This video is inspired by an example from "Professor Stewart's Casebook of Mathematical Mysteries". Please check my community post on March 3rd, 2024 for the photo of the particular page. Please don't take it too seriously.🤗
Suppose I define a set S of words by saying that a word in S is either the empty word, or it's obtained by appending an "x" to some other word in S. I can write this definition of S succinctly using the equation S = 1 | Sx where 1 denotes the empty word and | denotes "or". Now, "or" is kind of like addition, right? So let's subtract Sx from both sides to get S - Sx = 1. Factor out S: S(1 - x) = 1, so S = 1/(1 - x) = 1 | x | xx | xxx | xxxx | ... Lo and behold, this is correct: a word in S is either the empty word, or x, or xx, or xxx, or xxxx, or .... This crazy-looking calculation is one of the key ideas underlying the Kleene-Schützenberger theorem that regular languages can be represented by (noncommutative) rational functions. The * operation, indicating arbitrarily many repetitions of an expression, corresponds to taking an inverse: X* = 1/(1 - X).
You have successfully nerdsniped me. I go to your community tab, and I still haven't made it to this post because I keep stopping to do the quizzes. I even am trying to save one for later by saving the comments page (and hoping they don't spoil the answer). I'm just now at 10 days old. Let's see how long it takes for me to get to 12 or 13 days old to see your comment.
I've only taken calculus 101 and physics 101, and still I can completely comprehend why this is absurd to both tragic and comedic levels. I have a saying that physics research continues developing until it hits a wall, and that's when they go and knock on the doors of mathematics professors to see what they can borrow next.
Functional Analysis gives rigorous foundations to this and other seemingly non sensical manipulations, by treating these "symbols" as operators and giving necessary conditions for the convergence of such expressions in a given space.
@@jonathandawson3091 well, a more concrete view is that the integration operator is from 0 to some indeterminant x, and the integration operator on e^x is e^x-1 and is a linear operator
Guys, it is banter. Physicists use normal math but under physical constraints which make some mathematicians confused. No physicist would use or condone the argument in the video.
Okay but "With proper definitions everything i have done is legitimate" is just a Get Out of Jail Free card for doing whatever you want as long as it internally makes sense though lol
@@schokoladenjunge1 Yeah, but in context of this video it basically says "When everything I did is defined in a way that allows me to do the things I did I am allowed to do the things I did because of the way those things are defined" and I was mostly joking about that lol For example when dealing with PDEs you can sometimes get 0 in the denominator of the fraction and that's fine as long as you remember the proper way of interpreting it xd
This can be made rigorous using something called the holomorphic functional calculus. In fact, for any normal operator A on a Hilbert space, I can make sense of the expression f(A) for any bounded Borel measurable function on the spectrum of the operator.
I'm a physicist and this video gave me a migrain. I have never seen such horrible misstreatment of mathematics in any physics course I took, except the one we don't speak about (a.k.a physics of atoms and molecules, that prof was a chemist).
This manipulation can be made more rigorous through Functional Analysis, however I don't think that's something you would voluntarily decide to approach in your free time just for the sake of learning more about linear operators defined on higher spaces, and not to just show things like this as a party trick
@@ron-mathNot once in my life have I thought that I would hear a thanks after mentioning functional analysis to someone, but this time it's strangely nice to be proven wrong, so be my guest ^-^
@@healer1461I did take courses in FA since it connects to quantum mechanics and the derivation in this video is only on a very superficial level similar to what would be rigorous in those courses.
"As someone with a physics background, that constant *must* be 1." I recall working in 'natural units' where: * the speed of light * the mass of the electon * plank's constant * and the permittiviy of vaccum were all equal to 1, so yeah, that checks out.
Best unit system tbh, no need to carry on that garbage in each calculation. It makes dimensional analysis way more straight forward when every energy is just a frequency 1/s etc.
That's actually well-motivated, though. There's no physical reason that the only fundamentally meaningful speed in the universe should be exactly 299,792,458 meters per second where the "meter" is some fraction of the distance from the earth's equator to its north pole and the "second" is defined as some arbitrary number of hyperfine transitions in Cesium-133. It's FAR more justifiable just to say "that speed is 1" and go from there. The same thing is true for things like h-why should the fundamental unit of action be 6.626... x 10^-34 of some silly units we made up? Much better to just make that be 1, too. (Or the reduced Planck constant, as it were)
The most baffling thing with these physics acrobatics is that most of the times, they work very well. I remember a class on basic mecanical principles, and the teacher basically spent 4 hours violating derivatives and making integrals work with spit and spite.
Yeah, the way I like to think about it, physicists usually have some physical or intuitive justification for what they are doing. Obviously this video is a slight exaggeration of the egregious things physicists do, but nonetheless they usually have some justification of sorts. Then the mathematicians bend backwards and define theories so that the physicist's intuitive manipulation is correct. It kinda makes sense if you think of mathematicians as defining the math so that it accurately models whatever the physicists are looking at.
@@albertrichard3659That’s 100000% how mathematics is “discovered”. Einstein was not as good a mathematician as a physicist and sent his requested proofs be verified that his intuition was correct
@@superuser8636 Yeah, I wouldn't say 'discovered', but that's pretty much how all math progresses. There are, at any instant tons of things you could do with math but we only focus on the subset that captures the intuition behind whatever we're trying to solve because that's the whole point. Vladimir Arnold himself claimed that math is a subset of physics. It's a little bit too extreme a view for me but I do agree with the general sentiment.
A large part of my mathematics classes was dealing with various convoluted edge-cases that typically do not occur naturally. If we had the liberty to say "for most functions" or "for well-behaved functions" or something similar, my degree would've taken a lot less time to complete.
This actually has rigorous justifications if you work in an appropriate function space. All you need is a Banach space in which the operator norm is less than one so that expansion of the operator in Neumann series is justified.
You can also cheat and just note e^x's series is fully positive and converges for each value, so the operator is well-defined on series (ad hoc, and requires finding the radius of convergence, but who cares - every calc 1 class does this anyways :P).
