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

[@kaustavpal5510]

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

[@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!

[@JohnSmith-ol5kh]

Well, how can he use the power rule without the binomial theorem already being proven?

[@christopherwrogg6064]

@@JohnSmith-ol5kh Traditionally you prove integrability and then prove the derivative rules, namely ln and e^x from this it follows that you can prove the power rule without the binomial expansion.

[@lifeofphyraprun7601]

Fantastic!

[@aashsyed1277]

@@JohnSmith-ol5kh DISLIKED

[@siddanthvenkatesh2744]

@@JohnSmith-ol5kh y=x^n => ln(y)=nln(x) differentiate (1/y)*y'=(1/x)*n+ln(x)*0 y'=yn/x y'=x^n*n/x y'=x^(n-1)*n Reordering y'=nx^(n-1)

[@jaroddavid5933]

Eddie said "sus" before it was cool

[@JoaoPauluu]

lmfao

[@ripjawsquad]

timestamp

[@SoniSingh-yn3yx]

​@@ripjawsquad 4:15

[@ripjawsquad]

@@SoniSingh-yn3yx omg wow😮😅😢😂😂

[@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

[@anthonykula5169]

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

[@ElusiveEel]

And you see this again in the transition from power series to Taylor series

[@sumitkumaradhya2579]

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 !!!

[@siddanthvenkatesh2744]

No you don't have to use the binomial for the proof. Here is one of many without it. y=x^n => ln(y)=nln(x) differentiate (1/y)*y'=(1/x)*n+ln(x)*0 y'=yn/x y'=x^n*n/x y'=x^(n-1)*n Reordering y'=nx^(n-1)

[@janvisagie231]

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).

[@agarykane2127]

Did he say sus 😂 4:15

[@lifeofphyraprun7601]

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

[@prognosticii]

4:15 Eddie Woo predicted amogus

[@forthrightgambitia1032]

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.)

[@cheekyismymiddlename]

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

[@justacockypersonontheinter6889]

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)

[@evplayspiano]

when maths is sus 4:15

[@78anurag]

Ngl Collatz Conjecture acting kinda Sussy

[@docta2985]

You sir are probably the greatest teacher alive

[@gouranggehlot4896]

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

[@ma01rius76]

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..

[@joshuaronisjr]

Marius, look at my comment!

[@HeyItzMeDawg]

No, it's not circular. He's treating it as a variable. Think of that reasoning with any other variable: "The point is to find the f(x) formula, but it's being used here as f(0), f(1), f'(x), f'''(3)" Or "The point is to find the value of x, but it's being used here as x^2 + x" You can use a variable or function in a proof of its formula just fine. If you know some property of a function (such as its value at f(0)), you can use that in the proof of its formula. You're probably operating under the assumption that he has yet to prove that nCr _is_ the value of the coefficient r of a binomial expansion to the nth power, because the binomial theorem is typically first introduced in probabilities and combinatorics in order to evaluate the value of an n choose r problem. That is, nCr is _defined_ as the notation for combinations first and then its relation to the binomial coefficients is _proven._ But he's gone the other way around: he has first _defined_ the notation for nCr as the coefficients of the binomial expansion and will (presumably) later prove its connection to combinatorics. But since he has _defined_ the notation for nCr to be the coefficients of a binomial expansion, he can then use that property to prove its formula by using the notation in the manner in which it was defined. The notation could be n0, n1, n2, (with the numbers as a regular subscript) instead of nC0, nC1, nC2 and the proof would remain the same, it would just be _notated_ differently.

[@schrodingerbracat2927]

historically, the calculus power law is derived from the Binomial Theorem. however, what he presented is interesting. i would not call it a "proof" though.

[@ambroselam626]

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

[@OlayRivera]

Why i only heard this now... its amazing to know where did that notation came from...

[@KhanhTran-oj5sm]

It should be f '' (x) = ...+ 4.3. (n 4) x^2 +... instead of f''(x) = ...+ 4.3.2 (n 4) X^2 +... right?

[@fluoroantimonic9950]

I think that we are taking the factorial of that term so 2 is included

[@QYong-rq6iw]

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

[@bilalkhares9337]

oh right he's taken the same x term

[@sabitmahbub5510]

Thanks, it was bothering me so much

[@aryanmishra9994]

Sir just a doubt..is this result defined for x only equal to 0

[@michelef406]

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

[@krystofsedlacek]

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

[@凌彬翔]

i think it should just be 4*3 too!!

[@loukas7224]

What if (n-r)=0 ? Wouldn't that mean that 0^(0)=1 so the proof isn't correct for all values of (n-r) ?

[@divasv]

How??

[@HeyItzMeDawg]

0^0 is typically defined as equal to 1, particularly for things like the binomial expansion but also many other key formulas. In general though 0^0 is indeterminate and could be either 0 or 1 depending on context.

