Why We Care About Functions
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- เผยแพร่เมื่อ 24 ก.ค. 2024
- Made for 3Blue1Brown's Summer of Mathematical Exposition #SoME2
Many people are taught that functions are mathematical objects which take in an input, and somehow transform it into an output. This understanding only views what functions "do" to individual inputs, and while not wrong, we think we can do better!
In this video, we explore a global understanding of functions as objects which don't so much transform inputs into outputs, but instead simply connect things in the input space to things in the outputs space. Armed to the teeth with this new perspective, we show that mathematicians really care about functions because we can use their connections to transfer knowledge about one set into another set and vice versa. And you won't find a more fun example of that outside of our attention-getting puzzle about an impossible shape!
/ chillaxiommath
Chapters
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0:00 Intro
1:17 What is a function?
5:06 Back to the Grid problem
6:07 The function
10:49 Outro
Theres a way to do this using probability:
The expected number of lattice points you overlap is the same as the area You can show this by chopping it up into squares, and this doesn't change the expectation (because of linearity of expectation), and the expectation for a square can be shown to be its area.
But since the area is less than 1, the expected number of points is less than 1, so 0 must be possible.
Funnily enough my solution is a combination of 2 summer of math exposition videos i watched, one on buffon's noodle problem and another on proving things using probability.
I saw those videos too, but never put the connection together! That’s a great idea. This is one of those claims that’s so hard to prove directly, it almost requires a beautiful proof. Who knew would be two though?
This is the most mathematically beautiful TH-cam comment I have ever seen.
This works the other way too. If you have a shape with area greater than n, there's always a way to cover at least n+1 points with integer coordinates.
I'm not understanding the chopping up part and the expected value of a square part. Can you provide more detail please?
That was my thought too, because I saw it recently for a different problem. But the solution here lets you actually *find* the solution, not just proves that one exists somewhere, so I like it better.
I can't begin to tell ya how much I love the approach-ability of this concept because of how you attack it.
Me too! I never heard this before a few months ago and thought it was too good not to share
8:00 And as breaking up and rearranging the parts of a shape cannot magically add any area…"
Banach & Tarski would like to have a word with you…
Haha touche! I should have said you cannot non-magically change the area, because with Banach and Tarski’s axiom of choice witchcraft anything is possible! Luckily we are dealing with “measurable” parts of the shape for this proof, so we don’t have to worry too much about these issues here
That's what you get for taking the Axiom of Choice. Put that back, you don't know where it's been. :P
BT isn't a counterexample to this statement. indeed, you can't assign an area to the BT pieces
Man, I love proofs like these, where you transform something as mundane as a square on an infinite grid into something unexpected. Like a torus or a room with bouncy / mirror walls (i mean that billiards problem where you try to protect a target from getting hit, by placing obstacles in the path of the bullet / laser beam. PBS Infinite Series did a video on that). I love topology and I think these problems are often related to it, right?
Also I'm happy to see you are still making videos and with such high quality, you definitely deserve recognition, I hope SoME2 will help with that 😊
Always love the support and suggestions thank you so much! And say no more, Infinite series was absolutely epic and ended before it’s time. Channels like Kelsey’s infinite series were a big inspiration which led to me switching to a math undergraduate degree which has then led to a masters and now a PhD! I hope to make more videos and maybe inspire more people others!
Fantastic video. I like how the way you described the solution applies to the general setting of sets with lebesgue measure
Thank you! As some other commenters pointed out, it can be applied to even more situations than that, I just didn’t have time to think through the limit of its generality. Certainly any set contained in a Lebesgue Measurable set of measure less than 1 would also work. But also if we allow for rotations then things get really interesting and we can actually start to do this with areas great than 1 as well
Awesome video! Approachable, engaging, clear and accurate. Those are excellent qualities for a mathematics video. Keep up the good work!
This was so cool! Great video, define definetly sharing with all of my friends! Please make more videos like this one!
Great Video! Love the surprising but beautiful solution. I hope you continue doing works like this!!!!
What if the area is exactly one? It seems like there will always be a rotation where there's some overlap, but I'm not sure how to prove that.
Do you mean that for any shape area 1 there exists a rotation, without requiring a translation, that causes that shape to overlap a grid point?
