I Computed An Integral That Breaks Math
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- Опубліковано 23 вер 2024
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #integral
I'd like to address some of the comments on this video. Many have (correctly) pointed out that the method in the video is not exactly mathematically rigorous. I like to think of it in the same way first semester calculus students think of continuity: "a function is continuous if I can draw it without picking up my pencil". Is this approach rigorous? - No, but it's intuitive and leads to the correct answer. There are other ways to get the same answer in the video. (For example: using x^dx = e^(lnx*dx) and expanding the Taylor Series)
I'd probably need the opinon of an expert mathematical analysis (which I am certainly not), but I think the most dubious part of the solution is turning part of the integrand into a limit of h. I believe that it only works in this situation (and it might be coincidence) because we preserve the traditional dx at the end of the integrand. We certainly cannot do this is every circumstance since it would make most every integral equal 0. (for example: ∫xdx ≠ ∫ lim h->0 (x*h) = 0. )
I think, despite the lack of rigor, the method is interesting and worth thinking about. Regardless of how you feel about the video's methods, I hope you enjoyed it and I appreciate all of the comments! (especially the ones requiring us to to think critically)
Seriously, the only way in which this is useful is to animate students to think out-of-the-box, but thinking about the new interpretation of dx as e.g. a convector, a Volumeform, a generator of an orientation, or simply thinking about Lebesque measures, this thinking leads nowhere. It's not "wrong" what you are doing, it's simply not useful and more importantly it does not give rise to any new information. It's like solving a specific first order ODE in one coordinate... You can easily calculate, if you get a solution, but unless your solution is part of a bigger class solvable for a broader case, noone can use the result
This is surely mathematically legitimate. The main mathematical question is to find the "best definitions" to describe integrals of the form "integral f(x, dx)"
@@johnyboy3325 Correct, but f must at minimum tend to 0 as dx->0 for the integral to be finite. Therefore it's more practical to "factor" out that first order pole (dx), and we end up with normal integrals.
I would love to see how this could be made rigorous. The main problem I see is when you split this into a product of "(x^dx - 1)/dx" and "dx", and then take the limit of just the first part of this product as dx tends to zero, while holding dx in the second part as constant.
Having said that, I do believe this *can* be made rigorous. The history of math is full of the ridiculous being made rigorous, often with surprising implications. My favorite example is making sense of the geometric series for common ratios greater than 1. This can be used to develop the p-adic numbers, which allow you to use analysis to tackle questions in number theory.
@@martin.thogersen How would you assign a value to integral e^(-x/|dx|)? This satisfies your condition, but its not clear how to factor the dx out.
Mathematicians: "Physicist misuse differentials by using them as fractions"
Also BriTheMathGuy:
the whole introduction of differantials was made in order so that derivatives can be represented by fractions 🤦
@@ΚωνσταντινοςΔημητριου-τ4ε it's more like a ratio than a fraction
literally the first lesson in differential equations (variable separable) is making them like fractions
Differentials can be taken as fractions tho.
Coz dx literally means the change in x, a very small change, like ∆x but very small.
@@epikherolol8189 it's a ratio and not a fraction
3:01 "This is a good candidate to use Lobey Towel's Rule"
That doesn't sound like the right pronunciation but I don't know enough about L'Hôpital to argue
😂😂😂
So, is this how you phonetically perceive the french "l'Hôpital"? That's interesting !
@@yrden99 Attention, ne tombe pas dans l'erreur de penser que tous les anglophones le prononcent comme ça. La plupart le dissent de façon assez proche de la prononciation française.
That point where every french math student ends up cringing because "L'hôpital bad"
I love that face he makes while he's concentrated. He looks like he's disgusted by the math😂😂 love it
Best comment
lol so true
A lot of math solutions look like being plugged out of ass when being explained
Theoretical Mechanics Infinitesimal Math. It is disgusting and kinda works.
Haha
I do not know if it is just me, but I have always hated math, until I took Calculus. I am now fascinated. Everything just comes together and makes a lot more sense.
What is calculus?
I'm "french" and we separate our maths in algebra and analysis ( and probability and combinatorics( i hope it's the right term lol))
I suppose calculus is analysis?
@@roxazzino3115 calculus is anything to do with derivatives (dy/dx or f'(x) typically, which are about the rate of change of a variable or function with respect to another variable) or integrals (∫f(x) dx, which is the inverse of the derivative and can be thought of as a continuous sum of a variable over a given period). I would say it falls more under algebra out of those categories, as they are fundamentally algebraic tools and can be manipulated and derived algebraically, although in many ways it's in its own area of studying the change in, rather than the actual value of, whatever variables you're using. You would typically see it listed as its own branch of mathematics.
@@Fireball248 ah i see
We include all that in analysis, not algebra (here i mean).
Everything that has to do with real functions, differentiability, differential equation, squences of functions, taylors series and also topology ( the part about continuity and limits ) are included in analysis.
The rest is algebra for us ( even though there are some things that obviously overlap, like polynomials and topology )
@@Fireball248 Calculus is a part of analysis, because analysis governs all limit-related subjects. Among others, differentiation and integration are very closely connected with taking limits. Of course, there is some algebraic manipulation involved when working out integrals or derivatives, but fundamentally, it is part of analysis.
Very glad it ended up coming together for you! I for sure started to really enjoy math so much more when I got into calc. Thanks for watching!
