User:MichaelAngelosZolindakis

Prove 0.999... = 1

$0.999\ldots ={\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+\ldots$

${\frac {1}{10}}(0.999\ldots )={\frac {9}{100}}+{\frac {9}{1000}}+\ldots$

$0.999\ldots -{\frac {1}{10}}(0.999\ldots )={\frac {9}{10}}$

This equation is equivalent to

$(0.999\ldots )(1-{\frac {1}{10}})={\frac {9}{10}}$

which is equivalent to

$(0.999\ldots )({\frac {9}{10}})={\frac {9}{10}}$

which is equivalent to

$0.999\ldots ={\frac {9}{10}}{\frac {10}{9}}$

which is equivalent to

$0.999\ldots =1$

DONE

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The Thirteen Month Horoscope of Birds

01) {eagle, hawk, falcon, condor, vulture}

02) {dove, pigeon}

03) {penguin}

04) {parrot}

05) {ostrich}

06) {hummingbird}

07) {seagull}

08) {flamingo}

09)

10) {owl}

11) {sparrow, scissor-tail, oriole, cardinal}

12) {duck, goose, swan}

13) {crow, raven}

Notes:

 - 13 months for a lunar calendar

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In Gladiator (2000 film), in the beginning battle scene in which the Romans battle the Germans, the character Maximus leads a small band of cavalry around the Germans and attacks them from behind while the Roman infantry meets the Germans head-on. Maximus rides atop his horse with his dog by his side. Suppose this cavalry was comprised of only birds. Then the riders would be eagles or hawks or falcons or condors, maybe even seagulls. Their horses would be pigeons, and their dogs would be crows.

In Platoon (film) directed by Oliver Stone, in the scene in which the Americans enter the Vietnamese village, the character Barnes played by Tom Berenger shoots the wife of the village leader while trying to extract information on the NVA (North Vietnamese Army). He then points his gun at the village leader's daughter who is crying because she just witnessed her mother's death. Then Elias played by Willem Dafoe enters the scene preventing Barnes from shooting the young Vietnamese girl. Just before Elias hits Barnes with the but of his assault rifle, there is the sound of a rooster. Shortly afterward, Taylor, played by Charlie Sheen witnesses three US soldiers raping another young Vietnamese girl. All three US soldiers who were raping the young Vietnamese girl had necklaces with the cross, the Christian symbol. Taylor pulls the three US soldiers away from the Vietnamese girl, then one of the three spits in Taylor's face. Then Elias yells at them all, and to the left of Elias, there is a water-buffalo.

In Braveheart, when the character William Wallace portrayed by Mel Gibson rides into his village atop a horse, with his hands behind his head pretending to surrender to the English forces controlling his village just after the English magistrate killed his wife, he has a Nunchaku concealed in his long hair which he uses to hit the English soldier attempting to pull Wallace from his horse. The Scottish villagers band together and defeat the English forces. This is a "COWABUNGA" moment in movie history -- the weapon Wallace used, which marked the beginning of the Scottish rebellion from which the movie is based, the Nunchaku, is the weapon of Michelangelo of the Teenage Mutant Ninja Turtles.

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The Greek Orthodox Church began recognizing Saint Francis of Assisi as an Orthodox saint on November 19, 2019. His recognition as a saint of the Greek Orthodox church became official upon the founding of the first Saint Francis of Assisi Greek Orthodox Church. The founder of the church is Michael Angelo Zolindakis.
Iconography of the Saint Francis of Assisi Greek Orthodox Church:
The eight icons behind the altar of the church are as follows:

 - Saint Demetrius of Thessaloniki 
 - Saint Raphael the Archangel 
 - Saint Francis of Assisi
 - Mother Mary and child 
 - Jesus on the Crucifix 
 - Saint John the Baptist 
 - Saint Michael the Archangel 
 - Saint Catherine of Alexandria

Other Icons in the church:

