nyuuzyou/wikis huggingface dataset:
Magnitude of Relativistic Momentum

== Theorem ==
Let $B$ be a body in motion in space.
The linear momentum $\mathbf p$ of $B$ has a magnitude $p$ given by:
:$p = \dfrac {m_0 v} {\sqrt {1 = \dfrac {v^2} {c^2} } }$
where:
:$m_0$ is the rest mass of $B$
:$v$ is the speed of $v$
:$c$ is the speed of light.
== Proof ==
== Sources ==
* 
* 
Category:Linear Momentum
Category:Relativistic Mechanics
Magnitude of Scalar Triple Product equals Volume of Parallelepiped Contained by Vectors

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c$ be vectors in a vector space of $3$ dimensions:
Let $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ denote the scalar triple product of $\mathbf a, \mathbf b, \mathbf c$.
Then $\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ equals the volume of the parallelepiped contained by $\mathbf a, \mathbf b, \mathbf c$.
== Proof ==
Let us construct the parallelepiped $P$ contained by $\mathbf a, \mathbf b, \mathbf c$.
:
We have by Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors that:
:$\mathbf b \times \mathbf c$ is a vector area equal to and normal to the area of the bottom face $S$ of $P$.
The dot product $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ is equal to the product of this vector area and the projection of $\mathbf a$ along $\mathbf b \times \mathbf c$.
Depending on the relative orientations of $\mathbf a$, $\mathbf b$ and $\mathbf c$, $\mathbf a \cdot \paren {\mathbf b \times \mathbf c}$ may or may not be negative.
So, taking its absolute value, $\size {\mathbf a \cdot \paren {\mathbf b \times \mathbf c} }$ is the volume of the parallelepiped which has $\mathbf a$, $\mathbf b$ and $\mathbf c$ as edges.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors
* : Chapter $\text I$: Vector Algebra: $\S 4$
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.17$
* 
* 
Category:Scalar Triple Product
Category:Parallelepipeds
Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space of $3$ dimensions:
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Then $\norm {\mathbf a \times \mathbf b}$ equals the area of the parallelogram two of whose sides are $\mathbf a$ and $\mathbf b$.
== Proof ==
By definition of vector cross product:
:$\mathbf a \times \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \mathbf {\hat n}$
where:
:$\norm {\mathbf a}$ denotes the length of $\mathbf a$
:$\theta$ denotes the angle from $\mathbf a$ to $\mathbf b$, measured in the positive direction
:$\mathbf {\hat n}$ is the unit vector perpendicular to both $\mathbf a$ and $\mathbf b$ in the direction according to the right-hand rule.
As $\mathbf {\hat n}$ is the unit vector:
:$\norm {\paren {\norm {\mathbf a} \norm {\mathbf b} \sin \theta \, \hat {\mathbf n} } } = \norm {\mathbf a} \norm {\mathbf b} \sin \theta$
By Area of Parallelogram, the area of the parallelogram equals the product of one of its bases and the associated altitude.
Let $\mathbf a$ denote the base of the parallelogram.
Then its altitude is $\norm {\mathbf b} \sin \theta$.
The result follows.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $5$. Vector Area
* : $\S 22$: Cross or Vector Product: $22.15$
Category:Vector Cross Product
Magnitude of Vector Quantity in terms of Components

== Theorem ==
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf r$ be expressed in terms of its components:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
where $\mathbf i$, $\mathbf j$ and $\mathbf k$ denote the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Then:
:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2}$
where $\size {\mathbf r}$ denotes the magnitude of $\mathbf r$.
== Proof ==
Let the initial point of $\mathbf r$ be $\tuple {x_1, y_1, z_1}$.
Let the terminal point of $\mathbf r$ be $\tuple {x_2, y_2, z_2}$.
Thus, by definition of the components of $\mathbf r$, the magnitude of $\mathbf r$ equals the distance between $\tuple {x_1, y_1, z_1}$ and $\tuple {x_2, y_2, z_2}$.
The result follows from Distance Formula in 3 Dimensions.
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $7$. The unit vectors $\mathbf i$, $\mathbf j$, $\mathbf k$
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach: $(1.7 a)$
Category:Vectors
Magnitudes Proportional Compounded are Proportional Separated