Though I am not convinced by "dx" is "Iim Δx to 0" thing approach, 'cause in differential geometry, we know dx is a 1-form on the manifold(here is ℝ). But x^dx can be thought of as an element of C*(ℝ,ℝ) or Λ*(ℝ), the exterior algebra on ℝ. Like how we did to some characteristic class, x^dx can be expanded into Taylor series, with the multiplication of differential forms replaced by wedge product. Luckily, dxΛdx=0, lot of terms are 0, so x^dx-1 is equal to ln(x)dx. Furthermore, ℝ is one dimensional and contractable, so every 1-form is an exact form, we can find a 0-form(scalar field) F so that dF=(x^dx-1), and F is exactly the integral of ln(x), which gives out the answer in the video
Thanks for bringing the mathematical voodoo in the video back into a well-founded terrain! Very nice.
Taylor series is normally defined for real-valued functions, though I've seen some formulations for vector-valued ones. How do you know its well-defined for differential forms? Why should we assume that the x^2 element of the expansion corresponds to the exterior product?
@@mmoose3673 Forgive me for I do not know too much differential geometry, but when does @mcxzx_zhihao make the assumption that, in the Taylor Expansion, the "x^2" element correspond to the exterior product? I am more confused on how well-defined turning dx*dx into dx(wedge)dx is...?
@@mmoose3673 Well, it's not an assumption, but a definition in order for x^dx-1 and its integration to make sense. You want to stay in things somehow generated by differential forms like "dx" and able to be integral, the only choice(as far as I know) is the exterior algebra ring(like if you choose to use tensor product like dx⊗dx, then "integral" that we usually understand is not defined). And the Taylor series is more like a definition here to the exponential notation(same notation can also be seen from Todd class expressed from the Chern roots). Also, a converge Taylor series is obviously well defined for any ring. Btw(making sure we're talking about the same thing), the Taylor series here not trying to "approx" the behavior of the field on the manifold ℝ here, but "approx" the behavior of the "exp" defined on generally exterior algebra.
@@BradleyAndrew_TheVexis dx*dx is not that well-defined if you don't state that is a tensor product or that dx←Iim(Δx->0) applying here XD But I'm sure that wedge product in the Taylor series would make perfect sense and absolutely well-defined here on those rings(still like the wiki Todd class example). And again, these are all definitions to make the things here make sense.
It is an interesting question, but it feels like an abuse of notation.
Perhaps!
@BriTheMathGuy Indeed it is. Clearly you put a limit (dx->0) into the wrong place. By definition integral is to find a function, not to find the limit of a function.
molestation of notation, i think
It's just sneaky division by zero
@@L开门见山 integral computes a value, not finds a function. You are correct in asserting that what he did was claim the integral was finding the limit of a function, which is only an infinitesimal slice of what the integral is actually doing.
It's "Hospital rule" since that's where you end up everytime after using it and forgetting about Taylor.
Actually you need the L
@@KRYMauL maybe more than just an L kek
@@Cannongabang Yoy know what I meant. （＾＿＾）☆
You mean the theorem that sir De Hopital bought from Bernoulli and took all the credits?
@@CristianTraina You got to be careful when you sell anything for some money :) your credits could be stolen.
Also it is Bernoulli who gave the noble De l'Hôpital private lessons, thanks to which the latter published a book on mathematical analysis. So people started calling that theorem "De l'Hopital rule" because it was in his textbook.
I guess everything has to be put in perspective.
What the heck? I made my profile pic as a joke, but with very little awareness that it may make sense.
lmao
Your profile pic evaluates to x + C using the same method, kind of epic
I think it doesn't. He doesn't get any result really
∫ exp(dx) -1
makes sense;
@@rtfgx how do you call what he found "not a result"??
I *knew* there had to be a reason for that -1 in the original problem. The funny thing is, each individual step was something I had done before, but I'd never have considered stringing them together in this way to solve this problem.
Can you please explain how ln(x) got in there?
@@oo7362 d/dh x^h = x^h * ln(x). it's a standard derivative rule that's normally memorized, but there's a fuller explanation that involves the exponential function.
anyway, here's the expanation
d/dx r^x = d/dx (e^ln(r))^x
= d/dx e^(x*ln(r))
= d/dx f(x*ln(r)) where f(x)=e^x
= ln(r) * f'(x*ln(r))
= ln(r) * d/dx (e^ln(r))^x
= ln(r) * r^x
making the appropriate substitutions:
d/dh x^h = ln(x) * x^h
@@Pablo360able thanks alot. I undestand it now. It's my first year studying ln and integrals.
No way, this is amazing! Never expected such an elegant answer!!
Very glad you enjoyed it!
I tried computing the indefinite form of the integral numerically and I it behaves like the indefinite integral of lnx, so you are right
Very cool!
@Muhamad Kussai Alawad Sure! It was nothing fancy, I just wanted to convince myself that what was done in the video made sense. A regular indefinite integral can be though of as an inifite sum of infinitely many small rectangles, and even though this idea isn't very rigorous it's what the notation for integrals was made to represent. The integral ∫f(x)dx from a to b is the result of adding f(a)dx + f(a+dx)dx + ... + f(b)dx, where dx is the step between each rectangle (f(x)dx would be the signed area of each rectangle), so if you want a rough aproximation you can just make a computer iterate from a to b using a very small value of dx. If you don't really understand what I'm saying I highly recommend you to check out 3B1B's video on integrals (ua-cam.com/video/rfG8ce4nNh0/v-deo.html). With this in mind I made the computer do the same thing but instead of adding f(x)dx, I just made it compute (x^dx-1). This is some javascript code I wrote which you can just paste into your browser's console without needing to download anything.
function integral(a,b,n) {
let dx = (b-a)/n;
let I = 0;
for(let x = a; x > integral(1,2,100000);
0.38629183695806546 (which is aproximately ln(4)-1, it's acurate to 5 decimal places)
>> integral(1,Math.exp(1),100000);
0.9999975795854892 (which is almost 1, Math.exp(1) is just e to the first power)
As you can see the results are very similar to the values which we were expecting. If there's anything you don't understand or that you think might be wrong let me know.
