nyuuzyou/wikis huggingface dataset:
Sum of Squares of Hyperbolic Secant and Tangent

== Theorem ==
:$\sech^2 x + \tanh^2 x = 1$
where $\sech$ and $\tanh$ are hyperbolic secant and hyperbolic tangent.
== Proof ==
== Also presented as ==
== Also see ==
* Difference of Squares of Hyperbolic Cotangent and Cosecant
== Sources ==
* : $8.12$: Relationships among Hyperbolic Functions
* 
* 
Category:Sum of Squares of Hyperbolic Secant and Tangent
Category:Hyperbolic Secant Function
Category:Hyperbolic Tangent Function
Sum of Squares of Hyperbolic Secant and Tangent/Also presented as

== Sum of Squares of Hyperbolic Secant and Tangent: Also known as == 
Sum of Squares of Hyperbolic Secant and Tangent can also be reported as:
:$1 - \tanh^2 x = \sech^2 x$
or:
:$1 - \sech^2 x = \tanh^2 x$
== Sources ==
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $5$
* 
Category:Sum of Squares of Hyperbolic Secant and Tangent
Sum of Squares of Sine and Cosine

== Theorem ==
:$\cos^2 x + \sin^2 x = 1$
where $\sin$ and $\cos$ are sine and cosine.
=== Corollary 1 ===
=== Corollary 2 ===
== Algebraic Proof ==
== Geometric Proof ==
== Unit Circle Proof ==
== Euler's Formula Proof ==
== Proof using Euler's Identities ==
== Also known as ==
This result is often referred to as the Pythagorean Identity.
== Also see ==
* Difference of Squares of Hyperbolic Cosine and Sine
== Historical Note ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(1)$
* : $\S 5$: Trigonometric Functions: $5.19$
* 
::in which a mistake occurs
* 
* 
* 
* 
* : Chapter $5$: Eternal Triangles: The origins of trigonometry
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Sum of Squares of Sine and Cosine
Category:Sine Function
Category:Cosine Function
Sum of Squares of Sine and Cosine/Corollary 1

== Corollary to Sum of Squares of Sine and Cosine ==
For all $x \in \C$:
:$\sec^2 x - \tan^2 x = 1 \quad \text {(when $\cos x \ne 0$)}$
where $\sec$, $\tan$ and $\cos$ are secant, tangent and cosine respectively.
== Proof ==
When $\cos x \ne 0$:
== Also presented as ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(2)$
* : $\text V$. Trigonometry: Definitions of the ratios
* : $\S 5$: Trigonometric Functions: $5.20$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Difference of Squares of Secant and Tangent
Category:Sum of Squares of Sine and Cosine
Category:Tangent Function
Category:Secant Function
Sum of Squares of Sine and Cosine/Corollary 2

== Corollary to Sum of Squares of Sine and Cosine ==
For all $x \in \C$:
:$\csc^2 x - \cot^2 x = 1 \quad \text {(when $\sin x \ne 0$)}$
where $\csc$, $\cot$ and $\sin$ are cosecant, cotangent and sine respectively.
== Proof ==
When $\sin x \ne 0$:
== Also presented as ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(3)$
* : $\text V$. Trigonometry: Definitions of the ratios
* : $\S 5$: Trigonometric Functions: $5.21$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Difference of Squares of Cosecant and Cotangent
Category:Sum of Squares of Sine and Cosine
Category:Cosecant Function
Category:Cotangent Function
Sum of Squares of Sine and Cosine/Proof 1

== Theorem ==
== Proof ==
=== Warning ===
Note that we need to start from the algebraic definitions of sine and cosine:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
and then use the proofs of the Cosine of Sum that derive directly from these.
Otherwise these proofs are circular.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $(4.20)$
* : $\S 16.3 \ (3) \ \text{(i)}$
Category:Sum of Squares of Sine and Cosine
Sum of Squares of Sine and Cosine/Proof 3

== Theorem ==
== Proof ==
Let $P = \tuple {x, y}$ be a point on the circumference of a unit circle whose center is at the origin of a cartesian plane.
From Sine of Angle in Cartesian Plane and Cosine of Angle in Cartesian Plane:
:$P = \tuple {\cos \theta, \sin \theta}$
The graph of the unit circle is the locus of:
:$x^2 + y^2 = 1$
as given by Equation of Circle.
Substituting $x = \cos \theta$ and $y = \sin \theta$ yields:
:$\cos^2 \theta + \sin^2 \theta = 1$
== Notes ==
This proof requires that we start from the unit circle definitions of sine and cosine, otherwise the proof is circular.
Category:Sum of Squares of Sine and Cosine
Sum of Squares of Standard Normal Random Variables has Chi-Squared Distribution

