#the math fandom

doing numbers in the uni math discord for my top-notch memes

OH okay I actually have a really good explanation for this one, thanks to my number theory professor from college

so the Monty Hall problem is the setup of “there are three doors, behind one is a car and behind two are goats. you keep what is behind the door you pick and for the sake of the game you want a car not a goat, live your best goat life irl and not on my math problem. you pick one door. the announcer, who already knows what door the car is behind, reveals one of the doors that you haven’t picked as a goat. you have the opportunity to switch or not. should you?”

the instinctive answer is “eh why the fuck should I, it’s a 50/50 chance?”

but that’s the wrong answer, the mathematical answer is “it’s 2/3rds to get the car if you switch and 1/3rd if you don’t”

there are a lot of ways to mathematically convince yourself of that fact; probably the easiest way bc it’s a problem with a very small number of possibilities is just to chart out literally every single possibility, and then count number of times that a switch wins, and number of times that a stay wins. (ie, if the car is behind door 1, there are three possibilities of you pick door 1, 2, or 3, and then for each of those, either two or one possibilities of what the host picks.) you could plug it into a conditional probability formula. but my favorite intuitive solution is the following:

just re-imagine the problem with 100 doors. you pick one door out of 100 at random, 99 with goats behind them and one with a car behind it. your host opens 98 doors all with goats. you have the choice to keep the door that you picked, or choose the door that the host didn’t open. do you switch?

well, obviously. you had a 1/100 chance of getting it right in the first place, whereas the host knows where the car is, there is a 99% chance it was under one of the doors that you didn’t pick, and, well, now there’s only one option for “one of the doors that you didn’t pick” so there’s a 99% chance it was behind that one.

Binomial coefficients!

Binomial coefficients are pretty great, and there are essentially 3 ways of thinking about them? Which because this is math are all Equally Valid and which one you want to think about depends on what is the most useful situation! 

1. The One That Makes Sense Why That’s The Name:

We’re going to refer to the binomial coefficients as BC(n,k). For this, we want n and k to be integers with n>=k>=0. 

So a polynomial is just any expression in powers of ‘x’, right? like 5x^2+7x+1 or 3x^27-x^6. Blah blah blah. And bi means 2, because as you’re going to see we’re going to expand (x+1) and those are two things. And the coefficient is the number in front of the power of x. And this all very conveniently comes together to give you:

(x+1)^n=BC(n,n)x^n+BC(n,n-1)x^(n-1)+….+BC(n,1)x+BC(n,0)

so for example, if you know all the binomial coefficients when n=4 (which happen to be 1,4,6,4,1) without having to annoyingly foil something three times you just. immediately know that

(x+1)^4=x^4+4x^3+6x^2+4x+1

You can also do with with (x+y)^n, in which case your terms are going to be BC(n,k)x^(n-k)y^k. But that doesn’t look quite as pretty as using 1, and binomial coefficients are usually introduced with 1, the ‘y’ is just a generalization. but it also doesn’t have to be a variable – maybe you want to calculate (x+3)^4 – well, you do it like you’d be doing the (x+1)^4 above but make sure to multiply the coefficients by 1,3,9,27, and 81 respectively.

Super neat, right? FAST MATH.

2. Fuck It Just Give Me The Formula:

So the binomial coefficient (n,k) is calculated using factorials – which is when you multiply the number times all of the numbers smaller than it. Very specifically, the binomial coefficient BC(n,k) is just n!/k!(n-k)!. Note that 0!=1, by definition. Plug and chug. In our previous example, we knew that the binomial coefficients for n=4, well, we could have worked it out like this:

BC(4,4)=4!/4!(4-4)!=4!/4!=1BC(4,3)=4!/3!(4-3)!=(4*3*2*1)/(3*2*1*1)=4BC(4,2)=4!/2!(4-2)!=(4*3*2*1)/(2*1*2*1)=6BC(4,1)=4!/1!(4-1)!=(4*3*2*1)/(1*3*2*1)=4BC(4,0)=4!/0!(4-0)!=4!/4!=1

These actually come up a lot in probability, because let’s be real, how many times are you expanding (x+1)^4? The binary coefficient BC(n,k) is usually called n choose k. And it’s important because n choose k is the number of different ways you can pick k elements, order not mattering, from a list of n distinct elements. So basically let’s say there’s a class of 30 people and the professor is assigning a group project with groups of 3 pulled randomly. What are all of the possible potential groups that could be pulled randomly, at least for the first group? Well, 30 possibilities for the first person, 29 possibilities for the second, and 28 for the third, so the total possibilities are 30*29*28 – which is 30!/(30-3)!. But you only care about the members of the group, not what order they were pulled in. The set (1,2,3) has 3!=6 possible permutations: (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1). In general a set with k numbers has k! permutations. So we need to divide out by all the possibilities of the same three people having been pulled in a different order – which gives us 30!/3!(30-3)!. This turns out to be pretty big – 4060, actually. So the probability that you and your two best friends will all end up in the same group and that group will be pulled first out of all of the possible groups that can be pulled first is 1/4060, or approximately a 0.025% chance.

But you can see how in general, the question of “how many possible sets of k can I pull from a set of n” is hella useful? And it just. Is the binomial coefficient. And the explanation above is why. Honestly, I recommend memorizing the formula and memorizing what it’s used for, it comes up so much in combinatorics and just. Everything.

3. Oooh Pretty

So you know Pascal’s Triangle? That fun thing where you start with 1 // 1,1, and then each subsequent row you make by adding the two numbers immediately above it?

the rows are the binomial coefficients. 

…this is the divide?@ savrenim

I was so mad that I made a bad math pun back

I guess you can just say that gradient is basically the same as divergent only it produces a vector, but I’m wondering if you’ve got other explanations. (2/2)

so your prof is right in that this is just the product rule but in multiple dimensions. basically, you CANNOT take the divergence of a scalar field, only a vector field, because you need a vector entry? it’s like having a function f(x, y, z) and only supplying “x” and then going “calculate” and you’re like “I can’t that’s 3D and you gave me a 1D input???” similarly, gradient is “1D” and if you give it a “3D” vector entry it doesn’t make sense. (actually, funnily enough, gradient takes in 1D and outputs vector/3D which is going to be important). So once you agree that you gotta use gradients on scalar fields and produce vectors and divergences on vector fields and produce scalars, and that it’s dumb because the exact same symbol is used for both, it’s just a matter of writing everything out to see why it’s true: