This term we’d like to encourage class participation through use of this blog. Therefore, your first assignment is to:

- create an account, and
- leave a comment on this post.

To make things interesting, your comment should include a description of your favorite mathematical formula typeset in $\LaTeX$. If you don’t know how to use $\LaTeX$ this is a great opportunity to learn — a very basic introduction can be found here. (And if you don’t have a favorite mathematical formula, this is a great time to pick one!) Note that if you’d rather not use your real name in public you’re welcome to register under a pseudonym.

First!

Also, $f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta – z} d\zeta$

For holomorphic $f$, $C^1 \gamma$, and some other constraints.

I don’t know if it’s my

favoritebut it certainly blew my mind at first.I think the important condition here is that $\gamma$ is a closed loop, which is what makes this theorem so nice: it effectively says that a holomorphic function is completely determined by its values on the boundary! This fact will become useful when we talk about conformal surface parameterization. (Interestingly enough, it’s also a consequence of Stokes’ theorem, which Justin mentions below. )

A related thing which I also really really like is the Argument principle, which is highly related to the above.

Hmm… does it state that no matter what you say your TA will try to argue with you?

Not sure if getting trolled by TA….

If so…yup!

If not, the argument principle states that for meromorphic function $f$ and a closed curve $C$, with $f$ having no zeroes or poles on $C$:

$\begin{align}

\oint_C \frac{f’(z)}{f(z)} dz = 2\pi i(N – P)

\end{align}$

where $N$ is the number of zeroes and $P$ is the number of poles in the area enclosed by $C$ counted by multiplicity. I think this is almost equally mind-blowing to the Cauchy Integral Formula.

Stokes’ Theorem is pretty cool. Let $\mathbf{F}:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a vector field and let $\mathbf{\Sigma}$ be a surface in $\mathbb{R}^3$. Then

\[\int_{\Sigma} \nabla\times\mathbf{F}\cdot d\;\mathbf{\Sigma} = \oint_{\partial\Sigma}\mathbf{F}\cdot d\;\mathbf{r}\]

Great! It turns out Stokes’ theorem is pretty fundamental when talking about discretization. In class we’ll see an even more general version of Stokes’ using differential forms.

I kind of like $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

Cute.

How about a favorite theorem? My favorite theorem is Barrington’s Theorem from complexity theory (width-5 branching programs are equivalent in power to the class $NC_1$).

In terms of formulae, I’ve always liked the Biot-Savart law, for computing the magnetic field due to current in a wire:

$$\mathbf{B} = \int_C \frac{\mu_0 I (dL \times \hat{r})}{4 \pi |r|^2}$$

Interestingly enough, the Biot-Savart law also comes up in the context of the incompressible fluid flow mentioned below. There have been a couple nice papers recently which make use of Biot-Savart to numerically integrate a discrete smoke ring flow — see

hereandhere.I’m gonna have to go with Navier-Stokes, the momentum equations for viscous flow. For the sake of my carpal tunnel, only the x-component:

$\rho \frac{Du}{Dt}= – \frac{\partial p}{\partial x} + \frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z}$

$\rho$ is density of the fluid, u is the x-component of velocity, p is pressure, and $\tau_{ij}$ is tangential stress (and $\tau_{ii}$ is normal stress). Combined with the continuity and energy equations, they let you analyze incompressible unsteady 3D viscous flow. Without these equations, there’d be no modern CFD-designed planes!

Very nice. One way to enforce incompressibility numerically is to do a projection via

Helmholtz-Hodge decomposition— in class we’ll see how to do this kind of thing in the discrete setting.There’s always the simple yet ever so deep Euler’s Identity.

Let \(i=-1^{1/2}\) and \(\pi\) be the circumference of a circle divided by twice the radius, then

\(e^{i \pi}=-1\)

A surprising result.

Beautiful. Yes, this is the kind of thing that you’d be justified getting as a tattoo*.

(*The views expressed here do not necessarily reflect the opinion of the California Institute of Technology.)

I’m fond of: \(a^{\phi(n)} \equiv 1 (mod \ {n}) \)

Hmm… ok, this one has me stumped — definitely not from my neck of the woods. What is it?