9 Problem-Solving Principles — Full Breakdown & Transcript

[00:00] mean when they use the phrase problem solving. approaching puzzles that you've never seen before and still being

[00:13] able to systematically and creatively find some solution to them. I mean, you can teach someone how to solve one particular problem, But how do you teach someone how to approach a problem that

[00:29] Well, I honestly don't know how, but what I want to do for this I've enumerated nine of them in total, and each one we're

[00:41] And I think each one is kind of simple. yeah, of course, that's a thing that you should do. that you would be shocked at how often you can make very meaningful

[00:56] 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

[01:09] here on the spot, I want to ask you a question that we will be solving later a little thing to mull over, and a hard problem that will be fun to tackle when

[01:22] So the question asks, suppose the 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 0.58962 or something like that.

[01:36] 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 0 or it rounds down to 2 or to 4.

[01:50] And basically it's asking you to guess where is this probability, you know? But just intuitively as you hear the problem and you think looking at their ratio, what do you think that probability is going to be?

[02:06] And one thing I want to say is that this whole live quizzing software, 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 Cam Christensen.

[02:22] of what's actually a much deeper product at play. things that they've been working on that we'll be developing. So if any of you want to share some of the lectures here or go back and kind of go

[02:37] through the live quizzing experience, if you go to the link that it's in the description, but it's at itempool.com slash c slash 3b1b, but you can follow the description. along with it and your progress is tracked and you get scores and things like that.

[02:53] it will skip you to the right point in the video. there's just me talking through some problem, you'll also get to whatever the appropriate

[03:06] They often have explanations associated with them and hints and things like that. And ultimately, if you look back in just a couple days, So if you like challenging problem solving, contest math type stuff,

[03:21] which I think should be fun. It's not quite there yet, but definitely check back. there will be more item pool shenanigans going on.

[03:33] 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. They are looking for beta users, so feel free to reach out to them.

[03:47] Now let's dive into the actual content, shall we? And before we do, I want to specify that at some point today,

[03:59] in the next hour, I'm going to purposefully make a mistake. you're a little bit skeptical of each of the claims that I make. you know, throwing things at the screen, you're getting angry,

[04:14] you're clicking that unsubscribe button, you can at least quell a little bit of There will be one very purposeful mistake. Now before we get to the problem solving tip, I want to talk about geometry.

[04:26] This is one thing I was hoping to do a little bit more of in this series, And in particular, I want to talk about one of my favorite little bits of geometry. It's called the inscribed angle theorem.

[04:42] theorem like this to actually come up when you're solving problems. And basically an inscribed angle of a circle refers to if you have two lines that

[04:56] meet at a point of that circle, and what we care about is this little angle in here. And if you were just studying this, if you're an early mathematician trying

[05:08] 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

[05:20] lines from the center of the circle, it makes it a little bit clearer. here and mark that we have a different angle here. I'll call this one theta L for like the large angle,

[05:35] It's the small angle. And you might wonder is there a relationship between these two angles? And for our purposes today, the question is how

[05:48] Find a relationship between these two and prove, prove its existence. So problem solving tip, not problem sip, problem solving tip number one,

[06:01] 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

[06:14] asking very critically what is the definition of each term involved here, unraveling those definitions, and then just seeing how they piece together. In our context, what is defining the various points and intersections that we have here?

[06:34] A circle is by definition all of the points that are a common distance from the center. of the circle and this length is a radius of the circle.

[06:46] But moreover this other point P, that was not just chosen at random, And unraveling what that means, it means it is a common distance away

[06:58] and is the same distance away from the center as these other points. to the picture and to note that it's the same there. it comes down to adding something to the picture.

[07:14] left field and it just illuminates everything, you shift your perspective. Like again, I'm supposed to draw an extra circle here or put a, you know,

[07:27] 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. We can start giving a couple things names.

[07:43] but I've put it down as number two, that when you give things meaningful names, And in this context, it might seem like an innocuous thing.