If so, I don't think a circle with area 1 centered around the center of the grid cell, (1/2,1/2) could be rotated to touch the grid, as the points where the circle intersects the line between center and grid point, is roughly at (0.8989, 0.8989), for the upper right corner, likewise for the other 3, so within the boundary of the gridcell. Kinda has to be as the circle escapes the unit square at the parallels to the grid going through the circle's center, due to it's radius being about 0.56>0.5, so that area has to be missing somewhere in the square and being a circle that has to be around the edges of the square.
IIRC, you need at least area 2^2 in 2D and 2^3 in 3D to guarantee the shape touches the grid
This is a phenomenal question! This proof is a handsome way to show that any shape of area one can be moved to avoid the corners, but clearly if we allow for rotations we can allow for larger area shapes. As of now, I still don’t know what the max area is for which we can guarantee there is a translation and rotation which works, so if you have any ideas please let me know!
There are clearly area 1 shape that always hits a grid square(namely the grid square itself) under translation, but there are lots of arbitrarily horrible shapes of measure 1 that could be positioned to not hit any corner(in fact, any shape with any points in the reference cell not in the image can be made to do this, area
OMG, there videos are amazing! Super well-made, congrats!
Great Approach and Fantastic logic!
absolutly outstanding video! thank you
While functions may have been introduced in eighth grade, I don't think I fully grasped the definition until calculus, and even more, I don't think I mastered it until group theory.
Really nice video.
this is so cool. you need more subs
Lovely!
"It's all about finding X"
Good one!
Though really they lure you in with finding X and then you have to deal with sets of rules applies over things that are also sets of rules AAAAAAA
I think there is an even more fundamental function hidden around here: the one which associates to every measureable set A its area area(A). Without it we couldn't even STATE the problem, let alone solve it...
You should definitely make more videos like this
Wow. That’s actually an insanely clever proof.
Just wondered, how would one encounter the problem of identifying whether it is possible for shapes with areas larger than 1 as a function of internal properties of the shape?
EXCELLENT question. There may be some dusty paper out there in the archives somewhere explaining this, but at the moment it evades my googling abilities. From my perspective right now it’s an open problem that I would love to get to the bottom of!
Well, if passing it through the function used in this video yields a shape with area less than 1 (i.e. there is unused space) then there must be a way, using the same procedure. Not sure about what to do if it covers the whole box though...
7:01 "give me your lunch money" that was hilarious
Nice video! Thought this was minkowskis theorem from the thumbnail which uses a similar mapping
Cutting and rearranging things can't increase the area? What do you have against the Axiom of Choice? (see 'Banach-Tarski Paradox').
cutting and rearranging MEASURABLE sets doesn't increase the area. B-T paradox deals with non-measurable sets (you can't assign an area to the pieces)
8:00 [Laughing in Banach-Tarski]
great popularization work, don't mind me poking at the devil
B-T is not a counterexample of that statement. the culprit is that you cannot assign an area to the B-T pieces
i feel like there is one last step left: if you fill out the box completly, it will also have area one, but touch all four corners. but as you can rotate it by 45°, it doesnt break the proof
what about the graph of the function y = pi*x?
Is there an analogous proof for a shape of larger area when we also allow rotation?
EDIT: I suppose it can't be more than pi * (sqrt(2) / 2)^2 ≈ 1.57, because that would be a circle that can overlap 4 grid points and can thus never be taken off because of rotational symmetry... Not sure if thst is the lower boud
This is a really great question! Definitely the max area shape we can guarantee has a translation + rotation that avoids all the corners is somewhere between 1 and pi/2. The exact cut off though is a mystery to me, I would love to find out!
@@chillaxiommath a 1 by sqrt(2) rectangle will also intersect the lattice no matter how it's translated or rotated, so the answer is at most sqrt(2)
Once you have a 1 by √2 rectangle, i'm pretty sure you can cut of about 0.175 from the corners (as in you cut a right isoceles triangle with short edges 0.175). the actual optimal cut is some curve that i don't know yet.
so you can save 0.175^2/2 per corner so 0.06125 area saved, so about an area of 1.35.
The reason I think trimming off the corners works is that as long as parts of the old edges are involved, cutting this amount of corner is fine.
Here's a desmos graph showing the shape and why I think it works.
Click play on the "a" slider.
www.desmos.com/calculator/x6k9sodcgw
You can see that when a is out of the range of the slider the cut corners effectively don't matter.
So only one cut corner ever matters, and if thats the case then it definitely works.
