nyuuzyou/wikis huggingface dataset:
Particular Point Space is Weakly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is weakly locally compact.
== Proof ==
By Compact Space in Particular Point Space, $\set p$ is compact in $T$.
But by definition $\set p$ is open in $T$.
So $p$ is contained in a compact neighborhood, that is, $\set p$.
Now let $x \in S: x \ne p$.
Then $\set {x, p}$ is open in $T$.
We have that $\set {x, p}$ trivially has an open cover, that is, $\set {\set {x, p} }$ itself.
Hence any open cover of $\set {x, p}$ has a finite subcover: any one set that contains $x$ and $p$ is a cover for $\set {x, p}$.
So $\set {x, p}$ is a neighborhood of $x$ which is compact in $T$.
So $x$ is contained in a compact neighborhood.
Hence the result, by definition of weakly locally compact.
== Also see ==
* Particular Point Space is Locally Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $5$
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $15$
Category:Weakly Locally Compact Spaces
Category:Particular Point Topology
Particular Point Space is not Arc-Connected

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is not arc-connected.
== Proof ==
Let $q \in S$ be such that $q \ne p$.
Let $f: \closedint 0 1 \to T$ be an injection such that:
:$\map f 0 = q$
:$\map f 1 = p$
Because $f$ is an injection, it must be that:
:$\map {f^{-1} } {\set p} = \set 1$
where $f^{-1}$ denotes the preimage under $f$.
Now $\set p$ is open in $\tau_p$ by definition of particular point topology.
By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open.
Thus we have exhibited an open set whose preimage under $f$ is not open.
Hence $f$ is not continuous.
Since $f$ was an arbitrary injection, no arc between $p$ and $q$ can exist.
Hence $T$ is not arc-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $13$
Category:Connected Spaces
Category:Particular Point Topology
Particular Point Space is not Ultraconnected

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space with at least three points.
Then $T$ is not ultraconnected.
== Proof ==
Let $x, y \in S: x \ne p, y \ne p, x \ne y$.
Consider $\set x$ and $\set y$.
Neither are open as neither contain $p$.
So from Subset of Particular Point Space is either Open or Closed they are both closed.
We have that $\set x \cap \set y = \O$.
The result follows by definition of ultraconnected.
== Also see ==
* Sierpiński Space is Ultraconnected
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $10$
Category:Ultraconnected Spaces
Category:Particular Point Topology
Particular Point Space is not Weakly Countably Compact/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $8 \text { - } 10$: Particular Point Topology
:::Item $12$
== Mistake ==
:[A particular point space] $X$ is not weakly countably compact since any set which does not contain $p$ has no limit points.
However, this is true only of an infinite particular point space.
By definition, a space $X$ is weakly countably compact every infinite subset of $X$ has a limit point in $X$.
But a finite particular point space has no infinite subset.
So a finite particular point space is weakly countably compact vacuously.
== Also see ==
* Infinite Particular Point Space is not Weakly Countably Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $12$
Category:Mistakes/Counterexamples in Topology
Particular Point Space less Particular Point is Discrete

== Theorem ==
Let $T = \left({S, \tau_p}\right)$ be a particular point space, whose particular point is $p$.
Let $H = S \setminus \left\{ {p}\right\}$ where $\setminus$ denotes set difference.
Then the topological subspace $T_H = \left({H, \tau_H}\right)$ induced on $H$ by $\tau_p$ is a discrete space.
== Proof ==
Let $H = S \setminus \left\{{p}\right\}$.
Let $V \subseteq H$ be any subset of $H$.
As $p \notin V$, $V$ is a closed set of $T$.
Thus, by definition of closed set $S \setminus V$ is open in $T$.
By definition of subspace topology, $\left({S \setminus V}\right) \cap H$ is open in $T_H$.
From Intersection with Set Difference is Set Difference with Intersection:
:$\left({S \setminus V}\right) \cap H = \left({S \cap H}\right) \setminus V$
and so from Intersection with Subset is Subset:
:$\left({S \setminus V}\right) \cap H = H \setminus V$
Thus $H \setminus V$ is open in $H$.
But $H \setminus V$ is itself a subset of $H$, and so a closed set of $T$.
Thus, by the same argument, $V$ is open in $H$.
As $V$ is arbitrary, it follows that all subsets of $H$ are open in $H$.
The result follows by definition of discrete space.
Category:Particular Point Topology
Category:Discrete Topology
Particular Point Topology is Closed Extension Topology of Discrete Topology