@@aioia3885 how does the script sums up all the dx times f(x) , can you explain please?
@@wayneli4070 it doesn't actually sum f(x)dx, it would if this was a regular integral of the form ∫f(x)dx. Since this integral is ∫(x^dx-1) then instead of summing f(x)dx I just sum (x^dx-1).
@@aioia3885 woah bro you are awesome 👍
Very nice problem ! I actually got it in a different manner : x^dx = exp(dx*lnx) and then I used the mcLaurin series exp(x)= 1+x+xx/2... et said any power of dx was negligible compared to dx.
Nice, same here!
I thought this was going to be the way he solved it
You can do that but you don't have to approximate by saying some parts are neglitible, you can just use basic facts about differential forms to conclude that all powers of dx greater than 1 vanish
@@geomaggluckida8361 Just checked that up, I did not know, thanks ! I’m an engineering student and my math classes tend to gloss over the details, which is often quite a pain in calculus ! I’ll have to learn all of this properly some day. I guess I was honouring my trade by approximating tho haha.
Just something interesting that gives the same answer:
Re-write x^(dx) in terms of the taylor series of exp(t) around t=0 (where we substitute t=ln(x)dx). this transforms our integral into: n=1 to ∞ ∫∑(ln(x)dx)^n . the sum goes from n=1 rather than n=0 because the -1 in the original integrand cancels this term out. we can consider every term of order greater than or equal to 2 as equal to zero, which leaves us with just the first term in the expansion ∫ln(x)dx = xlnx - x + C
Just what i have thought
I tried it out before watching the rest of the video, and that's exactly what I came up with.
my first though when i saw this
Nice! Thanks for sharing!
3:59 - You’re a wizard Harry
I've been practicing my hand-waveiosa
@Itachi Uchiha and I read your comment in the voice of Itachi.
@Laura it's so confusing when people link to the END of the segment they're talking about... :-|
@@irrelevant_noob your watching a video on integrals, but this you find confusing? Lol 😂
@@Laurah847 in fact i do, i know the rules for integration and can follow along with the video, however there are some psychic skill that i lack... everything would be so much easier if i could understand how other peoples think. ^^
Another way, same answer:
x^dx = e^(ln x dx)
Taylor expand around dx = 0 (taking dx as just a variable, unrelated to x)
= 1 + ln x dx + 1/2 (ln x)^2 dx^2 + ...
Since we are integrating, we can take dx to be a first-order infinitesimal, i.e. assume dx^2 = 0.
= 1 + ln x dx
And the rest flows like warm butter.
How do you make sense of what you're doing, you make it look intuitive
A big step in this is having done problems like this a lot. After a while, things fall less into a necessity to rigorously prove things and moreso into being able to recognize identities which in fact do make things a lot easier once you know the tricks.
Also, when practicing to explain an issue to someone else, you begin to understand even better what the catches of that issue are.
In this case, he solved a limit which acts as an equivalent argument to the problem, which is just the natural log of x in disguise and then solves what's actually a very easy integral when you think about it. To get to the integral of lnx dx he took the differential "dx" at face value and just evaluated them wherever they were next to within the bounds of a limit to simplify the problem.
Most of the calculus here is not even integration itself but rather based purely on the understanding of limits, the underlying language of calculus (which in fairness defines the derivative and integral).
This manipulation works on many problems. The trick here is knowing limits and derivatives well. By merely accepting the differential and it's relation to the limit as being an important system of entities, unlike the impression many calc students have, he's allowed to solve the question more intuitively.
The example is really just a novelty. The only reason anyone knows about this integral is because of an unlikely mathematical coincidence:
lim (a^x − 1) / x as x → 0 = ln(a) .
Normally, you're not supposed to be able to solve an integral this way, and there's really no way to prove that it's even possible without using delta-epsilon form. This is one of those cases where it turns out to be true, and the intuition for it has been circulating in the math world as a fun little challenge problem.
Well math should be intuitive, otherwise you are not understanding what you are doing
I did it the same way as him, so it is pretty intuitive
Man, thank you. I saw this on tumblr a year ago and I tried to make sense of it for 2 hours. I tried to sneak a logarithm inside of it, but nothing came out of it. You just gained a subscriber.
I tested this in Python, and indeed it works. Here is the code if anyone is interested:
import time
import matplotlib.pyplot as plt
import numpy as np
a = 1
b = 2
def integrand(x):
return (x ** dx) - 1
def antiderivative(x):
return x * np.log(x) - x
dx_range = np.logspace(-6, -1, 1000)
S_list = list()
exact_list = list()
start = time.process_time()
for dx in dx_range:
x = np.arange(a, b, dx)
S = np.sum(integrand(x))
exact = antiderivative(b) - antiderivative(a)
S_list.append(S)
exact_list.append(exact)
plt.loglog(dx_range, S_list, label='Riemann sum')
plt.loglog(dx_range, exact_list, label='antiderivative xlog(x) - x')
end = time.process_time()
plt.xlabel('$dx$')
plt.legend()
print(f'ran in {end - start} seconds')
plt.show()
thank you!!
