This content introduces a series of "deceptively simple" problem-solving principles, illustrated through mathematical examples, with a focus on developing systematic approaches to novel challenges. It also highlights a new interactive platform for learning and practicing these concepts.
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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
going to 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 a 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, and a hard problem that will be fun to tackle when
the time comes.
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.
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 0 or it 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 and you think
of choosing two random numbers between 0 and 1,
looking at their ratio, what do you think that probability is going to be?
We'll return back to this later.
And one thing I want to say is that this whole live quizzing software,
it's something that's being 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 Cam Christensen.
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 we'll 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 itempool.com slash c slash 3b1b, 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 going to put things that are relevant to the lectures in there,
which I think should be fun.
It's not quite there yet, but definitely check back.
And I'm sure just over the next couple months
there will be more item pool shenanigans 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 going to 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 going to 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 problem solving tip, not problem sip, 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 piece 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 is the same distance away from the center as these other points.
So I am 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 it and you wonder like, how would you have known what to add?
Like again, 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 again, 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 going to 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, 3b1b.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, when I was going through it,
it's honestly pretty fun to just kind of click and like, oh,
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.
All right, so because there's such strong consistent- consensus,
I feel comfortable grading this.
Oh, I thought I got it right on 1 1 1 1.
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 walk through why that's the case, okay?
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 because I was just naming angles a and b,
but separate context, 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.
You 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 180 degrees, and there's two different alphas in there,
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.
Similarly, the one with all of the betas tells us that if
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
image is that alpha prime, beta prime, and theta L add up to 360 degrees, or 2 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, your 3Blue 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 canceling them out.
So I'm gonna, you know, add this top equation
and maybe I 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 2 times alpha plus beta.
You know, each of those gets subtracted off with the coefficient 2.
, 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 2 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 circumstances regarding,
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.
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 is 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 a not 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 one half of one plus cosine of two 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 it viscerally and 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 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.
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.
It's not great, but usually that comes out much more disastrous.
How pleasing.
I 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 vanilla 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 vanilla 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 2 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 did talk about in the trig 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 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,
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,
the idea of double projecting based on asking, you know,
is this line theta degrees away from that, or is that line theta degrees away from this?
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 2 theta.
We just were thinking about a context where we're able to relate an angle to 2 theta,
to 2 times that angle.
So somehow we want to realize this angle as an inscribed angle of some kind of triangle.
And I'll show you how you can do that.
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.
So let's say that that large angle we had, 2 theta, was actually 180 degrees, right?
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, if 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 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 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 through the center
is the same as rotating the whole image 90 degrees.
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
of the circle, but the other diagonal is also diameter of the circle,
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
relevant angles.
But I think that's a very beautiful way to think about Thale's 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 going to make that hypotenuse a diameter of the circle.
So this is still going to be an angle of theta, sitting right here.
The, uh, basically I flipped it around, so previously it was here,
but I flipped it around so that my 90 degree angle is sitting up and to the right,
rather than sitting down here.
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,
which now looks like this.
And what we care about is this long length, okay?
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.
Let me give you a little bit of time to think about this one,
because this is, uh, this is kind of a heart, part of the heart of this particular proof,
and I think it's very, very pleasing to see.
So it specifies that the hypotenuse of the large right triangle above has a length of one.
Our context is because it came from the hypotenuse of the triangle
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?
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.
So, 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 going to 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 I'm 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 center of my, uh,
of my new smaller circle that has radius only one half,
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 one half,
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.
And of course, that's exactly what we want it to be.
We've got the whole radius, which is one half,
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 exponentials to understand.
So I think that's quite beautiful.
I think that's just, um, one of the many many instances of where
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
can have one object described in two different ways,
very powerful in terms of showing non-obvious, um,
algebraic 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, uh, I will go ahead
and grade what we- we all know know the correct answer.
So for those of you- wait 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.
Um, so I might have like swapped around what the answers were.
So, uh, 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 red like,
oh no, but, uh, 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, um,
written the thing appropriately.
So just as a reminder of what the question is,
we're choosing two random numbers from the range zero through one,
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 zero is an even number, so that it rounds
down to zero or two or four or anything like that.
Now this is a tricky problem to think about and, uh, definitely no- no fault at all for,
uh, anyone who isn't immediately able to see roughly where it should be.
But I think, um, 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 zero and one.
I think that's a weird thing to think about, um,
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.
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,
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.
You know, so you might have in the back of your mind the idea that
we've chosen two points, they're each somewhere between zero and one.
And just to give an example of what we mean by uniform distribution,
the probability that x sits between 0.3 and like 0.5.
Because it's going to be somewhere between zero and one,
and the length of that range is 0.2, about a fifth of the entire length it
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 zero.
This is a thing many find very confusing.
How can a probability of an event that's possible,
it's definitely possible to hit 0.3, have probability zero?
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,
you know, 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 setup.
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, okay, 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 one as it is to be below one.
And that actually does tell you something, because if we're wondering how
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.
