nyuuzyou/wikis huggingface dataset:
Sum of Expectations of Independent Trials

== Theorem ==
Let $\EE_1, \EE_2, \ldots, \EE_n$ be a sequence of experiments whose outcomes are independent of each other.
Let $X_1, X_2, \ldots, X_n$ be discrete random variables on $\EE_1, \EE_2, \ldots, \EE_n$ respectively.
Let $\expect {X_j}$ denote the expectation of $X_j$ for $j \in \set {1, 2, \ldots, n}$.
Then we have, whenever both sides are defined:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
That is, the sum of the expectations equals the expectation of the sum.
== Proof 1 ==
== Proof 2 ==
Category:Expectation
Category:Sum of Expectations of Independent Trials
Sum of Expectations of Independent Trials/Proof 1

== Theorem ==
== Proof ==
The proof proceeds by induction on the number of terms $n$ in the sum.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \expect {\sum_{j \mathop = 1}^1 X_j} = \sum_{j \mathop = 1}^1 \expect {X_j}$
That is:
:$\expect {X_1} = \expect {X_1}$
which is tautologically true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
$\ds \expect {\sum_{j \mathop = 1}^k X_j} = \sum_{j \mathop = 1}^k \expect {X_j}$
from which it is to be shown that:
$\ds \expect {\sum_{j \mathop = 1}^{k + 1} X_j} = \sum_{j \mathop = 1}^{k + 1} \expect {X_j}$
=== Induction Step ===
This is our induction step:
Denote $Y = \ds \sum_{j \mathop = 1}^k X_j$
Then we compute:
The result follows by the Principle of Mathematical Induction.
Category:Sum of Expectations of Independent Trials
Sum of Expectations of Independent Trials/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction on the number of terms $n$ in the sum.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \expect {\sum_{j \mathop = 1}^n X_j} = \sum_{j \mathop = 1}^n \expect {X_j}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \expect {\sum_{j \mathop = 1}^1 X_j} = \sum_{j \mathop = 1}^1 \expect {X_j}$
That is:
:$\expect {X_1} = \expect {X_1}$
which is tautologically true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
$\ds \expect {\sum_{j \mathop = 1}^k X_j} = \sum_{j \mathop = 1}^k \expect {X_j}$
from which it is to be shown that:
$\ds \expect {\sum_{j \mathop = 1}^{k + 1} X_j} = \sum_{j \mathop = 1}^{k + 1} \expect {X_j}$
=== Induction Step ===
This is the induction step:
Denote the random variable $Y = \ds \sum_{j \mathop = 1}^k X_j$.
Then we compute:
The result follows by the Principle of Mathematical Induction.
Category:Sum of Expectations of Independent Trials
Sum of Exponential of i k x

== Theorem ==
:$\ds \sum_{k \mathop = 0}^n \map \exp {i k x} = \paren {i \sin \frac {n x} 2 + \cos \frac {n x} 2} \frac {\map \sin {\frac {\paren {n + 1} x} 2} } {\sin \frac x 2}$
where $x$ is a complex number that is not an integer multiple of $2 \pi$.
== Proof ==
Category:Exponential Function
Sum of External Angles of Polygon equals Four Right Angles

== Theorem ==
Let the external angles of a polygon be generated in the same direction going around the polygon.
Then the sum of all these external angles equals $4$ right angles, that is, $360 \degrees$.
== Proof ==
Let $V$ be an arbitrary vertex of a polygon $P$.
Let $T$ be the sum of all the external angles of $P$
Let $A$ be the size of the internal angle of $V$.
Let $E$ be the size of the external angle of $V$.
By definition, $E = 180 \degrees - A$.
Let $S$ denote the sum of all internal angles of $P$.
Hence we have:
:$T = n \times 180 \degrees - S$
From Sum of Internal Angles of Polygon:
:$S = \paren {n - 2} 180 \degrees$
Hence we have:
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.9$
* 
* 
Category:External Angles
Category:Polygons
Sum of Finite Set of Submodules is Supremum of Lattice of Submodules

