nyuuzyou/wikis huggingface dataset:
Legendre's Theorem

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime number.
Let $n$ be expressed in base $p$ representation.
Let $r$ be the digit sum of the representation of $n$ in base $p$.
Then $n!$ is divisible by $p^\mu$ but not by $p^{\mu + 1}$, where:
:$\mu = \dfrac {n - r} {p - 1}$
=== Corollary ===
== Proof ==
$n$ can be represented as:
Using De Polignac's Formula, we may extract all the powers of $p$ from $n!$.
:$\mu = \ds \sum_{k \mathop > 0} \floor {\dfrac n {p^k} }$
where $\mu$ is the multiplicity of $p$ in $n!$:
:$p^\mu \divides n!$
:$p^{\mu + 1} \nmid n!$
We have that:
Thus:
Hence the result.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $12$
Category:Factorials
Legendre's Theorem/Corollary

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $B$ be the binary representation of $n$.
Let $r$ be the number of unit digits in $B$.
Let $n!$ denote the factorial of $n$.
Then $2^{n - r}$ is a divisor of $n!$, but $2^{n - r + 1}$ is not.
== Proof ==
$n$ can be represented as:
where all of $e_1, e_2, \ldots, e_r$ are integers.
Then $r$ is the sum of the digits in $n$ as a binary number.
From Legendre's Theorem, we have:
:$\mu = \dfrac {n - r} {2 - 1}$
Hence the result.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $11$
Category:Factorials
Legendre Polynomial/Examples

== Examples of Legendre Polynomials ==
The first few Legendre polynomials are:
=== Legendre Polynomial $\map {P_0} x$ ===
=== Legendre Polynomial $\map {P_1} x$ ===
=== Legendre Polynomial $\map {P_2} x$ ===
=== Legendre Polynomial $\map {P_3} x$ ===
=== Legendre Polynomial: $\map {P_4} x$ ===
=== Legendre Polynomial: $\map {P_5} x$ ===
=== Legendre Polynomial: $\map {P_6} x$ ===
=== Legendre Polynomial: $\map {P_7} x$ ===
== Sources ==
* 
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P0

== Example of Legendre Polynomial ==
The zeroth Legendre polynomial is:
:$\map {P_0} x = 1$
== Proof ==
From Generating Function for Legendre Polynomials, $\map {P_0} x$ is the coefficient of the zeroth term of the generating function:
:$\map G t = \dfrac 1 {\sqrt {1 - 2 x t + t^2} }$
Setting $t = 0$ in $\map G t$:
Thus:
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.3$
* 
* 
* 
* 
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.3.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P1

== Example of Legendre Polynomial ==
The $1$st Legendre polynomial is:
:$\map {P_1} x = x$
== Proof ==
From Generating Function for Legendre Polynomials, $\map {P_1} x$ is the coefficient of the term in $t^1$ of the generating function:
:$\map G t = \dfrac 1 {\sqrt {1 - 2 x t + t^2} }$
Thus:
Further terms generate terms in $t$ of higher powers.
Hence the term in $t^1$ is seen to be $x t$.
Hence, by definition of generating function:
:$\map {P_1} x = x$
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.4$
* 
* 
* 
* 
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.4.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P2

== Example of Legendre Polynomial ==
The $2$nd Legendre polynomial is:
:$\map {P_2} x = \dfrac 1 2 \paren {3 x^2 - 1}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.5$
* 
* 
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.5.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P3

== Example of Legendre Polynomial ==
The $3$rd Legendre polynomial is:
:$\map {P_3} x = \dfrac 1 2 \paren {5 x^3 - 3 x}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.6$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.6.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P4

== Example of Legendre Polynomial ==
The $4$th Legendre polynomial is:
:$\map {P_4} x = \dfrac 1 8 \paren {35 x^4 - 30 x^2 + 3}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.7$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.7.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P5

== Example of Legendre Polynomial ==
The $5$th Legendre polynomial is:
:$\map {P_5} x = \dfrac 1 8 \paren {63 x^5 - 70 x^3 + 15 x}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.8$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.8.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P6

