Solved simply: the impossible integral
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- Опубліковано 15 кві 2021
- Yes, it can't be done using substitution, by parts or changing variables (and using the Jacobian); but there is a very clever trick to actually compute this integral, which is attributed to Crofton, an English mathematician.
This clever trick only requires the law of total expectation and some very simple algebraic manipulations, and is very elegant in solving this very complicated integral, and it is incredibly powerful in the sense that it can be used in much more general situations, not just this integral - when we want an average of some quantities (which needs to be a bounded symmetric function of n points), we can use the Crofton's differential equation already to convert the problem to the average quantity when 1 point is on the boundary of the domain. In this case, the differential equation is easy since we already know that the average distance is proportional to the radius.
The problem can be made even more interesting when we think of higher dimensions: what about the average distance in a unit ball, or an n-dimensional ball? The calculations might be a bit tedious, but doable, and it again simply relies on the Crofton's differential equation. The only difficult part would be to figure out the limits of integration and the Jacobian determinant when using higher-dimensional spherical polar coordinates, and you can see that in the sources below.
Even if you don't know the Jacobian, or multiple integrals, you can still at least understand the clever trick behind this, which is the more important message of this video.
I currently have plenty of video ideas, but none of them really forms fully into a plan yet, so if you do have any video ideas, drop a comment below!
SOURCE FOR THIS VIDEO SERIES:
www.tandfonline.com/doi/abs/1... [The paper that I'm following, with a bit of biological motivation of the problem]
www.jstor.org/stable/2589434?... (a more technical version for Crofton's differential equation)
If you want to know the higher-dimensional analog of spherical polar coordinates, and possibly want to derive the average distance in n-dimensional ball yourself, you can see the exercise 5.19 in Chapter 4 (Multiple integrals) on page 268 of C.H. Edwards, Jr., Advanced Calculus of Several Variables, Academic Press, San Diego, 1973
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Integral calculators hate him, see how he computed this integral with one simple trick.
Nice video!
Is (f(R)-f(R-dR))/dR as opposed to (f(R+dR)-f(R))/dR just to make the law of total expectation application a bit simpler?
It is just to make the expansion simpler - we can't really expand (R / R + dR)^4 that simply, or alternatively, you have to move (R + dR)^4 to the other side, so this is just more convenient for the explanation.
@@mathemaniac makes sense thanks.
Hey! This is a nice channel. I just love it. However I need some guidance.
Please tell me which video or playlist should I start from so that I can enjoy it fully.
@@jayantanayak4981 Can't really recommend as I don't know your maths level and interests, but for this video, it is actually the finale of a video series, so if you want, go check out the video series "Average distance between two points in a unit disc" and watch it in full.
Also done the "Essence of Group Theory" video series, which is about, you guessed it, group theory.
The differential equation part feels like too good to be true
Yes, indeed - the sheer simplicity is just mind-blowing, but it is indeed true :)
Like a magic trick out of thin air.
Mathemaniac: Your first instinct might be to use substitution, or integration by parts, or even change of variables
Me, an applied mathematician: haha Monte Carlo integration go brrrrrrrrrrr
Haha :)
lol hahahahaha, I was thinking the same thing. Fuckit, ENGINEERING TIME: g=10, pi=3 thus avg distance = r = 1
@@lucdegraaf5138 Me: Nope nope nope and nope!
I think the most unexpected part of that video is that the average distance in a unit ball is a rational number!
True, considering pi is ever present when dealing with circles.
However, using polar coordinates (r,theta,phi), it becomes a triple integral, where you look at the areas of circles of radius cos(phi)
it is rational for any odd dimension
This is how math should be shown. NO step skipping but at the same time making it fast so the user can stop the video and see the details for himself.
Now this is a great video.
Since the last video, I've learned to understand multi-integrals and the jacobian. I now understand the problem way more than before. This is certainly one of the coolest calculus problem's I've seen.
Also eyyyy, you have pi * 10^4 subscribers!
Yes, this is one of those integrals that look intimidating, but actually with a clever trick, it isn't that horrible at all!
That was really cool. I realized as soon as you introduced the functions g and h that they would just be simpler average distance problems as δ approached 0 and got exited for what was coming next.
Indeed, this is just such a cool trick that I need to share it on UA-cam!
Well done, another beautiful piece of mathematics. Keep it up man. We've got some quality content right here
Thanks for the appreciation!
I have been waiting on a digestible version of the more complete answer for this problem for years! Thank you for this!
Glad to help!
That is an incredible trick and a really well executed explanation!
Thanks for the compliment! This trick is so incredible that I just have to share it on the internet!
Incredible.
This method is too clever!
Yes, this is very clever, which is why I dedicated an entire video series to this :)
@@mathemaniac Thanks for sharing this with us!
To be fair, when I finally saw the equation of the radius of a circle I somewhat knew where the explanation was heading. Good video m8.
Thanks for the compliment!
The animated visualizations of formulas are awsome!
Thank you ... that was a fun series of videos.