== Theorem ==
Let $X_1, X_2, \ldots, X_n$ be independent random variables.
Let $X_i \sim \Gaussian 0 1$ for $1 \le i \le n$ where $\Gaussian 0 1$ is the standard normal distribution.
Then: 
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_n$
where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.
== Proof ==
By Square of Standard Normal Random Variable has Chi-Squared Distribution, we have: 
:$X^2_i \sim \chi^2_1$
for $1 \le i \le n$. 
So, by Sum of Chi-Squared Random Variables, we have: 
:$\ds \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_{1 + 1 + 1 \ldots} = \chi^2_n$ 
Category:Chi-Squared Distribution
Sum of Squares of Sum and Difference

== Theorem ==
:$\forall a, b \in \R: \paren {a + b}^2 + \paren {a - b}^2 = 2 \paren {a^2 + b^2}$
== Algebraic Proof ==
== Geometric Proof 1 ==
== Geometric Proof 2 ==
Category:Algebra
Category:Squares
Category:Sum of Squares of Sum and Difference
Sum of Squares of Sum and Difference/Geometric Proof 1

== Theorem ==
== Proof ==
:
That is, from the above diagram:
: $\left({AC + CD}\right)^2 + \left({AC - CD}\right)^2 = 2 \left({AC^2 + CD^2}\right)$
Let $AB$ be bisected at $C$, and let $D$ be another random point on $AB$.
Then the squares on $AD$ and $DB$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $DF$ parallel to $CE$ to intersect $BE$ at $F$, and join $AF$.
Construct $FG$ parallel to $AB$ to intersect $CE$ at $G$.
Since $AC = EC$ we have from Isosceles Triangle has Two Equal Angles that $\angle EAC = \angle AEC$.
Since $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles we have that $\angle EAC + \angle AEC$ also equals a right angle.
As they are equal, $\angle EAC$ and $\angle AEC$ are both equal to half a right angle.
For the same reason, $\angle EBC$ and $\angle BEC$ are also both equal to half a right angle.
We have that $\angle GEF = \angle BEC$, so $\angle GEF$ is also half a right angle.
From Parallelism implies Equal Corresponding Angles, $\angle EGF$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle EFG$ is half a right angle.
So $\angle GEF = \angle GFE$, and so from Triangle with Two Equal Angles is Isosceles, $EG = GF$.
From Parallelism implies Equal Corresponding Angles, $\angle FDB$ is a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle BFD$ is half a right angle.
So $\angle BFD = \angle FBD$, and so from Triangle with Two Equal Angles is Isosceles, $FD = DB$.
Since $AC = CE$, the square on $AC$ equals the square on $CE$.
So the squares on $AC$ and $CE$ together are twice the square on $AC$.
But from Pythagoras's Theorem, the square on $AE$ equals the squares on $AC$ and $CE$ because $\angle ACE$ is a right angle.
So the square on $AE$ is double the square on $AC$.
By a similar argument, the square on $EF$ is double the square on $GF$.
But from Opposite Sides and Angles of Parallelogram are Equal, $GF = CD$.
So the square on $EF$ is double the square on $CD$.
So the squares on $AE$ and $EF$ are double the squares on $AC$ and $CD$.
We have that $\angle AEF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AE$ and $EF$.
So the square on $AF$ is double the squares on $AC$ and $CD$.
We have that $\angle ADF$ is a right angle.
So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AD$ and $DF$.
So the squares on $AD$ and $DF$ are double the squares on $AC$ and $CD$.
But $DF = DB$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Sum of Squares of Sum and Difference
Sum of Squares of Sum and Difference/Geometric Proof 2

