comentario de fieles y arte matemático y Stowed-Nipon

Hay fieles de menos de 27 años,

que no los odia el mundo,

porque están glorificados.

Hay fieles de más de 27 años,

que los odia el mundo,

porque no están glorificados.

Arte:

[Ex][Ey][ a = x+y <==> a = 0 ]

Exposición:

x = z & y = (-z)

a = (x+y)+(-y) [ f(z) = z+(-y) ]

a = x+(-x) [ f(z) = z+(-x) ]

a = 0

Arte:

[Ex][ x = a <==> y = 0 ]

Exposición:

x = y+a

x = a

x = 0

0 = 0

y = 0

Arte:

[Ex][Ey][ a^{2} = x^{2}+y^{2} <==> ( a = 1 || a = (-1) ) ]

Exposición:

x = cos(t) & y = sin(t)

a^{2} = ( x^{2}+y^{2} )+(-1)·y^{2} [ f(z) = z+(-1)·y^{2} ]

a^{2} = x^{2}+(-1)·x^{2} [ f(z) = z+(-1)·x^{2} ]

a^{2} = 0

a^{2} = 1 [ f(z) = z+1 ]

Arte:

[Ex][ a^{2} = 2+x^{2} <==> ( a = 1 || a = (-1) ) ]

Exposición:

( x = i || x = (-i) )

a^{2} = 2+x^{2}

a^{2} = 2

a^{2} = 1

Arte:

[Ef(x)][ z = ( 1/(n+1) )·d_{x}[f(x)] <==> z = x^{n} ]

Exposición:

f(x) = x^{n+1}

z = (1/(n+1))·d_{x}[f(x)]

z = (1/(n+1)) [ f(z) = ( z/d_{x}[f(x)] ) ]

z = 1 [ f(z) = z·(n+1) ]

z = x^{n} [ f(z) = z·( x^{n} ) ]

Arte:

[Ef(x)][ int[ n·f(x) ]d[x] = nx+1 <==> x = 0 ]

Demostración:

f(x) = e^{nx}

e^{nx} = nx+1

( 1+nx+(1/2)·(nx)^{2}+... ) != 1+nx

nx = 0

Exposición:

int[ n·f(x) ]d[x] = nx+1

n·e^{nx} = nx+1 [ g(f(x)) = d_{x}[f(x)] ]

e^{nx} = nx+1 [ f(z) = (z/n) ]

e^{nx} = nx [ f(z) = z+(-1) ]

e^{nx} = x [ f(z) = (z/n) ]

e^{nx} = 0 [ f(z) = z+(-x) ]

x = 0 [ g(f(x)) = (1/n)·ln( f(x) ) ]

Arte:

[Ef(x)][ d_{x}[ (1/n)·f(x) ] = nx+1 <==> x = 0 ]

Demostración:

f(x) = e^{nx}

e^{nx} = nx+1

( 1+nx+(1/2)·(nx)^{2}+... ) != 1+nx

nx = 0

Exposición:

d_{x}[ (1/n)·f(x) ] = nx+1

(1/n)·e^{nx} = nx+1 [ g(f(x)) = int[f(x)]d[x] ]

e^{nx} = nx+1 [ f(z) = z·n ]

e^{nx} = nx [ f(z) = z+(-1) ]

e^{nx} = x [ f(z) = (z/n) ]

e^{nx} = 0 [ f(z) = z+(-x) ]

x = 0 [ g(f(x)) = (1/n)·ln( f(x) ) ]

Stowed-Nipon:

-maruto

-et-yuto-yamed

-et-yuto-yaming

-eretch-tet-yuto-yam

-koweretch-tet-yuto-yamed

I havere-kate-maruto drinket-yuto-yamed mutchet-muto.

I havere-kate-maruto drinket-yuto-yamed pocket-muto.

I gow-maruto to pisheretch-tet-yuto-yam,

becose I me stare-kate-maruto pishet-yuto-yaming.

I gow-maruto to shiteretch-tet-yuto-yam,

becose I me stare-kate-maruto shitet-yuto-yaming.

I me havere-kate-maruto pishet-yuto-yamed in sober-muto.

I me havere-kate-maruto shitet-yuto-yamed in sober-muto.

He stare-kate-maruto in sober-muto of the table-sosukni.

He stare-kate-maruto in sote-muto of the table-sosukni.

I take-maruto a coffi-sosukni wizh milki-sukni.

I take-maruto a coffi-sosukni wizhawt milki-sukni.

Medalla Fields del francés 2022:

Teorema:

[Ex][Ey][ x+y = ( b+b^{(1/2)} )^{(1/2)} ].