== Theorem ==
Let $S$ be a set and let $p \in S$.
Let $\tau_p$ be the particular point topology on $S$.
Let $T = \struct {S \setminus \set p, \vartheta}$ be the discrete topological space on $S \setminus \set p$.
Then $T^* = \struct {S, \tau_p}$ is a closed extension space of $T$.
== Proof ==
Directly apparent from the definitions of particular point topology, discrete topological space and closed extension space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $20$
Category:Closed Extension Topology
Category:Particular Point Topology
Category:Discrete Topology
Particular Point Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $\tau_p$ is a topology on $S$, and $T$ is a topological space.
== Proof ==
We have by definition that $\O \in \tau_p$, and as $p \in S$ we have that $S \in \tau_p$.
Now let $U_1, U_2 \in \tau_p$.
By definition $p \in U_1$ and $p \in U_2$, and so $p \in U_1 \cap U_2$ by definition of set intersection.
So $U_1 \cap U_2 \in \tau_p$.
Now let $\UU \subseteq \tau_p$.
We have that $\forall U \in \UU: p \in U$.
Hence from Subset of Union $p \in \bigcup \UU$.
So all the properties are fulfilled for $\tau_p$ to be a topology on $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology
Category:Particular Point Topology
Particular Point Topology with Three Points is not T4

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton or a doubleton.
That is, such that $S$ has more than two distinct elements.
Then $T$ is not a $T_4$ space.
== Proof ==
We have that there are at least three elements of $S$.
So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$.
Then $X = \set x, Y = \set y$ are closed in $T$ and $X \cap Y = \O$.
Suppose $U, V \in \tau_p$ are open sets in $T$ such that $X \subseteq U, Y \subseteq V$.
But as $p \in U, p \in V$ we have that $U \cap V \ne \O$.
So $T$ is not a $T_4$ space.
== Also see ==
* Sierpiński Space is $T_4$
* Sierpiński Space is $T_5$
== Mistakes in Sources ==
=== Non-Trivial Particular Point Topology is not T4 ===
See : Part $\text {II}$: Counterexamples: $8 \text { - } 10: \ 4$ where it is stated that:
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $4$
Category:T4 Spaces
Category:Particular Point Topology
Particular Point of Alexandroff Extension of Rational Number Space is Dispersion Point

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\left({\Q, \tau_d}\right)$.
Then $p$ is a dispersion point of $T^*$.
== Proof ==
By definition, $p$ is a dispersion point of $T^*$ :
:$\Q^*$ is a connected set in $T^*$
:$\Q^* \setminus \set p$ is totally disconnected in $T^*$.
From Alexandroff Extension of Rational Number Space is Connected, $\Q^*$ is a connected set in $T^*$.
It remains to be shown that $\Q^* \setminus \set p$ is totally disconnected in $T^*$.
By definition, $\Q^* \setminus \set p$ is the rational number space $\struct {\Q, \tau_d}$.
The result follows from Rational Numbers are Totally Disconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $5$
Category:Alexandroff Extensions
Category:Dispersion Points
Category:Rational Number Space
Particular Solution of Constant Coefficient Linear nth Order ODE

== Theorem ==
Consider the linear $n$th order ODE with constant coefficients:
:$(1): \quad \ds \sum_{k \mathop = 0}^n a_k \dfrac {\d^k y} {d x^k} = \map R x$
Let $(1)$ have the following $n$ initial conditions:
:$(2): \quad y = y_0, \dfrac {\d y} {\d x} = y_1, \dotsc, \dfrac {\d^{n - 1} y} {\d x^{n - 1} } = y_{n - 1}$
when $x = x_0$.
Then there exists exactly one particular solution of $(1)$ which satisfies $(2)$.
== Proof ==
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.3$ Arbitrary constants and initial conditions
Category:Linear ODEs
Particular Solution of System of Constant Coefficient Linear 1st Order ODEs