@@sarahseveringhaus357 no problem :)
the faces he makes while writing are so coooooooool
Thanks! Glad you think so!
Length of the video: 4:20
I applaud thee big brain sir.
😂😂
That was the clickbait for me tbh
Another way for this unambigous expression is:
x^{dx} = e^{ln(x) dx}
and then expand the exponential as a series:
1+ ln(x) dx + ln^2(x) d^2(x)/2 + ...
of course one has to be very careful at several steps in order for this to be well-defined. I personally find this series more interesting than the standard log(x) integral.
What other problems break math?
Usually math breaks me 😂
@@Laurah847 ha!
This triangle!
This: 1/0
integral from 0 to dx of some function (say, x^2 + 1 or something), but without the dx at the end?
Actually now that I think about it the result would simply be (unless my logic is wrong) the same as the integral (0 to dx) of (x^2 + 1) dx but all divided by dx, so basically ((dx)^3)/(3 dx) + dx/dx, or essentially infinitesimal + 1, so 1
which is a pretty boring result and a boring solution, so maybe not a great idea after all
Did I do it right? I am barely experienced with calculus so I'm not sure if what I did is legal
That "-1" have an importance there which is beyond the understanding of mortals.
I think it has to do with the lim(h->0): (x^h-1)/h, when h is very close to 0 we want both top and bottom part to be close to 0, otherwise we get infinity and it's boring.
The limit would be (x^h)/h which goes to 1/0. The limit diverges, and so does the integral.
In case anyone was wondering
I'm taking calc 3 right now, and I wish I paid more attention to calc 1 because it taught important concepts like Riemann sum integral, delta-epsilon proof, etc. This is so freaking cool. I still am unsure what dy/dx really is. Could you do a video on dy/dx? aka The Leibniz notation. This will really help a lot of calc students who are still weak on some of the fundamentals. Thanks!
Edit: So in calc class, we usually learn how to compute stuff but we never really get to learn what dy/dx really means and stuff like that.
I think that's a great idea for a video! I'll see what I can come up with. Thanks for watching and commenting!
Agreed. I wished my teachers explained more about the concepts in Calculus I. I am lost as I start Calculus III.
I don't know if this helps (or is even accurate because I'm in calc 2) but I think of the "d" in dx as a small delta, so basically a really really small change in x. The same goes for dy. Now think back to how slope is just (change in y)/(change in x). So if we have dy/dx, it's really just a slope (since "d" is a small delta). What goes on in my brain is that I imagine a horizontal line between two things, the things start moving very close together, so close that it looks like they are touching but not actually touching, that line represents "d", a very small delta or "distance" between two points. If it's infinitely small, it basically represents the "change in x for one point" (I think). Therefore the dy/dx is the "distance" between y values when there is an infinitely small distance between x values. So it's a slope. Just think of d as a very small distance. If you can, plot x^2 on a calculator and zoom in very very very closely. The line will look straight because it's pretty much showing the change in y/ change in x at that point, this helps to show that it's a slope. Hope this helps! I also hope I'm not wrong so I don't get roasted by Brian when he makes the video.
Second this
dy/dx is the slope of the tangent line ie the derivative.
2:46 here we have to notice that the limit is equal to the limit as h goes to zero of (x^h-1)/(h-0) which is the derivative of x^h at h=0. So using L‘Hôpital here is not great, because we are applying circular reasoning
That's the thing with math geeks. They end up solving problems I never realized I had.
True!
Math geek is literally in my name
For those who are worried this might not be rigourous, there seem to me to be two ways to go: the proposed integral can easily be viewed as the limit of a Riemann-like sum taking the limit Dx-> 0; the other way is by viewing it as the standard part of a hyperfinite sum where x takes values in a partition with interval size dx (which is infinitesimal), this is the route of nonstandard analysis and can also be made perfectly rigourous.
In both conceptions we have multiple ways of solving it.
The ln(exp(..)) trick works easily for both. The power series and L'Hôpital tricks work for nonstandard analysis. For a rigourous proof in this vein using limits you would have to show some limits/summations can be interchanged (I believe, I have not written it out).
One thing that will definitely not help is using differential forms. These just generalise integration / differentiation on R to other dimensions and curved spaces. For R itself they show us nothing new. (This can be seen from the definitions, e.g. the integral of a differential form is just a sum of integrals on patches small enough to be pretty much flat, so we learn nothing new about integrating flat space).
There is nothing about integration / differentiation on R that can be learnt from differential forms that cannot be seen more directly by some other means.
(Sorry pet peeve of mine)
"I computed an integral that breaks math"
I think you mean: "I broke math so that I could define a particular integral and convince people online"
2:06 ''I dunno but it kinda works out'' is my college physics course go-to mood
There is another way to prove this, and in my opinion it is an easier one. You can write x^dx as e^(dx*ln(x)) and taylor expand the exponential. You then drop the higher than first order terms and voila. You end up at the same integral.
I did this by using Taylor series expansion. x^dx can be rewritten as e^(dx*lnx). Then using the expansion of e^x, the first term is canceled by the -1. The terms with degree more than 1 are zero because having something like (dx)^2 in an integral is interpreted as multiplying the finite area of a regular integral by dx which is zero. The only term left in the series is lnx dx which is integrated to get the same answer. Cool thing about this way is that you see that the -1 is essential. Without the -1, you would to integrate 1 without a dx which is infinity as you are adding up an infinite amount of rectangles without the width approaching zero.
I think the "differential" is technically not the same as the limit towards zero of (delta x) (because that limit is exactly zero).
The differential is a one-form. A function that is two-point evaluation, first in a point and then in a vector of the tangent space.
