Very interesting use of complex numbers! One of my favorite unexpected applications is using generating functions (along with partial fractions involving roots of unity) to find an asymptotic solution to the McNugget problem. Some more whose concepts formed part of the foundation of modern math are using unique factorization is Z[sqrt(-2)] to solve x^3 = y^2 + 2 (the unique solution is x = 3), and using contour integration to solve real integrals. Another more whimsical example I'm fond of is investigating the Taylor series of 1/(2 - sin x). The signs of the coefficients appear to be periodic with the pattern +, +, +, +, -, -, -, -, - of period 9; however, the first discrepancy in this pattern occurs after almost 100 repetitions!
I was wondering why the double pendulum wasn’t stopping, and I looked at the differential equation and yep, didn’t see any first-order terms that would be damping it. This was a really neat case of your symbols-to-ideas argument, the math itself told me “this pendulum won’t lose any energy”
The proof I remember from The Book is more graph theoretical. Put the rectangle into a coordinate system so that the bottom left corner of the main rectangle is on integer coordinates, and the sides are parallel to the axes. Put a vertex at every corner of every small rectangle and at the center of every small rectangle. Put an edge from each corner-vertex with at least one integer coordinate to all the rectangle centers for which is it a corner. Now note that for each small rectangle center, it is connected to 0,2 or 4 corners. And each corner vertex that isn't a corner of the large rectangle is connected to 0,2 or 4 centers. The only vertices that could possibly have odd degree are the corners of the large rectangle. The handshaking lemma now finishes the proof.
Yeah I remember that! Amazing how different that proof was from the one in the video. The author also gave a “checkerboard” proof that was visually intuitive.
An alternate proof in the same flavor uses the idea of "rollability" of shapes in the plane. Thomas Kern has a good video about it that essentially outlines the proof.
A really great video! I am shocked you haven't gotten so many more likes. I haven't seen too many proofs where you can say either this part of the equation is 0 or this part of the equation is 0, so we know a property holds for one of the two dimensions.
If this is really in "proof from the book", i might need to give the book another try. My first glance thought it didn't explain things well, and had way too much graph theory
if you ask for weird, interesting things to explain - here is one: the Kasteleyn formlual for the number of domino tilings contains cos() and pi. where does the cos() come from in a combinatorial looking problem which is about 2*1 dominos on an integer sized rectangle? why is the result an integer?
Seems obvious to me. If one side or another has integer length then you can always trace out an "integer length path" so that the projection on one side is integer. Do not even need complex numbers. E.g., start at the corner lower left corner, U or R for which ever is an integer. Then you end up with some pattern of U's and R's all of which correspond to integer sides. When you end up at the "upper right corner" you will have traversed a number of edges that are integers in length. The pattern of U's and R's are precisely choosing the ones with integer lengths and is always possible by hypothesis. This problem is very basic by the fact that every inner triangle has at least one side that is an integer. You simply always choose that side as you work your way from one corner to the other and since it is always possible to choose an integer U or R it is always possible. The only snag is that you might not end up at the upper corner before you have exhausted the traversal but that is irrelevant since you will have ended up traversing at least one side when projected. Knowing this you could also look at the traversal in terms of the Euler exponential since we could represent it as a "winding number". One could do this in several ways such as transforming the shape in to a polar form and effectively transform integer sides in to multiples of 2pi. This does not really seems like a problem of any complexity. Am I missing something?
Thanks for the comment. I think the argument doesn’t quite work (though it took me a minute to see why). One clue that it’s flawed is that it “proves too much”; in this case it “proves” that both sides of the big rectangle must be integers! Pause the video at 0:03 (where it shows a sample tiling). If you try to traverse from the LL to the UR, you might go up “one rectangle” and then to the right “one rectangle”. At this point, you’re stuck. You can’t go right, and if you go up, there’s no guarantee that this length is an integer. (For one thing, this might not be the integer side, but also, you’ll only be traversing a portion of that side.) Does that make sense? -David (the voice on the video)
@@davidfinance If I understand you then no. The reason is that you are always guaranteed that you are traversing along integer edges. what you are saying is that "maybe you will then reach a point where you cannot go along an integer edge" by by hypothesis we always have the choice to go along some integer edge. Look at it this way. Every vertex except corners have either 3 or 4 edges coming from them. One edge will always be where we are coming from. In the 3 case then we have only 2 edges to leave and we are ALWAYS guaranteed that one is an integer(well, it is part of a side of a rectangle that is an integer). In the 4 case, it's just one extra path so we still have at least 1 possibility. What you will find in your argument is that, yes, if you just arbitrarily go through some paths you could have issues. This happens if BOTH sides of a rectangle are integers and we traverse an entire edge of the rectangle. But this will never actually happen if we do not skip vertices. While I could be wrong(I haven't technically proved it). I think I'm right. I think if you actually think through it that you will find that you can always make the right decision. It works out because there will always be a proper choice. The algorithm is extremely simple: if (R & U) are integers (try both) if R is integer go right if U is integer go up The complicated issue is when both paths are integers but this will always end up falling through eventually and then one will have one or the other. So this algorithm can only fail when you can't go right or up cause there is no integer edge, right? Which was your issue. If there are no rectangles with both sides integer then the first if never happens and we can just linearly walk through and we will just end up with some sequence of U's and R's. But if we do have such an integer rectangle then we will have to try both paths to see BUT it is easy to see that inductively it will return a valid path because the algorithm will always return U or R no matter what. Because we are guaranteed that we have either R or U to traverse the algorithm never fails.