== Theorem ==
Let $R$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Let $H_1, H_2, \ldots, H_n$ be submodules of $M$.
Then $H_1 + H_2 + \cdots + H_n$ is the supremum of $\set {H_1, H_2, \ldots, H_n}$ in the complete lattice of all submodules of $M$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$: Corollary
Category:Submodules
Sum of Floor and Floor of Negative

== Theorem ==
Let $x \in \R$. Then:
:$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$
where $\floor x$ denotes the floor of $x$.
== Proof ==
Let $x \in \Z$.
Then from Real Number is Integer iff equals Floor:
:$x = \floor x$
Now $x \in \Z \implies -x \in \Z$, so:
:$\floor {-x} = -x$
Thus:
:$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$
Now let $x \notin \Z$.
From Real Number is Floor plus Difference:
:$x = n + t$
where $n = \floor x$ and $t \in \hointr 0 1$.
Thus:
:$-x = -\paren {n + t} = -n - t = -n - 1 + \paren {1 - t}$
As $t \in \hointr 0 1$, we have:
:$1 - t \in \hointr 0 1$
Thus:
:$\floor {-x} = -n - 1$
So:
:$\floor x + \floor {-x} = n + \paren {-n - 1} = n - n - 1 = -1$
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.1$
Category:Floor Function
Sum of Floors not greater than Floor of Sum

== Theorem ==
Let $\floor x$ denote the floor function.
Then:
:$\floor x + \floor y \le \floor {x + y}$
The equality holds:
:$\floor x + \floor y = \floor {x + y}$
:
:$x \bmod 1 + y \bmod 1 < 1$
where $x \bmod 1$ denotes the modulo operation.
== Proof ==
From the definition of the modulo operation, we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which:
Hence the inequality.
The equality holds :
:$\floor {\paren {x \bmod 1} + \paren {y \bmod 1} } = 0$
that is, :
:$x \bmod 1 + y \bmod 1 < 1$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $7$
Category:Floor Function
Sum of Fourth Powers of Sine and Cosine

== Theorem ==
:$\cos^4 x + \sin^4 x = 1 - 2 \cos^2 x \sin^2 x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $4.$
Category:Trigonometric Identities
Category:Sine Function
Category:Cosine Function
Sum of Fourth Powers with Product of Squares

== Theorem ==
:$x^4 + x^2 y^2 + y^4 = \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
== Proof ==
Then:
Thus:
== Sources ==
* : $\S 2$: Special Products and Factors: $2.18$
* : $\S 2$: Special Products and Factors: $2.18.$
Category:Fourth Powers
Category:Square Function
Sum of Functions of Bounded Variation is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$. 
Then $f + g$ is of bounded variation on $\closedint a b$ with:
:$\map {V_{f + g} } {\closedint a b} \le \map {V_f} {\closedint a b} + \map {V_g} {\closedint a b}$
where $\map {V_{f + g} } {\closedint a b}$ denotes the total variation of $f + g$ on $\closedint a b$.
== Proof ==
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
Note that since $f$ and $g$ are of bounded variation, there exists $M, K \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le M$
:$\map {V_g} {P ; \closedint a b} \le K$
for all finite subdivisions $P$ of $\closedint a b$. 
We therefore have: 
:$\map {V_{f + g} } {P ; \closedint a b} \le M + K$
for all finite subdivisions $P$.
So $f + g$ is of bounded variation.
Note then that: 
== Sources ==
* : $\S 6.4$: Total Variation: Theorem $6.9$
Category:Total Variation of Real Function
Category:Bounded Variation
Sum of Functions of Exponential Order

== Theorem ==
Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $f + g: t \mapsto \map f t + \map g t$ is of exponential order $\max \set {a, b}$.
== Proof ==
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
Category:Exponential Order
Sum of General Harmonic Numbers in terms of Riemann Zeta Function

