nyuuzyou/wikis huggingface dataset:
Renaming Mapping from Set of Mappings on Single Element

== Theorem ==
Let $X$ and $Y$ be sets.
Let $E$ be the set of all mappings from $X$ to $Y$.
Let $b \in X$.
Let $\RR \subseteq E \times E$ be the relation on $E$ defined as:
:$\RR := \set {\tuple {f, g} \in \RR: \map f b = \map g b}$
Let $e_b: E / \RR \to Y$ be the renaming mapping induced by $\RR$.
Then $e_b$ is a bijection.
== Proof ==
This is an instance of Renaming Mapping is Bijection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $5$
Category:Mapping Theory
Category:Equivalence Relations
Renaming Mapping is Bijection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$.
The renaming mapping is a bijection.
== Proof 1 ==
== Proof 2 ==
== Different approaches ==
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : Chapter $\text{I}$: Sets and Functions: Factoring Functions
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $2$
Category:Equivalence Relations
Category:Bijections
Category:Quotient Sets
Category:Renaming Mapping is Bijection
Renaming Mapping is Bijection/Different approaches

== Different approaches to Renaming Mapping is Bijection ==
 considers the case where $r$ is an injection, but does not stress its bijective aspects from this particular perspective:
:This type of factorization of mappings ... is particularly useful when the set of inverse images $\map {\alpha^{-1} } {a'}$ coincides with $\overline S$; for, in this case, the mapping $\overline a$ is 1-1. Thus if $\overline a \overline \alpha = \overline b \overline \alpha$, then $a \alpha = b \alpha$ and $a \sim b$. Hence $\overline a = \overline b$. Thus we obtain here a factorization $\alpha = \nu \overline \alpha$ where $\overline \alpha$ is 1-1 onto $T$ and $\nu$ is the natural mapping.
Note that in the above, Jacobson uses:
:$\alpha$ for $f$
:$a'$ for the image of a representative element $a$ of $S$ under $\alpha$
:$\overline S$ for $S / \RR_f$
:$\nu$ for the quotient mapping $q_{\RR_f}: S \to S / \RR_f$
:$\overline a$ and $\overline b$ for representative elements of $\overline S$
:$\overline \alpha$ for the renaming mapping $r$.
 takes the approach of deducing the existence of the mapping $r$, and then determining under which conditions it is either injective or surjective.
From there, the surjective restriction of $r$ is taken, and $\RR$ is then identified with the equivalence induced by $f$.
Hence the bijective nature of $r$ is deduced rather than constructed.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.6$
Category:Renaming Mapping is Bijection
Renaming Mapping is Bijection/Proof 1

== Theorem ==
== Proof ==
=== Proof of Injectivity ===
To show that $r: S / \RR_f \to \Img f$ is an injection:
Thus $r: S / \RR_f \to \Img f$ is an injection.
=== Proof of Surjectivity ===
To show that $r: S / \RR_f \to \Img f$ is a surjection:
Note that for all mappings $f: S \to T$, $f: S \to \Img f$ is always a surjection from Surjection by Restriction of Codomain.
Thus by definition:
:$\forall y \in \Img f: \exists x \in S: \map f x = y$
So:
Thus $r: S / \RR_f \to \Img f$ is a surjection.
As $r: S / \RR_f \to \Img f$ is both an injection and a surjection, it is by definition a bijection.
Category:Renaming Mapping is Bijection
Renaming Mapping is Bijection/Proof 2

== Theorem ==
== Proof ==
From Renaming Mapping is Well-Defined, $r: S / \RR_f \to \Img f$ is a well-defined mapping.
By definition, $\RR_f$ is the equivalence relation induced by the mapping $f$.
Hence by definition:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
From Condition for Mapping from Quotient Set to be Injection, this is precisely the condition required for $r$ to be an injection.
Next it is noted that the codomain of $r$ is $\Img f$.
Then from Restriction of Mapping to Image is Surjection, we have that $f_{\restriction \Img f}$ is a surjection.
Hence from Condition for Mapping from Quotient Set to be Surjection it follows that $r$ is also a surjection.
Thus $r$ is shown to be both an injection and a surjection, and so by definition is a bijection.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.6$
Category:Renaming Mapping is Bijection
Renaming Mapping is Well-Defined

