an absurd approach to a simple mathematics problem
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There is a mathoverflow thread named "Awfully sophisticated proof for simple facts". It has some real gems in it. The top voted one is that the n-th root of 2 is irrational for n > 2, by way of Fermat's last theorem (if it is rational p/q, then p^n = q^n + q^n, which cannot happen). Unfortunately, Fermat's last theorem isn't strong enough to prove the irrationality of the square root.
How is a 2 lines proof "awfully sophisticated"?
@@ninck8992 Because it uses Fermat's last theorem. That's a pretty darn sophisticated piece of math.
@@ninck8992 because you would need to provide a brief outline of fermats lasts theorem for the proof to be rigorous
Yeah i find these sort of stuff fun. But , how can we be sure that Fermat's last theorem proof does not rely at some point on the irrationality of the nth-root of 2? Because if it did then you are using a circular argument.
@@gianpierocea I am not fluent in the theory of elliptic curves, but I'm familiar enough to be reasonably certain that this is a safe proof.
More videos on generating functions please
Have you seen the ones on the mathmajor number theory playlist?
Hell yeah!
I agree, sequences are my favorite
@@natepolidoro4565 "generatingfunctionology" was a great read on the subject. If you are interested, it dives into interesting aspects.
Agreed, this was cool
Killing a mosquito with a cannon!
It’s definitely dead now
That was spectacular. Reminds me of a recent calculus problem where I made a bunch of geometric constructions to derive this thing that I was then going to integrate and when I got the assignment back my tutor's comment said "this last 4 pages could be replaced by using Pythagoras' theorem on the diagonal of this triangle". Woopsie.
8:04 Directions unclear i just opened a portal to another dimension ! xD
gosh I wasn’t aware that math functions ever had ‘best friends’ LOL
Call T(n) the n-th triangular number and consider that the second differences are 1. Integrating twice gives T(n)=n^2/2 +an+b and using T(1)=1 we have a+b=1/2 and with T(2)=3 giving 2a+b=1. Solving this system of equations gives a=1/2 and b=0 so that T(n)=n^2/2 +n/2 = n(n+1)/2 which is the familiar formula for T(n).
More generally you can use generating functions to derive the Faulhaber's formula (sum of first n powers)
One way involves eulerian numbers and polynomials.
14:55
And to think Gauss did this all in his head as a schoolkid! 😉
I would use generative fonction to prove negative binomial does work...
Generating functions are nice to work with. If you think a thing might work on it, it will work on it. No need to worry about convergence or whatnot. It's just a way of storing numbers.
Awesome video, as always!
that was absurdly genius, ty
Fantastic, thank you for talking a bit about convergence and formal variables along the way. That's always something I got hung up on in generating functions, that is what we're allowed to do and not
I think you should make a video diving into the details of convergence vs. formal variables.
I once derived the sin and cos angle addition formulas using their complex forms, it was something like this:
sin(x+y) = 1/2i*(e^(x+y)i-e^-(x+y)i) = 1/2i(e^xi*e^yi-e^-xi*e^-yi)
and than I used eulers equasion on all the complex expenential, canceled out like terms and was left with cosx*siny+cosy*sinx
I also used an analogous method for cos(x+y)
this might seem overcomplicated, but as far as I'm aware this is the only method to derive the angle addition formulas for hyperbolic trig functions, which are
sinh(x+y) = coshx*sinhy+coshy*sinhx
cosh(x+y) = coshx*coshy+sinhx*sinhy
I love generating functions! One of my fav topics in discrete math courses I took.
I studied generating functions in a combinatorics class at university and loved them too. Everything just seemed to work like magic.
I always wondered haha. Also best catch phrase at the end.
That was massive!
I don't know how overkill this really is but the integral of 1/(x² + 1) from -infinity to infinity can be easily worked out using the arctan to be pi.
An alternative approach i saw (which imo is a really nice and simple example) is evaluating this via. complex integration:
You do the standard thing of choosing the contour to be a large semicircle enclosing the upper half plane and close it via the real line. Let it have radius R.
Now you can evaluate this just by looking at the residue at i, which will be simply 1/(2i). Now the integral's value will be pi by the residue theorem and a simple approximation will show that 1/(x²+1) goes to zero on the upper semi circular arc as R goes to infinity, which will be the result we expected.
Honorable mention:
Another non serious proof i saw (credits "An Overly Sophisticated Proof of a Disproportionately Simple Fact" by joseph newton absolutely hilarious) was proving that the cube root of 2 is irrational using fermats last theorem, which, famously, has a proof that is not completely easy to comprehend.
Not only does 1/(x^2 + 1) along the semicircle go to 0, but it goes to 0 _faster than_ 1/x. This is important, because you want the integral along the semicircle to go to 0, not just the integrand, and the length of the curve we integrate over increases proportionally with the radius.
How do you justify the binomial expansion for negative exponents?
As a student I came up with quite a few absurdly complicated ways to prove things, which I later realized were actually simple …. and sometimes wrong …
Thanks sir
For those curious, “formally” here means in relation to the form of the series, not “rigorously.” A formal power series is where we care about the structure, the form, of the terms and the series itself.
