nyuuzyou/wikis huggingface dataset:
Josephus Problem/Turks and Christians Variant/Mistake

== Source Work ==
: 
:The Puzzles:
::The Josephus Problem
:::$99$
== Mistake ==
:On board a ship, tossed in storms and in danger of shipwreck, are fifteen Christians and fifteen Turks. To lighten the load and save the ship, half are to be thrown overboard. One of the Christians suggests that all should stand in a circle and every ninth person counting around the circle should be chosen. How should the Christians arrange themselves in the circle to ensure that only the Turks die?
The solution is given as:
:The Christians and Turks should be placed in the following circular order, in which the first person follows the last, and the counting starts with the first person: $\text {CCCCTTTTTCCTCCCTCTTCCTTTCTTTCCTT}$.
== Correction ==
If you count them carefully, you will see that there are $32$ letters in the solution string.
The correct solution is:
:$\text {CCCCTTTTTCCTCCCTCTTCCTTTCTTCCT}$
== Sources ==
* : The Josephus Problem: $99$
Category:Mistakes/Curious and Interesting Puzzles
Julia Set/Examples/Filled for -0.75/Mistake

== Source Work ==
:
:Julia set
== Mistake ==
:(b) Filled Julia set for $c = -0.75$
::
== Correction ==
The screenshot given is actually the filled Julia set for the rational function $z \mapsto z^2 + c$ for the point $c = -1.0$.
This is the filled Julia set for the rational function $z \mapsto z^2 + c$ for the point $c = -0.75$:
::
Completely different beast.
== Julia Set Generator ==
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Jump Discontinuity/Examples/Example 1

== Example of Jump Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} 1 & : x < 2 \\ 2 & : x \ge 2 \end {cases}$
Then $f$ has a jump discontinuity at $x = 2$.
In this case, $\map f 2$ is defined, and equals the limit from the right.
:
== Sources ==
* 
Category:Examples of Jump Discontinuities
Jump Discontinuity/Examples/Example 2

== Example of Jump Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} 0 & : x > 1 \\ 1 & : x < 1 \\ \dfrac 1 2 & : x = 1 \end {cases}$
Then $f$ has a jump discontinuity at $x = 1$.
In this case, $\map f 1$ is defined, but equals neither limit.
== Sources ==
* 
* 
Category:Examples of Jump Discontinuities
Jump Discontinuity/Examples/Example 3

== Example of Jump Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} -1 & : x < 0 \\ 1 & : x > 0 \\ \text {undefined} & : x = 0 \end {cases}$
Then $f$ has a jump discontinuity at $x = 1$.
In this case, $\map f 0$ is not defined.
:
== Sources ==
* 
* 
Category:Examples of Jump Discontinuities
Jump Rule

== Theorem ==
Let $f : \R \to \R$ be a piecewise continuously differentiable real function with a discontinuity at $c \in \R$.
Suppose the limits $\map f {c^+}, \map f {c^-}, \map {f'} {c^+}, \map {f'} {c^-}$ exist.
Let $T \in \map {\DD'} \R$ be a distribution associated with $f$.
Then in the distributional sense we have that:
:$T_f' = T_{f'} + \paren {\map f {c^+} - \map f {c^-}} \delta_c$
where $\delta_c$ is the Dirac delta distribution.
== Proof ==
Let $\phi \in \map \DD {\R}$ be a test function with its support on $\closedint a b \subset \R$.
Let $c \in \closedint a b$.
Then:
== Also see ==
*Definition:Jump Discontinuity
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Category:Piecewise Continuously Differentiable Functions
Jump Rule/Finite Number of Discontinuities

== Theorem ==
Let $f : \R \to \R$ be a piecewise continuously differentiable real function with discontinuities at $\set {c_0, \ldots, c_n}$ where $c_k \in \R$.
Suppose for all $k \in \N_{\mathop \le n}$ the limits $\map f {c_k^+}, \map f {c_k^-}, \map {f'} {c_k^+}, \map {f'} {c_k^-}$ exist.
Let $T \in \map {\DD'} \R$ be a distribution associated with $f$.
Then in the distributional sense we have that:
:$\ds T_f' = T_{f'} + \sum_{k \mathop = 0}^n \paren {\map f {c_k^+} - \map f {c_k^-}} \delta_{c_k}$
where $\delta_c$ is the Dirac delta distribution.
== Proof ==
== Also see ==
*Definition:Jump Discontinuity
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense
Category:Examples of Distributional Derivatives
Category:Piecewise Continuously Differentiable Functions
Jung's Theorem

