+ y²)dxdy directly with the factor of 4 coming from the 4 cases (+,+), (+,-), (-,+),and (-,-) for the signs of the coord. differences.

[@Skandalos]

The channel mindyourdecisions had this problem a couple of years ago. A similar problem, the average distance of two points on a circle, was presented on 3blue1brown IIRC. Interesting problems, the math is over my head but I like visualizing them using geogebra or programming.

[@Boringpenguin]

It should be Mathemaniac who presented the circle version of the problem, he's also using manim so the style of his video is very similar to 3b1b's. Here's the link to the video series: kzbin.info/aero/PLDcSwjT2BF_XRsHboeAQ5esyUd6RdHtgC

[@Skandalos]

@@Boringpenguin I found the video: it was Numberphile in collab with 3b1b. kzbin.info/www/bejne/o4ulqKaja6dlbpI

[@thelightningwave]

That was a shit video, he glossed over the hard details because he's a hack and then went fully into the hard details. I watched that video so many times and each time I hate it even more. If anyone uses that video as a template to instruct they'll be a horrible professor. He's a great example of someone that just uses the algorithm for profit.

@@thelightningwave Why do you keep watching it, if you hate it?

[@thelightningwave]

@ I don't. I mean I kept watching that video to understand what he was talking about, but each time I watched I just realized that it's a horrible description/explanation of the problem. It wasn't until I watched it Soo many times and when flammable maths openly mocked him that I realized that mind your decisions is just math ClickBait and doesn't really explain things in a concise manner. I watch like at most three of his videos a year. I used to gobble his videos up like the end of the world was nire.

[@putputlawch6770]

A slight reframing: we first compute the distribution of X ~ Y ~ U(0,1) - U(0,1) ~ U(0,1) + U(-1,0) (random variable obtained by taking the difference of two independent uniforms on (0,1)) and then compute the expectation of sqrt(X² + Y²) The distribution of the difference U(0,1) - U(0,1) is the same as that of U(0,1) + U(-1,0) whose pdf is a triangle function (as convolution product of two rectangle functions) which is centered around 0. This is why we split the integral and get affine-linear terms like (1-x) and (1+x). The sqrt(X² + Y²) distribution reminds me of the chi-distribution, except that we swapped out the Gaussian random variables for our "triangle distributed" random variables X and Y, so I wonder if there's a connection between those.

[@Vladimir_Pavlov]

When calculating the integral (starting from 13:30), it is really better to go to the polar coordinates. 4*2*∫[0;π/4]∫[0;1/cosφ] (1-r*cosφ)*(1-r*sinφ)*r^2*dr*dφ= The multiplier 2 compensates for the fact that we integrate up to π/4. and not up to π/2. =... =∫[0; π/4] [(2/3)*(1/ cosφ^3)-(2/5)*(sinφ/cosφ^4)]dφ= ln(√2 +1)/3+(2+√2)/15.

[@leif1075]

Yea don't you agree that substitution was INSANE and should NOT BE ALLOWED I have NEVER seen a sub where you INCLUDETHE OTHER VARIABLE in the sub..what thenfuck y equals x times tan tbeta..how and why pit x in there . I don't see why or how ANYONE would EVER think of that..come on..that shouldn't eve. Be allowed why do that..maybe y equals cosine or just tangent x...why not just that

[@eyesyc]

Great calc3 refresher. Touched some points I was thinking about last night for diffforms.

[@steveodonnell3030]

I put the final answer in a calculator and got 0.52140543316. I ran one million simulations in excel and got 0.521539 Probably need a lot more simulations...

[@StarEclipse506]

That's actually amazingly good considering floating point side effects.

[@robshaw2639]

with a million samples i get 0.521395184807978, and the exact value is 0.521405433164720678, so the simulation checks out

[@michaelking7701]

“So I think you can already see why this very easy to state problem is actually quite gnarly.. Okay good!”

