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Mathematician here, treating (dx)²=0 just means we're doing smooth infinitesimal analysis, and this is just synthetic differential geometry. *places down Uno reverse card*
I figured that if the integrand is basically 0 if you set dx=0, then it should be fine. f(x) dx → f(x)×0=0 x^dx -1→ x^0 -1=1-1=0 sin(x) sin(dx) → sin(x) sin(0)=0 Furthermore, if you want a nontrivial solution, the strength of the differentials must be the same as the number of integrals. Each differential undoes an integral. ∫∫∫x dx² has not enough dx terms, so this "sum" would be ∞, if anything. ∫∫x³ dx⁴ has too many dx terms, so it just becomes 0. ∫cos(x) dx has just the right strength, so you'll get something not forcibly 0 or ∞.
I don't think so because while it's true those each of those individual functions go to zero the integral of those functions don't. This is basically because adding up an infinite amount of infinitesimals creates a real value. Even on your first example, f(x)dx may go to zero, but that doesn't mean the integral of it does. Integral(f(x)dx) ≠ integral(0). Two very different things. Infinitesimal often can be negligible to the point where they be considered zero, but once you start adding up infinite amounts of them, very different values. An infinite amount of infinitesimals added together could be anything, and infinite amount of zeros added together is zero.
@@briancoyle7770 that's not what I'm saying. Think about residues of poles at a point w. You're looking for the sweet spot, that is the order of the pole n, such that (z-w)^n gives you a non-zero finite result. If you use powers n, the limit would be stuck at 0. You're looking for the right balance. It's the same here. The same number of variable differentials for integrals gives you a normal function, too many differentials gives you 0, and not enough differentials gives you ∞. I say variable differentials because ∫∫dA looks like it has one, but dA=dx dy so it has two.
@@xinpingdonohoe3978 No I get what you mean there and I completely agree with you and the second have of your original comment. I just think that the first half I don't necessarily agree with. To me it just sounded like you were disagreeing with the video because you think that x^dx-1=0 which would mean that the answer to the video would be integral of 0 or C instead of his amswer.
@@briancoyle7770 oh right, okay. No, I just meant in a convergence sense. ∫sin(x+dx) could not succeed because dx→0 doesn't send the integrand to 0. ∫sin(x+dx)-sin(x) could succeed.
@@matthewe3813 The result can only be useful if you have shown that it is true, you can't show it to be true unless you apply proper rigor. Besides, a result doesn't need to be useful to be maths, it just need to be an actual result of proper rigor and not just a result of some vague hand-waving.
go back to the first explanation x^dx-1 = (x^dx-1)/dx * dx dx is small (approaches 0) so in a crude way we can take the limit as h -> 0 of (x^h-1)/h which is ln x so it becomes the integral of ln(x) dx
0.0000...01 is super small... 0.0000...01^2 is soooooo much smaller than that, so you ignore it. not really mathematically rigerous, but it's something us physicists do pretty often
@@henkolsonpietersen2242 It is called the first order perturbation, you could just as well have chosen the second order and kept the x^2 term of the expansion of e^x. I'm admittedly unsure how one would get the original video's answer to be equal in that case, however.
2:24 In stochastic calculus they define something called quadratic variation. A theorem is that if your function is deterministic then its quadratic variation is 0, if the function is stochastic then its first order variation is infinite. I think this is similar.
I showed this to one of my friends whose world was Mathematics. He was furious, and said to me, "If there's no dx multiplied then how do I know respect to what I integrate!?"
@@intellix7133 basically, Sf(x)dx (S is supposed to be an integral sign) means "plot the function f(x), then take the region between 0 and x and then do as follows: draw a rectangle which has the height from the x axis to f(0), and a width of dx, dx being a constant (draw the width in the direction of x). Then make another rectangle with the height from the x axis to f(dx) and a length of dx. Then repeat for f(2dx), f(3dx) and so on until you last rectangle reaches x. At the end, sum the areas of the rectangles. The value of the integral is the sum when dx approaches 0." This process basically gives you the area between the function and the x axis between x=0 and x=x, and the solution to the integral is some other function where you plug in a value for x and get said area. Technically that's for a proper integral where the area is defined (in this case between 0 and x), so because you can start from any point (doesn't have to be 0) you add a +C to the end function which symbolizes the added (or subtracted) area when changing from 0 to another number. Doesn't really matter if you got the +C part though, it's unimportant for this specifically, though it's worth mentioning. Anyway, imagine you have the integral Sf(x)(dx)ⁿ. It's equal to Sf(x)(dx)^(n-1)dx. Using the earlier definition, that means the area between the x axis and the function f(x)dx^(n-1). Since n>1 and dx approaches 0, the function itself is just 0. So the area will also be 0. Note: this is after splitting the integral in the vid into a bunch of seperate integrals. The area will be 0 so the solution will all be a constant, in other words the solution of all of the integrals together will be x(lnx-1) plus a bunch of constants. If we define another constant C to be the sum of those, then we have x(lnx-1)+C
In my opinion, it is neither correct nor incorrect; it is a choice. There are, to my knowledge, three different approaches to the `dx` symbol: it may be an "infinitesimal" value, it may be a value which becomes arbitrarily small in some context (this is the usual way calculus is taught), or it may be meaningless on its own, only used as a portion of some other syntax (integral ⟨...⟩ dx, or d ⟨...⟩/dx). The last approach is, in some way, the safest, but it makes this question strictly invalid. So we must choose one of the other options. The "limits" option is well-formalized, and partially used in this explanation, while an "infinitesimal" option may be easier to deal with intuitively. If we choose an infinitesimal approach, we once again have a choice about how to handle the expression `dx²`. We may go the way of the dual numbers, and say `dx² = 0` (which then implies `dx^(n + 2) = 0`, since `dx^(n + 2) = dx^n ⋅ dx^2 = dx^n ⋅ 0 = 0`), or we may say that `dx²` of a yet-smaller class of infinitesimals. However, if we do that, then the result of our integral is no longer a real number, like we would expect, but instead will come out as some `a + bε + cε² + ...`. Fortunately, all of these approaches are equivalent, I think (up to infinitesimal difference).