== Theorem ==
Let $S \subseteq \R^n$ be a compact subspace of an $n$-dimensional Euclidean space.
Let $d = \ds \max_{x, y \mathop \in S} \map d {x, y}$ be the diameter of $S$.
Then there exists a closed ball ${B_r}^-$ with radius $r$ such that:
:$r = d \sqrt {\dfrac n {2 \paren {n + 1} } }$
such that $S \subseteq {B_r}^-$.
=== Jung's Theorem in the Plane ===
== Proof ==
== Sources ==
* 
Category:Euclidean Space
Jung's Theorem in the Plane

== Theorem ==
Let $S \subseteq \R^2$ be a compact region in a Euclidean plane.
Let $d$ be the diameter of $S$.
Then there exists a circle $C$ with radius $r$ such that:
:$r = d \dfrac {\sqrt 3} 3$
such that $S \subseteq C$.
The parameter $\dfrac {\sqrt 3} 3$ can also be presented as $\dfrac 1 {\sqrt 3}$, and evaluates approximately as:
:$\dfrac {\sqrt 3} 3 \approx 0 \cdotp 57735 \, 02691 \ldots$
== Proof ==
This is an instance of Jung's Theorem, setting $n = 2$.
== Sources ==
* 
* : $0,57735 02691 \ldots$
Category:Euclidean Space
Justification for Geometrical Representation of Curl Operator

== Theorem ==
== Proof ==
According to the geometrical representation:
Let us take $3$ infinitesimal rectangular surfaces which intersect perpendicularly at a point $P$.
At $P$, let $\mathbf V$ be the value of a vector field.
:
Let the magnitudes of the components of $\mathbf V$ in the directions of the coordinate axes $x$, $y$ and $z$ be $V_x$, $V_y$ and $V_z$ respectively.
In the above diagram, the circular arrows indicate the directions in which contour integrals are to be taken according to the directions of the corresponding vector areas in question, according to the right-hand rule.
Take the surface $\Box abcd$ whose sides are $\d x$ and $\d y$ whose normal is in the direction of the $z$-axis.
Since $\Box abcd$ is very small, the magnitude of the component of $\mathbf V$ at the middle of each side can reasonably be taken to be the mean value along that side.
The arrows along the sides of the rectangular surfaces indicate the direction in which each component acts.
Because $V_x$, $V_y$ and $V_z$ are functions of the coordinates $\tuple {x, y, z}$ in the middle of the rectangle, the mean value along each of the $4$ sides $ab$, $bc$, $cd$ and $da$ will be respectively:
:$V_x - \dfrac 1 2 \dfrac {\partial V_x} {\partial y} \rd y$
:$V_y + \dfrac 1 2 \dfrac {\partial V_y} {\partial x} \rd x$
:$V_x + \dfrac 1 2 \dfrac {\partial V_x} {\partial y} \rd y$
:$V_y - \dfrac 1 2 \dfrac {\partial V_y} {\partial x} \rd x$
Hence the contour integral around $\Box abcd$ is:
:$\paren {\paren {V_x - \dfrac 1 2 \dfrac {\partial V_x} {\partial y} \rd y} - \paren {V_x + \dfrac 1 2 \dfrac {\partial V_x} {\partial y} \rd y} } \rd x + \paren {\paren {V_y + \dfrac 1 2 \dfrac {\partial V_y} {\partial x} \rd x} - \paren {V_y - \dfrac 1 2 \dfrac {\partial V_y} {\partial x} \rd x} } \rd y$
which works out as:
:$\paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \rd x \rd y$
Because the area of the rectangular surface is $\d x \rd y$, $\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y}$ is the magnitude of the component of the curl of $\mathbf V$ taken about the $z$-axis.
Hence we can write:
:$\curl_Z \mathbf V = \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k$
Doing the same thing with the other $2$ axes gives us:
Adding the $3$ components gives us:
:$\curl \mathbf V = \paren {\dfrac {\partial V_z} {\partial y} - \dfrac {\partial V_y} {\partial z} } \mathbf i + \paren {\dfrac {\partial V_x} {\partial z} - \dfrac {\partial V_z} {\partial x} } \mathbf j + \paren {\dfrac {\partial V_y} {\partial x} - \dfrac {\partial V_x} {\partial y} } \mathbf k$
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $4$. The Curl of a Vector Field
Category:Curl Operator
Justification for Geometrical Representation of Divergence Operator