== Example of Legendre Polynomial ==
The $6$th Legendre polynomial is:
:$\map {P_6} x = \dfrac 1 {16} \paren {231 x^6 - 315 x^4 + 105 x^2 - 5}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.9$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.9.$
* 
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples/P7

== Example of Legendre Polynomial ==
The $7$th Legendre polynomial is:
:$\map {P_7} x = \dfrac 1 {16} \paren {429 x^7 - 693 x^5 + 315 x^3 - 35 x}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Special Legendre Polynomials: $25.10$
* : $\S 28$: Legendre and Associated Legendre Functions: Special Legendre Polynomials: $28.10.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines

== Examples of Legendre Polynomials ==
The first few Legendre polynomials in terms of cosines are:
=== Legendre Polynomial $\map {P_0} {\cos \theta}$ ===
=== Legendre Polynomial $\map {P_1} {\cos \theta}$ ===
=== Legendre Polynomial $\map {P_2} {\cos \theta}$ ===
=== Legendre Polynomial $\map {P_3} {\cos \theta}$ ===
=== Legendre Polynomial: $\map {P_4} {\cos \theta}$ ===
=== Legendre Polynomial: $\map {P_5} {\cos \theta}$ ===
=== Legendre Polynomial: $\map {P_6} {\cos \theta}$ ===
=== Legendre Polynomial: $\map {P_7} {\cos \theta}$ ===
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P0

== Example of Legendre Polynomial ==
The zeroth Legendre polynomial of the cosine of $\theta$ is:
:$\map {P_0} {\cos \theta} = 1$
== Proof ==
From Legendre Polynomial $P_0$ of $x$:
:$\map {P_0} x = 1$
Substituting $\cos \theta$ for $x$ yields (trivially):
:$\map {P_0} {\cos \theta} = 1$
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.11$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.11.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P1

== Example of Legendre Polynomial ==
The $1$st Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_1} {\cos \theta} = \cos \theta$
== Proof ==
From Legendre Polynomial $P_1$ of $x$:
:$\map {P_1} x = x$
Substituting $\cos \theta$ for $x$ yields (trivially):
:$\map {P_1} {\cos \theta} = \cos \theta$
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.12$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.12.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P2

== Example of Legendre Polynomial ==
The $2$nd Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_2} {\cos \theta} = \dfrac 1 4 \paren {1 + 3 \cos 2 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.13$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.13.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P3

== Example of Legendre Polynomial ==
The $3$rd Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_3} {\cos \theta} = \dfrac 1 8 \paren {3 \cos \theta + 5 \cos 3 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.14$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.14.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P4

== Example of Legendre Polynomial ==
The $4$th Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_4} {\cos \theta} = \dfrac 1 {64} \paren {9 + 20 \cos 2 \theta + 35 \cos 4 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.15$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.15.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P5

== Example of Legendre Polynomial ==
The $5$th Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_5} {\cos \theta} = \dfrac 1 {128} \paren {30 \cos \theta + 35 \cos 3 \theta + 63 \cos 5 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.16$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.16.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P6

== Example of Legendre Polynomial ==
The $6$th Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_6} {\cos \theta} = \dfrac 1 {512} \paren {50 + 105 \cos 2 \theta + 126 \cos 4 \theta + 231 \cos 6 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.17$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.17.$
Category:Examples of Legendre Polynomials
Legendre Polynomial/Examples in Terms of Cosines/P7

== Example of Legendre Polynomial ==
The $7$th Legendre polynomial of the cosine of $\theta$ is given by:
:$\map {P_7} {\cos \theta} = \dfrac 1 {1024} \paren {175 \cos \theta + 189 \cos 3 \theta + 231 \cos 5 \theta + 429 \cos 7 \theta}$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $25.18$
* : $\S 28$: Legendre and Associated Legendre Functions: Legendre Polynomials in Terms of $\theta$ where $x = \cos \theta$: $28.18.$
Category:Examples of Legendre Polynomials
Legendre Polynomial Derivative Recursion Formula