@@davidfinance Oh, and it doesn't prove that both sides are integers. My suggestion is to try and find a counter example. it should be easy if I'm wrong. A computer program should easily be able to generate the squares to be used. In fact, if there is no path then the outer rect does not have integer sides.
@@MDNQ-ud1tySorry, I really don't follow. If you go to 0:03, start at the lower left, then go up the full height of that LL rectangle, and then to the right the full width of that same rectangle, you end up at a "3-junction". The left path is where you came from (so that's no good), and in your algorithm moving down is not an option. So you have to move up. But that's a partial rectangle edge so there's no guarantee that it's an integer. Right? I think you acknowledge that possibility but I don't understand how you address it.
@@davidfinance You would not do that. You are failing to understand that once you did that you would have to backtrack(which is what the recursion would do) and you can take a different path. The point isn't that you would end up at a "dead end" but that they exists a path that you can always take. You should give your specific path argument: URR is the path I think you mean which outlines the LL rectangle. If we assume UU is dead(else easy), URR is dead(assumption), and URU is dead(again easy) then there is a path that exists because either we could have went RR or the "dead end" is not dead because the intersection in URR is actually an integer as even though it hits in between the side of a rectangle it must hit it at an integer point if the enter side is an integer(since the difference of two integers is an integer). So no matter how you slice and dice it there will be some path that you can work your way through. If you ever get "trapped" in some dead in where you can't then there was some other path to take and if you traverse all paths then you will find one. In fact, if you want to allow for backtacking it too will work because the UDLR strings will represent integer addition and subtraction which is still an integer. I'm not saying the algorithm I'm explaining is perfect(it's pseudo code) but that if you have such an algorithm where it traces out the paths you can always find such a path because there is always integer sides you can follow. Even in the URR path, if it seams like a "dead" end, it is not because it will be equivalent to simply doing RU That is. URR = RU if the LL is an integer rect and so URRU = RUU too (which may not be an integer but if it is not then it means we should have went RRU or RRR. It is basically a maze that will always have a solution because because the paths are representing sums of integers. e.g., the outcome of any sequence represents an integer "solution" up to that point. If it is not the final UR corner and one cannot progress further then there is another path that would necessarily have to exist. Once simply has to setup the argument correctly and prove it using induction. It may not be that simple but it shouldn't be too complex.
This kinda feels similar as using complex variables in electronic circuits. Voltages and Amperes are not imaginary. But using imaginary arithmatic for capacitors and impedances works it out!
Very informative video thank you. In seconds 3:19 while expaining 2pi(x+y) equals to arc of circumference I would like ask this question. Let x=0 and y=1, then we should end up the complex number is i. But according to the formula 2pi(0+1)=2pi, we are ending up in 1. Am I missing something please help me.
h is a complex valued function. Input s are x,y which are in a plane, output h is in a different plane. The mapping is the function. So (0,1) point in input plane is mapped to (1,0) in the output plane
Very interesting and beautiful. A small bit of feedback - some people find music in the background really distracting. I am afraid I had to stop about 3/4 in once I could see where you were going. The music had become unbearable by that point. Most people won't notice the music. I suspect that you will lose more people than you gain by having music and would suggest that you don't have backing music. You are interesting enough not to need it.
Neither Sean nor I came up with this proof, but I think the idea is that you want to define a function that has a period of 1 in both the x and y directions and an average value of zero over its period. That way you can easily reason about its behavior over a range when either the width or height is a multiple of 1. The book has an alternate proof with a similar idea using a checkerboard where the squares have length 1/2.
Very interesting use of complex numbers! One of my favorite unexpected applications is using generating functions (along with partial fractions involving roots of unity) to find an asymptotic solution to the McNugget problem. Some more whose concepts formed part of the foundation of modern math are using unique factorization is Z[sqrt(-2)] to solve x^3 = y^2 + 2 (the unique solution is x = 3), and using contour integration to solve real integrals. Another more whimsical example I'm fond of is investigating the Taylor series of 1/(2 - sin x). The signs of the coefficients appear to be periodic with the pattern +, +, +, +, -, -, -, -, - of period 9; however, the first discrepancy in this pattern occurs after almost 100 repetitions!