[07:55] our radius alpha and this little angle beta. recognizing if alpha and beta show up elsewhere, rather than me telling you,

[08:08] I would like you to tell me in the context of our next quiz question. I'm going to pull up question number two for today. The dot in the middle is the center of the circle and I've labeled seven different angles.

[08:26] 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.

[08:40] 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. And then maybe it's all of those or maybe it's none of those.

[08:54] 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 And you know, if you if you want to be involved, the place to go, 3b1b.co live,

[09:10] that redirects you to an item pool page so that you can follow along. in the embedded page where the problem is just sitting right below you. to the live statistics, when I was going through it,

[09:24] I know I'm not a part of this, but it kind of feels like I am a part All right, so because there's such strong consistent- consensus, Oh, I thought I got it right on 1 1 1 1.

[09:40] and then none of the others are necessarily true. It comes down to exactly what we just highlighted, that these radii are common,

[09:54] Let's call this point a, and this point b. but separate context, separate picture, you know what I mean.

[10:07] 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, Likewise, the triangle up here is isosceles, and

[10:24] 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

[10:37] setups, it's super generalizable and it definitely will help you move forward. use that symmetry in some way, which is effectively what we've just done. I'll call it something related to the alphas, so maybe I just call it alpha prime,

[10:55] 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 have based on the triangles we're looking at.

[11:09] So the triangle with all of the alpha angles, the sum of those angles and then there's an alpha prime, and instead of writing 180 degrees, of course we like radians, so I'm going to say that equals pi radians.

[11:24] we take 2 times beta and we add beta prime, that's also pi. And then the other fact that we have that's just popping right out from the

[11:36] image is that alpha prime, beta prime, and theta L add up to 360 degrees, or 2 pi. now we have some objects that we can manipulate and work with to draw some kind

[11:51] of conclusion, which again, we're looking for a connection between theta small you know about the inscribed angle theorem, but I really want you to think about this

[12:03] If you were just approaching this and you didn't necessarily already know about it, and what principles can you take away as you do that? So in this context, once I have these three equations,

[12:18] there's a beta prime here and one here, I might think about canceling them out. and maybe I subtract off the other two equations.

[12:31] And to subtract these off, and what that means is the alpha prime gets cancelled, I'm subtracting off 2 times alpha plus beta.

[12:43] , 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

[12:58] It's 2 times the small angle. 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 circumstances regarding,

[13:14] you know, complex numbers or pure geometry situations, of course. Just anytime you want to relate an angle to 2 times that angle, realizing them in the context of a circle like this can be strangely useful.

[13:28] So this is just an image to have burned in your mind as you're solving problems. burned in the back of your mind, I want you to remember back to the lecture that we did on trigonometry, which conveniently is actually the example that I had pulled up here.

[13:43] 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, And you can get more exact about that where if we start with an initial cosine

[13:58] 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. but it's not at all obvious why these would be related.

[14:11] the other involves squaring the output. We proved it using complex numbers, but what I'd like to do is try to prove this

[14:23] 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.

[14:36] 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 one half of one plus cosine of two theta.

[14:50] 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 But let's say you didn't know that, we want to see it viscerally and geometrically,

[15:04] You just want a very direct understanding of this particular equation. two non-obvious things are related or even equal is to see if

[15:18] So we're going to look for one object that we have two different descriptions of, It might mean relating the left and right hand side.

[15:32] 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

[15:47] But on the other hand you can go iteratively and say well how many of them have no ones? Something that kind of seems harder and a more awkward way to count. this really not obvious fact from an algebraic standpoint.

[16:00] But again, just got to emphasize how like how general this ends up being. So what we want is some kind of object that each of these describes.

[16:12] 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. Oh, for the first time in my life I drew a quarter circle arc that wasn't terrible.

[16:26] It's not great, but usually that comes out much more disastrous. I should not be so pleased with a terrible quarter circle.