== Theorem ==
Consider the system of linear first order ordinary differential equations with constant coefficients:
Let $(1)$ and $(2)$ have the following $n$ initial conditions:
:$(3): \quad y = y_0, z = z_0$
when $x = x_0$.
Then there exists exactly one particular solution of $(1)$ and $(2)$ which satisfies $(3)$.
== Proof ==
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.3$ Arbitrary constants and initial conditions
Category:Linear First Order ODEs with Constant Coefficients‎
Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another

== Theorem ==
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
such that $y_1$ is not the trivial solution.
Then there exists a standard procedure to determine another particular solution $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are linearly independent.
== Proof ==
Let $\map {y_1} x$ be a non-trivial particular solution to $(1)$.
Thus, for all $C \in \R$, $C y_1$ is also a non-trivial particular solution to $(1)$.
Let $\map v x$ be a function of $x$ such that:
:$(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$
is a particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.
Thus:
:$(3): \quad {y_2}' ' + \map P x {y_2}' + \map Q x y = 0$
Differentiating $(2)$ twice $x$:
:${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$
and:
:${y_2}' ' = \map {v' '} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}' '} x$
Substituting these into $(3)$ and rearranging:
:$v \paren { {y_1}' ' + P {y_1}' + Q y_1} + v' ' y_1 + v' \paren {2 {y_1}' ' + P y_1} = 0$
Since $y_1$ is a particular solution of $(1)$, this reduces to:
:$v' ' y_1 + v' \paren {2 {y_1}' ' + P y_1} = 0$
which can be expressed as:
:$\dfrac {v' '} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$
Integration gives:
:$\ds \ln v' = - 2 \ln y_1 - \int P \rd x$
This leads to:
:$v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$
and so:
:$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
From Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution, $y_1$ and $v y_1$ are linearly independent.
== Sources ==
* : $\S 3.16$: The Use of a Known Solution to find Another
Category:Homogeneous LSOODEs
Particular Values of Binomial Coefficients

== Theorem ==
=== Binomial Coefficient $\dbinom 0 0$ ===
=== Binomial Coefficient $\dbinom 0 n$ ===
=== Binomial Coefficient $\dbinom 1 n$ ===
=== N Choose Negative Number is Zero ===
=== Binomial Coefficient with Zero ===
=== Binomial Coefficient with One ===
=== Binomial Coefficient with Self ===
=== Binomial Coefficient with Self minus One ===
=== Binomial Coefficient with Two ===
== Also see ==
* Properties of Binomial Coefficients
Category:Binomial Coefficients
Category:Examples of Binomial Coefficients‎
Particular Values of Cosecant Function

== Theorem ==
The following values of the cosecant function can be expressed as exact algebraic numbers.
This list is non-exhaustive.
 
=== Cosecant of Zero ===
=== Cosecant of 15 Degrees ===
=== Cosecant of 30 Degrees ===
=== Cosecant of 45 Degrees ===
=== Cosecant of 60 Degrees ===
=== Cosecant of 75 Degrees ===
=== Cosecant of Right Angle ===
=== Cosecant of 105 Degrees ===
=== Cosecant of 120 Degrees ===
=== Cosecant of 135 Degrees ===
=== Cosecant of 150 Degrees ===
=== Cosecant of 165 Degrees ===
=== Cosecant of Straight Angle ===
=== Cosecant of 195 Degrees ===
=== Cosecant of 210 Degrees ===
=== Cosecant of 225 Degrees ===
=== Cosecant of 240 Degrees ===
=== Cosecant of 255 Degrees ===
=== Cosecant of Three Right Angles ===
=== Cosecant of 285 Degrees ===
=== Cosecant of 300 Degrees ===
=== Cosecant of 315 Degrees ===
=== Cosecant of 330 Degrees ===
=== Cosecant of 345 Degrees ===
=== Cosecant of Full Angle ===
== Also see ==
* Particular Values of Sine Function
* Particular Values of Cosine Function
* Particular Values of Tangent Function
* Particular Values of Cotangent Function
* Particular Values of Secant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosecant Function
Particular Values of Cosine Function

== Theorem ==
The following values of the cosine function can be expressed as exact algebraic numbers.
This list is non-exhaustive.
 