You just blew my mind!
Never thought computations could surprise me more..XD
Glad you enjoyed it!
You can approach this a bit more rigorously: Consider dx as a differential 1-form on the real numbers and define (important bit: this is a definition) x^ω via the series sum_{n=0}^inf ω^n ln(x)^n / n! for any differential 1-form ω. Then ω^n is the n-fold exterior product of ω with itself. This means that ω^0 = 1, ω^1 = ω, ω^k = 0 for k > 1 (the product of every differential form with itself is 0). We thus have x^ω = 1 + ω ln(x). So our integral where ω=dx works out to be int (1+ ln(x) dx - 1) = int ln(x) dx which we know how to solve already.
This is still not fully sensible. What does it mean to exponentiate a differential form? In general, nothing. In this case though, ω is an element of the set of sections of the cotangent bundle T^*R=R\times R^*. By a choice of 1, we may identify R^* with R and thus we can exponentiate proj_2(ω(x)) for any x\in R. This is a special case of a phenomenon on all real Lie groups, or more generally parallelizable Riemannian manifolds. Outside of this context, the exponential map doesn’t exist!
@@jzcook77 I would've simply interpreted it as repeated multiplication in the algebra of differential forms (the same way that you'd define a^n for example in general groups) - the series definition is a *definition* either way (isn't it? I don't really know any lie theory sadly) so I don't really see a problem here in just defining ω^n in this context as ω∧ω∧...∧ω (n factors).
@@SVVV97 but is the original integrand in the video actually a 1-form? It seems that he is cheating by turning it into one the way it was done...
@@joefuentes2977 that's a good question. Basically the original problem is just badly posed and without contextual definitions doesn't make sense - but given that differential forms are the "things we integrate" (I think both Fortney as well as Bachmann use this wording in their books) in a quite general sense I think it's not entirely unreasonable to try this approach and assume it to be a differential form.
@@joefuentes2977 actually it is neither a 1-form nor any other n-form for fixed n, rather it is an element of the whole exterior Algebra which means it is a linear kombination of k-forms for different k
"Does it make sense? I don't know, but it works out." that sums up my math classes pretty well.
It's worth generalising it to consider Integral ( f(x) ^ dx ). I would define this as a limit of products of terms of the form f(x) ^ delta_x, with the values of x separated by delta_x. From here you can get the general rule that Integral ( f(x) ^ dx ) = exp (Integral ( log(f(x) dx ) and evaluate definite pseudointegrals of this type as the ratio of the indefinite pseudointegral evaluated at the two ends. It's also worth noting that the indefinite pseudointegral has an arbitrary multiplier constant, like the normal one has an arbitrary constant added.
to make sense of this you can also see the integral over a domain (divided by the length) as a weighted sum over that domain some sort of additive mean now this thing you are speaking about is actually a multiplicative mean or geometric mean think of sum{k=0 to k=n} f(a+k(b-a)/n)*1/n being replaced by the multiplicative version product {k=0 to k=1}f(a+k(b-a)/n)^(1/n) this limit would yield the integral. the exponential as a group morphism transforms additive means to multiplicative means so its exp ( (additive) integral of ln o f over that interval)
3:12 you can’t use lhopital here as it’s circular. In order to find the derivative of x^h differentiated by h you have to compute that exact limit, therefore you cannot use the derivative to compute the limit.
Same problem arises when we try to find the derivative of sin x. We arrive at the step where we have to solve the limit of sinx/x as x tends to zero. It would be solved in a single line using L'Hospital Rule, but we cannot do that because we are actually trying to find the derivative to sin x to begin with. We are supposed to proceed with Sandwich theorem to do that.
@@Bhuvan_MS yeah ik
Is this really an integral? Or should it be written as a limit of a sum? Although the latter interpretation doesn't allow an indefinite answer, since there is no equivalent to indefinite integrals in sums, or is there? (I don't know)
You *can* create a summation expression of an indefinite integral!
*NOTATION*:
\int_{a}^{b} ( f(t) dt ) is the integral of f(t) from t=a to t=b.
\sum_{i=0}^{\infty} ( f(i) ) is the sum of the expression f(i) from i=0 to infinity.
\lim_{n -> \infty} ( f(n) ) is the limit of the expression f(n) as n approaches infinity.
*CONSTRUCTION OF THE SUM*:
Let F(x) be an antiderivative of f(x):
J = \int_{a}^{x} ( f(t) dt ) = F(x) - F(a)
F(x) = J + F(a).
Note that here, F(a) is an arbitrary constant. Call it C.
J is a definite integral, so F(x) can be expressed as
F(x) = C + \lim_{n -> \infty} ( \sum_{i=0}^{n} ( f(a + (x-a)/n)*(x-a)/n ) ). (This is just the Riemann Sum of f from a to x plus a constant)
We defined F to be an antiderivative of f, so F(x) + C (which is equivalent to F(x) - c), is the general form of an antiderivative of f(x).
The accurate way to integrate x^dx is to rewrite it as e^(dxln(x))=e^t. And then integrate the powerseries of e^t = 1 + t + t²/2! + t³/3! + ...
This is interesting. It immediately brought to mind the fundamental theorem of calculus. The derivative of x lnx -x = lnx which makes me wonder about the relationship between x^dx -1 and lnx. Nothing is coming to mind so far though.
One observation is the solution is invariant under some kind of ln(x) transform (if you ln each part of the integrand, you get the same result)
An entire semester of Cal 2 failed to ever adequately explain just what the hell "dx" is doing in an integral, and you explained it in less than 3 minutes. I feel both elated for my new understanding and infinitely more annoyed than I already was at what a shit professor I had.
