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#1 Guest Join Date: Jul 2018 Location: The City Different Posts: 287

## Finding an Integral Using u-substitution

Please explain finding an integral by u-substitution to me like I am a 5-year old. I can do it, but I dont remember why it works. Someone explained it to me 25 or 30 years ago but I havent thought about it since and now I dont have a clue. Let me use the following example: After finding the appropriate u substitution you come up with the expression du/dx = 2x2. Then you are supposed to say that therefore du = 2x2 dx. What the heck? I know it works, but how do you justify this? du/dx is not a fraction! Neither du nor dx is a number! What is actually going on here?

I suppose I could google it or try to figure it out myself but why waste this opportunity to be teased and ridiculed by the brilliant minds on the SDMB. If a coherent discussion develops, I am pretty sure there will some talk about limits.
#2 Charter Member Join Date: Dec 2002 Location: Very east of Foggybog, WI Posts: 5,653
First you will seldom go wrong in treating a first derivative, du/dx, as a fraction (Just don't cancel the d's). Treating du and dx as separate and separable objects is pretty much the mathematics of infinitesimals.

But without going into detail think of the operation of integrating of f(x). We write that as (integral sign) f(x) dx. The dx here simply indicates the variable you are integrating with respect to so we can distinguish

int [ x2 y] dx = (1/3) x3 y
int [ x2 y] dy = (1/2) x2 y2

so think of your problem du(x)/dx = 2x2 as

define f(x) = du(x)/dx = x2

Then integrating f(x) is the same as integrating du(x)/dx nad you never have to worry about whether or not you're multiplying the dx out.

Last edited by OldGuy; 05-18-2020 at 11:21 PM.
#3 Guest Join Date: May 2016 Posts: 5,150
By "u-substitution", it seems that you just mean that instead of integrating f(x) dx, you just define u = u(x) to be some function that will end up making your integral simpler; practice will help a lot in finding a correct substitution.

As for the notation, it suffers from a bit of abuse in that du/dx is not a fraction, and you should not think of it as du divided by dx, as you point out, but on the other hand recall that dx and du are not numbers but rather differential 1-forms, so that you are able to integrate not f(x) but f(x) dx. The quantities with "d" in front of them are also variables you need to take account of and substitute for when changing coordinates.

The formula you need to understand is that if u = g(x), then du = g'(x) dx. In the above notation, that says du = (du/dx) dx, which looks tautological, but it's not, really, at least not due to elementary arithmetic.

As for your example, if du = 2x2 dx, what you need to do is write your original integral in terms of u and du instead of x and dx, possibly using the inverse dx = (1/2x2) du if you need it. Then, once the x's are gone, you can pretend u is the independent variable and integrate as usual.

#4 Guest Join Date: May 2016 Posts: 5,150
Quote:
 Originally Posted by OldGuy First you will seldom go wrong in treating a first derivative, du/dx, as a fraction (Just don't cancel the d's). Treating du and dx as separate and separable objects is pretty much the mathematics of infinitesimals. But without going into detail think of the operation of integrating of f(x). We write that as (integral sign) f(x) dx. The dx here simply indicates the variable you are integrating with respect to so we can distinguish int [ x2 y] dx = (1/3) x3 y int [ x2 y] dy = (1/2) x2 y2 so think of your problem du(x)/dx = 2x2 as define f(x) = du(x)/dx = x2 Then integrating f(x) is the same as integrating du(x)/dx nad you never have to worry about whether or not you're multiplying the dx out.
I would not go so far as to say dx "simply indicates the variable you are integrating", rather that f(x) is not something you can integrate over a 1-dimensional interval, while f(x)dx is something that you can integrate over an interval and get a number. (If we are going to make the connection to definite integrals...)