You can move the red polygon by dragging its center and changing a2 if you wanna check it.
I think that if you allow your shape to get very "large" (as in it only fits in a massive circle), you can get the area as close to 1 as you want while still being guaranteed to hit a point.
My reasoning is that giving some finite set of rotations, you can probably find a shape that has to hit a point at each of those angles. Like if you only allow 45 degree rotations, it's definitely possible to make something with an area of 1 that has to hit a point. (it is disconnected, but you can arbitrarily small connections between each part so they add basically no area)
It seems true that if you add more and more rotations, you'll be able to find a new shape with area 1 at each step.
Though this only covers a finite set of rotations, it seems true that you only need a small bit more area to make it cover all areas, hopefully some amount that you can get as small as you want by adding more rotations.
arent there space filling curves though?
2:34 is a bit misleading. yes a thing on the left connects to only one thing at the right, but two things left can connect to the same thing right (any polynom with even degree is a good example).
What a cutie of a proof that is~
I like that sentence for your Country. People forgot they have a team to play for.
Make a square of exactly size 1. Then remove one point at one of the corners. The area is then one infinitesimal less than 1, but the only available point is the corner, meaning you will not get the required offset.
Or is 1 - infinitesimal not less than 1 in this case?
if you remove a single point the area is still 1. where did you get such a strange idea?
what about the space filling Hilbert curve? a line has 0 area, yet the Hilbert curve allows it to map all of 2d space, including the corners. it has less than 1 unit of area, yet you can't find a modulo point P that is not on the Hilbert curve
The Hilbert curve, despite being a line, does not have 0 area. It's not alone in this: if you look up Osgood curves, you'll find they also have nonzero area, despite also not intersecting themselves, unlike the Hilbert curve. (You may be surprised to hear that the Hilbert curve intersects itself. In fact, the points 1/6, 1/2, and 5/6 along the Hilbert curve are all the same point, in the center of the square. Mathematically, H(1/6)=H(1/2)=H(5/6). Note that the Hilbert curve doesn't refer to any of the finite iterations but to the infinite limit. The finite iterations (made of line segments) don't intersect themselves, but the infinite limit does.)
Moral: despite what your intuition tells you, not _all_ lines have 0 area! _Smooth_ lines do, but many _fractal_ lines do not.
@@columbus8myhw i honestly didn't know the Hilbert curve has area or intersections. i was aware of the iterations thing. i did know fractals can do weird things to properties, even if constructed using iterations of integer dimensions, but didn't know they could give area to a line or stuff like that. i did know it could make 0s or infinities despite tied properties still being finite like the menger sponge having infinite surface area, 0 volume, and a non-zero finite bounding box
So how can I represent these steps with equations such I rigorously prove this?
*such that I can rigorously prove this
Oooo excellent question! So the actual proof proof will be a bit tedious, but will follow exactly what is outlined here. The first step is we will need to show that the function is “at most area preserving”. To do that we just need to show that the function restricted to each box in the grid is area preserving. Once we have that it is a famous result that the union of the outputs from each box, will have area less than or equal to the sum of the areas from the outputs of each box, and this remember will equal the area of the original shape.
Then call the total output shape S and the reference box B. As m(S) < m(B), where m is the Lebesgue measure which corresponds to area, S is a proper subset of B. This there exists a P = (u, v) in B that is not in S. Now consider f^-1(P), the pre image of P. By the algebraic definition of the function we can show that the pre image of P, i.e the set of all inputs that map to P, will be {(u + n, v + m) | n, m in Z}. Note if we translate every element of this set u down and v left you would get {(n, m) | n,m in Z}, so this is just an offset grid. Now let T be our original shape which when passes through the function outputs to S. If T intersected anything in the pre image that would mean there is a (u + n, v + m) in T. This, f(u + n, v + m) would be in f(T) = S. And remember by definition of pre image P = f(u + n, v + m), so this means P is in S! But this contradicts the fact that P is chosen not to be in S, so we must have that nothing in the pre image of P intersects T. Hence if we move the grid up u and left v each corner would avoid the shape T.
Now the BIG question from all of this is, can we show the function restricted to each box is area preserving. That takes some measure theory to show, but it’s very possible, and I would be happy to post a more in depth solution going over that too!
