nyuuzyou/wikis huggingface dataset:
Reciprocal of Square of 1 Less than Number Base

== Theorem ==
Let $b \in \Z$ be an integer such that $b > 2$.
Let $n = \paren {b - 1}^2$.
The reciprocal of $n$, expressed in base $b$, recurs with period $b - 1$:
:$\dfrac 1 n = \sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b = \sqbrk {0 \cdotp 012 \ldots cd012 \ldots}_b$
where:
:$c = b - 3$
:$d = b - 1$
== Proof ==
By Basis Representation Theorem, the number $\sqbrk {12 \ldots cd}_b$ can be written as:
Thus the recurring basis expansion $\sqbrk {0 \cdotp \dot 012 \ldots c \dot d}_b$ can be written as:
Hence the result.
== Examples ==
== Sources ==
* : $81$
* : $81$
Category:Reciprocals
Category:Reciprocal of Square of 1 Less than Number Base
Reciprocal of Square of 1 Less than Number Base/Examples/6

== Example of Reciprocal of Square of 1 Less than Number Base ==
Expressed in base $6$, the reciprocal of $5^2$ has the expansion:
:$\dfrac 1 {5^2} = \sqbrk {0 \cdotp \dot 0123 \dot 5}_6$
== Proof ==
From Square of $1$ Less than Number Base: $5$ we have:
:$5^2 = 41_6$
Performing the calculation using long division:
 0.0123501...
 ----------------
41)1.0000000000
 41
 --
 150
 122
 ---
 240
 203
 ---
 330
 325
 ---
 100
 41
 --
 ...
== Sources ==
* : $81$
* : $81$
Category:Reciprocal of Square of 1 Less than Number Base
Reciprocal of Strictly Negative Real Number is Strictly Negative

== Theorem ==
:$\forall x \in \R: x < 0 \implies \dfrac 1 x < 0$
== Proof ==
Let $x < 0$.
 $\dfrac 1 x > 0$.
Then:
But from Real Zero is Less than Real One:
:$1 > 0$
Therefore by Proof by Contradiction:
:$\dfrac 1 x < 0$
== Also see ==
* Reciprocal of Strictly Positive Real Number is Strictly Positive
Category:Real Numbers
Category:Reciprocals
Reciprocal of Strictly Positive Real Number is Strictly Positive

== Theorem ==
:$\forall x \in \R: x > 0 \implies \dfrac 1 x > 0$
== Proof ==
Let $x > 0$.
 $\dfrac 1 x < 0$.
Then:
But from Real Zero is Less than Real One:
:$1 > 0$
Therefore by Proof by Contradiction:
:$\dfrac 1 x > 0$
== Also see ==
* Reciprocal of Strictly Negative Real Number is Strictly Negative
== Sources ==
* : $\S 1$: Real Numbers: $\S 1.5$: Example
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(i)}$
Category:Reciprocals
Category:Real Numbers
Category:Inequalities
Reciprocal times Derivative of Gamma Function

== Theorem ==
Let $z \in \C \setminus \Z_{\le 0}$.
Then:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$
where:
:$\map \Gamma z$ denotes the Gamma function
:$\map {\Gamma'} z$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Also see ==
* Extension of Harmonic Number to Non-Integer Argument
== Sources ==
* : $\S 16$: Derivatives of the Gamma Function: $16.14$
* : $\S 25$: The Gamma Function: Derivatives of the Gamma Function: $25.14.$
Category:Reciprocal times Derivative of Gamma Function
Category:Digamma Function
Category:Gamma Function
Category:General Harmonic Numbers
Reciprocal times Derivative of Gamma Function/Corollary 1

== Corollary to Reciprocal times Derivative of Gamma Function ==
Let $z \in \C \setminus \Z_{\le 0}$.
Then:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \int_{\mathop \to 0}^{\mathop \to 1} \paren {\dfrac {1 - t^{z - 1} } {1 - t} } \rd t$
where:
:$\map \Gamma z$ denotes the Gamma function
:$\map {\Gamma'} z$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
Category:Reciprocal times Derivative of Gamma Function
Reciprocal times Derivative of Gamma Function/Corollary 2