The problem of finding antiderivatives can be formulated without integral signs and differentials, but on the other hand one must get used to the notation everybody uses.
#5 Guest Join Date: Jul 2018 Location: The City Different Posts: 287
Quote:
 Originally Posted by DPRK By "u-substitution", it seems that you just mean that instead of integrating f(x) dx, you just define u = u(x) to be some function that will end up making your integral simpler; practice will help a lot in finding a correct substitution. As for the notation, it suffers from a bit of abuse in that du/dx is not a fraction, and you should not think of it as du divided by dx, as you point out, but on the other hand recall that dx and du are not numbers but rather differential 1-forms, so that you are able to integrate not f(x) but f(x) dx. The quantities with "d" in front of them are also variables you need to take account of and substitute for when changing coordinates. The formula you need to understand is that if u = g(x), then du = g'(x) dx. In the above notation, that says du = (du/dx) dx, which looks tautological, but it's not, really, at least not due to elementary arithmetic. As for your example, if du = 2x2 dx, what you need to do is write your original integral in terms of u and du instead of x and dx, possibly using the inverse dx = (1/2x2) du if you need it. Then, once the x's are gone, you can pretend u is the independent variable and integrate as usual. Sorry for the terse explanation; please ask if anything needs elaboration.
What is a "differential 1-form"? Maybe that would help me understand.

The proposition that "if u = g(x), then du = g'(x) dx" is what is giving me trouble. Apparently it is true; people have been using it for hundreds of years. Can you give a simple proof that my tiny brain can comprehend?

I can use the fact that "if u = g(x), then du = g'(x) dx" to evaluate an integral. That's not the problem. I just want to know how we can be certain that it is true.
#6  Guest Join Date: Dec 2009 Location: the Land of Smiles Posts: 21,571
Quote:
 Originally Posted by OldGuy First you will seldom go wrong in treating a first derivative, du/dx, as a fraction (Just don't cancel the d's). ...
This is a key point I think. If you're not careful with your wording your work may be marked down by a pedantic teacher, but at least you get the right answer! I don't think the calculus stars 300 years ago had qualms about this.

And the great 12th-century Indian mathematician Bhaskara II has a memorable and relevant quote shown at the top of this page in Kim Plofker's book.
#7 Guest Join Date: May 2016 Posts: 5,150
Basically, the change-of-variables formula boils down to the chain rule for the derivative of a composite function: if u = g(x) and y = f(u), then dy/dx = dy/du ⋅ du/dx.

Quote:
 Originally Posted by Ynnad What is a "differential 1-form"? Maybe that would help me understand.
Without getting into a detailed geometrical explanation (unless that's what you want , dx represents a small displacement in the x-direction, and will satisfy (dx)(d/dx) = 1. If u = g(x) is another function, by how much will u increase if we again increase x by a unit infinitesimal amount? The rate of change is the derivative g'(x), so du = g'(x) dx. (Similarly, d/du = (d/dx) / g'(x).) Note that this is not mere work with fractions; with more variables you will have df = (∂f/∂x)dx + (∂f/∂y)dy , and changing coordinates in the volume element will factor in the determinant of the Jacobian matrix.)

Quote:
 The proposition that "if u = g(x), then du = g'(x) dx" is what is giving me trouble. Apparently it is true; people have been using it for hundreds of years. Can you give a simple proof that my tiny brain can comprehend? I can use the fact that "if u = g(x), then du = g'(x) dx" to evaluate an integral. That's not the problem. I just want to know how we can be certain that it is true.
Using only one variable may obscure some nuances, but for any function f we will have (df)(d/dx) = df/dx. On the other hand, (dx)(d/dx) = 1. Also note what we said above, about a change in x resulting in the change in u magnified by a factor of du/dx , and vice versa.
#8  Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 88,732
septimus, would you mind quoting that statement for us? Your Google Books link shows only a single sentence on the page you linked, in an Indian language that I can neither read nor recognize.
Quote:
 Quoth DPRK: Note that this is not mere work with fractions; with more variables you will have df = (∂f/∂x)dx + (∂f/∂y)dy , and changing coordinates in the volume element will factor in the determinant of the Jacobian matrix.
This is worth stressing. When you're dealing with total derivatives, you can in fact treat dy/dx as a fraction without getting into trouble, and there are even ways of formulating calculus where it really is a no-kidding genuine fraction. But that will absolutely not work once you eventually start working with partial derivatives (notated by those curvy "d"s that look like backwards 6s), and in fact you end up with some completely counterintuitive results like (∂x/∂y) * (∂y/∂z) * (∂z/∂x) = -1, rather than 1.
#9 Guest Join Date: Jun 2004 Posts: 480
Quote:
 Originally Posted by Chronos septimus, would you mind quoting that statement for us? Your Google Books link shows only a single sentence on the page you linked, in an Indian language that I can neither read nor recognize.