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as:
:$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$
where:
:$\RR_f$ is the equivalence induced by the mapping $f$
:$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$
:$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$.
The renaming mapping is always well-defined.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $2$
Category:Mapping Theory
Category:Quotient Sets
Category:Equivalence Relations
Category:Renaming Mapping is Well-Defined
Renaming Mapping is Well-Defined/Proof 1

== Theorem ==
== Proof ==
By Relation Induced by Mapping is Equivalence Relation, we have that $\RR_f$ is an equivalence relation.
To determine whether $r$ is well-defined, we have to determine whether $r: S / \RR_f \to \Img f$ actually defines a mapping at all.
Consider a typical element $\eqclass x {\RR_f}$ of $S / \RR_f$.
Suppose we were to choose another name for the class $\eqclass x {\RR_f}$.
Assume that $\eqclass x {\RR_f}$ is not a singleton.
For example, let us choose $y \in \eqclass x {\RR_f}, y \ne x$ such that: 
:$\eqclass x {\RR_f} = \eqclass y {\RR_f}$
then:
:$\map r {\eqclass x {\RR_f} } = \map r {\eqclass y {\RR_f} }$
Hence from the definition, we have:
Thus $r$ is well-defined.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : Chapter $\text{I}$: Sets and Functions: Factoring Functions: Theorem $10$
Category:Renaming Mapping is Well-Defined
Renaming Mapping is Well-Defined/Proof 2

== Theorem ==
== Proof ==
From Condition for Mapping from Quotient Set to be Well-Defined:
:there exists a mapping $\phi: S / \RR \to T$ such that $\phi \circ q_\RR = f$
:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$
But by definition of the equivalence induced by the mapping $f$:
:$\forall x, y \in S: \tuple {x, y} \in \RR_f \implies \map f x = \map f y$
The result follows directly.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.6$
Category:Renaming Mapping is Well-Defined
Reparameterization of Contour is Contour

== Theorem ==
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $\gamma: \closedint a b \to \C$ be a contour in $\C$.
That is, there exists a subdivision $a_0, a_1 , \ldots, a_n$ of $\closedint a b$ such that $\gamma \restriction_{I_i}$ is a smooth path for all $i \in \set {1, \ldots, n}$, where $I_i = \closedint {a_{i - 1} } {a_i}$.
Here, $\gamma \restriction_{I_i}$ denotes the restriction of $\gamma$ to $I_i$.
Let $\phi: \closedint c d \to \closedint a b$ be a differentiable real function with $\map \phi c = a$, and $\map \phi d = b$.
Suppose that $\phi$ is either bijective or strictly increasing.
Define $\sigma: \closedint c d \to \C$ as a composite function by $\sigma = \gamma \circ \phi$.
Then $\sigma$ is a contour.
For all $i \in \set {1, \ldots, n}$, let $J_i := \closedint {\map {\phi^{-1} } {a_{i - 1} } } {\map {\phi^{-1} } {a_i} }$.
Then:
:$\map {\sigma' \restriction_{J_i} } t = \map {\gamma' \restriction_{I_i} } {\map \phi t} \map {\phi'} t$
== Proof ==
$\phi$ is either bijective or strictly increasing.
From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, it follows that in both cases $\phi$ is both bijective and strictly monotone.
As $\map \phi c < \map \phi d$, $\phi$ must be strictly increasing.
As $\phi$ is strictly increasing, we have $c = \map {\phi^{-1} } {a_0} < \map {\phi^{-1} } {a_1} < \cdots < \map {\phi^{-1} } {a_n} = d$.
Hence, $\map {\phi^{-1} } {a_0}, \map {\phi^{-1} } {a_1}, \dotsc, \map {\phi^{-1} } {a_n}$ form a subdivision of $\closedint a b$.
Define $J_i := \closedint {\map {\phi^{-1} } {a_{i - 1} } } {\map {\phi^{-1} } {a_i} }$ as a closed real interval.
From Derivative of Complex Composite Function, it follows that $\sigma$ restricted to $J_i$ is complex differentiable.
Its derivative is given by:
:$\map {\sigma' \restriction_{J_i} } t = \map {\gamma' \restriction_{I_i} } {\map \phi t} \map {\phi'} t$
From Derivative of Monotone Function, it follows that for all $t \in \closedint c d$, we have $\map {\phi'} t > 0$.
As $\map {\gamma' \restriction_{I_i} } {\map \phi t} \ne 0$ by definition of contour, we have $\map {\sigma' \restriction_{J_i} } t \ne 0$.
Then $\sigma$ fulfils both requirement $(1)$ and $(2)$ from the definition of contour, so $\sigma$ is a contour.
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints

== Theorem ==
Let $\R^n$ be a real cartesian space of $n$ dimensions.
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $\gamma: \closedint a b \to \R^n$ be a smooth path in $\R^n$.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Suppose that $\sigma: \closedint c d \to \R^n$ is a reparameterization of $C$.
Then the start points and end points of $\gamma$ and $\sigma$ are identical:
:$\map \gamma a = \map \sigma c$
:$\map \gamma b = \map \sigma d$
=== Complex Plane ===
== Proof ==
By definition of reparameterization, there exists a bijective differentiable strictly increasing real function $\phi: \closedint c d \closedint a b$ such that $\sigma = \gamma \circ \phi$.
As $\map {\phi^{-1} }{a} \in \closedint c d$:
: $c \le \map {\phi^{-1} }{a}$
As $\phi$ is strictly increasing:
: $\map \phi c \le \map \phi { \map {\phi^{-1} } a } = a$
As $\map \phi c \in \closedint a b$:
:$\map \phi c = a$
Hence:
:$\map \sigma c = \map {\gamma \circ \phi} c = \map \gamma a$
As $\map {\phi^{-1} } b \in \closedint c d$:
: $d \ge \map {\phi^{-1} } b$
As $\phi$ is strictly increasing:
: $\map \phi d \ge \map \phi { \map {\phi^{-1} } b } = b$
As $\map \phi d \in \closedint a b$:
: $\map \phi d = b$
Hence:
:$\map \sigma d = \map {\gamma \circ \phi} d = \map \gamma b)$
== Sources ==
* $\S 2.2$
Category:Vector Analysis
Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints/Complex Plane

== Theorem ==
Let $\C$ denote the complex plane.
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $\gamma: \closedint a b \to \C$ be a smooth path in $\C$.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Let $\sigma: \closedint c d \to \C$ be a reparameterization of $C$.
Then the start points and end points of $\gamma$ and $\sigma$ are identical:
:$\map \gamma a = \map \sigma c$
:$\map \gamma b = \map \sigma d$
== Proof ==
By definition of reparameterization, there exists a bijective differentiable strictly increasing real function $\phi: \closedint c d \closedint a b$ such that $\sigma = \gamma \circ \phi$.
As $\map {\phi^{-1} }{a} \in \closedint c d$:
: $c \le \map {\phi^{-1} }{a}$
As $\phi$ is strictly increasing:
: $\map \phi c \le \map \phi { \map {\phi^{-1} } a } = a$
As $\map \phi c \in \closedint a b$:
:$\map \phi c = a$
Hence:
:$\map \sigma c = \map {\gamma \circ \phi} c = \map \gamma a$
As $\map {\phi^{-1} } b \in \closedint c d$:
: $d \ge \map {\phi^{-1} } b$
As $\phi$ is strictly increasing:
: $\map \phi d \ge \map \phi { \map {\phi^{-1} } b } = b$
As $\map \phi d \in \closedint a b$:
: $\map \phi d = b$
Hence:
:$\map \sigma d = \map {\gamma \circ \phi} d = \map \gamma b)$
== Sources ==
* $\S 2.2$
Category:Directed Smooth Curves (Complex Plane)
Reparameterization of Directed Smooth Curve Preserves Image