[@robwilkinson5682]

great solution, and a nice little lunch break project! import math import random data=[ ] lengths = [ ] trial_number = 10^6 for n in range(2*trial_number): point = [random.uniform(0,1), random.uniform(0,1)] data.append(point) for n in range(trial_number): length = math.sqrt((data[n][0]-data[n+trial_number][0])**2 + (data[n][1]-data[n+trial_number][1])**2) lengths.append(length) average = sum(lengths) / len(lengths) print(average)

[@rupen42]

If you like these kinds of simulations, I recommend looking into the numpy package and its random features. This simulation can be done with: ---------- import numpy as np TRIALS = 10**6 # Make 2 arrays with random 2D points. (the default is already between 0 and 1) points1 = np.random.uniform(size=(TRIALS, 2)) points2 = np.random.uniform(size=(TRIALS, 2)) # For each pair of points, calculate the difference of their coordinates. # Example: the pair (0.2, 0.8) - (0.3, 0.5) will give (-0.1, 0.3). # You can think of this as a vector that takes you from point2 to point1. diff_vectors = points1 - points2 # Now, for each of these differences, calculate the "size" of each of those vectors. # axis=1 makes it calculate the size of *each* vector, instead of the whole matrix. distances = np.linalg.norm(diff_vectors, axis=1) print(distances.mean()) # built-in average -------- This is shorter, faster to write, more readable, can scale to higher dimensions easily and runs *much* faster. I got 40x the speed.

[@tmpqtyutmpqty4733]

You won't be able to tell the exact answer from this

[@rupen42]

@@tmpqtyutmpqty4733 well, yes, of course. Good luck solving the 1000-D case analytically though!

@@lidarman2 You should start your avgdst with None, not 0.0. Zero is not a reasonable value. (And overriding all initial zeroes would even be wrong in general. Your solution can't tell whether avgdst has just not been set or whether all it currently just happens to sum to 0.) Numerically, it's better to just add all the samples, and divide them once at the end.

@@lidarman2 Who suggested creating the whole list? Yes, for your quick and dirty exercise, it's fine. You can keep a running count and a running sum, and at the end you divide them. No need to keep a whole list. That's simpler than working out the math like you did, and has fewer numerical problems.

[@The1RandomFool]

At 19:37 I noticed that log((1+sqrt(1+x^2))/x) = log(1/x+sqrt((1/x)^2+1)) = arcsinh(1/x). It makes integration by parts simpler. After that, I used a tangent substitution.

[@kartikchaturvedi7146]

why are we using a 4th dimension as a square is only within the 2 axes? kindly explain anyone

[@derekpoon5308]

The integral sums all distances. Should we explicitly divide it over square area (although it is equal to 1) to get average?

[@3x3-x3x-oXo]

It's defenitely there in the definition of average you are right. Expliciting it would depend on the public I guess.

[@ArminVollmer]

Btw, the average "Manhattan distance" between two points on a unit square is 2/3.

[@calcubite9298]

I was lost by about two minutes in. Why is a four-dimensional hypercube involved? What does it even mean to integrate the distance formula? Like I understand integrating a function over some interval, and I get (I think?) that the integral(1 dxdydz)= the volume of that bounded region, but.... yeah I'm stuck.

[@calcubite9298]

I am going to go learn about Jacobian matrices as I went fully cross-eyed by 4:44.

[@frotzecht3461]

Not that it matters in this case but your Jacobian is the wrong way round. It should have the new variables in the numerator.

[@GreenMeansGOF]

You should make a video on that integral at the end with the natural log.