@@mahdihasan6222 Yep. As someone who hasn't done a single derivative in his life, I have absolutely no idea what's wrong with it. I might need to take some Brilliant courses or something...
@@Lolwutdesu9000Not really, no. Yes, you can show "1=0" if youre working on a field where 0 has an inverse, but the only such field is the trivial field with one element, so it has zero applications. This on the other hand works when dealing with L^2, which is a far more relevant space. And this construction is quite silly, but the underlying reasoning of exploring polynomials of operators is also something you use in practise To put it more simply, yes, anyone can write a book, but it's more impressive if you wrote LoTR than if you wrote "Ernie and Bert's new kitten"
@@Lolwutdesu9000Your comment is stupid as it misses the essence of “application”. Physics is an applied science (despite the fact that it has gain a purely theoretical component over the years) which means the domain is there. We don’t just make up domains.
I don’t understand what’s so objectionable about it, should we all pretend to be babies who don’t understand differential operators on Banach spaces just because a physicist did it?
This video gave me a stroke, a migraine, 2 trivigintillion dollars in debt, a curse from a haunted mansion, a nightmare about Dirac's equation for the electron spin, a magic dog that solved the Collatz conjecture and absolute disgust
Technically, this methodology makes sense from an operator theory perspective. Call the integral operator I, and 1-I is a linear operator ((1-I)f=f-If) on the space of formal power series (who cares about convergence anyways :P - can also do on equivalence classes of functions differing by a constant, but this is the same thing). Knowing that f should be of that form, we get (1-I)f=(1-I)(a0+a1x+a2x^2+a3x^3+...)=(a0+a1x+a2x^2+a3x^3+...)-I(a1+a2x^2+a3x^3+...)=(a0-a1)+(a1-2a2)x+(a2-3a3)x^2+.... Since a positive series sums to 0 iff each term is 0, we have that an-(n+1)a_(n+1)=0 for all n. Since e^0=1, we must have a0=1, so 1-a1=0 and a1=1, so 1-2a2=0 and a2=1/2, so 1/2-3a3=0 and a3=1/6, and inductively, an=1/n!.
To be fair I can accept anything a physicist does except for writing the differential dx before the integrand like \int dx sinx. That is just disgusting...
@@lazarussevy2777I think it's just a constant, as lim dx -> 0 of sin(dx) = 0, so you are effectively summing up epsilons. I don't really have time for it now, but I think expanding it as the series gives (x - x^3/3! +...), such that the integral will be \int {(dx) - (dx)^3/3! +...}. Using the distributive property of integral you get = (dx + C) - something that is less than epsilon + something that is even more less than epsilon, and so on. That in the limiting case is equal to the C. Therefore, the unbounded integral of sin(dx) is just a constant. But please take it with a bit of salt, because this is not really rigorous
@@tenebrae711 I see what you did, but you still have to have a dx on the outside in order to integrate. To solve this issue, you have to multiply by dx/dx before you integrate. Then you have lim dx -> 0 of sin(dx)/dx, which we know is one, then the left over dx. This is ∫dx, which is x+C
It just depends on the length of the integral. When it's short use the dx as a bracket, when it's long it's clearer with dx in front and you use additional brackets.
@@Currywurst-zo8ooWhat do you mean by short integral or long integral? I've never heard of such a thing. Could you please explain how you would use the dx as a bracket?
I think the funniest part here is the division by (1- integral sign) Since e^x *(1-integral sign)=0, it follows that (1-integral sign) is 0 since e^x can’t be 0. We have a 0/0 situation here. I find it amusing
As a Math major, the feelings this video generates exactly how I felt when I was attending a class of Mathematical Methods in Physics for a Minor. We were taught many methods of solving Differential Equations, and also, Complex Analysis - in which I struggled to gasp a single thing as most explanations were as non-sensical as this. I wish I was joking. My heart finally felt at ease when I revisited all this concepts later on in courses meant for Math Majors, under completely different instructions for each discipline. Edit: Note:- for the people who argue this is fine if you view integral as a linear differential operator from a banach space to another (space of differentiable maps to space of continuous maps) and I do agree, the "so it must be 1" is still jack. I understand the video is a joke, and while I am still giggling at the joke, I certainly won't assert some statement for which is incomplete. Functional Analysis isn't trivial. The people who properly explained it in the comments have my gratitude though, as they simply didn't just say "Functional Analysis" and called it a day.
One of my math professors called physicists “idiot savants,” and I think that’s for good reason. Math owes a huge number of theorems (even if not their proofs) to physics. As a math guy, I think we get a little lost in the sauce
In engineering we learn D operator to solve differential equations where derivatives (d/dx) is treated as a algebraic element( more like x)..you can multiply them, factorize them also divide by them.Its really fascinating and wonderful !Though its not something ground breaking , it just reduces calculations.That integral reminded me that.
the key is to treat x as a vector; replacing 0 and 1 with 0(x) and 1(x) make this much more sensible; by treating the integrand as an operator on a functional object rather than on limit-depedent evaluations of said function, the dx limit is irrelevant bc we've abstracted away the evaluations, and we did not need to divide by zero. also, we can put everything on equal footing by virtue of full-rank tensors on the superposition of all R domain inputs; they transform like tensors after all. the identity operator and integrand are both linear w.r.t. any superposition of inputs of the function, and thus can be combined into a linear operator/tensor. this is not unique however because the integrand has a rowspace from -C(x). by linearity we can subtract both sides with -C(x) and pull the degree of freedom to the right side. there exists some -C(x) tensor for which (1-int) operator is unique and thus invertible, so by that virtue we can restrict ourselves to that case and continue by inverting it to the other side. since we make the assumption that C(x) is unique to invert the matrix, the right side is a unique operator, and it is valid to invoke division. now comes the fun part: we left-multiply each side with 1(x) to split off C(x) as the tensor to be transformed by an operator that can be expanded to its power series; things can be assumed to be smooth so Taylor's theorem is in range. the integral has lost its +C privileges so we don't have to worry about a residual polynomial blowout when computing them, and now WLOG we have a series that can be rewritten as e^c(x) by the definition of e^x. finally, the unique solution that makes e^(x) equal to e^c(x) is c(x) = 1(x), and we are done. i think that c=1 just happens to reflect the fixed point behind the system on which e is defined. im have a BA in CS and took physics classes on the side, hopefully i did a good attempt at justifying this madness even with the missing +C. it actually fits into quantum mechanics pretty well, literally, because x's norm here is restricted by 1, the radius of convergence for 1/(1-x). but it also just feels like mathematical anarchy.
