Difference between revisions of "User:Temperal/The Problem Solver's Resource11"

(include more)
(Holder's Inequality)
Line 14: Line 14:
 
===Holder's Inequality===
 
===Holder's Inequality===
 
For positive real numbers <math>a_{i_{j}}, 1\le i\le m, 1\le j\le n be</math>, the following holds:
 
For positive real numbers <math>a_{i_{j}}, 1\le i\le m, 1\le j\le n be</math>, the following holds:
<math>
+
 
 
<cmath>\prod_{i=1}^{m}\left(\sum_{j=1}^{n}a_{i_{j}}\right)\ge\left(\sum_{j=1}^{n}\sqrt[m]{\prod_{i=1}^{m}a_{i_{j}}}\right)^{m}</cmath>  
 
<cmath>\prod_{i=1}^{m}\left(\sum_{j=1}^{n}a_{i_{j}}\right)\ge\left(\sum_{j=1}^{n}\sqrt[m]{\prod_{i=1}^{m}a_{i_{j}}}\right)^{m}</cmath>  
 
===Muirhead's Inequality===
 
===Muirhead's Inequality===
For a sequence </math>A<math> that majorizes a sequence </math>B<math>, then given a set of positive integers </math>x_1,x_2,\ldots,x_n<math>, the following holds:
+
For a sequence <math>A</math> that majorizes a sequence <math>B</math>, then given a set of positive integers <math>x_1,x_2,\ldots,x_n</math>, the following holds:
  
 
<cmath>\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\ldots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</cmath>
 
<cmath>\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\ldots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}</cmath>
 
===Rearrangement Inequality===
 
===Rearrangement Inequality===
For any multi sets </math>{a_1,a_2,a_3\ldots,a_n}<math> and </math>{b_1,b_2,b_3\ldots,b_n}<math>, </math>a_1b_1+a_2b_2+\ldots+a_nb_n<math> is maximized when </math>a_k<math> is greater than or equal to exactly </math>i<math> of the other members of </math>A<math>, then </math>b_k<math> is also greater than or equal to exactly </math>i<math> of the other members of </math>B<math>.
+
For any multi sets <math>{a_1,a_2,a_3\ldots,a_n}</math> and <math>{b_1,b_2,b_3\ldots,b_n}</math>, <math>a_1b_1+a_2b_2+\ldots+a_nb_n</math> is maximized when <math>a_k</math> is greater than or equal to exactly <math>i</math> of the other members of <math>A</math>, then <math>b_k</math> is also greater than or equal to exactly <math>i</math> of the other members of <math>B</math>.
 
===Newton's Inequality===
 
===Newton's Inequality===
For non-negative real numbers </math>x_1,x_2,x_3\ldots,x_n<math> and </math>0 < k < n<math> the following holds:
+
For non-negative real numbers <math>x_1,x_2,x_3\ldots,x_n</math> and <math>0 < k < n</math> the following holds:
  
 
<cmath>d_k^2 \ge d_{k-1}d_{k+1}</cmath>,
 
<cmath>d_k^2 \ge d_{k-1}d_{k+1}</cmath>,
  
with equality exactly iff all </math>x_i<math> are equivalent.  
+
with equality exactly iff all <math>x_i</math> are equivalent.  
 
===Mauclarin's Inequality===
 
===Mauclarin's Inequality===
For non-negative real numbers </math>x_1,x_2,x_3 \ldots, x_n<math>, the following holds:
+
For non-negative real numbers <math>x_1,x_2,x_3 \ldots, x_n</math>, the following holds:
  
 
<cmath>x_1 \ge \sqrt[2]{x_2} \ge \sqrt[3]{x_3}\ldots \ge \sqrt[n]{x_n}</cmath>
 
<cmath>x_1 \ge \sqrt[2]{x_2} \ge \sqrt[3]{x_3}\ldots \ge \sqrt[n]{x_n}</cmath>
  
with equality iff all </math>x_i$ are equivalent.  
+
with equality iff all <math>x_i</math> are equivalent.
 
[[User:Temperal/The Problem Solver's Resource10|Back to page 10]] | Last page (But also see the  
 
[[User:Temperal/The Problem Solver's Resource10|Back to page 10]] | Last page (But also see the  
 
[[User:Temperal/The Problem Solver's Resource Tips and Tricks|tips and tricks page]], and the  
 
[[User:Temperal/The Problem Solver's Resource Tips and Tricks|tips and tricks page]], and the  
 
[[User:Temperal/The Problem Solver's Resource Competition|competition]]!
 
[[User:Temperal/The Problem Solver's Resource Competition|competition]]!
 
|}<br /><br />
 
|}<br /><br />

Revision as of 11:25, 13 October 2007



The Problem Solver's Resource
Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 11.

Advanced Number Theory

These are Olympiad-level theorems and properties of numbers that are routinely used on the IMO and other such competitions.

Jensen's Inequality

For a convex function $f(x)$ and real numbers $a_1,a_2,a_3,a_4\ldots,a_n$ and $x_1,x_2,x_3,x_4\ldots,x_n$, the following holds:

\[\sum_{i=1}^{n}a_i\cdot f(x_i)\ge f(\sum_{i=1}^{n}a_i\cdot x_i)\]

Holder's Inequality

For positive real numbers $a_{i_{j}}, 1\le i\le m, 1\le j\le n be$, the following holds:

\[\prod_{i=1}^{m}\left(\sum_{j=1}^{n}a_{i_{j}}\right)\ge\left(\sum_{j=1}^{n}\sqrt[m]{\prod_{i=1}^{m}a_{i_{j}}}\right)^{m}\]

Muirhead's Inequality

For a sequence $A$ that majorizes a sequence $B$, then given a set of positive integers $x_1,x_2,\ldots,x_n$, the following holds:

\[\sum_{sym} {x_1}^{a_1}{x_2}^{a_2}\ldots {x_n}^{a_n}\geq \sum_{sym} {x_1}^{b_1}{x_2}^{b_2}\cdots {x_n}^{b_n}\]

Rearrangement Inequality

For any multi sets ${a_1,a_2,a_3\ldots,a_n}$ and ${b_1,b_2,b_3\ldots,b_n}$, $a_1b_1+a_2b_2+\ldots+a_nb_n$ is maximized when $a_k$ is greater than or equal to exactly $i$ of the other members of $A$, then $b_k$ is also greater than or equal to exactly $i$ of the other members of $B$.

Newton's Inequality

For non-negative real numbers $x_1,x_2,x_3\ldots,x_n$ and $0 < k < n$ the following holds:

\[d_k^2 \ge d_{k-1}d_{k+1}\],

with equality exactly iff all $x_i$ are equivalent.

Mauclarin's Inequality

For non-negative real numbers $x_1,x_2,x_3 \ldots, x_n$, the following holds:

\[x_1 \ge \sqrt[2]{x_2} \ge \sqrt[3]{x_3}\ldots \ge \sqrt[n]{x_n}\]

with equality iff all $x_i$ are equivalent. Back to page 10 | Last page (But also see the tips and tricks page, and the competition!