Here's my proof:
Let's cut the shape into pieces based on whole parts of its coordinates in some grid. A shifted version will be what we need iff none of the parts have corners. Then, let's shift them all into one square. Since we shift them by whole amounts, a shifted version of the original shape will be what we need iff the union of the parts doesn't contain a prototype of a corner. But the only one it can contain is the one within the square we shifted them into. We can choose it arbitrarily from the square, which has more area than the original shape and therefore the union of the shifted parts, so we can choose one that isn't there
If you take the shape of the full box (Area =1) and remove exactly one corner from it, the area is less than 1, but I'm not seeing how you could satisfy the problem, as its touching 3 corners already and any movement away from them will cause the untouched corner to be touched.
A single point has zero area, and so removing one point doesn’t produce something with area less than 1.
the area is still 1... i think you should revisit some elementary math first
Hey, its Trevor! Just browsing SoME2 vids and found someone I knew, wow. Very cool problem.
Hey dude! What’s up?
@@chillaxiommath I, uh, had a mental breakdown and quit grad school, but other than that, excellent.
But so, what about Moore's Curve ??? It goes trought all points of a square space while having no area at all (like all fractals) 0 is less than 1 so it is like a counter example of this maybe ?
So, uh, fractals don't all have zero area (a trivial counter-example would be something like the interior of the Mandelbrot set). But I'm guessing you mean fractals 'constructed from lines'.
But you still need to be _very_ careful as to what you mean by 'area'. For the simplest argument I can think of: here we used Lebesgue measures to define area (it's noted briefly at the start of the video). This is a set-theoretic definition of area, and measures the area of 'a set of points'. I'm by no means an expert on the topic of the Moore curve, and I don't know _exactly_ what is meant by space filling. Firstly, I'll answer this according to the definition, I think you're using, and then I'll mention briefly why I'm uncertain, and why it might be argued in the opposite direction.
If we define the limit we use for the Moore curve in such a way that we say it hits every point (which feels non-trivial, in and of itself: sure, the distance of any point from the curve tends to zero, but for any finite curve length, the overall proportion of points in the unit square it contains is zero - you'll probably need some careful definition for how you're defining 'touching every point'), then every point our unit square in R^2 lies within the set of the points hit by the Moore curve. Since the Lebesgue measure describes area based on a set, what we've then argued is that the set of points hit by the Moore curve is the same as the set of points in our unit square. But if those sets are the same, both have Lebesgue measure of 1, and hence an area of 1.
As mentioned earlier, I'm not convinced that the Moore curve _does_ contain the same set of points as a unit square in R^2. In particular, it seems like it fills the unit cell in the sense that points on the Moore curve are _dense_ in the unit cell (i.e. your distance from such a point is always zero), but that's not the same as every point in the unit cell also being a point on the Moore curve.
In other words, the definition of area we're using is equivalent to discussing the number of points in our set. The only way to have a lower area is to leave out points. If you're saying the Moore curve has 0 area, then you're defining the set of points on it in such a way that it leaves out points (for instance, the point (0.5,0.5)). If you're saying it contains the same set of points than the unit square, then it _also_ has the same area.
I hope maybe that helps?
@@NinjaOfLU Oh wow! I didn't expect to get a comment this long about a dumb thing I said, I am no expert at all, I learn things on the internet here and there, I just happened to have seen a video about Moore's Curve which is a curve that loops back on itself while crossing all the possible points of a square (by getting the line longer and longer). I am in no way able to try to compete with any good mathematicians here.
And yes really, thank you for your response, you help me a lot with new concepts like this in maths.
Well-behaved Functions could map every input point to a singular output, because as long as there's just one answer to the operation we call it a function, all functions are automatically well-behaved, otherwise they'd not be functions.
x^2=y is a well behaved function but y=x^2 is not a function. As there are 2 solutions. As for y=4 x=-2 and x=+2 are both valid outputs.
Even though they are the same thing mathematically, however mapping wise, one maps x to y, and the other maps y to x.
Interesting, this is like a partial converse of Minkowski's theorem?
I never thought of it that way before, but you’re right! Would love to fill in the gaps to see what happens between areas of 1 and 4
See you next year.
Meander-ery, a fine word
This seems awfully similar to the inscribed square problem worded differently.
3:12 "As breaking up and rearranging [...] can't magically add any area..."
Banach and Tarski have something to say about that.
no they do not. Area in the plane has to be intended as Lebesgue measure, and the shape has to be measurable. then its intersection with the squares will be measurable too and sigma-additivity holds, from which you get the desired inequality.
