nyuuzyou/wikis huggingface dataset:
Cauchy's Inequality

== Theorem ==
:$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum {r_i s_i} }^2$
where all of $r_i, s_i \in \R$.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Vector Form ==
== Also known as ==
== Historical Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Cauchy's Inequality: $3.2.9$
* : 1. Functions on Euclidean Space: Norm and Inner Product
* : $\S 36$: Inequalities: $36.3$: Cauchy-Schwartz Inequality
* 
* : $(2)$
* : $(2)$
* 
Category:Cauchy's Inequality
Category:Analysis
Category:Inequalities
Category:Cauchy-Bunyakovsky-Schwarz Inequality
Cauchy's Inequality/Also known as

== Cauchy's Inequality: Also known as ==
Cauchy's Inequality is also known as the Cauchy-Schwarz Inequality.
However, that name is also given to the more general Cauchy-Bunyakovsky-Schwarz Inequality, and so will not be used in this context on .
Some sources refer to this as the Cauchy-Schwarz inequality for sums.
== Sources ==
* : $\S 36$: Inequalities: $36.3$: Cauchy-Schwartz Inequality
* 
* 
* 
Category:Cauchy's Inequality
Cauchy's Inequality/Also presented as

== Cauchy's Inequality: Also presented as ==
Cauchy's Inequality can also be expressed in the form:
:$\ds \sum_{i \mathop = 1}^n r_i s_i \le \sqrt {\paren {\sum_{i \mathop = 1}^n r_i} \paren {\sum_{i \mathop = 1}^n s_i} }$
where all of $r_i, s_i \in \R$.
== Sources ==
* 
* 
* 
Category:Cauchy's Inequality
Cauchy's Inequality/Proof 1

== Theorem ==
== Proof ==
For any $\lambda \in \R$, we define $f: \R \to \R$ as the function:
:$\ds \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$
Now:
:$\map f \lambda \ge 0$
because it is the sum of squares of real numbers.
Hence:
This is a quadratic equation in $\lambda$.
From Solution to Quadratic Equation:
:$\ds a \lambda^2 + b \lambda + c = 0: a = \sum { {s_i}^2}, b = 2 \sum {r_i s_i}, c = \sum { {r_i}^2}$
The discriminant of this equation (that is $b^2 - 4 a c$) is:
:$\ds D := 4 \paren {\sum {r_i s_i} }^2 - 4 \sum { {r_i}^2} \sum { {s_i}^2}$
 $D$ is (strictly) positive.
Then $\map f \lambda = 0$ has two distinct real roots, $\lambda_1 < \lambda_2$, say.
From Sign of Quadratic Function Between Roots, it follows that $f$ is (strictly) negative somewhere between $\lambda_1$ and $\lambda_2$.
But we have:
:$\forall \lambda \in \R: \map f \lambda \ge 0$
From this contradiction it follows that:
:$D \le 0$
which is the same thing as saying:
:$\ds \sum { {r_i}^2} \sum { {s_i}^2} \ge \paren {\sum {r_i s_i} }^2$
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.1$
* : $\S 1$: Real Numbers: $\S 1.11$: Example
Category:Cauchy's Inequality
Cauchy's Inequality/Proof 2

== Theorem ==
== Proof ==
From the Complex Number form of the Cauchy-Schwarz Inequality, we have:
As elements of $\R$ are also elements of $\C$, it follows that:
:$\ds \sum \size {r_i}^2 \sum \size {s_i}^2 \ge \size {\sum r_i s_i}^2$
where all of $r_i, s_i \in \R$.
But from the definition of modulus, it follows that:
:$\ds \forall r_i \in \R: \size {r_i}^2 = {r_i}^2$
Thus:
:$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum r_i s_i}^2$
where all of $r_i, s_i \in \R$.
Category:Cauchy's Inequality
Cauchy's Inequality/Vector Form