== Theorem ==
That is:
:$a : b = c : d \implies \left({a - b}\right) : b = \left({c - d}\right) : d$
== Proof ==
Let $AB, BE, CD, DF$ be magnitudes which are proportional componendo, so that $AB : BE = CD : DF$.
We need to show that they are proportional separando, that is, that $AE : EB = CF : DF$.
:
Let equimultiples $GH, HK, LM, MN$ be taken of $AE, EB, CF, FD$.
Let $KO, NP$ be other arbitrary equimultiples of $EB, FD$.
We have that $GH$ is the same multiple of $AE$ that $HK$ is of $EB$.
So from Multiplication of Numbers is Left Distributive over Addition $GH$ is the same multiple of $AE$ that $GK$ is of $AB$.
But $GH$ is the same multiple of $AE$ that $LM$ is of $CF$.
Therefore $GK$ is the same multiple of $AB$ that $LM$ is of $CF$.
Again, we have that $LM$ is the same multiple of $CF$ that $MN$ is of $FD$.
So from Multiplication of Numbers is Left Distributive over Addition $LM$ is the same multiple of $CF$ that $LN$ is of $CD$.
Therefore $GK, LN$ are equimultiples of $AB, CD$.
Again, we have that $HK$ is the same multiple of $EB$ that $MN$ is of $FD$.
Also, $KO$ is the same multiple of $EB$ that $NP$ is of $FD$.
So from Multiplication of Numbers is Right Distributive over Addition the sum $HO$ is also the same multiple of $EB$ that $MP$ is of $FD$.
We also have that $AB : BE = CD : DF$.
Also, we have that $GK, LN$ are equimultiples of $AB, CD$.
Also, we have that $EB, FD$ are equimultiples of $HO, MP$.
Therefore:
:$GK > HO \implies LN > MP$
:$GK = HO \implies LN = MP$
:$GK < HO \implies LN < MP$
Suppose $GK > HO$.
If we subtract $HK$ from each, we have that $GH > KO$.
But $GK > HO \implies LN > MP$.
Therefore $LN > MP$.
So if $MN$ is subtracted from each, we have that $LM > NP$.
So:
: $GH > KO \implies LM > NP$
Similarly we can prove that:
: $GH = KO \implies LM = NP$
: $GH < KO \implies LM < NP$
Also, we have that:
: $GH, LM$ are equimultiples of $AE, CF$
: $KO, NP$ are equimultiples of $EB, FD$
Therefore $AE : EB = CF : FD$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Magnitudes Proportional Separated are Proportional Compounded

== Theorem ==
That is:
:$a : b = c : d \implies \paren {a + b} : b = \paren {c + d} : d$
== Proof ==
Let $AE, EB, CF, FD$ be magnitudes which are proportional separando, that is:
:$AE : EB = CF : FD$
We need to show that they are also proportional componendo, that is:
:$AB : BE = CD : FD$
:
Suppose $CD : DF \ne AB : BE$.
Then as $AB : BE$ so will $CD$ be to some magnitude less than $DF$ or greater.
Suppose that it be in that ratio to a less magnitude $DG$.
Then since $AB : BE = CD : DG$ it follows from Magnitudes Proportional Compounded are Proportional Separated that:
:$AE : EB = CG : GD$
But by hypothesis:
:$AE : EB = CF : FD$
So by Equality of Ratios is Transitive we have that:
:$CG : GD = CF : FD$
But $CG > CF$.
Therefore $GD > FD$ from Relative Sizes of Components of Ratios.
But it is also less, which is impossible.
Therefore as $AB$ is to $BE$, so is not $CD$ to a lesser magnitude than $FD$.
Similarly we can show that neither is it in that ratio to a greater.
Hence the result.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Magnitudes with Irrational Ratio are Incommensurable