[16:38] And what does cosine mean in this context? Well, it tells us if we look at the x-coordinate of this point,

[16:51] And so now I want you to think about how can we Some geometric thing that we can point to in this image. let's draw a square off the side of this, you know,

[17:08] But then there's going to be a problem if the principle that we're Because if we describe this left-hand side as some kind of area,

[17:20] that would mean that we have to find another description of that same area, But that's going to be weird, because this doesn't involve squaring or anything like that. to describe that as some kind of ratio, or maybe some kind of length.

[17:39] 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.

[17:54] but I love it so much I'll talk about it again. the angle between the x-axis and the line, but on the other hand

[18:06] it's also telling us the angle between the line and the x-axis. And I know that sounds like the same thing, but it means when I say project down the point at the end of our radius, which was length 1,

[18:19] perpendicularly onto the x-axis, that length got scaled down by cosine theta. What if I say I want to project down in a perpendicular fashion onto this line?

[18:32] 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,

[18:44] 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.

[18:58] 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, the idea of double projecting based on asking, you know,

[19:13] is this line theta degrees away from that, or is that line theta degrees away from this? something of mathematical substance to recognize that symmetry. Well, we have a representation of cosine squared.

[19:27] We just were thinking about a context where we're able to relate an angle to 2 theta, So somehow we want to realize this angle as an inscribed angle of some kind of triangle.

[19:44] But before I do, just to mention another fact that if you've been puzzling around with these sorts of things, might be burning in your mind, is a specific instance of the inscribed angle theorem called Thales theorem.

[20:00] So let's say that that large angle we had, 2 theta, was actually 180 degrees, right? What would that mean in terms of the inscribed angle theorem? It means that if we take that diameter of the circle, if it's 180 degrees,

[20:15] 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 put a right triangle inside a circle,

[20:29] and whenever you do that, the hypotenuse of the right triangle is exactly the diameter If you wanted another proof of it, that's not just via the inscribed angle theorem,

[20:41] there's another very wonderful leveraging of symmetry that you can do, I'm going to reflect it through the origin, reflect it through the center of my circle, And recognize that reflecting through the center

[20:56] So if I rotated the image, not 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 the diameter

[21:11] which in particular means they have the same midpoint and they're the same distance apart, and you 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

[21:26] But I think that's a very beautiful way to think about Thale's theorem, it must be a rectangle, which means that this is a right angle. Well, we've got a right triangle sitting here that's from zero,

[21:41] 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, I'm going to make that hypotenuse a diameter of the circle.

[21:59] So this is still going to be an angle of theta, sitting right here. but I flipped it around so that my 90 degree angle is sitting up and to the right,

[22:12] The length that we care about is what happens when we project from the point at that 90 degree angle in a perpendicular fashion down onto the hypotenuse, And what we care about is this long length, okay?

[22:32] And actually, let me ask you as 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.

[22:47] because this is, uh, this is kind of a heart, part of the heart of this particular proof, So it specifies that the hypotenuse of the large right triangle above has a length of one.

[23:03] drawn in a unit circle, so the hypotenuse has a length of one. What is the length l in terms of theta? Okay, so can you find an expression for l in terms of theta?

[23:17] And I'll give you a little bit of, uh, a little bit of time for that, bring back our pause and ponder music, get myself a chance to take a drink.

[23:49] So while answers are rolling in, before I grade it, since it seems like a lot of you are already well ahead of me. That is the whole reason I'm bringing it up here.

[24:05] So in this context, I would draw some lines from the center of my, uh, of my new smaller circle that has radius only one half,

[24:17] And this is just lovely now, isn't it? Part of it, excuse me, part of it is the radius, which is one half,

[24:30] but then the projection down according to 2 theta onto this remaining leg ends up being that radius one half times the cosine of 2 theta.

[24:43] We've got the whole radius, which is one half, It's a radius of the circle times a scaled down version of that radius.