"You arrive at a page that cannot be viewed. Or the restrictions on viewing this book have been reached."

Neither memorable nor relevant, I'm afraid.
#10  Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 29,043
Quote:
 Originally Posted by Ynnad Please explain finding an integral by u-substitution to me like I am a 5-year old. I can do it, but I dont remember why it works.
Is this 5-year-old supposed to understand the Chain Rule for derivatives? Because u-substitution is more-or-less the Chain Rule in reverse.

The Chain Rule says, for example, that since the derivative of x3 (with respect to x) is 3x2, the derivative of u3 (with respect to x) is 3u2 * u'.

That is, the derivative of (something)3 is 3(something)2 times the derivative of the something.

The derivative of (10x5 + 7x + 93)3 would be 3(10x5 + 7x + 93)2 * (50x4 + 7).

So then the antiderivative of 3(10x5 + 7x + 93)2 * (50x4 + 7) would be (10x5 + 7x + 93)3. And u-substitution is really just a way of making this obvious.

There's a lot more I could say about how to actually do it, but that's the basic idea behind it.
#11 Guest Join Date: Mar 2012 Posts: 2,133
Quote:
 Originally Posted by Chronos septimus, would you mind quoting that statement for us? Your Google Books link shows only a single sentence on the page you linked, in an Indian language that I can neither read nor recognize.
Thats not an Indian language but I believe its Thai. Anyways, I have the book and here is the excerpt :

In other words, the at-that-time Sine-difference Δ Sin for a given arc α is considered simply proportional to the Cosine of α:

Δ Sin = Cos α * (225/R)

It has been noted that this and related statements reveal similarities be- tween Bh ̄askaras ideas of motion and concepts in differential calculus. (In fact, perhaps these ratios of small quantities are what he was referring to in his commentary on L ̄ıla ̄vat ̄ı 47 when he spoke of calculations with factors of 0/0 being useful in astronomy.) This analogy should not be stretched too far: for one thing, Bha ̄skara is dealing with particular increments of partic- ular trigonometric quantities, not with general functions or rates of change in the abstract. But it does bring out the conceptual boldness of the idea of an instantaneous speed, and of its derivation by means of ratios of small increments.
#12 Guest Join Date: Jul 2018 Location: The City Different Posts: 287
Quote:
 Originally Posted by Thudlow Boink Is this 5-year-old supposed to understand the Chain Rule for derivatives? Because u-substitution is more-or-less the Chain Rule in reverse. ... There's a lot more I could say about how to actually do it, but that's the basic idea behind it.
Using the Liebniz notation, the chain rule states that dy/dx = dy/du x du/dx. If dy, dx, and du were real numbers and dy/dx, dy/du, and du/dx were fractions then the proof of the chain rule would almost be trivial. However, they are not numbers or fractions. Nonetheless, I have seen several different proofs of the chain rule. Perhaps some sort of backwards proof of the chain rule could show that "if u = g(x), then du = g'(x) dx."

I know how u-substitution works. I just want to know why it works. In fact, I stumbled upon with this question when preparing training on solving differential equations using the separation of variables technique. It's really the same question but in a slightly different context. I just used the u-substitution context because I thought more people would be familiar with it. I wanted to be ready with a short simple answer in case some trainee asked the same question that I asked in this thread. Who knows, maybe one of the trainees could be a pedantic ***hole like me and just want to ask that question so that they could watch me squirm. (More likely though, I would get questions like "What's a differential?" or "Couldn't you just google it?".)

In looking for a simple answer, I have found several sources that resort to what I consider mere hand waving such as "using some 'informal' algebra", "treating dx as if it were a number even though it's not", etc. I have even seen some weasel words in this thread such as "you will seldom [emphasis added] go wrong ... ", "you can pretend ... ", and "... more-or-less."

One more thing. No one in their right mind wants to actually watch me physically squirm.
#13 Guest Join Date: May 2016 Posts: 5,150
Now, the slightly more modern point of view is that if you have a smooth space, which could be Euclidean space but also something curved like a differentiable manifold, you can zoom in to an infinitesimal neighborhood of a given point by considering the tangent space at that point, one way of looking at which is that curves passing through the point represent tangent vectors. A tangent vector applied to a function will give the corresponding directional derivative of the function at that point. Usually this kind of stuff is introduced when you study multi-variable calculus, not in a first-semester course.

