Difference between revisions of "User:Superagh"
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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. | 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, \dotsc, a_n, b_1, b_2, \dotsc, b_n, \dotsc, z_1, z_2, \dotsc, z_n</math> are nonnegative real numbers and <math>\lambda_a, \lambda_b, \dotsc, \lambda_z</math> are nonnegative reals with sum of 1, then: | ||
+ | <cmath>a_1^{\lambda_a}b_1^{\lambda_b} \dotsm z_1^{\lambda_z} + \cdots &+ a_n^{\lambda_a} b_n^{\lambda_b} \dotsm 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== | ==Combinatorics== |
Revision as of 19:58, 24 June 2020
Contents
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 ,
where
. Define the
as:
where
, and:
where
.
If , then
Power mean (weighted)
Statement: Let be positive real numbers. Let
be positive real numbers ("weights") such that
. For any
,
if ,
.
if ,
.
If , then
. Equality occurs if and only if all the
are equal.
Cauchy-Swartz Inequality
Let there be two sets of integers, and
, such that
is a positive integer, where all members of the sequences are real, then we have:
Equality holds if for all
, where
,
, or for all
, where
,
., or we have some constant
such that
for all
.
Bernoulli's Inequality
Given that ,
are real numbers such that
and
, we have:
Rearrangement Inequality
Given that and
We have:
is greater than any other pairings' sum.
Holder's Inequality
If are nonnegative real numbers and
are nonnegative reals with sum of 1, then:
\[a_1^{\lambda_a}b_1^{\lambda_b} \dotsm z_1^{\lambda_z} + \cdots &+ a_n^{\lambda_a} b_n^{\lambda_b} \dotsm z_n^{\lambda_z} \le{}& (a_1 + \cdots + a_n)^{\lambda_a} (b_1 + \cdots + b_n)^{\lambda_b} \cdots (z_1 + \cdots + z_n)^{\lambda_z} .\] (Error compiling LaTeX. Unknown error_msg)
This is a generalization of the Cauchy Swartz Inequality.