== Theorem ==
== Proof ==
According to the geometrical representation:
Let $\mathbf V$ be the value of a vector field in the middle of an infinitesimal volume element $P$ with edges $\d x$, $\d y$ and $\d z$ parallel to the coordinate axes $x$, $y$ and $z$ respectively.
:
The magnitudes of the components of $\mathbf V$ in the directions of the coordinate axes $x$, $y$ and $z$ are $V_x$, $V_y$ and $V_z$ respectively.
$\mathbf V$ can be considered to be a vector giving, for example, the velocity of a moving fluid through the volume element $P$.
Consider the $2$ faces of $P$ each with area $\d y \rd z$ perpendicular to the $x$-axis.
On one of these (the left-hand face in the above diagram), the value of the $x$ component of the vector at the middle of the face becomes:
:$V_x - \dfrac 1 2 \dfrac {\partial V_x} {\partial x} \rd x$
When $\d x$ becomes vanishingly small, this value can be taken to apply over the whole of that face.
Similarly, on the other face (the right-hand face in the above diagram), the value of the $x$ component of the vector at the middle of the face becomes:
:$V_x + \dfrac 1 2 \dfrac {\partial V_x} {\partial x} \rd x$
We can now define the flux through any face as the area of the face multiplied by the normal component of $\mathbf V$ acting upon it
That is, the dot product of the area of the face and $\mathbf V$.
This is positive when the flow is outwards, and negative when the flow is inwards.
Hence the excess of flux leaving the volume element over the flux entering it is given by:
By similar reasoning, the contributions parallel to the $y$-axis and $z$-axis are:
Hence the total flux leaving the volume element is:
:$\paren {\dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z} } \rd x \rd y \rd z$
It follows that the amount of flux per unit volume is:
:$\operatorname {div} \mathbf V := \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}$
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $3$. The Divergence of a Vector Field: $(4.6)$
Category:Divergence Operator
Justification for Geometrical Representation of Gradient Operator

== Theorem ==
== Proof ==
According to the geometrical representation:
:
Let $S$ be a constant value of $F$ giving rise to an equal surface of $F$.
Let $S + \d S$ be another constant value of $F$ giving rise to an equal surface of $F$ very close to $S$.
To economise on notation we refer to those equal surfaces themselves as $S$ and $S + \d S$ respectively.
Consider the point $A$ on $S$ whose position vector is $\mathbf r$.
Let $B$ be a point on $S + \d S$ whose position vector is $\mathbf r + \d \mathrm r$.
The smallest distance from $S$ to $S + \d S$ from $A$ is $AC$:
:whose direction is that of the unit normal $\mathbf {\hat n}$ at $A$
:whose length is $\norm {\d \mathbf n}$.
Let $\norm {\d \mathrm r}$ be the length of $AB$.
This is the magnitude of the rate of increase at $A$ of $S$ in the direction of $AB$ when $S$ and $S + \d S$ are infinitesimally close together.
The rate of increase is greatest in the direction is that of the unit normal, that is, along $AB$.
Hence at $A$ the rate of increase has value $\dfrac {\partial F} {\partial n}$.
Thus:
:$\dfrac {\partial F} {\partial r} = \dfrac {\partial F} {\partial n} \cos \theta$
where $\theta$ is the angle between $AB$ and $AC$.
Hence if $\mathbf {\hat n}$ is the unit normal to $S$ at any point in a scalar field, the vector quantity $\mathbf {\hat n} \dfrac {\partial F} {\partial n}$ gives the greatest rate of increase of $F$ at that point in magnitude and direction.
This is the gradient of $F$ at that point and can be written:
:$\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$
Thus the gradient of $F$ is a vector field whose direction is the fastest rate of increase of $F$ and whose magnitude is that rate of increase.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2$. The Gradient of a Scalar Field
Category:Gradient Operator
K-Connectivity Implies Lesser Connectivity