@@MCLooyverse I think you're overcomplicating it. Once I have Sf(x)*dx^n I can write it as Sf(x)dx^(n-1)dx, in other words I'm finding the area under the curve f(x)dx^(n-1), or y=0 since n>1 and dx->0. Btw, in both d/dx and S__dx the dx isn't just a symbol, but a multiplicative/divisor. In a derivitive you find a slope which is (f(x+h)-f(x))/((x+h)-x)=df(x)/dx, so dx is a divisor. In an integral you sum the areas of infinitely small rectangles with length dx, so the dx does multiply the function whereas the integral sign tells you to sum everything, but they are not actually connected in some way.
@@intellix7133 I wrote a comment explaining that and it literally disappeared lol. Look at what I told the other guy and tell me if you understand or want me to go into more detail
This is literally the solution I came up with too😳 I went back to watch your old video but before I played it, I tried solving it on my own and this is the solution I came up with! Although I looked at it through a slightly different perspective by incorporated elements of nonstandard calculus into my thought process. When I saw your old video I thought, oh cool a different solution but came up with same result. And then I see this video and you have the solution I came up with!
This actually does make sense on a conceptual level. You’re still taking an infinite sum, you can still approximate the integral in similar ways to normal ones. By taking small values of dx, and by using a finite sum.
The problem basically is: The summation of x^a-1 (with upper and lowers bounds not given), where a is equal to the step size that approaches 0. While he wasn’t rigorous with it, the actual problem makes sense.
Every time when we try to use different ways to solve those non-sense problems and always get the same result makes me wonder if there's something more fundamental that governs all theories and theorems
@@briancoyle7770 Not only this one, also something like "sum of all natural number equals -1/12", it just seems so nonsense that even a kindergarten child will tell you you're wrong, but all ways we could use to try evaluate tells us this is true
@@張謙-n3lNo I get what you mean. I also think that problems like these do have an underlying fundamental theory. Analytic continuation, p-adic numbers, multiple paradoxes, set theory, measure theory, infinity and infinitesimals. I've spent years trying to unify certain problems within these fields and I think it's possible.
@@張謙-n3l to slightly throw a wrench into this, you can evaluate 1+2+3+... to be other things. S=1+2+3+4+5+6+7+8+9+10+... =1+(2+3+4)+(5+6+7)+(8+9+10)+... =1+9+18+27+... =1+9S Therefore S=-1/8
It’s not really a nonsense problem in this case. You’re just expanding the definition of an integral a little bit. You’re taking the sum of values of x^a-1, where a is the step size that approaches 0. Using a definite integral then sets the upper and lower bounds of the summation.
Perhaps, if you convert the last step before integration to a Riemann sum, then dx^k approaching zero becomes more well defined. Also, minor nitpick, but maybe you could add parenthesis to the integral to include the minus one term.
d^2 =0 is used as the reason why every exact differential form is closed on wikipedia (en.m.wikipedia.org/wiki/Closed_and_exact_differential_forms) I wonder if someone could post an explanation on why that is, because "even closer to 0 than an infinitesimal" didn't really convince me yet.