There's a fancier version of what you just did called Neumann series, but you need to change the integral operator a little bit. Instead of just the integral, we define a much rigorous operator T, where the action of T to a function f(x) is T(f(x)) which is the integral of f(t) dt with the lower bound is some constant c (usually 0) and the upper bound is x. Therefore, the norm of the integral operator T would be less than 1, hence by the Neumann series, (1 - T)^(-1) = 1 + T + T^2 + ...
This is very creative and neat! I liked that you included the notion of operators and functionals, to show where you're coming from. Being creative in math is very important, most great physicists/mathematicians had their weird notational tricks and some of those even prevailed and became conventional. Unfortunately I think this video is largely misunderstood, which is a shame
I love doing math like this. Its really cool that despite playing fast and loose with rigor, if you can come up with a consistent logic in the math it still often agrees with reality.
This comment section is literally that IQ bell curve meme. :D The guy in the middle says "nooooo, you cannot terribly mistreat math like that, I need to wash my eyes", while the guys on the left and right say "this is totally fine, what an elegant derivation".
Generalising people with meme is such a childish and stupid idea. There is no math without formalism, you can't just do some unformalised tricks and hope everything will be okay, because after you build few theorems on those tricks it's impossible to tell if your proofs are valid. You are left with philosophy not math at this point
@@alang.2054 You are literally reiterating my point. If you have the necessary mathematical background in functional analysis and take enough care, then you are justified in making derivations like in the video (integration is just a bounded linear operator in some appropriate function space). If you don't have any such background and just do some unformalised tricks that seem fun to you, then you are not justified in doing so. All that basically corresponds exactly to the two guys on the sides of the bell curve meme (trained mathematician vs amateur). It's just a meme, don't take it too seriously, I'm not insulting anyone.
Ikr. Thanks to reading a bit of QM, I can appreciate the beauty of this. Makes me wonder why I chose to pursue computer science. Sorry, I'm gonna go cry somewhere :(
As a student this used to drive me nuts but as an older amateur just trying to occasionally do applications I've forgotten a lot of the 'proper' ways and love the sorts of abuses found in physics and engineering references
Let's try to do this properly. Let's define an operator S: C( [0,1] )->C( [0,1/2] ), S(f)(t)=\int_0^t f(x)dx. C( [0,1] ) is actually very restrictive, but it's good enough for exp. Since it is equipped with the uniform norm, we have ||S(f)||=|\int_0^{1/2} f(x)dx|
From 0:56 the video falls victim to the initial "suppression" of dx, which just a notational shorthand. In reality dx has been implied all along, it constantly accompanies the integral sign.
Greate video! But I got a question: why does the same procedure fails when substituting the intergral with differentiation? We also have (1-d/dx) e^x = 0, but expanding the series (1+d^n/dx^n) 0 does not give out e^x. Where is the problem?
What exactly do you mean by “operator” when referring to the integral sign? I know that you said to think in reverse, but what does an integral mean without a dx, if it means anything at all?
"What are you trying to tell me? That I can just choose whether to be an operator or an operand?" "No ∫. I'm trying to tell you that when you're ready, you won't have to."
Are there any other (actually useful) applications of this trick, similar to how generating functions seem to abuse formal series but turn out extremely useful? This video is just a re-derivation of exp(x) but if we alter the equation we should end up with a different function. Maybe if we extend our algebra with this integral sign some functions could start having closed-form expression representations?
e^x=D(e^x)=> e^x=(1/(1-D))(0) =>e^x=(1+D+D^2/2+D^3/6...)(0) (1-D)e^x=0 what does this mean? I remember learning a bit of it, it's called operational calculus started by Heaviside I believe. The solution I saw for differential equations like this used taylor expansions to solve for coefficients.
@@pauselab5569 correct me, but I believe it was made rigorous, just that it wasn't much better than Laplace transforms. It's honestly a lot more appealing to just understand the derivative operator itself rather than letting some transform do all the work. Neither will work in all cases anyways.
you cant expand the inverse of (1-D) since it isnt a bounded operator afaik. rudin treats this subject in chapter 10 of functional analysis. the section is called "symbolic calculus" but it is also known as functional calculus
The recurring theme in my physics class is that they make it work somehow but dont explain why they can do that. I cannot understand how they keep getting away with those shenanigans. It's probably easier to see and understand the tricks when you delve deeper and get some more global knowledge but for now I'm just thinking like a medieval peasant and want to burn the witch
For what it's worth, if you keep integrating and adding 1, you get e^x Int(1,x) = x Int(Int(1,x)+1,x) = x+x^2/2 etc and then you add 1 to get the result of the taylor series of e^x
The arbitrary choice of letting the integral of 0 be 1 makes it work for e^x. If you make other choices, it works for the other functions that are equal to their own derivative.
but like, this is (mostly) valid if you treat everything as formal symbols!! there is the issue of why integrating 0 should give 1 rather than another constant and why repeatedly integrating shouldn't add more constants. However, we can just take this as a definition (i.e. integrating sends 0 to 1 and 1 to x and so on) and, doing this, the final answer will be off at most by a constant!
This is actually pretty much okay to do if you know functional analysis. Look at some of the definitions of Bernoulli polynomials on the wikipedia page to see stuff where derivative operators are used like this.
Next time I'm called to the drawing board, I'll just say "As someone with a physics background, I can do something like..." and proceed to write total nonsense on the board, hoping that it'll work.