Holy cow that's a heck of a method... basically imposing a geometric scaling constraint and then magic!
Enjoyed the video series, keep up the great work!
Thanks for the appreciation!
I watched this video when it came out, and I got complete lost. Now I tried again, and it makes sense! This is such a clever method!
Yes indeed, which is why I wanted to make a video series about it in the first place!
Nice use of topological properties - {have to rewatch the other videos in this series}...
This video is so so so well made yet few people watch it. Nevertheless, please keep sharing your ideas and I believe everyone here will enjoy this process!
Thanks so much for the kind words!
Wow that was actually a fun problem to solve for the average distance between two points in the unit ball. Amazing approach and beautiful solution.
I wonder if the same thing can be done to find the average distances between two points in a unit square
Glad you found this problem fun! The use of symmetry means that computing g(R,R) in the square case is not quite straightforward. However, the square case has been covered somewhere else before on UA-cam.
@@mathemaniac symmetry can still somewhat be exploited by restricting one of the points to move along one (or even half a side). Although it still seems difficult
Even though I hadn't seen and knew the 'c' of calculus , but very glad that I watched it anyhow 😍
Nice video! Interesting that the average distance in n-dimensional space has π in it when n is even but not when n is odd. Wonder if there is an intuitive explanation for this.
I don' think there is an "intuitive" explanation in the sense of "oh, obviously!" because we can't even visualise higher dimensions. The reason seems to be that when actually doing the integral, you will encounter a term cos^(n+1) theta (just like here, we have cos^3 theta), which provides the extra pi if n is odd; but rational when n is even.
This looks like more physics than mathematics yet this was a great video to watch. I wonder how the limit arguments can be stated properly to arrive at the differential equation. I guess you have to assume or show that the functions are C1 or at least differentiable and that there is some kind of uniform convergence maybe
Very nice. This has some implications for what unit balls look like in high dimensions, and how our intuition about the shape of space in 2 or 3 dimensions fails in higher dimensions.
Yes! That's a great point!
Is there any intuitive explanation for why the limit n to infinity would go to the square root of 2? What conclusions can we draw from that about infinite dimensional space?
An intuitive but not a very explanatory answer is that in higher dimensions, the sphere behave more and more like lots of spikes. You can even compute the volume of an n-dimensional ball, and the volume shrinks and shrinks down to nothing as n goes to infinity. So eventually, when you sample the points, it will be at the edge of the "spheres" along the coordinate axes, and so the average distance is that between (1,0,0,...,0) and (0,1,0,...,0), which is sqrt(2).
Yes, this is not a very satisfactory answer, but probably a more "intuitive" one, simply because it is extremely difficult if not impossible to visualise higher dimensions.
Wow thanks so much! A lot! It helped a lot!
Glad it helped!
How could you calculate the average distance between two random points in a semicircle or in any circular sector?
that is amazingly cool way of solving it
Glad that you enjoyed it! Can't really take credit for this though - I just followed the paper in the description.
@@mathemaniac still, we would probably never learn that without you!
Also how do you go about evaluating the g(R,R) in an n-dimensional ball? The 3D ball was already a bit of a mess for me to deal with but I got it after correcting a few mistakes
Read the description to find out how to use n-dimensional spherical polars, and hopefully that should make sense, albeit very tedious.
You just hacked the problem .
Nice
I wonder: does the uniform PDF for random points result in the maximum average distance? If not, what PDF would maximize this average?
EDIT: Sampling points exclusively from the edge of the disc gives a higher average distance.
I don't even know what a integral is, I'm still going to watch it tho, I'm curious.
The trick where we reduce the problem into a simpler one is the very clever bit, and doesn't really require any integration at all - and that's the most important bit of the entire video, not the actual integration that we do at the end.
an*
Integrals are another set of functions that are piecewise related.
I'm not good enough to try the n dimensional case, I'm just here to enjoy the show!
That's cool! I didn't expect everyone to try, but just that hopefully someone might see this as some stimulating exercise, and try it.
This is amazing.
Thanks!
Thank you sir finally!
0:0 This is the part where you scare away any eventual non-mathematicians.
I tried the sphere version... first I got that g(R,R)=3R but I noticed an error so I tried fixing it and now I'm stuck because an integral turns either to infinity or zero, can somebody tell me if rho+=2R.sin(theta)/cos(phi) ?
There shouldn't be any fractions here. In the spherical polars, we use, say, the south pole as our origin. By the way, g(R,R) should be 6R/5 in the three-dimensional case.
@@mathemaniac thanks! I was placing the origin at the "west" of the sphere, maybe that was it
I dont understand how the average Distance in the unit Ball tends to sqrt(2) if the Volume of an n-dim. ball tends to 0?
You can imagine the unit ball in higher dimensions are "pointier", and so even if the volume tends to 0, distances between points can still grow.
@@mathemaniac yeah, I think that makes sense to me. Thanks!
„I solved it SIMPLY“
Impressive
This is great
Thanks!
