Tips to be a better problem solver [Last live lecture] | Lockdown math ep. 10

welcome back everybody it's hard to define exactly what mathematicians mean when they use the phrase problem-solving however you go about it it's going to involve some notion of approaching puzzles that you've never seen before and still being able to systematically and creatively find some solution to them but that's a weird thing when you think about it because it makes it a very hard thing to teach I mean you can teach someone how to solve one particular problem maybe even a class of problems and teach them how to solve another problem but how do you teach someone how to approach a problem that they've never seen before and still make progress on it well I honestly don't know how but what I want to do for this lecture is to talk through a couple different problem-solving principles I've enumerated nine of them in total and each one we're gonna walk through in the context of a specific example and I think each one is kind of simple you'll look at it and you'll nod along thinking yeah yeah of course that's the thing that you should do but I would argue that each one is deceptively simple that you would be shocked at how often you can make very meaningful progress in very hard problems just by keeping some of these tips in the back of your mind and to give you a little flavor for where we're going to be going today I want to ask not as a quiz that I expect you to necessarily be able to solve here on the spot I want to ask you a question that we will be solving later on in the lecture just so you can have it in the back of your mind a little thing to mull over any a hard problem that will be fun to tackle when the time comes so the question asks suppose that two numbers are chosen at random from the range 0 through 1 and it's done according to a uniform distribution so maybe you pick like 0.385 and point 5 8 9 6 2 or something like that each one is chosen at random suppose P is the probability that the ratio of the first number to the second rounds down to an even number so you know maybe it rounds down to zero or rounds down to 2 or to 4 and basically it's asking you to guess where is this probability you know we'll solve it exactly we'll get an exact expression but just intuitively as you hear the problem when you think of choosing two random numbers between 0 & 1 looking at their ratio what do you think that probability is going to be well return back to this later and one thing I want to say is that this whole live quizzing software it's something that's built by some friends of mine Ben eater who many of you may know because of his YouTube fame and then another person who used to work with us at Khan Academy named kam Kristensen and this is just one small little outcropping of what's actually a much deeper product at play and I want to give you a little preview of some of the other things that they've been working on that will be developing so if any of you want to share some of the lectures here or go back and kind of go through the live quizzing experience if you go to the link that it's in the description but it's at item pool comm /c / 3 b 1b but you can follow the description you can basically watch the lectures in a way where you can do these live quizzes along with it and your progress is tracked and you get scores and things like that so as you skip ahead to various different questions it will skip you to the right point in the video and as the video plays forward you'll also get to yeah there's just me talking through some problem you'll also get to whatever the appropriate part of the problem is they often have explanations associated with them and hints and things like that and ultimately if you look back in just a couple days I'm going to fill out like homework and challenge problems so if you like challenging problem solving contest math type stuff I'm gonna put things that are relevant to the lectures in there which I think should be fun it's not not quite there yet but definitely check back and I'm sure just over the next couple months there will be more item portion Anakin's going on if any of you want to use this maybe to do some live polling in your own live classes for teachers who are dealing with the whole remote landscape or if you have your own streams anything like that they are looking for beta users so feel free to reach out to them it should be available on the website now let's dive into the actual content shall we some of the nine deceptively simple problem-solving tricks and before we do I want to specify that at some point today in the next hour I'm going to purposefully make a mistake okay and I tell you that for two reasons one so that you keep your eye out and you you know you're a little bit skeptical of each of the claims that I make and two so that when I make that mistake and you notice it and you're just you know throwing things at the screen you're getting angry you're clicking that unsubscribe button you can at least quell a little bit of what you're thinking so just keep that in mind there will be one very purposeful mistake now before we get to the problem solving tip I want to talk about geometry this is one thing I was hoping to do a little bit more of in this series but this would be just a fun time to give a little example of it and in particular I want to talk about one of my favorite little bits of geometry it's not too simple but it's also not too hard it's called the inscribed angle theorem it comes up way more than is reasonable for a simple little theorem like this to actually come up when you're solving problems okay and basically an inscribed angle of a circle refers to if you have two lines that meet at a point of that circle and what we care about is this little angle in here okay and if you were just studying this if you're an early mathematician trying to make sense out of this you might make a guess that something about this angle is going to be related to the arc of the circle that those lines hit I mean that arc can also be described as an angle of some kind if we draw lines from the center of the circle it makes it a little bit clearer so I'm just gonna draw a couple green lines here and Mark that we have a different angle here and maybe I give them names I'll call this one theta L for like the large angle and this one I'm gonna call theta s it's the small angle and you might wonder is there a relationship between these two angles and if so what is it and for our purposes today the question is how do you systematically approach a puzzle like this find a relationship between these two and prove prove its existence so problems of problem solving tip it's not problems if problem solving tip number one which is just just so useful in a way that again is kind of deceptively simple make sure you're always using the defining features of whatever your setup is I swear a good 70% of the problem sets that I did as an undergraduate math major essentially came down to looking at what was given asking very critically what is the definition of each term involved here unraveling those definitions and then just seeing how they pieced together so use whatever's the defining feature in our context what is defining the various points and intersections that we have here well it's the idea that they're on a circle a circle is by definition all of the points that are a common distance from the center so in particular we know that this length is a radius of the circle and this length is a radius of the circle we should use the fact that those are the same but moreover this other point P that was not just chosen at random it's defined to be on the circle and unraveling what that means it means it is a common distance away and it's the same distance away from the center as these other points so I and then inspired to draw a line to add something to the picture and to note that it's the same there you know quite often when you see people solve hard geometry problems it comes down to adding something to the picture and the most beautiful geometry involves adding something that seems out of left field and it just illuminates everything you shift your perspective and sometimes you look at him you wonder like how would you have known what to add like I get I'm supposed to draw an extra circle here or put a you know put a rectangle around this triangle whatever it is you're doing but how do you how do you systematically know what you should add so in this context if it relates to the definition of your objects probably a good idea and that line actually will be helpful to us we can start giving