== Theorem ==
== Proof ==
Let $A$ and $B$ be magnitudes which do not have to one another the ratio which a number has to a number.
Suppose $A$ and $B$ are commensurable.
Then from Ratio of Commensurable Magnitudes it follows that $A$ and $B$ have to one another the ratio which a number has to a number.
From this contradiction follows the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Magnitudes with Rational Ratio are Commensurable

== Theorem ==
=== Porism ===
== Proof ==
Let $A$ and $B$ be magnitudes which have to one another the ratio which the number $D$ has to the number $E$.
Let $A$ be divided into as many equal parts as there are units in $D$.
Let $C$ be equal to one of those parts.
Let $F$ be made up of as many magnitudes equal to $C$ as there are units in $E$.
Since:
: there are in $A$ as many magnitudes equal to $C$ as there are units in $D$
then:
:whatever part the unit is of $D$, the same part is $C$ of $A$.
From Ratios of Fractions in Lowest Terms:
:$\dfrac C A = \dfrac 1 D$
and so from the porism to Ratios of Equal Magnitudes:
:$\dfrac A C = \dfrac D 1 = D$
Since:
: there are in $F$ as many magnitudes equal to $C$ as there are units in $E$
then from Ratios of Fractions in Lowest Terms:
:$\dfrac C F = \dfrac 1 E$
From Equality of Ratios Ex Aequali:
:$\dfrac A F = \dfrac D E$
But:
:$\dfrac D E = \dfrac A B$
and so from Equality of Ratios is Transitive:
:$\dfrac A B = \dfrac A F$
Therefore $A$ has the same ratio to each of the magnitudes $B$ and $F$.
Therefore from Magnitudes with Same Ratios are Equal:
:$B = F$
But $C$ measures $F$.
Therefore $C$ also measures $B$.
Further, $C$ also measures $A$.
Therefore $C$ measures both $A$ and $B$.
Therefore $A$ is commensurable with $B$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Magnitudes with Same Ratios are Equal

== Theorem ==
That is:
:$a : c = b : c \implies a = b$
:$c : a = c : b \implies a = b$
== Proof ==
:
Let each of $A$ and $B$ have the same ratio to $C$, i.e. $A : C = B : C$.
Suppose $A \ne B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : C \ne B : C$.
But $A : C = B : C$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
Again, let $C$ have the same ratio to each of $A$ and $B$, i.e. $C : A = C : B$.
Suppose $A \ne B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $C : A \ne C : B$.
But $C : A = C : B$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Mahaviracharya/Ganita Sara Samgraha

== Problems by : ==
== Chapter $\text {II}$: Arithmetical Operations ==
== Chapter $\text {III}$: Fractions ==
== Chapter $\text {IV}$: Miscellaneous Problems ==
== Chapter $\text V$: Rule of Three ==
== Chapter $\text {VI}$: Mixed Problems ==
== Chapter $\text {VII}$: Calculations relating to the Measurement of Areas ==
== Chapter $\text {VIII}$: Calculations regarding Excavations ==
== Chapter $\text {IX}$ Calculations relating to Shadows ==
Category:Examples of Algebra Problems
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter IV/17-22

== : Chapter $\text {IV}$: Miscellaneous Problems (on Fractions): Problem $17$ to $22$ ==
One night, in a month of the spring season, a certain young lady ... was lovingly happy along with her husband ... on the floor of a big mansion, white like the moon, and situated in a pleasure-garden with trees bent down with the load of bunches of flowers and fruits, and resonant with the sweet sounds of parrots, cuckoos and bees which were all intoxicated with the honey obtained from the flowers therein.
Then on a love-quarrel arising between the husband and the wife, that lady's necklace made up of pearls became sundered and fell on the floor.
:One-third of that necklace reached the maid-servant there;
:One-sixth fell on the bed;
:then one-half of what remained
::(and one-half of what remained thereafter and again one-half of what remained thereafter)
::and so on, counting six times [in all] fell all of them everywhere;
:and there were found to remain (unscattered) $1,161$ pearls;
and if you know ... give out the measure of the pearls.
The translation by remarks:
:Certain epithets here have not been considered fit for translation.
I don't know what he means.
== Solution ==
Improbably, there were in fact $148, 608$ pearls in that necklace.
== Proof ==
Let $n$ be the number of pearls in the necklace.
After $\frac 1 6$ and $\frac 1 3$ have fallen to the maid-servant and on the bed, there are $\frac 1 2$ remaining.
One-half and half again, and so on, counting $6$ times in all, means $\dfrac 1 {2^6} \times \dfrac 1 2$ remain.
Hence $n \times 1 {128} = 1,161$ and the result follows.
== Sources ==
* : Chapter $\text {IV}$: Miscellaneous Problems (on Fractions): Problem $17$ to $22$
* : Chapter $\text {IV}$: Miscellaneous Problems (on Fractions): Problem $17$ to $22$
* : Indian Puzzles: $50$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VI

