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The Problem Solver's Resource

Introduction \| Other Tips and Tricks \| Methods of Proof \| You are currently viewing page 8.

Intermediate Number Theory

These are more complex number theory theorems that may turn up on the USAMO or Pre-Olympiad tests. This will also cover diverging and converging series, and other such calculus-related topics.

General Mean Inequality

Take a set of functions $m_j(a) = \left({\frac{\sum a_i^j}{n}}\right)^{1/j}$ .

Note that $m_0$ does not exist. The geometric mean is $m_0 = \lim_{k \to 0} m_k$ . For non-negative real numbers $a_1,a_2,\ldots,a_n$ , the following holds:

$m_x \le m_y$ for reals $x<y$ .

I $m_2$ is the quadratic mean, $m_1$ is the arithmetic mean, $m_0$ the geometric mean, and $m_{-1}$ the harmonic mean.

Chebyshev's Inequality

Given real numbers $a_1 \ge a_2 \ge ... \ge a_n \ge 0$ and $b_1 \ge b_2 \ge ... \ge b_n$ , we have

${\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}$ .

Minkowsky's Inequality

Given real numbers $a_1,a_2,...,a_n$ and $b_1,b_2,\ldots,b_n$ , the following holds:

$\sqrt{\sum a_i^2} + \sqrt{\sum b_i^2} \ge \sqrt{\sum (a_i+b_i)^2}$

Nesbitt's Inequality

For all positive real numbers $a$ , $b$ and $c$ , the following holds:

${\frac{a}{b+c}} + {\frac{b}{c+a}} + {\frac{c}{a+b}} \ge {\frac{3}{2}}$ .

Schur's inequality

Given positive real numbers $a,b,c$ and real $r$ , the following holds:

$a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0$ .

Fermat-Euler Identitity

If $gcd(a,m)=1$ , then $a^{\phi{m}}\equiv1\pmod{m}$ , where $\phi{m}$ is the number of relatively prime numbers lower than $m$ .

Gauss's Theorem

If $a|bc$ and $(a,b) = 1$ , then $a|c$ .

Power Mean Inequality

For a real number $k$ and positive real numbers $a_1, a_2, ..., a_n$ , the $k$ th power mean of the $a_i$ is

$M(k) = \left( \frac{\sum_{i=1}^n a_{i}^k}{n} \right) ^ {\frac{1}{k}}$ when $k \neq 0$ and is given by the geometric mean]] of the $a_i$ when $k = 0$ .

Diverging-Converging Theorem

A series $\sum_{i=0}^{\infty}S_i$ converges iff $\lim S_i=0$ .

Errata

All quadratic residues are $0$ or $1\pmod{4}$ and $0$ , $1$ , or $4$ $\pmod{8}$ .

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@@ Line 24: / Line 24: @@
 Given real numbers <math>a_1 \ge a_2 \ge ... \ge a_n \ge 0</math> and <math>b_1 \ge b_2 \ge ... \ge b_n</math>, we have
-%{\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}%.
+<math>{\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}</math>.
 ===Minkowsky's Inequality===

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