a couple things names that's something that a hint maybe it shouldn't even be described as a tip but I've put it down as number two that when you give things meaningful names that actually helps you move forward in your problem and in this context it might seem like an innocuous thing I'll call this little angle that we formed with our radius alpha and this little angle beta and just to see if that helps us move forward you know recognizing if alpha and beta show up elsewhere rather than me telling you I would like you to tell me in the context of our next quiz question so before I answer that probability one for you I'm gonna pull up question number two for today we've got a diagram essentially what we just drew the dot in the middle is the center of the circle and I've labeled seven different angles I've just labeled them a through G and what I want you to do is tell me which of these equations is true one possibility is that four of the angles a B D and E have a sum that's the same as C plus F plus G which just looking at C F and G that would be 360 degrees right the other is that a is equal to B and D is equal to e another possibility is that C is equal to F and then maybe it's all of those or maybe it's none of those and it looks like we already have strong strong consensus on this one with 360 400 just a strong large number of people coming in agreeing what they believe the right answer is and you know if you if you want to be involved the place to go 3 B 1 B Co /live that redirects you to an item pool page so that you can follow along and for those who are watching in the future maybe you're watching it in the embedded page where the problem is just sitting right below you and even if you're not technically participating in the data contributing to the live statistics what I was going through it it's honestly pretty fun to just kind of looking like I know I'm not a part of this but it kind of feels like I am a part of what was happening so I just love it alright so because there's such strong consistence consensus I feel comfortable grading this I thought I got it right on one one one one so the correct answer is that a is equal to B D is equal to E and then none of the others are necessarily true and let's let's walk through why that's the case yeah it comes down to exactly what we just highlighted that these radii are common which means you know let me just give more things names let's call this point a and this point B maybe that's confusing cuz I was just naming angles and B but separate contacts separate picture you know what I mean the triangle a P and then the center of the circle that's isosceles you know there's this symmetry about I could even draw the little axis of symmetry and that tells us that this is also alpha likewise the triangle up here is isosceles and that tells us that this has an angle of beta and what we just did in effect is leverage a little bit of symmetry in this case it was innocuous but quite often looking for symmetry in much harder setups it's super generalizable and it definitely will help you move forward if you recognize that there's something symmetric use that symmetry in some way which is effectively what we've just done giving things a couple more names I might want to call this angle I'll call it something related to the alphas so maybe I just call it alpha Prime and similarly this angle over here is beta Prime and by noting those facts I have everything I need to do to draw a connection between this small angle and this large angle we essentially just write down the facts that we have based on the triangles we're looking at so the triangle with all of the Alpha angles the sum of those angles has to be a hundred eighty degrees and there's two different alphas in there and then there's an alpha Prime and instead of writing a hundred eighty degrees of course we like radians so I'm going to say that equals PI radians similarly the one with all of the betas tells us that if we take two times beta and we add beta prime that's also PI and then the other fact that we have that just popping right out from the image is that alpha prime beta Prime and theta L add up to 360 degrees or two pi so just writing down all of the relevant facts that we have now we have some objects that we can manipulate and work with to draw some kind of conclusion which again we're looking for a connection between theta small and theta large and I know a lot of you you're three blue and brown audience members you know about the inscribed angle theorem but I really want you to think about this from a beginner's mind right if you were just approaching this and you didn't necessarily already know about it what would you have done to find that solution and what principles can you take away as you do that because that helps us as we start to get to harder and harder geometry setups so in this context once I have these three equations recognizing that there's an alpha prime here and one here there's a beta prime here and one here I might think about cancelling them out so I'm gonna you know add this top equation and maybe I've subtract off the other two equations and to subtract these off and what that means is the Alpha prime gets cancelled so does the beta prime I'm left with my large angle I'm subtracting off two times alpha plus beta you know each of those get subtracted off with the coefficient two and then we have two pi minus two copies of Pi so that's all equal to zero which is saying the same thing as theta L is equal to two times well rather than writing alpha plus beta I'll just recognize that that is the small little angle that we had it's two times the small angle again I can't emphasize enough just what a weirdly useful fact this turns out to be in various geometry puzzles you might do it's definitely come up on the channel a number of times in circumstance regarding you know complex numbers or pure geometry situations of course just anytime you want to relate an angle to two times that angle realizing them in the context of a circle like this can be strangely useful so this is just an image to have burned in your mind as you're solving problems and to give you one example of a problem that you can solve once this is sitting there burned in the back of your mind I want you to remember back to the lecture that we did on trigonometry which conveniently it's actually the example that I had pulled up here so one of the central things we were talking about was how just playing with graphs you can get this bizarre-looking fact that if you square the cosine function you get something that looks again just like a cosine graph and you can get more exact about that where if we start with an initial cosine graph and you manipulate it a little you know we shift it up we scale it down we say we need to double the frequency you get the exact same graph so we have two different expressions for the same thing but it's not at all obvious why these would be related one of them involves doubling the angle it's a reference to the other involves squaring the output okay so it's an obvious fact we proved it using complex numbers but what I'd like to do is try to prove this using geometry to kind of viscerally see the fact pop out right in front of us and to do that let's go ahead and write down what the fact is again so that we can start thinking about it we want to find that the cosine squared of theta which is just saying take the cosine of theta and square it it's that awkward notation is equal to 1/2 of 1 plus cosine of 2 theta you've got the strained relationship between squaring things and doubling the angle which as we've talked about in the whole series is really a reflection of the fact that a cosine is a shadow of an exponential function but let's say you didn't know that we want to see if this relating geometrically so you don't already know the double angle identities or anything like that you just want a very direct understanding of this particular equation well one common thing that comes up is that a way to show the two non obvious things are related or even equal is to see if you can find one object that you can describe in two different ways so we're going to look for one object that we have two different descriptions of and this often can give you nice equations in this context it might mean relating the left and right hand side this comes up in combinatorics all the time when you have a counting puzzle where you know you do something like count how many ways you can have a string of five bits that are either zeros or ones and on the one hand you can count it multiplicatively kind of