@@julianbruns7459 That's because of how you define the "d" operator. In vector calculus and differential geometry there are the so called "differential 1-forms" which kinda sorta represent the distribution of a certain vector field over each curve in each point, and they are defined (in R³) as f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz and they can be integrated over a line. The so called 2-forms are f(x,y,z)dy/\dz+ g(x,y,z)dz/\dx and h(x,y,z)dx/\dy they kind of rapresent the distribution of a pseudovector (vectors that rapresent rotation and are ment to be used in a cross product) field over a space (or a bivector field if you know what these are) can be integrated over a surface (also the /\ is the grassman product, it's kind of like the cross product but takes vectors and outputs bivectors), and 3-forms are f(x,y,z)dx/\dy/\dz and kind of rapresent the distribution of a pseudoscalar (a kind of flux, or trivector) over a volume. Now if you take the exterior derivative of a scalar function, it becomes a 1-form and so on, a function (now I will skip the x,y,z part) on R³ f has exterior derivative f[x]dx+f[y]dy+f[z]dz (the [] brackets rapresent a partial derivative over a certain variable). Now the exterior derivative of a 1-form is the 2-form (h[y]-g[z])dy/\dz+ (f[z]-h[x])dz/\dx+ (g[x]+f[y])dx/\dy (all the dx/\dx etc terms go to 0 since a/\a is 0 for everything), but if all of these function are the derivative of a scalar function you get f[z,x]-f[x,z] etc., except you know these partials are equal to each other for the Schwartz theorem so they go to 0. Now an "exact" form is a form that is the exterior derivative of something else, while a "closed" one is one that has 0 derivative. It's clear that an exact form is always closed, the opposite is true only on simply connected spaces. Now in R1 how does it make sense? Well, the only form that exists is the 1-form f(x)dx, but the exterior derivative of that is f'(x)dx/\dx, exept dx/\dx Is zero since a/\a=0 for every a. I hope that this explains something to you.
Except it was only a single integral in the first place, so whilst you could *do* a double integral or something, that's now actually what is going on.
We’re only taking the single integral. dx can be considered a constant that is the step size of the summation, that approaches 0. So when taking the integral of f(x)dx^2, you can move the one dx into the front (since it acts as a constant). A function that converges everywhere multiplied by a constant that approaches 0, just approaches 0.
I don't buy your argument, that you can just drop infinity elements of the series, because they are close to zero. In my mind you only found integral of a lower bound of the expression.
@@Alians0108the ∫ deals with a dx because you're doing the limit of summing up an infinite number of infinitesimals Kinda like multiplying 100000×0.001=100, the ∫ works as the 100000 and the dx works as the .001 But now if you have dx², then that's gonna be much smaller and only bring you back to having a dx but in the final answer, where it's infinitesimal and doesn't matter (so like 100000×[0.001]² = 0.1) --------- That's how I'm thinking about it, might be wrong, dunno :P
Think about the product rule: the derivative of "f times g" is ("f" times dg/dx) + (df/dx times g). Now picture it as a rectangle with sides "f" and "g." You're adding infinitesimally thin slices to the edges, one slice with length "f" and thickness "dg/dx" and one slice with length "g" and thickness "df/dx." There should be another corner piece with dimensions "df/dx" by "dg/dx", but as "dx" becomes thinner and thinner, that extra corner piece becomes more and more overwhelmed by the two extra side slices.
I have a better way, its like the one with h The derivative of a^x is a^x*ln(a) But it also a^x*(a^dx-1)/dx So ln(a)=(a^dx-1)/dx So x^dx-1=(x^dx-1)/dx*dx=ln(x)dx
I didn't see your first video, so maybe it's what was done, but the question can be answered using riemann sums. You just need to define indefinite integral with it (a definite from a to x + constant, for example) and it become clear, because a^x/x -> 1+ ln(a) for x->0 Very interesting
Is it possible to make a substitution such that the higher power dx's are just regular differentials of different variables, and then then evaluate the integral of each section independently? For some arbitrary choice of n
The limit of Riemann sums of the original integral exactly match the integral of ln(x), so the conclusion of this video is definitely accurate. Although the way it is presented is certainly a bit unrigorous.
I thought of that integral from your old video randomly today and you happen to make a new video about it hah! I swear we have some sort of a connection Also please do more interesting integrals like what you were doing while drawing the screen! They're so fun
That's still some interesting abuse of notation you got there. Do tell us more about raising x to an infinitesimal power, and then lecture us on Errett Bishop's paper on schizophrenia in mathematics. Give a Reimann integral version of xᵈˣ - 1 as well.
If the previous method was dubious, this was extremely dubious. But it sort of make sense if you treat dx as a nilpotent infinitesimal (which is different from how Leibniz viewed infinitesimals!) Cool nonetheless! :D
I don’t think I’d really call it “right” or call it exactly “wrong”? There’s some sense to it, but, well, I think one could define some things in a way that would allow making this rigorous, but, I see little motivation for why one would want to assign a meaning to the original expression? ... though... I guess if you wanted to allow \int f(x,dx) for arbitrary smooth functions f of two variables then, I suppose that could probably be done.. Would there be any real mathematical merit in it? I think it would mostly just boil down to “\int (f(x,dx) - f(x,0))” = \int \frac{\partial f(x,y)}{\partial y} |_{y=0} dx Which, idk, is there much use in having such a notation? Maayybe it could be occasionally helpful for some trick in a heuristic calculator in physics. But, I think usually, it’s probably better to use a clearer notation
This integral can be considered the summation of x^a-1 (with upper and lowers bounds not given), where a is equal to the step size that approaches 0. While he wasn’t rigorous with it, the actual problem makes sense.