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space $V$.
Then:
:$\paren {\mathbf a \cdot \mathbf b}^2 \le \paren {\mathbf a \cdot \mathbf a} \paren {\mathbf b \cdot \mathbf b}$
where $\cdot$ denotes dot product.
== Proof ==
Let us express $\mathbf a$ and $\mathbf b$ in component form:
where the ordered $n$-tuple $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $V$.
Then we have:
== Sources ==
* : $(2)$
* : $(2)$
* 
Category:Cauchy's Inequality
Category:Vector Analysis
Cauchy's Integral Formula

== Theorem ==
Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.
Let $f: U \to \C$ be holomorphic on some open set containing $D$.
Then for each $a$ in the interior of $D$:
:$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {z - a} \rd z$
where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.
=== General Result ===
== Proof ==
Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $\map f z$ is holomorphic on $R$.
Let $z_0$ be any point in the region $R$ such that:
:$\dfrac {\map f z} {z - z_0}$ is holomorphic on $R \setminus \set {z_0}$
We draw a circle $C_r$ with center at $z_0$ and radius $r$ such that $r \to 0$.
This makes $C$ and $C_r$ a multiply connected region for a sufficiently small $r > 0$.
According to Cauchy's Integral Theorem for a multiply connected region:
Let:
Now:
:$\ds I = 2 \pi i \map f {z_0} + \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z$
From Complex-Differentiable Function is Continuous, $f$ is continous.
According to Epsilon-Delta definition, for every $\epsilon \in \R_{>0}$ there exists a $\delta \in \openint 0 r$ such that:
:$\forall z \in \C: \cmod {z - z_0} < \delta \implies \cmod {\map f z - \map f {z_0} } < \epsilon$
Hence:
As $\epsilon \to 0$:
:$\ds \oint_{C_r} \frac {\map f z - \map f {z_0} } {z - z_0} \rd z = 0$
So:
== Also see ==
* Definition:Complex Contour Integral
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.26$: Cauchy ()
* 
Category:Complex Analysis
Category:Cauchy's Integral Formula
Cauchy's Integral Formula/General Result

== Theorem ==
Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.
Let $f: U \to \C$ be holomorphic on some open set $U$ such that $D \subseteq U$.
Let $n \in \N$ be a natural number.
Then for each $a$ in the interior of $D$, the $n$-th derivative of $f$ at $a$ exists and can be written as:
:$\ds \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.
=== Corollary ===
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
=== Basis for the Induction ===
$\map P 0$ holds, as this is:
:$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}} \rd z$
which is Cauchy's Integral Formula.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map {f^{\paren k} } a = \frac {k!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z$
Then we need to show:
:$\ds \map {f^{\paren {k + 1} } } a = \frac {\paren {k + 1}!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z$
=== Induction Step ===
This is our induction step:
To justify the passage of the derivative under the integral sign in line 1, note that for $\alpha$ near $a$, we have the following bound for all $z \in \partial D$:
:$\ds \cmod {\frac {\paren {k + 1} \map f z} {\paren {z - \alpha}^{k + 2} } } = \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2} }$
Since $f$ is continuous on the compact set $\partial D$, it follows that $\ds \sup_{z \mathop \in \partial D} \cmod {\map f z} < \infty$.
Thus the arc length integral:
:$\ds \oint_{\partial D} \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2}} ds(z) \le \paren {k + 1} \map \ell {\partial D} \sup_{z \mathop \in \partial D}\frac{\cmod {\map f z} }{r^{k + 2} } < \infty$
Hence an argument involving the Fundamental Theorem of Calculus followed by the dominated convergence theorem justifies the swap.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N: \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
== Also known as ==
== Sources ==
* 
Category:Cauchy's Integral Formula
Cauchy's Integral Formula/General Result/Corollary

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:
:$\sqbrk {z^n} \map G z := a_n$
Let $\map G z$ be convergent for $z = z_0$ and $0 < r < \cmod {z_0}$.
Then:
:$\sqbrk {z^n} \map G z = \ds \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }$
== Proof ==
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(32)$
Category:Generating Functions
Category:Cauchy's Integral Formula
Cauchy's Integral Formula/Historical Note

