nyuuzyou/wikis huggingface dataset:
Radon-Nikodym Theorem/Lemma 1

== Lemma ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$ such that: 
:$\nu$ is absolutely continuous with respect to $\mu$.
Define $\mathcal F$ to be the set of $\Sigma$-measurable functions $f : X \to \overline \R_{\ge 0}$ with: 
:$\ds \int_A f \rd \mu \le \map \nu A$
for each $A \in \Sigma$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $\mathcal F$. 
Then:
:for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$. 
== Proof ==
Let $A \in \Sigma$. 
We first prove that for each $f, g \in \mathcal F$, we have: 
:$\max \set {f, g} \in \mathcal F$
From Measurable Functions Determine Measurable Sets, we have: 
:$\set {x \in X : \map f x < \map g x}$ is $\Sigma$-measurable
and:
:$\set {x \in X : \map f x \ge \map g x}$ is $\Sigma$-measurable.
Define: 
:$A_1 = \set {x \in X : \map f x < \map g x} \cap A$
and:
:$A_2 = \set {x \in X : \map f x \ge \map g x} \cap A$
For $x \in A_1$, we have:
:$\map {\max \set {f, g} } x = \map g x$
and for $x \in A_1$, we have:
:$\map {\max \set {f, g} } x = \map f x$ 
From Sigma-Algebra Closed under Countable Intersection, we have: 
:$A_1$ and $A_2$ are $\Sigma$-measurable. 
Clearly we also have:
:$A = A_1 \cup A_2$
with $A_1$ and $A_2$ disjoint.
From Pointwise Maximum of Measurable Functions is Measurable, we have: 
:$\max \set {f, g}$ is $\Sigma$-measurable.
Then:
So, we have: 
:$\ds \int_A \max \set {f, g} \rd \mu \le \map \nu A$
for each $A \in \Sigma$. 
So:
:$\max \set {f, g} \in \mathcal F$.
We now prove that:
:for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$. 
by induction. 
For all $n \in \N$ let $\map P n$ be the proposition:
:$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$ 
=== Basis for Induction ===
We have: 
:$\max \set {f_1} = f_1$
so:
:$\max \set {f_1} \in \mathcal F$ 
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n \ge 1$, then it logically follows that $\map P {n + 1}$ is true.
Our induction hypothesis is:
:$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$ 
We aim to show that: 
:$\max \set {f_1, f_2, \ldots, f_{n + 1} } \in \mathcal F$
=== Induction Step ===
We have: 
:$f_{n + 1} \in \mathcal F$
and:
:$\max \set {f_1, f_2, \ldots, f_n} \in \mathcal F$
so: 
as required.
We therefore obtain:
:for each $n \in \N$, the pointwise maximum $\max \set {f_1, f_2, \ldots, f_n}$ is contained in $\mathcal F$.
Category:Radon-Nikodym Theorem
Radon-Nikodym Theorem/Lemma 2

== Lemma ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ and $\nu$ be $\sigma$-finite measures on $\struct {X, \Sigma}$.
Then tThere exists a pairwise disjoint sequence of $\Sigma$-measurable sets $\sequence {X_n}_{n \in \mathop \N}$ such that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$ 
where: 
:$\map \mu {X_n} < \infty$ and $\map \nu {X_n} < \infty$ for each $n$.
== Proof ==
Since $\mu$ is $\sigma$-finite, there exists a sequence of $\Sigma$-measurable sets $\sequence {A_n}_{n \mathop \in \N}$ such that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty A_n$
with: 
:$\map \mu {A_n} < \infty$ for each $n$.
From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {E_n}_{n \mathop \in \N}$ with:
:$\ds X = \bigcup_{n \mathop = 1}^\infty E_n$
with:
:$E_n \subseteq A_n$ for each $n$. 
From Measure is Monotone, we have that: 
:$\map {\mu} {E_n} \le \map {\mu} {A_n}$ for each $n$.
So:
:$\map {\mu} {E_n}$ is finite for each $n$.
Since $\nu$ is $\sigma$-finite, there exists a sequence of $\Sigma$-measurable sets $\sequence {B_n}_{n \mathop \in \N}$ such that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty B_n$
with: 
:$\map \nu {B_n} < \infty$ for each $n$.
From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {F_n}_{n \mathop \in \N}$ with:
:$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$
with:
:$F_n \subseteq B_n$ for each $n$. 
From Measure is Monotone, we have that: 
:$\map \nu {F_n} \le \map \nu {B_n}$ for each $n$.
So:
:$\map \nu {F_n}$ is finite for each $n$.
Note that for $\tuple {i_1, j_1} \ne \tuple {i_2, j_2}$, we have: 
So $\set {E_i \cap F_j : \tuple {i, j} \in \N^2}$ is a pairwise disjoint family of sets.
We have: 
For each $i, j$, we now have: 
and:
From Cartesian Product of Countable Sets is Countable, we have: 
:$\set {E_i \cap F_j : \tuple {i, j} \in \N^2}$ is countable.
So we can find a bijection: 
:$f : \N \to \set {E_i \cap F_j : \tuple {i, j} \in \N^2}$
Writing: 
:$X_n = \map f n$ 
for each $n \in \N$, we have: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty X_n$
with $\sequence {X_n}_{n \mathop \in \N}$ pairwise disjoint: 
:$\map \mu {X_n} < \infty$ and $\map \nu {X_n} < \infty$ for each $n$.
So we are done.
Category:Radon-Nikodym Theorem
Raising Exponential Order