== Corollary to Reciprocal times Derivative of Gamma Function ==
Let $a$ and $b \in \C$ such that $\paren {\dfrac a {2 b} + 1} \in \C \setminus \Z_{\le 0}$ and $\paren {\dfrac a {2 b} + \dfrac 1 2 } \in \C \setminus \Z_{\le 0}$
Then:
:$\ds \map \psi {\dfrac a {2 b} + 1} - \map \psi {\dfrac a {2 b} + \dfrac 1 2} = 2 b \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {a + b k} $
where:
:$\psi$ is the digamma function
== Proof ==
== Sources ==
* : Chapter $8$. Analogues of the Gamma Function
Category:Reciprocal times Derivative of Gamma Function
Category:Digamma Function
Reciprocal times Derivative of Gamma Function/Examples

== Examples of Use of Reciprocal times Derivative of Gamma Function==
=== Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + k}$ ===
=== Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 2 k}$ ===
=== Example: $\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 3 k}$ ===
=== Example: $\map {H^{\paren 1} } {\dfrac 1 2}$ ===
=== Example: $\map {H^{\paren 1} } {-\dfrac 1 2}$ ===
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Reciprocal times Derivative of Gamma Function/Examples/Sum of Reciprocal of 1 + 2k Alternating in Sign

== Example of Use of Reciprocal times Derivative of Gamma Function ==
:$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 2 k} = 1 - \dfrac \pi 4$
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Reciprocal times Derivative of Gamma Function/Examples/Sum of Reciprocal of 1 + 3k Alternating in Sign

== Example of Use of Reciprocal times Derivative of Gamma Function ==
:$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + 3 k} = 1 - \dfrac 1 3 \ln 2 - \dfrac {\pi \sqrt 3} 9$
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Reciprocal times Derivative of Gamma Function/Examples/Sum of Reciprocal of 1 + k Alternating in Sign

== Example of Use of Reciprocal times Derivative of Gamma Function ==
:$\ds \sum_{k \mathop = 1}^\infty \dfrac {\paren {-1}^{k + 1} } {1 + k} = 1 - \ln 2$
== Proof ==
Category:Examples of Use of Reciprocal times Derivative of Gamma Function
Reciprocals of Odd Numbers adding to 1

== Theorem ==
$105$ is the smallest positive integer $n$ such that $1$ can be expressed as the sum of reciprocals of distinct odd integers such that none are less than $\dfrac 1 n$:
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {33} + \dfrac 1 {35} + \dfrac 1 {45} + \dfrac 1 {55} + \dfrac 1 {77} + \dfrac 1 {105}$
There are $5$ ways of expressing $1$ as the sum of reciprocals of only $9$ distinct odd integers, but then the least term is less than $\dfrac 1 {105}$.
The $9$-term solutions are as follows:
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {35} + \dfrac 1 {45} + \dfrac 1 {231}$
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {231} + \dfrac 1 {315}$
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {33} + \dfrac 1 {45} + \dfrac 1 {385}$
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {165} + \dfrac 1 {693}$
:$1 = \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} + \dfrac 1 {15} + \dfrac 1 {21} + \dfrac 1 {135} + \dfrac 1 {10 \, 395}$
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $105$
* : Egyptian Fractions: $4$
* : $105$
Category:Reciprocals
Category:Unit Fractions
Reciprocals of Odd Numbers adding to 1/Historical Note

== Theorem ==
 reports in his of $1986$ that this result was reported by and in Volume $10$ of the , but no information has so far been found to back this up.
It is reported in the various relevant pages in the On-Line Encyclopedia of Integer Sequences that this set was apparently discovered by S. Yamashita in $1976$, but once more corroborative evidence is hard to come by.
Category:Historical Notes
Reciprocals of Odd Numbers adding to 1/Mistake

== Source Work ==
:
:The Dictionary
::$105$
:
:The Dictionary
::$105$
== Mistake ==
:There are $4$ ways of representing $1$ as the sum of odd reciprocals, using only $9$ of them ...
There are in fact $5$ such ways, as can be seen in Reciprocals of Odd Numbers adding to 1.
== Sources ==
* : $105$
* : $105$
Category:Mistakes/Curious and Interesting Numbers
Reciprocals of Prime Numbers