[@bakaneko39]

Solving this problem with Corfton formula works really nicely (easier for the average distance on a disk)

[@stephenhamer8192]

In case you were wondering [spoiler alert] [ ( 2 + sqrt 2 ) / 15] + [ ln (1 + sqrt 2)) / 15 ] = 0.5214 to 4 dp, so if you guessed "a half" you were close drive.google.com/file/d/1antIBxzFTw5Cfvw6nxMKLn_634byUxDA/view?usp=sharing I was slightly puzzled by the speedy intro of 4D space at the start, and by the near-absence of any mention of the probability function. We are calculating the "Expectation Value" of the distance, d(P, Q) between points P, Q of a unit-square. The distance function is a function of 4 parameters , the (x,y) co-ords of points P, Q, which can each be chosen independently in the set [0, 1]; hence a 4D probability space. OK, got that. Call this the 4D unit-cube, 4U. The required EV is then the integral over all points of 4U of prob(P, Q) x d(P, Q), where prob (P, Q) is the probability of choosing the 4 parameters of (P, Q). Since the choices are all distinct and independent, we have prob(P, Q) = 1 / total number of (P, Q)s. But what is the total number of (P, Q)s? I took me a minute to twig that the needed "counting" function is the function that "clocks" (takes the value) "1" for each choice of (P, Q), but that's just the "characteristic function", χ on 4U, which when integrated over 4U yields the "vol" of 4U (I think the Prof mentioned this) Thus prob (P, Q) = 1 / vol 4U, which is just a number which can be factored-out of our original integral; and, in fact, it's just the number 1 (which the prof also mentioned), so it, in effect, disappears Sorry, rather laboured all that

[@kartiksunaad]

I really thought there would be a way to do this without integration, using some clever trick. By the way, you sure of your answer? I put the integral that we had at the start ( the integral of Sqrt[(x2-x1)^2+(y2-y1)^2] ) in Mathematica and it spit out this as the answer: 1/120 (16 + 8√2 + 80π - 70 Log[1 + √2] - 15 Log[2] - 40 Log[3 - 2√2] + 30 Log[2 + √2]) In particular there is a π in the answer 🤔

[@random19911004]

It's seems like the answer should have been much more clean, and there would be a way to get it just by arguing about symmetry, but unfortunately not

[@cernejr]

Approx. 0.52 ?? For the 1D problem (a unit line segment) the answer is apparently 1/3 . I like this stuff.

[@teeweezeven]

Can confirm it is 1/3. Was fun calculating that!

[@lPlanetarizado]

i wonde what happens if you let n tends to infinity for a "n-integral", the value converge?

[@user-nb6zu3rk4f]

@IPlanetarizado I’m sure it converges to 0

[@ulrichs.3228]

@@teeweezeven Huh, I wonder what happens with those values if we generalize to n-dimensional hypercubes. But not enough to calculate it formally, maybe I'll run a little sim.

[@teeweezeven]

@@ulrichs.3228 I mean, going from 1 to 2 dimensions already complicates things a whole lot. All I can confidently say is that as n gets bigger, the distance also gets bigger, probably tending towards infinity.

[@sldimaf]

~0.521

[@YellowBunny]

I gave this a try without watching the video and thought this might be the solution: integral from 0 to 1 integral from 0 to 1 (2-x-y)*(sqrt(x^2+y^2)) dx dy But that ended up being something that I can't solve and that is equal to around 0.651757 according to WolframAlpha. I don't really know why my idea is wrong but I definitely wouldn't have tried to use a quadruple integral after already being unable to solve a double integral.

[@rumbleinthejungle3358]

Thanks for the upload

[@sharpnova2]

jesus this is a hell of a headache of a problem. i worked it out in my own way using probability density functions and got a somewhat similar answer with nested radicals that i *think* is equivalent to your answer. but it's late and i'm too tired to prove they are the same. but it seems likely. incidentally, my method really easily extends itself to n-dimensions

> but it's late and i'm too tired to prove they are the same. I'd plug both of them in a calculator and see whether they agree numerically, and only if they do, start proving.

[@johnchristian5027]

What a brilliant use of mathematics, you'd expect a simpler solution to this problem!

[@ethandavis7310]

The simple solution is just running a simulation

[@natepolidoro4565]

The number at the end is about 0.521405433165

[@mimzim7141]

could do a discret version with an nxn lattice and compute all the distances. then increase n and see if it converges towards your value.

