# Homework 0 Assigned

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

1. create an account, and
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.

### 17 Responses to “Homework 0 Assigned”

1. Ilya Nepomnyashchiy says:

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 favorite but it certainly blew my mind at first.

• Keenan says:

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. )

• Ilya Nepomnyashchiy says:

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

• Keenan says:

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

• Ilya Nepomnyashchiy says:

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.

2. justinj says:

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

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

• Keenan says:

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.

3. awein says:

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

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

• Keenan says:

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 here and here.

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!

• Keenan says:

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.

6. Matanya says:

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.

• Keenan says:

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.)

7. rlawler says:

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

• Keenan says:

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