nyuuzyou/wikis huggingface dataset:
Morphism from Multiplicative Group of Complex Numbers to Unit Circle

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ denote the multiplicative group of complex numbers.
Let $f: \C_{\ne 0} \to \C_{\ne 0}$ be the mapping defined as:
:$\forall z \in \C_{\ne 0}: \map f z = \dfrac z {\cmod z}$
where $\cmod z$ denotes the modulus of $z$.
Then $f$ is an endomorphism on $\struct {\C_{\ne 0}, \times}$ whose kernel is the positive real axis:
:$\set {z \in \C: z = x + 0 i, x \in \R_{>0} }$
and whose image is the unit circle:
:$\set {z \in \C: \cmod z = 1}$
== Proof ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{M}$
Category:Multiplicative Group of Complex Numbers
Category:Examples of Group Homomorphisms
Morphism from Ring with Unity to Module

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be an $R$-module.
Then for every $m \in M$ there exists a unique $R$-module morphism:
:$\psi: R \to M$
that sends $1$ to $m$.
== Proof ==
=== Existence ===
Let $r \in R$.
Let $\map \psi r := r m$.
This map is $R$-linear by definition of a module.
=== Uniqueness ===
Let $\psi_1$ and $\psi_2$ be two such morphisms.
Then $\psi_1 - \psi_2$ is an $R$-module morphism whose kernel contains $1$.
Thus:
:$\map \ker {\psi_1 - \psi_2} = R$
and:
:$\psi_1 = \psi_2$
== Also see ==
*Universal Property of Free Module on Set
Category:Rings with Unity
Category:Module Theory
Morphism in Preorder Category is Epic

== Theorem ==
Let $\mathbf P$ be a preorder category.
Let $f \in \mathbf P_1$ be a morphism.
Then $f$ is epic.
== Proof ==
Suppose that $g, h \in \mathbf P_1$ are morphisms such that:
:$g \circ f = h \circ f$
In particular then, $g$ and $h$ have equal domain and codomain.
Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects.
Thus necessarily $g = h$, and hence $f$ is epic.
== Sources ==
* : $\S 2.1$: Example $2.4$
* : $\S 2.8$: Exercise $2.2$
Category:Preorder Categories
Morphism in Preorder Category is Monic

== Theorem ==
Let $\mathbf P$ be a preorder category.
Let $f \in \mathbf P_1$ be a morphism.
Then $f$ is monic.
== Proof ==
Suppose that $g, h \in \mathbf P_1$ are morphisms such that:
:$f \circ g = f \circ h$
In particular then, $g$ and $h$ have equal domain and codomain.
Since $\mathbf P$ is a preorder category, there is at most one morphism between any two objects.
Thus necessarily $g = h$, and hence $f$ is monic.
== Sources ==
* : $\S 2.1$: Example $2.4$
* : $\S 2.8$: Exercise $2.2$
Category:Preorder Categories
Moscow Anthology/Amusing Problems/1

== Problem ==
:Two school students are observing a steam locomotive, the smoke from whose smokestack appears to be rising in a vertical plume.
:As they have been taking meteorological measurements, they know that the wind is steady and gust-free, and its speed is $20$ miles per hour.
:How do they know the train's speed?
== Solution ==
As the smoke was rising vertically upwards, the speed of the smoke must have been $0$ relative to the wind.
Hence the smoke was travelling at the same speed as the wind.
Thus it must have been rising from a smokestack that is travelling at that same speed.
It follows that the train itself was travelling at $20$ miles per hour.
== Sources ==
* : $\text I$: Amusing Problems: $1$. Observant Children
Category:Moscow Anthology
Motion of Body Falling through Air

== Theorem ==
The motion of a body $B$ falling through air can be described using the following differential equation:
:$m \dfrac {\d^2 y} {\d t^2} = m g - k \dfrac {d y} {d t}$
where:
:$m$ denotes mass of $B$
:$y$ denotes the height of $B$ from an arbitrary reference
:$t$ denotes time elapsed from an arbitrary reference
:$g$ denotes the local gravitational constant acting on $B$
:$k$ denotes the coefficient of resistive force exerted on $B$ by the air (assumed to be proportional to the speed of $B$)
== Proof ==
From Newton's Second Law of Motion, the force on $B$ equals its mass multiplied by its acceleration.
Thus the force $F$ on $B$ is given by:
:$F = m \dfrac {\d^2 y} {\d t^2}$
where it is assumed that the acceleration is in a downward direction.
The force on $B$ due to gravity is $m g$.
The force on $B$ due to the air it is passing through is $k$ multiplied by the speed of $B$, in the opposite direction to its travel.
That is::
:$k \dfrac {d y} {d t}$
Hence the required differential equation:
:$m \dfrac {\d^2 y} {\d t^2} = m g - k \dfrac {d y} {d t}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(3)$
Category:Gravity
Category:Examples of Differential Equations
Motion of Body with Constant Mass