I would feel more comfortable doing this with the Laplace transform, although probably it's just using a different and fancier name for the same and simpler/more fundamental thing.
I'm unironically amazed. The "integral" is essentially now an operator on the function space..moreover, it's actually an "inverse" (unfortunately there's no one to one mapping for differetiation, so no actually inverse) of a linear transformation.. the d/dx . Since the kernel of the map d/dx is set of al constant functions, this inverse is supposed to give a whole family of functions. Now, we can represent the whole family by a unique representative.. namely the function with value 0 at x=0, and define that as the output of the inverse, making it a linear transformation on the function space. And thus, the 1/(1-int) thing makes sense, since int is just an infinite dimensional matrix and so by Taylor expansion, we can do what he did. Thus, with proper rigor, all of it does make sense. Damn.. I'm really amazed.
I studied mathematics in college and graduate school and my god my analysis professors, especially functional analysis, would have lost their fucking minds lmao
From a comment above: "the correct way goes like this: define F(f) = int_0^x f. Then F(e^x) = e^x - 1 and rearranging we get (id - F)^-1(1) = e^x, where id is the identity and id - F is invertible since ||F|| < 1 in L2([0,1]). so we are calculating iterated integrals of the constant function 1, and not 0 like the video suggests"
The assumption that the inverse for your operator exists would require justification but the assumption about the constants being 1 is good enough for qualification i feel, especially since in physics you hopefully can test these equations through experiment.
One should be very careful. This is true in the case if exp^(x), but other functions i don't believe this will work, since this proof hinges on the point that you know e^x is its anti-derivative (but its interesting to see the behaviour of the operator). Thanks for sharing!
This video is inspired by an example from "Professor Stewart's Casebook of Mathematical Mysteries". Please check my community post on March 3rd, 2024 for the photo of the particular page.
Please don't take it too seriously.🤗
Suppose I define a set S of words by saying that a word in S is either the empty word, or it's obtained by appending an "x" to some other word in S. I can write this definition of S succinctly using the equation S = 1 | Sx where 1 denotes the empty word and | denotes "or". Now, "or" is kind of like addition, right? So let's subtract Sx from both sides to get S - Sx = 1. Factor out S: S(1 - x) = 1, so S = 1/(1 - x) = 1 | x | xx | xxx | xxxx | ... Lo and behold, this is correct: a word in S is either the empty word, or x, or xx, or xxx, or xxxx, or .... This crazy-looking calculation is one of the key ideas underlying the Kleene-Schützenberger theorem that regular languages can be represented by (noncommutative) rational functions. The * operation, indicating arbitrarily many repetitions of an expression, corresponds to taking an inverse: X* = 1/(1 - X).
You have successfully nerdsniped me. I go to your community tab, and I still haven't made it to this post because I keep stopping to do the quizzes. I even am trying to save one for later by saving the comments page (and hoping they don't spoil the answer).
I'm just now at 10 days old. Let's see how long it takes for me to get to 12 or 13 days old to see your comment.
Some comments do spoil the answer :)@@ZipplyZane
Dud u post on community tab as if it was Twitter
@@26IME 🤣
The real joke is that a physicist forgot how to do a Taylor expansion
Fr we do that shit every time we get the chance
Nah, physicists don't usually venture beyond the second degree of Taylor expansion. That's why he forgot to expand it beyond the third degree.
Need to wash my eyes after watching this
I've only taken calculus 101 and physics 101, and still I can completely comprehend why this is absurd to both tragic and comedic levels. I have a saying that physics research continues developing until it hits a wall, and that's when they go and knock on the doors of mathematics professors to see what they can borrow next.
@@brianjones9780 Lol the last bit is truly spoken like someone who has only taken calc and physics 101
@@seacucumber6768 ...am I wrong though?
Functional Analysis gives rigorous foundations to this and other seemingly non sensical manipulations, by treating these "symbols" as operators and giving necessary conditions for the convergence of such expressions in a given space.
@@brianjones9780 Yes
bro violated that dx
In a lot of places in math there is no dx either
What is a meaning on an integral without dx..
@@kyspace1024then calculus got violating nothing is variable so integration doesnt exist if theres no variable parameter
@@dipankarhowladar9949The variable of integration is often omitted, you can see that quite often in measure theory texts for example
@@derp_gamer7596 A linear operator acting on a functional space
"its a constant so it equals 1"
Wow, just learned an important lesson when solving for x
Yes, downvoted the video for now. If OP can explain that I will upvote. Without explanation it is not even a joke, and is very dangerous.
@@jonathandawson3091 well, a more concrete view is that the integration operator is from 0 to some indeterminant x, and the integration operator on e^x is e^x-1 and is a linear operator
@@lfx2407 Wow yeah that actually makes sense. I feel a bit stupid now. Thanks
more jokingly, its because the constant must be 1 in natural units ;)
Actually, for any C, C*e^x is also its own antiderivative. In this case C was 1 and so that's why 1 is there.
Y'all physics people are crazy💀
I work with a few and yep they are.
Eventualmente, los matemáticos terminan realizando la demostración rigurosa de nuestros cálculos
@claudiuionel5196 it is true.
Not me. This really blew the fuse in my brain
Guys, it is banter. Physicists use normal math but under physical constraints which make some mathematicians confused. No physicist would use or condone the argument in the video.
Okay but "With proper definitions everything i have done is legitimate" is just a Get Out of Jail Free card for doing whatever you want as long as it internally makes sense though lol
Thats how any math intuition is turned into publishable material tbh
@@schokoladenjunge1 Yeah, but in context of this video it basically says "When everything I did is defined in a way that allows me to do the things I did I am allowed to do the things I did because of the way those things are defined" and I was mostly joking about that lol
For example when dealing with PDEs you can sometimes get 0 in the denominator of the fraction and that's fine as long as you remember the proper way of interpreting it xd
@@sidlestone4575The first rule of Tautology Club is the first rule of Tautology Club.
