Guifré del Bergadà:
Doctorat en análisis matemátic.
Capítol I:
Teorema:
[Ek][ k >] 2 & cos((1/pi)·i) [< ( k/(k+(-1)) ) ]
[Ek][ k >] 2 & sin((1/pi)·i) [< ( i/(k+(-1)) ) ]
Desmostració:
Destrocter ponens:
[Ap][ (1/(2p)!)·(1/pi)^{2p} > 1 ]
Destrocter ponens:
[Ap][ (1/(2p+1)!)·(1/pi)^{2p+1} > (1/k) ]
Teorema:
[Ek][ k >] 2 & cosh(1/pi) [< ( k/(k+(-1)) ) ]
[Ek][ k >] 2 & sinh(1/pi) [< ( 1/(k+(-1)) ) ]
Teorema:
(1/3) [< e^{(-1)·(1/pi)} [< 1
1 [< e^{(1/pi)} [< 3
Teorema:
( sin(x) & cos(x) ) convergeish la serie a [0i,i]_{C}
( sinh(x) & cosh(x) ) convergeish la serie a [0,1]_{C}
Capítol II:
Teorema:
[Ek][ k >] 2 & cos((1/pi)·i) >] ( k/(k+1) ) ]
[Ek][ k >] 2 & sin((1/pi)·i) >] ( i/(k+1) ) ]
Desmostració:
Destrocter ponens:
[Ap][ (1/(2p)!)·(1/pi)^{2p} < (1/(2p)!)·(1/pi)^{2p}( (-1)·(1/k) )^{p} ]
[Ap][ (1/(2p)!)·(1/pi)^{2p}·( (-1)·(1/k) )^{p} < ( (-1)·(1/k) )^{p} ]
Destrocter ponens:
[Ap][ (1/(2p+1)!)·(1/pi)^{2p+1} < (1/(2p+1)!)·(1/pi)^{2p+1}·(1/k)·( (-1)·(1/k) )^{p} ]
[Ap][ (1/(2p+1)!)·(1/pi)^{2p+1}·(1/k)·( (-1)·(1/k) )^{p} < (1/k)·( (-1)·(1/k) )^{p} ]
Teorema:
[Ek][ k >] 2 & cosh(1/pi) >] ( k/(k+1) ) ]
[Ek][ k >] 2 & sinh(1/pi) >] ( 1/(k+1) ) ]
Teorema:
(1/3) [< e^{(-1)·(1/pi)} [< 1
1 [< e^{(1/pi)} [< 3
Teorema:
( sin(x) & cos(x) ) convergeish la serie a [0i,i]_{C}
( sinh(x) & cosh(x) ) convergeish la serie a [0,1]_{C}
Hugo de Portugal:
Doctorat en lógica algebraica.
[p(x)] = Binari concret.
]q(x)[ = Borrós semblant-abstracte.
Teorema:
min{[p(x)],]q(x)[} = [p(x)] <==> max{[p(x)],]q(x)[} = ]q(x)[
max{¬[p(x)],¬]q(x)[} = ¬[p(x)] <==> min{¬[p(x)],¬]q(x)[} = ¬]q(x)[
Teorema:
min{[f(1)],]f(n)[} = [f(1)] <==> max{[f(1)],]f(n)[} = ]f(n)[
max{[f(-1)],]f(-n)[} = [f(-1)] <==> min{[f(-1)],]f(-n)[} = ]f(-n)[
Don Casasayas de Euskal-Herria:
Doctorat en análisis matemátic.
Capítol I:
Teorema:
Si a [< b ==> [Au][Eq][En][ a [< q+(u/n) [< b ]
Demostració:
a [< ( (a+(-u))+(b+u) )/2 [< b
q = ( (a+(-u)+b)/2 ) & n = 2
Teorema:
Si a [< b ==> [Au][Av][Eq][En][Em][ a [< q+(u/n)+(v/m) [< b ]
Demostració:
a [< ( (5a+(-5)·u+(-2)·v)+(5b+5u+2v) )/10 [< b
q = ( (5a+(-5)·u+(-2)·v+5b)/10 ) & n = 2 & m = 5
Teorema:
[Ax][Ea_{n}][Eb_{n}][ a_{n} [< x [< b_{n} ...
