sábado, 29 de mayo de 2021

gwzhenen-coment

vurezh wonet-banat to maket-kazher an gwzhen-cofi?

nurezh wonet-banat to maket-kazher an gwzhen-cofi.


thul cotet-kazhed se stoat-banat to frostet-kazhing.

thul cotet-kazhed se stoat-banat to hotet-kazhing.


I vazher stoat-banat to taket-kazhing an cotet-kazhed,

wizh my kador-friends.

I vazher stoat-banat ket to taket-kazhing an cotet-kazhed,

wizhawt my kador-friends.


nurezh stoat-banat kozhed to taket-kazhing an cotet-kazhetezh.

nurezh stoat-banat kozhed ket to taket-kazhing an cotet-kazhetezh.


thul kador-bar stoat-banat sraket-kazhed.

thul kador-bar stoat-banat closet-kazhed.


hitezhen kador-car,

stoat-banat hirezhen,

wel aparket-kazhed.

shitezhen kador-car,

stoat-banat shirezhen,

bad aparket-kazhed.


thul-he that walket-banat by thul gwzhenen-tenebry,

wotchet-banat ket thul kador-way.

thul-he that walket-banat by thul gwzhenen-light,

wotchet-banat thul kador-way.


he stoat-banat hirezhen,

in my right costet-kazhed.

she stoat-banat hirezhen,

in my left costet-kazhed.

she stoat-banat shirezhen,

in yur left costet-kazhed.

he stoat-banat shirezhen,

in yur right costet-kazhed.


Teoría del Gaélical-Irish:


thul [o] an

hitezhen [o] shitezhen

hirezhen [o] shirezhen


-kazher

-kazhed

-kazhing


present:

-banat [o] -banat ket


I

yu

he

she


nurezh

vurezh

hiezh

shiezh


kador-plat.

kador-plate.

error: métodes numérics

f(x_{i}) = | x+(-1)·x_{i} |


f(x_{i}+y_{i}) [< f(x_{i})+f(y_{i})

| ( x+y )+(-1)·( x_{i}+y_{i} ) | [< | x+(-1)·x_{i} |+| y+(-1)·y_{i} |


1+1 = 2

0.75+1.25 = 2

0 = f(0.75+1.25) [< f(0.75)+f(1.25) = 0.50


f(x_{i}·y_{i}) [< |y_{i}|·f(x_{i})+f(x_{i})·f(y_{i})+|x_{i}|·f(y_{i})

| ( x·y )+(-1)·( x_{i}·y_{i} ) | [< |y|| x+(-1)·x_{i} |+|x_{i}|| y+(-1)·y_{i} |

| ( x·y )+(-1)·( x_{i}·y_{i} ) | [< |y_{i}|| x+(-1)·x_{i} |+|x|| y+(-1)·y_{i} |


si ( x [< (-1) || 1 [< x ) ==>

f(1/x_{i}) [< (1/|x_{i}|)·f(x_{i})

si (-1) < x < 1 ==>

f(1/x_{i}) >] (1/|x_{i}|)·f(x_{i})


| (1/x)+(-1)·(1/x_{i}) | = |x_{i}+(-x)|/( |x|·|x_{i}| ) [< ...

... |x_{i}+(-x)|/( |x_{i}+(-x)|·|x_{i}|+(-1)·|x_{i}|^{2} )

f(1/x_{i}) [< (1/|x_{i}|)·f(x_{i})/( f(x_{i})+(-1)·|x_{i}| )


f(y_{i}/x_{i}) [< (1/|x_{i}|)·( ( |x_{i}|·f(y_{i})+|y_{i}|·f(x_{n}) )/( f(x_{i})+(-1)·|x_{i}| ) )

| (y/x)+(-1)·(y_{i}/x_{i}) | [< ( |x_{i}||y+(-1)·y_{i}|+|y_{i}||x_{i}+(-x)| )/( |x|·|x_{i}| )


si x [< (-1) || 1 [< x ==>

f(y_{i}/x_{i}) [< (1/|x_{i}|)·( |x_{i}|·f(y_{i})+|y_{i}|·f(x_{n}) )

si (-1) < x || x < 1 ==>

f(y_{i}/x_{i}) >] (1/|x_{i}|)·( |x_{i}|·f(y_{i})+|y_{i}|·f(x_{n}) )


f( (x_{i})^{2} ) [< ( f(x_{i}) )^{2}+2·|x_{i}|·f(x_{i})

| x^{2}+(-1)·(x_{i})^{2} | = |x+(-1)·x_{i}|·|x+x_{i}|


Si x >] x_{i} ==>

f( ln(x_{i}) ) [< ln( f(x_{i})/|x_{i}|+1 )

| ln(x)+(-1)·ln(x_{i}) | = |ln(x/x_{i})|


f( e^{x_{i}} ) [< ( e^{f(x_{i})}+1 )·e^{|x_{i}|}

| e^{x}+(-1)·e^{x_{i}} | [< |e^{x+(-1)·x_{i}+x_{i}}|+|e^{x_{i}}|

imatge y sonit

A^{j}_{i}·( cos(at)+i·sin(at) )

B^{j}_{i}·( cosh(i·bt)+sinh(i·bt) )


a = frecuencia de la ona portadora [ fibra óptica en color ]-[ ona hertzaria ]

b = to de la escala musical.

A^{j}_{i} = matriu de imatge.

B^{j}_{i} = matriu de sonit.


( cos(at)+i·sin(at) )

( cosh(i·bt)+sinh(i·bt) )


led monocromàtic.

speaker monocromàtic.

º

8 colors visibles.

