Difference between revisions of "User:Superagh"

Revision as of 19:05, 24 June 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)

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.

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 ====Holder's Inequality====
-If <math>a_1, a_2, \cdots, a_n, b_1, b_2, \cdots, b_n, \cdots, z_1, z_2, \cdots, z_n</math> are nonnegative real numbers and <math>\lambda_a, \lambda_b, \cdots, \lambda_z</math> are nonnegative reals with sum of 1, then:
+If <math>a_1, a_2, \cdots, a_n</math>, <math>b_1, b_2, \cdots, b_n</math>, <math>\cdots</math>, <math>z_1, z_2, \cdots, z_n</math> are nonnegative real numbers and <math>\lambda_a, \lambda_b, \cdots, \lambda_z</math> are nonnegative reals with sum of <math>1</math>, then:
-<cmath>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} .</cmath> This is a generalization of the Cauchy Swartz Inequality.
+<cmath>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} .</cmath> This is a generalization of the Cauchy Swartz Inequality.
 ==Combinatorics==

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