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Intuition for Change of Variables Theorem

by mathmonkey
Tags: intuition, theorem, variables
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mathmonkey
#1
Jun14-12, 08:19 PM
P: 33
Hi,

The change of variables theorem states that given a diffeomorphism [itex]g:A \rightarrow B[/itex] between open sets, and a continuous function [itex]f:A \rightarrow R[/itex], then [itex]
\int _A f = \int _B f \circ g |Det Dg|[/itex] given that either one of the integrals exist.

I was wondering if anyone here could help explain to me the intuition behind the change of variables theorem. More specifically, I'm interested in the intuition (though formal arguments are welcome as well!) behind how the determinant function creeps into the equation and why the determinant of Dg? I'd be grateful for any insight into the theorem. Thanks!
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mathwonk
#2
Jun14-12, 09:10 PM
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the simplest case is when the change of variables map is linear and the domain is a rectangle. then the theorem just says that the volume of a cube is expanded by exactly the determinant of the linear map. since every smooth map is locally linear and every domain is approximable by rectangles, the theorem follows from that case. (yeah right...)

i.e. to get the intuition, compute the area of the image of the unit square in the first quadrant of R^2 under a matrix with entries a,b,c,d, and check that it equals |ad-bc|.
algebrat
#3
Jun15-12, 02:24 AM
P: 428
Quote Quote by mathmonkey View Post
Hi,

The change of variables theorem states that given a diffeomorphism [itex]g:A \rightarrow B[/itex] between open sets, and a continuous function [itex]f:A \rightarrow R[/itex], then [itex]
\int _A f = \int _B f \circ g |Det Dg|[/itex] given that either one of the integrals exist.
This doesn't look right, guess I'll have to prove that by using intuition.

Let's call the area elements dA and dB. At x, or y=g(x), we want to multiply the height f by the area dA. But dB isn't the right size, so we need [itex]dA=\frac{dA}{dB}dB[/itex]. But [itex]Dg[/itex] is more like [itex]\frac{dB}{dA}[/itex], then [itex]\frac{dA}{dB}[/itex], so you need to fix something in your statement.

Maybe you can fix it if you say g:B->A.

EDIT: That won't do it, there's more wrong, you'll have to do an overhaul

But Mathwonk is right, try rounding out your intuition with a linear change of variable, and maybe check what happens in the 1-d case too. (It should reduce to u-substitution.)

chiro
#4
Jun15-12, 03:11 AM
P: 4,572
Intuition for Change of Variables Theorem

Quote Quote by mathmonkey View Post
Hi,

The change of variables theorem states that given a diffeomorphism [itex]g:A \rightarrow B[/itex] between open sets, and a continuous function [itex]f:A \rightarrow R[/itex], then [itex]
\int _A f = \int _B f \circ g |Det Dg|[/itex] given that either one of the integrals exist.

I was wondering if anyone here could help explain to me the intuition behind the change of variables theorem. More specifically, I'm interested in the intuition (though formal arguments are welcome as well!) behind how the determinant function creeps into the equation and why the determinant of Dg? I'd be grateful for any insight into the theorem. Thanks!
Hey mathmonkey and welcome to the forums.

Do you understand what the Jacobian refers to with respect to the derivative in a multi-variable scenario (i.e. generalizing to multiple variables)?

Are you also aware that the determinant for a given operator represents the volumetric scaling given a linear operator?

Finally, are you aware of how this linear algebra relates to orthogonal cartesian like systems and how the determinant represents overall volumetric 'differential' change given the fact of the nature of R^n?
Muphrid
#5
Jun15-12, 09:35 AM
P: 834
Let [itex]g(x) = x'[/itex], and let there be two functions [itex]f,f'[/itex] such that [itex]f(x) = f'(x')[/itex]. The Jacobian operator [itex]\underline g[/itex] relates derivatives of this function. For some vector [itex]a[/itex], we have, by the chain rule,

[itex]a \cdot \nabla f(x) = a \cdot \nabla f'(x') = [a \cdot \nabla g(x)] \cdot \nabla' f'(x')[/itex]

And we define the Jacobian as [itex]a \cdot \nabla g(x) \equiv \underline g(a)[/itex].

When [itex]x = x(\tau)[/itex] (i.e. [itex]x[/itex] is parameterized by some scalar and traces out a path), the chain rule leads to the following result:

[tex]\frac{dx}{d\tau} = \underline g^{-1} \left( \frac{dx'}{d\tau} \right)[/tex]

Or, since this has to be true regardless of the parameterization, we can say

[tex]dx = \underline g^{-1}(dx')[/tex]

Now, [itex]dx,dx'[/itex] are vectors, and even when one is not using their vector parts, the effect of the Jacobian on them must be accounted for, and it must be used on every factor making up the differential in question. That is, if one has a 2D area, then the inverse Jacobian has to be applied to both [itex]dx'[/itex] and [itex]dy'[/itex]. When this is done on every dimension in the space (the volume element), the correct way to do this yields the determinant of the inverse Jacobian. I can write this much more formally, but I don't think the "correct way" of doing this can be understood completely without multivectors. If you're interested, I'd be happy to elaborate, however.