Furthermore, at any point you can also consider the cotangent space. Eg a smooth function equal to 0 at a given point represents a cotangent vector. So now if you start with any smooth function f, its differential df will give a cotangent vector at any point, therefore it is a differential 1-form.

Anyway, the advantage of this point of view is that it provides a rigorous way of the interpreting intuitive calculations with infinitesimals which is independent of a particular choice of coordinates, and can be generalized.
#14 Guest Join Date: May 2016 Posts: 5,150
Quote:
 Originally Posted by Ynnad Using the Liebniz notation, the chain rule states that dy/dx = dy/du x du/dx. If dy, dx, and du were real numbers and dy/dx, dy/du, and du/dx were fractions then the proof of the chain rule would almost be trivial. However, they are not numbers or fractions. Nonetheless, I have seen several different proofs of the chain rule. Perhaps some sort of backwards proof of the chain rule could show that "if u = g(x), then du = g'(x) dx." I know how u-substitution works. I just want to know why it works. In fact, I stumbled upon with this question when preparing training on solving differential equations using the separation of variables technique. It's really the same question but in a slightly different context. I just used the u-substitution context because I thought more people would be familiar with it. I wanted to be ready with a short simple answer in case some trainee asked the same question that I asked in this thread. Who knows, maybe one of the trainees could be a pedantic ***hole like me and just want to ask that question so that they could watch me squirm. (More likely though, I would get questions like "What's a differential?" or "Couldn't you just google it?".) In looking for a simple answer, I have found several sources that resort to what I consider mere hand waving such as "using some 'informal' algebra", "treating dx as if it were a number even though it's not", etc. I have even seen some weasel words in this thread such as "you will seldom [emphasis added] go wrong ... ", "you can pretend ... ", and "... more-or-less." One more thing. No one in their right mind wants to actually watch me physically squirm.
The Leibniz notation may be obscuring what is going on. If you think of the Chain Rule in terms of the total derivative of a function, the total derivative of the composite of two functions is just the composite of the total derivatives. The composition of two linear functions is calculated via matrix multiplication; in one variable the derivative at any point is given by a single number, and you are multiplying those numbers together.

Anyway, back to the beginning, when you write "dx" it is not a "number" (then again your coordinate function x is not a mere "number" either, rather at any point you may evaluate it to get a number). Like x, dx may be evaluated at any point and gives a cotangent vector, which is not a "number" but some linear map, but in terms of your fixed coordinate x (which gives a resulting coordinate dx on the cotangent space) the linear map is represented by a number, namely the derivative of the function at that point. Think of f(x) = f(a) + f'(a) (x - a) + ... df (evaluated at a) = f'(a) dx (evaluated at a) is the infinitesimal/differential version of that, and this is abbreviated by the formula df = f'(x) dx
#15 Guest Join Date: May 2016 Posts: 5,150
Quote:
 Originally Posted by Ynnad In looking for a simple answer, I have found several sources that resort to what I consider mere hand waving such as "using some 'informal' algebra", "treating dx as if it were a number even though it's not", etc. I have even seen some weasel words in this thread such as "you will seldom [emphasis added] go wrong ... ", "you can pretend ... ", and "... more-or-less."
Mathematics is not a place for lies or informality or weaselling, so you should discount those sources Even people who give informal "physics proofs" know how to make them completely mathematically rigorous, assuming they know what they are talking about.

The short answer you can tell people seems to be that both "x" and "dx" are coordinates, so that when you change coordinates and write your integral in terms of the new coordinates, you need to substitute for both of them. Or, more basically, when finding antiderivatives you are using the Chain Rule--- the trainees probably remember that name.
#16 Guest Join Date: Jul 2018 Location: The City Different Posts: 287
I just found this video that does not answer my question but at least it kinda, sorta, maybe asks my question.

#17 Guest Join Date: May 2016 Posts: 5,150
Quote:
 Originally Posted by Ynnad I just found this video that does not answer my question but at least it kinda, sorta, maybe asks my question. https://www.khanacademy.org/math/ap-...-algebraically
dx and dy are "algebraic expressions" just like x and y, and are "super mathematically rigorous" no less than x and y, so I am not convinced that narrator knows what they are talking about.