This can be made rigorous using something called the holomorphic functional calculus. In fact, for any normal operator A on a Hilbert space, I can make sense of the expression f(A) for any bounded Borel measurable function on the spectrum of the operator.
@@THEEVANTHETOONThanks for tonight’s reading material
You committed about 5 different sins in two minutes, and I love it.
I see no sins-in fact, I see no trig functions at all!
@@SeanCMonahan Of course, you don't see any sins - they are imaginable!
I'm a physicist and this video gave me a migrain. I have never seen such horrible misstreatment of mathematics in any physics course I took, except the one we don't speak about (a.k.a physics of atoms and molecules, that prof was a chemist).
👻 "the one we don't speak about". Don't take this one too seriously :).
This manipulation can be made more rigorous through Functional Analysis, however I don't think that's something you would voluntarily decide to approach in your free time just for the sake of learning more about linear operators defined on higher spaces, and not to just show things like this as a party trick
I have been waiting for the key word functional analysis in the comment. Thanks @@healer1461
@@ron-mathNot once in my life have I thought that I would hear a thanks after mentioning functional analysis to someone, but this time it's strangely nice to be proven wrong, so be my guest ^-^
@@healer1461I did take courses in FA since it connects to quantum mechanics and the derivation in this video is only on a very superficial level similar to what would be rigorous in those courses.
"As someone with a physics background, that constant *must* be 1."
I recall working in 'natural units' where:
* the speed of light
* the mass of the electon
* plank's constant
* and the permittiviy of vaccum
were all equal to 1, so yeah, that checks out.
b r u h
Best unit system tbh, no need to carry on that garbage in each calculation. It makes dimensional analysis way more straight forward when every energy is just a frequency 1/s etc.
That's actually well-motivated, though. There's no physical reason that the only fundamentally meaningful speed in the universe should be exactly 299,792,458 meters per second where the "meter" is some fraction of the distance from the earth's equator to its north pole and the "second" is defined as some arbitrary number of hyperfine transitions in Cesium-133. It's FAR more justifiable just to say "that speed is 1" and go from there.
The same thing is true for things like h-why should the fundamental unit of action be 6.626... x 10^-34 of some silly units we made up? Much better to just make that be 1, too. (Or the reduced Planck constant, as it were)
The constants of nature are arbitrary. Mathematician when we choose that arbitrary to be 1: "YoU ViOlAtEd MaThEmAtIcS!!!!"
The most baffling thing with these physics acrobatics is that most of the times, they work very well. I remember a class on basic mecanical principles, and the teacher basically spent 4 hours violating derivatives and making integrals work with spit and spite.
Yeah, the way I like to think about it, physicists usually have some physical or intuitive justification for what they are doing. Obviously this video is a slight exaggeration of the egregious things physicists do, but nonetheless they usually have some justification of sorts. Then the mathematicians bend backwards and define theories so that the physicist's intuitive manipulation is correct. It kinda makes sense if you think of mathematicians as defining the math so that it accurately models whatever the physicists are looking at.
@@albertrichard3659lol
@@albertrichard3659That’s 100000% how mathematics is “discovered”. Einstein was not as good a mathematician as a physicist and sent his requested proofs be verified that his intuition was correct
@@superuser8636 Yeah, I wouldn't say 'discovered', but that's pretty much how all math progresses. There are, at any instant tons of things you could do with math but we only focus on the subset that captures the intuition behind whatever we're trying to solve because that's the whole point. Vladimir Arnold himself claimed that math is a subset of physics. It's a little bit too extreme a view for me but I do agree with the general sentiment.
A large part of my mathematics classes was dealing with various convoluted edge-cases that typically do not occur naturally. If we had the liberty to say "for most functions" or "for well-behaved functions" or something similar, my degree would've taken a lot less time to complete.
This actually has rigorous justifications if you work in an appropriate function space. All you need is a Banach space in which the operator norm is less than one so that expansion of the operator in Neumann series is justified.
I was hoping to find this comment
Thanks, I was wondering how you could use that for a function since if the series diverges then you can’t apply the trick.
You can also cheat and just note e^x's series is fully positive and converges for each value, so the operator is well-defined on series (ad hoc, and requires finding the radius of convergence, but who cares - every calc 1 class does this anyways :P).
Thanks for that!! It was the cure to the headache this video gave me.
Thanks for that!! It was the cure to the headache this video gave me.
This might just be the most cursed thing I've seen in a year
For those afraid of calculus
@@DistortedV12*for those who understand calculus
@@mahdihasan6222 for those not understanding functional analysis
@@mahdihasan6222 Yep. As someone who hasn't done a single derivative in his life, I have absolutely no idea what's wrong with it. I might need to take some Brilliant courses or something...
I had a stroke watching this, this is not okay.
Relax bro.
the real crime is the lack of \left( and
ight)
why are you talking in latex
why arent you
bro put \left(
ight) for the outer brackets but not the inner ones i can’t xD
Sloppy mathematics is part of life;
Bad notation also is;
But not using your \left's and
ight's when you need them is just prepostreous;
The most beautiful thing about this is that with some little ajusts and dealing with the right space this process becomes correct
That's a stupid statement. Everything in mathematics can be made to be correct in the right domain.
That's a stupid statement. Everything in mathematics can be made to be correct in the right domain.
@@Lolwutdesu9000Not really, no. Yes, you can show "1=0" if youre working on a field where 0 has an inverse, but the only such field is the trivial field with one element, so it has zero applications. This on the other hand works when dealing with L^2, which is a far more relevant space. And this construction is quite silly, but the underlying reasoning of exploring polynomials of operators is also something you use in practise
To put it more simply, yes, anyone can write a book, but it's more impressive if you wrote LoTR than if you wrote "Ernie and Bert's new kitten"
@@Lolwutdesu9000Your comment is stupid as it misses the essence of “application”. Physics is an applied science (despite the fact that it has gain a purely theoretical component over the years) which means the domain is there. We don’t just make up domains.
@@Lolwutdesu9000 its kind of funny calling a statement stupid, when you actually have no idea what you are talking about
Students using chatGPT be like...
