## Calculus: Theorems about continuous functions

— 1. Theorems about continuous functions —

— 1.1. Infimum, supremum —

Definition 1 (least upper bound, greatest lower bound). A set ${A}$ of real numbers is bounded above if there exists a number ${x}$ such that ${x \ge a}$ for all ${a \in A}$. Such ${x}$ is called an upper bound of ${A}$.

A number ${x}$ is callead least upper bound of ${A}$ if ${x}$ is an upper bound of ${A}$ and ${x \le y}$ for all upper bound ${y}$ of ${A}$.

Lower bound and greatest lower bound are defined similarly.

Before introducing the theorems, it is necessary to note that the set of real numbers has an important property, call least-upper-bound property, as this will be used in the proofs.

Theorem 2 (least-upper-bound property of ${\mathbf{R}}$). For every nonempty subset ${S \subset \mathbf{R}}$ that is bounded above has a least upper bound, i.e. ${\sup S}$ is in ${\mathbf{R}}$.

## Calculus: Limits

My writing about Calculus when I was/am taking MATH1051, MATH2400 and reading Spivak Calculus book.

— 1. Definition of limits —

— 1.1. Definition of limits of function —

For me, the precise definition of limits is a bit hard to understand so I decided to reconstruct the whole thing by writing some of my guesses on why did mathematicians write the definition in this way. By the way, French mathematician, Augustin-Louis Cauchy (1789-1857) is the one who introduced ${\delta, \varepsilon}$ into definition of limits.

Limit, the notation ${\lim_{x \rightarrow a}f(x)=\ell}$ is understood vaguely as: as ${x}$ gets closer to ${a}$, ${f(x)}$ gets closer to a number ${\ell}$. What does it mean by ‘get closer’? It is easier to understand if we imagine ${a}$ lies on a number line and suppose we pick an arbitrary point ${x}$ on the line. If we apply the term ‘${x}$ gets closer to ${a}$‘, we can understand it as distance from ${x}$ to ${a}$ on the number line decreases. Its distance is ${|x-a|}$ so saying ${x}$ gets closer to ${a}$ is the same as saying ${|x-a|}$ gets closer to ${0}$. Note that for any distance ${|x-a|}$, there are two possible ${x}$ on the number line that can satisfy that distance. What we care about is the distance from ${x}$ to ${a}$ not the position of ${x}$, so using ‘${|x-a|}$ gets closer to ${0}$‘ is better than ‘${x}$ gets closer to ${a}$‘. Similarly to ${|f(x)-\ell|}$ and we get a new definition:

Definition 1. The symbol ${\lim_{x \rightarrow a}f(x)=\ell}$ means that as ${|x-a|}$ gets closer to ${0}$ then ${|f(x)-\ell|}$ gets closer to ${0}$.

Note that we can get ${|x-a|}$ as close to ${0}$ as we want by choosing appropriate ${x}$, but we can’t guarantee the same thing for ${|f(x)-\ell|}$ since ${f(x)}$ depends on ${x}$. Thus, firsly, in order for ${|f(x)-\ell|}$ to be able to approach ${0}$ we need ${|f(x)-\ell|}$ to be as small as we want. It is equivalent to saying:

For any ${\varepsilon>0}$ there exists some ${x}$ so ${\varepsilon>|f(x)-\ell|}$.