[24:55] which is what we get when we take this right triangle and we And it gives us this non-trivial relationship in trigonometry The fact that otherwise we were going into like

[25:11] So I think that's quite beautiful. the inscribed angle theorem suspiciously slos- suspiciously shows up. Um, and again, what I want you to take away is this principle that if you

[25:26] very powerful in terms of showing non-obvious, um, algebraic relations or anything that's kind of written down symbolically So with all of that, let's actually turn to the probability

[25:41] question that I asked at the beginning of the lecture. So going back to our live quiz, uh, I will go ahead and grade what we- we all know know the correct answer.

[25:56] So for those of you- wait a minute. No, I just slipped this one in like last minute before the lesson today. So, uh, well, it looks like 1380 if we were absolutely

[26:13] 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 red like, So congratulations to those of you who got that.

[26:28] 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

[26:44] written the thing appropriately. we're choosing two random numbers from the range zero through one, what the probability that they round down- the ratio of these numbers

[27:01] Remember zero is an even number, so that it rounds Now this is a tricky problem to think about and, uh, definitely no- no fault at all for,

[27:13] uh, anyone who isn't immediately able to see roughly where it should be. 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.

[27:28] Choosing two random numbers between zero and one. 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.

[27:43] Um, but essentially, uh, it's what you would expect where you're choosing some random point on this line, and the idea is that each point is as likely as another, should have a given probability that's independent of where that range showed up.

[27:59] You know, so you might have in the back of your mind the idea that And just to give an example of what we mean by uniform distribution,

[28:12] the probability that x sits between 0.3 and like 0.5. and the length of that range is 0.2, about a fifth of the entire length it

[28:25] What it means to be uniform is that that probability is actually just the length of the segment that it came from. what about the probability that it's precisely 0.3 or precisely 0.5?

[28:40] 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 zero. How can a probability of an event that's possible,

[28:55] Made a whole video about it trying to describe this, but really what it comes down to is that the things that have probability, Those are the fundamental objects, and it doesn't really matter

[29:10] how we treat the boundary and just think in terms of ranges. which is if we take the ratio of x and y and we round that down, which sometimes,

[29:22] saying we find the greatest integer smaller than that, That's a weird thing to think about, it's a hard problem in that way.

[29:34] it's kind of like use the defining features of the setup. I guess we'll be using lengths in some way to yield probabilities,

[29:46] Give things meaningful names, you know, maybe x and y Symmetry, okay, maybe, you know, the idea that choosing x and then y is

[30:00] 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 one as it is to be below one. And that actually does tell you something, because if we're wondering how

[30:13] often do you round down to be zero, 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. This would be a probability of 50 percent.

[30:30] but it's not clear how you would apply that to things like even numbers, same object, So principle number five here is where we're going to come in.

[30:42] It's something that you can not be like, yeah, yeah, 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,

[30:56] but just having some kind of sketch, uh, to give meaning to your terms can be very 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.

[31:08] we'll want to think about a single point with xy coordinates. whole problem two-dimensional in a very helpful way. which, where have you gone little straight edge?

[31:25] Oh, here we go. Even your objects sometimes get tired of math class and want to play truant now and then,

[31:38] but he has to stay whether he likes to or not. So let's say this is our x coordinate. x can fall anywhere between 0 and 1 with uniform probability.

[31:53] So when we have a pair of numbers, you know, something like 0.2, what is that, maybe like 0.8, pair of numbers, it's just a single point in this diagram.

[32:05] we're choosing a random point inside a square, a square with side length 1. because it's going to come down to some view of what's going on in this square.

[32:21] 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 0, so this would end up being even.

[32:37] But let's see if we can try to find a way of thinking And again, one very useful thing, if you get stuck,

[32:50] that I have enumerated down here as principle number six, You're solving something, it's hard. actually can solve and get some kind of foothold.

[33:04] 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, but I want you to answer it in a geometric way,

[33:19] because the next simpler question might be probability that it rounds down to two, We're going to do it as a live quiz because rather than me answering things,

[33:33] So jumping up to question number four at this point, Okay, we've got a, b, c, and d, and it's asking us which of these four regions

[33:46] corresponds to values of x and y, where 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?

