16 October, 2013

Erdős' Anti-Calculus Proposition


Proposition

Erdös contends that in Complex Analysis, versus Single Variable Calculus, a function will not have a derivative of zero when it reaches a maximum value.

In Single Variable Calculus:

A function \(f(x)\), which is smooth (continuous and differentiable) on a open domain \(D\), will have a derivative of zero when it attains a maximum on \(D\).

However, for a complex function \(g(z)\), on a closed domain \(U\) with border \(B\), the maximum (Guaranteed by the Extreme Value Theorem), may not have a derivative of zero.

Proof

Let there exist a complex \(\epsilon\) such that \(|\epsilon| > 0\) and \(\delta\) such that \(|\delta| > 0\) where \(\epsilon\) and \(\delta\) are arbitrarily small.

Furthermore, let the complex function \(f(z)\) be analytic and non-constant on an open domain \(D\), bounded by a contour \(B\) along which \(f(z)\) is continuous and differentiable. Thus, \(f\) is continuous and differentiable on the closed region \(U = D \cup B\).

By the Maximum Modulus Theorem, the maximum of \(f\) occurs on \(B\) at \(z_0\).

Now, suppose that \(f'(z_0)\) is indeed zero. This would mean that for some \(\epsilon\), there is are points \(w = z_0 + \delta\) where \(\delta < \epsilon\) for which \(f(w)=f(z_0)\), since \(f(z)\) is continuous. \(z_0 + \delta\) is a maximum point for \(f(z)\) on \(D\). By the Maximum Modulus Theorem, this means that \(f\) must be constant on \(D\).

Since \(f(z)\) is non-constant on \(D\), this is a contradiction. \(#\)

14 October, 2013

Some Complex Math--the Relation Between Analyticity and Antidifferentiability

In Complex Analysis, the field of mathematics relying on doing almost all types of regular mathematics, but in the Complex Number plane, as opposed to the Real Number line, the concept of Analyticity, the state of a function having the property of being Analytic, is central.

For a function \(f(x)\) to be analytic in the Real Numbers, it must have some power series representation \(\sum{c_n {(x-x_0)}^n}\). A function satisfies this property if it is infinitely differentiable, meaning that \(f'(x), f''(x),\frac{d^3 f}{d x^3}, ..., \frac{d^\infty f}{d x^\infty}\) all exist, and are continuous. Most continuously differentiable functions are analytic in \(R^1\), but there are some functions that are continuous and differentiable that do not satisfy this property, for example, the non-analytic, yet smooth function:
$$f(x)=\begin{cases}\exp(-1/x)&\text{if }x>0,\\ 0&\text{if }x\le0,\end{cases}$$

The proof of Non-analyticity is here.

In the Complex Field, Analyticity is a slightly different animal. In general, if a function, \(f(z)\) is differentiable on some open domain \(D\), it is analytic on that domain. It can then be proved that if a function \(f(z)\) is analytic, it is infinitely differentiable on the same open domain \(D\).

Now, my quandry is thus:

In contour integration of complex functions, one of the more important concepts is the idea of path-independence, or conservativity. A function \(f\) is said to be conservative iff (if and only if) \(\int_a^b {f(z(t)) dt}\) evaluates to the same value no matter what simple smooth path \(z(t)\) represents, so long as \(a\) and \(b\) are the same.

If a function is conservative, then \(\int_\Gamma f(z) dz\) is simple. First, find the antiderivative of \(f\) (its existence is guaranteed based upon its conservativity), next, find the endpoints of \(\Gamma\), and finally, integrate \(f\) from the beginning of  \(\Gamma\) to the end using the Fundamental Theorem of  Calculus.

$$\int_\Gamma f(z) dz = \int\limits_a ^b f(z) dz = (F(b)-F(a))$$

Now, the guarantee of the existence of an antiderivative for \(f\) is bidirectional, meaning that, if \(f\) is continuous, and it has an antiderivative, it must therefore be conservative. This turns out to be extremely powerful, as the analyticity of \(f\) implies that \(f\) is both continuous and has an antiderivative over the domain of analyticity.

My issue is that while it seems obvious that if \(f\) is infinitely differentiable, there must also be a function, \(g\) for which \(g'=f\), the proof for that statement is very non-obvious for me. I believe the answer is Green's theorem, but it sadly uses a very, very long proof that goes through the classification of all possible types of Analytic functions.

Anyways, I hope that my little rant about analyticity has entertained those whose knowledge of complex calculus is greater than mine, and educated those whose knowledge of complex of Calculus was increased by this.

13 October, 2013

Present Tense Verbs, Their Meanings and Their Conjugations

Verb Conjugations



A note on pronunciation:
For the most part, all consonants are pronounced as they would be in English, with one notable exception: v is the Bilabial voiced fricative, rather than the labial voiced fricative it is in English.

The vowels are as follows:

  • a: the ah sound in far
  • e: the eh sound them
  • i: the ee sound in deem
  • o: the o sound in bone
  • u: the oo sound in boon
  • ä: the ay sound in hay
  • á,é,í,ó,ú,'ä: This indicates that there is a stop before the beginning of the vowel. It is used when there are two vowels right next to each other, but the second is in a separate syllable from the first.