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\dfrac \d {\d x} \map {P_{n + 1} } x - x \dfrac \d {\d x} \map {P_n} x = \paren {n + 1} \map {P_n} x$
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.21$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.21.$
Category:Legendre Polynomial Derivative Recursion Formula
Category:Legendre Polynomials
Legendre Polynomial Derivative Recursion Formula/Corollary 1

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$x \dfrac \d {\d x} \map {P_n} x - \dfrac \d {\d x} \map {P_{n - 1} } x = n \map {P_n} x$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.22$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.22.$
Category:Legendre Polynomial Derivative Recursion Formula
Legendre Polynomial Derivative Recursion Formula/Corollary 2

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\dfrac \d {\d x} \map {P_{n + 1} } x - \dfrac \d {\d x} \map {P_{n - 1} } x = \paren {2 n + 1} \map {P_n} x$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.23$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.23.$
Category:Legendre Polynomial Derivative Recursion Formula
Legendre Polynomial Derivative Recursion Formula/Corollary 3

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then:
:$\paren {x^2 - 1} \dfrac \d {\d x} \map {P_n} x = n x \map {P_n} x - n \map {P_{n - 1} } x$
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Recurrence Formulas for Legendre Polynomials: $25.24$
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.24.$
Category:Legendre Polynomial Derivative Recursion Formula
Legendre Polynomial Derivative Recursion Formula/Corollary 3/Mistake

== Source Work ==
:Chapter $28$: Legendre and Associated Legendre Functions
::Recurrence Formulas for Legendre Polynomials: $28.24.$
== Mistake ==
:$\paren {x^2 - 1} \dfrac \d {\d x} \map {P_n} x - n x \map {P_n} x - n \map {P_{n - 1} } x$
== Correction ==
An equals sign has been replaced by a minus sign.
This is the correct expression:
:$\paren {x^2 - 1} \dfrac \d {\d x} \map {P_n} x = n x \map {P_n} x - n \map {P_{n - 1} } x$
The mistake appears to have crept in since the $1$st edition, which presented this correctly.
== Sources ==
* : $\S 28$: Legendre and Associated Legendre Functions: Recurrence Formulas for Legendre Polynomials: $28.24.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Legendre Polynomial of Even Index is Even Function

== Theorem ==
Let $n \in \N$ be an even natural number: $n \in \set {0, 2, 4, \ldots}$
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then $\map {P_n} x$ is an even function.
== Proof ==
From Legendre Polynomial of $-x$:
:$\map {P_n} x = \paren {-1}^n \map {P_n} {-x}$
When $n$ is even we have:
:$\paren {-1}^n = 1$
and so:
:$\map {P_n} x = \map {P_n} {-x}$
Hence the result by definition of even function.
== Also see ==
* Legendre Polynomial of Odd Index is Odd Function
Category:Legendre Polynomials
Category:Examples of Even Functions
Legendre Polynomial of Odd Index is Odd Function

== Theorem ==
Let $n \in \N$ be an odd natural number: $n \in \set {1, 3, 5, \ldots}$
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
Then $\map {P_n} x$ is an odd function.
== Proof ==
From Legendre Polynomial of $-x$:
:$\map {P_n} x = \paren {-1}^n \map {P_n} {-x}$
When $n$ is odd we have:
:$\paren {-1}^n = -1$
and so:
:$\map {P_n} x = -\map {P_n} {-x}$
Hence the result by definition of odd function.
== Also see ==
* Legendre Polynomial of Even Index is Even Function
Category:Legendre Polynomials
Category:Examples of Odd Functions
Legendre Symbol/Examples/2 over 7

== Example of Use of Legendre Symbol ==
:$\paren {\dfrac 2 7} = 1$
== Proof ==
From Quadratic Residue Examples: $7$:
That is, $2$ is a quadratic residue of $7$:
:$2^2 \equiv 4 \pmod 7$
Hence the result by definition of Legendre symbol.
== Sources ==
* 
* 
Category:Examples of Use of Legendre Symbol
Legendre Symbol/Examples/3 over 7

== Example of Use of Legendre Symbol ==
:$\paren {\dfrac 3 7} = -1$
== Proof ==
From Quadratic Residue Examples: $7$:
That is, $3$ is not a quadratic residue of $7$.
Hence the result by definition of Legendre symbol.
== Sources ==
* 
* 
Category:Examples of Use of Legendre Symbol
Legendre Symbol is Multiplicative

