# The Unapologetic Mathematician

## Examples and Notation

Okay, let’s do some simple examples of differentials, which will lead to some notational “syntactic sugar”.

First of all, if we pick an orthonormal basis $\left\{e_i\right\}_{i=1}^n$ we can write any point as $x=x^ie_i$. This gives us $n$ nice functions to consider: $x^i:\mathbb{R}^n\rightarrow\mathbb{R}$ is the function that takes a point and returns its $i$th coordinate. This is actually a sort of subtle point that’s important to consider deeply. We’re used to thinking of $x^i$ as a variable, which stands in for some real number. I’m saying that we want to consider it as a function in its own right. In a way, this is just extending what we did when we considered polynomials as functions and we can do everything algebraically with abstract “variables” as we can with specific “functions” as our $x^i$.

Analytically, though, we can ask how the function $x^i$ behaves as we move our input point around. It’s easy to find the partial derivatives. If $k\neq i$ then

$\displaystyle\left[D_kx^i\right](x)=\lim\limits_{t\to0}\frac{x^i(x+te_k)-x^i(x)}{t}=\lim\limits_{t\to0}\frac{0}{t}=0$

since moving in the $e_k$ direction doesn’t change the $i$th component. On the other hand, if $k=i$ then

$\displaystyle\left[D_kx^i\right](x)=\lim\limits_{t\to0}\frac{x^i(x+te_k)-x^i(x)}{t}=\lim\limits_{t\to0}\frac{t}{t}=1$

since moving a distance $t$ in the $e_k$ direction adds exactly $t$ to the $i$th component. That is, we can write $D_kx^i=\delta_k^i$ — the Kronecker delta.

Of course, since ${0}$ and ${1}$ are both constant, they’re clearly continuous everywhere. Thus by the condition we worked out yesterday the differential of $x^i$ exists, and we find

$\displaystyle dx^i(x;t)=\delta_k^it^k=t^i$

We can also write the differential as a linear functional $dx^i(x)$. Since this takes a vector $t$ and returns its $i$th component, it is exactly the dual basis element $\eta^i$. That is, once we pick an orthonormal basis for our vector space of displacements, we can actually write the dual basis of linear functionals as the differentials $dx^i$. And from now on that’s exactly what we’ll do.

So, for example, let’s say we’ve got a differentiable function $f:\mathbb{R}^n\rightarrow\mathbb{R}$. Then we can write its differential as a linear functional

$df(x)=\left[D_1f\right](x)dx^1+\dots+\left[D_nf\right](x)dx^n=\left[D_if\right](x)dx^i$

In the one-dimensional case, we write $df(x)=f'(x)dx$, leading us to the standard Leibniz notation

$\displaystyle\frac{df}{dx}=f'$

If we have to evaluate this function, we use an “evaluation bar” $\frac{df}{dx}\bigr\vert_{x}=f'(x)$, or $\frac{df}{dx}\bigr\vert_{x=a}=f'(a)$ telling us to substitute $a$ for $x$ in the formula for $\frac{df}{dx}$. We also can write the operator that takes in a function and returns its derivative by simply removing the function from this Leibniz notation: $\frac{d}{dx}$.

Now when it comes to more than one variable, we can’t just “divide” by one of the differentials $dx^i$, but we’re going to use something like this notation to read off the coefficient anyway. In order to remind us that we’re not really dividing and that there are other variables floating around, we replace the $d$ with a curly version: $\partial$. Then we can write the partial derivative

$\displaystyle\frac{\partial f}{\partial x^i}=D_if$

and the whole differential as

$\displaystyle df=\frac{\partial f}{\partial x^1}dx^1+\dots+\frac{\partial f}{\partial x^n}dx^n=\frac{\partial f}{\partial x^i}dx^i$

Notice here that when we see an upper index in the denominator of this notation, we consider it to be a lower index. Similarly, if we find a lower index in the denominator, we’ll consider it to be like an upper index for the purposes of the summation convention. We can even incorporate evaluation bars

$\displaystyle df(a)=\frac{\partial f}{\partial x^1}\biggr\vert_{x=a}dx^1+\dots+\frac{\partial f}{\partial x^n}\biggr\vert_{x=a}dx^n=\frac{\partial f}{\partial x^i}\biggr\vert_{x=a}dx^i$

or strip out the function altogether to write the “differential operator”

$\displaystyle d=\frac{\partial}{\partial x^1}dx^1+\dots+\frac{\partial}{\partial x^n}dx^n=\frac{\partial}{\partial x^i}dx^i$

October 2, 2009 - Posted by | Analysis, Calculus