I was wondering why the double pendulum wasn’t stopping, and I looked at the differential equation and yep, didn’t see any first-order terms that would be damping it. This was a really neat case of your symbols-to-ideas argument, the math itself told me “this pendulum won’t lose any energy”
The proof I remember from The Book is more graph theoretical. Put the rectangle into a coordinate system so that the bottom left corner of the main rectangle is on integer coordinates, and the sides are parallel to the axes.
Put a vertex at every corner of every small rectangle and at the center of every small rectangle. Put an edge from each corner-vertex with at least one integer coordinate to all the rectangle centers for which is it a corner.
Now note that for each small rectangle center, it is connected to 0,2 or 4 corners. And each corner vertex that isn't a corner of the large rectangle is connected to 0,2 or 4 centers. The only vertices that could possibly have odd degree are the corners of the large rectangle. The handshaking lemma now finishes the proof.
Yeah I remember that! Amazing how different that proof was from the one in the video. The author also gave a “checkerboard” proof that was visually intuitive.
This is really beautiful, and very well explained. Thanks for sharing it.
Incredibly elegant! Love your approach. 😉😉
An alternate proof in the same flavor uses the idea of "rollability" of shapes in the plane. Thomas Kern has a good video about it that essentially outlines the proof.
A really great video! I am shocked you haven't gotten so many more likes. I haven't seen too many proofs where you can say either this part of the equation is 0 or this part of the equation is 0, so we know a property holds for one of the two dimensions.
Great working with you, Sean!
Beautiful!
If this is really in "proof from the book", i might need to give the book another try. My first glance thought it didn't explain things well, and had way too much graph theory
if you ask for weird, interesting things to explain - here is one: the Kasteleyn formlual for the number of domino tilings contains cos() and pi. where does the cos() come from in a combinatorial looking problem which is about 2*1 dominos on an integer sized rectangle? why is the result an integer?
Geometry is usually more visual appealing than its underlying algebraic equations.
Seems obvious to me. If one side or another has integer length then you can always trace out an "integer length path" so that the projection on one side is integer. Do not even need complex numbers. E.g., start at the corner lower left corner, U or R for which ever is an integer. Then you end up with some pattern of U's and R's all of which correspond to integer sides. When you end up at the "upper right corner" you will have traversed a number of edges that are integers in length. The pattern of U's and R's are precisely choosing the ones with integer lengths and is always possible by hypothesis. This problem is very basic by the fact that every inner triangle has at least one side that is an integer. You simply always choose that side as you work your way from one corner to the other and since it is always possible to choose an integer U or R it is always possible. The only snag is that you might not end up at the upper corner before you have exhausted the traversal but that is irrelevant since you will have ended up traversing at least one side when projected.
Knowing this you could also look at the traversal in terms of the Euler exponential since we could represent it as a "winding number". One could do this in several ways such as transforming the shape in to a polar form and effectively transform integer sides in to multiples of 2pi.
This does not really seems like a problem of any complexity. Am I missing something?
Thanks for the comment. I think the argument doesn’t quite work (though it took me a minute to see why). One clue that it’s flawed is that it “proves too much”; in this case it “proves” that both sides of the big rectangle must be integers!
Pause the video at 0:03 (where it shows a sample tiling). If you try to traverse from the LL to the UR, you might go up “one rectangle” and then to the right “one rectangle”. At this point, you’re stuck. You can’t go right, and if you go up, there’s no guarantee that this length is an integer. (For one thing, this might not be the integer side, but also, you’ll only be traversing a portion of that side.)
Does that make sense?
-David (the voice on the video)
@@davidfinance If I understand you then no.
The reason is that you are always guaranteed that you are traversing along integer edges.
what you are saying is that "maybe you will then reach a point where you cannot go along an integer edge" by by hypothesis we always have the choice to go along some integer edge.
Look at it this way.
Every vertex except corners have either 3 or 4 edges coming from them. One edge will always be where we are coming from.
In the 3 case then we have only 2 edges to leave and we are ALWAYS guaranteed that one is an integer(well, it is part of a side of a rectangle that is an integer).
In the 4 case, it's just one extra path so we still have at least 1 possibility.
What you will find in your argument is that, yes, if you just arbitrarily go through some paths you could have issues.
This happens if BOTH sides of a rectangle are integers and we traverse an entire edge of the rectangle. But this will never actually happen if we do not skip vertices.
While I could be wrong(I haven't technically proved it). I think I'm right. I think if you actually think through it that you will find that you can always make the right decision. It works out because there will always be a proper choice.
