1
$\begingroup$

I am currently reading an old math book which contains the following unexplained notation:

Let $f(x,y)$ and $g(x,y)$ be functions $\mathbb{R}^2\rightarrow \mathbb{R}$. The notation

$$\frac{\partial(f,g)}{\partial(x,y)}$$

apparently refers to a function of the form $\mathbb{R}^2\rightarrow \mathbb{R}$. I am not sure, but I suspect it may be defined as

$$\frac{\partial(f,g)}{\partial(x,y)} \equiv \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}.$$

Is this notation/definition common in any particular field?

And especially if so, could there be an obvious interpretation of the following notation, which this book also uses without explanation?

$$\frac{\partial[f,g]}{\partial(x,y)}$$

Thanks for your help.


Edit: I believe @Fred is correct that the parentheses are used to denote the Jacobian. Here is the notation, as used in a simplified excerpt of Calculating Curves by Ron Doerfler and others:

$$ \left\{ \begin{array}{c} 0 = \frac{\partial u}{\partial y} f_1(x) + \frac{\partial v}{\partial y}\\ 0 = \frac{\partial u}{\partial x} f_2(y) + \frac{\partial v}{\partial x}\\ \end{array}\right.$$

Let us assume that $\frac{\partial(u,v)}{\partial(x,y)}=0$. Then the above equations yield that

$$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}[f_1(x) - f_2(y)] = 0.$$

Thus we can posit that

$$\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial x}\frac{\partial u}{\partial y} = e^\theta.$$

I am still unsure about the meaning of the square bracket notation. The square brackets are a little more difficult to place in context, but here is an attempt:

$$g_3(z) = u f_3(z) + v$$

We clearly have $\frac{\partial[g_3(z), z]}{\partial(x,y)} = 0$. By substituting the above equation and observing that $\frac{\partial[f_3(z), z]}{\partial(x,y)} = 0$, we obtain

$$f_3(z)\frac{\partial(u,z)}{\partial(x,y)} + \frac{\partial(v,z)}{\partial(x,y)} = 0.$$

Possibly the square brackets are simply an alias for round brackets which are used to avoid potentially visually-noisy nested round brackets(?).

$\endgroup$
  • $\begingroup$ Never used the above notation. It might be the Lie bracket $[f,g] = fg - gf$. $\endgroup$ – Wuestenfux Jun 12 '17 at 7:51
  • $\begingroup$ What book? And how does the book use this notation? Even a single example should clarify a lot. $\endgroup$ – Chris Culter Jun 12 '17 at 8:01
3
$\begingroup$

In vector calculus, we said, for the generalization of the chain rule, that $$\frac{\partial \mathbf f}{\partial \mathbf x}=\frac{\partial \mathbf f}{\partial \mathbf u}\frac{\partial \mathbf u}{\partial \mathbf x}$$ Where $\mathbf x, \mathbf f, \mathbf u$ are vectors (of functions).

Thus, $$\frac{\partial (f, g)}{\partial (x, y)}= \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial g}{\partial x} \end{bmatrix} \begin{bmatrix} \frac{\partial x}{\partial x}& \frac{\partial x}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{bmatrix}$$

Well, it is true that sometimes books use this notation for the determinant of this matrix (However I didn't). Then we get, $$ \begin{vmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{vmatrix} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}- \frac{\partial f}{\partial y} \frac{\partial g}{\partial x} $$ Which is the Jacobian of $h(x,y)=(f(x,y), g(x,y))$.

$\endgroup$
1
$\begingroup$

Let $h: \mathbb R^2 \to \mathbb R^2 $ be defined by $h(x,y) =(f(x,y),g(x,y))$.

Then $\frac{\partial(f,g)}{\partial(x,y)} $ is the Jacobian of $h$.

$\endgroup$
1
$\begingroup$

$$\frac{\partial(f,g)}{\partial(x,y)} = \left(\begin{matrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{matrix}\right)$$

However, I think that sometimes the notations is used for the determinant if this matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.