Difference between revisions of "User:Superagh"

(Power mean (weighted))
(Power mean (weighted))
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====Power mean (weighted)====
 
====Power mean (weighted)====
For all positive reals <math>a_1, \ldots, a_n</math> and nonnegative reals <math>p_1, \ldots, p_n</math> with <math>p_1+p_2+\cdots + p_n=1</math>, we have
 
<cmath>\prod_{i=1}^n a_i^{p_i} \leq \sum_{i=1}^n p_ia_i.</cmath>
 
  
 
==Combinatorics==
 
==Combinatorics==

Revision as of 15:26, 24 June 2020

Introduction

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)

Combinatorics

Number Theory

Geometry