== Theorem ==
:$\ds \harm r x - \dfrac 1 {n^r} \sum_{k \mathop = 0}^{n - 1} \harm r {x - \dfrac k n} = \paren {1 - n^{1 - r} } \map \zeta r$
where:
:$\harm r x$ and $\harm r {x - \dfrac k n}$ denotes the general harmonic number of order $r$ evaluated at $x$ and $\paren {\dfrac {x - k} n}$, respectively.
:$\map \zeta r$ is the Completed Riemann zeta function
:$r$ and $x$ are complex numbers with $\paren {\dfrac {x - k} n} \notin \Z_{<0}$
:$n \in \Z_{>0}$
=== Corollary ===
== Proof ==
=== Lemma ===
== Sources ==
* : Chapter $7$. Sums of Powers, Bernoulli Numbers, and the Gamma Function
Category:Sum of General Harmonic Numbers in terms of Riemann Zeta Function
Category:General Harmonic Numbers
Category:Riemann Zeta Function
Sum of General Harmonic Numbers in terms of Riemann Zeta Function/Corollary

== Corollary to Sum of General Harmonic Numbers in terms of Riemann Zeta Function ==
:$\ds \sum_{k \mathop = 1}^{n - 1} \harm r {-\dfrac k n} = \paren {n - n^r} \map \zeta r$
where:
:$\harm r x$ and $\harm r {-\dfrac k n}$ denotes the general harmonic number of order $r$ evaluated at $x$ and $\paren {\dfrac {-k} n}$, respectively.
:$\map \zeta r$ is the Completed Riemann zeta function
:$r$ and $x$ are complex numbers
:$n \in \Z_{>0}$.
== Proof ==
== Sources ==
* : Chapter $7$. Sums of Powers, Bernoulli Numbers, and the Gamma Function
Category:Riemann Zeta Function
Category:General Harmonic Numbers
Category:Sum of General Harmonic Numbers in terms of Riemann Zeta Function
Sum of General Harmonic Numbers in terms of Riemann Zeta Function/Lemma

== Lemma for Sum of General Harmonic Numbers in terms of Riemann Zeta Function ==
:$\ds \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + x}^r} = \sum_{k \mathop = 0}^{n - 1} \sum_{j \mathop = 1}^\infty \frac 1 {\paren {n j - k + x }^r}$
where:
:$r$ and $x$ are complex numbers
:$n \in \Z_{>0}$.
== Proof ==
Category:Sum of General Harmonic Numbers in terms of Riemann Zeta Function
Sum of Geometric Sequence

== Theorem ==
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also presented as ==
Note that when $x < 1$ the result is usually given as:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {1 - x^n} {1 - x}$
Some sources give it as:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {1 - x^{n + 1} } {1 - x}$
and likewise its corollary:
:$\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
== Examples ==
== Also see ==
* Sum of Infinite Geometric Sequence
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Theorem $\text{1-2}$
* : $\S 1.2.1$: The Geometric Series
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : Chapter $\text {B}.5$: Fermat's Calculation of $\int_0^b x^n \rd x$ for Positive Rational $n$
Category:Sums of Sequences
Category:Geometric Sequences
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Corollary 1

== Corollary to Sum of Geometric Sequence ==
Let $a, a r, a r^2, \ldots, a r^{n - 1}$ be a geometric sequence.
Then:
:$\ds \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {r^n - 1} } {r - 1}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This result is also seen presented as:
:$\ds \sum_{j \mathop = 0}^{n - 1} a r^j = \frac {a \paren {1 - r^n} } {1 - r}$
which is usually more manageable when $r < 1$.
== Sources ==
* : Book $\text{IX}$. Propositions
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Sum of Geometric Progression to $n$ Terms: $3.1.10$
* : $\S 19$: Geometric Series: $19.4$
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $2$
* 
* 
* 
* : $\S 21$: Series of Constants: Geometric Series: $21.4.$
* 
Category:Geometric Sequences
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Corollary 1/Proof 1

== Corollary to Sum of Geometric Sequence ==
== Proof ==
By the Distributive Property:
:$a + a r + a r^2 + \cdots + a r^{n - 1} = a \paren {1 + r + r^2 + \cdots + r^{n - 1} }$
The result follows from Sum of Geometric Sequence.
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Corollary 2

== Theorem ==
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Let $n \in \N_{>0}$.
Then: 
:$\ds \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$
== Proof ==
 
Category:Geometric Sequences
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Examples/Common Ratio 1