== Theorem ==
The decimal representations of the reciprocals of the first few prime numbers are as follows:
:
$n$ 	$1 / n$ 	Period	Also see
$2$ 	$0 \cdotp 5$ 	$0$ 	Reciprocal of $2$
$3$ 	$0 \cdotp \dot 3$ 	$1$ 	Reciprocal of $3$
$5$ 	$0 \cdotp 2$ 	$0$ 	Reciprocal of $5$
$7$ 	$0 \cdotp \dot 14285 \, \dot 7$ 	$6$ 	Reciprocal of $7$
$11$	$0 \cdotp \dot 0 \dot 9$ 	$2$ 	Reciprocal of $11$
$13$	$0 \cdotp \dot 07692 \, \dot 3$ 	$6$ 	Reciprocal of $13$
$17$	$0 \cdotp \dot 05882 \, 35294 \, 11764 \, \dot 7$ 	$16$ 	Reciprocal of $17$
$19$	$0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$ 	$18$ 	Reciprocal of $19$
$23$	$0 \cdotp \dot 04347 \, 82608 \, 69565 \, 21739 \, 1 \dot 3$ 	$22$ 	Reciprocal of $23$
$29$	$0 \cdotp \dot 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 93 \dot 1$ 	$28$ 	Reciprocal of $29$
$31$	$0 \cdotp \dot 03225 \, 80645 \, 1612 \dot 9$ 	$15$ 	Reciprocal of $31$
$37$	$0 \cdotp \dot 02 \dot 7$ 	$3$ 	Reciprocal of $37$
$41$	$0 \cdotp \dot 0243 \dot 9$ 	$5$ 	Reciprocal of $41$
$43$	$0 \cdotp \dot 02325 \, 58139 \, 53488 \, 37209 \, \dot 3$ 	$21$ 	Reciprocal of $43$
$47$	$0 \cdotp \dot 02127 \, 65957 \, 44680 \, 85106 \, 38297 \, 87234 \, 04255 \, 31914 \, 89361 \, \dot 7$ 	$46$ 	Reciprocal of $47$
$53$	$0 \cdotp \dot 01886 \, 79245 \, 28 \dot 3$ 	$13$ 	Reciprocal of $53$
$59$	$0 \cdotp \dot 01694 \, 91525 \, 42372 \, 88135 \, 59322 \, 03389 \, 83050 \, 84745 \, 76271 \, 18644 \, 06779 \, 66 \dot 1$ 	$58$ 	Reciprocal of $59$
$61$	$0 \cdotp \dot 01639 \, 34426 \, 22950 \, 81967 \, 21311 \, 47540 \, 98360 \, 65573 \, 77049 \, 18032 \, 78688 \, 5245 \dot 9$ 	$60$ 	Reciprocal of $61$
$67$	$0 \cdotp \dot 01492 \, 53731 \, 34328 \, 35820 \, 89552 \, 23880 \, 59 \dot 7$ 	$33$ 	Reciprocal of $67$
$71$	$0 \cdotp \dot 01408 \, 45070 \, 42253 \, 52112 \, 67605 \, 63380 \, 2816 \dot 9$ 	$35$ 	Reciprocal of $71$
$73$	$0 \cdotp \dot 01369 \, 86 \dot 3$ 	$8$ 	Reciprocal of $73$
$79$	$0 \cdotp \dot 01265 \, 82278 \, 48 \dot 1$ 	$13$ 	Reciprocal of $79$
$83$	$0 \cdotp \dot 01204 \, 81927 \, 71084 \, 33734 \, 93975 \, 90361 \, 44578 \, 31325 \, \dot 3$ 	$41$ 	Reciprocal of $83$
$89$	$0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$ 	$44$ 	Reciprocal of $89$
$97$	$0 \cdotp \dot 01030 \, 92783 \, 50515 \, 46391 \, 75257 \, 73195 \, 87628 \, 86597 \, 93814 \, 43298 \, 96907 \, 21649 \, 48453 \, 60824 \, 74226 \, 80412 \, 37113 \, 40206 \, 18556 \, \dot 7$	$96$ 	Reciprocal of $97$
== Sources ==
* : Tables: $6$ The Decimal Reciprocals of the Primes from $7$ to $97$
* : Tables: $6$ The Decimal Reciprocals of the Primes from $7$ to $97$
Category:Prime Numbers
Category:Examples of Reciprocals
Reciprocals whose Decimal Expansion contain Equal Numbers of Digits from 0 to 9