=== Cosine of Zero ===
=== Cosine of 15 Degrees ===
=== Cosine of 30 Degrees ===
=== Cosine of 45 Degrees ===
=== Cosine of 60 Degrees ===
=== Cosine of 75 Degrees ===
=== Cosine of Right Angle ===
=== Cosine of 105 Degrees ===
=== Cosine of 120 Degrees ===
=== Cosine of 135 Degrees ===
=== Cosine of 150 Degrees ===
=== Cosine of 165 Degrees ===
=== Cosine of Straight Angle ===
=== Cosine of 195 Degrees ===
=== Cosine of 210 Degrees ===
=== Cosine of 225 Degrees ===
=== Cosine of 240 Degrees ===
=== Cosine of 255 Degrees ===
=== Cosine of Three Right Angles ===
=== Cosine of 285 Degrees ===
=== Cosine of 300 Degrees ===
=== Cosine of 315 Degrees ===
=== Cosine of 330 Degrees ===
=== Cosine of 345 Degrees ===
=== Cosine of Full Angle ===
== Also see ==
* Particular Values of Sine Function
* Particular Values of Tangent Function
* Particular Values of Cotangent Function
* Particular Values of Secant Function
* Particular Values of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Particular Values of Cotangent Function

== Theorem ==
The following values of the cotangent function can be expressed as exact algebraic numbers.
This list is non-exhaustive.
 
=== Cotangent of Zero ===
=== Cotangent of 15 Degrees ===
=== Cotangent of 30 Degrees ===
=== Cotangent of 45 Degrees ===
=== Cotangent of 60 Degrees ===
=== Cotangent of 75 Degrees ===
=== Cotangent of Right Angle ===
=== Cotangent of 105 Degrees ===
=== Cotangent of 120 Degrees ===
=== Cotangent of 135 Degrees ===
=== Cotangent of 150 Degrees ===
=== Cotangent of 165 Degrees ===
=== Cotangent of Straight Angle ===
=== Cotangent of 195 Degrees ===
=== Cotangent of 210 Degrees ===
=== Cotangent of 225 Degrees ===
=== Cotangent of 240 Degrees ===
=== Cotangent of 255 Degrees ===
=== Cotangent of Three Right Angles ===
=== Cotangent of 285 Degrees ===
=== Cotangent of 300 Degrees ===
=== Cotangent of 315 Degrees ===
=== Cotangent of 330 Degrees ===
=== Cotangent of 345 Degrees ===
=== Cotangent of Full Angle ===
== Also see ==
* Particular Values of Sine Function
* Particular Values of Cosine Function
* Particular Values of Tangent Function
* Particular Values of Secant Function
* Particular Values of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Particular Values of Legendre Polynomials/-1

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\map {P_n} {-1} = \paren {-1}^n$ 
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {P_n} {-1} = \paren {-1}^n$
=== Basis for the Induction ===
From Legendre Polynomial $P_0$:
:$\map {P_0} {-1} = 1 = \paren {-1}^0$
Thus $\map P 0$ is seen to hold.
From Legendre Polynomial $P_1$ we have:
:$\map {P_1} x = x$
Substituting $x = -1$ yields 
:$\map {P_1} {-1} = -1 = \paren {-1}^1$
Thus $\map P 1$ is seen to hold.
$\map P 0$ and $\map P 1$ form the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\map {P_k} 1 = \paren {-1}^k$
from which it is to be shown that:
:$\map {P_{k + 1} } 1 = \paren {-1}^{k + 1}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \map {P_n} {-1} = {-1}$
== Sources ==
* : $\S 25$: Legendre Functions: Special Results Involving Legendre Polynomials: $25.30$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Results Involving Legendre Polynomials: $28.30.$
Category:Particular Values of Legendre Polynomials
Category:Proofs by Induction
Particular Values of Legendre Polynomials/-x