@@APaleDot Nope he's correct. You can chose to ignore whatever powers you like to simplify a calculation a priori, but if for different seemingly possible choices you get different contradictory results you have done something that isn't justified. Some limit you did wasn't exchangeable or the expression and the algebraic manipulations were ill-defined/ dubious. What is done here just isn't correct. Maybe with definitions it could be made correct, but not like this. In this sense it "isn't even wrong" as it is barely math at all.
@@APaleDot Well for example ignoring lower powers or less powers. As he didn't justify WHY he neglects higher powers there's a priori no reason not to and the problem begins as his expression he starts with isnt defined in the first place. In standard analysis there are no infitesimals. So he's kind of doing non-standard analysis on thr hyper reals, but i don't think here it is clear what he is doing either at multiple points (i have next to zero knowledge in none standard tho - i know that this isnt how you define an integral there either). The way i would make it sensible is interpreting dx as a 1-form, but there are still problems and it isn't at all clear why we should write the integral of the 1-form "ln(x)dx" in such a convoluted way. What he's essentially doing is introducing new notation defining integral of lnxdx with a new symbol and then adhoc reasoning why it could make sense.
@@mamaoforever1786 What he's doing isn't well-defined. It isn't even clear what "dx" is. Is it a 1-form basis vector? Well then exp(dx) does make sense, but together with the integral it makes no sense again. Also expanding dx^n makes nonsense again, but we could fix it by interpreting it as an outer product. In this case indeed dx^n for n>1 is zero (not close to it; its just zero and thats not because dx is small - its because of an algebraic property). So we have arrived with assumptions from an undefined object at one that could be interpreted in a clear way. Yeah and that's bullshit. We can't make secure deductions if we start with something dubious. What we conclude may be correct but it also may be false. The problem is why should we do any of this or not? It isn't clear what we are doing here. Well ... what we are doing here is just defining a notation - just a symbol - the integral of lnx ... and potentially for ln(f(x)) for some appropriate function and then reasoning adhoc why this new notation kind of makes sense, but why on earth should we do that? I don't see any use in writing this integral with such a symbol.
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5 days ago but realeased one hour ago??
engineers are smiling and this video while mathematicians are dying inside
So true 🤣
I'm not, but I started as a physics major before switching to math, so I'm cool with infinitesimals.
Physicists are thinking it was obvious from the start.
That’s a third degree abuse of notation here
Mathematician here, treating (dx)²=0 just means we're doing smooth infinitesimal analysis, and this is just synthetic differential geometry.
*places down Uno reverse card*
I figured that if the integrand is basically 0 if you set dx=0, then it should be fine.
f(x) dx → f(x)×0=0
x^dx -1→ x^0 -1=1-1=0
sin(x) sin(dx) → sin(x) sin(0)=0
Furthermore, if you want a nontrivial solution, the strength of the differentials must be the same as the number of integrals. Each differential undoes an integral.
∫∫∫x dx² has not enough dx terms, so this "sum" would be ∞, if anything.
∫∫x³ dx⁴ has too many dx terms, so it just becomes 0.
∫cos(x) dx has just the right strength, so you'll get something not forcibly 0 or ∞.
Wow, how you got that symbols?
I don't think so because while it's true those each of those individual functions go to zero the integral of those functions don't. This is basically because adding up an infinite amount of infinitesimals creates a real value. Even on your first example, f(x)dx may go to zero, but that doesn't mean the integral of it does. Integral(f(x)dx) ≠ integral(0). Two very different things. Infinitesimal often can be negligible to the point where they be considered zero, but once you start adding up infinite amounts of them, very different values. An infinite amount of infinitesimals added together could be anything, and infinite amount of zeros added together is zero.
@@briancoyle7770 that's not what I'm saying. Think about residues of poles at a point w. You're looking for the sweet spot, that is the order of the pole n, such that (z-w)^n gives you a non-zero finite result. If you use powers n, the limit would be stuck at 0. You're looking for the right balance.
It's the same here. The same number of variable differentials for integrals gives you a normal function, too many differentials gives you 0, and not enough differentials gives you ∞.
I say variable differentials because ∫∫dA looks like it has one, but dA=dx dy so it has two.
@@xinpingdonohoe3978 No I get what you mean there and I completely agree with you and the second have of your original comment. I just think that the first half I don't necessarily agree with. To me it just sounded like you were disagreeing with the video because you think that x^dx-1=0 which would mean that the answer to the video would be integral of 0 or C instead of his amswer.
@@briancoyle7770 oh right, okay. No, I just meant in a convergence sense.
∫sin(x+dx) could not succeed because dx→0 doesn't send the integrand to 0.
∫sin(x+dx)-sin(x) could succeed.
dubious maths is dubious, but its still maths.
No it isn't.