== Theorem ==
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $\gamma : \closedint a b \to \C$ be a smooth path.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Suppose that $\sigma : \closedint c d \to \C$ is a reparameterization of $C$.
Then $\Img \gamma = \Img \sigma$, where $\Img \gamma$ denotes the image of $\gamma$.
== Proof ==
By definition of directed smooth curve, there exists a bijective differentiable strictly increasing function $\phi: \closedint c d \to \closedint a b$ such that $\sigma = \gamma \circ \phi$.
From Surjection by Restriction of Codomain, it follows that there exists a function $\tilde \gamma: \closedint a b \to \Img \gamma$ that is the surjective restriction of $\gamma$ to $\Img \gamma$.
Put $\tilde \sigma := \tilde \gamma \circ \phi : \closedint c d \to \Img \gamma$.
From Composite of Surjections is Surjection, it follows that $\tilde \sigma$ is surjective.
Then:
:$\Img {\tilde \sigma} = \Img \gamma$
Now:
:$\map \sigma t = \map {\tilde \sigma} t$ for all $t \in \closedint c d = \Dom \sigma = \Dom {\tilde \sigma}$
where $\Dom \sigma$ denotes the domain of $\sigma$.
It follows that:
:$\Img \sigma = \Img {\tilde \sigma}$
Hence:
:$\Img {\tilde \gamma} = \Img \sigma$
== Sources ==
* $\S 2.2$
Category:Directed Smooth Curves (Complex Plane)
Reparameterization of Directed Smooth Curve with Given Domain

== Theorem ==
Let $\gamma: \closedint a b \to \C$ be a smooth path in the complex plane.
Let $C$ be a directed smooth curve with parameterization $\gamma$.
Let $\closedint { a_0 }{ b_0 }$ be a closed real interval, where $a_0 < b_0$.
Then there exists a smooth path
:$\gamma_0 : \closedint { a_0 }{ b_0 } \to \C$
that is a reparameterization of $C$.
== Proof ==
Define $\phi : \closedint a b \to \closedint { a_0 }{ b_0 }$ by:
:$\map{ \phi }{ t } = \dfrac{ b_0 - a_0 }{ b - a } \paren{ t - a } + a_0$
Power Rule for Derivatives shows that
:$\map{ \phi' }{ t } = \dfrac{ b_0 - a_0 }{ b - a }$
Real Function with Strictly Positive Derivative is Strictly Increasing shows that $\phi$ is strictly increasing, as $\map{ \phi' }{ t } > 0$ for all $t \in \closedint a b$.
Strictly Monotone Real Function is Bijective shows that $\phi$ is bijective.
Set $\gamma_0 = \gamma \circ \phi : \closedint a b \to \C$.
By definition of directed smooth curve, it follows that $\gamma_0$ is a reparameterization of $C$.
== Sources ==
* $\S 2.2$
Category:Directed Smooth Curves (Complex Plane)
Repdigit Number consisting of Instances of 9 is Kaprekar

== Theorem ==
A repdigit number that consists entirely of the digit $9$ is a Kaprekar number.
== Proof ==
We note as examples:
Now we consider:
So $\paren {10^{n + 1} - 1}^2$ can be split as:
:$\paren {10^{n + 1} - 2} + 1 = 10^{n + 1} - 1 = \ds \sum_{k \mathop = 0}^n 9 \times 10^k$
Thus $\ds \sum_{k \mathop = 0}^n 9 \times 10^k$ is a Kaprekar number.
== Sources ==
* : $99$
* : $999$
* : $99$
* : $999$
Category:Kaprekar Numbers
Category:Repdigit Numbers
Repdigit Triangular Numbers

== Theorem ==
The only repdigit numbers which are also triangular are:
:$55, 66, 666$
== Proof ==
Let $\dfrac {d \paren {10^j - 1}} 9$ be a $j$-digit repdigit number.
Suppose it is triangular.
For $d = 1$, write:
Since $3 k + 1 \perp 3 k + 2$, one of them is equal to $2^{j + 1}$ while the other is equal to $5^j$.
This gives $2^{j + 1} \pm 1 = 5^j$.
Since $j \ge 1$, taking modulo $4$:
:$0 \pm 1 \equiv 1^j \equiv 1 \pmod 4$
so we have:
:$2^{j + 1} + 1 = 5^j$
By 1 plus Power of 2 is not Perfect Power except 9, the above equation has no solutions for $j > 1$.
For $j = 1$ we have equality.
Hence $1$ is the only triangular repunit.
By Odd Square is Eight Triangles Plus One:
:$x = 1 + \dfrac {8 d \paren {10^j - 1}} 9$ is a perfect square.
For $d = 2, 4, 7, 9$, $x$ ends in $2$ or $3$, which is not a perfect square by Square Modulo 5/Corollary.
For $d = 3, 8$, $x$ ends in $5$.
Since it is a perfect square, $x$ must end in $25$.
This is satisfied only when $d = 3, j = 1$.
For $d = 3, j > 1$ and $d = 8, j = 1$, $x$ ends in $65$.
For $d = 8, j > 1$, $x$ ends in $05$.
This corresponds to $\dfrac {x - 1} 8 = 3$.
For $d = 5$, $x$ is in the form $44 \dots 41$.
We show that $x$ is square only when $x = 441$.
This corresponds to $\dfrac {x - 1} 8 = 55$.
For $d = 6$, $x$ is either:
:$49$ for $j = 1$
:$53 \dots 329$ for $j > 1$, with $j - 1$ $3$'s
we show that $x$ is square only when $x = 49, 529, 5329$.
These correspond to $\dfrac {x - 1} 8 = 6$, $66$ and $666$.
We have hence demonstrated that the only triangular repdigit numbers are:
:$1, 3, 6, 55, 66, 666$
== Sources ==
* : $55$
* : $55$
Category:Repdigit Numbers
Category:Triangular Numbers
Repeated Composition of Injection is Injection