== : Chapter $\text {VI}$: Mixed Problems ==
=== Problems $95 \frac 1 2$ and $96 \frac 1 2$ ===
=== Problem $219$ ===
=== Problem $236 \text - 237$ ===
=== Problem $289$ ===
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VI/219

== : Chapter $\text {VI}$: Mixed Problems: Problem $219$ ==
In how many ways can different numbers of flavours be used in combination together, being selected from:
:the astringent
:the bitter
:the sour
:the pungent
:and the saline
:together with the sweet taste?
== Solution ==
:$64$
== Proof ==
A simple application of Cardinality of Power Set of Finite Set.
There are $6$ different flavours.
The number of different combinations is equal to the number of different subsets of this set of $6$.
Hence the number is:
:$2^6 = 64$
== Sources ==
* : Chapter $\text {VI}$: Mixed Problems: $219$
* : Chapter $\text {VI}$: Mixed Problems: $219$
* : Indian Puzzles: $51$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VI/236-237

== : Chapter $\text {VI}$: Mixed Problems: Problem $236 \text - 237$ ==
$3$ merchants saw in the road a purse containing money.
One said:
:If I secure this purse, I shall become twice as rich as both of you together.
Then the second said:
:I shall become $3$ times as rich.
Then the third said:
:I shall become $5$ times as rich.
What is the value of the money in the purse, as also the money on hand with each of the $3$ merchants?
== Solution ==
The solution is not unique.
:The second has $3$ times as much as the first.
:The third has $5$ times as much as the first.
:The purse holds $15$ times as much as the first owns.
== Proof ==
Let $x$, $y$ and $z$ denote the amount of money owned by the $1$st, $2$nd and $3$rd merchant respectively.
Let $u$ denote the amount of money in the purse.
Let $a$, $b$ and $c$ denote the factor by which $x$, $y$ and $z$ respectively will be richer, were they to get the purse, than the other two combined.
We have:
Let $T = u + x + y + z$.
Then:
Setting $a = 2$, $b = 3$, $c = 5$ we have:
Hence:
and:
from which it is seen that:
:$x : y : z = 1 : 3 : 5$
and the solution in smallest integers is:
== Sources ==
* : Chapter $\text {VI}$: Mixed Problems: $236 \text - 237$
* : Chapter $\text {VI}$: Mixed Problems: $236 \text - 237$
* : Indian Puzzles: $52$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VI/289

== : Chapter $\text {VI}$: Mixed Problems: Problem $289$ ==
Arrows, if they are thin cylinders, circular in cross-section, can be packed in hexagonal bundles.
Let the arrows around the circumference be $18$ in number.
How many in all are the arrows to be found in the bundle in the quiver?
== Solution ==
:$37$
== Proof ==
The arrows are packed as centered hexagonal numbers.
The answer is immediately apparent.
== Sources ==
* : Chapter $\text {VI}$: Mixed Problems: $289$
* : Chapter $\text {VI}$: Mixed Problems: $289$
* : Indian Puzzles: $53$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VI/95-96