going through each one and saying well you're multiplying the possibilities by two but on the other hand you can go iteratively and say well how many of them have no ones how many of them have one one two ones something that kind of seems harder and a more awkward way to count but by describing the same thing twice you end up with this really not obvious fact from an algebraic standpoint and we're going to do the same thing more geometrically here but again just gotta emphasize how like how general this ends up being so what we want is some kind of object that each of these describes and to do that maybe we just think okay let's let's draw a unit circle which is where something like a cosine typically comes up well for the first time in my life I drew a quarter circle arc that wasn't terrible it's not great but usually that comes out much more disastrous how pleasing it should not be so pleased with a terrible quarter circle great okay so that's our angle theta right and what does cosine mean in this context not cosine squared but just plain vanila cosine well it tells us if we look at the x-coordinate of this point that's the cosine of our angle and so now I want you to think about how can we represent the square of the cosine as some kind of object some geometric thing that we can point to in this image and the first instinct might be something like well let's draw a square off the side of this you know something with area to it and interpret it like that but then there's going to be a problem if the principle that we're trying to apply is describing the same object in two different ways because if we describe this left-hand side as some kind of area that would mean that we have to find another description of that same area that same object with the right-hand side but that's going to be weird because this doesn't involve squaring or anything like that if it's just a plain vanila cosine term it seems much more natural to describe that as some kind of ratio or maybe some kind of length it's just that we need the two theta to pop up somehow so instead let's seek a way to describe cosine squared that does not involve area but is instead something more of the flavor of a ratio or a length and in this context the key comes down to leveraging symmetry again and it's a sneaky bit of symmetry it's something we didn't talk about in the trek lecture but I love it so much I'll talk about it again if we think of this angle theta on the one hand it's telling us the angle between the x-axis and the line but on the other hand it's also telling us the angle between the line and the x-axis and I know that see it sounds like the same thing but it means when I say project down the point at the end of our radius which was length one perpendicularly onto the x-axis that length got scaled down by cosine theta but now what if I do things the other way what if I say I want to project down in a perpendicular fashion onto this line well again I just have two lines separated by an angle theta I'm doing a projection which means it gets scaled down by the cosine of that angle so now I'm taking the cosine of theta scaled down by the cosine of theta and it gets me cosine squared of theta so cosine squared refers to this length a portion of the hypotenuse of our right triangle if that hypotenuse had a length of 1 which in our unit circle it always does and incidentally you can show very similar reasoning that the sine of theta is this other other portion of that hypotenuse and this gives you a nice sort of a clever proof of the Pythagorean theorem baby of double projecting based on asking you know is is this line theta degrees away from that or is that line theta degrees away from this it sounds like you're saying the same thing but it gets you something of mathematical substance to recognize that symmetry now why do I say this well we have a representation of cosine squared what we want now is to do something in terms of two theta we just were thinking about a context where we're able to relate an angle to two theta two to two times that angle so somehow we want to realize this this angle as an inscribed angle of some kind of triangle and I'll show you okay how you can do that but before I do just to mention another fact that if you been puzzlingly around with these sorts of things might be burning in your mind is a specific instance of the inscribed angle theorem called Thales theorem so let's say that that large angle we had two theta was actually a hundred eighty degrees and it was pi radians what would that mean in terms of the inscribed angle theorem it means that if we take that diameter of the circle it's 180 degrees it's just drawing out a diameter and we have an inscribed angle with lines that hit either end of that diameter then this angle is necessarily half of that so what it means is that we can we can put a right triangle inside a circle and whenever you do that the hypotenuse of the right triangle is exactly the diameter of that circle it's a very cute fact if you wanted another proof of it that's not just via the the inscribed angle theorem there's another very wonderful leveraging of symmetry that you can do where basically you take this point and you say I'm going to reflect it through the origin reflect it through the center of my circle and see where I get and recognize that reflecting to the center is the same as rotating the whole image 90 degrees so if I have rotated the image 90 180 degrees my other vertex would end up at that same point but now what we have is a quadrilateral and one of the diagonals is a diameter of the circle but the other diagonal is also a diameter of the circle which in particular means they have the same midpoint under the same distance apart and it can convince yourself a little that implies it must be a rectangle that could also be a little side homework problem if you wanted to chase around the relevant angles but I think that's a very beautiful way to think about fales theorem that you reflect everything 180 degrees and you necessarily conclude it must be a rectangle which means that this is a right angle now for our purposes what does that mean well we've got a right triangle sitting here that's from zero we've got one of the points here another point on the circle let's inscribe that in a separate circle okay so I'm going to take a copy of that triangle but I'm going to instead of making the hypotenuse a radius of the circle I'm gonna make that hypotenuse a diameter of the circle so this is still going to be an angle of theta sitting right here the basically I flipped it around so previously I was here but I flipped it around so that my ninety degree angle is sitting up unto the right rather than sitting down here the length that we care about is what what happens when we project from the point at that ninety degree angle in a perpendicular fashion down onto the hypotenuse which now looks like this and what we care about is this long length okay and actually let me ask you as a is a live quiz to see if you can come up with an expression for that length in the context of the diagram that we're now looking at so pulling up our quiz again congratulations to everyone who got this one correct let me give you a little bit of time to think about this one because this is this is kind of a heart part of the heart of this particulate proof and I think it's very very pleasing to see so it specifies that the hypotenuse of the large bright triangle above has a length of 1 in our context B is because it came from the hypotenuse of the triangle drawn in a unit circle so that potenuse has a length of 1 what is the length L in terms of theta yeah so can you find an expression for L in terms of theta and I'll give you a little bit of a little bit of time for that bring back our pause and ponder music get myself a chance to take a drink once again it looks like we have some strong consensus for today so while answers are rolling in before I grade it I'm just gonna go ahead and start describing how it goes since it seems like a lot of you are already well ahead of me it's fun to have though so of course we're going to use the inscribed angle theorem that is the whole reason and bringing it up here it's a way to relate a single angle to twice that angle so in this context I would draw some lines from the centre of my of my new smaller circle that has radius only 1/2 and recognize that this is 2 theta and this is just lovely now isn't it because what is the length we care about part of it excuse me part of it is the radius which is 1/2 but then the projection down according to 2 theta onto this remaining