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.

the whole introduction of differantials was made in order so that derivatives can be represented by fractions 🤦

​@@user-qh9yn3gk8u 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

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"

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

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!

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

@@user-er8ro4hv1e 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.

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.

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"??

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.

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.

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 (kzfaq.info/get/bejne/qMx3a8aWmdSymGQ.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 👍

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.

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. ^^

😂😂

That was the clickbait for me tbh

Very glad you enjoyed it!

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

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

Thanks! Glad you think so!

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!

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

True!

Math geek is literally in my name

Glad you enjoyed it!

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.

perect

Glad you liked it!

thank you!!

@@sarahseveringhaus357 no problem :)

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).

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!

Glad you enjoyed it!

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)

He may have, I'm not sure. Thanks for watching!

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.

I love this definition! 😂

​@@BriTheMathGuy you replied me...hihi subscribed :D

@@tronskywalker3633 Thanks a ton!

Glad you enjoyed it! Thanks for watching!

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

Very glad you enjoyed it!

That's because professors know the question itself is nonsense.

I might still be a little confused myself!

@@BriTheMathGuy What is x^dx?

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

Thanks! Very glad to hear it!

😂

Thanks! and thanks for watching!

😂

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

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.

Crazy right?!

Sorry :(

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

Thank you! Cheers!

Glad you think so!

Your assumption is correct. X is constant with respect to h. It is an independent variable and not a function of anything. Implicit differentiation is for expressions where the value of one variable depends on the other.

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.