== Theorem ==
The following positive integers $p$ have reciprocals whose decimal expansions:
:$(1): \quad$ have the maximum period, that is: $p - 1$
:$(2): \quad$ have an equal number, $\dfrac {p - 1} {10}$, of each of the digits from $0$ to $9$:
::$61$, $131$, $\ldots$
== Proof ==
From Reciprocal of $61$:
From Reciprocal of $131$:
== Sources ==
* : $61$
Category:Reciprocals
Rectangle Contained by Medial Straight Lines Commensurable in Length is Medial

== Theorem ==
== Proof ==
Let $x$ and $\lambda x$ be two medials such that:
:$x \frown \lambda x$
where $\frown$ denotes that $x$ and $\lambda x$ are commensurable in length.
By Areas of Triangles and Parallelograms Proportional to Base:
:$x^2 : x \cdot \lambda x = x : \lambda x$
From Commensurability of Elements of Proportional Magnitudes: 
:$x^2 \frown x \cdot \lambda x$
From Medial is Irrational we have that $x^2$ is medial.
Therefore by Straight Line Commensurable with Medial Straight Line is Medial: Porism:
:$x \cdot \lambda x$ is medial.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Rectangle Divided into Differently Shaped Equal Area Subrectangles

== Theorem ==
Let $R$ be a rectangle.
Let $R$ be divided into $n$ smaller rectangles which are of equal area but with different lengths of sides.
Then $n \ge 7$.
== Proof ==
== Sources ==
* : $7$
* : $7$
Category:Recreational Mathematics
Category:Rectangles
Category:7
Category:Dissections
Rectangle Divided into Incomparable Subrectangles

== Theorem ==
Let $R$ be a rectangle.
Let $R$ be divided into $n$ smaller rectangles which are pairwise incomparable.
Then $n \ge 7$.
The smallest rectangle with integer sides that can be so divided into rectangles with integer sides is $13 \times 22$.
:
== Proof ==
== Sources ==
* 
* : $7$
* : $7$
Category:Recreational Mathematics
Category:Rectangles
Category:7
Category:Dissections
Rectangle is Parallelogram

== Theorem ==
Let $ABCD$ be a rectangle.
Then $ABCD$ is a parallelogram.
== Proof ==
Let $ABCD$ be a rectangle.
 $ABCD$ is not a parallelogram.
, let line segments $AB$ and $CD$ not be parallel.
By , let us produce $AB$ and $CD$ into two infinite straight lines.
By the Parallel Postulate, the $AD$ and $BC$ will eventually meet at one side or the other.
Let their point of intersection be $E$.
There are two possibilities:
:$ADE$ is a triangle
:$CBE$ is a triangle.
, let $\triangle ADE$ be a triangle:
:
By Sum of Angles of Triangle equals Two Right Angles, $\angle ADE + \angle DEA + \angle EAD$ add to two right angles.
But by the definition of a rectangle, $\angle ADE + \angle EAD$ add to two right angles.
As $\angle DEA$ is not a zero angle (else $\triangle ADE$ would not be a triangle), this cannot be so.
Hence $ABCD$ is not a rectangle.
But , $ABCD$ is a rectangle.
This is a contradiction.
Hence by Proof by Contradiction, $AB$ and $CD$ are parallel.
The same argument can be used to demonstrate that $AD$ and $BC$ are also parallel.
Hence the result by definition of parallelogram.
Category:Rectangles
Category:Parallelograms
Rectangle is Sum of Square and Rectangle