== Theorem ==
Let $G = \struct {V, E}$ be a $k$-connected graph.
Then $G$ is $l$-connected for all $l \in \Z : 0 < l < k$.
== Proof ==
Suppose that $G$ is $k$-connected.
Then:
:$\card V > k > l$
:$G$ is connected
:If $W$ is a vertex cut of $G$, then $\card W \ge k > l$ so $\card W \ge l$.
Category:Connectedness (Graph Theory)
K-Cycle can be Factored into Transpositions

== Theorem ==
Every $k$-cycle can be factorised into the product of $k - 1$ transpositions.
This factorisation is not unique.
== Proof ==
The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ has the factorisation:
:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & k \end{pmatrix} \ldots \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$
Therefore, the general $k$-cycle $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ has the factorisation:
:$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_k \end{pmatrix} \ldots \begin{pmatrix} i_1 & i_3 \end{pmatrix} \begin{pmatrix} i_1 & i_2 \end{pmatrix}$
The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ also has the factorisation:
:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} \ldots \begin{pmatrix} k - 1 & k \end{pmatrix}$
Therefore, the general $k$-cycle $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ also has the factorisation:
:$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_2 \end{pmatrix} \begin{pmatrix} i_2 & i_3 \end{pmatrix} \ldots \begin{pmatrix} i_{k - 1} & i_k \end{pmatrix}$
== Sources ==
* : Appendix: Elementary set and number theory
* : Chapter $9$: Permutations $9.10$
Category:Cyclic Permutations
Kakeya Problem

== Problem ==
The Kakeya Problem is the question:
:What is the smallest possible area of a set in the plane inside which a needle of length $1$ can be moved continuously in order to reverse its direction?
== Solution ==
There is no such smallest area.
That is, let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then there exists a plane figure which fulfils the conditions of the Kakeya Problem whose area is less than $\epsilon$.
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
Category:Kakeya Problem
Category:Plane Geometry
Kakutani's Fixed Point Theorem

== Theorem ==
Let $S \subset \R^n$ be nonempty, compact, and convex.
Let $\Phi : S \to 2^S$ be a correspondence.
Let the following conditions be satisfied:
:$(1): \quad \map \Phi x$ is nonempty and convex for all $x$
:$(2): \quad \map \Phi \cdot$ is upper hemi-continuous
Then $\Phi$ has a fixed point.
== Proof ==
== Sources ==
* : $\text I$ Strategic Games: Chapter $2$ Nash Equilibrium: $2.4$ Existence of a Nash Equilibrium: Lemma $20.1$ (Kakutani's fixed point theorem)
Kakutani's Theorem

== Theorem ==
Let $X$ be a normed vector space.
Let $w$ be the weak topology on $X$. 
Let $B_X^-$ be the closed unit ball in $X$. 
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Then $X$ is reflexive $\struct {B_X^-, w}$ is compact.
== Proof ==
Let $B_{X^{\ast \ast} }^-$ be the closed unit ball in $X^{\ast \ast}$. 
Let $w^\ast$ be the weak-* topology on $X^{\ast \ast}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation. 
=== Necessary Condition ===
Suppose that $X$ is reflexive.
By Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual, $\iota B_X^- = B_{X^{\ast \ast} }^-$.
From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Homeomorphism onto Image, we have that: 
:$\iota : \struct {X, w} \to \struct {\iota X, w^\ast}$ is a homeomorphism.
From Restriction of Homeomorphism is Homeomorphism, we have that: 
:$\iota : \struct {B_X^-, w} \to \struct {\iota B_X^-, w^\ast} = \struct {B_{X^{\ast \ast} }^-, w^\ast}$ is a homeomorphism.
From the Banach-Alaoglu Theorem, $\struct {B_{X^{\ast \ast} }^-, w^\ast}$ is compact.
Since $\struct {B_X^-, w}$ is homeomorphic to $\struct {B_{X^{\ast \ast} }^-, w^\ast}$, we have that $\struct {B_X^-, w}$ is compact.
=== Sufficient Condition ===
Suppose that $\struct {B_X^-, w}$ is compact.
From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual, $\iota : \struct {X, w} \to \struct {X^{\ast \ast}, w^\ast}$ is continuous.
From Continuous Image of Compact Space is Compact, $\iota B_X^-$ is a compact subset of $\struct {X^{\ast \ast}, w^\ast}$.
In particular, $\iota B_X^-$ is closed in $\struct {X^{\ast \ast}, w^\ast}$.
So:
:$\map {\cl_{w^\ast} } {\iota B_X^-} = \iota B_X^-$ 
from Set is Closed iff Equals Topological Closure.
From Goldstine's Theorem, we have that: 
:$\map {\cl_{w^\ast} } {\iota B_X^-} = B_{X^{\ast \ast} }^-$
So, we have: 
:$\iota B_X^- = B_{X^{\ast \ast} }^-$
From Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual, $X$ is reflexive.
Category:Normed Vector Spaces
Category:Reflexive Spaces
Category:Weak Topologies on Topological Vector Spaces
Kappa Curve has Double Cusp at Origin