@@Nolys-bk4kd 1 + 1 = 3 is maths, but maybe not good maths. But still maths
@@matthewe3813 No it isn't, it's just pointless symbols with no attempt at rigor whatsoever.
@@Nolys-bk4kd But when the result is useful, there is some use and it makes it maths
@@matthewe3813 The result can only be useful if you have shown that it is true, you can't show it to be true unless you apply proper rigor.
Besides, a result doesn't need to be useful to be maths, it just need to be an actual result of proper rigor and not just a result of some vague hand-waving.
Is there any explanation for why dx^2 should be treated like 0? The "its super small trust me" explanation doesn't quite do it for me.
go back to the first explanation
x^dx-1 = (x^dx-1)/dx * dx
dx is small (approaches 0) so in a crude way we can take the limit as h -> 0 of (x^h-1)/h which is ln x
so it becomes the integral of ln(x) dx
0.0000...01 is super small... 0.0000...01^2 is soooooo much smaller than that, so you ignore it. not really mathematically rigerous, but it's something us physicists do pretty often
@@rewazza Yes, but we're considering a mathematical expression. dx is also really really close to zero, but we do not ignore its multiplicants.
Well, that assumption is why the product rule works, after all. Should be enough for you.
@@henkolsonpietersen2242 It is called the first order perturbation, you could just as well have chosen the second order and kept the x^2 term of the expansion of e^x. I'm admittedly unsure how one would get the original video's answer to be equal in that case, however.
2:24 In stochastic calculus they define something called quadratic variation. A theorem is that if your function is deterministic then its quadratic variation is 0, if the function is stochastic then its first order variation is infinite. I think this is similar.
I showed this to one of my friends whose world was Mathematics. He was furious, and said to me, "If there's no dx multiplied then how do I know respect to what I integrate!?"
bbbut rigor!
The (dx)ⁿ=0 while n>1 is actually correct (sort of), but that explanation is skipping a ton of steps.
Interesting, I'd actually like to hear the explanation if it isn't too much complicated
@@intellix7133 basically, Sf(x)dx (S is supposed to be an integral sign) means "plot the function f(x), then take the region between 0 and x and then do as follows: draw a rectangle which has the height from the x axis to f(0), and a width of dx, dx being a constant (draw the width in the direction of x). Then make another rectangle with the height from the x axis to f(dx) and a length of dx. Then repeat for f(2dx), f(3dx) and so on until you last rectangle reaches x. At the end, sum the areas of the rectangles. The value of the integral is the sum when dx approaches 0." This process basically gives you the area between the function and the x axis between x=0 and x=x, and the solution to the integral is some other function where you plug in a value for x and get said area. Technically that's for a proper integral where the area is defined (in this case between 0 and x), so because you can start from any point (doesn't have to be 0) you add a +C to the end function which symbolizes the added (or subtracted) area when changing from 0 to another number. Doesn't really matter if you got the +C part though, it's unimportant for this specifically, though it's worth mentioning. Anyway, imagine you have the integral Sf(x)(dx)ⁿ. It's equal to Sf(x)(dx)^(n-1)dx. Using the earlier definition, that means the area between the x axis and the function f(x)dx^(n-1). Since n>1 and dx approaches 0, the function itself is just 0. So the area will also be 0.
Note: this is after splitting the integral in the vid into a bunch of seperate integrals. The area will be 0 so the solution will all be a constant, in other words the solution of all of the integrals together will be x(lnx-1) plus a bunch of constants. If we define another constant C to be the sum of those, then we have x(lnx-1)+C
In my opinion, it is neither correct nor incorrect; it is a choice. There are, to my knowledge, three different approaches to the `dx` symbol: it may be an "infinitesimal" value, it may be a value which becomes arbitrarily small in some context (this is the usual way calculus is taught), or it may be meaningless on its own, only used as a portion of some other syntax (integral ⟨...⟩ dx, or d ⟨...⟩/dx).
The last approach is, in some way, the safest, but it makes this question strictly invalid. So we must choose one of the other options. The "limits" option is well-formalized, and partially used in this explanation, while an "infinitesimal" option may be easier to deal with intuitively.
If we choose an infinitesimal approach, we once again have a choice about how to handle the expression `dx²`. We may go the way of the dual numbers, and say `dx² = 0` (which then implies `dx^(n + 2) = 0`, since `dx^(n + 2) = dx^n ⋅ dx^2 = dx^n ⋅ 0 = 0`), or we may say that `dx²` of a yet-smaller class of infinitesimals. However, if we do that, then the result of our integral is no longer a real number, like we would expect, but instead will come out as some `a + bε + cε² + ...`.
Fortunately, all of these approaches are equivalent, I think (up to infinitesimal difference).
@@MCLooyverse I think you're overcomplicating it. Once I have Sf(x)*dx^n I can write it as Sf(x)dx^(n-1)dx, in other words I'm finding the area under the curve f(x)dx^(n-1), or y=0 since n>1 and dx->0. Btw, in both d/dx and S__dx the dx isn't just a symbol, but a multiplicative/divisor. In a derivitive you find a slope which is (f(x+h)-f(x))/((x+h)-x)=df(x)/dx, so dx is a divisor. In an integral you sum the areas of infinitely small rectangles with length dx, so the dx does multiply the function whereas the integral sign tells you to sum everything, but they are not actually connected in some way.
