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The Problem Solver's Resource

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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 $a_{i_{j}}, 1\le i\le m, 1\le j\le n$ , 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 $x_1,x_2,x_3 \ldots, x_n$ , and $d_1,d_2,d_3 \ldots, d_n$ such that $d_k = \frac{\displaystyle \sum_{ 1\leq i_1 < i_2 < \cdots < i_k \leq n}x_{i_1} x_{i_2} \cdots x_{i_k}}{\displaystyle {n \choose k}}$ , for $k\subset [1,n]$ the following holds:

$d_1 \ge \sqrt[2]{d_2} \ge \sqrt[3]{d_3}\ldots \ge \sqrt[n]{d_n}$

with equality iff all $x_i$ are equivalent.

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@@ Line 13: / Line 13: @@
 ===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</math>, the following holds:
+<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===
 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:

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