 - Saint George, note: The icon will not be of Saint George killing the dragon, which references the Golden Legend of Saint George and the Dragon.  Michael Angelo is working on an icon to depict a new modified version of the legend in which Saint George befriends the beast, similar to the Wolf of Gubbio Saint Francis legend. 
 - Saint Elias
 - Saint Constantine and Helen

Prayer_to_Saint_Michael

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THE MAYOR

 Dedication: Michael Angelos to the Magnificent Eric Garcetti of Los Angeles

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Programming Exercises:

- Write 2 programs, in which each program draws a circle and a simple cliché Greek line pattern outlining the circle with only lines for the pattern 
   (i)  For the first program use Turtle_graphics

Do not be distracted by the image of non-digital turtle graphics

turtle graphics

   (ii) For the second program, use a single segment of the pattern and then use Givens_rotation and translation matrix transformations to plot the segment many times around the outline.  
          Do not use 4 segments as that would produce an image that resembles the Nazi Swastika.
 An example output is in this file: https://wiki.riteme.site/wiki/File:Circular_greek_pattern_python_code_output.jpg
 Another example with slightly different segment designs is in this file: https://commons.wikimedia.org/wiki/File:Ace.20200718.09.greek.circular.patterns.jpg

- Write code to return the annual Chinese animal using the integer year as input, i.e. if input is 1936 then return "mouse", if input is 1182 then return "cat".  
   Hint: the 12 animals in order are dragon for the year 2000 ace, then snake for 2001, then horse, then goat, monkey, rooster, dog, boar, mouse, ox, cat, rabbit
   Hint: if the year is in bce then use a negative number as the input so that for example if the input is 428 bce then return "mouse"

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Math Exercise

Let B be a 4x4 matrix. Apply the following matrix operations: (i) double column 1, (ii) halve row 3, (iii) add row 3 to row 1, (iv) interchange columns 1 and 4, (v) subtract row 2 from each of the other rows, (vi) replace column 4 by column 3, (vii) delete column 1. (a) Write the result as a product of 8 matrixes and (b) write the result again as a product of 3 matrixes.

Suppose $\mathbf {B} ={\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}\in \mathbb {R} ^{4\times 4}$

(i) double column 1:

Observe that

${\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}{\begin{pmatrix}2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}={\begin{pmatrix}2&2&3&4\\10&6&7&8\\18&10&11&12\\26&14&15&16\\\end{pmatrix}}$

Let $T_{1}={\begin{pmatrix}2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}$ be the right-side transformation matrix that doubles column 1.

(ii) halve row 3:

Observe that

${\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&{\frac {1}{2}}&0\\0&0&0&1\\\end{pmatrix}}{\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}={\begin{pmatrix}1&2&3&4\\5&6&7&8\\{\frac {9}{2}}&5&{\frac {11}{2}}&6\\26&14&15&16\\\end{pmatrix}}$

Let $T_{2}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&{\frac {1}{2}}&0\\0&0&0&1\\\end{pmatrix}}$ be the left-side transformation matrix that halves row 3.

(iii) add row 3 to row 1:

Observe that

${\begin{pmatrix}1&0&1&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}{\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}={\begin{pmatrix}10&12&14&16\\5&6&7&8\\9&10&11&12\\26&14&15&16\\\end{pmatrix}}$

Let $T_{3}={\begin{pmatrix}1&0&1&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}$ be the left-side transformation matrix that add row 3 to row 1.

(iv) interchange columns 1 and 4:

Observe that

${\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}{\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\\\end{pmatrix}}={\begin{pmatrix}4&2&3&1\\8&6&7&5\\12&10&11&9\\16&14&15&13\\\end{pmatrix}}$

Let $T_{4}={\begin{pmatrix}0&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\\\end{pmatrix}}$ be the right-side transformation matrix that interchanges column 1 and 4.