[@phee4174]

8:54 I'm confused why are the bounds of the third integral sign thingy 0 to -1, shouldn't they be 0 to 1 as those are the values x is ranging between in the purple triangle?

[@RecreationallyCynical]

I think you're right. The bounds should be 0 to 1. On the next board Michael corrected the bounds on the second integral and said the editor will fix it in post for the previous board. I think maybe the editor thought both needed correction.

[@digxx]

The integral is actually Int_{-1}^0 dx Int_{-x}^1 dz + Int_0^1 dx Int_0^{1-x} dz = Int_{-1}^0 dx (1+x) + Int_0^1 dx (1-x) = 2*Int_0^1 dx (1-x) upon change of variables x->-x in the first integral.

[@alexandresoutonogueira7675]

#ruby 2.5.3 def dist ((rand()-rand())**2.0+(rand()-rand())**2.0)**0.5 end size = 1000.0 v = [] 1000.times do v

[@ethandavis7310]

Approximated the answer numerically with about 30 seconds of writing a python script

[@hel13ay]

Please, make the video for derivation of Probability of distance between two points being less than x.

[@leif1075]

I'm sorry y equals x tan theta is even a Valid substitution..you don't know if y will.ever equal x times tangent theta because x is a point in the square but some dummy variable..

[@MurshidIslam]

Why is the average distance = ithe ntegral of the distance formula over all possible P and Q divided by the 4D volume of the unit cube?

[@Patapom3]

At one point you pose y=x tan(theta), like... you can really do that??? Use another variable as substitute for a variable? 🤯

[@werquast]

I think this was your best video so far!

[@ludomine7746]

tried doing it with expectation of uniform random variables, got that the distance squared is expected to be 1, which i suspect is also wrong😅

[@grandmartinvandam]

0.57 ish? The farthest apart they could be is about 1.41... seems reasonable.

[@nibn4r]

So what is the final answer? Something to compare to the intuitive guesses we all made at the start. Something close to 0.5 I think?

[@falconhawker]

Even by intuition 2 point on a diagonal will cause the final average to EXCEED 0.5 , so intuition should say the avg must be >0.5

[@coreyyanofsky]

being a statistician i'd have done it with probability theory: - the components of P and Q are IID uniform - the absolute difference of two uniform RVs is a triangle distribution - squaring is a change of variables - summing two such RVs is convolution, probably not tractable so we leave it as is - law of the unconscious statistician turns this into a double integral, even more ugly but the fact that this video exists at all indicates it's tractable

[@3x3-x3x-oXo]

You'd end up with the exact same computations in the end.

[@coreyyanofsky]

​@@3x3-x3x-oXo yes, it's exactly the same in the end; the point of the comment is that probability theory provides conceptual chunks that get you to the final integral faster, e.g., it makes it clear that you can handle the two dimensions separately so you don't have to write down a quadruple integral

[@ddiq47]

It's nice to be able to find a closed form solution for this, but monte carlo sims got me 3 sig fig accuracy very fast

[@cicik57]

maybe there can be some geometrical trick to calculate it easier?

[@josephmartos]

Can pls some1 explain why the expected value has that 4dim volume cube thing?

Michael explains it in the video?

[@leif1075]

Why not just write x1 minus x2 as delta x and the difference in y as delta y then it becomes a much easier 2d integral?

That's more or less what he's doing in a roundabout way in the video.

[@BethKjos]

Gee this problem is getting tedious; let me just apply the Feynman method.

[@eldarallakhverdiiev2211]

it can go much easier if using polar coordinate system instead of rectangular, can't it?

[@romajimamulo]

On one level yes, but your boundaries become really awkward and difficult to work with

[@lucachiesura5191]

Something about the universal parabolic constant...good

[@spoolster64]

Thank you! I was trying to remember where I saw ln(1+sqrt(2)) before xD

[@binaryblade2]

Now show how that varies with a p-adic metric

[@Allyballybean]

I admit I was lost at 4 minutes.