I just had an aneurysm
Please no :( Don't take it too seriously bro.
I don’t understand what’s so objectionable about it, should we all pretend to be babies who don’t understand differential operators on Banach spaces just because a physicist did it?
This video gave me a stroke, a migraine, 2 trivigintillion dollars in debt, a curse from a haunted mansion, a nightmare about Dirac's equation for the electron spin, a magic dog that solved the Collatz conjecture and absolute disgust
It's fine if you consider integration to be an operator over functions. It can be cleaned up via Functional Analysis.
Technically, this methodology makes sense from an operator theory perspective. Call the integral operator I, and 1-I is a linear operator ((1-I)f=f-If) on the space of formal power series (who cares about convergence anyways :P - can also do on equivalence classes of functions differing by a constant, but this is the same thing). Knowing that f should be of that form, we get (1-I)f=(1-I)(a0+a1x+a2x^2+a3x^3+...)=(a0+a1x+a2x^2+a3x^3+...)-I(a1+a2x^2+a3x^3+...)=(a0-a1)+(a1-2a2)x+(a2-3a3)x^2+.... Since a positive series sums to 0 iff each term is 0, we have that an-(n+1)a_(n+1)=0 for all n. Since e^0=1, we must have a0=1, so 1-a1=0 and a1=1, so 1-2a2=0 and a2=1/2, so 1/2-3a3=0 and a3=1/6, and inductively, an=1/n!.
To be fair I can accept anything a physicist does except for writing the differential dx before the integrand like \int dx sinx. That is just disgusting...
What about ∫(sin(dx))? 🙃
@@lazarussevy2777I think it's just a constant, as lim dx -> 0 of sin(dx) = 0, so you are effectively summing up epsilons. I don't really have time for it now, but I think expanding it as the series gives (x - x^3/3! +...), such that the integral will be \int {(dx) - (dx)^3/3! +...}. Using the distributive property of integral you get = (dx + C) - something that is less than epsilon + something that is even more less than epsilon, and so on. That in the limiting case is equal to the C. Therefore, the unbounded integral of sin(dx) is just a constant. But please take it with a bit of salt, because this is not really rigorous
@@tenebrae711 I see what you did, but you still have to have a dx on the outside in order to integrate. To solve this issue, you have to multiply by dx/dx before you integrate. Then you have lim dx -> 0 of sin(dx)/dx, which we know is one, then the left over dx. This is ∫dx, which is x+C
It just depends on the length of the integral. When it's short use the dx as a bracket, when it's long it's clearer with dx in front and you use additional brackets.
@@Currywurst-zo8ooWhat do you mean by short integral or long integral? I've never heard of such a thing. Could you please explain how you would use the dx as a bracket?
This works because differential operators can be converted into powers via an integral transform: Fourier, Laplace, mellin, etc.
Wow that's crazy that it works! Very cool
Exactly what I thought as well. Can’t resist sharing with you.
It's just physics \s
I think the funniest part here is the division by (1- integral sign)
Since e^x *(1-integral sign)=0, it follows that (1-integral sign) is 0 since e^x can’t be 0. We have a 0/0 situation here.
I find it amusing
This video brings me great joy. I shall impart this forbidden wisdom to unsuspecting victims. Thank you.
As a Math major, the feelings this video generates exactly how I felt when I was attending a class of Mathematical Methods in Physics for a Minor. We were taught many methods of solving Differential Equations, and also, Complex Analysis - in which I struggled to gasp a single thing as most explanations were as non-sensical as this. I wish I was joking.
My heart finally felt at ease when I revisited all this concepts later on in courses meant for Math Majors, under completely different instructions for each discipline.
Edit: Note:- for the people who argue this is fine if you view integral as a linear differential operator from a banach space to another (space of differentiable maps to space of continuous maps) and I do agree, the "so it must be 1" is still jack. I understand the video is a joke, and while I am still giggling at the joke, I certainly won't assert some statement for which is incomplete. Functional Analysis isn't trivial. The people who properly explained it in the comments have my gratitude though, as they simply didn't just say "Functional Analysis" and called it a day.
One of my math professors called physicists “idiot savants,” and I think that’s for good reason. Math owes a huge number of theorems (even if not their proofs) to physics. As a math guy, I think we get a little lost in the sauce
as a mathematician I find it elegant
In engineering we learn D operator to solve differential equations where derivatives (d/dx) is treated as a algebraic element( more like x)..you can multiply them, factorize them also divide by them.Its really fascinating and wonderful !Though its not something ground breaking , it just reduces calculations.That integral reminded me that.
Bro appreciate all your vids it always helps me sleep keep up the good work
🤣
I didn't thought it would go smoother than the slope of my marks
EDIT: These are the most likes a comment of mine has gotten since I joined TH-cam.
This video fired up some old memories of QFT classes at my uni way back when. I still can't fathom how this math voodoo gave us such good predictions.
the key is to treat x as a vector; replacing 0 and 1 with 0(x) and 1(x) make this much more sensible; by treating the integrand as an operator on a functional object rather than on limit-depedent evaluations of said function, the dx limit is irrelevant bc we've abstracted away the evaluations, and we did not need to divide by zero. also, we can put everything on equal footing by virtue of full-rank tensors on the superposition of all R domain inputs; they transform like tensors after all.
the identity operator and integrand are both linear w.r.t. any superposition of inputs of the function, and thus can be combined into a linear operator/tensor. this is not unique however because the integrand has a rowspace from -C(x). by linearity we can subtract both sides with -C(x) and pull the degree of freedom to the right side. there exists some -C(x) tensor for which (1-int) operator is unique and thus invertible, so by that virtue we can restrict ourselves to that case and continue by inverting it to the other side. since we make the assumption that C(x) is unique to invert the matrix, the right side is a unique operator, and it is valid to invoke division.
now comes the fun part: we left-multiply each side with 1(x) to split off C(x) as the tensor to be transformed by an operator that can be expanded to its power series; things can be assumed to be smooth so Taylor's theorem is in range. the integral has lost its +C privileges so we don't have to worry about a residual polynomial blowout when computing them, and now WLOG we have a series that can be rewritten as e^c(x) by the definition of e^x. finally, the unique solution that makes e^(x) equal to e^c(x) is c(x) = 1(x), and we are done. i think that c=1 just happens to reflect the fixed point behind the system on which e is defined.
im have a BA in CS and took physics classes on the side, hopefully i did a good attempt at justifying this madness even with the missing +C. it actually fits into quantum mechanics pretty well, literally, because x's norm here is restricted by 1, the radius of convergence for 1/(1-x). but it also just feels like mathematical anarchy.