== Historical Note on Cauchy-Goursat Theorem ==
Cauchy's Integral Formula was developed by during his work to establish the groundwork of the discipline of complex analysis.
 independently discovered it during his own exercise to rebuild the theory from first principles.
== Sources ==
* : Chapter $\text {A}.26$: Cauchy ()
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Cauchy's Lemma (Group Theory)

== Theorem ==
Let $\struct {G, \circ}$ be a group of finite order whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $\struct {G, \circ}$ has an element of order $p$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cauchy's Group Theorem, which establishes that, given the same conditions, $G$ also has a subgroup of order $p$.
== Sources ==
* 
* 
Category:Group Theory
Category:Cauchy's Lemma (Group Theory)
Cauchy's Lemma (Group Theory)/Proof 1

== Theorem ==
== Proof ==
Let $\order G = n$ such that $p \divides n$.
Let:
:$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$
where $G^p$ is the cartesian product $\underbrace {G \times G \times \cdots \times G}_p$.
The first $p - 1$ coordinates of an element of $X$ can be chosen arbitrarily.
The last coordinate is determined by the fact that:
:$a_1 a_2 \cdots a_{p - 1} = a_p^{-1}$
So from the Product Rule for Counting, it follows that:
:$\card X = n^{p - 1}$
Let $C_p$ be a cyclic group of order $p$ generated by the element $c$.
Let $C_p$ act on the set $X$ by the rule:
:$c * \tuple {a_1, a_2, \ldots, a_p} = \tuple {a_2, a_3, \ldots, a_p, a_1}$
By the Orbit-Stabilizer Theorem, the number of elements in any orbit is a divisor of the order of $C_p$, which is $p$.
As $p$ is prime, an orbit has either $p$ elements or $1$ element by definition.
Let $r$ be the number of orbits with one element.
Let $s$ be the number of orbits with $p$ elements.
Then by the Partition Equation:
:$r + s p = n^{p - 1} = \card X$
By hypothesis, $p \divides n$, so:
:$r + s p = n^{p - 1} \implies p \divides r$
We know that $r \ne 0$ because, for example, the orbit of $\tuple {e, e, \ldots, e} \in X$ has only one element.
So there must be at least $p$ orbits with only one element.
Each such element has the form $\tuple {a, a, \ldots, a} \in X$ so $a^p = e$.
So $G$ contains at least $p$ elements $x$ satisfying $x^p = e$.
So $G$ contains an element $a \ne e$ such that $a^p = e$.
That is, $a$ must have order $p$.
== Sources ==
* 
* : Chapter $2$: The Sylow Theorems: $\S 55$
Category:Cauchy's Lemma (Group Theory)
Cauchy's Lemma (Group Theory)/Proof 2

== Theorem ==
== Proof ==
By the corollary to the First Sylow Theorem, $G$ has subgroups of order $p^r$ for all $r$ such that $p^r \divides \order G$.
Thus $G$ has at least one subgroup $H$ of order $p$.
As a Prime Group is Cyclic, $H$ is a cyclic group.
Thus by definition $H$ has an element of order $p$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.19$
* : Chapter $2$: The Sylow Theorems: $\S 56$. First Sylow Theorem
Category:Cauchy's Lemma (Group Theory)
Cauchy's Lemma (Number Theory)