I'm trying to picture some sort of fractal shape, as a way to thwart your logic.
Give us your worst! The great news about the proof is that we just need to find that one little point that sneaks through the fractal in the output shape and it all still works. And one thing to notice is that if the fractal shape ended up covering everything so there is no point that sneaks through then I think we would find the shape we started with would still have to have area >= 1. I thought the same thing when I first saw the proof though, it feels like magic
@@chillaxiommath I think I can cover every rational point with 0 area, but not every real point. Now I'm trying to use Banach-Tarski to double the area without doubling the area.
@@PhilipSmolen good luck with that
What about a point with an infinite number of infinitely thin spokes radiating out of it? That would technically be a shape with zero area that touches every possible point
The problem is that when you multiply infinity with 0 you can really get anything. You would really have to get into the details to prove the area is less than one.
Since the proof is quite clear, my guess is that as you make the branches longer, they get thinner in time to "dodge" the lattice points.
If it has every point in the plane it has infinite area. If you just mean all the lattice points, then you can translate it so that it doesn’t hit any of them anymore.
10:23 Did anyone else think this was a segue to a sponsor?
I'm pretty sure the 'intersection preserving' part isn't a thing. Consider that ANY function acting on the points of the plane would be 'intersection preserving.'
The Shape is at least a set, call it *S* . If f( *S* ) is simply all points *a* such that *a* = f( *b* ) for some *b* in *S* , well then there is a *q* in f( *S* ) such that f( *p* ) = *q* whenever *p* is a point in *S* (in other words, *S* 'intersects' with *p* ), and there you have it: *p* is in *S* , and f( *p* ) is in f( *S* ).
This entire line of reasoning has got nothing to do with the nature of the function - it's just one or two definitions from Set Theory.
This is so true! Any function defined from R^2 to R^2 like this will clearly be “intersection preserving” as I called it, so it’s not super special. BUT, it is important and in my experience sometimes it’s the simplest ideas that are the easiest to overlook. What I want to do with this channel is shine a light on the simple but brilliant observations that line the history of math, and hopefully make more people curious about what other hard problems are just one simple perspective shift away.
@@chillaxiommath So, perhaps a good tweak would be to say 'a nice fact about any function like this...' instead of 'an essential property of *this* function...'
No big deal, still a cool video overall.
A tiny bit disappointed that you did not define the image of a function, would have been a great fit while taking about shapes.
If you teach her to master your body and she gets your soul is she a compiler then? Is that the origin of the word compiler? Consciousness exists in the electrical and so the better question is did you traverse a wormhole? Broke da game bro Luv you.
I may have been using different rules when coming up with a shape of area less than 1 that will always intersect a grid square. My answer was an infinite line with an irrational slope. The line is infinitely thin, so its area is less than one, and since the slope is irrational, extending it long enough will eventually hit a grid square. When mapping this shape to the single box, the result would've been a solid square that seemingly has both 0 area and 1 area.
This is what I get by trying to solve this problem using infinity.
Why would the line intersect with a grid point just because its slope is not rational? In fact, it should not be hard to prove that any line with irrational slope can be placed in the grid while completely avoiding every intersection point (just make the grid finer and prove that the line can pass through at most one gridpoint in the refined grid, etc.)
"since the slope is irrational, extending it long enough will eventually hit a grid square". No: lines with irrational slope are exactly the ones which DO NOT interesect integer points apart from (0,0)! Indeed, if a line y=cx contains the nonzero integer point (m,n), then n=cm, which means that c=n/m is rational.
2:40 “What would we say f(G) is?” f(G) = {1, 9}
This is indeed often a useful thing to do, it just generalizes the notion of a function so isnt a function anymore. Or thinking another way, its a function but it doesnt map to the set S of integers 1 through 10, but to the set of all _subsets_ of S
That could be described by another mathematical concept called binary relation. Functions are just a subset of binary relations. The notation there would be more like this: Gf1, Gf7. It reads as "G is f-related to 1 and G is f-related to 7".
then that would be called a "multivalued function". it's a generalization of functions. people have enought troubles understanding functions, let alone multivalued ones...
Nice hair
🥹😊🥹😊🥹😊 Thank you! (if you run your eyes across those emojis really fast I’m supposed to be blinking away tears of gratitude)
U look like the Indian singer Sonu Nigam. Nice video though.