Demostración:

x^{2}+2xy+y^{2} = b+b^{(1/2)}

x·(x+y)+(x+y)·y = b+b^{(1/2)}

x^{1+[...(y)...[1]...(y)...]}+y^{1+[...(x)...[1]...(x)...]} = b+b^{(1/2)}

x = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( 1+[...(y)...[1]...(y)...] ) )}

y = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( 1+[...(x)...[1]...(x)...] ) )}

Si x = (b/2)+(b^{(1/2)}/2) ==>

( (b/2)+(b^{(1/2)}/2) )·( ( (b/2)+(b^{(1/2)}/2) )+y ) = (b/2)+(b^{(1/2)}/2)

(b/2)+(b^{(1/2)}/2) = (b/2)+(b^{(1/2)}/2)

Teorema: [ de medalla fields por plantear esta ecuación aunque no de la solución ]

c^{( n/( n+[...(x)...[n]...(x)...] ) )} = c^{( 1/( 1+[...(x)...[1]...(x)...] ) )}

c·( c+c^{( 1/( 1+[...(c)...[1]...(c)...] ) )} ) = c

Teorema:

[Ex][Ey][ x^{n}+y^{n} = ( b+b^{(1/2)} )^{(1/2)} ].

Demostración:

x^{2n}+2(xy)^{n}+y^{2n} = b+b^{(1/2)}

x^{n}·(x^{n}+y^{n})+(x^{n}+y^{n})·y^{n} = b+b^{(1/2)}

x^{n+[...(y^{n})...[n]...(y^{n})...]}+y^{n+[...(x^{n})...[n]...(x^{n})...]} = b+b^{(1/2)}

x = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( n+[...(y^{n})...[n]...(y^{n})...] ) )}

y = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( n+[...(x^{n})...[n]...(x^{n})...] ) )}

Si x = ( (b/2)+(b^{(1/2)}/2) )^{(1/n)} ==>

( (b/2)+(b^{(1/2)}/2) )·( ( (b/2)+(b^{(1/2)}/2) )+y^{n} ) = (b/2)+(b^{(1/2)}/2)

(b/2)+(b^{(1/2)}/2) = (b/2)+(b^{(1/2)}/2)

Teorema:

[Ex][Ey][ x+y^{3} = ( b+b^{(1/2)} )^{(1/2)} ].

Demostración:

x^{2}+2xy^{3}+y^{6} = b+b^{(1/2)}

x·(x+y^{3})+(x+y^{3})·y^{3} = b+b^{(1/2)}

x^{1+[...(y^{3})...[1]...(y^{3})...]}+y^{3+[...(x)...[3]...(x)...]} = b+b^{(1/2)}

x = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( 1+[...(y^{3})...[1]...(y^{3})...] ) )}

y = ( (b/2)+(b^{(1/2)}/2) )^{( 1/( 3+[...(x)...[3]...(x)...] ) )}

Si x = ( (b/2)+(b^{(1/2)}/2) ) ==>

( (b/2)+(b^{(1/2)}/2) )·( ( (b/2)+(b^{(1/2)}/2) )+y^{3} ) = (b/2)+(b^{(1/2)}/2)

(b/2)+(b^{(1/2)}/2) = (b/2)+(b^{(1/2)}/2)

Medalla Fields del Inglés 2022:

Teorema:

[An][ p_{n+1}+(-1)·p_{n} [< 240 ].

Demostración:

< 1,3,7,9 > [< 4!·0!·10

< 1,3,7,9 > [< 3!·1!·40

< 1,3,7,9 > [< 2!·2!·60

Teorema:

[An][ p_{n+1}+(-1)·p_{n} [< 480 ].

Demostración:

< 1,3,7,9 > [< 4!·0!·20

< 1,3,7,9 > [< 3!·1!·80

< 1,3,7,9 > [< 2!·2!·120

Teorema:

[Ec][An][ ( 1+(1/n) )^{n} [< e^{c·e^{e^{e^{(-n)}}}} ]

Demostración:

e^{(-n)} >] 0

e^{e^{(-n)}} >] e^{0} = 1

e^{e^{e^{(-n)}}} >] e^{1} = e

Sea M >] 1 ==>

Se define un c >] ( M/e )

( 1+(1/n) )^{n} [< e [< e^{M} [< e^{(M/e)·e^{e^{e^{(-n)}}}} [< e^{c·e^{e^{e^{(-n)}}}}

Teorema:

Si ( (`p(x) || q(x) ) ==> ( a(x) & b(x) ) ) ==> ( [1] & [2] )...

[1] ( p(x) ==> a(x) )

[2] ( q(x) ==> b(x) )

Negad en murciano y valenciano para no perder:

pink records, contraseña records, moon records y ferpas records.

Murciano:

Entreshko a casa

Ishko de casa

Valencià:

Entreshkû a casa

Ishkû de casa

Perifrástico:

Valenciano-y-Murciano:

vore ishkir

varas ishkir

vara ishkir

( várem || váramos ) ishkir

( váreu || várais ) ishkir

varan ishkir