== Theorem ==
Let $B$ be a body with constant mass $m$ undergoing a force $\mathbf F$.
Then the equation of motion of $B$ is given by:
:$\mathbf F = m \mathbf a$
where $\mathbf a$ is the acceleration of $B$.
== Proof ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Definitions: $3$. Definitions of terms
Category:Dynamics
Motion of Body with Variable Mass

== Theorem ==
Let $B$ be a body undergoing a force $\mathbf F$.
Let $B$ be travelling at a velocity $\mathbf v$ at time $t$.
Let mass travelling at a velocity $\mathbf v + \mathbf w$ be added to $B$ at a rate of $\dfrac {\d m} {\d t}$.
Let $m$ be the mass of $B$ at time $t$.
Then the equation of motion of $B$ is given by:
:$\mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$
== Proof ==
From Newton's Second Law of Motion:
:$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$
Then the added momentum being added to $B$ by the mass being added to it is given by:
:$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$
Hence:
== Sources ==
* : Miscellaneous Problems for Chapter $2$
Category:Dynamics
Motion of Cart attached to Wall by Spring

== Theorem ==
=== Problem Definition ===
Then the motion of $C$ is described by the second order ODE:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac k m \mathbf x = 0$
== Proof ==
By Newton's Second Law of Motion, the force on $C$ equals its mass times its acceleration:
:$\mathbf F = m \mathbf a$
By Acceleration is Second Derivative of Displacement with respect to Time:
:$\mathbf a = \dfrac {\d^2 \mathbf x} {\d t^2}$
By Hooke's Law:
:$\mathbf F = -k \mathbf x$
So:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(2)$
Category:Mechanics
Motion of Cart attached to Wall by Spring/Problem Definition

== Definition of Problem ==
:
Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.
Let $C$ be free to move along a horizontal straight line with zero friction.
Let the force constant of $S$ be $k$.
Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(4)$
Category:Mechanics
Motion of Cart attached to Wall by Spring under Damping

== Theorem ==
=== Problem Definition ===
Then the motion of $C$ is described by the second order ODE:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$
== Proof ==
Let $\mathbf F_s$ be the force on $C$ exerted by the spring $S$.
Let $\mathbf F_d$ be the damping force.
By Newton's Second Law of Motion, the total force on $C$ equals its mass times its acceleration:
:$\mathbf F_s + \mathbf F_d = m \mathbf a$
By Acceleration is Second Derivative of Displacement with respect to Time:
:$\mathbf a = \dfrac {\d^2 \mathbf x} {\d t^2}$
By definition, the velocity $\mathbf v$ is defined as:
:$\mathbf v = \dfrac {\d \mathbf x} {\d t}$
By Hooke's Law:
:$\mathbf F = -k \mathbf x$
By definition of the damping force:
:$\mathbf F_d = - c \mathbf v$
So:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(10)$
Category:Mechanics
Motion of Cart attached to Wall by Spring under Damping/Problem Definition

== Definition of Problem ==
:
Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.
Let $C$ be free to move along a straight line in a medium which applies a damping force $F_d$ whose magnitude is proportional to the speed of $C$.
Let the force constant of $S$ be $k$.
Let the constant of proportion of the damping force $F_d$ be $c$.
Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems$
Category:Mechanics
Motion of Cart attached to Wall by Spring with no Damping

== Theorem ==
=== Problem Definition ===
Suppose the damping force $c$ is zero.
Then the motion of $C$ can be described by the second order ODE:
:$m \dfrac {\d^2 \mathbf x} {\d t^2} = -k \mathbf x = 0$
== Proof ==
From Motion of Cart attached to Wall by Spring under Damping:
:$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$
The result follows by setting $c = 0$.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(5)$
Category:Mechanics
Motion of Particle in Polar Coordinates