== Theorem ==
Let $S$ be a set.
Let $f: S \to S$ be an injection.
Let the sequence of mappings:
:$f^0, f^1, f^2, \ldots, f^n, \ldots$
be defined as:
:$\forall n \in \N: \map {f^n} x = \begin {cases}
x & : n = 0 \\
\map f x & : n = 1 \\
\map f {\map {f^{n - 1} } x} & : n > 1
\end{cases}$
Then for all $n \in \N$, $f^n$ is an injection.
== Proof ==
Proof by induction:
For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition:
:$f^n$ is an injection.
$\map P 0$ is true, as this is the case Identity Mapping is Injection.
$\map P 1$ is true, as this is the assertion that $f$ is an injection.
=== Basis for the Induction ===
$\map P 2$ is the assertion that $f^2$ is an injection.
But $f \circ f$ is an injection by Composite of Injections is Injection.
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:
:$f^k$ is an injection.
Then we need to show:
:$f^{k + 1}$ is an injection.
=== Induction Step ===
This is our induction step:
By the induction hypothesis $f^k$ is an injection.
By Composite of Injections is Injection it follows that $f^{k + 1}$ is an injection.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore for all $n \in \N$:
:$f^n$ is an injection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 7$: Relations: Exercise $3$
Category:Injections
Category:Composite Mappings
Repeated Fourier Transform of Even Function

== Theorem ==
Let $f: \R \to \R$ be an even real function which is Lebesgue integrable.
Let $\ds \map \FF {\map f t} = \map F s = \int_{-\infty}^\infty e^{-2 \pi i s t} \map f t \rd t$ be the Fourier transform of $f$.
Let $\ds \map \FF {\map F s} = \map g t = \int_{-\infty}^\infty e^{-2 \pi i t s} \map F s \rd s$ be the Fourier transform of $F$.
Then:
:$\map g t = \map f t$
== Proof ==
== Sources ==
* : Chapter $2$: Groundwork: The Fourier transform and Fourier's integral theorem
Category:Fourier Transforms
Repeated Fourier Transform of Odd Function

== Theorem ==
Let $f: \R \to \R$ be an odd real function which is Lebesgue integrable.
Let $\ds \map \FF {\map f t} = \map F s = \int_{-\infty}^\infty e^{-2 \pi i s t} \map f t \rd t$ be the Fourier transform of $f$.
Let $\ds \map \FF {\map F s} = \map g t = \int_{-\infty}^\infty e^{-2 \pi i t s} \map F s \rd s$ be the Fourier transform of $F$.
Then:
:$\map g t = -\map f t$
== Proof ==
== Sources ==
* : Chapter $2$: Groundwork: The Fourier transform and Fourier's integral theorem
Category:Fourier Transforms
Repeated Fourier Transform of Real Function

== Theorem ==
Let $f: \R \to \R$ be a real function which is Lebesgue integrable.
Let $\ds \map \FF {\map f t} = \map F s = \int_{-\infty}^\infty e^{-2 \pi i s t} \map f t \rd t$ be the Fourier transform of $f$.
Let $\ds \map \FF {\map F s} = \map g t = \int_{-\infty}^\infty e^{-2 \pi i t s} \map F s \rd s$ be the Fourier transform of $F$.
Then:
:$\map g t = \map f {-t}$
== Proof ==
== Sources ==
* : Chapter $2$: Groundwork: The Fourier transform and Fourier's integral theorem
Category:Fourier Transforms
Repeated Simpson's Rule/Also known as