@@intellix7133 I wrote a comment explaining that and it literally disappeared lol. Look at what I told the other guy and tell me if you understand or want me to go into more detail
This is literally the solution I came up with too😳
I went back to watch your old video but before I played it, I tried solving it on my own and this is the solution I came up with! Although I looked at it through a slightly different perspective by incorporated elements of nonstandard calculus into my thought process. When I saw your old video I thought, oh cool a different solution but came up with same result. And then I see this video and you have the solution I came up with!
This actually does make sense on a conceptual level. You’re still taking an infinite sum, you can still approximate the integral in similar ways to normal ones. By taking small values of dx, and by using a finite sum.
The problem basically is:
The summation of x^a-1 (with upper and lowers bounds not given), where a is equal to the step size that approaches 0. While he wasn’t rigorous with it, the actual problem makes sense.
I think this can be made rigorous with a broader definition of an integral, using nonstandard analysis.
wierd wierstrauss and cauchy delta episilon may help
why are dubious methods so unreasonably effective wtf
Why not just derive it from the limit of the finite sum? This seems like the most natural way to find this imo
Everyone talking about advanced stuff while me not understanding a single thing and getting confused by weird symbols: 😅😅😅😅🤣
learn math then
@@trinityy-7Leave him alone.
SAME
Every time when we try to use different ways to solve those non-sense problems and always get the same result makes me wonder if there's something more fundamental that governs all theories and theorems
With problems like this, nonstandard Calculus makes it way more intuitive and clear in my opinion
@@briancoyle7770 Not only this one, also something like "sum of all natural number equals -1/12", it just seems so nonsense that even a kindergarten child will tell you you're wrong, but all ways we could use to try evaluate tells us this is true
@@張謙-n3lNo I get what you mean. I also think that problems like these do have an underlying fundamental theory. Analytic continuation, p-adic numbers, multiple paradoxes, set theory, measure theory, infinity and infinitesimals. I've spent years trying to unify certain problems within these fields and I think it's possible.
@@張謙-n3l to slightly throw a wrench into this, you can evaluate 1+2+3+... to be other things.
S=1+2+3+4+5+6+7+8+9+10+...
=1+(2+3+4)+(5+6+7)+(8+9+10)+...
=1+9+18+27+...
=1+9S
Therefore S=-1/8
It’s not really a nonsense problem in this case. You’re just expanding the definition of an integral a little bit.
You’re taking the sum of values of x^a-1, where a is the step size that approaches 0. Using a definite integral then sets the upper and lower bounds of the summation.
Perhaps, if you convert the last step before integration to a Riemann sum, then dx^k approaching zero becomes more well defined. Also, minor nitpick, but maybe you could add parenthesis to the integral to include the minus one term.
Can you please do a video about the partial differential
Make a video about deriving the cubic formula
Bri comeback with a quality stuff❤️
if you consider d as the exterior derivative of x then i kind of guess that it makes sense that d²=0 (maybe)
d^2 =0 is used as the reason why every exact differential form is closed on wikipedia (en.m.wikipedia.org/wiki/Closed_and_exact_differential_forms)
I wonder if someone could post an explanation on why that is, because "even closer to 0 than an infinitesimal" didn't really convince me yet.
@@julianbruns7459 That's because of how you define the "d" operator. In vector calculus and differential geometry there are the so called "differential 1-forms" which kinda sorta represent the distribution of a certain vector field over each curve in each point, and they are defined (in R³) as f(x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz and they can be integrated over a line. The so called 2-forms are f(x,y,z)dy/\dz+ g(x,y,z)dz/\dx and h(x,y,z)dx/\dy they kind of rapresent the distribution of a pseudovector (vectors that rapresent rotation and are ment to be used in a cross product) field over a space (or a bivector field if you know what these are) can be integrated over a surface (also the /\ is the grassman product, it's kind of like the cross product but takes vectors and outputs bivectors), and 3-forms are f(x,y,z)dx/\dy/\dz and kind of rapresent the distribution of a pseudoscalar (a kind of flux, or trivector) over a volume. Now if you take the exterior derivative of a scalar function, it becomes a 1-form and so on, a function (now I will skip the x,y,z part) on R³ f has exterior derivative f[x]dx+f[y]dy+f[z]dz (the [] brackets rapresent a partial derivative over a certain variable). Now the exterior derivative of a 1-form is the 2-form (h[y]-g[z])dy/\dz+ (f[z]-h[x])dz/\dx+ (g[x]+f[y])dx/\dy (all the dx/\dx etc terms go to 0 since a/\a is 0 for everything), but if all of these function are the derivative of a scalar function you get f[z,x]-f[x,z] etc., except you know these partials are equal to each other for the Schwartz theorem so they go to 0. Now an "exact" form is a form that is the exterior derivative of something else, while a "closed" one is one that has 0 derivative. It's clear that an exact form is always closed, the opposite is true only on simply connected spaces.
