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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 21:08, 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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