The Unapologetic Mathematician

Mathematics for the interested outsider

Higher Differentials and Composite Functions

Last time we saw an example of what can go wrong when we try to translate higher differentials the way we did the first-order differential. Today I want to identify exactly what goes wrong, and I’ll make use of the summation convention to greatly simplify the process.

So, let’s take a function f of n variables \left\{y^j\right\}_{j=1}^n and a collection of n functions \left\{g^j\right\}_{j=1}^n, each depending on m variables \left\{x^i\right\}_{j=1}^m. We can think of these as the components of a vector-valued function g:X\rightarrow\mathbb{R}^n which has continuous second partial derivatives on some region X\subseteq\mathbb{R}^m. If the function f:Y\rightarrow\mathbb{R} has continuous second partial derivatives on some region Y\subseteq\mathbb{R}^n containing the image g(X), then we can compose the two functions to give a single function h=f\circ g:X\rightarrow\mathbb{R}, and we’re going to investigate the second differential of h with respect to the variables x^i.

To that end, we want to calculate the second partial derivative

\displaystyle\frac{\partial^2h}{\partial x^b\partial x^a}=\frac{\partial}{\partial x^b}\frac{\partial}{\partial x^a}h

First, we take the derivative in terms of x^a, and we use the chain rule to write

\displaystyle\frac{\partial h}{\partial x^a}=\frac{\partial g^j}{\partial x^a}\frac{\partial f}{\partial y^j}

Now we have to take the derivative in terms of x^b. Luckily, this operation is linear, so we don’t have to worry about the hidden summations in the notation. We do, however, have to use the product rule to handle the multiplications

\displaystyle\begin{aligned}\frac{\partial}{\partial x^b}\frac{\partial h}{\partial x^a}&=\frac{\partial}{\partial x^b}\left(\frac{\partial g^j}{\partial x^a}\frac{\partial f}{\partial y^j}\right)\\&=\frac{\partial^2g^j}{\partial x^b\partial x^a}\frac{\partial f}{\partial y^j}+\frac{\partial g^j}{\partial x^a}\frac{\partial^2f}{\partial x^b\partial y^j}\\&=\frac{\partial^2g^j}{\partial x^b\partial x^a}\frac{\partial f}{\partial y^j}+\frac{\partial g^j}{\partial x^a}\frac{\partial g^k}{\partial x^b}\frac{\partial^2f}{\partial y^k\partial y^j}\end{aligned}

where we’ve used the chain rule again to convert a derivative in terms of x^b into one in terms of y^k.

And here we’ve come to the problem itself. For we can write out the second differential in terms of the x^i

\displaystyle\begin{aligned}d^2h&=\frac{\partial^2h}{\partial x^b\partial x^a}dx^adx^b\\&=\frac{\partial^2f}{\partial y^k\partial y^j}\left(\frac{\partial g^j}{\partial x^a}dx^a\right)\left(\frac{\partial g^k}{\partial x^b}dx^b\right)+\frac{\partial f}{\partial y^j}\frac{\partial g^j}{\partial x^b\partial x^a}dx^adx^b\\&=\frac{\partial^2f}{\partial y^k\partial y^j}dg^jdg^k+\frac{\partial f}{\partial y^j}d^2g^j\end{aligned}

The first term here is the second differential in terms of the y^j. If there were an analogue of Cauchy’s invariant rule, this would be all there is to the formula. But we’ve got another term — one due to the product rule — based on the second differentials of the functions g^j themselves. This is the term that ruins the nice transformation properties of higher differentials, and which makes them unsuitable for many of our purposes.

Notice, though, that we have not contradicted Clairaut’s theorem here. Indeed, as long as f and all the g^j have continuous second partial derivatives, then so will h. Further, the formula we derived for the second partial derivatives of h is manifestly symmetric between the two derivatives, and so the mixed partials commute.

October 19, 2009 Posted by | Analysis, Calculus | 2 Comments

   

Follow

Get every new post delivered to your Inbox.

Join 366 other followers