== Theorem ==
== Proof ==
:
That is, from the above diagram:
:$\paren {AC + CD}^2 + \paren {CD - AC}^2 = 2 \paren {AC^2 + CD^2}$
Let $AB$ be bisected at $C$, and let $AB$ be produced to some point $D$.
Then the squares on $AD$ and $BD$ are double the squares on $AC$ and $CD$.
The proof is as follows.
Construct $CE$ perpendicular to $AB$ and construct $CE = AC$.
Join $EA$ and $EB$.
Construct $EF$ parallel to $AB$, and Construct $DF$ parallel to $CE$.
From Parallelism implies Supplementary Interior Angles, $\angle CEF + \angle EFD$ equal two right angles.
Therefore $\angle FEB + \angle EFD$ is less than two right angles.
So by the Parallel Postulate, $EB$ and $FB$ must meet when produced in the direction $B$ and $D$.
Let this point be $G$, and let $AG$ be joined.
From Isosceles Triangle has Two Equal Angles, as $AC = AE$, $\angle EAC = \angle AEC$.
Because $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles, each of $\angle EAC$ and $\angle AEC$ are each half a right angle.
For the same reason, each of $\angle CEB$ and $\angle EBC$ are each half a right angle.
Since $\angle EBC$ is half a right angle, from Two Straight Lines make Equal Opposite Angles, $\angle DBG$ is also half a right angle.
From Parallelism implies Equal Alternate Angles, $\angle BDG = \angle DCE$, so $\angle BDG$ is also a right angle.
So from Sum of Angles of Triangle Equals Two Right Angles, $\angle DGB$ is half a right angle.
As $\angle DGB = \angle DBG$, from Triangle with Two Equal Angles is Isosceles, we have that $BD = BG$.
From Opposite Sides and Angles of Parallelogram are Equal, $\angle EFD = \angle ECD$ which is a right angle, 
Since $\angle EGF$ is half a right angle, and $\angle EFD$ is a right angle, $\angle FEG$ is half a right angle from Sum of Angles of Triangle Equals Two Right Angles.
So $\angle EGF = \angle FEG$, and so from Triangle with Two Equal Angles is Isosceles, $GF = EF$.
Since $EC = CA$, the square on $EC$ equals the square on $CA$.
So the squares on $EC$ and $CA$ are together twice the square on $CA$.
From Pythagoras's Theorem, the square on $EA$ is equal to the the squares on $EC$ and $CA$, and so twice the square on $AC$.
Again, since $FG = FE$, the square on $FG$ equals the square on $FE$.
So the squares on $FG$ and $EF$ are together twice the square on $FE$.
From Pythagoras's Theorem, the square on $EG$ is equal to the the squares on $GF$ and $FE$, and so twice the square on $EF$.
From Opposite Sides and Angles of Parallelogram are Equal, $EF = CD$.
But the square on $EA$ has been shown to be double the square on $AC$.
So the squares on $AE$ and $EG$ are twice the squares on $AC$ and $CD$.
From Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AE$ and $EG$.
So the square on $AG$is double the squares on $AC$ and $CD$.
But by Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AD$ and $DG$.
As $DG = DB$, the result follows.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Sum of Squares of Sum and Difference
Sum of Squares of Two Odd Integers is not Square

== Theorem ==
Let $m$ and $n$ be odd integers.
Then $m^2 + n^2$ is not a square number.
== Proof ==
 $m^2 + n^2$ is a square number.
Because $m$ and $n$ are both odd, we have:
From Parity of Integer equals Parity of its Square, $m^2$ and $n^2$ are both odd integers.
Hence $m^2 + n^2$ is an even integer.
But from Square Modulo 4, $m^2 + n^2$ is even $x^2 \equiv 0 \pmod 4$.
This contradicts the deduction that $m^2 + n^2 \equiv 2 \pmod 4$.
Hence by Proof by Contradiction it follows that $m^2 + n^2$ cannot be square.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $8 \ \text {(a)}$
Category:Odd Squares
Category:Sums of Squares
Sum of Squares on Pairs of Rows and Columns of Moessner's Order 4 Magic Square

== Theorem ==
The sums of the squares of the entries are equal on the following pairs of rows and columns of Moessner's order $4$ magic square:
:Rows $1$ and $4$
:Rows $2$ and $3$
:Columns $1$ and $4$
:Columns $2$ and $3$.
== Proof ==
Recall Moessner's order $4$ magic square:
=== Rows $1$ and $4$ ===
=== Rows $2$ and $3$ ===
=== Columns $1$ and $4$ ===
=== Columns $2$ and $3$ ===
== Sources ==
* : $16$
* : $16$
Category:Magic Squares
Sum of Stopping Times is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ and $S$ be stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Then the pointwise sum $T + S$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
Let $t \in \Z_{\ge 0}$. 
Then, if for $\omega \in \Omega$ we have $\map T \omega + \map S \omega = t$, we have: 
:$\map T \omega \le t$
and:
:$\map S \omega \le t$ 
If we have: 
:$\map S \omega = s \le t$ 
and:
:$\map S \omega + \map T \omega = t$ 
we have: 
:$\map T \omega = t - s$
So, we have: 
:$\ds \set {\omega \in \Omega : \map S \omega + \map T \omega \le t} = \bigcup_{s \in \Z_{\ge 0}, \, 0 \le s \le t} \set {\omega \in \Omega : \map S \omega = s} \cap \set {\omega \in \Omega : \map T \omega = t - s}$
Since $S$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
:$\set {\omega \in \Omega : \map S \omega = s} \in \FF_s$
Since $s \le t$ and $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have: 
:$\FF_s \subseteq \FF_t$
and so:
:$\set {\omega \in \Omega : \map S \omega = s} \in \FF_t$
Similarly, we have:
:$\set {\omega \in \Omega : \map T \omega = t - s} \in \FF_{t - s}$
and so:
:$\set {\omega \in \Omega : \map T \omega = t - s} \in \FF_t$
Since $\FF_t$ is closed under finite union, we have: 
:$\ds \set {\omega \in \Omega : \map S \omega + \map T \omega \le t} \in \FF_t$
So $S + T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Sum of Strictly Negative Elements is Strictly Negative