Now in R1 how does it make sense?
Well, the only form that exists is the 1-form f(x)dx, but the exterior derivative of that is f'(x)dx/\dx, exept dx/\dx Is zero since a/\a=0 for every a. I hope that this explains something to you.
why did you make them go to 0? Couldn't you do a double integral, then a triple integral, and so on until it converges to a specific function?
Except it was only a single integral in the first place, so whilst you could *do* a double integral or something, that's now actually what is going on.
We’re only taking the single integral. dx can be considered a constant that is the step size of the summation, that approaches 0. So when taking the integral of f(x)dx^2, you can move the one dx into the front (since it acts as a constant). A function that converges everywhere multiplied by a constant that approaches 0, just approaches 0.
I don't buy your argument, that you can just drop infinity elements of the series, because they are close to zero. In my mind you only found integral of a lower bound of the expression.
In integral calculus, we say that dx ^ n = 0
n > 1
@@duckymomo7935But why specifically from n = 2 ?
@@duckymomo7935 But on what basis you do that?
@@Alians0108the ∫ deals with a dx because you're doing the limit of summing up an infinite number of infinitesimals
Kinda like multiplying 100000×0.001=100, the ∫ works as the 100000 and the dx works as the .001
But now if you have dx², then that's gonna be much smaller and only bring you back to having a dx but in the final answer, where it's infinitesimal and doesn't matter (so like 100000×[0.001]² = 0.1)
---------
That's how I'm thinking about it, might be wrong, dunno :P
Think about the product rule: the derivative of "f times g" is ("f" times dg/dx) + (df/dx times g).
Now picture it as a rectangle with sides "f" and "g." You're adding infinitesimally thin slices to the edges, one slice with length "f" and thickness "dg/dx" and one slice with length "g" and thickness "df/dx."
There should be another corner piece with dimensions "df/dx" by "dg/dx", but as "dx" becomes thinner and thinner, that extra corner piece becomes more and more overwhelmed by the two extra side slices.
If dx is so small that you treat it like 0 with the rest of the exponents, shouldn't (ln(x) * dx) ^ 0 be similar to 0^0? Why 1?
e^0 = 1
The 0^0 in the sum defining exp(0) , is equal to 1.
0^0 is pretty much only undefined if the 0 in the exponent isn’t specifically an integer.
Isn't that a Feynman technique?
0:10 ok and?
Its probaly equal to x using a formula that i dont feel like citing
Plot Twist: the dx was d * x where d and x are variables
I have a better way, its like the one with h
The derivative of a^x is a^x*ln(a)
But it also a^x*(a^dx-1)/dx
So ln(a)=(a^dx-1)/dx
So x^dx-1=(x^dx-1)/dx*dx=ln(x)dx
I didn't see your first video, so maybe it's what was done, but the question can be answered using riemann sums. You just need to define indefinite integral with it (a definite from a to x + constant, for example) and it become clear, because a^x/x -> 1+ ln(a) for x->0
Very interesting
I suppose placing the dx in the exponent is abuse of notation for some less compact expression. I'd rather have see the full form.
Is it possible to make a substitution such that the higher power dx's are just regular differentials of different variables, and then then evaluate the integral of each section independently? For some arbitrary choice of n
i made this integral
integral from 1 to infinity of (sin(cos(x^3)))/(x^2)
The limit of Riemann sums of the original integral exactly match the integral of ln(x), so the conclusion of this video is definitely accurate. Although the way it is presented is certainly a bit unrigorous.
This is not physics this is maths!!!!!
calculus equivalent of evaluating 64/16 by canceling out the 6s and getting 4/1
what is the program that you are using to do this math videos?
Higher order differential terms don't exist. They can't hurt you.
-Physicist
I thought of that integral from your old video randomly today and you happen to make a new video about it hah! I swear we have some sort of a connection
Also please do more interesting integrals like what you were doing while drawing the screen! They're so fun
that's not very controverstial if most of us doesn't even know how to solve that 💀
That's still some interesting abuse of notation you got there. Do tell us more about raising x to an infinitesimal power, and then lecture us on Errett Bishop's paper on schizophrenia in mathematics. Give a Reimann integral version of xᵈˣ - 1 as well.
Non newtonian calculi?
this is equal to (1/dx+1) * x^(dx+1) - ∫
∫xdx
= ∫x(0) , dx=0 by the fundemental theorem of engineering
= ∫ (the integral of nothing with respect to nothing)
= ???????
underated comment
What is i^^i?
The title is wrong genius
Solve subfactorial of i
Isn't this what Newton did -- treating the infinitesimal seriously, but ignoring its higher powers?