== Theorem ==
== Proof ==
:
Let $AB$ be the given straight line cut at random at the point $C$.
Construct the square $CDEB$ on $AB$.
Produce $ED$ to $F$ and construct $AF$ parallel to $CD$.
Then $\Box ABEF = \Box ACDF + \Box CBED$.
Now $\Box ABEF$ is the rectangle contained by $AB$ and $BC$, as $BC = CD = AF$.
Similarly, from Opposite Sides and Angles of Parallelogram are Equal:
: $\Box ACDF$ is the rectangle contained by $AC$ and $BC$, as $BC = AF$
: $\Box CBED$ is the square on $BC$.
So the rectangle contained by $AB$ and $BC$ equals the rectangle contained by $AC$ and $BC$ together with the square on $BC$.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Rectangles
Category:Squares
Rectangles Contained by Proportional Straight Lines

== Theorem ==
Note: in the above, equal is to be taken to mean of equal area.
== Proof ==
Let the four straight lines $AB, CD, E, F$ be proportional, that is, $AB : CD = E : F$.
What we need to show is that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.
:
Let $AG, CH$ be drawn perpendicular to $AB$ and $CD$.
Let $AG = F$, $CH = E$.
Complete the parallelograms $BG$ and $DH$.
We have that $AB : CD = E : F$, while $E = CH$ and $F = AG$.
So in $\Box BG$ and $\Box DH$ the sides about the equal angles are reciprocally proportional.
But from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional‎:
: $\Box BG = \Box DH$ (in area).
We also have that:
: $\Box BG$ is the rectangle contained by $AB$ and $F$
: $\Box DH$ is the rectangle contained by $CD$ and $E$
Hence the result.
Now suppose that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.
We use the same construction, and note that $\Box BG = \Box DH$ (in area).
But they are equiangular, as all angles are equal to a right angle.
So from Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional‎:
: $AB : CD = CH : AG$
But $E = CH$ and $F = AG$.
So:
: $AB : CD = E : F$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Rectangles
Rectangles Contained by Three Proportional Straight Lines

== Theorem ==
== Proof ==
Let the three straight lines $A, B, C$ be proportional, that is: 
:$A : B = B : C$
Then we need to show that the rectangle contained by $A$ and $C$ equals the square on $B$.
:
Let $D = B$.
Then $A : B = D : C$
By Rectangles Contained by Proportional Straight Lines, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$.
But as $B = D$, the rectangle contained by $B$ and $D$ equals the square on $B$.
So the rectangle contained by $A$ and $C$ equals the square on $B$.
Now let the rectangle contained by $A$ and $C$ be equal to the square on $B$.
Using the same construction, the rectangle contained by $A$ and $C$ equals the rectangle contained by $B$ and $D$ because $B = D$.
So by Rectangles Contained by Proportional Straight Lines:
: $A : B = D : C$
But as $B = D$:
: $A : B = B : C$
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Rectangles
Rectangles with Equal Bases and Equal Altitudes are Congruent

== Theorem ==
Let $ABCD$ and $EFGH$ be rectangles.
Then $ABCD$ and $EFGH$ are congruent if:
: the base of $ABCD$ equals the base of $EFGH$
: the altitude of $ABCD$ equals the altitude of $EFGH$.
== Proof ==
A rectangle is a parallelogram whose vertices are right angles.
Thus the altitudes of $ABCD$ and of $EFGH$ coincide with the sides of $ABCD$ and $EFGH$ which are adjacent to the bases.
The result then follows from Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.24$: Corollary
Category:Rectangles
Rectangular Delta Sequence

== Theorem ==
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\map {\delta_n} x := \begin{cases}
0 & : x < - \frac 1 {2n} \\
n & : - \frac 1 {2n} \le x \le \frac 1 {2n} \\
0 & : x > \frac 1 {2n} \end{cases}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function.
Then:
== Sources ==
: Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
Rectangular Formula for Definite Integrals

== Theorem ==
Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, n}: x_r - x_{r - 1} = \dfrac {b - a} n$
Then the definite integral of $f$ $x$ from $a$ to $b$ can be approximated as:
:$\ds \int_a^b \map f x \rd x \approx \sum_{r \mathop = 0}^{n - 1} h \map f {x_r}$
where $h = \dfrac {b - a} n$.
== Proof ==
== Also see ==
* Trapezium Rule for Definite Integrals
* Simpson's Rule
== Sources ==
* : $\S 15$: Approximate Formulas for Definite Integrals: $15.15$
* : $\S 18$: Definite Integrals: Approximate Formulas for Definite Integrals: $18.15$
Category:Definite Integrals
Recurrence Formula for Associated Legendre Function of the First Kind/Formulation 1