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $N$ be the (strict) negativity property on $D$:
:$\forall a \in D: \map N a \iff \map P {-a}$
Then for all $a \in D$:
:$\map N a, \map N b \implies \map N {a + b}$
== Proof ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $10 \ \text {(iii)}$
Category:Ordered Integral Domains
Sum of Submodules is Submodule

== Theorem ==
Let $R$ be a ring.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Let $H$ and $K$ be submodules of $M$.
Then $H + K$ is also a submodule of $M$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$
Category:Submodules
Sum of Summations equals Summation of Sum

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
=== Infinite Sequence ===
== Proof ==
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(8)$
Category:Summations
Sum of Summations equals Summation of Sum/Infinite Sequence

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers $\Z$.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be infinite.
Let $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ be convergent.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.3$
Category:Summations
Sum of Summations over Overlapping Domains

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ and $S: \Z \to \set {\T, \F}$ be propositional functions on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of both $R$ and $S$ be finite.
Then:
:$\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j = \sum_{\map R j \mathop \lor \map S j} a_j + \sum_{\map R j \mathop \land \map S j} a_j$
where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively.
=== Infinite Series ===
== Proof ==
== Example ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(11)$
Category:Summations
Category:Sum of Summations over Overlapping Domains
Sum of Summations over Overlapping Domains/Example

== Example of Sum of Summations over Overlapping Domains ==
:$\ds \sum_{1 \mathop \le j \mathop \le m} a_j + \sum_{m \mathop \le j \mathop \le n} a_j = \paren {\sum_{1 \mathop \le j \mathop \le n} a_j} + a_m$
== Proof ==
Let $\map R j$ be the propositional function $1 \mathop \le j \mathop \le m$.
Let $\map S j$ be the propositional function $m \mathop \le j \mathop \le n$.
Then we have:
and:
The result follows from Sum of Summations over Overlapping Domains.
== Sources ==
* : $\S 1.2.3$: Sums and Products: $(12)$
Category:Sum of Summations over Overlapping Domains
Sum of Summations over Overlapping Domains/Infinite Series

== Theorem ==
Let $R: \Z \to \set {\T, \F}$ and $S: \Z \to \set {\T, \F}$ be propositional functions on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of both $R$ and $S$ be infinite.
Then provided that any $3$ of the $4$ summations converge:
:$\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j = \sum_{\map R j \mathop \lor \map S J} a_j + \sum_{\map R j \mathop \land \map S j} a_j$
where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively.
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $6$
Category:Summations
Category:Sum of Summations over Overlapping Domains
Sum of Summations over Overlapping Domains/Proof

== Theorem ==
== Proof ==
Let:
:$A := \set {j \in \Z: \map R j}$
:$B := \set {j \in \Z: \map R j}$
The result then follows from Cardinality of Set Union.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $21$
Category:Sum of Summations over Overlapping Domains
Sum of Tangents of Angles in Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\tan A + \tan B + \tan C = \tan A \tan B \tan C$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(22)$
Category:Sum of Tangents of Angles in Triangle
Category:Triangles
Category:Tangent Function
Sum of Terms of Magic Cube

== Theorem ==
The total of all the entries in a magic cube of order $n$ is given by:
:$T_n = \dfrac {n^3 \paren {n^3 + 1} } 2$
== Proof ==
Let $M_n$ denote a magic cube of order $n$.
$M_n$ is by definition an arrangement of the first $n^3$ (strictly) positive integers into an $n \times n \times n$ cubic array containing the positive integers from $1$ upwards.
Thus there are $n^3$ entries in $M_n$, going from $1$ to $n^3$.
Thus:
=== Sequence ===
Category:Magic Cubes
Sum of Terms of Magic Cube/Sequence