In the end, just by requiring a simple notion that the value of the function [itex]f(x) = f'(x')[/itex], we're able to derive several rules for the invariance of the derivative and integrals, as long as the Jacobian is used where prescribed.

On the intuitive level, the Jacobian helps keep track of how lengths, areas, and so on are distorted through the transformation, and using the inverse Jacobian is necessary to make sure that the area or volume or whatever used in the integration is the same as the original one. Even though we change coordinates, the region we're integrating over is not actually changing, and the Jacobian's inverse helps ensure that.
mathmonkey
#6
Jun15-12, 12:48 PM
P: 33
Quote Quote by algebrat View Post
This doesn't look right, guess I'll have to prove that by using intuition.
Ah, of course! Sorry for the typo, I meant [itex] F:B \rightarrow R [/itex]. That should be the right statement of the theorem now. I'm not sure how to edit my original post to reflect that unfortunately.

To everyone else:

Thanks for the help. I now understand that the determinant of the linear map is used for scaling volume just from the example of the parallelopiped formed by the standard bases under a linear map. Now I think the source of my problem is understanding why it is the determinant of the Jacobian that is used for preserving volume, not, for example, the determinant of [itex]g[/itex] itself (Though I suppose that would only make sense if [itex]g[/itex] were a linear map)?
algebrat
#7
Jun15-12, 02:04 PM
P: 428
Quote Quote by mathmonkey View Post
Now I think the source of my problem is understanding why it is the determinant of the Jacobian that is used for preserving volume, not, for example, the determinant of [itex]g[/itex] itself (Though I suppose that would only make sense if [itex]g[/itex] were a linear map)?
g is a multivariable function, that is, from a mulit-input to a multi-output of datas. But it is not a system of linear equations, so it is not a matrix, so we cannot take the determinant. Luckily, the integral only needs local data, so we take the linear approximation of g near each point, that is, the system of differentials. This is the Jacobian. Then the Jacobian maps the unit n-cube to an n-prism which has volume the determinant of the jacobian.

g would have mapped a very small n-cube with volume dV, to approximately a very small n-prism with volume dV'=det(Dg)dV
algebrat
#8
Jun15-12, 02:10 PM
P: 428
Quote Quote by mathmonkey View Post
Ah, of course! Sorry for the typo, I meant [itex] F:B \rightarrow R [/itex]. That should be the right statement of the theorem now. I'm not sure how to edit my original post to reflect that unfortunately.
It's still not correct, what is fg doing in the integral over the region B? Why is f being integrated over A?
mathmonkey
#9
Jun15-12, 02:35 PM
P: 33
Yeah, sorry, getting my letters mixed up. I'll just restate it:

Given a diffeomorphism [itex]g: A \rightarrow B[/itex] between open sets, and a continuous function [itex]f:B \rightarrow R[/itex], then [itex]\int _B f = \int _A f \circ g |DetDg| [/itex] given that either one of the integrals exist.

Anyway, I think I understand what you're saying. In order for the determinant function to make sense, we need a linear map from [itex]A[/itex] to [itex]B[/itex]. Although [itex]g[/itex] may not necessarily be a linear map, if we take its derivative, we can then make sense of the determinant function for [itex]Dg[/itex]. Now this value of the determinant will scale the volumes of infinitesimally small cubes from [itex]A [/itex] to [itex]B[/itex]. Is that along the lines of what you mean?
algebrat
#10
Jun15-12, 04:32 PM
P: 428
Yes, that is exactly what I'm saying (though you sort of began to say it above first, I was just rewording it),

We have (don't ask me precisely where did I get this notation, I'm mixing it up to highlight things)
[tex]dB(y)=|Dg(x)|dA(x)[/tex]
so now we are free to integrate f over either region, and still be multiplying by the same "base" value, like in 1-d riemann sums df=f(y)*dy=f(g(x))*(g'(x))dx

In other words, dA is mapped to dB, which is different by the factor |Dg|. But we really wanted dB. The directions are a little bit confusing, which is why I have often made the same notational mistake you originally did, so to remember the form it helps to have this understanding and think it through. I think I usually draw a picture to help me, but even then it is a little tricky to remember everything quickly (for mere mortals like me).


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