An ingenious & unexpected proof of the Binomial Theorem (2 of 2: Proof)

[kaustavpal]

The way he unpacks the story with drama and suspense is astounding, more so, considering the story is a mathematical proof.

[jaroddavid]

Eddie said "sus" before it was cool

[joshuaronisjr]

Awesome video Eddie! Thanks!

For people saying its circular...not really. Here's why:

First, look at the left side. There's no (nCr) there, that's just taking derivatives. No assumptions there.

Now, look at the right side. Although Eddie is writing that f(x) = (nC0) + (nC1)x + (nC2)x^2 +...+(nCn)x^n, at this point he doesn't actually know what any of those coefficients are!

He may as well have said f(x) = k + Lx + Mx^2 + ... + Px^n, or whatever!

However, he DOES know that f(x) CAN be written in that form - as a sum of a bunch of Xs, each raised to some power, and some coefficient in front. You can take a moment to convince yourself of that. He just doesn't know what those coefficients ARE.

Now, the reason that CALLING them (nC0), (nC1), ...(nCn) makes sense comes from the very nature of the function itself.

Take for example: f(x) = (1+x)^3

f(x) = (1+x)(1+x)(1+x)

The coefficient associated with the first x, the X^0, (so the 1) is like asking: "How many ways are there for me to not choose any of the X_s in here to multiply together?" Obviously, the answer is 1, but we don't actually need to know that yet. All we know is that (nC0) is a valid name for whatever that constant is.

The coefficient associated with the second x, with the X^1, (so n) is like asking: "How many ways are there for me to choose a single of the X_s in this function to multiply together, and then multiply it with two of the 1_s?" Obviously, the answer is 3, since we can choose the first X and multiply it with two 1s, the second X and multiply it with two 1s, or the third X and multiply it with two 1s. But we don't actually need to know that yet. All we know is that (nC1) is a valid name for whatever that constant is.

And so on and so on...

Then, he goes on to prove that r!(nCr) (which remember, we DON'T YET KNOW WHAT (nCr) IS! ITS JUST A PLACE HOLDER FOR A COEFFICIENT) is equal to ( n!/(n-r)! ). Which means that (nCr), which we previously didn't know, is ( n!/r!(n-r)! ).

And thus, no circularity.

The main thing to remember is that even though he's writing (nC0 nC1 nC2...nCn) as the coefficients, they don't actually mean anything until he derives what they actually are! He didn't assume the binomial expansion to begin with, he just chose those as place holders for constants that he DID know were there, but he didn't know what the constants actually were.


And, he chose (nCm) (which is read as n CHOOSE m) because as a name, it makes sense, since for the x^m term we're trying to find the number of ways to multiply m of the n X_s together!

[jasonbarrylambeth]

I know this video is years old and likely this will never reach you... but... currently I am studying my masters of teaching and your videos have inspired me more than the 18 month degree (with 3 placements). The engagement you extract from students is absolutely insane... you are a gift to those kids and they probably don't even understand why. Keep being amazing, you will likely birth the next Einstein or Cantor :p

[anthonykula]

Eddie,
I teach calculus in high school. What a great introduction to discrete math. Nicely put together.

[lifeofphyraprun]

Mind blowing!Very very unexpected!And your narration style is just the best of the best Mr.Woo!

[sumitkumaradhya]

The formula of differentiation used here is d/dx (x^n) = nx^n-1 which we derive from the first principles using the binomial theorem itself. Now you can't use the calculus formula again to show a proof for the binomial theorem. That will be nothing but going round the circle.


Basically calculus originates from the Taylor's theorem which inturn comes from the binomial theorem. Therefore Calculus originates from the binomial theorem. Now you can't use calculus to get back the binomial theorem again !!!

[janvisagie]

I always remember using induction to prove this theorem and I love the alternative given here, though I have one big worry: the terms inside the brackets were chosen such that it only has one variable, x. Therefore if you have an equation with 2 variables you get a function that is dependent on both f(x, y). How should I go about this, just set y=1 and work with f(x, 1) which is the same equation as was used in this video. Secondly, it appears that this proof only focuses on the coefficients that are generated by the binomial theorem and not the pattern for the variables and their powers (this may have been part of a previous vid, I'm unsure).

[QYong-rqiw]

Just to remind there's a minor mistake at 12:34

[marius]

9:21 isn't this a circular argument?
the whole point is to find the nCr formula, but it's being used here as nC(0), nC(1) etc..

[fluoroantimonic]

4:15 it's so sus that he predicted sus

[krystofsedlacek]

At 12:30 shouldnt it just be times 4 and 3?

[forthrightgambitia]

I am presuming that for this proof to be rigorous we'd need to prove the uniqueness of the Taylor series? (Which if the two function definitions have all the same derivatives will obviously be the same.)

[凌彬翔]

At 12:30 shouldnt it just be times 4 and

[ambroselam]

“And that’s fricking badass I think.” ~Eddie Woo 2014

[michelef]

Is the connection between apparently unrelated things called duality or is it plain wrong to use this word in such a loose way?

[cheekyismymiddlename]

this is slightly mind blowing...wow. never would have thought

[justacockypersonontheinter]

hi. i like the video, and the proof. the first part strongly reminds me taylor approximation and the second part is finite differences shewn in book by Euler foundations of diff. calculus chapter 1.
for exponent taking integers perhaps we can prove it by this method, but how about integers and negative exponent?
does the proof stay exact the same? and it might sound weird, but do you know anyone who recorded "original problem" where it have appear the first time? i mean retrospective look, often times helpful. because again, Euler in his books widely used binominal theorem but actually didn't prove it, except by induction. thank you, i will apreciate any help)

[gouranggehlot]

It's just the maclaurin expansion performed in a different fashion, (genius don't do different things they do the same thing differently.),

[epsilia]

4:14 If only they knew at that time ...