== : Chapter $\text {VI}$: Mixed Problems: Problem $95 \frac 1 2$ and $96 \frac 1 2$ ==
Three puranas formed the pay of one man who is a mounted soldier.
At that rate there were $65$ soldiers in all.
Some among them broke down, and the amount of their pay was given to those who remained in the field.
Of this, each man obtained $10$ puranas.
You tell me, after thinking well, how many remained in the field and how many broke down.
== Solution ==
:$50$ soldiers broke down, while $15$ remained in the field.
== Proof ==
Let $x$ denote the number of soldiers who remained in the field.
Let $y$ denote the number of soldiers who broke down.
We have:
which leaves $65 - 15 = 50$ soldiers who broke down.
== Sources ==
* : Chapter $\text {VI}$: Mixed Problems: $95 \frac 1 2$ and $96 \frac 1 2$
* : Chapter $\text {VI}$: Mixed Problems: $95 \frac 1 2$ and $96 \frac 1 2$
* : Indian Puzzles: $48$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahaviracharya/Ganita Sara Samgraha/Chapter VII/180

== : Chapter $\text {VII}$: Measurement of Areas: Problem $180$ ==
Two pillars are of known height: $p$ and $q$.
Two strings are tied, one to the top of each.
Each of these strings is stretched so as to touch the foot of the other pillar.
From the point where the two strings meet, another string is suspended vertically till it touches the ground.
What is the length $r$ of this suspended string?
== Solution ==
:$r = \dfrac {p q} {p + q}$
that is:
:$\dfrac 1 r = \dfrac 1 p + \dfrac 1 q$
== Proof ==
:
Let $AB$ and $CD$ be the pillars, standing vertically on the ground at $A$ and $C$.
Let $EF$ be the string hanging to the ground.
Let the height of $AB$ be $p$, and the height of $CD$ be $q$.
Let the length of $EF$ be $r$.
Let the distance of $AC$ be $d$.
Let the distance of $F$ from $C$ be $y$.
We have that:
:$\triangle AEF$ and $\triangle ACD$ are similar
:$\triangle ABC$ and $\triangle FEC$ are similar.
Thus:
That is:
== Sources ==
* : Chapter $\text {VII}$: Measurement of Areas: $180 \frac 1 2$
* : Chapter $\text {VII}$: Measurement of Areas: $180 \frac 1 2$
* : Indian Puzzles: $54$
Category:Mahaviracharya/Ganita Sara Samgraha
Mahler's Inequality

== Theorem ==
The geometric mean of the termwise sum of two finite sequences of positive real numbers is never less than the sum of their two separate geometric means:
:$\ds \prod_{k \mathop = 1}^n \paren {x_k + y_k}^{1/n} \ge \prod_{k \mathop = 1}^n x_k^{1/n} + \prod_{k \mathop = 1}^n y_k^{1/n}$
where $x_k, y_k > 0$ for all $k$.
== Proof ==
This leads to:
Category:Algebra
Category:Measures of Central Tendency
Main Lemma of Propositional Tableaux

== Lemma ==
Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Either $\mathbf H$ has a tableau confutation or $\mathbf H$ has a model.
== Proof ==
If $\mathbf H$ is finite, then the Finite Main Lemma of Propositional Tableaux applies.
So, assume that $\mathbf H$ is countably infinite.
Suppose $\mathbf H$ does not have a tableau confutation.
By Countable Hypothesis Set has Finished Tableau, there is a finished tableau $T$ with hypothesis set $\mathbf H$.
By Finished Propositional Tableau has Finished Branch or is Confutation‎, as $T$ is () not a confutation, it must have a finished branch; call it $\Gamma$.
By Finished Set Lemma, the set of WFFs of propositional logic on $\Gamma$ has a model $\MM$.
Finally, because all the WFFs in the hypothesis set occur on $\Gamma$, $\MM \models \mathbf H$.
== Sources ==
* : $\S 1.11$: Compactness: Lemma $1.11.1$
Category:Propositional Tableaux
Major Straight Line is Divisible Uniquely

== Theorem ==
== Proof ==
Let $AB$ be a major straight line.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
:$AC$ and $CB$ are incommensurable in square
:$AC^2 + CB^2$ is rational
:$AC$ and $CB$ contain a medial rectangle.
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
From :
:$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
and so:
:$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right) = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
But $AB^2 + CB^2$ and $AD^2 + DB^2$ are rational.
So $\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right)$ is rational.
Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational.
But $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are medial.
From this cannot be the case.
So there can be no such $D$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Major is Irrational