Secondly, we need the definition to say ${|f(x)-\ell|}$ approaches ${0}$ as ${|x-a|}$ approaches ${0}$. This means for small enough ${\delta}$, ${|f(x)-\ell|}$ must be bounded by some ${\varepsilon}$ for all ${|x-a|<\delta}$. For better understanding more about this argument, let’s assume the contrary, then as ${\delta}$ gets closer to ${0}$, ${|f(x)-\ell|}$ still does not have a bound for all ${|x-a|<\delta}$. Say we find at ${|x-a|=\delta_1<\delta}$ has ${|f(x)-\ell |=M_1}$. We consider all ${x}$ in ${|x-a|<\delta_1}$ and follow that there must be an ${x}$ so ${|x-a|=\delta_2<\delta_1}$ and ${|f(x)-\ell|=M_2>M_1}$ according to the assumption. Next consider all ${x}$ in ${|x-a| < \delta_2}$ and from our assumption, there exists ${x}$ so ${|f(x)-\ell |=M_3>M_2}$ for ${|x-a|=\delta_3<\delta_2}$. This keeps going and we can see that as ${|x-a|}$ gets closer to ${0}$, ${|f(x)-\ell |}$ gets larger, which contradicts to our aim that ${f(x)}$ must approaches ${\ell}$. Thus, we got the second part of the definition: (more…)

## Reading a solution using algebraic integers of a HMMT number theory problem

I tried to attack the following problem but eventually gave up.

Problem 1 (HMMT Ferb 2015 Team P9) Let ${z=e^{\tfrac{2\pi i}{201}}}$ and ${\omega= e^{\tfrac{2 \pi i}{10}}}$. Prove that

$\displaystyle A=\displaystyle \prod_{a=0}^{9}\prod_{b=0}^{100}\prod_{c=0}^{100}(\omega^a+z^b+z^c)$

is an integer and find its remainder upon division by ${101}$

The only idea I have is to expand this expression, but it was too complex to do. So I decided to read the solution using algebraic integers. I’m new at algebraic number theory and I don’t know much about algebraic integers. My only reference is Problems from the book (PFTB), Chapter 9.

## Yellowstone Permutation – St. Petersburg MO 2015, grade 9, 2nd round

This problem was given to the St. Petersburg Olympiad 2015, 2nd round, grade 9 and is known as Yellowstone Permutation. You can see here for the article about this sequence and here for the OEIS entry.

Problem. (Yellowstone Permutation) A sequence of integers is defined as follows: $a_1=1,a_2=2,a_3=3$ and for $n>3$, $a_n$ is the smallest integer not occurring earlier, which is relatively prime to $a_{n-1}$ but not relatively prime to $a_{n-2}.$ Prove that every natural number occurs exactly once in this sequence.

M. Ivanov

Here is my proof for this interesting problem. Another proof can be found in the article given link above.

## The normal approximation to the binomial distribution

A coin is flipped $n$ times with probability of getting heads is $p$. This is a binomial approximation $X \sim B(n,p)$. The Year 12 MATH B textbook gives an normal approximation to this: $B(n,p) \sim N(np,np(1-p))=N(\mu, \sigma^2)$ where $\mu=E[X]$ is the mean (or expected value) and $\sigma$ is the standard deviation. Since the textbook doesn’t give a proof for this so I will go and prove.

## IMO 2016

IMO 2016 was held from 6/7/2016 to 16/07/2016. Although I didn’t attend this, I just want to post my solutions and how I think about the difficulty of the problems, which can be found here. I will post my solutions on this post. Many awesome solutions can be found on AoPS.

## Bài hình học Tuần 4 tháng 12 – 2015 của thầy Trần Quang Hùng

Thầy Trần Quang Hùng có một chuyên mục Mỗi tuần một bài toán trên blog của thầy. Dưới đây là bài toán của thầy Hùng đưa lên trong Tuần 4 tháng 12.

PROBLEM. Cho tam giác $ABC$ nhọn với đường cao $AH$ và tâm ngoại tiếp $O$. Đường thẳng qua $H$ vuông góc với $OH$ lần lượt cắt $CA,AB$ tại $E,F$. Gọi $M,N$ theo thứ tự là trực tâm của tam giác $OFB$$OHB$$P,Q$ theo thứ tự là trực tâm tam giác $OEC, OHC$. Chứng minh rằng $MN,PQ,EF$ đồng quy.

Sau đây là lời giải của tôi được đưa lên trong topic Mỗi tuần một bài toán trên DDTH.

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