# A month, you say?

Much has changed in the last month.

I moved to Rhode Island and began grad school. That’s a pretty big change. I’m suddenly much more focused in my studies again (undeniably a good thing), figuring out what I will do. Solid.

And I’m struggling to get acquainted with the curious structure of classes at Brown. There are many, many calculus classes here, for example. As a tutor, I’m somewhat expected to know these things. Classes on differentiation, integration, fast-paced and (maybe) slow-paced versions, calc I but with vectors incorporated – all of these fell under the blanket heading of Calc I at Georgia Tech. And my feelings are mixed. It’s an interesting idea. The general freedom to make mistakes at Brown is something that I firmly stand behind, though.

But there is one thing that I think is very poorly done – why is there not more interdepartmental cooperation? Brown is an ivy, and we’re close to many other schools that are excellent at many things in math. Why is there no form of cooperation between these universities? This is something that I absolutely must change. Somehow. I’ll work on that.

Ok, let’s actually do some math.

I recently came across a fun paper, The Fundamental Theorem of Algebra: A Most Elementary Proof, by Oswaldo Rio Branco de Oliveira on proving the Fundamental Theorem of Algebra with no bells, whistles, or ballyhoo in general. All that is assumed is the Bolzano-Weierstrass Theorem and that polynomials are continuous. Here is the gist of the proof.

THEOREM: Let $P(z) = a_0 + a_1 z + … + a_n z^n, a_n \not = 0,$ be a complex polynomial with degree $n geq 1$. Then P has a zero.

Proof: We have that $|P| \geq | |a_n| |z|^n – |a_{n-1}||z|^{n-1} – … – |a_0||z||$, and so $\lim_{|z| \to \infty} |P(z)| = \infty$. By continuity, $|P|$ has a global minimum at some $z_0 \in \mathbb{C}$. We suppose wlog that $z_0 = 0$. Then $|P(z)|^2 – |P(0)|^2 \geq 0 \forall z \in \mathbb{C}$. Then we may write $P(z) = P(0) + z^k Q(z)$ for some $k \in \{1, …, n \}$, and where $Q$ is a polynomial and $Q(0) \neq 0$ (the idea being that one factored that part out already).

Pick some $\zeta \in \mathbb{C}$ and substitute $z = r \zeta, r \geq 0$ into the above inequality and dividing by $r^k$, we get: $2 \mathrm{Re} [ \overline{P(0)} \zeta ^k Q(r \zeta)] + r^k |\zeta ^k Q(r \zeta )|^2 \geq 0 \forall r > 0, \forall \zeta$. The left side is a continuous function of r for nonnegative r, and so taking the limit as $r \to 0$, one finds $2 \mathrm{Re} [ \overline{P(0)} \zeta ^k Q(0)] \geq 0, \forall \zeta$.

Now suppose $\alpha := \overline{P(0)}Q(0) = a + b i$. For $k$ odd, setting $\zeta = \pm 1$ and $\zeta = \pm i$ in this inequality lets us conclude that $a = b = 0$. So then we have $P(0) = 0$, and the odd case is complete. Now before I go on, I give a brief lemma, which I’ll not prove here. But it just requires using binomial expansions and keeping track of lots of exponents and factorials.

Lemma (credit given to Estermann): For $\zeta = \left( 1 + \frac{i}{k} \right)^2$ and $k \geq 2$, even, we have that $\mathrm{Re}[\zeta ^k] < 0 < \mathrm{Im} [\zeta ^k]$

For k even, we don’t have the handy cancellation that we used above. But let us choose $\zeta$ as in this lemma, and write $\zeta ^k = x + iy; \quad x < 0, y > o$. Then we can substitute $\zeta ^ k$ and $\overline{ \zeta ^k}$ in the inequality, and a little work shows that $\mathrm{Re}[\alpha (x \pm iy)] = ax \mp by \geq 0$. So $ax \geq 0$ and since $x < 0$, we see $a \leq 0$. But then $a = 0$. Similarly, we get $b = 0$ after considering $\mp by \geq 0$. Then we again see that $P(0) = 0$, and the theorem is proved as long as you believe Etermann’s Lemma.

I like it when such results have relatively simple proofs. The first time I came across the FTA, we used lots of machinery to prove it. Some integration and differentiation on series, in particular.

And now that I’m vaguely settled and now that I see new things routinely, perhaps I’ll update this more. Not at Tao-pace, but something.

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