So there's a 50-50 chance that this should happen.
This would be a probability of 50 percent.
So that gets you somewhere, which is kind of nice,
but it's not clear how you would apply that to things like 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 can not 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, uh, 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 as 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 straight edge,
which, where have you gone little straight edge?
I can only throw you so far.
Oh, here we go.
Things, they run away from you.
Even your objects sometimes get tired of math class and want to play truant now and then,
but he has to stay whether he likes to or not.
All right. All right, absurd.
So let's say this is our x coordinate.
x can fall anywhere between 0 and 1 with uniform probability.
y can fall between 0 and 1.
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.
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 1.
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 0, 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,
that 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,
something that will generalize to rounding to other things,
because the next simpler question might be probability that it rounds down to two,
and things like that.
You know where this is going to go.
We're going to 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.
Okay, we've got a, b, c, and d, and it's asking us which of these four regions
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?
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
the video and do so.
I don't want to rush anyone.
So it looks like 1265 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.
Okay, you can do so just with a pile of examples and just see which one
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.
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 diagonally.
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, y 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 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?
And again, I don't want to answer it.
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?
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 thinking 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 want it 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.
So so Okay, so once again, I'm going to 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 or equal to x,
which is less than or equal to 3 times y.
Now quite often we don't think of y as 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 2y 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, 3y 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 equalities associated with each of these.
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.
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,
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.
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
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.
No, yeah, sorry.
x is more than two times what y is, but it's not three times more than what y is.
It's not, not, no, 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, it's very easy to get confuddled 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 possibilities already is one,
so this, 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 does x divided by
y sit between four and five, we're going to be drawing lines that
intersect at x over four.
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
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.
And we're going to 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?
Right, 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 going to look like one half 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 one half times a base of some kind.
So I'm going to take, maybe I'll write this out on a
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.
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.
What is that length?
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,
so we don't want to collapse things too soon.
That could maybe be another problem solving tip, don't collapse things too soon.
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 a fourth minus a fifth.
That's the distance between these given how they were defined.
So we have a fourth minus a fifth.
And in general we have this kind of oscillating sum, a sixth minus a seventh,
where we have all the reciprocals of the natural numbers,
but we're adding that up infinitely many different times, okay,
and we want to know what that sum happens to be.
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.
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,
actually equals the natural log of two.
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 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,
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 I 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 maybe more explicit, if we evaluate the integral from zero to one,
that's the same as taking this whole expression and evaluating
it at one and subtracting at zero.
So that will give us what happens when we plug in at one.
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 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.
And think about what has to go on there in order to be able
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
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 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 merely 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 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.
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, wow, 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
instinct that makes them really good at this stuff.
But I think the truth of the matter is that the people who are
showing that kind of ingenuity, they've just exposed 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 formulaically 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 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,
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, is 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 silly mistakes that come in here or there.
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.
Okay, because there's going to 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 one half times the natural log of two.
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,
we're around 0.346, around 0.35, okay?
So if we write that down as one of the, as 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 a half,
because that's the region where it rounds down to zero.
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 applies 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 gut check your answer, have like two different
perspectives that can give you a reasonability 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 one half,
and it's only after that that we start alternating.
It was plus plus, then 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 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.
Which in turn implies that that remainder of the
sum looks like 1 minus the natural log of 2.
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 2.
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,
which is just two minus the natural log of two.
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, no,
we don't want the natural log of 2, that was wrong.
2 minus that, maybe 0.653.
Does that pass our basic reasonability test?
Yeah, I think so, right?
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,
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.
So that's about a 12th, or 0.083.
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 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.
All right, 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 going to 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 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.
Really, though, I also just miss 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, 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 derangements and the principles
of inclusion and exclusion, just like combinatorics 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 would be a perfect example for 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 why e pops out.
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.
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
if you wanted to, because the very last 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 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 when those mismatch, 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, what's the word I'm looking for?
Cheating!
Seeing what it turns out to be.
So I'm going to 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 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.
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,
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.
This next one is taking 0.66 divided by 0.56 on and on,
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,
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.
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 zero,
and some that would round down to zero, some that would round down to two,
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 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,
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 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
region here, when it's in that green triangle,
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,
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 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 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,
what proportion of them give me true, and yeah, 0.65,
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,
natural log of two.
So we have this wonderful way to kind of empirically verify.
And of course, and most of the times with in-lockdown math that I've pulled up 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 pyplot.
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 0 and 20.
So my number of bins is 20.
I was about to sneeze, but I held it in.
Don't you love when that happens when 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 the sneeze.
Right.
Again, I'm more pleased with myself than I should be there.
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,
I have something that has 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 rounded down to one, rounded down to two.
And you can get this, um, this 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 Desmos 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 a company's 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
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,
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,
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 Katsini, is a Bézier Knight.
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 Bézier curve, very useful for computer graphics.
It's a kind of cubic parametric term, and they just recreated
a starry knight 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, it'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 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 and actual 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 forums 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 going to 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.
So Thank you.
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