leg ends up being that radius 1/2 times the cosine of 2 theta and of course that's exactly what we want it to be we've got the whole radius which is 1/2 and then we're multiplying that by 1 plus the cosine of 2 theta it's a radius of the circle times a scaled down version of that radius so that's two different ways of viewing the same object which is what we get when we take this right triangle and we project down and look at what part of the hypotenuse that cuts off and it gives us this non-trivial relationship in trigonometry between the cosine squared and the cosine of 2 theta the fact that otherwise we were going into like complex numbers and Exponential's to understand so I think that's quite beautiful I think that's just one of the many many instances of where the inscribed angle theorem suspiciously lotion it's just suspiciously shows up and again what I want you to take away is this principle that if you can have one object described in two different ways very powerful in terms of showing non obvious how to break relations or anything that's kind of written down symbolically without immediate intuition on top of it so with all of that let's actually turn to the probability question that I asked at the beginning of the lecture so going back to our live quiz I will go ahead and grade what we we all now know the correct answer so for those of you in a minute oh it was marked incorrectly oh that's my bad no I just slipped this one in like last minute before the lesson today so I I might have like swapped around what the answers were so well it looks like 1380 if we were absolutely wrong according to whatever jerk wrote this quiz so I don't know how that shows up on the user interface if it's like shocking read like oh no but D was the actual correct answer here so congratulations to those of you who got that now back to our probability question I'm curious to see what people said on this one just in terms of their instincts so this one we had a little bit more of a spread and ah interesting here the actual correct answer does show up quite a bit quite a bit lower than and now I'm questioning myself to make sure that I've actually written the thing appropriately so just as a reminder of what the question is we're choosing two random numbers from the range 0 through 1 each according to a uniform distribution and we're we're guessing what the probability that they round down the ratio of these numbers rounds down to an even number remember 0 is an even number so that it rounds down to 0 where 2 or 4 or anything like that now this is a tricky problem to think about and definitely no no fault at all for anyone who isn't immediately able to see roughly where it should be but I think with a with a little bit of progress on our way even before we get the exact solution we can get to a point where you might be able to intuitively give some kind of ballpark estimate so what have we got here choosing two random numbers between 0 & 1 I think that's a weird thing to think about especially if you're not familiar with probability that well or when the phrase uniform distribution is thrown up if it's not clear what that means but essentially it's what you would expect when you're choosing some random point on this line and the idea is that each point is as likely as another or more specifically a given range of points of a certain size should have a given probability that's independent of where that range showed up it's only dependent on its size now so you might have in the back of your mind the idea that we've chosen two points there each somewhere between 0 and 1 and just to give an example of what we mean by uniform distribution the probability that X sits between point three and like zero point five because it's going to be somewhere between zero and one and the length of that range is zero point two about a fifth of the entire length that could have come from what it means to be uniform is that that probability is actually just the length of the segment that it came from now one thing you might ask is well what if what about the probability that it's precisely 0.3 or precisely 0.5 would it matter if we made these less than or equal to signs and the answer is it doesn't actually matter because the probability of hitting any specific value on a real number line ends up being 0 this is a thing that I find very confusing how can a probability of an event that's possible it's definitely possible to hit 0.3 have probability 0 made a whole video about it trying to describe this but really what it comes down to is that the things that have probability you should think of ranges those are the fundamental objects and it doesn't really matter how we treat the boundary and just think in terms of ranges even still though what we're asking is a very bizarre question which is if we take the ratio of x and y and we round that down which sometimes no mathematicians write using this floor function saying we find the greatest integer smaller than that how do we know if that's an even number that's a weird thing to think about it's a hard problem in that way so you know if you think through the principles here it's kind of like use the defining features of the set up well not clear how to use the fact that it's a uniform distribution I guess we'll be using lengths in some way to yield probabilities so maybe that gives us some geometry but that's not really helpful give things meaningful names you know maybe x and y are meaningful or some something suggestive like that symmetry ok maybe you know the idea that choosing X and then Y is as likely as choosing Y than X you could use that to conclude that this ratio x over Y is as likely to be above 1 as it is to be below 1 and that actually does tell you something because if we're wondering how often do you round down to be 0 right you can say well x over Y is as likely to be x is as likely to be bigger than Y as Y is likely to be bigger than X so there's a 50/50 chance that this should happen this would be probability of si percent so let gets you somewhere which is kind of nice but it's not clear how you would apply that to things like the even numbers same object two different ways unclear so principle number five here is where we're going to come in again it seems simple it's something that you cannot along be like yeah yeah drawing pictures it helps to think through what I'm looking at but really when you find yourself struggling with some setup that's not already visual or pictorial you know it doesn't have to be making a geometric but just having some kind of sketch to give meaning to your terms can be very helpful and as a more specific problem-solving tip when you have some numbers multiple different numbers see if you can make them coordinates in some space so rather than thinking about x and y is separate things here we'll want to think about a single point with XY coordinates and what that does for us is it actually turns the whole problem two-dimensional in a very helpful way so first of all I've lost track of my straightedge which what have you gone little straightedge I can only throw you so far here we go things they run away from you even your object sometimes get tired of math class and want to play true into now and then but he has to stay whether he likes to or not I'm sorry so let's say this is our x coordinate X can fall anywhere between 0 & 1 with uniform probability Y can fall between 0 & 1 so when we have a pair of numbers you know something like zero point two what does that maybe like zero point eight pair of numbers is just a single point in this diagram and now to choose both of those numbers uniformly at random means that we're choosing a random point inside a square a square with side length one and now maybe we can make a little bit of progress because it's going to come down to some view of what's going on in this square now with this specific example if we're thinking about the ratio x over y and taking its floor taking just rounding it down well that's 0.2 over 0.8 that's going to round down to zero so this would end up being even and like I just said that that happens with 50% probability but let's see if we can try to find a way of thinking about that that generalizes up to other examples and again one very useful thing if you get stuck but I have enumerated down here as principle number six is to ask a simpler variant of the problem you're solving something it's hard it's too hard see if you can make it simpler in a way that you actually can solve and get some kind of foothold maybe that means loosening the constraints of the problem in some setups