Difference between revisions of "Superagh's Olympiad Notes"

Latest revision as of 12:59, 22 July 2020

Introduction

SINCE MY COMPUTER WON'T LOAD THIS FOR SOME REASON, I'LL BE UPDATING THIS AS I GO THOUGH :)

Ok, so inspired by master math solver Lcz, I have decided to take Oly notes (for me) online! I'll probably be yelled at even more for staring at the computer, but I know that this is for my good. (Also this thing is almost the exact same format as Lcz's :P ). (Ok, actually, a LOT of credits to Lcz)

NEVERYMIND I"M DOING THIS ON MY BLOG SINCE IT"LL LOAD

Algebra

Problems worth noting/reviewing I'll leave this empty for now, I want to start on HARD stuff yeah!

Inequalities

We shall begin with INEQUALITIES! They should be fun enough. I should probably begin with some theorems.

Power mean (special case)

Statement: Given that $a_1, a_2, a_3, ... a_n > 0$ , $a_{i} \in \mathbb{R}$ where $1 \le i \le n$ . Define the $pm_x(a_1, a_2, \cdots , a_n)$ as: $(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^{\frac{1}{x}},$ where $x\neq0$ , and: $\sqrt[n]{a_{1}a_{2}a_{3} \cdots a_{n}}.$ where $x=0$ .

If $x \ge y$ , then $pm_x(a_1, a_2, \cdots , a_n) \ge pm_y(a_1, a_2, \cdots , a_n).$

Power mean (weighted)

Statement: Let $a_1, a_2, a_3, . . . a_n$ be positive real numbers. Let $w_1, w_2, w_3, . . . w_n$ be positive real numbers ("weights") such that $w_1+w_2+w_3+ . . . w_n=1$ . For any $r \in \mathbb{R}$ ,

if $r=0$ ,

$P(r)=a_1^{w_1} a_2^{w_2} a_3^{w_3} . . . a_n^{w_n}$ .

if $r \neq 0$ ,

$P(r)=(w_1a_1^r+w_2a_2^r+w_3a_3^r . . . +w_na_n^r)^{\frac{1}{r}}$ .

If $r>s$ , then $P(r) \geq P(s)$ . Equality occurs if and only if all the $a_i$ are equal.

Cauchy-Swartz Inequality

Let there be two sets of integers, $a_1, a_2, \cdots a_n$ and $b_1, b_2, \cdots b_n$ , such that $n$ is a positive integer, where all members of the sequences are real, then we have: $(a_1^2+a_2^2+\cdots +a_n^2)(b_1^2+b_2^2+ \cdots +b_n^2)\ge (a_1b_1 + a_2b_2 + \cdots +a_nb_n)^2.$ Equality holds if for all $a_i$ , where $1\le i \le n$ , $a_i=0$ , or for all $b_i$ , where $1\le i \le n$ , $b_i=0$ ., or we have some constant $k$ such that $b_i=ka_i$ for all $i$ .

Bernoulli's Inequality

Given that $n$ , $x$ are real numbers such that $n\ge 0$ and $x \ge -1$ , we have: $(1+x)^n \ge 1+nx.$

Rearrangement Inequality

Given that $x_1 \ge x_2 \ge x_3 \cdots x_n$ and $y_1 \ge y_2 \ge y_3 \cdots y_n.$ We have: $x_1y_1+x_2y_2 + \cdots + x_ny_n$ is greater than any other pairings' sum.

Holder's Inequality

If $a_1, a_2, \cdots, a_n$ , $b_1, b_2, \cdots, b_n$ , $\cdots$ , $z_1, z_2, \cdots, z_n$ are nonnegative real numbers and $\lambda_a, \lambda_b, \cdots, \lambda_z$ are nonnegative reals with sum of $1$ , then: $a_1^{\lambda_a}b_1^{\lambda_b} \cdots z_1^{\lambda_z} + \cdots + a_n^{\lambda_a} b_n^{\lambda_b} \cdots z_n^{\lambda_z} \le (a_1 + \cdots + a_n)^{\lambda_a} (b_1 + \cdots + b_n)^{\lambda_b} \cdots (z_1 + \cdots + z_n)^{\lambda_z} .$ This is a generalization of the Cauchy Swartz Inequality.

Combinatorics

ok look bro

$a^2+b^2=c^2$

lol

$\phi{n} \equiv 1 \pmod{asdf}$ .

Number Theory

Geometry

hahaha

geometry sucks

~Lcz 5:02 PM CST, 6/25/2020

@@ Line 3: / Line 3: @@
 Ok, so inspired by master math solver Lcz, I have decided to take Oly notes (for me) online! I'll probably be yelled at even more for staring at the computer, but I know that this is for my good. (Also this thing is almost the exact same format as Lcz's :P ). (Ok, actually, a LOT of credits to Lcz)
+NEVERYMIND I"M DOING THIS ON MY BLOG SINCE IT"LL LOAD
 ==Algebra==
@@ Line 15: / Line 17: @@
 If <math>x \ge y</math>, then<cmath>pm_x(a_1, a_2, \cdots , a_n) \ge pm_y(a_1, a_2, \cdots , a_n).</cmath>
-Power mean (weighted)
+====Power mean (weighted)====
 Statement: Let <math>a_1, a_2, a_3, . . . a_n</math> be positive real numbers. Let <math>w_1, w_2, w_3, . . . w_n</math> be positive real numbers ("weights") such that <math>w_1+w_2+w_3+ . . . w_n=1</math>. For any <math>r \in \mathbb{R}</math>,
@@ Line 33: / Line 35: @@
 ====Bernoulli's Inequality====
 Given that <math>n</math>, <math>x</math> are real numbers such that <math>n\ge 0</math> and <math>x \ge -1</math>, we have:<cmath>(1+x)^n \ge 1+nx.</cmath>
-Rearrangement Inequality
+====Rearrangement Inequality====
 Given that<cmath>x_1 \ge x_2 \ge x_3 \cdots x_n</cmath>and<cmath>y_1 \ge y_2 \ge y_3 \cdots y_n.</cmath>We have:<cmath>x_1y_1+x_2y_2 + \cdots + x_ny_n</cmath>is greater than any other pairings' sum.
@@ Line 40: / Line 42: @@
 ==Combinatorics==
+ok look bro
+<math>a^2+b^2=c^2</math>
+lol
+<math>\phi{n} \equiv 1 \pmod{asdf}</math>.
 ==Number Theory==
 ==Geometry==
+hahaha
+geometry sucks
+~Lcz 5:02 PM CST, 6/25/2020

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