== Theorem ==
Let $p$ be a odd prime.
Let $a, b \in \Z$.
Then:
:$\paren {\dfrac {a b} p} = \paren {\dfrac a p} \paren {\dfrac b p}$
where $\paren {\dfrac a p}$ is the Legendre symbol.
== Proof ==
We have: 
Category:Legendre Symbol
Legendre Symbol of Congruent Integers

== Theorem ==
Let $p$ be a odd prime.
Let $a, b \in \Z$ be such that $a \equiv b \pmod p$.
Then:
:$\paren {\dfrac a p} = \paren {\dfrac b p}$
where $\paren {\dfrac a p}$ is the Legendre symbol.
== Proof ==
Category:Legendre Symbol
Legendre Transform is Involution

== Theorem ==
The Legendre transform is an Involution.
== Proof ==
Let $\map f x$ be a strictly convex real function.
Let $p = \map {f'} x$.
By definition of the Legendre transform, the transformed real function is of the form:
:$\map {f^*} p = - \map f {\map x p} + p \map x p$
By Legendre Transform of Strictly Convex Real Function is Strictly Convex, $f^*$ is strictly convex.
Let $t = \map { {f^*}'} p$.
Let $\map {\paren {f^*}^*} t = - \map {f^*} {\map p t} + t \map p t$.
Then:
$t$ is an arbitrary independent variable, hence can be renamed.
Set $t = x$.
Then:
:$\tuple {t,f^{**} } = \tuple {x, f}$
which is the original pair of function and its variable.
Hence, by definition, the Legendre transform is an involution.
==Sources==
* : $\S 4.18$: The Legendre Tranformation
Category:Calculus of Variations
Legendre Transform of Strictly Convex Real Function is Strictly Convex

== Theorem ==
Let $\map f x$ be a strictly convex real function.
Then the function $\map {f^*} p$ acquired through the Legendre Transform is also strictly convex.
== Proof ==
We have that $\map f x$ is real and strictly convex.
Hence, by Real Function is Strictly Convex iff Derivative is Strictly Increasing, $\map {f'} x$ is strictly increasing.
Then:
:$\map {f} x > 0$
Therefore, the first derivative of $f^*$ is strictly increasing.
By Real Function is Strictly Convex iff Derivative is Strictly Increasing, $f^*$ is strictly convex.
== Sources ==
* : $\S 4.18$: The Legendre Tranformation
Category:Calculus of Variations
Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle

== Theorem ==
Let $a$ and $b$ be the legs of a Pythagorean triangle $P_1$.
Let $\tuple {a, b}$ be used as the generator for a new Pythagorean triangle $P_2$.
Then the hypotenuse of $P_2$ is the square of the hypotenuse of $P_1$.
== Proof ==
By Pythagoras's Theorem, the square of the hypotenuse of $P_1$ is $a^2 + b^2$.
By Solutions of Pythagorean Equation, the sides of $P_2$ can be expressed as $\tuple {2 a b, a^2 - b^2, a^2 + b^2}$, where the hypotenuse is $a^2 + b^2$.
== Examples ==
== Historical Note ==
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Category:Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle
Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle/Examples/3-4-5

== Example of Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle ==
The $3-4-5$ triangle has hypotenuse $5$.
:
By Solutions of Pythagorean Equation, the Pythagorean triangle $P_2$ formed by the generator $\tuple {4, 3}$ is:
It is observed that $25 = 5^2$.
:
== Sources ==
* : $13$
* : $13$
Category:Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle
Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle/Historical Note

== Historical Note on Legs of Pythagorean Triangle used as Generator for another Pythagorean Triangle ==
According to in of $1997$, this observation was made by , although no details are given of where or when.
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Leibniz's Formula for Pi