[34:16] So, okay, I'm going to go ahead and lock in answers, but if you want to keep thinking about it, definitely feel free to pause

[34:39] I don't want to rush anyone. So it looks like 1265 of you, 1273, always answers rolling in at the end, And let's take a moment to think about why that's the case.

[34:56] seemed to fall in the region or not, but let's see if we can understand this in a way that lets us make progress onto the other even numbers. saying is that that ratio sits somewhere between zero and one.

[35:13] 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

[35:28] or another, I multiply everything by y, and we're basically asking when is x less than y? that region is going to be described by the equality. Well, that's just a straight line that goes diagonally.

[35:45] If we draw our line y equals x, that's what we get. look to the left of it or to the right of it, either you can think very directly and say,

[36:00] 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

[36:14] x divided by 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 divided by y rounds down to be two?

[36:30] I want you to answer it. When is it that x divided by y inside our unit square of points x comma y rounds down to two?

[36:43] And we've got four possible geometric regions that this could correspond to, And really, you know, rather than just thinking about which of these is it, really try to think through why it's the case and how you can prove

[36:57] For example, I want it to be the case that if I didn't show the correct answer here, you would be able to confidently come and say like, no,

[37:09] I'm quite positive that the correct answer is nothing that you've shown here. So again, I'll give you give you a little moment to think about that.

[37:24] So so Okay, so once again, I'm going to lock in answers potentially

[37:47] earlier than you want me to, but keep the lesson moving forward.

[38:10] because hopefully the explanation will make them do so. and let's go ahead and think through why that's the case.

[38:22] 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 It's saying that x divided by y is greater than or equal to 2 if it's

[38:37] 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 or equal to x,

[38:52] which is less than or equal to 3 times y. 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 2y less than or equal to x.

[39:07] Well, that's the same thing as saying y is less than or equal to x halves. And same deal, 3y being greater than x, that's the same thing as saying y is greater than x divided by 3.

[39:21] The line y equals x halves, which has a slope of one half, you can think of it as intersecting at the point where y equals one half when x equals 1.

[39:34] Right, 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,

[39:48] where one of these represents x over 2, and one of these represents x over 3. And then part of the part of the intrusion into the space of my last inequality.

[40:00] 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 where rounding down gets to 2. And I want you to appreciate how this is a kind of complicated thing to

[40:14] 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.

[40:27] x is more than two times what y is, but it's not three times more than what y is. I said it wrong. When y is more than two times what x is, but it's not three times more than x is.

[40:42] And it's even harder to try to give some sort of probability to that, It is the area of this region because the full area of possibilities already is one,

[40:56] so this, the probability of something happening should be one, and we just need to look at the area of that. because for the next term when we want to know when does x divided by

[41:12] y sit between four and five, we're going to be drawing lines that Maybe I'll go to a different color for this one. x over four and x over five, which is going to require very small

[41:27] handwriting at this point, but I'm going to 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.

[41:45] so that gives us sort of another phase of challenge to the problem. 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,

[42:00] 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. I only went out to like a hundred or something like that,

[42:12] So we've made progress, but this is still hard. Right, that's not necessarily an obvious thing.

[42:24] 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 going to look like one half base times height. direction as being their height, every one of them has a height of one.

[42:39] So each one is going to look like one half times a base of some kind. different piece of paper actually so I can keep it up close. I'll take one half times whatever the base of the triangle is times the height.

[42:53] So our first triangle, that base is, that base has a length one. So that's going to correspond to the one half probability of going to zero. The next triangle, we have to look at this length here between one third and one half.

[43:07] Well, actually I'm just going to write it out as a half minus a third. It equals a sixth, but writing it out like that kind of reminds us where it came from, That could maybe be another problem solving tip, don't collapse things too soon.

[43:23] came from because sometimes that helps see overall patterns. Well, it's a fourth minus a fifth. So we have a fourth minus a fifth.