(v) subtract row 2 from each of the other rows:

Observe that

${\begin{pmatrix}1&-1&0&0\\0&1&0&0\\0&-1&1&0\\0&-1&0&1\\\end{pmatrix}}{\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}={\begin{pmatrix}-4&-4&-4&-4\\5&6&7&8\\4&4&4&4\\8&8&8&8\\\end{pmatrix}}$

Let $T_{5}={\begin{pmatrix}1&-1&0&0\\0&1&0&0\\0&-1&1&0\\0&-1&0&1\\\end{pmatrix}}$ be the left-side transformation matrix that subtracts row 2 from each of the other rows.

(vi) replace column 4 by column 3:

Observe that

${\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}{\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&0\\\end{pmatrix}}={\begin{pmatrix}1&2&3&3\\5&6&7&7\\9&10&11&11\\13&14&15&15\\\end{pmatrix}}$

Let $T_{6}={\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&1\\0&0&0&0\\\end{pmatrix}}$ be the right-side transformation matrix that replaces column 4 by column 3.

(vii) delete column 1:

Observe that

${\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&12\\13&14&15&16\\\end{pmatrix}}{\begin{pmatrix}0&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}={\begin{pmatrix}0&2&3&4\\0&6&7&8\\0&10&11&12\\0&14&15&16\\\end{pmatrix}}$

Let $T_{7}={\begin{pmatrix}0&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}}$ be the right-side transformation matrix that deletes column 1.

(a) Given a 4x4 matrix B, the result of all the transformations as a product of 8 matrixes is

$T_{5}*T_{3}*T_{2}*B*T_{1}*T_{4}*T_{6}*T_{7}$.

(b) The result as a product of 3 matrixes is

$A*B*C$, where $A=T_{5}*T_{3}*T_{2}$ and $C=T_{1}*T_{4}*T_{6}*T_{7}$.

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Math Exercise -- derivation of the coefficients for linear regression line $y=mx+k$

Derive the matrix equation ${\begin{pmatrix}\sum _{i=1}^{n}1&\sum _{i=1}^{n}x_{i}\\\sum _{i=1}^{n}x_{i}&\sum _{i=1}^{n}x_{i}^{2}\\\end{pmatrix}}{\begin{pmatrix}k\\m\\\end{pmatrix}}={\begin{pmatrix}\sum _{i=1}^{n}y_{i}\\\sum _{i=1}^{n}x_{i}y_{i}\\\end{pmatrix}}$ for $n$ data points $(x_{i},y_{i})$, $i={1,...,n},k\in \mathbb {R} ,m\in \mathbb {R}$

Want to minimize the expression $\sum _{i=1}^{n}(y_{i}-(mx_{i}+k))^{2}$ with respect to $m,k$.

$\sum _{i=1}^{n}(y_{i}-(mx_{i}+k))^{2}=\sum (y_{i}^{2}-2(mx_{i}+k)y_{i}+(mx_{i}+k)^{2})$

$=\sum y_{i}^{2}+2\sum (mx_{i}+k)y_{i}+\sum (mx_{i}+k)^{2}$

$=\sum y_{i}^{2}-2\sum mx_{i}y_{i}-2\sum ky_{i}+\sum (m^{2}x_{i}^{2}+2mx_{i}k+k^{2})$

$=\sum y_{i}^{2}-2m\sum x_{i}y_{i}-2k\sum y_{i}+m^{2}\sum x_{i}^{2}+2mk\sum x_{i}+k^{2}\sum 1$

$=F(m,k)$

To minimize the expression, set the partial derivatives of $F(m,k)$ to zero.