== Sequence of Sums of Terms of Magic Cubes ==
The sequence of the sum totals of all the entries in magic cubes of order $n$ begins:
:$1, \paren {36,} \, 378, 2080, 7875, 23 \, 436, 58 \, 996, 131 \, 328, \ldots$
However, note that while $36 = \dfrac {2^3 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.
Category:Magic Cubes
Sum of Terms of Magic Square

== Theorem ==
The total of all the entries in a magic square of order $n$ is given by:
:$T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$
== Proof ==
Let $M_n$ denote a magic square of order $n$.
$M_n$ is by definition a square matrix of order $n$ containing the positive integers from $1$ upwards.
Thus there are $n^2$ entries in $M_n$, going from $1$ to $n^2$.
Thus:
=== Sequence ===
Category:Magic Squares
Sum of Terms of Magic Square/Sequence

== Sequence of Sums of Terms of Magic Squares ==
The sequence of the sum totals of all the entries in magic squares of order $n$ begins:
:$1, \paren {10,} \, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10 \, 440, 14 \, 365, 19 \, 306, 25 \, 425, 32 \, 896, \ldots$
However, note that while $10 = \dfrac {2^2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist.
Category:Magic Squares
Sum of Triangular Matrices

== Theorem ==
Let $\mathbf A = \left[{a}\right]_{n}, \mathbf B = \left[{b}\right]_{n}$ be square matrices of order $n$.
Let $\mathbf C = \mathbf A + \mathbf B$ be the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, then so is $\mathbf C$.
If $\mathbf A$ and $\mathbf B$ are lower triangular matrices, then so is $\mathbf C$.
== Proof ==
From the definition of matrix addition, we have:
:$\forall i, j \in \left[{1 .. n}\right]: c_{ij} = a_{ij} + b_{ij}$
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, we have:
:$\forall i > j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i > j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself upper triangular.
Similarly, if $\mathbf A$ and $\mathbf B$ are lower triangular matrices, we have:
:$\forall i < j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i < j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself lower triangular.
Category:Triangular Matrices
Category:Matrix Entrywise Addition
Sum of Trigamma of n plus 3 over n plus 2

== Theorem ==
where:
:$\map {\psi_1} n$ is the trigamma function of $n$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.
== Proof ==
== Also see ==
* Sum of Digamma of n plus 2 over n plus 2 squared
== Sources ==
* https://www.instagram.com/p/C03-ZRTBdfl/
Category:Trigamma Function
Category:Riemann Zeta Function
Category:Apéry's Constant
Sum of Trigonometric Functions over Power

== Theorem ==
== Proof ==
First, let:
Now, consider:
:$\mathbf A + i \, \mathbf B$
where $i$ is the imaginary unit:
So equating real and imaginary parts:
This proves $(1)$ and $(2)$ respectively.
$(3)$ is proved by finding the quotient of the two results:
Category:Trigonometric Identities
Sum of Two Angles of Three containing Solid Angle is Greater than Other Angle

== Theorem ==
== Proof ==
:
Let $A$ be a solid angle which is contained by the three plane angles $\angle BAC$, $\angle CAD$ and $\angle DAB$.
It is to be demonstrated that any two of these plane angles together are greater than the third.
Suppose $\angle BAC$, $\angle CAD$ and $\angle DAB$ are all equal.
Then any two together are greater than the remaining one and the result follows.
Suppose otherwise.
, suppose $\angle BAC$ is greater than $\angle BAD$.
Let $E$ be constructed on $AB$ such that $\angle BAE$, in the plane through $BA$ and $AC$, is equal to $\angle DAB$.
Let $AE = AD$.
Let $BEC$ be drawn across through $E$ such that the straight lines $AB$ and $AC$ are cut at $B$ and $C$.
Let $DB$ and $DC$ be joined.
We have that:
:$DA = AE$
and $AB$ is common.
So we have two sides equal to two sides, while $\angle DAB = \angle DAE$.
Therefore from :
:$DB = BE$
We have from :
:$BD + DC > BC$
But we have that $BD = BE$.
Therefore $DC > EC$.
We have that:
:$DA = AE$
and $AC$ is common.
We also have that $DC > DE$.
Therefore from :
:$\angle DAC > \angle EAC$
But:
:$\angle DAB = \angle BAE$
and so:
:$\angle DAB + \angle DAC > \angle BAC$
Similarly it can be shown that the remaining plane angles, taken two together, are greater than the remaining one.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Solid Angles