But we must admit that the notation IS potentially confusing: dy/dx is not a fraction, and yet we have these things dx as well as d/dx defined, not to mention the curly d's, stuff like |dx| and (dx)1/2, dx dy as well as dx ^ dy, you name it...
#18  Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 29,043
Quote:
 Originally Posted by Ynnad I know how u-substitution works. I just want to know why it works. In fact, I stumbled upon with this question when preparing training on solving differential equations using the separation of variables technique. It's really the same question but in a slightly different context. I just used the u-substitution context because I thought more people would be familiar with it.
I guess there are different ways one could answer the question, but one way is "It works by definition."

We define ∫ f(x) dx to mean "the general antiderivative of the function f(x) with respect to the variable x"; and we define du to be "du/dx * dx" (that is, the derivative of u with respect to x, times dx) without ever really defining what dx by itself means. These definitions then give us a convenient way of doing what we need to do with integrals or separable differential equations.
#19  Guest Join Date: Dec 2010 Posts: 8,548
Just calling it another name for the chain rule seems incomplete to me. Otherwise, I couldn't do more complicated substitutions, like this (admittedly contrived) example:
Say I want to compute:
∫3x5 dx

u2 = x3
Take the derivative of both sides:
2u du = 3x2 dx

Note that the original integral is one part of each:
∫3x5 dx = ∫x3 * 3x2 dx = ∫u2 * 2u du = ∫2u3 du
Integrate the normal way (ignoring the +C for convenience):
u4/2
And substitute back to x:
(x3)2/2 = x6/2

That is of course the same as what we'd get if we did it the normal way in the beginning.
#20  Guest Join Date: Mar 2010 Location: MD outside DC Posts: 6,527
Quote:
 Originally Posted by Ynnad What is a "differential 1-form"? Maybe that would help me understand. The proposition that "if u = g(x), then du = g'(x) dx" is what is giving me trouble. Apparently it is true; people have been using it for hundreds of years. Can you give a simple proof that my tiny brain can comprehend? I can use the fact that "if u = g(x), then du = g'(x) dx" to evaluate an integral. That's not the problem. I just want to know how we can be certain that it is true.
Lets start with the definition of the derivative g'(x) = limit( (g(x+h)-g(x))/h), if we let dx=h be an infinitesimally small change in x and du=g(x+h)-g(h) be the resulting change in u=g(x) then we get g'(x)=du/dx, or du=g'(x)dx

Now on to integration. Remember that when I write Int(f'(x)dx)=f(x)+c I am calculating the area under the curve f'(x) by taking little tiny slices of width dx and height f'(x) and adding their areas together. to make thing concrete I can chop the region a to b up into 10^1000 equal slices of width dx=10^-1000 resulting in f(b)-f(a)

Now suppose I am interested in calculating Int(f(g(x))g'(x)dx) from a to b. I can take dx=10^1000 equal width bars along the x axis and add their areas together. This will be the sum of all (b-a)*10^1000 regions each with area f(g(x))g'(x)x10^-1000

Next I am going to consider a different graph where the height is f(g(x)) with the same set of bars, but now I am going to relabel the x-axis from x to u=g(x), so that the curve now has height f(u) and extends from u=g(a) to u=g(b) so if for example g(x)=2*x^2 then the point at x=0.1 will be replaced with the value u=0.02 and the value x=0.2 with be replaced with the value u=0.08 and the region will extend from 0 to 2. If I use the same bars as before, they will have height f'(g(x)) and width in terms of u equal to du=g'(x)dx=g'(x)x10^-1000, so their total area will be the sum of the (b-a)10^1000 regions with area f(g(x))g'(x)x10^-1000 the same as my original graph. But it is also the graph of f'(u) vs u from the limits g(a) to g(b), and has area equal to Int(f(u)du) from g(a) to g(b).