== Theorem ==
Let $\map f t: \R \to \mathbb F$ a function, where $\mathbb F \in \set {\R, \C}$.
Let $f$ be continuous on the real interval $\hointr 0 \to$, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\hointr 0 \to$.
Let $f$ be of exponential order $a$.
Let $b > a$.
Then $f$ is of exponential order $b$.
== Proof ==
From the definition of exponential order, there exist strictly positive real numbers $M$ and $K$ such that:
:$\forall t \ge M: \size {\map f t} < K e^{a t}$
From Exponential is Strictly Increasing, we have:
:$K e^{a t} < K e^{b t}$
Therefore:
:$\forall t \ge M: \size {\map f t} < K e^{b t}$
The result follows from the definition of exponential order.
Category:Exponential Order
Ramanujan's Approximations to Pi

== 's Approximations to Pi ==
=== $\left({9^2 + \dfrac {19^2} {22} }\right)^{1/4}$ ===
=== $\dfrac {63} {25} \left({17 + 15 \sqrt 5}\right) \left({7 + 15 \sqrt 5}\right)$ ===
=== Approximation involving $\dfrac {99^2} {1103}$ ===
Category:Pi
Category:Approximations to Pi
Ramanujan's Approximations to Pi/2 pi root 2 as 99 squared over 1103

== Approximation to Pi by ==
:$2 \pi \sqrt 2 \approx \dfrac {99^2} {1103}$
== Proof ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Approximations to Pi
Ramanujan's Approximations to Pi/4th Root of 9 Squared plus 19 Squared over 22

== Approximation to Pi by ==
:$\pi \approx \paren {9^2 + \dfrac {19^2} {22} }^{1/4} = 3 \cdotp 14159 \, 26526 \, 2$
== Proof ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Approximations to Pi
Ramanujan's Approximations to Pi/63 over 25 by Product of Sums involving 15 Root 5

== Approximation to Pi by ==
:$\pi \approx \dfrac {63 } {25 } \dfrac {\paren {17 + 15 \sqrt 5} } {\paren {7 + 15 \sqrt 5} }$
== Proof ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Approximations to Pi
Ramanujan's Arctangent Sum

== Theorem ==
Let $n > 0$ and $r \in \N$.
Then:
:$\ds \sum_{k \mathop = 0}^{r - 1} \map \arctan {\dfrac 2 {\paren {n + 2 k + 1}^2} } = \map \arctan {\dfrac {2 r} {\paren {n^2 + 2 n r + 1} } }$
=== Corollary ===
== Proof ==
Recall the Difference of Arctangents:
We first observe:
We now sum both sides on $k$, from $0 \le k \le r - 1$. 
On the , we obtain:
Therefore:
:$\ds \map \arctan {\frac 1 n} - \map \arctan {\frac 1 {\paren {n + 2 r} } } = \sum_{k \mathop = 0}^{r - 1} \map \arctan {\dfrac 2 {\paren {n + 2 k + 1}^2} } $
Finally:
== Examples ==
== Sources ==
* : Chapter $\text {2}$. Sums Related to the Harmonic Series or the Inverse Tangent Function
Category:Ramanujan's Arctangent Sum
Category:Arctangent Function
Ramanujan's Arctangent Sum/Corollary

