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Difference between revisions of "User:Temperal/The Problem Solver's Resource8"

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This will also cover diverging and converging series, and other such calculus-related topics.
 
This will also cover diverging and converging series, and other such calculus-related topics.
  
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==Useful facts and Formulas==
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===General Mean Inequality===
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All quadratic resiues are 0 or 1<math>\pmod{4}</math>and  0,1, or 4 <math>\pmod{8}</math>.
 
  
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Fermat-Euler Identitity-If <math>gcd(a,m)=1</math>, then <math>a^{\phi{m}}\equiv1\pmod{m}</math>, where <math>\phi{m}</math> is the number of relitvely prime  numbers lower than <math>m</math>.
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Take a set of functions <math>m_j(a) = \left({\frac{\sum a_i^j}{n}}\right)^{1/j}</math>.
  
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Gauss's Theorem-If <math>a|bc</math> and <math>(a,b) = 1</math>, then <math>a|c</math>.
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Note that <math>m_0</math> does not exist. The geometric mean is <math>\displaystyle m_0 = \lim_{k \to 0} m_k</math>.
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For non-negative real numbers <math>a_1,a_2,\ldots,a_n</math>, the following holds:
  
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<math>m_x \le m_y</math> for reals <math>x<y</math>.
  
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I<math>m_2</math> is the quadratic mean, <math>m_1</math> is the arithmetic mean, <math>m_0</math> the geometric mean, and <math>m_{-1}</math> the harmonic mean.
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===Chebyshev's Inequality===
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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
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%{\frac{\sum a_ib_i}{n}} \ge {\frac{\sum a_i}{n}}{\frac{\sum b_i}{n}}%.
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===Minkowsky's Inequality===
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Given real numbers <math>a_1,a_2,...,a_n</math> and <math>b_1,b_2,\ldots,b_n</math>, the following holds:
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<math>\sqrt{\sum a_i^2} + \sqrt{\sum b_i^2} \ge \sqrt{\sum (a_i+b_i)^2}</math>
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===Nesbitt's Inequality===
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For all positive real numbers <math>a</math>, <math>b</math> and <math>c</math>, the following holds:
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<math>\displaystyle{\frac{a}{b+c}} + {\frac{b}{c+a}} + {\frac{c}{a+b}} \ge {\frac{3}{2}}</math>.
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===Schur's inequality===
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Given positive real numbers <math>a,b,c</math> and real <math>r</math>, the following holds:
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<math>a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0</math>.
  
 
[[User:Temperal/The Problem Solver's Resource7|Back to page 7]] | [[User:Temperal/The Problem Solver's Resource9|Continue to page 9]]
 
[[User:Temperal/The Problem Solver's Resource7|Back to page 7]] | [[User:Temperal/The Problem Solver's Resource9|Continue to page 9]]
 
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Revision as of 22:08, 5 October 2007



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 $\displaystyle m_0 = \lim_{k \to 0} m_k$. 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:

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

Schur's inequality

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

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

Back to page 7 | Continue to page 9



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