So Int(f(g(x))g'(x)dx) from a to b = Int(f(u)du) from g(a) to g(b)

Last edited by Buck Godot; 05-19-2020 at 05:02 PM.
#21 Guest Join Date: Mar 2012 Posts: 2,133
Quote:
 Originally Posted by Dr. Strangelove Start with this substitution: u2 = x3 Take the derivative of both sides: 2u du = 3x2 dx ....
Sometimes substitutions like this can get you into trouble :
U^2 = X^3 => U = +/- X^ (3/2)
#22 Guest Join Date: Jul 2018 Location: The City Different Posts: 287
Quote:
 Originally Posted by Buck Godot ... then we get g'(x)=du/dx, or du=g'(x)dx ...
This is the part that I don't know how to explain. How did we get from g'(x)=du/dx to du=g'(x)dx? Is there a rigorous proof for this? Is there a simple explanation that could be used for someone that knows little or nothing about derivatives and differentials that doesn't involve "handwaving"? DPRK has had some suggestions but I'm afraid my audience would just return a blank stare. I have all summer to work on this.
#23  Guest Join Date: Dec 2010 Posts: 8,548
Are you familiar with infinitesimals? Willing to use them?

They provide a perfectly rigorous explanation for your questions, and backed the original intuition behind calculus for both Newton and Leibniz. However, the existence of the infinitesimals was not rigorously established until the 60s. This is why most calculus pedagogy uses the (ε, δ) limit form, with all the baggage that entails (such as the claim that dy/dx is not a fraction, even though Leibniz certainly thought it was).

You aren't going to prove the existence of the infinitesimals in an introductory class--but you aren't going to establish the real numbers either. So that doesn't seem like a big deal to me.
#24 Guest Join Date: Jul 2018 Location: The City Different Posts: 287
Quote:
 Originally Posted by Dr. Strangelove Are you familiar with infinitesimals?
I know that they are rather small. Actually they are really, really, really, really small. They may be even smaller than that.
#25  Guest Join Date: Dec 2010 Posts: 8,548
Quote:
 Originally Posted by Ynnad I know that they are rather small. Actually they are really, really, really, really small. They may be even smaller than that.
Then you are halfway there. The only other thing to remember is that they are not quite zero.
#26  Charter Member Join Date: May 2000 Location: Lincoln, IL Posts: 29,043
Quote:
 Originally Posted by Ynnad This is the part that I don't know how to explain. How did we get from g'(x)=du/dx to du=g'(x)dx? Is there a rigorous proof for this?
As I said above: it's true by definition. At least, that's the approach taken by all the standard Calculus textbooks I can remember seeing.

You can either define "du" to mean "the derivative of u with respect to x times dx," or you can define it to mean something else and then prove/explain why that something else is equivalent to g'(x)dx.
#27  Guest Join Date: Mar 2010 Location: MD outside DC Posts: 6,527
Quote:
 Originally Posted by Ynnad This is the part that I don't know how to explain. How did we get from g'(x)=du/dx to du=g'(x)dx? Is there a rigorous proof for this? Is there a simple explanation that could be used for someone that knows little or nothing about derivatives and differentials that doesn't involve "handwaving"? DPRK has had some suggestions but I'm afraid my audience would just return a blank stare. I have all summer to work on this.

I think it may be easier to think of a concrete example where there are actual physical quantities associated with x and u. Let us suppose that I am looking at an accelerating car and I want to know how far its gone after a particular time. In this case x is time in seconds and u is distance in meters and it happens that u=g(x)=x^2. Suppose I want to know how fast the car is going at a particular instant in time x.

One way I can do this is to at a particular time, x, I advance the time just one little tiny tiny smidgen, dx, and see how much the distance changes, lets call this change in distance du.

Well the distance at time=x was g(x)=x^2, and the distance of time (x+dx) was g(x+dx). So du defined as the change in distance was g(x+dx)-g(x).

Now we can work this out and find that
du=g(x+dx)-g(x)=(x+dx)^2-x^2 = x+2xdx+dx^2-2x = 2x dx+dx^2

Now dx is really really small, so dx^2 is really really really small, so small that you might as well ignore it. So we compute that du=2x dx, and in particular at that instant in time the distance it travels per unit time is du/dx=2x, which by amazing coincidence is g'(x).

Now suppose I have a wonky car that is doing all sorts of wonky things. So that its position over time is.

u=g(x) = x^3-5/x+ sin(x) -(e^x). I start out the same and get as far as,

du=g(x+dx)-g(x), and then divide both side by dx. So we get du/dx=(g(x+dx)-g(x))/dx, now computing all those differences in a pain in the ass so I'm not going to do that right now. Instead I'm going to just keep things in terms of functions.