[@凌彬翔]

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

[@scoringwolf]

where do you teach,please give online live lessons

[@charlesbowyer]

Many problems - as noted in previous comments. Just McLaurins (not unexpected) and circularity to do with differentiating x^n. However - it is only intended as a teaching video - and shows the diversity of seeing different angles of the same problem.

[@siddanthvenkatesh2744]

You don't have to use the binomial for the proof. Here is one of many without it. y=x^n => ln(y)=nln(x) differentiate (1/y)*y'=(1/x)*n+ln(x)*0 y'=yn/x y'=x^n*n/x y'=x^(n-1)*n Reordering y'=nx^(n-1)

[@siddanthvenkatesh2744]

I don't know what other problem there may be.

[@noelgeorgi3313]

​@@siddanthvenkatesh2744This still posses a problem For d/dx(lnx)=1/x the func e^x must have itself as it's derivative. Which can only be established by maclurin series(which includes d/dx(x^n)). Only by that we would be able to prove that only a func related to power can have it as it's own derivative (we termed this power func as the exponential function) Therefore the only possibly non circular proof would be first proving d/dx(x^n) for whole numbers through normal binomial theorem (proved by principal of multiplication), establishing the maclurin series and thereby establishing the exponential function to have itself as its derivative, using the above proof extending d/dx(x^n) for all real numbers and finally proving the Newton's binomial theorem

[@fluoroantimonic9950]

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

[@erickjian7025]

But for the constant term in the brackets, isn't 1 a special case? what if it's a number larger than 1 (say a)?

[@alvinlepik5265]

The proof is quite mindblowing, actually - I prefer the sigma-notation proof (induction), because it clearly indicates What we are proving. This proof is valid, but it's not obvious :/

[@JohnSmith-ol5kh]

It seems a bit circular. How can you use the power rule for derivatives if you haven't proved that thebinomial theorem is true?

[@nin10dorox]

You dont need all the coefficients for the derivative of a power. You just need the first two, which we can understand as special cases, and all the rest will vanish.

[@HeyItzMeDawg]

The power rule is a convenient way of evaluating derivatives but you can just manually compute derivatives if you have to using limits. Lots of mathematics feels "circular"... insofar as lots of things in mathematics are heavily interrelated and proving one thing often allows you to prove a bunch of other things and vice versa. One of the biggest unsolved problems in mathematics is known as "P vs NP" - people have tried hundreds of different ways to do it and discovered thousands of seemingly unrelated problems that are actually connected, such that if any single one were somehow solved, all could be solved, including "P vs NP". In this case, you can use the power rule to prove the binomial theorem or the binomial theorem to prove the power rule. If you haven't proven either you can use limit theory as a substitute for the power rule to prove either the binomial theorem or the power rule, or if easier use limits to prove the binomial theorem and use the binomial theorem to prove the power rule.

[@NotYourAverageNothing]

9:13 Now you’re begging the question (kinda).

[@0mar.ali.khorshed12]

You are my hero

[@epsilia3611]

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

[@sadiqsheikh9546]

Where does he teach and which grade is he teaching in this video?

[@LG-dm3ts]

in Sydney Australia probably final year of High school

[@bojanrobev4302]

Should it be ( n-r-1) ?

[@richardguyver6676]

I had to look twice, but I agree with Eddie. When r=3, the last term was (n-2) ie (n-(3-1)).

[@eggtimer2]

Is it "permissable" to differentiate n! Which is only defined for integers? I E., It's not continuous. No gamma function in sight ...

[@ElusiveEel]

n is a constant, x is the variable differentiated with respect to

[@chirayubetkekar5692]

well just because the rth derivative of 2 functions is same...doesnt mean the functions are same right?

[@fredlurch3690]

Chirayu Betkekar that‘s right. Consider 2 polynomials of degree 2. Then every derivative higher than 3 is 0 for both of these polynomials but that - of course- does not imply that the polynomials are equal. However here we are only interested in the coefficents of these functions and there the derivatives can help you a lot. Just google taylor expansion. That‘s basically what he did in this proof

[@ckzombie1130]

Nicely proven, haven't seen it done this way before. How did you come up with this? Also, what is the reason for taking the derivative ( besides that it works for the proof). Thank you =)

[@skylorbrown4374]

What he did is very similar to what is done when finding the Maclaurin series of a binomial.

[@jennicawilton4322]

Can someone enlighten me, please? It seems he's only proved it for the case x=0, like proving the gradients of y=x^2 + 1 and y=x^4 + 1 both have the same derivative at x=0, so their gradients are equal for all x. What have I missed? Not trolling. Genuine question.