== Theorem ==
Let $a$ and $b$ be odd positive integers. 
Suppose $a$ and $b$ satisfy: 
Then there exist non-negative integers $s, t, u, v$ such that:
== Proof ==
Because $a$ is odd, we can write:
:$a = 2 k + 1$
for some positive integer $k$.
Then:
From $(1)$, we have that $4 a - b^2$ is a positive integer. 
By Integer as Sum of Three Odd Squares, there exist $3$ odd positive integers $x, y, z$ such that:
:$(3): \quad 4 a - b^2 = x^2 + y^2 + z^2$
Because $b, x, y, z$ are all odd integers, $b + x + y + z$ must be even.
It is now to be shown that $b + x + y \pm z$ is divisible by $4$.
Suppose that $b + x + y + z$ is not divisible by $4$.
Because $b + x + y + z$ is even:
:$b + x + y + z \equiv 2 \pmod 4$
Writing $z = 2 l + 1$:
That is:
:if $b + x + y + z$ is not divisible by $4$
:then $b + x + y - z$ is divisible by $4$.
Let us choose the case such that $b + x + y \pm z$ is divisible by $4$.
We define:
We are to show that $s, t, u, v$ are non-negative, and will satisfy:
First we show that $s, t, u, v$ are non-negative.
Because $x, y, z$ are positive:
:$s, t, u, v \ge \dfrac {b - x - y - z} 4$
So we need to show:
:$\dfrac {b - x - y - z} 4 \ge 0$
or equivalently:
:$\dfrac {b - x - y - z} 4 > -1$
Now:
showing that $s, t, u, v$ are non-negative.
Now we check $(1)$:
Now we check $(2)$:
Therefore $s, t, u, v$ as defined are the integers we are looking for.
== Sources ==
* 
Category:Integer as Sum of Polygonal Numbers
Cauchy's Mean Theorem

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R$ be real numbers which are all positive.
Let $A_n$ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.
Let $G_n$ be the geometric mean of $x_1, x_2, \ldots, x_n$.
Then:
:$A_n \ge G_n$
with equality holding :
:$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
That is, all terms are equal.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof of Equality Condition ==
== Also known as ==
== Also see ==
* AM-HM Inequality
* GM-HM Inequality
== Sources ==
* : $3$: Elementary Analytic Methods: $3.2$ Inequalities: Relation Between Arithmetic, Geometric, Harmonic and Generalized Means: $3.2.1$
* : $\S 36$: Inequalities: Inequalities involving Arithmetic, Geometric and Harmonic Means: $36.4$
* 
* 
* 
* 
* 
* 
* 
* 
Category:Cauchy's Mean Theorem
Category:Arithmetic Mean
Category:Geometric Mean
Category:Inequalities
Cauchy's Mean Theorem/Also known as

== Theorem ==
Cauchy's Mean Theorem is widely known as:
* the Arithmetic-Geometric Mean Inequality
* the AM-GM Inequality.
Some sources give this as Cauchy's formula, but this is too imprecise to be useful.
== Sources ==
* 
* 
* 
* 
* 
Category:Cauchy's Mean Theorem
Cauchy's Mean Theorem/Proof 1

== Theorem ==
== Proof ==
The arithmetic mean of $x_1, x_2, \ldots, x_n$ is defined as:
:$\ds A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$
The geometric mean of $x_1, x_2, \ldots, x_n$ is defined as:
:$\ds G_n = \paren {\prod_{k \mathop = 1}^n x_k}^{1/n}$
We prove the result by induction:
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:For all positive real numbers $x_1, x_2, \ldots, x_n: A_n \ge G_n$.
$\map P 1$ is true, as this just says:
:$\dfrac {x_1} 1 \ge {x_1}^{1/1}$
which is trivially true.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\dfrac {x_1 + x_2} 2 \ge \sqrt {x_1 x_2}$
As $x_1, x_2 > 0$ we can take their square roots and do the following:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we show that:
:$(1): \quad$ If $\map P {2^k}$ is true, where $k \ge 1$, then it logically follows that $\map P {2^{k + 1} }$ is true
:$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true.
The result will follow by Forward-Backward Induction.
This is our first induction hypothesis:
:$A_{2^k} \ge G_{2^k}$
Then we need to show:
:$A_{2^{k + 1} } \ge G_{2^{k + 1} }$
=== Induction Step ===
This is our induction step:
Let $m = 2^k$.
Then $2^{k + 1} = 2 m$.
Because $\map P m$ is true:
:$\paren {x_1 x_2 \dotsm x_m}^{1/m} \le \dfrac 1 m \paren {x_1 + x_2 + \dotsb + x_m}$
Also:
:$\paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} \le \dfrac 1 m \paren {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} }$
But we have $\map P 2$, so:
:$\paren {\paren {x_1 x_2 \dotsm x_m}^{1/m} \paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} }^{1/2} \le \dfrac 1 2 \paren {\dfrac {x_1 + x_2 + \cdots + x_m} m + \dfrac {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} } m}$
So:
:$\paren {x_1 x_2 \dotsm x_{2 m} }^{1/2m} \le \dfrac {x_1 + x_2 + \dotsb + x_{2 m} } {2 m}$
So $\map P {2 m} = \map P {2^{k + 1} }$ holds.
So $\map P {2^n}$ holds for all $n$ by induction.
Now suppose $\map P k$ holds.
Then:
So $\map P k \implies \map P {k - 1}$ and the result follows by Forward-Backward Induction.
Therefore $A_n \ge G_n$ for all $n$.
== Proof of Equality Condition ==
== Sources ==
* : $\S 3$: Natural Numbers: $\S 3.10$: Example
Category:Cauchy's Mean Theorem
Category:Proofs by Induction
Cauchy's Mean Theorem/Proof 2