== Theorem ==
Consider the Sum of Geometric Sequence defined on the standard number fields for all $x \ne 1$.
:$\ds \sum_{j \mathop = 0}^n a x^j = a \paren {\frac {1 - x^{n + 1} } {1 - x} }$
When $x = 1$, the formula reduces to:
:$\ds \sum_{j \mathop = 0}^n a 1^j = a \paren {n + 1}$
== Proof ==
When $x = 1$, the is undefined:
:$a \paren {\dfrac {1 - 1^{n + 1} } {1 - 1} } = a \dfrac 0 0$
However, the degenerates to:
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $11$
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Examples/Index to Minus 1

== Theorem ==
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$
== Proof ==
The summation on the is vacuous:
:$\ds \sum_{j \mathop = 0}^{-1} x^j = 0$
while on the we have:
as long as $x \ne 1$.
However, the theorem itself is based on the assumption that $n \ge 0$, so while the result is correct, the derivation to achieve it is not.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $9$
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Examples/Index to Minus 2

== Theorem ==
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
:$\ds \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$
== Proof ==
The summation on the is vacuous:
:$\ds \sum_{j \mathop = 0}^{-2} x^j = 0$
while on the we have:
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $10$
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Examples/One Seventh from 1 to n

== Theorem ==
:$\ds \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \paren {1 - \frac 1 {7^{n + 1} } }$
== Proof ==
Hence the result.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $12$
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Proof 1

== Theorem ==
== Proof ==
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
=== Basis for the Induction ===
$\map P 1$ is the case:
so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{j \mathop = 0}^{k - 1} x^j = \frac {x^k - 1} {x - 1}$
Then we need to show:
:$\ds \sum_{j \mathop = 0}^k x^j = \frac {x^{k + 1} - 1} {x - 1}$
=== 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:
:$\ds \forall n \in \N_{>0}: \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Theorem $\text {1-2}$
* : $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (2)$
Category:Sum of Geometric Sequence
Category:Proofs by Induction
Sum of Geometric Sequence/Proof 2

== Theorem ==
== Proof ==
Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.
Then:
The result follows.
== Sources ==
* : $\S 1.2.4$: Factorials and binomial coefficients: Exercise $1.2: 1$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes: Footnote $2$
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Proof 3

== Theorem ==
== Proof ==
From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Set $a = x$ and $b = 1$:
:$\ds x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = \paren {x - 1} \sum_{j \mathop = 0}^{n - 1} x^j$
from which the result follows directly.
Category:Sum of Geometric Sequence
Sum of Geometric Sequence/Proof 4/Lemma

== Lemma ==
Let $n \in \N_{>0}$.
Then:
:$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
== Proof ==
Proof by induction on $n$:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
=== Basis for the Induction ===
$\map P 1$ is the case:
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:
:$\ds \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i = 1 - x^k$
Then we need to show:
:$\ds \paren {1 - x} \sum_{i \mathop = 0}^k x^i = 1 - x^{k + 1}$
=== 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:
:$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
Category:Sum of Geometric Sequence
Sum of Harmonic Numbers approaches Harmonic Number of Product of Indices

== Theorem ==
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let:
:$\map T {m, n} := H_m + H_n - H_{m n}$
where $H_n$ denotes the $n$th harmonic number.
Then as $m$ or $n$ increases, $\map T {m, n}$ never increases, and reaches its minimum when $m$ and $n$ approach infinity.
== Proof ==
Similarly:
:$\map T {m, n + 1} - \map T {m, n} \le 0$
From Approximate Size of Sum of Harmonic Series, the limiting value of $\map T {m, n}$ is the Euler-Mascheroni constant:
:$\ds \lim_{m, n \mathop \to \infty} H_m + H_n - H_{m n} = \gamma$
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $7$
Category:Harmonic Numbers
Sum of Hyperbolic Sine and Cosine equals Exponential

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$e^z = \cosh z + \sinh z$
where:
:$e^z$ denotes the complex exponential function
:$\cosh z$ denotes the cosine function
:$\sinh z$ denotes sine function
== Proof ==
== Also see ==
* Euler's Formula
== Sources ==
* 
* 
Category:Exponential Function
Category:Hyperbolic Sine Function
Category:Hyperbolic Cosine Function
Sum of Ideals is Ideal