If the previous method was dubious, this was extremely dubious. But it sort of make sense if you treat dx as a nilpotent infinitesimal (which is different from how Leibniz viewed infinitesimals!)
Cool nonetheless! :D
What does “contraverstial” mean?
It's something arguable, something uncertain, thus leading to a heated discussion.
@@wynneve Learn to write
@@trumpgaming5998 I think you have replied to a wrong person.
@@wynneve Both of you are idiots there is no word "contraverstial" in the English language that has a definition.
@@trumpgaming5998
Shut up
Fascinating. Does the math community agree with you? Are you a brilliant solo amateur?
I don’t think I’d really call it “right” or call it exactly “wrong”?
There’s some sense to it, but,
well,
I think one could define some things in a way that would allow making this rigorous,
but, I see little motivation for why one would want to assign a meaning to the original expression?
... though...
I guess if you wanted to allow \int f(x,dx) for arbitrary smooth functions f of two variables
then, I suppose that could probably be done..
Would there be any real mathematical merit in it?
I think it would mostly just boil down to “\int (f(x,dx) - f(x,0))” = \int \frac{\partial f(x,y)}{\partial y} |_{y=0} dx
Which,
idk, is there much use in having such a notation?
Maayybe it could be occasionally helpful for some trick in a heuristic calculator in physics.
But, I think usually, it’s probably better to use a clearer notation
You killed math 🥺
That makes no sense.
According to that logic
Integral(e^x dx) = integral(0) = +c
Since dx approaches zero
Only higher powers of dx approach zero faster than the summation of the integral approaches infinity.
This integral can be considered the summation of x^a-1 (with upper and lowers bounds not given), where a is equal to the step size that approaches 0. While he wasn’t rigorous with it, the actual problem makes sense.
@@APaleDot Nope he's correct. You can chose to ignore whatever powers you like to simplify a calculation a priori, but if for different seemingly possible choices you get different contradictory results you have done something that isn't justified. Some limit you did wasn't exchangeable or the expression and the algebraic manipulations were ill-defined/ dubious. What is done here just isn't correct. Maybe with definitions it could be made correct, but not like this. In this sense it "isn't even wrong" as it is barely math at all.
@@IsomerSoma
What different choices are possible here? What contradictory result can you obtain?
@@APaleDot Well for example ignoring lower powers or less powers. As he didn't justify WHY he neglects higher powers there's a priori no reason not to and the problem begins as his expression he starts with isnt defined in the first place. In standard analysis there are no infitesimals. So he's kind of doing non-standard analysis on thr hyper reals, but i don't think here it is clear what he is doing either at multiple points (i have next to zero knowledge in none standard tho - i know that this isnt how you define an integral there either).
The way i would make it sensible is interpreting dx as a 1-form, but there are still problems and it isn't at all clear why we should write the integral of the 1-form "ln(x)dx" in such a convoluted way.
What he's essentially doing is introducing new notation defining integral of lnxdx with a new symbol and then adhoc reasoning why it could make sense.
xlogx - x +c
very skibidi
integrate x!-x+dx+x^dx+xdx!+Γ(dx)^x+ψ^(0)(dx^x)-1 next
ψ is polygamma
nice!
No. This is not rigorous and it is unacceptable.
☝️🤓
asnwer=1x dx
Bruh
No
Taylor series?
asnwer=1x log x
Interesting
you hurt math in a way that was not hurt before! what a waste of time this was!
Большего бреда в жизни не видел
Ayo ! How did he pinned the comment 5 days ago while the video was released 7 minutes ago 😂😂😂
IKR!?!?!? why is nobody talking bout that
That's what integrating with dx in the exponent does to the space-time continuum. I warned y'all!
It may not be uploaded to everyone, but for sponsors only. Or pre released
It was released early for people with channel membership
rubbish!
Why?
@@mamaoforever1786 That's my first response. Show me I'm wrong and I'll be intrigued. What do you get when you differentiate it?
@@rob876 Most places i've looked show the exact same result. And from watching the video it seems mostly correct. What do you think is wrong?
so which steps of the video are wrong?
@@mamaoforever1786 What he's doing isn't well-defined. It isn't even clear what "dx" is. Is it a 1-form basis vector? Well then exp(dx) does make sense, but together with the integral it makes no sense again. Also expanding dx^n makes nonsense again, but we could fix it by interpreting it as an outer product. In this case indeed dx^n for n>1 is zero (not close to it; its just zero and thats not because dx is small - its because of an algebraic property). So we have arrived with assumptions from an undefined object at one that could be interpreted in a clear way. Yeah and that's bullshit. We can't make secure deductions if we start with something dubious. What we conclude may be correct but it also may be false.
The problem is why should we do any of this or not? It isn't clear what we are doing here. Well ... what we are doing here is just defining a notation - just a symbol - the integral of lnx ... and potentially for ln(f(x)) for some appropriate function and then reasoning adhoc why this new notation kind of makes sense, but why on earth should we do that? I don't see any use in writing this integral with such a symbol.