0:02 what kind of unholy function is this😬😬
3:02 actually, we shouldn’t use L’Hôpital’s Rule here, because the fact that the derivative of eˣ is eˣ relies on the fact that the limit as h → 0 of (eʰ-1)/h is 1.
The length of the video is 4:20 :))
Wow you fit in a lot of information into such a short video! Impressive!
Glad you liked it!
Well at least give that type of function some proper definition: A function f is called ambiguous if and only if it can be described by handwaving on youtube to show your apparent mathematical dominance. claim: f:x->\int x^{dx}-1 is ambiguous. proof: 0:00
I love this definition! 😂
@@BriTheMathGuy you replied me...hihi subscribed :D
@@tronskywalker3633 Thanks a ton!
The face impression at 0:30 is how I think a mathematician mind look like from inside when performing integration.
What would the geometric interpretation be of what you just did? Anyone got any ideas?
I dont know maths but heres my thought.
So u have a function with value f(x) at point x. Imagine that value being a line of length x. If we put our line to some power of integer, we get a shape: 2 = square, 3 = cube, 4 = tesseract etc.
But putting it to dx is like taking an infinite root of it. dx~0=1/infinity = infinite root.
So instead of going up in dimensions, you go below the 1st dimension.
What you just computed (x^dx) is the descriptor for a 0 dimensional object, whatever that means :D AS in a line can be described by its length, a square can be described by its area, cube by volume etc. x^dx would be the descriptor for that object.
If we have normal functions x^2, x^3 etc. and we differentiate them we get the rate at which the descriptor would grow if we increased the descriptor for the 1D line:
a square would increase in area 2x proportionally to the current x.
So differentiating x^dx would give the rate of growth of the 0th dimensional object with respect to 1d line x. I got this to be 0. So something that would have 0 dimensions wouldn't change if its 1D counterpart changed size. Makes sense kindof... if 1D is a line, and 0D is a dot. ofcourse the size of the dot wont change if we increase the size of the line. Unlike for the Cube or square vs line, where the area of the square would increase and the volume of the cube would increase if the line got longer.
So calculating the integral of x^dx would give... idk my head starting to hurt from this thinkering
Well as far as I can tell this is just nonsensical abuse of notation, so it's difficult to make any real sense of it. This much can be said however, if we consider the characterization of the f(x)*dx term in the integral as an infinitesimal area determined by the infinitesimal length dx times the height f(x), then f(x)^dx is the same thing just on an exponential scale. That is to say that ln(f(x)^dx)=ln(f(x))dx. In some sense you could simply define the integral of f(x)^dx this way, which would perhaps be more appropriate.
@@asynesthesickid I don’t think we HAVE to stick to the old area of a rectangle intuition. We could scratch it for something more general and redefine a new type of Riemann sum in favour of giving meaning to exponentials. It doesn’t even have to spit out area and yet perhaps in a nice coincidence (due to perspectives) give us the same numerical final answer. I wouldn’t know where to start so don’t ask lol.
That infinite root sounds promising. Maybe you could end up with some interesting connection with factors and prime numbers by having (kx)^dx where k is an integer
@@asdfasdf71865 Perhaps finally give more insight into that stubborn Riemann hypothesis 😂. I didn’t understand though, what infinite root you talking about? I presume it is another comment
"I don't know but it kind of works out" This is my motto
😂
Amazing approach towards this ambiguous problem
Glad you enjoyed it! Thanks for watching!
Man, I i love using Lowbie-Towel's rule!
I feel like what you've done breaks laws in at least 7 countries.
But I don't have the slightest clue *what* laws you broke.
😂
I think the key to making sense out of this without going into more abstract math is to simply think of it as lim n->inf \sum_n{x^d -1}, where d=L/n and L the integration interval. Then, write x^d = exp{d log x} and approximate that as 1 + d log x, from its Taylor series. That immediately gives the Riemann sum: lim d ->0 \sum{ d log x} = int dx log x. It's interesting to see that you can write something that looks meaningless and find out that it can have a well defined meaning in some context, but then that context should also be stated, so that you may anticipate what other crazy looking "integrals" may not make sense the same way.
Hasn't papa flammy already done this integral? Good video btw. I've just discovered your channel.
He may have, I'm not sure. Thanks for watching!
When you evaluate a right-Darboux integral on the interval [a, b] of f(x)·dx, what you are actually doing is evaluating the limit as Δx -> 0+ of the sum starting at n = 1 and ending at n = (b - a)/Δx of f[x(n)]·Δx, with x(n) = a + n·Δx. In fact, this is the definition of such an integral.
So if we take the usage of symbols at face value, then the integral of x^dx - 1 on [a, b] should, by definition, be equal to the limit as Δx -> 0 of the sum starting at n = 1 and ending at n = (b - a)/Δx of x(n)^Δx - 1 = ([x(n)^Δx - 1]/Δx)·Δx.
The video claimed, without rigorous proof, that the integral of interest is actually equal to the integral of ln(x)·dx on [a, b], which by definition, is the limit as Δx -> 0 of the sum starting at n = 1 ad ending at n = (b - a)/Δx of ln[x(n)]·Δx. Therefore, the video is making the implicit claim that the limit as Δx -> 0 of the sum starting at n = 1 and ending at n = (b - a)/Δx of ([x(n)^Δx - 1]/Δx)·Δx is equal to the limit as Δx -> 0 of the sum starting at n = 1 ad ending at n = (b - a)/Δx of ln[x(n)]·Δx, which amounts to replacing [x(n)^Δx - 1]/Δx with ln[x(n)] in the formula. This is the key observation to rigorously justify this result. So now you may ask, is this equality true? Can it be rigorously proven that the above key equality is true?