${\frac {\partial F}{\partial m}}=-2\sum x_{i}y_{i}+2m\sum x_{i}^{2}+2k\sum x_{i}=0$

${\frac {\partial F}{\partial k}}=-2\sum y_{i}+2m\sum x_{i}+2k\sum 1=0$

$\iff$

$(2\sum x_{i}^{2})m+(2\sum x_{i})k=2\sum x_{i}y_{i}$

$(2\sum x_{i})m+(2\sum 1)k=2\sum y_{i}$

$\iff$

$(\sum x_{i}^{2})m+(\sum x_{i})k=\sum x_{i}y_{i}$

$(\sum x_{i})m+(\sum 1)k=\sum y_{i}$

$\iff$

$(\sum 1)k+(\sum x_{i})m=\sum y_{i}$

$(\sum x_{i})k+(\sum x_{i}^{2})m=\sum x_{i}y_{i}$

$\iff$

${\begin{pmatrix}\sum _{i=1}^{n}1&\sum _{i=1}^{n}x_{i}\\\sum _{i=1}^{n}x_{i}&\sum _{i=1}^{n}x_{i}^{2}\\\end{pmatrix}}{\begin{pmatrix}k\\m\\\end{pmatrix}}={\begin{pmatrix}\sum _{i=1}^{n}y_{i}\\\sum _{i=1}^{n}x_{i}y_{i}\\\end{pmatrix}}$ $\blacksquare$

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Math Exercise

Let $A\in \mathbb {C} ^{m\times m}$ and $b\in \mathbb {C} ^{m}$ be arbitrary. Show that any $x\in K_{n}$ is equal to $p(A)b$ for some polynomial $p$ of degree $\leq n-1$.

A polynomial $p(A)$ of degree $k$ is $p(A)=a_{0}+a_{1}A+a_{2}A^{2}+\ldots +a_{k}A^{k}$, where $a_{0},\ldots ,a_{k}\in \mathbb {C}$.

Given matrix $A$, vector $b$, the Krylov Sequence is the set of vectors $b,Ab,A^{2}b,A^{3}b,\ldots$. The Krylov_subspaces are the spaces spanned by successively larger groups of these vectors.

In the Arnoldi_iteration Algorithm, considering $AQ=QH,H$ is a Hessenberg_matrix, $Q$ orthogonal, the columns of $Q_{n}\in \mathbb {C} ^{m\times n}$ are the first n columns of $Q$. So $Q_{n}=[q_{1}q_{2}\ldots q_{n},H_{n}\in \mathbb {C} ^{(n+1)\times n}]$ is the upper left section of $H$ which is also Hessenberg,

$H_{n}={\begin{pmatrix}h_{1,1}&\ldots &&&h_{1,n}\\h_{2,1}&h_{2,2}&&&\vdots \\&\ddots &\ddots &&\\&&&h_{n,n-1}&h_{n,n}\\&&&&h_{n+1,n}\\\end{pmatrix}}$ then $AQ_{n}=Q_{n+1}H_{n}$

$\iff$

$[A_{n}][q_{1}q_{2}\ldots q_{n}]=[q_{1}q_{2}\ldots q_{n+1}]{\begin{pmatrix}h_{1,1}&\ldots &h_{1,n}\\h_{2,1}&&\vdots \\&\ddots &\\&&h_{n+1,n}\\\end{pmatrix}}$

$\iff$

$Aq_{n}=h_{1,n}q_{1}+\ldots +h_{n,n}q_{n}+h_{n+1,n}q_{n+1}$

so that $q_{n+1}$ satisfies an (n+1) term recurrence relation involving itself and the previous Krylov vectors. Thus the vectors ${q_{j}}$ form bases of successive Krylov Subspaces generated by $A$ and $b$, where

$K_{n}={b,Ab,A^{2}b,\ldots ,A^{n-1}b}={q_{1},q_{2},\ldots ,q_{n}}\subset \mathbb {C} ^{m}$.

Therefore any $x\in K_{n}$ satisfies

$x=p(A)b$, where $p$ is a polynomial of degree $n-1$

$=(a_{0}+a_{1}A+a_{2}A^{2}+\ldots +a_{n-1}A^{n-1})b$

$=a_{0}b+a_{1}Ab+a_{2}A^{2}b+\ldots +a_{n-1}A^{n-1}b$ $\blacksquare$

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Miscellaneous