== Theorem ==
Let $\KK$ be a kappa curve expressed in Cartesian coordinates as:
:$x^4 + x^2 y^2 = a^2 y^2$
$\KK$ has a double cusp at the origin.
This double cusp is a double cusp of the first kind.
:
== Proof ==
== Sources ==
* 
* 
Category:Kappa Curve
Category:Double Cusps
Kaprekar's Process for 2-Digit Numbers

== Theorem ==
Kaprekar's process, when applied to a non-repdigit $2$-digit positive integer leads to the cycle:
:$09 \to 81 \to 63 \to 27 \to 45 \to 09$
Note that it is important to retain the leading zero on the $9$, or the process trivially terminates in $0$.
== Proof ==
Let $n \in \Z_{>0}$ be a $2$-digit positive integer.
, let $n$ be expressed in decimal notation as:
:$n = 10 x + y$
where:
: $x > y$
:$0 \le x \le 9, 0 \le y \le 9$
The reversal of $n$ is $10 y + x$.
We have:
Thus after the first iteration of Kaprekar's process, $n$ is one of:
:$09$
:$18$
:$27$
:$36$
:$45$
or their reversals.
Applying Kaprekar's process to $09$ gives:
All those multiples of $9$ can be seen to be in (or end up in) this loop.
Hence the result.
== Sources ==
* : $63$
* : $63$
Category:Kaprekar's Process
Category:Reversals
Kaprekar's Process for 2-Digit Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$63$
:
:The Dictionary
::$63$
== Mistake ==
:Kaprekar's process for $2$-digit numbers leads to the cycle $63 - 27 - 45 - 9 - 81 \ldots$
The statement needs to specify that the $2$-digit numbers do not include repdigit numbers, as for example:
:$\map K {11} = 11 - 11 = 0$
trivially.
== Sources ==
* : $63$
* : $63$
Category:Mistakes/Curious and Interesting Numbers
Kaprekar's Process on 3 Digit Number ends in 495

== Theorem ==
Let $n$ be a $3$-digit integer whose digits are not all the same.
Kaprekar's process, when applied to $n$, results in $495$ after no more than $6$ iterations.
== Proof ==
Let $n = \sqbrk {abc}_{10}$ denote a $3$-digit integer whose digits are $a, b, c$.
If $a = b = c$ then Kaprekar's process trivially results in $0$ after the first iteration.
, let $a \ge b \ge c$ but such that $a \ne c$.
By the Basis Representation Theorem:
:$n = 10^2 a + 10 b + c$
Let $n' = 10^2 a' + 10 b' + c'$ be the result of Kaprekar's process after the $1$st iteration on $n$.
We have:
Thus we have that:
Hence we have that:
There are in fact only the following possibilities for $n'$:
 we inspect Kaprekar's process on $198$:
and the result is seen to follow.
== Sources ==
* : $495$
* : $495$
Category:Kaprekar's Process
Category:495
Kaprekar's Process on 3 Digit Number ends in 495/Mistake

== Source Work ==
:
:The Dictionary
::$495$
== Mistake ==
:Take any $3$-digit number whose digits are not all the same and is not a palindrome. Arrange its digits into ascending and descending order and subtract. Repeat. This is called Kaprekar's process. All $3$-digit numbers eventually end up with $495$.
The $3$-digit number in question needs only to have its digits not all the same. A palindrome ends up at $495$ in the same way as any other number, for example:
== Sources ==
* : $495$
Category:Mistakes/Curious and Interesting Numbers
Kaprekar's Process on 5 Digit Number