== Corollary to Ramanujan's Arctangent Sum ==
Let $n > 0$
Then:
:$\ds \sum_{k \mathop = 0}^\infty \map \arctan {\dfrac 2 {\paren {n + 2 k + 1}^2} } = \map \arctan {\dfrac 1 n }$
== Proof ==
Letting $r \to \infty$, we have:
== Sources ==
* : Chapter $\text {2}$. Sums Related to the Harmonic Series or the Inverse Tangent Function
Category:Ramanujan's Arctangent Sum
Category:Arctangent Function
Ramanujan's Arctangent Sum/Examples

== Examples of Use of Ramanujan's Arctangent Sum==
=== Example: $\ds \sum_{k \mathop = 1}^\infty \map \arctan {\dfrac 1 {2 \paren {n + k}^2} }$ ===
=== Example: $\ds \sum_{k \mathop = 1}^\infty \map \arctan {\dfrac 1 {2 k^2} }$ ===
=== Example: $\ds \sum_{k \mathop = 0}^\infty \map \arctan {\dfrac 2 {\paren {2 k + 1}^2} }$ ===
Category:Examples of Use of Ramanujan's Arctangent Sum
Ramanujan's Arctangent Sum/Examples/Sum over k from 0 to Infinity of Arctan of Two Over the Square of 2K Plus 1

== Example of Use of Ramanujan's Arctangent Sum ==
== Proof ==
== Sources ==
* : Chapter $\text {2}$. Sums Related to the Harmonic Series or the Inverse Tangent Function
Category:Examples of Use of Ramanujan's Arctangent Sum
Ramanujan's Arctangent Sum/Examples/Sum over k from 1 to Infinity of Arctan of Reciprocal of Twice the Square of K

== Example of Use of Ramanujan's Arctangent Sum ==
== Proof ==
== Sources ==
* : Chapter $\text {2}$. Sums Related to the Harmonic Series or the Inverse Tangent Function
Category:Examples of Use of Ramanujan's Arctangent Sum
Ramanujan's Arctangent Sum/Examples/Sum over k from 1 to Infinity of Arctan of Reciprocal of Twice the Square of N Plus K

== Example of Use of Ramanujan's Arctangent Sum ==
:$\ds \sum_{k \mathop = 1}^\infty \map \arctan {\dfrac 1 {2 \paren {n + k}^2} } = \map \arctan {\dfrac 1 {2 n + 1} } $
== Proof ==
== Sources ==
* : Chapter $\text {2}$. Sums Related to the Harmonic Series or the Inverse Tangent Function
Category:Examples of Use of Ramanujan's Arctangent Sum
Ramanujan's Continued Fraction of One

== Theorem ==
Let $x \notin \Z_{<0}$.
Then: 
:$1 = \cfrac {x + 1} {x + \cfrac {x + 2} {x + 1 + \cfrac {x + 3 } {x + 2 + \ddots} } }$
=== Corollary ===
== Proof ==
We have:
In order for this sequence to continue, it needs to be shown that:
:$x + n = x + \paren {n - 1} + \dfrac {x + n + 1} {x + n + 1}$
Thus:
The result follows.
== Sources ==
* : Chapter $\text {12}$. Continued Fractions
Category:Ramanujan's Continued Fraction of One
Category:Continued Fractions
Category:Number Theory
Ramanujan's Continued Fraction of One/Corollary

== Corollary to Ramanujan's Continued Fraction of One == 
:$1 = \cfrac 2 {1 + \cfrac 3 {2 + \cfrac 4 {3 + \ddots} } }$
== Proof ==
From Ramanujan's Continued Fraction of One, we have:
:$1 = \cfrac {x + 1} {x + \cfrac {x + 2} {x + 1 + \cfrac {x + 3 } {x + 2 + \ddots} } }$
Set x := 1, then...
:$1 = \cfrac 2 {1 + \cfrac 3 {2 + \cfrac 4 {3 + \ddots} } }$
Category:Ramanujan's Continued Fraction of One
Ramanujan's Infinite Nested Roots

== Theorem ==
:$3 = \sqrt {1 + 2 \sqrt {1 + 3 \sqrt { 1 + \cdots} } }$
== Proof ==
We have:
In order for this sequence to continue, it needs to be shown that:
:$n + 1 = \sqrt {1 + n \left({n + 2}\right)}$
Thus:
The result follows.
== Sources ==
* : Chapter $\text {12}$. Continued Fractions
* 
Category:Number Theory
Ramanujan Phi Function in terms of Digamma Function