8 octaves audibles.


infra-rojos

infra-so

ultra-violeta

ultra-so

jueves, 27 de mayo de 2021

música

[00+01][00+04][00+04][00+04] = 13k

[00+04][00+07][00+07][00+07] = 25k = 5^{2}k

[00+07][00+10][00+10][00+10] = 37k

[00+10][12+01][12+01][12+01] = 49k = 7^{2}k

49 = 37+12 = (25+12)+12 = ( (13+12)+12 )+12


[00+01][00+05][00+08][00+05] = 19k

[00+01][00+08][00+08][00+08] = 25k = 5^{2}k

[00+07][00+11][12+02][00+11] = 43k

[00+07][12+02][12+02][12+02] = 49k = 7^{2}k

49 = 43+6 = ( 25+18 )+6 = ( (19+6)+18 )+6


[00+01][00+04][00+08][00+04] = 17k

[00+02][00+07][00+07][00+07] = 23k

[00+07][00+10][12+02][00+10] = 41k

[00+08][12+01][12+01][12+01] = 47k

47 = 41+6 = ( 23+18 )+6 = ( (17+6)+18 )+6

ecuacions diferencials

d_{x}[ ln( plot[(-n)]-[o(x)o]-e(x) ) ] = ( plot[(-n)]-[o(x)o]-e(x) )^{(-n)}

d_{x}[ plot[(-n)]-[o(x)o]-e(x) ] = ( plot[(-n)]-[o(x)o]-e(x) )^{(-n)+1}


d_{x}[ ln( plot[n]-[o(x)o]-e(x) ) ] = ( plot[n]-[o(x)o]-e(x) )^{n}

d_{x}[ plot[n]-[o(x)o]-e(x) ] = ( plot[n]-[o(x)o]-e(x) )^{n+1}


d_{x}[ ln( plot[(-n)]-[o(x)o]-e(x) ) ] = ...

... ( plot[(-n)]-[o(x)o]-e(x) )^{(-1)}·d_{x}[ plot[(-n)]-[o(x)o]-e(x) ]

d_{x}[ ln( plot[n]-[o(x)o]-e(x) ) ] = ...

... ( plot[n]-[o(x)o]-e(x) )^{(-1)}·d_{x}[ plot[n]-[o(x)o]-e(x) ]


y(x) [o(x)o] ln( d_{x}[y(x)] ) = cx

plot[1]-[o(x)o]-ln( d_{x}[y(x)] ) = cx

y(x) = int[ plot[(-1)]-[o(x)o]-e(cx) ] d[x]

y(x) = (1/2)·cx^{2}

plot[(-1)]-[o(x)o]-e(x) = x


d_{x}[y(x)]^{n}·d_{xx}^{2}[y(x)] = d_{x}[y(x)]

( y(x) )^{[o(x)o]n} [o(x)o] ln( d_{x}[y(x)] ) = x

plot[n]-[o(x)o]-ln( d_{x}[y(x)] ) = x

y(x) = int[ plot[(-n)]-[o(x)o]-e(x) ] d[x]


d_{xx}^{2}[y(x)] = d_{x}[y(x)]^{n+1}

( y(x) )^{[o(x)o](-n)} [o(x)o] ln( d_{x}[y(x)] ) = x

plot[(-n)]-[o(x)o]-ln( d_{x}[y(x)] ) = x

y(x) = int[ plot[n]-[o(x)o]-e(x) ] d[x]


( y(x) )^{[o(x)o]n} [o(x)o] e^{d_{x}[y(x)]} = x

plov[n]-[o(x)o]-e( d_{x}[y(x)] ) = x

y(x) = int[ plov[(-n)]-[o(x)o]-ln(x) ] d[x]


d_{x}[ plov[(-n)]-[o(x)o]-ln(x) ] = ...

... ( plov[(-n)]-[o(x)o]-ln(x) )^{(-n)}·e^{(-1)·plov[(-n)]-[o(x)o]-ln(x)}

d_{x}[ e^{plov[(-n)]-[o(x)o]-ln(x)} ] = ( plov[(-n)]-[o(x)o]-ln(x) )^{(-n)}

miércoles, 26 de mayo de 2021

Elecciones

Madrid-Cásteldor:

PP = 65

Más-Madrid = 25

PSOE = 25

UP = 10

precio del metro = 1.05€

precio de cercanías 2 zonas = 1.55€

precio del peaje = 2.05€


Aragó-Càteldor:

PSOE = 33

ERC = 33

Junts = 32

CUP = 11

UP = 8

Cs = 5

PP = 3

precio del metro = 2€

precio de cercanías 2 zonas = 3€

precio del peaje = 4€


Naffarrotzak-Éuskaldor:

PNV = 16

EH-Bildu =12

PSOE = 6

UP = 4

PP = 2


Paisotzok Astur-Nort-koashek:

PRC = 5

PP = 3

PSOE = 2


Galicialdor:

PP = 11

BNG = 6

PSOE = 3


Portugale-y:

PS = 15

PSD = 13

BE = 2

analisis matemàtic

x [< y+x <==> 0 [< y

x >] y+x <==> 0 >] y


x < y+x <==> 0 < y

x > y+x <==> 0 > y


(n+1)x [< y+x <==> nx [< y

(n+1)x >] y+x <==> nx >] y


(n+1)x < y+x <==> nx < y

(n+1)x > y+x <==> nx > y


x+y [< f(x)+y <==> x [< f(x)

x+y >] f(x)+y <==> x >] f(x)


x+y < f(x)+y <==> x < f(x)

x+y > f(x)+y <==> x > f(x)