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\map {P_n} x = \paren {-1}^n \map {P_n} {-x}$ 
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {P_n} x = \paren {-1}^n \map {P_n} {-x}$
First we note that:
That is:
:$(1): \quad \map {P_n} x = \paren {-1}^n \map {P_n} {-x} \iff \map {P_n} {-x} = \paren {-1}^n \map {P_n} x$
=== Basis for the Induction ===
From Legendre Polynomial $P_0$:
Thus $\map P 0$ is seen to hold.
From Legendre Polynomial $P_1$ we have:
Thus $\map P 1$ is seen to hold.
$\map P 0$ and $\map P 1$ form the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\map {P_k} x = \paren {-1}^k \map {P_k} {-x}$
which from $(1)$ can be expressed as:
:$\map {P_k} {-x} = \paren {-1}^k \map {P_k} x$
from which it is to be shown that:
:$\map {P_{k + 1} } x = \paren {-1}^{k + 1} \map {P_{k + 1} } {-x}$
which from $(1)$ can be expressed as:
:$\map {P_{k + 1} } {-x} = \paren {-1}^{k + 1} \map {P_{k + 1} } x$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \map {P_n} x = \paren {-1}^n \map {P_n} {-x}$
== Sources ==
* : $\S 25$: Legendre Functions: Special Results Involving Legendre Polynomials: $25.31$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Results Involving Legendre Polynomials: $28.31.$
Category:Particular Values of Legendre Polynomials
Category:Proofs by Induction
Particular Values of Legendre Polynomials/0

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\map {P_n} 0 = \begin {cases}
0 & : \text {$n$ odd} \\ \\
\paren {-1}^r \dfrac {\paren {2 r}!} {\paren {2^r r!}^2} & : n = 2 r: r \in \N
\end {cases}$
That is, for even $n$:
:$\map {P_n} 0 = \paren {-1}^{n / 2} \dfrac {1 \cdot 3 \cdot 5 \cdots \paren {n - 1} } {2 \cdot 4 \cdot 6 \cdots n}$
== Proof ==
First let $n \in \N$ be odd.
From Legendre Polynomial of Odd Index is Odd Function:
:$\map {P_n} x = -\map {P_n} {-x}$
So from Odd Function of Zero is Zero:
:$\map {P_n} 0 = 0$
Now let $n \in \N$ be even such that $n = 2 r$.
The proof proceeds by mathematical induction.
For all $r \in \Z_{\ge 0}$, let $\map P r$ be the proposition:
:$\map {P_{2 r} } 0 = \paren {-1}^r \dfrac {\paren {2 r}!} {\paren {2^r r!}^2}$
First we check the case for $r = 0$.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
From Legendre Polynomial $P_2$ we have:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\map {P_{2 k} } 0 = \paren {-1}^k \dfrac {\paren {2 k}!} {\paren {2^k k!}^2} = \paren {-1}^k \dfrac {1 \cdot 3 \cdot 5 \cdots \paren {2 k - 1} } {2 \cdot 4 \cdot 6 \cdots 2 k}$
from which it is to be shown that:
:$\map {P_{2 \paren {k + 1} } } 0 = \map {P_{2 k + 2} } 0 = \paren {-1}^{k + 1} \dfrac {\paren {2 \paren {k + 1} }!} {\paren {2^{k + 1} \paren {k + 1}!}^2} = \paren {-1}^{k + 1} \dfrac {1 \cdot 3 \cdot 5 \cdots \paren {2 k - 1} \cdot \paren {2 k + 1} } {2 \cdot 4 \cdot 6 \cdots 2 k \cdot \paren {2 k + 2} }$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \map {P_n} 0 = \begin {cases}
0 & : \text {$n$ odd} \\ \\
\paren {-1}^r \dfrac {\paren {2 r}!} {2^{r + 1} r!} & : n = 2 r: r \in \N
\end {cases}$
== Sources ==
* : $\S 25$: Legendre Functions: Special Results Involving Legendre Polynomials: $25.32$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Results Involving Legendre Polynomials: $28.32.$
Category:Particular Values of Legendre Polynomials
Particular Values of Legendre Polynomials/1

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\map {P_n} 1 = 1$
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\map {P_n} 1 = 1$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\map {P_0} 1 = 1$
which is demonstrated in Legendre Polynomial $P_0$.
Thus $\map P 0$ is seen to hold.
From Legendre Polynomial $P_1$ we have:
:$\map {P_1} x = x$
Substituting $x = 1$ yields 
:$\map {P_1} 1 = 1$
Thus $\map P 1$ is seen to hold.
$\map P 0$ and $\map P 1$ form the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$\map {P_k} 1 = 1$
from which it is to be shown that:
:$\map {P_{k + 1} } 1 = 1$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \map {P_n} 1 = 1$
== Sources ==
* : $\S 25$: Legendre Functions: Special Results Involving Legendre Polynomials: $25.29$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Results Involving Legendre Polynomials: $28.29.$
Category:Particular Values of Legendre Polynomials