The answer is yes. It is a well known fact that ln[x(n)] = lim [x(n)^Δx - 1]/Δx (Δx -> 0). In fact, this is often taken to be the definition of the natural logarithm to begin with. Another well-known fact is not as commonly known, however, is that this convergence is not merely pointwise, but also uniform. What does this imply? It implies that if you replace every summand by the limit of the summands, leaving all other parts of the expression unchanged, then you have equality. Since uniform convergence satisfied, the replacement done in the video does yield an equality.
Ahh, thx this makes a lot more sense.
Are you a mathematician? That was detailed.
@@joefuentes2977 Real analysis is taught in undergraduate mathematics, so it is possible to not be a mathematician and still be equipped to solve this problem. That being said, you would never encounter an integral exercise like this one unless you are doing higher mathematics.
What you are actually doing when evaluating a Darboux integral on the interval [a, b] of f where f is bounded function on [a, b] is taking the supremum of the set of all possible P-lower sums of f and the infimum of the set of all possible P-upper sums of f. If they are equal, then f is integrable on [a, b]. There is no reason to divide into n equal subintervals as you have done with x. This is not in the definition of integral. You certainly can evaluate an integral this way, but it relies on the fact that the function is integrable in the first place. To determine if it is integrable, we must consider all possible partitions of [a, b].
I have read about this new "integral" and I agree that it is well-defined. Although I don't find your argument particularly convincing given how you have used the definition of integral incorrectly.
@@willassad8670 you killed it bro
Really good videos! I think spending a couple more seconds contemplating the result and saying how pretty it looks at the end of your video would make the process feel more rewarding! I often get kind of frustrated when we get the result and the video cuts right away 😋
Nice result, but I have a slight bit of criticism if you'll allow me... using L'Hopital's rule to evaluate the derivative of an exponential the way you did isn't valid, since it involves going through this exact limit as the exponent goes to 0. There are other ways to define this result (which is correct) but L'Hopital is unfortunately not one of them. Again, great video otherwise!
I used the L'Hopital to destroy the L'Hopital
I dont see the problem there
@@NutziHD circular reasoning is the problem. You can't use a result to prove itself.
@@AlexandreRibeiroXRV7 you can't use this to proof this rule, but when proof is done you can use it as a shorthand for calculations, not as rigorous proof.
@@AlexandreRibeiroXRV7 Ah, now I understand your point. We actually did not define the natural logarithm with this limit in my course. We showed that exp : R -> R+ is bijective and called its inverse ln : R+ -> R. His reasoning is only circular, if instead the limit was your definition of the ln.
Now try to compute ∫x/dx
I did not regret clicking on this video. Really liked it!
Thanks! Very glad to hear it!
it seems like other methods in the comments found the same answer, but the thing as treating dx like its just any limit to 0 seems a bit wishy-washy to me
Love from India 🇮🇳
Thanks! and thanks for watching!
this is actually a real type of integral known as the product integral
Hey just wanna say I`m glad you still make videos...please keep going
Thank you so much!! I'll do my best!
@@BriTheMathGuy omg thanks for the response 😭 ps- you helped me with my math assignment today thanks for that too!
I remember integrating the area of an n sided regular polygon circumscribed about a circle of radius R, and let n approach infinity. I ended up with an integral from theta = 0 to 2 pi of something times the sin(d-theta). Didn't know how to treat the sine of d-theta, but then I remembered that the limit as x -> 0 of sin(x)/x = 1. So i thought maybe this is justification to simply replace sin(d-theta) with just d-theta. After making this substitution, low and behold, out popped pi*R^2 as expected. This seems very similar to your approach in this video.
When i see dx as an exponent.. I just skipped.. Wahahaha... Thank you for sharing this... Not all profs are sharing this to anyone...this helps a lot
Very glad you enjoyed it!
That's because professors know the question itself is nonsense.
My gut reaction was to do e^(ln(x)dx) and then expand that via its taylor expansion/defining sum. The 0! term cancels out the -1, the 1! term gives you ln(x)dx and so you get the same answer as if you set higher order terms of dx to 0. Which is cool i think.
Felt amazing
(still kinda confused if any of that actually makes any physical sense or not 😅)
I am just a math enthusiast.😅
I might still be a little confused myself!
@@BriTheMathGuy What is x^dx?
I thought of this differently: essentially since dx is some value approaching 0, x^dx is the same as the limit (as h -> 0) of x^h - which is well defined and equals 1 for all x ≠ 0. Therefore the integrand is just 0, and we can just chuck in a dx on the end since multiplying the zero is the same as not having it there at all. So the integral is just 0.
Can you please make a video of all the classes that you have to take for an applied mathematics degree and those you should expect to see in college (please)
If you pick a school I'm sure you can find the necessary coursework for their applied mathematics program on their site. If not you could find an advisor for the department on the directory and ask them for it. Often there will be course descriptions, text books used, and sometimes past exams. You can get a very real look at what kind of stuff is done during college.
Me absolutely forgetting I’ve already watched this already
Please stop!, integral is defined and denoted as $ \int_a^b f $ or $ \int_a^b f dx $. The dx there is just a notation, not an actual variable for you to manipulate. Furthermore, nothing is broken other than the way you approach this.