The algorithm is extremely simple:
if (R & U) are integers (try both)
if R is integer go right
if U is integer go up
The complicated issue is when both paths are integers but this will always end up falling through eventually and then one will have one or the other.
So this algorithm can only fail when you can't go right or up cause there is no integer edge, right? Which was your issue.
If there are no rectangles with both sides integer then the first if never happens and we can just linearly walk through and we will just end up with some sequence of U's and R's.
But if we do have such an integer rectangle then we will have to try both paths to see BUT it is easy to see that inductively it will return a valid path because the algorithm will always return U or R no matter what.
Because we are guaranteed that we have either R or U to traverse the algorithm never fails.
@@davidfinance Oh, and it doesn't prove that both sides are integers.
My suggestion is to try and find a counter example. it should be easy if I'm wrong. A computer program should easily be able to generate the squares to be used.
In fact, if there is no path then the outer rect does not have integer sides.
@@MDNQ-ud1tySorry, I really don't follow. If you go to 0:03, start at the lower left, then go up the full height of that LL rectangle, and then to the right the full width of that same rectangle, you end up at a "3-junction". The left path is where you came from (so that's no good), and in your algorithm moving down is not an option. So you have to move up. But that's a partial rectangle edge so there's no guarantee that it's an integer. Right? I think you acknowledge that possibility but I don't understand how you address it.
@@davidfinance
You would not do that.
You are failing to understand that once you did that you would have to backtrack(which is what the recursion would do) and you can take a different path.
The point isn't that you would end up at a "dead end" but that they exists a path that you can always take.
You should give your specific path argument:
URR is the path I think you mean which outlines the LL rectangle.
If we assume UU is dead(else easy), URR is dead(assumption), and URU is dead(again easy) then there is a path that exists because either we could have went RR or the "dead end" is not dead because the intersection in URR is actually an integer as even though it hits in between the side of a rectangle it must hit it at an integer point if the enter side is an integer(since the difference of two integers is an integer).
So no matter how you slice and dice it there will be some path that you can work your way through.
If you ever get "trapped" in some dead in where you can't then there was some other path to take and if you traverse all paths then you will find one.
In fact, if you want to allow for backtacking it too will work because the UDLR strings will represent integer addition and subtraction which is still an integer.
I'm not saying the algorithm I'm explaining is perfect(it's pseudo code) but that if you have such an algorithm where it traces out the paths you can always find such a path because there is always integer sides you can follow.
Even in the URR path, if it seams like a "dead" end, it is not because it will be equivalent to simply doing RU
That is. URR = RU if the LL is an integer rect and so URRU = RUU too (which may not be an integer but if it is not then it means we should have went RRU or RRR.
It is basically a maze that will always have a solution because because the paths are representing sums of integers. e.g., the outcome of any sequence represents an integer "solution" up to that point. If it is not the final UR corner and one cannot progress further then there is another path that would necessarily have to exist.
Once simply has to setup the argument correctly and prove it using induction. It may not be that simple but it shouldn't be too complex.
Very nice!
Why? Why some amazing youtubers and channels just stop making great videos? And do many bad youtubers and channels keep going??? Why, God? Why???
Do you not see the #SoME2 hashtag on the title? Many people just made a video or two for the competition
This kinda feels similar as using complex variables in electronic circuits. Voltages and Amperes are not imaginary. But using imaginary arithmatic for capacitors and impedances works it out!
sir how you came up with the idea of defining such a function and then using it in such an ingenious way
Very informative video thank you. In seconds 3:19 while expaining 2pi(x+y) equals to arc of circumference I would like ask this question. Let x=0 and y=1, then we should end up the complex number is i. But according to the formula 2pi(0+1)=2pi, we are ending up in 1. Am I missing something please help me.
h is a complex valued function. Input s are x,y which are in a plane, output h is in a different plane. The mapping is the function. So (0,1) point in input plane is mapped to (1,0) in the output plane
Very interesting and beautiful. A small bit of feedback - some people find music in the background really distracting. I am afraid I had to stop about 3/4 in once I could see where you were going. The music had become unbearable by that point. Most people won't notice the music. I suspect that you will lose more people than you gain by having music and would suggest that you don't have backing music. You are interesting enough not to need it.
Thanks for the feedback about the music. I'm starting to think that most math videos should be music-free.
Can you share video code?
Cool effects, Great looking animator, soothing voice, nice idea.
for the next video maybe more explosions and animations and less math.
Do another
Beautiful!
sir how you came up with the idea of defining such a function and then using it in such an ingenious way
Pls reply
Neither Sean nor I came up with this proof, but I think the idea is that you want to define a function that has a period of 1 in both the x and y directions and an average value of zero over its period. That way you can easily reason about its behavior over a range when either the width or height is a multiple of 1.
The book has an alternate proof with a similar idea using a checkerboard where the squares have length 1/2.