2014-07-02

Back to the old way

\(
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\r}{\rightarrow}
\newcommand{\d}{\mathbin{\cdot}}

\)Leibniz' way of expressing the derivative of one implicit function with respect to another was to write \(\frac{dy}{dx}\) – intuitively the infinitesimal change in \(y\) divided by the infinitesimal change in \(x\), giving a gradient. Of course, when division is flipped, this becomes \(\frac{dx}{dy}\), putting the variables in the familiar \((x,y)\) order.

So, now imagine that \(x\) isn't the independent variable, and we instead have an unnamed independent variable that can be used via \(\large\iota\). The derivative of \(x\) is just known as \(Dx\), and the derivative of \(y\), with respect to the independent variable, is \(Dy\).

Let's do something easy. Let \(y = x + \sin x\). Notice that \(\sin x\) becomes \(\text{sin}\circ x\) when we evaluate the expression, since \(x\) is an \(\R\r\R\) function (or \(\C\r\C\) or whatever). Also, the whole thing is forked over \(+\). We can rewrite the equation as \(y = ({\large\iota} + \sin)\ x\). Then we can differentiate, using the chain rule: \(Dy = (1 + \cos)\ x\d Dx\). Magically, we're left with:
\[
\frac{Dx}{Dy} = 1 + \cos x
\]Whereas it's a sin in normal mathematics to throw around solitary \(dx\)s, \(Dx\)s are just ordinary functions. You can do whatever you want with them.

Let's try something a little more interesting. Consider:
\[\begin{aligned}
{}_\d^2x+{}_\d^2y&{}=1&& \\
2\d x\d Dx+2\d y\d Dy&{}=0&& \quad \text{applied }D\text{ to both sides} \\
\frac{y}{x}+\frac{Dx}{Dy}&{}=0&& \quad \text{applied }(2\d y\d Dx/)\text{ to both sides} \\
\frac{Dx}{Dy}&{}=\frac{y}{x}-0&& \quad \text{applied }\left(\frac{y}{x}-\right)\text{ to both sides}
\end{aligned}\]See that curried functions are useful for documentation. They give us the option to be more symbolic, rather than using verbose phrases like “divide through by ...” which have to be translated before we can see them interacting with the equation at hand.

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