== Theorem ==
Let $n$ be a $5$-digit integer whose digits are not all the same.
Kaprekar's process, when applied to $n$, results in one of the following $3$ cycles:
:$53 \, 955 \to 59 \, 994 \to 53 \, 955$
:$61 \, 974 \to 82 \, 962 \to 75 \, 933 \to 63 \, 954 \to 61 \, 974$
:$62 \, 964 \to 71 \, 973 \to 83 \, 952 \to 74 \, 943 \to 62 \, 964$
== Proof ==
We have:
== Sources ==
* : $\text {XIII}$: Numbers Curious and Serious: $350$. A Persistent Difference
* : $99,954$
* : $99,954$
Category:Kaprekar's Process
Kaprekar's Process on 5 Digit Number/Mistake

== Source Work ==
:
:The Dictionary
::$99,954$
:
:The Dictionary
::$99,954$
== Mistake ==
:Kaprekar's process for all $5$-digit numbers whose digits are not all equal leads to one of $3$ separate cycles. The smallest cycle is $99,954 - 95,553$. The other two cycles are $98,532 - 97,443 - 96,642 - 97,731$ and $98,622 - 97,533 - 96,543 - 97,641$.
== Correction ==
Those are not the numbers in the cycles themselves, but those after arranging them in descending order of their digits.
The cycles themselves are:
:$53 \, 955 \to 59 \, 994 \to 53 \, 955$
:$61 \, 974 \to 82 \, 962 \to 75 \, 933 \to 63 \, 954 \to 61 \, 974$
:$62 \, 964 \to 71 \, 973 \to 83 \, 952 \to 74 \, 943 \to 62 \, 964$
== Sources ==
* : $99,954$
* : $99,954$
Category:Mistakes/Curious and Interesting Numbers
Kaprekar's Symmetry

== Theorem ==
Let $n$ be a Kaprekar number with $D$ digits.
Then $10^D - n$ is also a Kaprekar number.
== Examples ==
== Proof ==
Since $n$ is a Kaprekar number of $D$ digits:
:$\begin {cases} n^2 = a \times 10^D + b \\ n = a + b \end {cases}$
for some positive integers $a$ and $b$, $b < 10^D$.
Hence:
and we have:
:$\paren {10^D - a - 2 b} + b = 10^D - a - b = 10^D - n$
Finally we check that $10^D - a - 2 b \ge 0$:
 $10^D - a - 2 b \le -1$.
Then:
but squares are positive, a contradiction.
Hence $10^D - n$ is also a Kaprekar number of $D$ digits.
== Sources ==
* : $297$
* : $297$
Category:Kaprekar Numbers
Kaprekar Number/Examples/22,222,222,222,222

== Example of Kaprekar Number ==
$22 \, 222 \, 222 \, 222 \, 222$ is a Kaprekar number:
== Sources ==
* : $22,222,222,222,222$
* : $22,222,222,222,222$
Category:Kaprekar Numbers
Category:22,222,222,222,222
Kasteleyn's Formula

== Theorem ==
The number of perfect covers of a chessboard of dimensions $m \times n$ is given by the formula:
:$\ds \prod_{j \mathop = 1}^{\ceiling {\frac m 2} } \prod_{k \mathop = 1}^{\ceiling {\frac n 2} } \paren {4 \cos^2 \frac {\pi j} {m + 1} + 4 \cos^2 \frac {\pi k} {n + 1} }$
== Proof ==
== Sources ==
* 
* : Chapter $1$: $1.1$ Chessboard Problems
Category:Chessboard Tilings
Category:Chessboard Puzzles
Category:Dominoes
Keith.U/Whatever/Definition:Exponential/Real

==Definition ==
For all definitions of the real exponential function:
:The domain of $\exp$ is $\R$
:The codomain of $\exp$ is $\R_{>0}$
For $x \in \R$, the real number $\exp x$ is called the exponential of $x$.
=== As a Power Series Expansion ===
=== As a Limit of a Sequence ===
=== As an Extension of the Rational Exponential ===
=== As the Inverse to the Natural Logarithm ===
=== As the Solution of a Differential Equation ===
== Equivalence of Definitions ==
As shown in Equivalence of Definitions of Exponential Function the definitions above are equivalent.
== Notation ==
The exponential of $x$, $\exp x$, is frequently written as $e^x$. The consistency of this power notation is demonstrated in Equivalence of Definitions of Real Exponential Functions.
== Also see ==
* Properties of Exponential Function
Category:Definitions/Real Analysis