[@TheOnlyAndreySotnikov]

Can we generalize it for an arbitrary dimension n?

[@toes161616]

The answer almost looks like the golden ratio

[@AndrejPanjkov]

Michael's result is about 0.52. Are there bounds we can put on the result before we work out the integrals? Certainly the average is in (0,sqrt(2)).

[@alexpotts6520]

It's very quick to work out that the result is exactly 1/3 in one dimension, and then using the triangle inequality gives you an upper bound of n/3 in n dimensions.

You can use en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities for some bounds. Or more generally, artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality

[@dobridimitrov6688]

so the average distance is ~ 0.5214... does it mean that: the chance of random distance being > 0.5 is more than 50%?

[@Vladimir_Pavlov]

Based only on the average value (Expected value) of a random variable, it is impossible to draw a conclusion about the distribution of probabilities of its values. Nevertheless, in this case P[sqrt(x^2+y^2)>1/2] =125/96 - π/4 =0.517...

[@falconhawker]

the distance of 2 points on a diagonal when x,y exceed 0.5

[@nixlarfs1002]

Now do this with Manhattan distance:)

[@schweinmachtbree1013]

I think a hyperbolic substitution might be cleaner than the trigonometric substitution

[@leif1075]

I don't see why anyone would.think of x tan theta.is thst even allowed you dot know that y will ever equal x tsn theta since x is a point in the square. So yea thinking about it morenkw.i ealize that is not a valid substitution

[@plislegalineu3005]

*the editor corrected the wrong thing*

[@2kreskimatmy]

but why 4 dimensions

[@Fine_Mouche]

min distance = 0, max istance = sqrt(2) ~1.4142, the average distance = (2 + sqrt(2) / 15) + ln(1 + sqrt(2)) / 3 ~ 0,521405

[@PunmasterSTP]

Damn. That's really all I can say.

[@frentz7]

Jacobian is there, because an ordinary "substitution" such as u = 3x will "inflate" the region by a factor of 3 .. and so the same quantity (such as 9x^2 + 7, or aka u^2 + 7) placed into a new domain (such as u : 0 to 30, instead of x : 0 to 10) ... well it just aint right. Without some sort of Jacobian factor (such as a 1/3 = dx/du). Otherwise you're getting your quantity accumulated across three times as much territory ...

[@aidanokeeffe7928]

Now the variance please

[@Vladimir_Pavlov]

Some knowledge from probability theory allows to carry out transformations 4-dimensional integral to two-dimensional faster. A random variable A is the difference between two independent random variables x1 and x2, uniformly distributed over the interval [0,1] : A=x1-x2. Its distribution density is constructed as a convolution of the distribution densities x1 and x2. Result: pA(x)=1-∣x∣ if x∈[-1,1] ;pA(x)=0 if x∉[-1,1]. The same result for the distribution density pB(y) of a random variable B=y1-y2. According to the average calculation rule: Avg=∫[-1,1]∫[-1,1] sqrt(x^2+y^2)*pA(x)*pB(y)dy*dx= =4∫[0,1]∫[0,1] sqrt (x^2 +y^2)*(1-x)*(1-y)dy*dx. It is clear that for an n-dimensional unit cube, calculations can be performed using the formula Avg=2^n*∫[0,1]∫[0,1]∫[0,1]...... (1- x)*(1-y)*(1-z)......*sqrt (x^2+y^2+z^2......)dx*dy*dz....