This was a lot of fun. Gonna check out the channel
This is an almost very sound linear algebra derivation. It basically says that e^x-(e^x)' = 0 up to an additive constant
this is the funniest video i have seen in a long while, thanks lmao
i'm pretty sure ive never seen any physicist explain it this way
only with operators, and then once they introduce those the huge shortcuts happen
That picture at the end makes me realise how perfectly Jack Quaid is cast as Feynman in Oppenheimer
There's a fancier version of what you just did called Neumann series, but you need to change the integral operator a little bit. Instead of just the integral, we define a much rigorous operator T, where the action of T to a function f(x) is T(f(x)) which is the integral of f(t) dt with the lower bound is some constant c (usually 0) and the upper bound is x. Therefore, the norm of the integral operator T would be less than 1, hence by the Neumann series, (1 - T)^(-1) = 1 + T + T^2 + ...
I hope I can learn more about that soon, I'm going to take a functional analysis course next semester
This is the comment I was looking for.
I'm having high blood pressure, stroke, heart attack and panick attack watching this.
I showed this video to my calculus lecturer and received a letter of expulsion from my college the next day
This is very creative and neat! I liked that you included the notion of operators and functionals, to show where you're coming from. Being creative in math is very important, most great physicists/mathematicians had their weird notational tricks and some of those even prevailed and became conventional. Unfortunately I think this video is largely misunderstood, which is a shame
I am fine with it. Glad you enjoyed it!
You sure notational tricks is the correct word
This is so cursed
This usage perspective on operations reminds me of how putting e to the power of the derivative operator creates a shift operator
I love doing math like this. Its really cool that despite playing fast and loose with rigor, if you can come up with a consistent logic in the math it still often agrees with reality.
Acrobatics Calculus is an extremely accurate description for this.
This comment section is literally that IQ bell curve meme. :D The guy in the middle says "nooooo, you cannot terribly mistreat math like that, I need to wash my eyes", while the guys on the left and right say "this is totally fine, what an elegant derivation".
Generalising people with meme is such a childish and stupid idea. There is no math without formalism, you can't just do some unformalised tricks and hope everything will be okay, because after you build few theorems on those tricks it's impossible to tell if your proofs are valid. You are left with philosophy not math at this point
@@alang.2054 You are literally reiterating my point. If you have the necessary mathematical background in functional analysis and take enough care, then you are justified in making derivations like in the video (integration is just a bounded linear operator in some appropriate function space). If you don't have any such background and just do some unformalised tricks that seem fun to you, then you are not justified in doing so. All that basically corresponds exactly to the two guys on the sides of the bell curve meme (trained mathematician vs amateur). It's just a meme, don't take it too seriously, I'm not insulting anyone.
@@tsevasa bro he just literally proved your point 😭😭 i wish the joke didn't write itself this clearly
Ikr. Thanks to reading a bit of QM, I can appreciate the beauty of this. Makes me wonder why I chose to pursue computer science.
Sorry, I'm gonna go cry somewhere :(
@@iknowsomestuff7131 Because maybe you want to actually make money?
This is one of the weirdest things I've ever seen in my life. Brilliant !! :D I don't know if these steps are legit but it looks fantastic.
My eyes just melted from disbelief
This has to be some sort of world record in abuse of notation.
That is a smart way to get the Taylor expansion of ex, considering the integration as a normal operator.
man's forgot the +c 💀😭
Haha. Not really but very good that you remembered the lesson!
the c-herry on top hh
"let's start by assuming c=0"
@@DatBoi_TheGudBIASfuck yeah
Man murdered the +C and stashed the body in the basement
I want a version of this vid where everything is properly defined, and the defined mathematics can be applied in other cases.
It is a good idea. However, I would definitely need collaborators on this direction.
As a student this used to drive me nuts but as an older amateur just trying to occasionally do applications I've forgotten a lot of the 'proper' ways and love the sorts of abuses found in physics and engineering references
this is exactly what my physics teachers did in physics I and II at uni
Thank you for helping me reaffirm that I made the right decision to choose pure math over physics.
:)
This is really slick. I love it
Let's try to do this properly.
Let's define an operator S: C( [0,1] )->C( [0,1/2] ), S(f)(t)=\int_0^t f(x)dx. C( [0,1] ) is actually very restrictive, but it's good enough for exp. Since it is equipped with the uniform norm, we have ||S(f)||=|\int_0^{1/2} f(x)dx|
Nice work!
The essence of Mathematics lies in its freedom.
- Georg Cantor
From 0:56 the video falls victim to the initial "suppression" of dx, which just a notational shorthand. In reality dx has been implied all along, it constantly accompanies the integral sign.
How does MatLab respond to this syntax?
1:15 As someone without a physic background I can't stop to question myself why you choose all those seemingly random therm 😂
It's acrobatics haha.
psychic?
next time i take an exam ill write at the end "with the proper definitions everything i just did is ok"
Greate video! But I got a question: why does the same procedure fails when substituting the intergral with differentiation? We also have (1-d/dx) e^x = 0, but expanding the series (1+d^n/dx^n) 0 does not give out e^x. Where is the problem?
Nice question but I don't know the answer in my mind right now...
What exactly do you mean by “operator” when referring to the integral sign? I know that you said to think in reverse, but what does an integral mean without a dx, if it means anything at all?
"What are you trying to tell me? That I can just choose whether to be an operator or an operand?"