or maybe it means looking at a sub-problem so in our context rather than asking the probability that it rounds down to be even let me just ask the probability that it becomes one but I want you to answer it in a geometric way is something that will generalize to rounding to other things because the next simpler question might be probability that arounds down to two and things like that you know where this is gonna go we're gonna do it as a live quiz because rather than me answering things I want you guys to answer things for me so jumping up to question number four at this point it's going to give us a couple different diagrams yeah we've got a B C and E and it's asking us which of these four regions corresponds to values of x and y where X divided taking the floor of X divided by Y is equal to zero which is to say you take the ratio you round it down you get zero which region corresponds to that fact [Music] you [Music] you okay I'm gonna go ahead and lock in answers but if you want to keep thinking about it definitely feel free to pause the video and do so I don't want to rush anyone so looks like twelve hundred and sixty five of you 1273 always answers rolling in at the end correctly answered that it's C and let's take a moment to think about why that's the case yeah you can do so just with a pile of examples and just see which ones seem to fall in the region or not but let's see if we can understand this in a way that lets us make progress on to the other even numbers so when we say that it rounds down to zero what we're basically saying is that that ratio sits somewhere between zero and one and it's awkward to think of X divided by Y we kind of like to think of Y in terms of X so if I multiply everything by Y which is okay to do with these inequalities because Y is always positive so that's not going to affect whether the inequality flips one way or another I multiply everything by Y and we're basically asking when is X less than Y and whenever you see an inequality the boundary of that region is going to be described by the equality so we're going to wonder when is it the case that y equals x well that's just a straight line that goes to agonal if we draw our line y equals x that's what we get now that's the boundary of our region and to know whether we should look to the left of it or to the right of it either you can think very directly and say well you know why should be greater than X so at a given point we want to move upward in the positive y direction you could look at a specific example like this but however you do it you'll draw the conclusion that geometrically the region of points such that x / y rounds down to zero is this sort of grilled-cheese cut of our diagram so with that maybe you can start to think about the harder variant which is when is it that x / y rounds down to be - and again I don't want to answer it I want you to answer it when is it that x / y inside our unit square of points x comma y rounds down to two and we've got four possible geometric regions that this could correspond to a B C and D and really you know rather than just think about which of these is it really try to think through why it's the case and how you can prove that one of these boundaries is actually what it's supposed to be for example I wanted to be the case that if I didn't show the correct answer here let's say I was just trolling with everyone and I didn't show the correct answer you would be able to confidently come and say like no I'm quite positive that the correct answer is nothing that you've shown here see if that's the level of reasoning that you can put behind it so again I'll give you give you a little moment to think about that [Music] you [Music] okay so once again I'm gonna lock in answers potentially earlier than you want me to but keep the lesson moving forward no hard feelings if things haven't clicked yet because hopefully the explanation will make them do so so the correct answer is C which it looks like most of you got and let's go ahead and think through why that's the case very similar reasoning to what we were just doing the idea is that rather than thinking about this ratio and a floor which is kind of a kind of an awkward thing let's explicitly write out the inequality this is referring to it's saying that X divided by Y is greater than or equal to 2 if it's rounding down to that but it's not greater than 3 so it's less than 3 and you know again it's a little bit awkward to think of this ratio so let's write that as 2 times y is less than equal to X which is less than or equal to 3 times y now quite often we don't think of Y is a function of X we think of X as a function of Y if that makes you feel more comfortable so if you want in the back of your mind you can kind of think 2 y less than or equal to X well that's the same thing as saying Y is less than or equal to X halves and same deal 3 y being greater than X that's the same thing as saying Y is greater than X divided by 3 because that way we can look at the equality's associated with each of these the line y equals x halves which has a slope of 1/2 you can think of it as intersecting at the point where y equals 1/2 when x equals 1 so it'll be a line like this that describes part of the boundary of our region and the other line is when y is equal to X thirds so we know we actually have to be above this line that I'm about to draw or one of these represents x over 2 and one of these represents x over 3 and then part of the pardon the intrusion into the space of my last inequality all right so we want to be above the x equals x over 3 below the X divided by 2 this region here shows us everything we're rounding down gets to 2 now what you appreciate how this is a kind of complicated thing to think about if we hadn't gone into a picture that involves two dimensions if you were just thinking of x and y varying along this line and wondering when is it the case that X is more than twice or Y is more than two times what X is now yeah sorry X is more than two times or Y is but it's not three times more than what Y is it's not no no I said it wrong I said it wrong when y is more than two times what X is but it's not three times more than X is it's very easy to get confront fuddled in that way and it's even harder to try to give some sort of probability to that whereas in our diagram it has a very clear meaning it is the area of this region because the full area of the possibilities already is one so these the probability of something happening should be one and we just need to look at the area of that and now maybe you can see where this is going to go because for the next term when we want to know when is X divided by Y sit between four and five we're gonna be drawing lines that intersect at x over four and you'll go to a different color for this one x over four and x over five which is going to require a very small handwriting at this point but I'm gonna give it a try nevertheless x fourths y equals x fifths and this little sliver of area gives us all of the times that our ratio x over y rounds to be four and we're gonna have to add infinitely many of these so that gives us sort of another phase of challenge to the problem I've drawn this all out in desmos by the way if you want to just sort of see what some of these regions look like where we've got our top region of places where it rounds to zero then we've got another region corresponding to rounding to two rounding to four rounding to six and just on and on each one of these regions rounding and rounding rounding I only went out to like a hundred or something like that but that gives you a sense of what we're trying to do so we've made progress but this is still hard what is the area of all of those triangles added together that's not necessarily an obvious thing so let's just start by writing it out and seeing what help that can give us so every one of these is a triangle it's gonna look like 1/2 base times height and in fact every one of them if we think of the left right direction as being their height every one of them has a height of one so each one is going to look like 1/2 times a base of some kind so I'm gonna take don't write this out on a different piece of paper actually so I can keep it up close I'll think 1/2 times whatever the base of the triangle is times the height so our first triangle that bases that base has a length 1 so that's going to correspond to the 1/2 probability of going to 0 the next triangle we have to look at this length here between 1/3 and 1/2 what is that length well actually I'm just gonna write it out as 1/2 minus 1/3 it equals 1/6 but writing