== Theorem ==
Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.
Then:
: $J = J_1 + J_2$ is an ideal of $R$
where $J_1 + J_2$ is as defined in subset product with respect to $\struct{R, +}$.
=== General Result ===
=== Corollary ===
== Proof ==
By definition, $\struct {R, +}$ is an abelian group.
So from Subset Product of Abelian Subgroups, we have that:
:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
is itself a subgroup of $R$.
Now consider $a \circ b$ where $a \in J, b \in R$.
Then:
Similarly, $b \circ a \in J_1 + J_2$
So by definition $J_1 + J_2$ is an ideal of $R$.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $40$
* : $\S 6$: Rings and fields
Category:Ideal Theory
Sum of Ideals is Ideal/Corollary

== Corollary to Sum of Ideals is Ideal ==
Let $J_1$ and $J_2$ be ideals of a ring $\struct {R, +, \circ}$.
Let $J = J_1 + J_2$ be an ideal of $R$ where $J_1 + J_2$ is as defined in subset product.
Then:
:$J_1 \subseteq J_1 + J_2$
:$J_2 \subseteq J_1 + J_2$
== Proof ==
From Sum of Ideals is Ideal we have that $j$ is an ideal of $R$.
Then:
:$0_R \in J_1 + J_2$
and so:
:$\forall x \in J_1: x + 0_R = x \in J_1 + J_2$
Similarly for $J_2$.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $40$
Category:Ideal Theory
Sum of Ideals is Ideal/General Result

== Theorem ==
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Then:
:$J = J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product with respect to $\struct{R, +}$.
=== Corollary ===
== Proof ==
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
$\map P 1$ is true, as this just says $J_1$ is an ideal of $R$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$J_1 + J_2$ is an ideal of $R$
which is proved in Sum of Ideals is Ideal.
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:
:$J_1 + J_2 + \cdots + J_k$ is an ideal of $R$.
Then we need to show:
:$J_1 + J_2 + \cdots + J_k + J_{k + 1}$ is an ideal of $R$.
=== Induction Step ===
This is our induction step:
Let $J = J_1 + J_2 + \cdots + J_k$.
From the induction hypothesis, $J$ is an ideal.
From the base case, $J + J_{k + 1}$ is an ideal.
That is:
:$J_1 + J_2 + \cdots + J_k + J_{k + 1}$ is an ideal of $R$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: J_1 + J_2 + \cdots + J_n$ is an ideal of $R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
* : $\S 2.3$: Some properties of subrings and ideals: Lemma $2.16$
Category:Ideal Theory
Sum of Ideals is Ideal/General Result/Corollary

== Corollary to Sum of Ideals is Ideal/General Result ==
Let $J_1, J_2, \ldots, J_n$ be ideals of a ring $\struct {R, +, \circ}$.
Let $J = J_1 + J_2 + \cdots + J_n$ be an ideal of $R$ where $J_1 + J_2 + \cdots + J_n$ is as defined in subset product.
$J$ is contained in every subring of $R$ containing $\ds \bigcup_{k \mathop = 1}^n {J_k}$.
== Proof ==
From Sum of Ideals is Ideal/General Result we have that $J$ is an ideal of $R$.
The result follows directly from the definition of join of subgroups.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ideal Theory
Sum of Independent Binomial Random Variables

== Theorem ==
Let $X$ and $Y$ be discrete random variables with a binomial distribution:
:$X \sim \Binomial m p$
and
:$Y \sim \Binomial n p$
Let $X$ and $Y$ be independent.
Then their sum $Z = X + Y$ is distributed as:
:$Z \sim \Binomial {m + n} p$
== Proof ==
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = \paren {q + p s}^m$
:$\map {\Pi_Y} s = \paren {q + p s}^n$
respectively.
Now because of their independence, we have:
This is the probability generating function for a discrete random variable with a binomial distribution:
:$\Binomial {m + n} p$
Therefore:
:$Z = X + Y \sim \Binomial {m + n} p$
== Sources ==
* : $\S 4.4$: Sums of independent random variables: Exercise $8$
Category:Binomial Distribution
Category:Probability Generating Functions