Now remember that I wanted dx to be really really small, in fact I am going to take the limit as dx goes to zero.

du/dx = limit (g(x+dx)-g(x))/dx)

now to make thigs possibly more familiar I have going to change notation and say h=dx

so du/dx=limit (g(x+h)-g(x)/h) this right side is the definition of g'(x). So we see that for dx really really small du/dx=g'(x). Note you can't argue with this step since without agreeing that its true we can't discuss g'(x) at all.

now I use what I know about differentiation to compute that
g'(x)=3x^2-5ln(x)+cos(x)-e^x so if I move time ahead a little tiny smidgen dx the distance will change by du=(3x^2-5ln(x)+cos(x)-e^x)dx
#28 Guest Join Date: Mar 2012 Posts: 2,133
Quote:
 Originally Posted by Buck Godot I think it may be easier to think of a concrete example where there are actual physical quantities associated with x and...
I liked your idea of concrete example 

Quote:
 Originally Posted by Buck Godot So that its position over time is. u=g(x) = x^3-5/x+ sin(x) -(e^x). I start out the same and get....
Not sure if the limit even exists as x>0 and not sure if you can integrate it. Are you sure this is a concrete example ?
#29  Guest Join Date: Dec 2009 Location: the Land of Smiles Posts: 21,571
How does Google Books decide which pages to let some user see and which pages produce the "That page is not in this preview" message?

Quote:
 Originally Posted by septimus ... the great 12th-century Indian mathematician Bhaskara II has a memorable and relevant quote shown at the top of this page in Kim Plofker's book.

Quote:
 Originally Posted by Chronos septimus, would you mind quoting that statement for us? Your Google Books link shows only a single sentence on the page you linked, in an Indian language that I can neither read nor recognize.
I've added color codes to the following excerpt. The beginning of the paragraph is on page 197 and shown in Gray. (The URL links to page 198.) The 12th-century comment alluding to quantities like dy/dx is shown in Red.
Quote:
 Originally Posted by Kim Plofker's Book It has been noted that this and related statements reveal similarities between Bhaskara's ideas of motion and concepts in differential calculus. (In fact, perhaps these ratios of small quantities are what he was referring to in his commentary on Lilavati 47 when he spoke of calculations with factors of 0/0 being "useful in astronomy.")
As to the mysterious "Indian sentence" you saw: I'll guess you saw "That page is not in this preview" in Thai. (Try the new URL above.) I don't know why Google Books lets me preview that page but not you. (Over half of the book's pages are blocked to me also, but not 197-198.)

Blame me for the Thai error message (via the 'co.th' URL). Sometimes I manually edit those URLs to be '.com' but Google changes them back to 'co.th' in the browser's locator bar (unless I stay logged in to Google). I guess I figured that since Google changes English to Thai automatically for me, it would do the vice versa for you. (But that's not how Google works. For me it leaves the English 'co.uk' intact; it just changes '.com'.)
#30 Guest Join Date: May 2016 Posts: 5,150
Something to consider:

To the real numbers ℝ you may adjoin an infinitesimal element, let's call it dx, which satisfies (dx)² = 0. This is a function on a formal infinitesimal line segment, a fattened point if you will, just as functions at a single point are given by a real number. dx is a coordinate function, just like x is on the real line.

Differential 1-forms on the real line now take the form g(x) dx, where g(x) is an ordinary smooth function. g(x) dx is not a function on the real line, rather a function on the space of infinitesimal paths in ℝ, i.e., something of the form f(x) + g(x) dx, with the f(x) = 0 so you just have g(x) dx.

Now let f(x) be an ordinary smooth function we wish to differentiate. Define a new function h(x,ε) = f(x+ε) (on the infinitesimal path space, so ε² = 0).

The derivative of f will now be obtained via taking h(x,ε) - f(x) = f(x+ε) - f(x). This will be a linear function of ε, namely εf'(x).

The corresponding 1-form is
(df)(x) = f(x + dx) - f(x) = f'(x) dx.

In summary, nothing that has not been said already in this thread, including that df = f'(x) dx very easily and that 1-forms are functions on an appropriate space (nothing non-rigorous about it). It still remains to make the connection to antidifferentiation and to definite integrals, but the differential forms that appear should not be confusing anybody.

Last edited by DPRK; 05-20-2020 at 01:19 AM.

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