[43:40] And in general we have this kind of oscillating sum, a sixth minus a seventh, but we're adding that up infinitely many different times, okay,

[43:52] And from here, 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.

[44:07] You might recognize, let's say if you were watching a particular lockdown math lecture a week or two ago, that this alternating sum, one minus a half plus a third minus fourth plus a fifth, on and on,

[44:20] actually equals the natural log of two. 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

[44:32] formula and then immediately see that this is how you're going to solve it, unless it's something that you've seen before, which can make it seem all the more opaque. We did this strange thing where we made it seem like a harder question at first,

[44:47] 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.

[45:01] 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,

[45:13] each one of those terms I can nicely express as an integral of a much more x cubed minus, now let's see, plus x to the fourth.

[45:25] that increases and then we divide by what the exponent is. And the reason that you would want to do this is So just to make this maybe more explicit, if we evaluate the integral from zero to one,

[45:44] it at one and subtracting at zero. And again, this is just such a bizarre thing that if you hadn't recognized the sum,

[45:56] like something that you could have found, which is frustrating in the context But I'll keep walking through it just to give a little bit of closure to this.

[46:11] We've got this infinite sum that if you had been familiar with geometric sums, you would be able to write this as 1 over 1 plus x, So you always take 1 over 1 minus the thing you're multiplying by, which again,

[46:29] it's kind of one of these things where it's relying on you recognizing it in some way. 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.

[46:45] And we're evaluating this between 0 and 1, which is to say we're taking the natural log of 1 plus 1, or 2, minus the natural log of 1, which is 0. And that's why all of these things are the natural log of 2.

[46:58] to take this and then apply it to our 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

[47:12] that can be collapsed because of geometric series, And then thinking about like, what is what is the thing that you can teach

[47:24] someone to say come away and be able to solve problems in the same way? but I actually think it's maybe the most potent one and the most honest one.

[47:36] Read as much as you can. You know, watch YouTube videos on math if they're substantive, things like that. because what you want is to be able to say like, well,

[47:49] how could I have come to this on my own without having merely recognized it? is really just pattern recognition, but wearing a little bit of added clothing.

[48:01] And sometimes it's patterns not so much that you're directly recognizing exactly this, Like geometric series, that comes up a lot. you will recognize geometric series, even if it's in a context that you've

[48:16] never actually seen before, or a specific geometric series that you've never seen before. knowing that you can have the natural log pop out of an integral like this, And I think recognizing the truth of this as being the key

[48:30] to a lot of problem solving is actually pretty inspiring. Because oftentimes you find yourself in a situation where somebody is, they're just faster, they're just better, they just recognize things more so than you do.

[48:43] 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

[48:56] wondering like, wow, I just, you know, I'm just not in the same league at all, right? That person, they just have some sort of innate But I think the truth of the matter is that the people who are

[49:11] showing that kind of ingenuity, they've just exposed themselves There is a path towards that which takes the form of practice. you're looking at and trying to digest the deeper principles behind it.

[49:28] okay, I can say leverage symmetry, but what does that actually mean? and turn that into something that's formulaically useful? Don't just be satisfied with the answer, see if

[49:44] So in that way, this number seven, like, it's the most frustrating, but it's 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,

[49:59] You just have to do a lot of practice and expose yourself to a lot. you wouldn't have thought that that's a thing that you're going to

[50:11] They show up in unexpected places. So what you can do then, is say, well, we've got our whole expression that involves this very alternating sum, and you'd say, okay,

[50:24] 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,

[50:38] all of these areas. Now at this point, there is one thing that I think separates really good problem solvers, to ones who are like merely good, who, you know, they aren't necessarily perfect scores.

[50:54] 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.

[51:10] 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. numerically what my answer turns out to be, right?

[51:27] So let's just see what does that end up being? And natural log of two is around 0.69, so maybe it's not too surprising, So if we write that down as one of the, as the answer that we just got,

[51:44] I told you I would purposefully make a mistake, so hopefully you're not yelling too loud at this point, does that make sense? And if you look at our picture, well, the probability has to be at least a half,

[51:59] So certainly it couldn't be the case that the whole thing adds up to be only 0.35, Everybody makes silly mistakes, everybody drops a minus sign

[52:11] or applies some rule that doesn't quite apply in a circumstance. 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.