== Theorem ==
== Proof ==
Let:
:$\map f x = \ln x$
for $x > 0$.
With a view to apply Jensen's Inequality: Real Analysis: Corollary, we can show that $f$ is concave on $\openint 0 \infty$. 
By Second Derivative of Concave Real Function is Non-Positive, it is sufficient to show that $\map {f} x \le 0$ for all $x > 0$. 
We have, by Derivative of Natural Logarithm:
:$\map {f'} x = \dfrac 1 x$
We then have, by Derivative of Power:
:$\map {f} x = -\dfrac 1 {x^2}$ 
As $x^2 > 0$ for all $x > 0$, we have:
:$\dfrac 1 {x^2} > 0$
Therefore: 
:$-\dfrac 1 {x^2} = \map {f} x < 0$
so $f$ is indeed concave on $\openint 0 \infty$.
As $x_1, x_2, \ldots, x_n$ are all positive, they all lie in the interval $\openint 0 \infty$. 
We therefore have, by Jensen's Inequality: Real Analysis: Corollary: 
:$\ds \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
for real $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$, with at least one of which being non-zero. 
As $n$ is a positive integer, we have: 
:$\dfrac 1 n > 0$
We can therefore set: 
:$\lambda_i = \dfrac 1 n$
for $1 \le i \le n$.
This gives: 
and:
We therefore have: 
:$\map \ln {A_n} \ge \map \ln {G_n}$
Note that for $x > 0$: 
:$\dfrac 1 x = \map {f'} x > 0$
Therefore, by Derivative of Monotone Function, $f$ is increasing on $\openint 0 \infty$. 
We therefore have: 
:$A_n \ge G_n$
== Proof of Equality Condition ==
Category:Cauchy's Mean Theorem
Cauchy's Mean Theorem/Proof 3

== Theorem ==
== Proof ==
For $p \in \R$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
By definition of Hölder Mean with $p = 0$:
:$\map {M_0} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$
From Hölder Mean for Exponent 1 is Arithmetic Mean:
:$\map {M_1} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$
The result follows from Inequality of Hölder Means:
:$0 < 1 \implies \map {M_0} {x_1, x_2, \ldots, x_n} < \map {M_1} {x_1, x_2, \ldots, x_n}$
unless $x_1 = x_2 = \cdots = x_n$, in which case:
:$\map {M_0} {x_1, x_2, \ldots, x_n} = \map {M_1} {x_1, x_2, \ldots, x_n}$
Hence the result.
Category:Cauchy's Mean Theorem
Cauchy's Mean Theorem/Proof of Equality Condition

