|
|
| Line 1: |
Line 1: |
| − | ==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 <math>a_1, a_2, a_3, ... a_n > 0</math>, <math>a_{i} \in \mathbb{R}</math> where <math>1 \le i \le n</math>. Define the <math>pm_x(a_1, a_2, \cdots , a_n)</math> as:<cmath>(\frac{a_1^x+a_2^x+\cdots+a_n^x}{n})^{\frac{1}{x}},</cmath>where <math>x\neq0</math>, and:<cmath>\sqrt[n]{a_{1}a_{2}a_{3} \cdots a_{n}}.</cmath>where <math>x=0</math>.
| |
| − |
| |
| − | 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)
| |
| − | 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>,
| |
| − |
| |
| − | if <math>r=0</math>,
| |
| − |
| |
| − | <cmath>P(r)=a_1^{w_1} a_2^{w_2} a_3^{w_3} . . . a_n^{w_n}</cmath>.
| |
| − |
| |
| − | if <math>r \neq 0</math>,
| |
| − |
| |
| − | <cmath>P(r)=(w_1a_1^r+w_2a_2^r+w_3a_3^r . . . +w_na_n^r)^{\frac{1}{r}}</cmath>.
| |
| − |
| |
| − | If <math>r>s</math>, then <math>P(r) \geq P(s)</math>. Equality occurs if and only if all the <math>a_i</math> are equal.
| |
| − |
| |
| − | ====Cauchy-Swartz Inequality====
| |
| − | Let there be two sets of integers, <math>a_1, a_2, \cdots a_n</math> and <math>b_1, b_2, \cdots b_n</math>, such that <math>n</math> is a positive integer, where all members of the sequences are real, then we have:<cmath>(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.</cmath>Equality holds if for all <math>a_i</math>, where <math>1\le i \le n</math>, <math>a_i=0</math>, or for all <math>b_i</math>, where <math>1\le i \le n</math>, <math>b_i=0</math>., or we have some constant <math>k</math> such that <math>b_i=ka_i</math> for all <math>i</math>.
| |
| − |
| |
| − | ====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
| |
| − | 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.
| |
| − |
| |
| − | ====Holder's Inequality====
| |
| − | 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.
| |
| − |
| |
| − | ==Combinatorics==
| |
| − | ==Number Theory==
| |
| − | ==Geometry==
| |