== Theorem ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Then:
:$\paren {n + 1 - m} \map { {P_{n + 1} }^m} x - \paren {2 n + 1} x \map { {P_n}^m} x + \paren {n + m} \map { {P_{n - 1} }^m} x = 0$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Recurrence Formulas: $26.12$
* : $\S 28$: Legendre and Associated Legendre Functions: Generating Function for $\map { {P_n}^m} x$: $28.60.$
Category:Recurrence Formula for Associated Legendre Function of the First Kind
Recurrence Formula for Bessel Function of the First Kind

== Theorem ==
Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
:$\map {J_{n + 1} } x = \dfrac {2 n} x \map {J_n} x - \map {J_{n - 1} } x$
And: 
:$\map {J_{n + 1} } x = -2 \map {J_n'} x + \map {J_{n - 1} } x$
== Proof ==
From Generating Function for Bessel Function of the First Kind of Order n of x we have:
:$\ds \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} } = \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n$
Differentiating both sides of the equation with respect to $t$:
This is the first recurrence formula.
We prove the second recurrence formula by differentiating both sides of the original equation with respect to $x$:
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {II}$. Bessel functions: $2$
Category:Bessel Functions
Category:Recurrence Relations
Recurrence Formula for Sum of Sequence of Fibonacci Numbers

== Theorem ==
Let $g_n$ be the sum of the first $n$ Fibonacci numbers.
Then:
:$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$
== Proof ==
Let $F_n$ be the $n$th Fibonacci number.
By definition:
:$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$
Hence, $g_n$, the sum of the first $n$ Fibonacci numbers is
:$g_0 = 0, g_1 = 1, g_2 = 2, g_3 = 4, g_4 = 7, \ldots$
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^n F_j = g_n = g_{n - 1} + g_{n - 2} + 1$
=== Basis for the Induction ===
$\map P 2$ is the case $g_2 = g_1 + g_0 + 1$, which holds:
:$2 = 1 + 0 + 1$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{j \mathop = 0}^k F_j = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1$
Then we need to show:
:$\ds \sum_{j \mathop = 0}^{k + 1} F_j = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$
where $g_n$ is the sum of the first $n$ Fibonacci numbers.
Category:Fibonacci Numbers
Category:Recurrence Relations
Recurrence Relation for Bell Numbers

== Theorem ==
Let $B_n$ be the Bell number for $n \in \Z_{\ge 0}$.
Then:
:$B_{n + 1} = \ds \sum_{k \mathop = 0}^n \dbinom n k B_k$
where $\dbinom n k$ are binomial coefficients.
== Proof ==
By definition of Bell numbers:
:$B_{n + 1}$ is the number of partitions of a (finite) set whose cardinality is $n + 1$.
Let $k \in \set {k \in \Z: 0 \le k \le n}$.
Let us form a partition of a (finite) set $S$ with cardinality $n + 1$ such that one component has $n + 1 - k > 0$ elements.
We can do this by first choosing $1$ element from $S$. We put this element into that single component.
Then choose $k$ more elements from $S$, and let the remaining $n - k$ elements be put into the same component as the first element.
From Cardinality of Set of Subsets and the definition of binomial coefficient, there are $\dbinom n k$ ways to do this.
For the chosen $k$ elements, there are $B_k$ ways to partition them.
Thus there are $\dbinom n k B_k$ possible partitions for $S$:
:$\dbinom n k$ of selecting $n - k$ elements to form one component with the one singled-out element
:for each of these, $B_k$ ways to partition the remaining $k$ elements.
Summing the number of ways to do this over all possible $k$:
:$\ds B_{n + 1} = \sum_{k \mathop = 0}^n \dbinom n k B_k$
=== Illustration ===
Category:Recurrence Relation for Bell Numbers
Category:Bell Numbers
Category:Recurrence Relations