[52:26] perspectives that can give you a reasonability check. more carefully at how we were applying things, this sum isn't quite the alternating sum that converges to natural log of 2.

[52:40] and it's only after that that we start alternating. And you might see this by recognizing we were subtracting all the even numbers here,

[52:53] So it's similar, but it's not the same. 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,

[53:08] so that would mean we're subtracting the one half, then we're adding the one third, then we're subtracting the one fourth, we're flipping all of the signs of everything beyond that 1, that is the thing that equals the natural log of 2.

[53:20] Which in turn implies that that remainder of the sum looks like 1 minus the natural log of 2. When we plug it into our original expression, it's saying that

[53:34] That didn't even pass our basic reasonability test. Instead, it'll be one half of one plus one minus the natural log of two,

[53:46] which is just two minus the natural log of two. What does this actually equal numerically? or if you want to see more precisely, we can pull up a calculator.

[54:02] we don't want the natural log of 2, that was wrong. Does that pass our basic reasonability test?

[54:14] 0.65, that looks like a reasonable answer to what the area 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,

[54:30] well, about one sixth, half of a sixth, basically, because this length was a half minus a third, which makes it a sixth, and then it's a triangle, so one half base times height.

[54:42] So that's about a 12th, or 0.083. Something around 0.65 seems pretty reasonable. that this is probably the correct answer.

[54:57] because probability questions very often you can kind of cheat and just and seeing what happens. So I want to do that actually on this one, just to give us a little bit of confidence

[55:12] 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. any questions from the audience, and it certainly looks like there are.

[55:27] So let's see if we can address some of these, get myself out of the way of the questions. an intentional mistake in his long speech to test his son. Every time I make a mistake, it was always purposeful,

[55:44] All right, keep making more episodes. You know, the main one, honestly, if I go much longer in this lockdown the style of like a 1980s music video, just to keep stylistic consistency.

[56:02] You know, I think that's just going to be a little bit too much effort, so I'll have to wait until whenever it's possible to get a haircut again, then we can go. I've got a long pile of things I want to get to, and you know, the lectures take time.

[56:17] I might spin up something like this again, might do it on a separate channel. 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.

[56:32] And lastly, can you help us understand derangements and the principles Yeah, boy. Tell you what, maybe just a separate video.

[56:46] If I do the combinatorics course, actually, that would be a perfect example for it. but very naturally what you end up with is taking 1 over 1 factorial minus 1 over 2

[57:00] factorial plus 1 over 3 factorial minus, and you're doing this alternating sum with And if you think of e as 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.

[57:15] 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 gut check this programmatically for mathematical problem solving, is to learn at least a little bit of programming.

[57:31] 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.

[57:45] Sometimes you can go about it mathematically, but then when you try to make it computational, 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?

[57:59] 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. when it's not computationally viable to give a perfect illustration of what

[58:13] And I actually think that does make for better problem solving in general, So let's just try this out, see if we can understand the probability question we

[58:25] were just looking at in terms of just, you know, what's the word I'm looking for? So I'm going to import numpy, which evidently some heathens refer to as numpy,

[58:40] which just, I just can't, that seems awful to me. this first random refers to the library, the second one is a function.

[58:52] It will return a number between 0 and 1 according to a uniform distribution. So it is as likely to choose something between, you know, 0.1 and 0.2 as it is to choose something between 0.8 and 0.9.

[59:04] And what's nice is I can get a list of them, so in this case I get a list of 10 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,

[59:16] and if I take x divided by y, it does it term by term. So for example this first number that we see in here, that's 0.739, that's taking the 0.52 divided by the 0.7.