== Theorem ==
:$\map \psi {\dfrac 1 z} + \map \psi {1 - \dfrac 1 z} = -2 \gamma - z \map \phi z$
where:
:$\map \psi z$ denotes the digamma function
:$\map \phi z$ denotes the Ramanujan phi function
:$z \in \C$, where $\dfrac 1 z \notin \Z_{\le 0}$ and $\paren {1 - \dfrac 1 z} \notin \Z_{\le 0}$
== Proof ==
== Sources ==
* : Chapter $8$. Analogues of the Gamma Function
Category:Digamma Function
Category:Ramanujan Phi Function
Ramanujan Sum is Multiplicative

== Theorem ==
Let $q \in \N_{>0}$, $n \in \N$.
Let $\map {c_q} n$ be the Ramanujan sum.
Then $\map {c_q} n$ is multiplicative in $q$.
== Proof ==
Let $q, r \in \N$ such that:
:$\gcd \set {q, r} = 1$
where $\gcd \set {q, r}$ denotes the greatest common divisor of $q$ and $r$.
Then:
This completes the proof.
Category:Analytic Number Theory
Ramus's Identity

== Theorem ==
Let $k, m, n \in \Z_{\ge 0}$ be positive integers such that $0 \le k < m$.
Then:
== Proof ==
Let $\omega := e^{2 \pi i / m}$.
Then by the Binomial Theorem:
:$(1): \quad \ds \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k} = \sum_t \sum_{0 \mathop \le j \mathop < m} \binom n t \omega^{j \paren {t - k} }$
By Sum of Geometric Sequence:
:$\ds \sum_{0 \mathop \ge j \mathop < m} \omega^{j \paren {t - k} } = \begin{cases} \dfrac {1 - \omega^{m \paren {t - k} } } {1 - \omega^{t - k} } & : t - k \not \equiv 0 \pmod m \\ m & : t \equiv k \pmod m \end{cases}$
We have that:
:$\omega = \exp \dfrac {2 \pi i} m \implies \omega^m = 1$
and so:
Thus the summation on the of $(1)$ is:
:$m \ds \sum_{t \bmod m \mathop = k} \binom n t$
The summation on the of $(1)$ is:
Because the is wholly real, so must the be.
So, taking the real parts of the and equating it to the reveals the result.
== Examples ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: Exercise $38$
Category:Binomial Coefficients
Category:Ramus's Identity
Random Variable has Zero Variance iff Almost Surely Constant

== Theorem ==
Let $X$ be a random variable such that $\expect {X^2}$ exists.
Then $\var X = 0$ there exists $c \in \R$ with $\map \Pr {X = c} = 1$.
That is, $X$ is almost surely constant.
== Proof ==
=== Sufficient Condition ===
Suppose that there exists some $c \in \R$ with $\map \Pr {X = c} = 1$.
From Expectation of Almost Surely Constant Random Variable: 
:$\expect X = c$
Let $\map \supp X$ be the support of $X$. 
Since $\map \Pr {X = c} = 1$, we have: 
:$\map \supp X = \set c$
We therefore have: 
Hence, from Variance as Expectation of Square minus Square of Expectation:
=== Necessary Condition ===
Suppose now that $\var X = 0$. 
By definition of variance: 
:$\expect {\paren {X - \expect X}^2} = 0$
From Condition for Expectation of Non-Negative Random Variable to be Zero: 
:$\map \Pr {\paren {X - \expect X}^2 = 0} = 1$
That is: 
:$\map \Pr {X - \expect X = 0} = 1$
giving: 
:$\map \Pr {X = \expect X} = 1$
Setting $c = \expect X \in \R$, we see that we have $c \in \R$ such that: 
:$\map \Pr {X = c} = 1$ 
Category:Variance
Random Vector/Examples/Throwing 6-sided Die 4 Times

== Example of Random Vector ==
Let $\EE$ be the experiment of throwing a standard $6$-sided die $4$ times.
Then the outcome of $\EE$ can be expressed as a random vector $\tuple {x_1, x_2, x_3, x_4}$, where each $x_i$ is uniformly distributed over the sample space $\set {1, 2, 3, 4, 5, 6}$.
If $\EE$ is performed with the results $5$, $2$, $5$ and $6$, then the outcome would be reported as $\tuple {5, 2, 5, 6}$.
== Sources ==
* 
Category:Random Variables
Random Vector is Random Variable