it out like that kind of reminds us where it came from so we don't want to collapse things too soon that could maybe be another problem solving tip don't collapse things too soon and try to let your notation have a memory for where things came from because sometimes that helps see overall patterns this next one what's the distance between these two points well it's 1/4 minus 1/5 that's the distance between these given how they were defined so we have 1/4 minus 1/5 and in general we have this kind of oscillating some 1/6 minus 1/7 where we have all the reciprocals of the natural numbers but we're adding in that up infinitely many different times and we want to know what that sound happens to be and from here the the problem-solving tip associated with this will seem a little bit strange but it might be the case that you recognize this fact from somewhere else you might recognize let's say if you were watching a particular lock-down math lecture week or two ago this alternating sum one minus 1/2 plus 1/3 minus fourth plus 1/5 on and on actually equals the natural log of 2 okay and the way this actually came about it's such a weird procedure it's worth just like walking through again really quickly because it's a bizarre thing that you're not gonna you're not gonna be able to just stare at this formula and then immediately see that this how you're gonna solve it unless it's something that you've seen before which can make it seem all the more opaque we did the strange thing where we made it seem like a harder question at first where rather than asking about one particular sum we turned it into a function which is effectively asking about infinitely many different sums like this so X to the fourth over four then we're adding X to the fifth over five and the reason for doing this is that this plays nicely in calculus land because those denominators are now related to the exponents in a way that we can kind of cancel out by doing an integration trick each one of those terms that can nicely express as an integral of a much more simple monomial term X cubed minus now let's see plus X to the fourth if I integrate this thing each one of them has an exponent that increases and then we divide by what the exponent is and the reason that you would want to do this is that this now has a nice way to collapse itself so just to make this may be more explicit if we evaluate the integral from 0 to 1 that's the same as taking this whole expression and evaluating it at 1 and subtracting at 0 so that will give us what happens when we plug in at 1 and again this is just such a bizarre thing that if you hadn't recognized the sum seeing someone prove it to you like this doesn't necessarily make it feel like something that you could have found which is frustrating in the context of trying to develop problem-solving tips that are generalizable but I'll keep walking through it just to give a little bit of closure to this we've got this infinite sum that if you had been familiar with geometric sums where each term looks like a certain product from the last you would be able to write this as 1 over 1 plus X because we're always multiplying by negative x so you always take 1 over 1 minus the thing you're multiplying by which again it's kind of one of these things where it's relying on you recognizing it in some way and then the last bit of recognition is knowing how to take integrals of 1 divided by a thing and in this context it works out very nicely to just be the natural log and we're evaluating this between 0 & 1 which is to say we're taking the natural log of one plus one or two minus the natural log of one which is zero and that's why all of these things are the natural log of two and think about what has to go on there in order to be able to take this and then apply it to a probability question you would have to recognize the alternating sum as something that you had seen from another context or if you didn't you would have to be aware of this trick to somehow turn it into a polynomial that can be nicely expressed as an integral that can be collapsed because of geometric series which can be integrated because of the natural log of X and then thinking about like what is what is the thing that you can teach someone to take say come away and be able to solve problems in the same way I have what might seem like kind of a facetious tip but I actually think it's maybe the most potent one and the most honest one the way that you can get to this sort of point just read a lot read as much as you can you know watch youtube videos on math if they're substantive things like that and think a lot about problems which is maybe frustrating because what you want is to be able to say like well how could I have come to this on my own without having nearly recognized it but I think the truth of the matter is a lot of what looks like insight and ingenuity is really just pattern recognition but wearing a little bit of added clothing and sometimes it's patterns not so much that you're directly recognizing exactly this but maybe you had seen a series like this or a tactic like this before like geometric series that comes up a lot if you read a lot and if you think a lot about problems you will recognize geometric series even if it's in a context that you've never actually seen before or a specific geometric series that you've never seen before similarly if you read a lot and you think a lot about calculus knowing that you can have the natural log pop out of an integral like this it becomes second nature and I think recognizing the truth of this as being the key to a lot of problem solving it's actually pretty inspiring because often times you find yourself in a situation where somebody is they're just faster they're just better they're just recognize things more so than you do and that can be a little bit intimidating right to look at one problem think of yourself as pretty savvy with math and knowing what's going on and then just having someone burn through with a wonderful bit of cleverness this super beautiful argument that leaves you sitting there in the dust wondering like why I just you know I'm just not in the same league at all right and you sometimes even think well he just has a math gene right that person they just have some sort of innate instincts that makes them really good at this stuff but I think the truth of the matter is the the people who are showing that kind of ingenuity they Virts expose themselves to a huge number of patterns and you too could get there right there is a path towards that which takes the form of practice but not just practice practice where you're taking each problem that you're looking at and trying to digest the deeper principles behind it maybe some of the ones I'm trying to talk through today like okay I can say leverage symmetry but what does that actually mean you know what does it mean to look at a symmetry of a problem and turn that into something that's formulaic ly useful you just have to see it a lot and then be pensive when you do don't just be satisfied with the answer see if you can understand why that answer came out so in that way this number seven like it's it's the most frustrating but it's the the most real of all the problem-solving tips that there can be which is that true problem-solving comes down to a kind of pattern recognition and there's no two ways around it you just have to do a lot of practice and expose yourself to a lot now I would bet that when we did talk about this infinite series a couple lectures back you wouldn't have thought that that's a thing that you're going to be using in a probability question one day but that's just how these things go they show up in unexpected places so what you can do then and say well we've got our whole expression that involves this very alternating sum and you'd say okay that means that the answer to our final question is you know one half of the natural log of two it's one half of what that alternating sum came out to be which is very nice you know it involves this natural log expression and it's very clean and we've got this picture for where it came from adding up all of these all of these areas now at this point there is one thing that I think separates really good problem solvers the ones who get like nearly perfect scores all the time to ones who are like merely good who you know they aren't necessarily perfect scores there's some like sealing mistakes they come in here they're at this point when you've done the problem and you've got your nice elegant solution you want to draw a box around it you're not done just always always principle