I remember learning generating functions and it was quite fun.
Me divierten mucho estos laburos totalmente laboriosos y complejos para hacer algo sencillo.
Abrazos
I'd love to see a compilation of every time Michael says "switch the order of summation/integration" in his videos.
Sounds like a drinking game.
This is incredible and brilliant.
It's hard to imagine that using derivatives, geometric series formulas, binomial theorem, etc to prove the arithmetic series formula is "circular reasoning"-free.
How I missed the overkill series!
Imagined this: exams have finished, still a few days of term time left so what to do to fill in teaching hours that are entertaining and learning at the same time?
Give the students that as a step by step process making sure everyone has completed step m before going on to step m+1 with tailored guidance to make sure everyone is keeping up?
It may not be pedagogically helpful but it is math after all - and there are plenty of math lessons in that exposition that should motivate learners everywhere ?.!
Pause: at this point I do not know whether to end that sentence (points at the sentence) with a full stop, question mark or an exclamation mark.
After all: math is math 🙂and a scenic route is just as good as a fast route (audible puns intended 🙂)
EDIT: added a full stop, question mark and exclamation mark to end od said sentence not necessarily in that order )
Doesn't the binomial formula already imply n(n+1)/2? Or is there a circular argument here?
Not absurd. Beautiful!
I like the idea of presenting a proof using formal power series, but the example is maybe not the best, and the way the proof is presented seems aimed at scaring the viewer rather then edifying her; it definitely does not give the impression that this is a technique that one could use easily to solve (easy or difficult) problems. It also leaves a lot of question of whether this constitutes a proof at all (why are differentiation formulas derived for real functions valid in the formal poser series world, for instance).
I think it would be much easier to understand if you first studied what multiplying by 1-X means for formal power series, namely: for each position (except that of X^0) the coefficient of X^(n-1) gets subtracted from the coefficient of X^n. Doing this once turns you power series into the sum of terms nX^n, and doing it again turns that series into the one with all coefficients 1, except the constant coefficient which remains 0. A third application turns the series into X (the coefficients of all powers other than X^1 are zero). That shows (without filling a blockboard) that (1-X)^3 times the initial series is X, so that the initial series is X/(1-X)^3.
Cool. I found out about this method looking up ways to find closed forms. There is one thing leaving me confused, though. You said we do not worry about convergence and we work with formal variables. Then I wonder, why does this work? Why mere formal manipulations lead to the right answer?
Suppose you have a random variable with known probability density function (pdf) and you want an equation for the n-th moment. You could compute it directly, but you could also take the complex conjugate of the Fourier transform of the pdf. The result is called the characteristic function. Take the n-th derivative of that function, set the frequency to zero and the n-th moment is 1/i^n times the result. (That actually can be useful if you have a sum of random variables since the resulting pdf is the convolution of the individual pdf's, which becomes multiplication in the Fourier domain.)
Neat! Another fun proof is to use least-squares to fit a polynomial curve to an appropriate number of sums. Turns out the best fit is a quadratic with coefficients 1/2, 1/2, and 0.
Excellent! Had me laughing... 🙂
Fantastic!! 😀
would be interesting to prove the same for the sum of first n squares and cubes
this is a great video! one thing: is the choose function defined for negative numbers? could you not just keep the (1-x)^3 in the denominator and expand it from there?
whoa I didn't expect that
I wonder what Gauss would have to say about this?
#overkill
Love it!
i missed this series
What does “everything is happening formally” mean?
I see that a lot for generating functions, but what does it even mean to manipulate infinite sums if we don’t have concept of convergence. What manipulations are okay and what aren’t. We exchanged the order of summation, why are we allowed to if there’s no concept of convergence and we just take it “formally”?
Arguably ridiculous; objectively beautiful.
No matter how ridiculous it may look you still can derieve formulas for power sums using the same method. Unlike the usual ones
I enjoyed that!
That was fun!
This is intense, it is like hitting a thumbtack with a big hammer.
I will stick to my easy and cute way of proving it:
Let S= 1+2+3+.........+n
and also S=n+(n-1)+(n-2)+.........+1
Then adding the two sums gives
2S=(n+1)+(n+1)+..........+(n+1) [n times of (n+1)]
2S=n(n+1)
S=n(n+1)/2
Therefore 1+2+3+.........+n=n(n+1)/2
generating function for triangular numbers
but does it work for finding the sum of the first n squares
I don't know good ways to do things absurdly, but something I've been working on is solving recurrence relations algebraically from the very basics without mathematical induction, so solving a(n+1) - a(n) = C which has closed form a(0) + C n. The question is how to get this closed form using only the rules of algebra and of relations a(n) where a(n) defines the members of a set, and its inverse defined as n = a_index(a(n)) (but a(a_index(k)) can be not equal to k). Do tell.
this was crazy
Perhaps this is the true way young Gauss did it 😂
Here's a problem to solve the hard way:
Two trains start 20 miles apart, and travel towards each other at 10 miles per hour. Just as they start, a fly takes off from the front of one train, flies at 15mph directly to the other, turns around, flies back to the first… and zigzags back and forth until the trains meet. How far does the fly fly?