[59:28] so we can see all of our ratios like that. And in general I might just define a ratios list where I'm going to take my,

[59:40] a bunch of random numbers of size n, and then divide it by another bunch of random numbers of the same size. n is not defined, but I'll give it a definition, some nice and big number like a million.

[59:53] And now I'm holding on to a million ratios, a million examples of x divided by y, which is kind of cool if you think about it.

[1:00:05] and some that would round down to zero, some that would round down to two, And now I can start asking questions, like, you know,

[1:00:17] when is it that I take these ratios, I take the floor function of them, not the four, 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,

[1:00:30] 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.

[1:00:44] We could have also asked when is it about two, okay, and it looks like 0.83. And remember from our diagram what we were looking for is when when it's in this green

[1:00:56] which has an area that was a half minus a third, but all times a half, because it's one half base times height for a triangle. So if we pop back over to our terminal and say, okay,

[1:01:10] if we were looking for one half times the base of that triangle, which was a half minus a third, we would expect that proportion to have been 0.83. And we could even answer our actual question, which is to take the floor,

[1:01:26] and then if I say I want to divide by two and ask when the remainder is zero, 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,

[1:01:38] which is about the answer that we were looking for. You know, we were looking for something that was too half of two minus the natural log, So we have this wonderful way to kind of empirically verify.

[1:01:55] And of course, and most of the times with in-lockdown math that I've pulled up Python, If you wanted to, you could try to visualize stuff. So something like matplotlib is definitely a great pyplot.

[1:02:10] This is a very good library for just like simple data visualization that you can pull up pretty quickly. So I'm going to take a histogram of all of my data, and I have to specify a range.

[1:02:27] those values get bucketed in between like 0 and 20. So my number of bins is 20.

[1:02:39] Don't you love when that happens when like a sneeze It just recedes into the darkness because it knows that you're better than the sneeze. Again, I'm more pleased with myself than I should be there.

[1:02:54] And um, I always, you always have to add a relative width on these sorts of histograms because they just look ugly if they're all like side by side, I think sometimes. So if I do this, um, and then I show what the plot ends up being,

[1:03:06] I have something that has shown up on my screen, but not your screen. So you can get a sense of, you know, this bar represents all of the ones that All the ones that rounded down to one, rounded down to two.

[1:03:22] And you can get this, um, this nice sense for what all of your data is. on top of whatever you find more analytically. for more like pure mathematical problem solving.

[1:03:36] 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 Desmos a lot to like show graphs of things because I just love Desmos.

[1:03:50] And I'm actually friends with some of the people who work there because quite often a company's people are as delightful as its products. were doing like an art contest among some students.

[1:04:04] And I just wanted to showcase some of, I'm not sure if these are the winners or the finalists of the art contest, but basically in various different categories of students who are, I think it was like 12 through 14, 15 through 17,

[1:04:16] 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 are mathematical graphs that someone wrote just with an analytic description and they were,

[1:04:28] 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.

[1:04:43] 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.

[1:04:55] This is by Katsini, is a Bézier Knight. where each one of the curves here is described according to something

[1:05:10] It's a kind of cubic parametric term, and they just recreated a starry knight in a way that's completely beautiful, I think.

[1:05:23] 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. up with little pictures of like a smiley face and things like that.

[1:05:40] 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. thank you to Ben Eater and to Cam, who've been extremely helpful with

[1:05:56] 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

[1:06:08] 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, it's simply named Ben Eater, like 100% check it out, definitely subscribe to it, try some of the projects.

[1:06:21] I think it's absolutely great. So cannot emphasize enough how grateful I am in the direction of both of those two. you kind of relive the lockdown math experience and have like

[1:06:38] So stay tuned on that. And if you're interested in being a beta user of it, there will be forums for how you can how you can reach out to both of them. This has been very fun for me, very different for me.

[1:06:52] And I will shortly get back to the more usual videos, which I'm very excited about. especially if you like problem solving stuff like this, That's some of what's on the horizon.

[1:07:08] And with that I will simply say keep loving math and enjoy the rest of your day.

[1:07:22] So Thank you.