number eight here always gut check your answer hey because there's gonna be some little mistake that happens all the time the great problem solvers aren't the ones who just never make little mistakes they're the ones who have some way of recognizing what those mistakes are so in this context let's say I I just want to see numerically what my answer turns out to be right we said that it was 1/2 times the natural log of 2 so let's just see what does that end up being and that 4 log of 2 is around point 6 9 so maybe it's not too surprising where around zero point 3 4 6 around 0.35 okay so if we write that down as one of the is the answer that we just got I told you I would purposefully make a mistake so hopefully you're not yelling too loud at this point does that make sense does that pass a basic reasonability test and if you look at our picture well the probability has to be at least 1/2 because that's the region where it rounds down to 0 so certainly it couldn't be the case that the whole thing adds up to be only 0.35 so there must have been some mistake we had along the way everybody makes silly mistakes everybody drops a minus sign or apply some rule that doesn't quite apply in a circumstance you're not going to make you're not going to approach perfection by avoiding silly mistakes the way to do it is to be able to systematically know when you've when you've made them so always go check your answer have like 2 different perspectives that can give you a reason ability check in this context if it inspired us to go and look a little bit more carefully at how we were applying things this sum isn't quite the alternating sum that converges to natural log of 2 in particular we added the 1 but then we also add 1/2 and it's only after that that we start alternating it was plus plus than minus plus minus plus on and on and you might see this by recognizing we were subtracting all the even numbers here but we're subtracting all the odd numbers here so it's similar but it's not the same we will be able to use our knowledge but let me just give this part where things actually start alternating a name something like s what this bottom equation is telling us is that when we take 1 minus s so that would mean we're subtracting in 1/2 then we're adding the 1/3 then we're subtracting the 1/4 we're flipping all of the signs of everything beyond that 1 that is the thing that equals the natural log of 2 which in turn implies that that remainder of the sum looks like one minus the natural log of two okay so all of that what does that tell us when we plug it into our original expression it's saying that the actual answer should not be one half the natural log of two that didn't even pay past our basic reasonability test instead it'll be one half of one plus one minus the natural log of 2 which is just 2 minus the natural log of 2 so does that pass our reasonability check what does this actually equal numerically you can get a loose approximation in your head if you want or if you want to see more precisely we can pull up a calculator so if we go over here and we're saying no no we don't want the natural log of two that was wrong too - that may be 0.65 three does that pass our reason our basic reasonability test yeah I think so all right 0.65 that looks like a reasonable answer to what the area in our in our diagram was because if we looked at that diagram which was you know we've got one wedge here that's covering 0.5 and then this other one covers well about one-sixth half of a sixth basically because this length was 1/2 minus 1/3 which makes it a sixth and then it's a triangle so 1/2 base times height so that's about a twelfth or point zero eight three and then the rest of it you know it's not going to fill a huge amount something around 0.65 seems pretty reasonable and so that you know we could call that good on our gut check that this is probably the correct answer but we could go one level further if we wanted because probability questions very often you can kind of cheat and just see the answer by simulating it and actually taking a bunch of samples and seeing what happens so I want to do that actually on this one just to give us a little bit of confidence that our answer of 0.65 that there wasn't some other silly mistake that we made along the way because everyone knows there certainly are silly mistakes that we can make before I jump to the programming though I just want to see if there's any questions from the audience and it certainly looks like there are so let's see if we can address some of these get myself out of the way of the questions this is like from the book of The Chosen where the rabbi makes an intentional mistake in his long speech to test his son it is exactly like that my children I just want to test you every time I make a mistake it was always purposeful and I was always doing it just to test you I keep making more episodes why am i stopping two reasons you know the main one honestly if I go much longer in this lockdown without getting a haircut we're gonna have to start filming these in the style of like a 1980s music video just to keep stylistic consistency I'm not up for that you know I think that's just gonna be a little bit too much effort so I'll have to wait until whenever it's possible to get a haircut again then then we can go really there I also just missed my old content I love visualizing stuff I've got a long pile of things I want to get to and you know the lectures take time I might spin up something like this again I might do it on a separate channel we'll see what plays out but I would love to do like a full course where it's a little bit more clear exactly what we're going to talk about from beginning to end loosely I'm thinking like combinatorics would be fun but don't hold me to that and lastly can you help us understand arrangements and the principles of inclusion and exclusion just like combinatory intuitive yeah boy how much time do we have today tell you what maybe just a separate video if I do the combinatorics course actually that that would be a perfect example for it it it takes a little bit too much to describe here but very naturally what you end up with is taking 1 over 1 factorial minus 1 over 2 factorial plus 1 over 3 factorial minus and you're doing this alternating sum with factorials which is ye pops out and if you think of E is fundamentally being the sum of the reciprocals of factorials it doesn't seem crazy surprising that it's related to counting problems in that way but for another day I'll say because I don't know if we have time today because I would love to just show how you could maybe get check this programmatically if you wanted to because the very last I have principle I have for problem solving for mathematical problem solving is to learn at least a little bit of programming okay and the reason here one for what we're about to do you can sometimes basically cheat and see what an answer is numerically to see that that verifies how you're thinking about it analytically but also importantly it forces you to think about things in two separate ways sometimes you can go about it mathematically but then when you try to make it computational you you run into certain walls about like how are things actually defined or I don't have infinity available to me how can I do this in a more approximate way very often when I'm animating things for usual videos the piece of math that I'm describing you know it lines up with some way that I'm programming it to give the illustrations and windows mismatched that's actually what's most interesting when it's not computationally viable to give a perfect illustration of what I'm describing and I actually think that does make for better problem solving in general where you're coming at it from two different angles so let's just try this out see if we can understand the probability question we were just looking at in terms of just you know yeah what's the word I'm looking for cheating seeing seeing what it turns out to be so I'm gonna import numpy which evidently some heathens refer to as numpy which just I just can't that seems awful to me so we can find random numbers if you just call random this first random refers to the library the second one is a function it will return a number between 0 & 1 according to a uniform distribution so it is as likely to choose something between you know zero point one and zero point two as it is to choose something between zero point eight and zero point nine and what's nice is I can get a list of them so in this case I get a list of ten random numbers and maybe I call that something like X and