Mathspp: mathematics and programming

Pt En In this post I will share with you my favourite proof that the series of the inverses diverges: $\sum_{i=1}^\infty \frac1i = \infty $. Claim : the series $\sum_i \frac1i$ diverges. Twitter proof : consider the series $$ \begin{align} &\frac12 + \frac12 + \frac12 + \cdots = \\ &\frac12 + 2 \times\frac14 + 4\times \frac18 + \cdots = \\ &\frac12 + \frac14 + \frac14 + \frac18 + \cdots \leq \\ &\frac12 + \frac13 + \frac14 + \frac15 + \cdots \end{align}$$ that clearly diverges because it is a series of a constant nonzero term. By the comparison test, the series of the inverses also diverges. Comment with your favourite way to prove this fact!! Neste post quero partilhar com todos a minha prova preferida de que a série dos inversos dos naturais diverge: $\sum_{i=1}^\infty \frac1i = \infty $. Proposição : a série $\sum_i \frac1i$ diverge. Prova num tweet : considere-se a série$$ \begin{align} &\frac12 + \frac12 + \frac12 + \cdots = \\ &...

Pt En In this post we prove what the minimum number of moves to solve the problem of the Tower of Hanoi is! Claim: let $T(n)$ denote the number of moves it takes to solve the Tower of Hanoi with $n$ disks; then $T(n) = 2^{n}-1$. Twitter proof: note that to solve the problem with $n$ disks, we first have to move the top $n-1$ disks to one of the two poles, move the bottom disk (the bigger one) to the remaining pole, and then move the top $n-1$ disks to the top of the bigger disk. Each time we move the top $n-1$ disks to another pole we must take, at least, $T(n-1)$ moves (by definition of $T$) hence we clearly have $T(n) = 2T(n-1) + 1$. Just notice that $B(n) = 2^n - 1$ satisfies the recurrence relation and that $T(0) = B(0) = 0$. If you are having trouble understanding what I mean by to solve the problem with $n$ disks, we first have to move the top $n-1$ disks to one of the two poles, move the bottom disk (the bigger one) to the remaining pole, and then move t...

Pt En This post is going to follow up what was discussed here , and I will show you how changing the value of the discount factor may affect what the optimal policy is. For that, consider the MDP defined below: This MDP is inspired in the example I have been using; I added an extra state and stripped the MDP of the majority of the transitions so we can focus on what is essential here. From the MDP above there is a clear contender for the title of optimal policy: $$\begin{align} &\pi(H) = E\\ &\pi(T) = D\\ &\pi(TU) = D\end{align}$$ Assume that when we enter the MDP there is a $50\%$ chance we start Hungry and a $50\%$ chance we start Thirsty. If we use the policy $\pi$ then the expected reward is $$E[R | \pi] = 0.5E[R_T | \pi] + 0.5E[R_H | \pi]$$ where $E[R_s | \pi]$ is the expected reward we get by following policy $\pi$ starting from state $s$. From my previous post we know that $E[R_T | \pi] = E[R_H | \pi] = \frac{1}{1-\gamma}$ and hence $$E[R | \pi] ...