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R$ be real numbers which are all positive.
Let $A_n$ be the arithmetic mean of $x_1, x_2, \ldots, x_n$.
Let $G_n$ be the geometric mean of $x_1, x_2, \ldots, x_n$.
Then:
:$A_n = G_n$
:
:$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
That is, all terms are equal.
Then:
== Proof ==
=== Necessary Condition ===
Let:
:$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j = x$
Then:
So:
:$A_n = G_n = x$
=== Sufficient Condition ===
Let $A_n = G_n$.
We prove the result by induction:
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
$\map P 1$ is true, as this just says:
which is trivially true.
==== Basis for the Induction ====
$\map P 2$ is the case:
This is our basis for the induction.
==== Induction Hypothesis ====
Now we show that:
:$(1): \quad$ If $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {2 k}$ is true
:$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true.
The result will follow by Forward-Backward Induction.
This is our first induction hypothesis:
:$A_k = G_k \implies \forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$
Also, let:
:$A_k' := \ds \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
and:
:$G_k' := \ds \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
By the induction hypothesis:
:$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$
We need to show:
:$A_{2 k} = G_{2 k} \implies \forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$
==== Induction Step ====
This is our induction step:
Suppose:
:$A_{2 k} = G_{2 k}$
Let:
:$A_k' := \ds \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
and:
:$G_k' := \ds \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
By the induction hypothesis:
:$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$
Then:
Then:
That is:
:$\forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$
:$\forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = x$
Hence:
:$\forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$
Now suppose $\map P k$ holds.
Then:
But:
:$\paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ is the geometric mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
:$\dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ is the arithmetic mean of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
We have that $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ has $k$ elements.
Hence by the induction hypothesis:
:$\forall i, j \in \set {1, 2, \ldots, k - 1}: x_i = x_j = G_{k - 1}$
So $\map P k \implies \map P {k - 1}$ and the result follows by Forward-Backward Induction.
Therefore $A_n \ge G_n$ for all $n$.
:$\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
Category:Cauchy's Mean Theorem
Cauchy's Residue Theorem

== Theorem ==
Let $U$ be a simply connected open subset of the complex plane $\C$.
Let $a_1, a_2, \dots, a_n$ be finitely many points of $U$.
Let $f: U \to \C$ be analytic in $U \setminus \set {a_1, a_2, \dots, a_n}$. 
Let $L$ be a contour in $\C$ oriented anticlockwise.
Let $\partial U_k$ denote the closed contour bounding $U_k$.
Then:
:$\ds \oint_L \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
where $\Res f {a_k}$ denotes the residue at $a_k$ of $f$.
== Proof ==
Let $\set {U_1, \dotsc, U_n}$ be a set of open subsets of $U$ such that $a_i \in U_i$, and $a_i \notin U_j$ for $i \ne j$.
Let $U_i \cap U_j = \O$ for all $i \ne j$.
By Existence of Laurent Series, around each $a_k$ there is an expansion:
:$\ds \map f z = \sum_{j \mathop = -\infty}^\infty c_j \paren {z - a_k}^j$ 
convergent in $U_k$.
Write: 
:$\ds X = \bigcup_{i \mathop = 1}^n U_i$
Then, by Contour Integral of Concatenation of Contours: 
:$\ds \oint_L \map f z \rd z = \oint_{\partial \paren {U \setminus X} } \map f z \rd z + \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$ 
As all poles of $f$ in $U$ are contained in $X$, $f$ is holomorphic on $U \setminus X$.
So by the Cauchy-Goursat Theorem: 
:$\ds \oint_{\partial \paren {U \setminus X} } \map f z \rd z = 0$
giving: 
:$\ds \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z$
we have: 
So:
:$\ds \oint_L \map f z \rd z = \sum_{k \mathop = 1}^n \oint_{\partial U_k} \map f z \rd z = 2 \pi i \sum_{k \mathop = 1}^n \Res f {a_k}$
== Examples ==
== Also known as ==
== Sources ==
* 
* 
* 
Category:Cauchy's Residue Theorem
Category:Complex Analysis
Cauchy's Residue Theorem/Examples/Arbitrary Example 1

== Example of Use of Cauchy's Residue Theorem ==
:$\ds \int_\Gamma \dfrac c {z - a} = 2 \pi i c$
where $\Gamma$ is a simple closed curve enclosing the point $z = a$
== Sources ==
* 
Category:Examples of Use of Cauchy's Residue Theorem