Parker Glynn-Adey

Public Talks for UTSC

Posted in Math by pgadey on 2018/07/05

From Colourings to Fixed Points

The notes for the talk are available here.

Uniform Convergence

The notes for the talk are available here.

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Keeping Sets Different

Posted in Math by pgadey on 2011/08/19

Consider \mathcal{B} the Borel \sigma-algebra of Lebesgue measurable subsets of [0,1]. We then define a pseudometric d(A,B) = \mu(A \Delta B) on \mathcal{B} where A \Delta B = A \setminus B \cup B \setminus A is the symmetric difference of sets. Ignoring sets of zero measure, we have that d is indeed a metric. We wish to show that \mathcal{B} is not sequentially compact in its metric.

Write x = d_0.d_1d_2\dots for the binary expansion of x \in [0,1]. Consider E_i = \{x = d_0.d_1d_2\dots\ :\ d_i = 1\}. We compute E_i \Delta E_j = \{x : d_i \neq d_j\}. If we think probabilistically, where the digits d_1 and d_j represent independent coin tosses, we get: \mu(E_i \Delta E_j) = 1/2 for i \neq j. Thus d(E_i,E_j) is constant for i \neq j and hence \{E_i\} can have no convergent subsequences

This came from the September 2005 UoT Analysis comprehensive. The solution is due to Dror Bar-Natan.

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BPM 1.4.2 (Sp 84)

Posted in Math by pgadey on 2011/03/10

Let f(x) = x \log(1 + 1/x) for 0 < x < \infty. We show that f is monotonically increasing. First we rewrite f(x) = x \log \left( \frac{x + 1}{x} \right) = x [ \log(x+1) - \log(x)]. Taking derivatives, and doing some algebra, we get f'(x) = [\log(x + 1) - \log(x)] - 1/(x+1). We wish to show that 0 < f'(x) everywhere. This would follow from 1/(x+1) < \log(x + 1) - \log(x). However, this follows from the convexity of \log(x). Which says that 1/(x + \epsilon) < [\log(x + \epsilon) - \log(x)] / \epsilon for all 0 < x, \epsilon.

We compute \lim_{x \rightarrow 0} f(x). Taking x_n = 1/(e^n - 1) we see that f(x_n) = 1/(e^n - 1) \log(e^n) = n/(e^n - 1). We then have that f(x_n) \rightarrow 0. Since f is monotonically increasing, and positive, we have that \lim f(x) = 0 as x \rightarrow 0. We compute \lim_{x \rightarrow \infty} f(x). Applying concavity, we have f(e^n) = e^n \log(1 + e^{-n}) \geq e^n \epsilon e^{-n} = \epsilon for all 0 < \epsilon < 1. We then check that f(x) = x \log(1 + 1/x) \stackrel{\star}{\leq} x (1/x) \leq 1, where \star is by applying \log(1 + y) \leq y. It follows that \lim_{x \rightarrow \infty} f(x) = 1.

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