== Repeated Simpson's Rule: Also known as == 
The repeated Simpson's rule is also known as just Simpson's rule.
It is also known as the parabolic formula.
It can also be seen as Simpson's formula, but this may be confused with Simpson's formulas, which are a set of completely different results.
== Sources ==
* : $\S 15$: Approximate Formulas for Definite Integrals: $15.17$
* : $\S 18$: Definite Integrals: Approximate Formulas for Definite Integrals: $18.17$
Category:Simpson's Rule
Repeated Sum of Cubes of Digits of Multiple of 3

== Theorem ==
Let $k \in \Z_{>0}$ be a positive integer.
Let $f: \Z_{>0} \to \Z_{>0}$ be the mapping defined as:
:$\forall m \in \Z_{>0}: \map f m = $ the sum of the cubes of the digits of $n$.
Let $n_0 \in \Z_{>0}$ be a (strictly) positive integer which is a multiple of $3$.
Consider the sequence:
:$s_n = \begin{cases} n_0 & : n = 0 \\
\map f {s_{n - 1} } & : n > 0 \end{cases}$
Then:
:$\exists r \in \N_{>0}: s_r = 153$
That is, by performing $f$ repeatedly on a multiple of $3$ eventually results in the pluperfect digital invariant $153$.
== Proof ==
We verify by brute force:
:Starting on $n_0 \le 2916 = 4 \times 9^3$, we will end on $153$.
First we prove that if $3 \divides n_0$, then $3 \divides \map f {n_0}$.
From Divisibility by 3:
The sum of digits of $n_0$ is divisible by $3$.
By Fermat's Little Theorem:
:$\forall x \in \Z: x^3 \equiv x \pmod 3$
Therefore sum of the cubes of the digits of $n_0$ is also divisible by $3$.
That is the definition of $\map f {n_0}$.
Thus $3 \divides \map f {n_0}$.
It remains to be shown that for every $n_0 > 2916$, $\map f {n_0} < n_0$.
For $n_0 \le 9999$:
:$\map f {n_0} \le 9^3 + 9^3 + 9^3 + 9^3 = 2916 < n_0$
Now suppose $n_0$ is a $k$-digit number, where $k \ge 5$.
Then:
This shows that $\map f {n_0} < n_0$ for all $n_0 > 2916$.
Thus for any $n_0 > 2916$, $s_n$ is eventually less than $2916$.
Therefore we will eventually reach $153$.
== Sources ==
* : $153$
* 
* : $153$
Category:Recreational Mathematics
Category:Pluperfect Digital Invariants
Category:Proofs by Induction
Repellor/Examples/Unit Circle under Complex Square Function

== Example of Repellor ==
Consider the complex function $f: \C \to \C$ defined as:
:$\forall z \in \C: \map f z = z^2$
Then the unit circle $\set {z \in \C: \size z = 1}$ of the Argand plane is a repellor.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Attractors
Replicative Function of x minus Floor of x is Replicative

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $f$ be a replicative function.
Let $g: \R \to \R$ be the real function defined as:
:$\map g x = \map f {x - \floor x}$
Then $g$ is also a replicative function.
== Proof ==
=== Lemma ===
First observe:
:$\ds \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n} = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }$
We need to show that the two final sums are equal.
Suppose:
:$\dfrac j n \le x - \floor x < \dfrac {j + 1} n$
for some integer $j$.
By Real Number minus Floor:
:$0 \le x - \floor x < 1$
This gives:
:$0 \le j \le n - 1$
Let $y = x - \dfrac j n$.
Then:
:$\floor x \le x - \dfrac j n = y$
By Number not less than Integer iff Floor not less than Integer:
:$\floor x \le \floor y$
On the other hand:
:$y \le x$
which gives
:$\floor y \le \floor x$
So $\floor y = \floor x$.
We now have:
:$\dfrac j n \le x - \floor x = y + \dfrac j n - \floor y < \dfrac {j + 1} n$
which gives:
:$0 \le y - \floor y < \dfrac 1 n$
which satisfies the condition for the lemma.
Hence:
We also have:
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{i)}$
Category:Replicative Functions
Category:Replicative Function of x minus Floor of x is Replicative