Seems to work for any sum of first n kth powers. I just did it to find the sum of first n squares.
Can we change the other summation without justification? The sum clearly diverges.
For the sum order change, Iverson brackets are much easier than reading the bounds from a picture:
sum_n=0^infty sum_m=0^n mx^n = sum_n=0^infty sum_m=0^infty [m
That's hilarious. Like solving the harmonic oscillator via the Hamilton-Jacobi equation :)
I liked the part where he opened a portal
Can't you use re-indexing as n -> m = n + 1? Using n again is inconsistent. Doing this and then replacing m with n again means: n = n + 1 or 0 = 1 which is false, so something inconsistent must have happened. In which case the proof does not work. For the proof to work we need to set m = n on LHS and m = n + 1 on the RHS, which seems inconsistent! So "m = n" is a false assumption, but it does not lead to a contradiction!
However since n on LHS is a dummy variable, m can replace n to make the proof work. Then we proved that the assumption "m = n" follows after the assumption "m = n + 1" got discharged. In this case we need to find an elimination rule for discharging the assumption. The reason for this rule in this case is: "because we wish so." - seems illegal.
Actually we need something stronger than this: we need a rule to delete "m = n + 1" from the proof sequence.
I don’t remember this being the best friend
I thought it was just 1/(1-x) for the geometric series.
3:13 proof is on the board, that's a good place to stop
9:30 What do you mean by "best friends"?
I believe @blackpenredpen coined the term:
ua-cam.com/video/Ux7vl6zXxj0/v-deo.html
blackpenredpen actually refers to 1/(1-x) as your best friend because it comes up time and time again when doing sums and stuff
@@AbstractNoesis Also it has a really easy derivative, the square of itself! d/dx 1/1(1-x) = 1/(1-x)^2
Nice!
Possibly a newbie question, but how can we take the derivative of a discrete sum when differentiation is only defined for continuous functions?
The derivative is with respect to x. The sum is continuous wrt x. The discreteness is wrt m or n.
@@landsgevaer oh duh, that’s what I was missing, thanks. Generating functions are funny because that x just gets inserted out of nowhere to create the generator, so I tend to forget what its actual nature is.
Thank you Penn. Thank you for helping me escape from my life.
Terrible application of Tonnelli thm to counting measure on the non negative integers .Bon courage !!!
generating functions seem to have a lot of similarities with the z-transform in discrete systems used in engineering. the z variable represents a right shift in the index of a value in a sequence (i.e. x_n -> x_{n+1}) so a sequence can be represented as an infinite sum across z^n, where each coefficient is the value of the sequence at that z-shifted time/index. partial fraction decomposition with z as a formal parameter is used to show that sequences generated by recursive linear equations are a sum of geometric series
Make a series of videos where u use overkill theorems to prove well known math statements. Next step could be proving that the square root of 2 is irrational over Q using galois theory 🤣.
Would love that
Another ridiculus way of finding this formula is to use the graph with points in the video containing 1 and then 2 an then 3 up to n points. Then we can calculate the area of this triangle
(base x height/2 )
n(n+1)/2
☺
why sum to infinity, why not do the entire thing summing to N. That way you are not starting off with a sum to nowhere.
Fucking cool
❤ I love it ❤
how many problems will give me that v shaped back ??
Bouldering problems
@@sjswitzer1 quantify your answer !
V10 at least.
'ridiculous proof'
Proof the Pythagorean theorem as a 'limit' of it's spherical form: cos a * cos b = cos c.
Did we just become best friends!?
the scientific method thrives on alternate proofs...
In Watership Down, rabbits use 5 and many and infinity interchangeably.
Michael is obvs a rabbit
I was expecting -1/12 as answer
I think we can also give the reason for taking out the derivative outside the summation at 8:58 as that the derivative is a linear operator and is easily distributive over a sum.
This reasoning is not enough for an infinite sum, only finite sums; you need the sum of derivatives to uniformly converge (which here of course it does).
@@matthew-m You refering to Hilbert-Schmidt norm?
I wouldn't call it "ridiculous", this is just a very "interesting" approach to the problem!
Who called it absurd?
@@samueldeandrade8535 Michael Penn. "absurd" in the title, "ridiculous" in the video!
@@rainerzufall42 oh really? I think it is interesting too. I usually don't refuse info like that. Especially in this case that gives an example of what happens if we apply some theory to get what we already know. Such info is not just interesting, but important.
crazy :)
I have laid down simple principles and fundamentals that underlay all formal linear manipulations akin to generating functions in the concept of an arithmetic space, in my work Foundations of Formal Arithmetic.
This deserves a Rube Goldberg Field's Medal.
omg genfuncs
Yeah absurd approach then how you
can find formula for Catalan numbers , Bell numbers
or formula for orthogonal polynomials such as Legendre or Hermite
Let me guess you use characterictic equations