== Theorem ==
Consider a particle $p$ of mass $m$ moving in the plane under the influence of a force $\mathbf F$.
Let the position of $p$ at time $t$ be given in polar coordinates as $\polar {r, \theta}$.
Let $\mathbf F$ be expressed as:
:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
where:
:$\mathbf u_r$ is the unit vector in the direction of the radial coordinate of $p$
:$\mathbf u_\theta$ is the unit vector in the direction of the angular coordinate of $p$
:$F_r$ and $F_\theta$ are the magnitudes of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.
Then the second order ordinary differential equations governing the motion of $m$ under the force $\mathbf F$ are:
== Proof ==
Let the radius vector $\mathbf r$ from the origin to $p$ be expressed as:
:$(1): \quad \mathbf r = r \mathbf u_r$
From Velocity Vector in Polar Coordinates, the velocity $\mathbf v$ of $p$ can be expressed in vector form as:
:$\mathbf v = r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r$
From Acceleration Vector in Polar Coordinates, the the acceleration $\mathbf a$ of $p$ can be expressed as:
:$\mathbf a = \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta + \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r$
We have:
:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
and from Newton's Second Law of Motion:
:$\mathbf F = m \mathbf a$
Hence:
:$\mathbf F = m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} } \mathbf u_\theta + m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2} \mathbf u_r$
Equating components:
Hence the result:
== Sources ==
* : $\S 3.21$: Newton's Law of Gravitation: $(7)$
* : Chapter $\text {B}.25$: Kepler's Laws and Newton's Law of Gravitation
Category:Mathematical Physics
Motion of Rocket in Outer Space

== Theorem ==
Let $B$ be a rocket travelling in outer space.
Let the velocity of $B$ at time $t$ be $\mathbf v$.
Let the mass of $B$ at time $t$ be $m$.
Let the exhaust velocity of $B$ be constant at $\mathbf b$.
Then the equation of motion of $B$ is given by:
:$m \dfrac {\d \mathbf v} {\d t} = - \mathbf b \dfrac {\d m} {\d t}$
== Proof 1 ==
== Proof 2 ==
Category:Rocket Science
Motion of Rocket in Outer Space/Proof 1

== Theorem ==
== Proof ==
From Motion of Body with Variable Mass:
:$(1): \quad \mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$
where:
:$\mathbf F$ is the external force being applied
:$\mathbf w$ is the velocity of the added mass relative to $B$.
In this scenario:
:there is no external force and so $\mathbf F = \mathbf 0$
:the velocity of the added mass relative to $B$ is $-\mathbf b$.
Thus $(1)$ becomes:
:$-\mathbf b \dfrac {\d m} {\d t} = m \dfrac {\d \mathbf v} {\d t}$
Category:Rocket Science
Motion of Rocket in Outer Space/Proof 2

== Theorem ==
== Proof ==
From Newton's Second Law of Motion:
:$\mathbf F = \dfrac \d {\d t} \paren {m \mathbf v}$
At time $t + \Delta t$, let:
: the mass of $B$ be $m + \Delta m$
: the velocity of $B$ be $\mathbf v + \Delta \mathbf v$.
The fuel is being consumed, so the increase in mass of the fuel during time $\Delta t$ is $-\Delta m$.
Thus the exhaust products, therefore of mass $-\Delta m$, are expelled at a velocity $-\mathbf b$ relative to $B$.
Thus this material is actually moving at a velocity $\mathbf v - \mathbf b$.
By Conservation of Momentum, the total linear momentum is constant.
Thus:
== Sources ==
* : Chapter $\text {B}.8$: Rocket Propulsion in Outer Space
Category:Rocket Science
Motion of Simple Pendulum