Why is the avg. distance doubled when the radius is doubled? This is intuitive, but there is no solid proof or explanation.
That's one sexy result.
Indeed!
Essence of ring and field please
Thanks for the request! Currently, I couldn't find a unique enough perspective on rings and fields to make videos on them (I have also avoided symmetric groups for my series on group theory for this reason). I would make videos on those if I could find a good enough perspective that nobody has talked about.
@@mathemaniac ok, thanks
Seeing this integral made me cry. Plz no. Be gentle
But the content of the video is hopefully simple - we are not computing it in the usual way, and there is a very clever trick to do this!
I love your video, sorry im late :(
There is no need to be sorry :)
genius.
"You won't really be able to do the whole integral"
aight bet
⭐️
Hey buddy , I actually tried solving the exercise you gave , that is to try and find the average distance between 2 points on a sphere/ball , I got a triple integral because of the presence of 3 variables in the spherical coordinates , and I am very very close to the correct answer , in fact I am getting 9R/35 as my answer which is precisely 1/4th of the correct answer . Now I don't really know what's the source of my error , it can be either a calculation error (I checked my calculations at least 3 times) or something else . So umm can you help me rectify my error ? And the hardest part for me is that forget multivariable calculus , our school haven't even yet started integral calculus ! (I am self-taught by UA-cam and books)
SO SORRY FOR BOTHERING YOU , it was indeed a calculation error , except this everything was smooth sailing thanks to you :) Also , subscribed !
I got lost at the cos(theta)d(theta) = d(sin(theta)) step. What logic is this?
It is the usual u-substitution.
idk what this is but pretty cool
Glad that you enjoyed it!
Explain how scientists calculate π value
I made a video about that - "How is pi calculated to trillions of digits" a year ago. Check that out if you want.
Please I need a help in solving integration problem can u help me
Elegant
Yes, this trick is really elegant, which is why I have to share on UA-cam!
この動画は作戦実行当日動画だと考えてます。
NICEEEEEE!!!!!
Yes, the trick is so nice!
Was anyone able to solve the full integral?
So many midroll ads :(
MindYourDecisions has a similar video on finding the distance between two points in a s̶p̶h̶e̶r̶e̶ square (youtube[dot]com/watch?v=i4VqXRRXi68). But so far these are the only two videos I could find solving this exact problem. All others seem to be about distances from the origin and the square of distances between two points since they're easier.
I didn't fully watch the previous videos, but I'm not sure if the shape being a circle makes the problem easier or harder. I found another video too that deals with the same problem, but with a square.
I'll try solving this problem for a sphere myself :)
MindYourDecisions made a video on the *square*, not a sphere, right? I do know that video, and it is because of the fact that Presh kind of assumed the knowledge of pdf and the Jacobian seemingly out of nowhere, that I made a video series addressing those concepts in the previous videos so that everyone is on the same page. If you have already learnt all these, you don't need to watch the previous videos.
I would say that the shape of the circle is much more difficult at first glance, because the limits of integration are more complicated, and even if we change to polar coordinates, there would be cosine inside a square root, and so there is really no hope of tackling the definite integral directly.
However, because of the rotational symmetry of the setup, the method used here is more suited to the circular case than the square case, but it is, I would argue, much more difficult to come up with this sneaky trick. I mean, at least, I can't think of this method at all.
The paper listed in the description also addressed the problem for a square, and they also said "and you thought the integral for the square was tough to evaluate directly!", so the circle case being more difficult is not only my opinion haha :)
@@mathemaniac yeah I meant *square. I was watching an MIT Lecture on the average distance on a sphere and that's when I remembered your video and started looking for others who tried to do the same problem.
Yeah Presh did assume some knowledge out of nowhere so props to you for including them. And yes, it does seem like circles are more difficult despite the symmetry.
But a wild guess - since you're integration for a circle, couldn't you try integrating with root of a cosine with ellipting integrals, for theta from -pi/2 to pi/2? Looks like there can be an interesting solution to me. Like how Presh did, by reducing the independent variables to the distances between points instead of their coordinates.
HOLY SHIT
This channel scares me. I'm scared.
So, a complex 15 minute video is called "Solved Simply"?
Or is it because I am not interested in circles and all the information I have on circles is just from school(which is very less, they don't teach anything useful)? I've never watched a maths video so helplessly, my brain is feeling insulted. I'm learning about circles now!(or after a month, haha).
"Simply" is obviously in the eye of the beholder - a toddler can't see these maths as simple. What I mean is that for people who have learnt integral calculus, and therefore understand the problem at hand, would find this approach a lot easier than expected.
All you need to know about circles is the area of it, and how to derive the length of a chord. Nothing more!
@@mathemaniac How to derive the length of a chord, what does that mean? Deriving the length function?
@@shambhav9534 No I mean just finding the length of a chord. It honestly shouldn't be too difficult.
I said no more reaserch ....bye
Wrong! My first instinct is to jump off of a building.
I couldn't get the documents.I dont have a subscription. Can someone please send them to me ? Respond and Ill give you my email.