I'm trying to understand Taylor's Theorem for functions of $n$ variables, but all this higher dimensionality is causing me trouble. One of my problems is understanding the higher order differentials. For example, if I have a function $f(x, y)$, then it's first differential is:
$$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy.$$
To me this quantity is saying that:
A differential change in the value of function $f(x,y)$ is equal to how fast function $f(x,y)$ is changing
with respect to $x$ multiplied by a differential change in the
$x$-coordinate plus how fast function $f(x,y)$ is changing with
respect to $y$ multiplied by a differential change in the
$y$-coordinate.
This seems intuitive. But when we get into higher order differentials I get confused:
$$d^2f= \frac{\partial^2 f}{\partial y ^2}dy^2 + 2\frac{\partial^2 f}{\partial y \partial x}dy\:dx + \frac{\partial^2 f}{\partial x ^2}dx^2$$
How would one interpret this quantity? What about even higher order differentials? say $d^3f$ or $d^{1500 }f$ =)
Thank you for any help! =)
Best Answer
I think all of this makes a lot more sense when you approach it from a multilinear set up.
If $f: \mathbb{R}^2 \to \mathbb{R}$ is a function, then its differential $df$ gives a different linear map at each point. Fundamentally we have
$$ f(\mathbf{p}+\vec{v})\approx f(\mathbf{p})+df(\mathbf{p}, \vec{v}) $$
Now the second differential ($d^2f$ in your notation), should be something which records how $df$ changes from one point to the next. In other words, we should like
$$ df(\mathbf{p}+\vec{w}, \vec{v}) \approx df(\mathbf{p},\vec{v})+d^2f(\mathbf{p},\vec{w},\vec{v}) $$
Slogan: "$d^2 f$ is the gadget which takes two vectors $\vec{v}$ and $\vec{w}$ and spits out the approximate change in $df$ from $p$ to $p+\vec{w}$ when it is evaluated in the direction $\vec{v}$ "
At a given point $p$, this map gadget $d^2f$ should be linear in both $\vec{w}$ and $\vec{v}$. So it is a multilinear function which varies from point to point, aka a $2$-tensor field!
We can figure out an expression for $d^2f$ as follows:
$$ \frac{\partial ^2 f}{\partial x^2} dx \otimes dx + \frac{\partial ^2 f}{\partial x \partial y} dx \otimes dy + \frac{\partial ^2 f}{\partial y \partial x} dy \otimes dx + \frac{\partial ^2 f}{\partial y^2} dy \otimes dy $$
Taylor's theorem comes about when you try to approximate changes not just in $df$, but carry those through to changes in $f$. You do that, basically, by starting from one point and restricting your method of approximation to a line segment. So you only ever plug the same vector into both arguments of the second differential, which means you are really working with the associated quadratic form.
If you would like to come to an understanding of Taylor's theorem along the lines suggested here, I recommend you check out my online course here:
http://ximera.osu.edu/course/kisonecat/m2o2c2/course/
It will guide you through a selection of exercises which gradually build in difficulty, with you developing most of the mathematics yourself. A copious hint system (which more often than not prompts you with a simpler or related question) should ensure that you can get through it.