== Theorem ==
== Proof ==
From :
:Let $AB$ and $BC$ be straight lines which are incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial.
We have that the rectangle contained by $AB$ and $BC$ is medial.
From:
: 
and:
: :
it follows that:
:$2 \cdot AB \cdot BC$ is medial.
But $AB^2 + BC^2$ is rational.
Therefore $2 \cdot AB \cdot BC$ is incommensurable with $AB^2 + BC^2$.
So by :
:$AC^2 = \left({AB + BC}\right)^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$ is also incommensurable with $AB^2 + BC^2$.
Therefore $AC^2$ is irrational.
Hence $AC$ is irrational.
Such a straight line is called major.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Man looking South in All Directions

== Problem ==
Where can a man look south in all directions?
=== Variant ===
== Solution ==
When he is standing at the North pole.
== Also see ==
* Place where Wind always Blows from North, a variant of this.
== Historical Note ==
== Sources ==
* 
* : Henry van Etten: $119$
Category:Geography Puzzles
Category:Classic Problems
Category:Man looking South in All Directions
Mandelbrot Set/Graphics/Complete Set

== Graphic of Mandelbrot Set ==
The following is a depiction of the complete Mandelbrot set.
The Mandelbrot set itself is white.
:
The region of the complex plane depicted is the rectangle $\closedint {-2.2} {2.2} \times \closedint {-1.65 i} {1.65 i}$.
== Online Fractal Generator ==
== Sources ==
* 
* 
Category:Graphics of Mandelbrot Set
Manipulation of Absolutely Convergent Series/Characteristic Function

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a real or complex series that is absolutely convergent.
Let $A \subseteq \N$.
Then:
:$\ds \sum_{n \mathop = 1}^\infty a_n \map {\chi_A} n = \sum_{n \mathop \in A} a_n$
where $\chi_A$ is the characteristic function of $A$.
== Proof ==
For all $N \in \N$, we have:
:$\ds \sum_{n \mathop = 1}^N \size {a_n \map {\chi_A} n} \le \sum_{n \mathop = 1}^N \size {a_n} \le \sum_{n \mathop = 1}^\infty \size {a_n}$
It follows that:
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n \map {\chi_A} n} \le \sum_{n \mathop = 1}^\infty \size {a_n}$
Then $\ds \sum_{n \mathop = 1}^\infty a_n \map {\chi_A} n$ is absolutely convergent.
It follows from Manipulation of Absolutely Convergent Series: Permutation that the order of the terms in the series does not matter.
As $\map {\chi_A} n = 0$ for all $n \notin A$, we have:
:$\ds \sum_{n \mathop = 1}^\infty a_n \map {\chi_A} n = \sum_{n \mathop \in A} a_n$
Category:Series
Category:Convergence
Manipulation of Absolutely Convergent Series/Permutation

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a real or complex series that is absolutely convergent.
If $\pi: \N \to \N$ is a permutation of $N$, then:
:$\ds \sum_{n \mathop = 1}^\infty a_n = \sum_{n \mathop = 1}^\infty a_{\map \pi n}$
== Proof ==
Let $\epsilon > 0$.
From Tail of Convergent Series tends to Zero, it follows that there exists $N \in \N$ such that:
:$\ds \sum_{n \mathop = N}^\infty \size {a_n} < \epsilon$
By definition, a permutation is bijective.
Hence we can find $M \in \N$ such that:
:$\set {1, \ldots, N - 1} \subseteq \set {\map \pi 1, \ldots, \map \pi M}$
Let $m \in \N$, and put $B = \set {n \in N: \map {\pi^{-1} } n > m}$.
For all $m \ge M$, it follows that:
By definition of convergent series, it follows that:
:$\ds \sum_{n \mathop = 1}^\infty a_n = \lim_{m \mathop \to \infty} \sum_{k \mathop = 1}^m a_{\map \pi k} = \sum_{k \mathop = 1}^\infty a_{\map \pi k}$
== Sources ==
* : $\S 1.2.3$: Operations with series
Category:Convergence
Category:Series