[@lPlanetarizado]

do you know how to prove that it diverge for n->inf? i have been thinking and i cant do it

[@Vladimir_Pavlov]

@@lPlanetarizado Avg(n)= 2^n*∫[0;1]...∫[0;1](1-t(1))*...(1-t(n))*sqrt [t(1)^2+...+t(n)^2]dt(1)*...dt(n). (1) The inequality about the arithmetic mean and geometric mean gives [ t(1)^2 +...t(n)^2]/n ≥[t(1)^2*...*t(n)^2]^(1/n)=[t(1)*...*t(n)]^(2/n). So sqrt [t(1)^2+...+t(n)^2]≥ sqrt(n)*[t(1)*...*t(n)]^(1/n). (2) Of (1) and (2) Avg(n) ≥(2^n)*sqrt(n)*{∫[0;1] (1- z)*z^(1/n) dz}^n=... ...= (2^n)*sqrt(n)*{1/(1/n +1)- 1/(1/n +2)}^n = = (2^n)*sqrt(n)*{n^2/[(n+1)*(2n+1)]^n = =sqrt (n)/[(1+1/n)^n*(1+1/2*n)^n] It turns out that for n ≫1 Avg(n)≥sqrt(n)/[e*sqrt(e)]=sqrt(n/e^3). The estimate is rough, but shows that Avg(n) is proportional to sqrt(n).

@@lPlanetarizado You can use en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities to prove divergence in the original formulation. You don't need to solve the problem to a closed form to prove divergence.

[@lPlanetarizado]

@ i see ....i knew there has to be an easier way...i tried the integral - just for fun- with spherical coordinate substitution, but you get a complicated expression too

[@simonwillover4175]

Does michael have a calc 3 course play list? Im sorry but i havent done calc 3 (just 1 and 2) so I don't know where he's getting these conclusions from.

[@groa4]

Would there be a problem with finding the average of the square of the distance to simplify the integration and then come back and get the actual average distance?

[@diffusegd]

Fun fact: the average distance between 2 points in a unit cube also has a closed form, and yes, you do get 6 nested integrals. I don't know whether there's a closed form for 4+ dimensions

[@ranshen1486]

(1) What is the average distance of 2 points on the edges of a unit square? (2) How does the ratio of question (1) to the original average in the video change as dimension grows towards infinity?

[@Fine_Mouche]

In a cube ? in a N-cube ?

[@johnalley8397]

And I thought the 4D quadruple integral was the tricksie bit! NOPE!

[@mokouf3]

I remember that one other math channel posted same question, but using a little bit different concept about probability density function in the middle.

[@samlsne]

Using conditional probability makes the problem easy

[@MrEliteXXL]

I'm overwhelmed

[@colincolenso]

By geometrical estimation, I would have guessed the average distance of any two points is somewhere between ([1/2 of root 2 on 4] and [1/2 of 1/2]) = 0.35355 and 0.25, with less weight to the 0.25 due to geometry. As the answer is 1/3, the weighting of the lesser number 0.2022... 0.00833 (differential) the weighting is about 0.412 or 1 in 2.426.

[@mansoormohamedali]

Hello Mr. Penn. I have a question that has been keeping me all night these past few days. Hope you can help. Question: Let have any area on earth ( a city or a region). That we can calculate or know its area. We have n (a natural number) number of fixed spot we can put on the area.How can we calculate the minimal distance we can place n number of fixed spot such that we have the same distance between any point on the edge of the boundary to the next fixed spot or from fixed spot to fixed spot. For example: if we have n =1, so the only logical place to put the fixed point is at the center. From the the center ( fixed spot) to any point in the boundary are the same distance. What is the reasoning in the case n = 2,or n = 3, … n = inf?

[@rozpiotr]

craziness 🙃

[@MrCoffeypaul]

I like lamb.

[@byronwilliams7977]

I saw this on Presh Talwalker's channel a while back, loved it then, still love it now.

[@BachelorChowFlavour]

A bit disappointing that a clever trick could not be found...

[@AndrejPanjkov]

some simpler problems: = average distance between a point and all points on a line segment = average distance between points on two line segments

[@joeo3377]

It's been a while since I did an integral this involved, but when you did the change of variables y = xtanθ, why isn't the differential that results xsec²θdθ+tanθdx? Don't the differentials follow a Leibniz rule?

[@quintonpierre]

What about if the sides of the unit square are connected in a pacman like manner ?