"No ∫. I'm trying to tell you that when you're ready, you won't have to."
So pictorial, I like it.
saw the von neumann series coming from the thumbnail. love it
Are there any other (actually useful) applications of this trick, similar to how generating functions seem to abuse formal series but turn out extremely useful? This video is just a re-derivation of exp(x) but if we alter the equation we should end up with a different function. Maybe if we extend our algebra with this integral sign some functions could start having closed-form expression representations?
e^x=D(e^x)=> e^x=(1/(1-D))(0)
=>e^x=(1+D+D^2/2+D^3/6...)(0)
(1-D)e^x=0
what does this mean? I remember learning a bit of it, it's called operational calculus started by Heaviside I believe. The solution I saw for differential equations like this used taylor expansions to solve for coefficients.
Don’t learn that it is full of paradoxes and most concepts are defined loosely without any rigor involved. Laplace is wayyyyy better.
@@pauselab5569 correct me, but I believe it was made rigorous, just that it wasn't much better than Laplace transforms. It's honestly a lot more appealing to just understand the derivative operator itself rather than letting some transform do all the work. Neither will work in all cases anyways.
you cant expand the inverse of (1-D) since it isnt a bounded operator afaik.
rudin treats this subject in chapter 10 of functional analysis. the section is called "symbolic calculus" but it is also known as functional calculus
@@er4795 No. Viewing D as an unbounded operator on L2(R), it has spectrum(D)=iR. In particular, (1-D)^{-1} is bounded.
Can you do similar video but about math proofs? Y'know, the upside-down A, mirrored E, and that stuff
You mean acrobatics of proofs? That would be a real crime lol 🤣
The recurring theme in my physics class is that they make it work somehow but dont explain why they can do that. I cannot understand how they keep getting away with those shenanigans.
It's probably easier to see and understand the tricks when you delve deeper and get some more global knowledge but for now I'm just thinking like a medieval peasant and want to burn the witch
For what it's worth, if you keep integrating and adding 1, you get e^x
Int(1,x) = x
Int(Int(1,x)+1,x) = x+x^2/2
etc
and then you add 1 to get the result of the taylor series of e^x
Ahh, functional analysis! Indeed, sir! Have a great day!
e^x is not the only function with equal derivative and integral, this works for 0 too
The arbitrary choice of letting the integral of 0 be 1 makes it work for e^x. If you make other choices, it works for the other functions that are equal to their own derivative.
@@chadwick3593 what branch of math are you talking about? I never heard of someone equating the integral of 0 to 1
@@chadwick3593 The integral of 0 is 0x which is still 0
Love the channel
but like, this is (mostly) valid if you treat everything as formal symbols!! there is the issue of why integrating 0 should give 1 rather than another constant and why repeatedly integrating shouldn't add more constants. However, we can just take this as a definition (i.e. integrating sends 0 to 1 and 1 to x and so on) and, doing this, the final answer will be off at most by a constant!
This is actually pretty much okay to do if you know functional analysis. Look at some of the definitions of Bernoulli polynomials on the wikipedia page to see stuff where derivative operators are used like this.
Reminds me of my first year mathematics course in engg.
Where all of us did only acrobatics.
I still have epsilon delta trauma
Next time I'm called to the drawing board, I'll just say "As someone with a physics background, I can do something like..." and proceed to write total nonsense on the board, hoping that it'll work.
This is my new favorite video on YT not gonna lie. I was laughing the whole way through out-loud. Yes, I am weird
This channel should be called Rum & Math because there’s no way a sober person could come up with this
there’s a sick, twisted part of me that enjoys anything that is cursed, (music, programming, images, etc) but this on a whole new plane of cursed
This appears on page 65, 68 of "Game, Set and Math" by Ian Stewart, first published in 1989.
I would feel more comfortable doing this with the Laplace transform, although probably it's just using a different and fancier name for the same and simpler/more fundamental thing.
Isn't this actually valid if you treat integration as an operator on appropriate function space?
I'm unironically amazed. The "integral" is essentially now an operator on the function space..moreover, it's actually an "inverse" (unfortunately there's no one to one mapping for differetiation, so no actually inverse) of a linear transformation.. the d/dx . Since the kernel of the map d/dx is set of al constant functions, this inverse is supposed to give a whole family of functions. Now, we can represent the whole family by a unique representative.. namely the function with value 0 at x=0, and define that as the output of the inverse, making it a linear transformation on the function space. And thus, the 1/(1-int) thing makes sense, since int is just an infinite dimensional matrix and so by Taylor expansion, we can do what he did.
Thus, with proper rigor, all of it does make sense.
Damn.. I'm really amazed.
Is that ʃe^x without dx a 0-form? And is that standalone ʃ a 0-form too and means ʃ1?
It's true, all constants are equal to 1, if they're not change the units until they are
I studied mathematics in college and graduate school and my god my analysis professors, especially functional analysis, would have lost their fucking minds lmao
I took functional analysis this semester. The Neumann series would like to have a word with you.
Doesnt this work for any smooth function?
Can you explain this in more detail?
Specifically, why those specific constants of integration?
From a comment above: "the correct way goes like this: define F(f) = int_0^x f. Then F(e^x) = e^x - 1 and rearranging we get (id - F)^-1(1) = e^x, where id is the identity and id - F is invertible since ||F|| < 1 in L2([0,1]). so we are calculating iterated integrals of the constant function 1, and not 0 like the video suggests"
This looks like operational calculus, and linear operators can be transformed to be used like an algebraic variable
The assumption that the inverse for your operator exists would require justification but the assumption about the constants being 1 is good enough for qualification i feel, especially since in physics you hopefully can test these equations through experiment.
Wait according to this you could divide by e^x at the start and get that integral of one equals one
Hell was made for you specifically.
>S-Tier Acrobatics
Brilliant work.
One should be very careful. This is true in the case if exp^(x), but other functions i don't believe this will work, since this proof hinges on the point that you know e^x is its anti-derivative (but its interesting to see the behaviour of the operator). Thanks for sharing!