# Interpreting higher order differentials

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! =)

• Can you tell me the interpretation of the 1501-st derivative of a real function of a single real variable? – Siminore Jul 3 '14 at 11:26
• @Siminore no =) – jjepsuomi Jul 3 '14 at 16:24

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.

• +1 Appreciate it, thank you! =) Need to check out your course ;) – jjepsuomi Jul 3 '14 at 16:29
• @Steven Gubkin : Is the material that you linked to at Ximera the same as what is now at ximera.osu.edu/mooculus/calculus3/… ? – Toby Bartels Dec 4 '17 at 22:49
• @TobyBartels Nope. I will see if I can take my old files and at least compile them into a pdf. – Steven Gubkin Dec 4 '17 at 23:02
• It might be a week or so: the whole ximera'' system was completely revamped from this time. I do remember reading some of your stuff on sections of iterated tangent bundles on n-category cafe? And how a connection is really just a section of a map from $T^2M \to TM \times TM$? Did that stuff ever get resolved? I would be interested in thinking more about this kind of thing. – Steven Gubkin Dec 4 '17 at 23:18
• I suppose you mean the stuff at ncatlab.org/nlab/show/cogerm+differential+form and the forum posts listed in the references there? I'd say that that stuff is pretty much resolved, but what's been resolved goes beyond what's at that nLab page; it's in the forum posts but not in a very accessible form. – Toby Bartels Dec 5 '17 at 6:04

Lets denote $$df = \left(dx\frac{\partial}{\partial x} + dy\frac{\partial}{\partial y}\right)f$$ then $$d^2f = \left(dx\frac{\partial}{\partial x} + dy\frac{\partial}{\partial y}\right)\left(dx\frac{\partial}{\partial x} + dy\frac{\partial}{\partial y}\right)f = \left(dx\frac{\partial}{\partial x} + dy\frac{\partial}{\partial y}\right)^2f$$ this in the loose sense but with expressions of the form we can expand like $$(a+b)^2 = a^2 + ab + ba + b^2$$ I kept the middle terms order dependent since I am not making the assumption that they commute at this stage, though for functions I have come across (physicist here) they do, but I do not know if this generally true always.

Anyway, using the expression for a and b I am highlighting that it is simply a binomal expansion of the operators so for $$(a+b)^3 = a^3+3a^2b + 3ab^2 + b^3$$ or equivalently $$d^3f = \left(dx\frac{\partial}{\partial x} + dy\frac{\partial}{\partial y}\right)^3f,\\ =\left(dx^3\frac{\partial^3}{\partial x^3} +3dx^2dy\frac{\partial^3}{\partial x^2\partial y} + 3dxdy^2\frac{\partial^3}{\partial x\partial y^2} + dy^3\frac{\partial^3}{\partial y^3}\right)f$$ and so on and so on..try generating the summation rule for $d^nf$ ;)

This is my humble opinion as always

• +1 Thank you for your help @Chinny84 appreciate the help =) It's still difficult though to grasp on the meaning of e.g. $d^3 f$ :) I'm looking for an intuition for these formulas so that I won't just blindly follow the algebra :) – jjepsuomi Jul 3 '14 at 11:06

In $$d^2f= \frac{\partial^2 f}{\partial y ^2}dy^2 + 2\frac{\partial^2 f}{\partial y \partial x}dy\:dx + \frac{\partial^2 f}{\partial x ^2}dx^2$$, $\frac{\partial^2 f}{\partial y ^2}dy^2$ is the rate of change of $\frac{\partial f}{\partial y}$ on $y$, same for $\frac{\partial^2 f}{\partial x ^2}dx^2$. $2\frac{\partial^2 f}{\partial y \partial x}dy\:dx$ is the rate of change of $\frac{\partial f}{\partial y}$ on $x$ and rate of change of $\frac{\partial f}{\partial x}$ on $y$.

You can similarly interpret the terms appearing in higher order differentials.

Hint

Let us consider first the case of a function $f(x)$ and you expand it by a first order Taylor series, you represent locally the function by a straight line. If you go to the second order expansion, you represent locally the function by a parabola.

When you have a function $f(x,y)$ and you expand it by a first order Taylor series, you represent locally the function by a plane. If you go to the second order expansion, you represent locally the function by a paraboloid and so on.

• You are very welcome. I hope and wish that my explanation was not too simplistic (but this is really how I always percieved the problem. Cheers :) – Claude Leibovici Jul 3 '14 at 18:26
• +1 your answer was very good! Thank you very much :) No problem at all ;D – jjepsuomi Jul 4 '14 at 5:12

The "quantity" $$d^rf({\bf z}):=\sum_{k=0}^r{r\choose k}{\partial ^rf({\bf z})\over\partial x^k\partial y^{r-k}}dx^k\,dy^{r-k}$$ is a homogeneous polynomial of degree $r$ in the variables $dx$, $dy$ with coefficients the various $r$th order partial derivatives of $f$ at the given point ${\bf z}$. (Originally the polynomial had $2^r$ terms, but only $r+1$ of them really different.) This polynomial collects all terms of total degree $r$ in the Taylor expansion of $f$ at ${\bf z}$: \eqalign{j^n_{\bf z}f(d{\bf z})&=\sum_{r=0}^n {1\over r!}d^rf({\bf z}) \cr&= f({\bf z})+ \bigl(f_x({\bf z})dx+ f_y({\bf z}) dy\bigr)+{1\over2}\bigl(f_{xx}({\bf z})dx^2+2 f_{xy}({\bf z})dx\,dy+ f_{yy}({\bf z})dy^2\bigr)\cr&\ \ \ +{1\over6}\bigl(f_{xxx}({\bf z})dx^3+3 f_{xxy}({\bf z})dx^2\,dy+3 f_{xyy}({\bf z})dx\,dy^2+ f_{yyy}({\bf z})dy^3\bigr)+{1\over24}\ldots\cr}

• +1 Thank you for your help! Appreciate it – jjepsuomi Jul 3 '14 at 16:28