== Theorem ==
Let $n \in \N$. 
Let $\struct {X, \Sigma, \Pr}$ be a probability space.
Let $\struct {S_1, \Sigma_1}, \struct {S_2, \Sigma_2}, \ldots, \struct {S_n, \Sigma_n}$ be measurable spaces.
Let: 
:$\ds S = \prod_{i \mathop = 1}^n S_i$
Let: 
:$\ds \Sigma' = \bigotimes_{i \mathop = 1}^n \Sigma_i$ 
where $\ds \bigotimes_{i \mathop = 1}^n \Sigma_i$ denotes the product $\sigma$-algebra of $\Sigma_1, \Sigma_2, \ldots, \Sigma_n$. 
For each $i \in \set {1, 2, \ldots, n}$:
:let $X_i$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S_i, \Sigma_i}$. 
Define a function $\mathbf X : \Omega \to S$ by: 
:$\map {\mathbf X} \omega = \tuple {\map {X_1} \omega, \map {X_2} \omega, \ldots, \map {X_n} \omega}$
for each $\omega \in \Omega$.
Then $\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$.
== Proof ==
Note that from the definition of product $\sigma$-algebra: finite case, we have:
:$\ds \Sigma' = \map \sigma {\set {\prod_{i \mathop = 1}^n S_i : S_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} } }$
where $\sigma$ denotes the $\sigma$-algebra generated by a collection of subsets. 
We aim to apply Mapping Measurable iff Measurable on Generator.
We therefore aim to show that: 
:${\mathbf X}^{-1} \sqbrk B \in \Sigma$ 
for each: 
:$\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$
Let: 
:$\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$
Then, for each $i \in \set {1, 2, \ldots, n}$ there exists $E_i \in \Sigma_i$ such that: 
:$\ds B = \prod_{i \mathop = 1}^n E_i$ 
Then, for $\omega \in \Omega$, we have $\map {\mathbf X} \omega \in B$ :
:$\ds \map {X_i} \omega \in E_i$ for each $i \in \set {1, 2, \ldots, n}$.
That is, $\omega \in {\mathbf X}^{-1} \sqbrk B$ : 
:$\omega \in {X_i}^{-1} \sqbrk {E_i}$ for each $i \in \set {1, 2, \ldots, n}$.
So:
:$\ds {\mathbf X}^{-1} \sqbrk B = \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i}$
Since $X_i$ is $\Sigma/\Sigma_i$-measurable, we have: 
:${X_i}^{-1} \sqbrk {E_i} \in \Sigma$ 
for each $i \in \set {1, 2, \ldots, n}$. 
From Sigma-Algebra Closed under Finite Intersection, we therefore have: 
:$\ds \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i} \in \Sigma$
so:
:$\ds {\mathbf X}^{-1} \sqbrk B \in \Sigma$ 
which was the demand. 
From Mapping Measurable iff Measurable on Generator, we then have: 
:$\mathbf X$ is $\Sigma/\Sigma'$-measurable.
So:
:$\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$
as required.
== Also see ==
* Vector is Random Vector iff Components are Random Variables shows that the converse is also true: if $\tuple {X_1, X_2, \ldots, X_n}$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$ then each $X_i$ is a random variable taking values in $\struct {S_i, \Sigma_i}$.
Category:Random Vectors
Category:Random Variables
Range and Kernel of Idempotent are Algebraically Complementary

== Theorem ==
Let $H$ be a Hilbert space.
Let $A$ be an idempotent operator.
Then $\ker A$ and $\Rng A$ are algebraically complementary, that is:
:$\ker A \cap \Rng A = \left({0}\right)$, the zero subspace
:$\ker A + \Rng A = H$, where $+$ signifies setwise addition.
== Proof ==
== Sources ==
* $\text {II}.3.2 \ \text {(c)}$
Category:Linear Transformations on Hilbert Spaces
Range of Characters

== Theorem ==
Let $G$ be a finite abelian group of order $m$.
Let $\chi: G \to \C^\times$ be a character on $G$.
Then for any $g \in G$, $\map \chi g$ is an $m$th root of unity.
If $e$ is the identity of $G$ then $\map \chi g = 1$.
== Proof ==
The claim that $\map \chi e = 1$ is shown by Group Homomorphism Preserves Identity.
Let $g \in G$ be arbitrary.
Let $k$ be the order of $g$.
By Order of Element Divides Order of Finite Group:
:$\exists l \in \Z: m = k l$
Therefore:
:$g^m = \paren {g^k}^l = e^l = e$
By the homomorphism property:
:$1 = \map \chi e = \map \chi {g^m} = \map \chi g^m$
Hence the result.
Category:Analytic Number Theory
Range of Common Logarithm of Number between 1 and 10