== Theorem ==
Consider a simple pendulum consisting of a bob whose mass is $m$, at the end of a rod of negligible mass of length $a$.
Let the bob be pulled to one side so that the rod is at an angle $\alpha$ from the vertical and then released.
Let $T$ be the time period of the simple pendulum, that is, the time through which the bob takes to travel from one end of its path to the other, and back again.
Then:
:$T = 4 \sqrt {\dfrac a g} \map K k$
where:
:$k = \map \sin {\dfrac \alpha 2}$
:$\map K k$ is the complete elliptic integral of the first kind.
== Proof ==
At a time $t$, let:
:the rod be at an angle $\theta$ to the vertical
:the bob be travelling at a speed $v$
:the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $s$.
:
At its maximum displacement, the speed of the bob is $0$, so its kinetic energy is $0$.
By the Principle of Conservation of Energy:
:$\dfrac 1 2 m v^2 = m g \paren {a \cos \theta - a \cos \alpha}$
We have that:
:$s = a \theta$
:$v = \dfrac {\d s} {\d t} = a \dfrac {\d \theta} {\d t}$
The rate of change of $s$ at time $t$ is the speed of the bob.
So:
Substituting:
:$\cos \theta = 1 - 2 \sin^2 \dfrac \theta 2$
:$\cos \alpha = 1 - 2 \sin^2 \dfrac \alpha 2$
we get:
:$\ds T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {1 - 2 \sin^2 \frac \theta 2 - 1 + 2 \sin^2 \frac \alpha 2} } } = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {\sin^2 \frac \alpha 2 - \sin^2 \frac \theta 2} } }$
We now put $k = \sin \dfrac \alpha 2$:
:$\ds T = 2 \sqrt {\frac a g} \int_0^\alpha {\frac {\d \theta} {\sqrt {k^2 - \sin^2 \frac \theta 2} } }$
Next, let us introduce the variable $\phi$, such that:
:$\sin \dfrac \theta 2 = k \sin \phi$
and where $\phi$ goes from $0 \to \pi / 2$ as $\theta$ goes from $0 \to \alpha$.
Differentiating with respect to $\phi$ we have:
:$\dfrac 1 2 \cos \dfrac \theta 2 \dfrac {\d \theta} {\d {\phi}} = k \cos \phi$
Thus:
Then:
The integral:
:$\ds \int_0^{\pi / 2} {\frac {\d \phi} {\sqrt {1 - k^2 \sin^2 \phi} } }$
is the complete elliptic integral of the first kind and is a function of $k$, defined on the interval $0 < k < 1$.
Hence the result.
== Also see ==
* Approximate Motion of Simple Pendulum
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
* 
* : Chapter $9$: Patterns in Nature: Differential equations
Category:Mathematical Physics
Category:Simple Pendulums
Category:Complete Elliptic Integral of the First Kind
Moving Average/Examples/Three-Point

== Examples of Moving Average ==
An example of a moving average is as follows:
Let $T = \sequence {y_n}$ be a time series whose observations are equally spaced in time.
Let $y_s$ be a representative observation pf $T$.
A three-point moving average on $T$ would replace $y_s$ with $y_s'$ thus:
:$y_s' := \dfrac {y_{s - 1} + 2 y_s + y_{s + 1} } 4$
for all $s \ge 2$.
== Sources ==
* 
* 
Category:Examples of Moving Averages
Multibrot Set/Graphics/Multibrot 4

== Graphic of Multibrot Set ==
The following is a depiction of the complete multibrot $4$ set.
The multibrot $4$ set itself is white.
:
The region of the complex plane depicted is the rectangle $\closedint {-2.2} {2.2} \times \closedint {-1.65 i} {1.65 i}$.
The iterated mapping here is $z \mapsto z^4 + c$
== Online Fractal Generator ==
== Sources ==
* 
* 
Category:Graphics of Multibrot Sets
Multifunction/Examples/Arbitrary Multifunction 1

== Example of Multifunction ==
Consider the implicit function:
:$y^2 = x + 2$
For $x > 2$, there are $2$ values of $y$ for every $x$.
Hence on that domain $y$ is a two-valued (multi)function of $x$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.3$ Functions of a Real Variable: $\text {(a)}$ Many-valued Functions
Category:Examples of Multifunctions
Multifunction/Examples/Unit Circle

== Example of Multifunction ==
Consider the equation for the unit circle whose center is at the origin of the Cartesian plane:
 
:$x^2 + y^2 = 1$
This can be considered as the graph of a multifunction whose domain is the closed real interval $\closedint {-1} 1$ and whose branches are:
== Sources ==
* 
* 
Category:Examples of Multifunctions
Multiindices under Addition form Commutative Monoid

== Theorem ==
Let $Z$ be the set of multiindices.
Let $+$ denote the addition of multiindices.
Then $\left({Z, +}\right)$ is a commutative monoid.
== Proof ==
We check each of the axioms in turn.
Let $k = \left \langle {k_j}\right \rangle_{j \in J}$, $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ and $m = \left \langle {m_j}\right \rangle_{j \in J}$ be multiindices.
=== Closure ===
Trivially we have that:
:$j \mapsto \left({ k_j + \ell_j }\right)$
is a sequence of integers indexed by $J$.
We know that finitely many of the $k_j$ are non-zero, and finitely many of the $\ell_j$ are non-zero.
Therefore finitely many of the $k_j + \ell_j$ are non-zero.
This shows that $k + \ell$ is a multiindex and so $+$ is closed.
=== Associativity ===
For all $j \in J$, we have:
Thus $+$ is shown to be associative.
=== Commutativity ===
For all $j \in J$, we have:
Thus $+$ is shown to be commutative.
=== Identity Element ===
Let $0_Z$ be the multiindex defined by:
:$\left(0_Z\right)_j = 0$
for all $j \in J$.
Then we have, for all $j \in J$:
Since we have seen that $\left({Z, +}\right)$ is commutative, this shows that $0_Z$ is an identity element for $Z$.
Category:Polynomial Theory
Multilinear Mapping from Free Modules is Determined by Bases