[@aryamansingha]

Ok first think about the how any function can be written as some infinite sum of x's of different powers and coefficients such as f(x)=ax +bx^2+cx^3.... Now that you have convinced yourself that that is possible. On the right hand side he is using nC0, nC1, nC2... to represent coefficients. Think of them like a,b,c... what that means is he is trying to generalize those coefficents in terms of the position(r) that coefficient is in and the total number(n) of coeficients. He does this by using the rth derivative at 0 since it removes x from the question. See a,b,c... (nC0,nC1, nC2) are not dependent on x even if the rth derivative f{r}(0) is. It is quite different from your example in which you are comparing the derivatives/ gradient themselves which are dependent on x and/or the The coefficient and/or the power of X. Hope this helps.

[@jennicawilton4322]

@@aryamansingha Oh, I think I get it. So by taking the rth derivative we can work out the coefficient for rth term, because it becomes a constant. And you can equate it for x=0 because that particular coefficient isn't affected by x at that level of derivative. Then you could take different derivatives and repeat the process to get all the separate coefficients. Have I understood correctly?

[@shiveshtuli7608]

Wait a minute did he just proofed that there is a relationship between fractals and calculus ....... ........ ....... .....

[@janamohammad8745]

What a really amazing young teacher I'm already falling in love with you!

[@m.kamalmkhallalati119]

Thanks hagi

[@MarekDz60]

Well, depends how one derives the derivative of x^n. If in a standard way then the demonstrated proof is circular. One other way of getting the formula of (x^n)' is by induction, but Eddie's leading idea was to avoid induction, so he cannot use this route. The question is then how Eddie gets the formula for (x^n)'. Razzle dazzle them and they'll never get wise kzbin.info/www/bejne/j4iWfnyfraqjqZY

[@phagunshah1754]

Excellent

[@andrewhemowitz4326]

are u in Australia here???

[@justacockypersonontheinter6889]

i am australian too by the way, Victoria)))

[@targaryen3671]

This is awesome 👍

[@krystofsedlacek]

Oh no 4:15

[@starinsky2873]

Is this high school math

[@raj-m]

Any Indians here? Just wondering about the silence of the class?

[@joshuawalsh6968]

Very entertaining

[@veenakirale1848]

You must be paid for lacks for your explanation.. Seriously

[@agarykane2127]

If he explained everything , the video would be for more than an hour and don’t forget that he teaches real students which don’t have infinite amount of time

[@sevarchy]

What exactly are you proving? You have just shown (nCr) = [n! / r! (n-r)!] which is the definition of combinations. The binomial theorem is not (nCr) = [n! / r! (n-r)!]. The binomial theorem is simply an expansion of (x+y)^n explained by combinations which are called binomial coefficients. I am confused because you state you want to prove the binomial theorem. Meanwhile, the end result isn't the binomial theorem. You go to calculus and use the power rule to justify the steps, only to validate the definition of combinations. Further, mathematical induction is needed for binomial theorem. You think you are proving the binomial theorem (while you aren't), and then you use the power rule (which requires binomial theorem to get it proved). Then when you write "But f(x) = (nC0) + (nC1)x + ..." that's basically the binomial expansion of (x+1)^n. You now have also used the binomial theorem in the proof of the binomial theorem... what exactly are you even doing?

[@kcwilliamson50]

I agree. He didn't prove the binomial theorem. He simply demonstrated the definition of nCr. It's a cool demonstration but it didn't accomplish the goal.

[@Swan584]

This is badass

[@someone229]

This is just a proof that n choose r = n!/r!(n-r)! not the binomial theorem

[@johnwroblewski6458]

actually it's a proof of the particular case of the binomial theorem (x + y)^n, where y = 1. I agree that the phrasing is very misleading and incorrect.

[@christopherwrogg6064]

@@johnwroblewski6458 It's not really that hard to extrapolate this to the general case...

[@aashsyed1277]

HOLY COW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@prakharchaurasiya8107]

Proving the derivative of two things equal doesn't make them equal.

[@kshin6489]

What do you mean? He exactly proved it correctly.

[@berserker8884]

That is not what the problem was about. It was using derivatives to show an equality of coeficients. I assume he defined the binomial coeficient as the expansion of the binomial series and then he tried to prove what the coeficients must be using the derivatives. All that he assumed is that the derivative is equal to itself, aside from that you can actually differentiate the function the way he did, but thats just tedius real analysis you have to go through one day. So he got that the derivative is equal to sth and also sth else, so those two things must be equal. He never did anything like proving two functions are equal by showing their derivatives are equal.

[@basiltom5414]

👏👏👏👏👏👏👏

[@exhalerwolf1272]

I don't see how this is a proof for binomial theorem. More like an alternate unnecessarily weird proof of nCr