I would like to use this idea to say that all integrals are 0 because we are multiplying by a dx which is basically 0
Voodo math
Crazy right?!
Take d as constant and then make the integral for x, it is obviously for x as a variable 😊
I accidentally deleted my comment : (
It's not like it was particularly important, but still I'm a bit sad
Sorry :(
∫ x^(dx) - 1 =
∫ exp( dx * ln(x) ) - 1 =
(to first order using exp(z)-1 ~z as z->0, which is cool for x>0 and dx very small, which it is)
∫ ln(x) dx
= x ln(x) - x + C and GG
So this is doable with taylor expansions to the first order and you can make sense of expressions like this anytime:
∫ f(x, dx)
Lovely indeed
If Calculus students don't believe that the differential is vital to the integral, that tells me that they were not taught calculus, or at least not with any care to what's happening. If the "dx" wasn't vital, why would changing "dx" to "dy" completely change the answer?
In respect to what is something changing if there is no reference upon which a measurement is made? That by definition means nothing is changing, leaving us with the same f(x).
It scares me that there are people wasting time studying Calculus without knowing what a differential is. It's even scarier to realize they likely got past differential calculus before seeing their first integral while still not knowing what differentiation is.
This can only mean that either students are not paying the slightest attention whatsoever or that our education system is abstractly failing our society and passing potential future engineers that build our society from this math. Imagine having someone build the wing to an aircraft without intuitively knowing what a differential is.
Wtf. Who said any of this. You're twisting this guys words to mean that nobody knows calc 2 subjects and that's plainly not the case
@@ethansmusic9898 I don't recall stating he said anything of the sort.
I'm speaking on something that does actually happen with a lot of students.
@@ethansmusic9898 0:16 - 0:22
Actually, the ability to multiply and divide them is actually why we use dx in the first place, you even have to to calculate a derivative, I’ve even coded a derivative before, and from what I’ve learned, dx is just a really small number
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"cos(i)" breaks math. Because it's a real number.
That was pretty cool and you explained it very very well. You may very well be right on this honestly
I swear it feels like every other video I watch from this channel feels like an April Fools joke that’s just plausible enough to lead us on until the end where it actually ends up making perfect sense.
This is the best issue I've seen today
“lobi-tal” lmao 😂 love it
Every time that I see someone saying "break physics", "break chemistry", "break mathematics" it pisses me off because it is cheap clickbait.
The hacky way to solve this is to write x^dx as x^(lnx dx) and then do a first order taylor expansion.
Although it has a meaning in terms of maths as an infinite sum of small pieces, I've never met such problems in physics.
without the -1 the integral is undefined because integral sign multiplies by infinity and dx = 0
Also you are correct in doing the limit thing
Not what I was expecting, but intriguing nonetheless.
Some decades ago, I was toying with the following idea. An integral is a kind of "continuous sum." Can we define an analogous "continuous product"?
My motivation was to approximate the factorial function, g(x) = x!, in a way that doesn't require x to be an integer.
So for the integral, ∫ [a,b] f(x) dx, take the (simplest form of) Riemann sum: ∑ᵢ₌₁ⁿ f(xᵢ) ∆x, with ∆x = (b-a)/n, and xᵢ = a + i∆x
Then in place of that sum, take the product: ∏ᵢ₌₁ⁿ f(xᵢ)^∆x, and write the limit as n→∞, as:
∫P [a,b] f(x)^dx
[Writing the "∫" sign by closing the top curl, to make the top half look like a capital "P"]
By rewriting the product using exponentials and logs, this can be converted to:
e^[ ∫ [a,b] ln(f(x)) dx ]
But it's more 'fun' to write it with that new symbol, and with "dx" as an exponent.
And taking f(x) = x, you can approximate n! with (a, b) = (½, n+½)
Fred
it is not allowed to use l’hopital’s rule in this case because it literally is a definition of ln(x)
So this is called multiplicative calculus. The multiplicative integral can be defined as e to the power of the integral from a to b of ln composed with the function on x dx.
Yeah that’s what I’d call “abuse of notation”
Probably true!
It is only abuse of notation if you do not define the terms properly. This definitely can be made rigorous. I wrote a post about it in the comments section.
The units do also add up!
Let's consider your final integral, and say that dx is in A units and f(x) = ln(x) is in B units. This would make the final value of the integral be in A*B units (imagine m^2, m/s, whatever).
Considering the limit you used to reach f(x), we can reverse-engineer the units of the original integrand. Since h is the variable you used to signify dx, then h is also in A units. As such, we have the following equation with differentials, where U are the unknown units of x^h - 1:
((d(x^h - 1) U)/(dh A)) / ((dh A)/(dh A)) = (x^h * ln(x)) U / A
We then know that by substituting h=0, we must get to f(x) in B units, which leads to the equation:
ln(x) U / A = ln(x) B
By isolating U, we can conclude that the unknow units are A*B, which in turn means that the integral over the original integrand must also in A*B units, so both integrals actually give the same units!
tbh this kinds of integrals exist in maths called Fourier Mukai type transformation
this a weird one, rules seem arbitrarily enforced to move the problem along in a desired manner
Not sure if you'll see this but the part where you turned the dx into a limit as h-->0 makes sense intuitively, however I'm not sure if it can be rigorously shown to be equivalent.
It is equivalent. The definition of dx is a limit. The definition of integrals is in terms of limits.
@@angelmendez-rivera351 it isn't actually. by taking the limit as dx->0, he should have ln x * dx = 0.
I think you should've pointed out that if you compute the sum it does seem to approach the integral of lnx, because the approach is very sus indeed