== Theorem ==
Let $x \in \R$ be a real number such that:
:$1 \le x < 10$
Then:
:$0 \le \log_{10} x \le 1$
where $\log_{10}$ denotes the common logarithm function.
== Proof ==
We have:
The result follows from Logarithm is Strictly Increasing.
== Sources ==
* : Chapter $1$: Logarithms
Category:Common Logarithms
Range of Idempotent is Kernel of Complementary Idempotent

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $A$ be an idempotent operator.
Then:
:$\Rng A = \map \ker {I - A}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
If $h \in \map \ker {I - A}$, we have:
:$\map {\paren {I - A} } h = {\mathbf 0}_\HH$
That is:
:$h - A h = {\mathbf 0}_\HH$ 
so that:
:$A h = h$ 
and so:
:$h \in \Rng A$ 
So we have:
:$\map \ker {I - A} \subseteq \Rng A$ 
Now let $h \in \Rng A$.
Then there exists $k \in \HH$ such that $h = A k$. 
Then we have $h - A h = h - A^2 k$.
Since $A$ is an idempotent operator, we have:
:$A^2 k = A k = h$
Hence:
:$h - A h = {\mathbf 0}_\HH$
So:
:$h \in \map \ker {I - A}$
We therefore obtain:
:$\Rng A \subseteq \map \ker {I - A}$
So:
:$\Rng A = \map \ker {I - A}$
== Sources ==
* $\text {II}.3.2 \ \text {(b)}$
Category:Range of Idempotent is Kernel of Complementary Idempotent
Category:Linear Transformations on Hilbert Spaces
Range of Idempotent is Kernel of Complementary Idempotent/Corollary 1

== Corollary to Range of Idempotent is Kernel of Complementary Idempotent ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $A$ be an idempotent operator.
Then:
:$\ker A = \Rng {I - A}$
== Proof ==
From Complementary Idempotent is Idempotent, we have that:
:$I - A$ is an idempotent operator.
Hence by Range of Idempotent is Kernel of Complementary Idempotent, we have:
:$\Rng {I - A} = \map \ker {I - \paren {I - A} } = \ker A$
== Sources ==
* $\text {II}.3.2 \ \text {(b)}$
Category:Range of Idempotent is Kernel of Complementary Idempotent
Range of Idempotent is Kernel of Complementary Idempotent/Corollary 2

== Corollary to Range of Idempotent is Kernel of Complementary Idempotent ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $A$ be an idempotent operator.
Then:
:$\Rng A$ is a closed linear subspace of $\HH$.
== Proof ==
From Range of Idempotent is Kernel of Complementary Idempotent, we have:
:$\Rng A = \map \ker {I - A}$
From Space of Bounded Linear Transformations forms Vector Space:
:$I - A$ is a Definition:Bounded Linear Transformation.
From Kernel of Bounded Linear Transformation is Closed Linear Subspace, we have:
:$\map \ker {I - A}$ is a closed linear subspace of $\HH$.
Hence $\Rng A$ is a closed linear subspace of $\HH$.
== Sources ==
* $\text {II}.3.2 \ \text {(b)}$
Category:Range of Idempotent is Kernel of Complementary Idempotent
Range of Infinite Sequence may be Finite

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence.
Then it is possible for the range of $\sequence {x_n}$ to be finite.
== Proof ==
Consider the infinite sequence $\sequence {x_n}_{n \mathop \in \N}$ defined as:
:$\forall n \in \N: x_n = \dfrac {1 + \paren {-1}^n} 2$
Thus:
:$\sequence {x_n}_{n \mathop \in \N} = 1, 0, 1, 0, \dotsc$
Hence the range of $\sequence {x_n}$ is $\set {0, 1}$, which is finite.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences
Category:Sequences
Range of Modulo Operation for Negative Modulus

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y < 0$.
Then:
:$0 \ge x \bmod y > y$
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(2) \ \text{b)}$
Category:Modulo Arithmetic
Range of Modulo Operation for Positive Modulus

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y > 0$.
Then:
:$0 \le x \bmod y < y$
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: $(2) \ \text{a)}$
Category:Modulo Arithmetic