maybe I create another list of Y so X some random numbers Y some random numbers and if I take X divided by Y it does it term by term so for example this this first number that we see in here that's point seven three nine that's taking the point five two divided by the point seven this next one is taking point six six divided by point five six on and on so we can see all of our ratios like that and in general I might just define the ratios list where I'm gonna take my a bunch of random numbers of size N and then divide it by another bunch of random numbers of the same size and is not too fine but I'll give it a definition some nice and big number like a million okay and now I'm holding on to a million ratios a million examples of X divided by y where x and y are chosen according to our constraints which is kind of cool if you think about it so you know we can see some examples here some that have come up to 0 and some that would round down to zero sum that would round down to to some that would round down to one so that's nice and now I can start asking questions like you know when is it that I take these ratios I take the floor function of them not the floor the floor and I want to know when is that equal to zero this gives me a list of trues and falses basically saying it is or it isn't zero and if I take the mean of that which is treating the trues and falses as ones and zeros this tells me the proportion in total that will actually be zeros so we expect it to be about a half and we can verify yeah okay it's about a half we could have also asked when is it about to okay and it looks like 0.8 3 and remember from our diagram what we were looking for is when when it's in this green region here when it's in that green triangle which has an area that was 1/2 minus 1/3 but all times 1/2 because it's 1/2 base times height for a triangle so if we pop back over to our terminal and say ok if we were looking for 1/2 times the base of that triangle which was 1/2 minus 1/3 we would expect that proportion to have been point 8 3 and yeah it looks like it was about that and we could even answer our actual question which is to take the floor and then if I say I want to divide by 2 and ask when the remainder is 0 that's a way of asking when it's even and then taking that whole list and taking a mean of it is a way of asking how often that ends up looking like true what proportion of them give me true and yet point 6 5 which is about the answer that we were looking for you know we were looking for something that was 2 half of 2 minus the natural log natural log of 2 so we have this this wonderful way to kind of empirically verify and of course in most of the times with in lockdown map that I pulled a Python I'm just doing relatively simple things numerically if you wanted to you could try to visualize stuff so something like matplotlib is definitely a great pie plot I'll import it as PLT this is a very good library for just like simple data visualization that you can pull up pretty quickly so in our context I can pull up a histogram so I'm going to take a histogram of all of my data and I have to specify a range so maybe we just want to look at when the when those values get bucketed in between like zero and 20 so my number of bins is 20 that was about to sneeze but I held it in don't you love when that happens when you're like a sneeze is coming and you don't have to catch it it just recedes into the darkness because it knows that you're better than this needs right again I'm more pleased with myself than I should be there and I always you always have to add a relative width on these sorts of histograms because they just look ugly if they're all excited by side I think sometimes so if I do this and then I show what the plot ends up being I have something that is shown up on my screen but not your screen so let's pop it on over yeah there we go so you can get a sense of you know this bar represents all of the ones that rounded down to zero and you see it's about half of them about half a million all the ones that round it down to one rounded down to two and you can get this nice nice sense for what all of your data is again just adding that kind of empirical validation on top of whatever you find more analytically and that back and forth I really do think is helpful for more like pure mathematical problem solving so with that that's actually all that I have for the lesson today just two quick things that I want to go over before we end things here I've been using decimals a lot to like show graphs of things because I just love desmos and I'm actually friends with some of the people who work there because quite often accompanies people are as delightful as its products and the CEO Eli shared with me the fact that they were doing like an art contest among some students and I just wanted to showcase some of I'm not sure if these are the winners of the finalists of the art contest but basically in various different categories of students who were I think it was like 12 to 14 15 to 17 or something like that using like mathematical graphs to try to draw pictures okay so keep in mind what I'm about to show you our mathematical graphs did someone wrote just with an analytic description and they were they were just prompted to create something artistic from that okay so one of my favorites and I think this one was from someone named Carrie is two giraffes that just from an artistic standpoint it's actually quite lovely and then to think through like actually mathematically describing everything involved here it's such a beautiful blend of well like the creative side of things the artistic side with aesthetics and everything and the analytic side another that was this is just genuinely insane this is by a cat simi is a Bayesian night so I definitely wanted to highlight this on behalf of the team over at desmos where each one of the curves here is described according to something that's called a Bezier curve very useful for computer graphics it's a kind of cubic parametric term and they just recreated a starry night in a way that's completely beautiful I think and then the very last one which is genuinely shocking that you can do with mathematical graphs in any way was a self-portrait by Jared that's just like genuinely insane I mean I remember playing around with like a ti-84 and trying to come up with little pictures of like a smiley face and things like that so the amount that things have changed in terms of when someone's noodling off with their graphing calculator in class and what they can do truly next level and then at the very end I just want to say again like a highlighted thank you to Ben eater and to cam who've been extremely helpful with the whole series in ways that's like hard to even articulate properly eater in particular I mean he's let me borrow a lot of his equipment and helped out with like figuring out live footage type stuff because that's not something I usually do which is not even to mention the work on the live stats and live quizzes so if you aren't already familiar with his channel that's simply named Ben eater like 100% check it out definitely subscribe to it try some of the projects he has a very project oriented way of teaching I think it's absolutely great so cannot emphasize enough how how grateful I am in the direction of both of those two and check out the the item pool site where we are going to let you kind of relive the lockdown math experience and have like homework all challenges associated with each one of them so stay tuned on that and if you're interested in being a beta user of it there will be forms for how you can how you can reach out to both of them thank you everyone for joining in the whole series this has been very fun for me very different for me and I will shortly get back to the more usual videos which I'm very excited about there's a couple topics that I just think you're thoroughly gonna enjoy especially if you like problem-solving stuff like this like really getting into good meaty problem-solving that's some of what's on the horizon and with that I will simply say keep loving math and enjoy the rest of your day [Music] [Music]

Channel: 3Blue1Brown

Views: 575,781

Rating: 4.9596491 out of 5

Keywords: math, problem-solving

Id: QvuQH4_05LI

Length: 68min 19sec (4099 seconds)

Published: Fri May 22 2020

Reddit Comments

[u/theadamabrams]

I swear I've seen a video before about the specific ratio-rounds-to-even-number puzzle before, solved the same way with shading parts of a square (and also mentioning variants like rounding up), but I can't find it now. Am I imagining this??