The Unapologetic Mathematician

Mathematics for the interested outsider

Cauchy’s Invariant Rule

An immediate corollary of the chain rule is another piece of “syntactic sugar”.

If we have functions g:X\rightarrow\mathbb{R}^n and f:Y\rightarrow\mathbb{R}^p for some open regions X\subseteq\mathbb{R}^m and Y\subseteq\mathbb{R}^n so that the image g(X) is contained in Y, we can compose the two functions to get a new function f\circ g:X\rightarrow\mathbb{R}^p. In terms of formulas, we can choose coordinates y^i on \mathbb{R}^n and write out both the function f(y^1,\dots,y^n) and the component functions g^1(x),\dots,g^n(x). We get a formula for \left[f\circ g\right](x) by substituting g^i(x) for y^i in the formula for f and write y^i=g^i(x).

The language there seems a little convoluted, so I’d like to give an example. We might define a function f(x,y)=e^{x^2+y^2} for all points (x,y) in the plane \mathbb{R}^2. This is all well and good, but we might want to talk about the function in polar coordinates. To this end, we may define x=r\cos(\theta) and y=r\sin(\theta). These are the component functions describing a transformation g from the region (r,\theta)\in(0,\infty)\times(-\pi,\pi)\subseteq\mathbb{R}^2 to the region where (x,y)\neq(0,0). We can substitute r\cos(\theta) for x and r\sin(\theta) for y in our formula for f to get a new function f\circ g with formula

\displaystyle f(g(r,\theta))=e^{r^2\cos(\theta)^2+r^2\sin(\theta)^2}=e^{r^2}

This much is straightforward. The thing is, now we want to take differentials. What Cauchy’s invariant rule tells us is that we can calculate the differential of f\circ g by not only substituting g^i(x) for y^i, but also substituting dg^i(x;t) for s^i in the formula for df(y;s). That is, if h=f\circ g then we have the equivalence

\displaystyle dh(x;t)=df(g^1(x),\dots,g^n(x);dg^1(x;t),\dots,dg^n(x;t))

In our particular example, we can easily calculate the differential of f using our first formula:

df(x,y)=2xe^{x^2+y^2}dx+2ye^{x^2+y^2}dy

or using our second formula:

df(r,\theta)=2re^{r^2}dr

We want to call both of these simply df. But can we do so unambiguously? Indeed, if x=r\cos(\theta) then we find

\displaystyle dx=\cos(\theta)dr-r\sin(\theta)d\theta

and if y=r\sin(\theta) then we find

\displaystyle dy=\sin(\theta)dr+r\cos(\theta)d\theta

We substitute these into our formula for df(x,y) to find

\displaystyle\begin{aligned}df(r,\theta)&=2r\cos(\theta)e^{r^2\cos(\theta)^2+r^2\sin(\theta)^2}\left(\cos(\theta)dr-r\sin(\theta)d\theta\right)+2r\sin(\theta)e^{r^2\cos(\theta)^2+r^2\sin(\theta)^2}\left(\sin(\theta)dr+r\cos(\theta)d\theta\right)\\&=2r\cos(\theta)e^{r^2}\left(\cos(\theta)dr-r\sin(\theta)d\theta\right)+2r\sin(\theta)e^{r^2}\left(\sin(\theta)dr+r\cos(\theta)d\theta\right)\\&=2r\cos(\theta)e^{r^2}\cos(\theta)dr+2r\sin(\theta)e^{r^2}\sin(\theta)dr-2r\cos(\theta)e^{r^2}r\sin(\theta)d\theta+2r\sin(\theta)e^{r^2}r\cos(\theta)d\theta\\&=\left(2r\cos(\theta)^2e^{r^2}+2r\sin^2(\theta)e^{r^2}\right)dr+\left(2r^2\cos(\theta)\sin(\theta)e^{r^2}-2r^2\cos(\theta)\sin(\theta)e^{r^2}\right)d\theta\\&=2re^{r^2}dr\end{aligned}

just the same as if we calculated directly from the formula in terms of r and \theta.

That is, we can substitute our formulæ for the coordinate functions y^i=g^i(x) before taking the differential in terms of x, or we can take the differential in terms of y and then substitute our formulæ for the coordinate functions y^i=g^i(x) and their differentials dy^i=dg^i(x) into the result. Either way, we end up in the same place, so we don’t have to worry about ending up with two (or more!) “different” differentials of f.

So, how do we verify this using the chain rule? Just write out the differentials out using partial derivatives. For example, we know that

\displaystyle df(y;s^1,\dots,s^n)=\frac{\partial f}{\partial y^i}\biggr\vert_ys^i

and so on. So, performing our substitutions we can find:

\displaystyle\begin{aligned}df(g(x);dg^1(x;t),\dots,dg^n(x;t))&=\frac{\partial f}{\partial y^i}\biggr\vert_{y=g(x)}dg^i(x;t)\\&=\frac{\partial f}{\partial y^i}\biggr\vert_{y=g(x)}\frac{\partial g^i}{\partial x^j}\biggr\vert_xt^j\\&=\frac{\partial\left[f\circ g\right]}{\partial x^j}\biggr\vert_xt^j\\&=d\left[f\circ g\right](x;t)\end{aligned}

The important part here is the passage from products of two partial derivatives to single partial derivatives of f\circ g. This works out because when we consider differentials as linear transformations, the matrix entries are the partial derivatives. The composition of the linear transformations df(g(x)) and dg(x) is given by the product of these matrices, and the entries of the resulting matrix must (by uniqueness) be the partial derivatives of the composite function.

October 8, 2009 Posted by | Analysis, Calculus | 5 Comments

   

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