== Theorem ==
:$\dfrac \pi 4 = 1 - \dfrac 1 3 + \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 - \cdots \approx 0 \cdotp 78539 \, 81633 \, 9744 \ldots$
That is:
:$\ds \pi = 4 \sum_{k \mathop \ge 0} \paren {-1}^k \frac 1 {2 k + 1}$
== Elementary Proof ==
== Leibniz's Proof ==
== Proof by Taylor Expansion ==
== Proof by Dirichlet Beta Function ==
== Proof by Digamma Function ==
== Also known as ==
== Also see ==
* Gregory Series
== Historical Note ==
== Sources ==
* : Chapter $\text{VII}$: Master of All Trades
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.15$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $0 \cdotp 78539 \, 81 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.15.$
Category:Leibniz's Formula for Pi
Category:Formulas for Pi
Leibniz's Formula for Pi/Elementary Proof

== Theorem ==
== Proof ==
First we note that:
:$(1): \quad \dfrac 1 {1 + t^2} = 1 - t^2 + t^4 - t^6 + \cdots + t^{4 n} - \dfrac {t^{4 n + 2} } {1 + t^2}$
which is demonstrated here.
Now consider the real number $x \in \R: 0 \le x \le 1$.
We can integrate expression $(1)$ $t$ from $0$ to $x$:
:$(2): \quad \ds \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots + \frac {x^{4 n + 1} } {4 n + 1} - \map {R_n} x$
where:
:$\ds \map {R_n} x = \int_0^x \frac {t^{4 n + 2} } {1 + t^2} \rd t$
From Square of Real Number is Non-Negative we have that:
:$t^2 \ge 0$
and so:
:$1 \le 1 + t^2$
From Relative Sizes of Definite Integrals, we have:
:$\ds 0 \le \map {R_n} x \le \int_0^x t^{4 n + 2} \rd t$
that is:
:$0 \le \map {R_n} x \le \dfrac {x^{4n + 3} } {4 n + 3}$
But as $0 \le x \le 1$ it is clear that:
:$\dfrac {x^{4 n + 3} } {4 n + 3} \le \dfrac 1 {4 n + 3}$
So:
:$0 \le \map {R_n} x \le \dfrac 1 {4 n + 3}$
From Basic Null Sequences and the Squeeze Theorem, $\dfrac 1 {4 n + 3} \to 0$ as $n \to \infty$.
This leads us directly to:
:$(3): \quad \ds \int_0^x \frac {\d t} {1 + t^2} = x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \frac {x^9} 9 \cdots$
But from Derivative of Arctangent Function, we also have that:
:$\dfrac \d {\d x} \arctan t = \dfrac 1 {1 + t^2}$
and thence from the Fundamental Theorem of Calculus we have:
:$\ds \arctan x = \int_0^x \frac {\d t} {1 + t^2}$
From $(3)$ it follows immediately that:
:$(4): \quad \arctan x = x - \dfrac {x^3} 3 + \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 \cdots$
Now all we need to do is plug $x = 1$ into $(4)$.
=== Comment ===
Note that we did not just take the Sum of Infinite Geometric Sequence:
:$\dfrac 1 {1 - \paren {-t^2} } = 1 + \paren {-t^2} + \paren {-t^2}^2 + \paren {-t^2}^3 + \cdots$
and integrate it term by term, as we have not at this stage proved that this is permissible.
== Sources ==
* : Chapter $\text {B}.10$: An elementary proof of Leibniz's Formula $\pi / 4 = 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \ldots$
Category:Leibniz's Formula for Pi
Category:Formulas for Pi
Leibniz's Formula for Pi/Historical Note

== Historical Note on Leibniz's Formula for Pi ==
 discovered his formula for $\pi$ in $1673$.
He took great pleasure and pride in this discovery.
:It's as if, by this expansion, the veil which hung over that strange number had been drawn aside.
:::: -- 
Simple as it is, Leibniz's Formula for Pi is inefficient, in that it needs hundreds of terms in order to calculate a few decimal places.
Some sources ascribe this formula to .
== Sources ==
* : Chapter $\text{VII}$: Master of All Trades
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : Chapter $\text {A}.19$: Leibniz ()
* : Chapter $\text {B}.13$: How Leibniz Discovered his Formula $\pi / 4 = 1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \cdots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Historical Notes