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $M_1,\ldots,M_n$ be free $R$-modules.
Let $B_1,\ldots,B_n$ be bases of $M_1,\ldots,M_n$.
Let $N$ be an $R$-module.
Let $f:B_1\times\cdots\times B_n\to N$ be a function.
Then there exists a unique multilinear map $\phi:M_1\times\cdots\times M_n\to N$ such that $\phi(b)=f(b)$ for all $b\in B_1\times\cdots\times B_n$.
== Proof ==
== Also see ==
* Universal Property of Free Modules
Category:Abstract Algebra
Category:Module Theory
Multinomial Coefficient/Examples/2, 3, 2 from 7

== Example of Multinomial Coefficients ==
The multinomial coefficient $\dbinom 7 {2, 3, 2}$ evaluates as:
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.16.$ Example
Category:Examples of Multinomial Coefficients
Multinomial Coefficient/Examples/4, 2, 2, 0 from 8

== Example of Multinomial Coefficients ==
The multinomial coefficient $\dbinom 8 {4, 2, 2, 0}$ evaluates as:
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.16.$ Example
Category:Examples of Multinomial Coefficients
Multinomial Coefficient expressed as Product of Binomial Coefficients

== Theorem ==
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
where:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a multinomial coefficient
:$\dbinom {k_1 + k_2} {k_1}$ etc. denotes binomial coefficients.
== Proof ==
The proof proceeds by induction.
For all $m \in \Z_{> 1}$, let $\map P m$ be the proposition:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
=== Basis for the Induction ===
$\map P 2$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r} {k_1 + k_2 + \cdots + k_{r - 1} }$
from which it is to be shown that:
:$\dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1, k_2, \ldots, k_r, k_{r + 1} } = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m \in \Z_{>1}: \dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(43)$
Category:Binomial Coefficients
Category:Multinomial Coefficients
Multinomial Theorem

== Theorem ==
Let $x_1, x_2, \ldots, x_k \in F$, where $F$ is a field.
Then:
:$\ds \paren {x_1 + x_2 + \cdots + x_m}^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_m \mathop = n} \binom n {k_1, k_2, \ldots, k_m} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_m}^{k_m}$
where:
:$m \in \Z_{> 0}$ is a positive integer
:$n \in \Z_{\ge 0}$ is a non-negative integer
:$\dbinom n {k_1, k_2, \ldots, k_m} = \dfrac {n!} {k_1! \, k_2! \, \cdots k_m!}$ denotes a multinomial coefficient.
The sum is taken for all non-negative integers $k_1, k_2, \ldots, k_m$ such that $k_1 + k_2 + \cdots + k_m = n$, and with the understanding that wherever $0^0$ may appear it shall be considered to have a value of $1$.
== Proof ==
The proof proceeds by induction on $m$.
For each $m \in \N_{\ge 1}$, let $\map P m$ be the proposition:
:$\ds \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_m}^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_m \mathop = n} \binom n {k_1, k_2, \ldots, k_m} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_m}^{k_m}$
=== Basis for the Induction ===
Trivially, for all $n \in \N$:
and so it is seen that $\map P 1$ holds.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_r}^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop = n} \binom n {k_1, k_2, \ldots, k_r} {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r}$
from which it is to be shown that:
:$\ds \forall n \in \N: \paren {x_1 + x_2 + \cdots + x_r + x_{r + 1} }^n = \sum_{k_1 \mathop + k_2 \mathop + \mathop \cdots \mathop + k_r \mathop + k_{r + 1} \mathop = n} \binom n {k_1, k_2, \ldots, k_r, k_{r + 1} } {x_1}^{k_1} {x_2}^{k_2} \cdots {x_r}^{k_r} {x_{r + 1} }^{k_{r + 1} }$
=== Induction Step ===
Now:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
== Also known as ==
== Also see ==
* Binomial Theorem: the multinomial theorem is a generalization
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.16$
* : $\S 1.2.6$: Binomial Coefficients: $(42)$
* 
* 
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.17.$
Category:Multinomial Theorem
Category:Multinomial Coefficients
Category:Binomial Coefficients
Category:Algebra
Category:Discrete Mathematics
Category:Proofs by Induction
Multinomial Theorem/Also known as

== Multinomial Theorem: Also known as == 
The multinomial theorem is also known as the multinomial formula.
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.16$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Multinomial Formula: $3.17.$
Category:Multinomial Theorem