[@stomio1164]

I think this exsample shows how it is different between the simplicity for human understanding and world of math. math is made in order to solve many human-like problems easier, but as always, we can’t produce a “perfect” tool. Maybe we have to be super-surprised of the fact that this easy method is so versatile.

[@1.4142]

Presh did this

[@natepolidoro4565]

2:17 lol

[@Wielorybkek]

Now what happens when we use different metrics? :D

[@christopherdyson1158]

Seeing the thumbnail I interpreted the question different, but it seems like it might still be interesting... for two specified points what is the average path length between the two... which I feel like will just be infinite, so maybe add the restrictions that the path must not cross itself and has a certain width.

[@SuperYoonHo]

Awesome sir!!!

[@farfa2937]

Really interesting problem, tho it's quite the L that the result is not a cute number like 1/4 or something like that.

[@herbertschulz4713]

Do you get the same result in a polar coordinate system? (Random variable with phi,r)

[@robertholder]

In the easy to ask, but surprisingly tricky category: If rolling a die 100 times, what is the probability of getting five in a row of the same number at least once? Was helping a student with that, and there are some subtle traps to fall into.

[@lucasm4299]

Instead solve: what is the probability of never getting five in a row of the same number in 100 rolls?

[@robertholder]

@@lucasm4299 Yes, the student got that far. The tricky part comes with managing the dependencies between rolls and the windows of five rolls.

[@daaa2299]

Isn't it somehow related to bertrand paradox? We can't say that every point has the same probability of appearing if we are talking about infinite sets, we need some probability distribution

[@trustnoone81]

Whenever I see a problem that starts with "choose two points at random in this 2D region" I hear bells and whistles going on somewhere in my subconscious.

[@eu7059]

Assume uniform distribution

[@replicaacliper]

No not really, a uniform distribution over the square is the really only natural "random" distribution for a point. Bertrand's paradox is more about situations where the probability distribution is ambiguous. Multivariate continuous distributions exist and there is no issues with doing statistics with them.

[@mrpenguin815]

It probably should state the distribution, but if its not then I think it's safe to assume it is uniform. HW: solve for all cases of the beta distribution in terms of parameters a and b.

[@farfa2937]

You'd think if he had a non-uniform distribution he'd have said it. The most "natural" thing to do is assume uniformity, since he didn't say (for example) "this corner of the square has more points than the others".

[@DividingLimits]

Layman person computer programmer here. Given two tuples (Cartesian points- x1,y1, x2,y2), what’s the algorithm I can implement in code? I didn’t follow the simplification and all of the substitutions. The final values do not seem to link back to the original inputs at all.

[@3x3-x3x-oXo]

In code you want to compute the average value over many random experiments (where x1, x_2, y_1,y_2 are random floats between 0 and 1) of the value sqrt((x_1-x_2)^2 + (y_1,y_2)^2). So it's a simple for loop. What he computes is the expectation, that is the idealized average where the number of trials becomes infinite.

[@jerry3790]

I came up with a solution to this problem that ended up being incorrect, but I’m still not entirely sure what is fundamentally wrong with it. It is well known that the average distance between two points on a line of length 1 is 1/3. Therefore the expected values for (x1-x2) and (y1-y2) should also be 1/3. Plugging this into the distance formula gives a result of sqrt2/3

[@bscutajar]

That's because in general avg(F(x,y)) is not equal to F(avg(x),avg(y)). In fact that is a very wrong assumption in most cases. Take for example kill death ratios of a videogame scoreboard. The average kill death ratio is not necessarily 1, but the ratio of average kills and average deaths is always necessarily 1, since kills=deaths.

[@byronwatkins2565]

Did you multiply by 4? This result looks a bit small...

[@indocesare14]

It can be interesting to ask the same question for a square with non-constant density. Depending on the density function there should be parts of the square with "more points" than others, so the average distance should change. However analytically solving the integral may really become challenging if not impossible