... & [Eq][ b_{n}+(-1)·a_{n} = q ] & lim[a_{n}] = lim[b_{n}] = x ]
Demostració:
a_{n} = x+(-1)·(1/n)
b_{n} = x+(1/n)
q = (2/n)
Teorema:
[Ax][Ay][Ea_{n}][Eb_{n}][ x [< a_{n}+b_{n} [< y...
... & [Eq][ b_{n}+(-1)·a_{n} = q ] & lim[a_{n}] = lim[b_{n}] ]
Demostració:
a_{n} = ( (x+y)/4 )+(-1)·(1/n)
b_{n} = ( (x+y)/4 )+(1/n)
q = (2/n)
Capítol II:
Teorema:
max{x,y} = ( ( (x+y)+|y+(-x)| )/2 )
min{x,y} = ( ( (x+y)+|i(y+(-x))| )/2 )
Demostració:
x [< y <==> 0 [< y+(-x)
x >] y <==> 0 >] y+(-x)
Teorema:
x^{2} = ( ( x+|x| )/2 )^{2}+( ( x+(-1)·|x| )/2 )^{2}
x^{2} = ( ( x+|ix| )/2 )^{2}+( ( x+(-1)·|ix| )/2 )^{2}
Teorema:
|x_{1}+...+x_{n}| [< |x_{1}|+...+|x_{n}|
|i(x_{1}+...+x_{n})| >] |ix_{1}|+...+|ix_{n}|
Pla d'estudis de la universitat de Stroniken:
Nota de tall = 7.50
1r semestre:
Guifré del Bergadà:
Análisis matemátic I
Análisis matemátic III
Don Casasayas:
Álgebra
Probabilitats
Hugo de Portugal:
Lógica
Topología
Jûan Garriga:
Informática
Especies combinatories
2n semestre:
Guifré del Bergadà:
Análisis matemátic II
Análisis matemátic IV
Don Casasayas:
Ecuacions Diferencials
Análisis Complex y Borrós
Hugo de Portugal:
Teoria de conjunts
Lógica algebraica y Dualogía
Jûan Garriga:
Álgebra lineal y geometría lineal
Geometría Diferencial
Hidrogen verd:
Aigua:
4·H_{2}+O_{4} <==> 4·H_{2}O
[4·H_{2}]·[O_{4}] <==> [4e]·[4·H_{2}O]
Aigua oxigenada:
2·H_{2}+O_{6} <==> 2·H_{2}O_{3}
[2·H_{2}]·[O_{6}] <==> [2e]·[2·H_{2}O_{3}]
Oxigen:
2·O_{4}+O_{4} <==> 2·O_{6}
[2·O_{4}]·[O_{4}] <==> [2e]·[2·O_{6}]
Ley que no es del mundo:
Si no es tu dinero:
Si enseñas el DNI en el banco robas.
Si no enseñas el DNI en el banco no robas.
Si es tu dinero:
Si enseñas el DNI en el banco no te roban.
Si no enseñas el DNI en el banco te roban.
menjar [o] menjjar [o] menjjate [o] menjjet-kazhe
pujar [o] pujjar [o] pujjate [o] pujjet-kazhe
bajar [o] baishar [o] bashate [o] bashet-kazhe
dejar [o] deishar [o] deshate [o] deshet-kazhe
yo havere-po encontratered una miravilem demostraciorum,
apud en el marginis non sere-po capered la demostraciorum.
A + B = A [ || ] B & ¬A + ¬B = ¬A [&] ¬B
A = < {a_{1},...(n)...,a_{n}},{a_{1}} > & S[A] = (n+1)·x^{n}
¬A = < }a_{1},...(n)...,a_{n}{,}a_{1}{ > & S[A] = ((-n)+(-1))·x^{n}
A = {a_{1},...(n)...,a_{n}} [&] {a_{1},...,a_{k}} & S[A] = kx^{k}
¬A = }a_{1},...(n)...,a_{n}{ [&] }a_{1},...,a_{k}{ & S[A] = (-k)·x^{k}
A = }a_{1},...(n)...,a_{n}{ [ \ ] }a_{1},...,a_{k}{ & S[A] = ((-n)+k)·x^{k}
¬A = {a_{1},...(n)...,a_{n}} [ \ ] {a_{1},...,a_{k}} & S[A] = (n+(-k))·x^{k}
Teoría de ingeniería y de economía:
Definició:
[ n // k ]+[ n // (k+1) ] = [ (n+1) // (k+1) ]
sum[k = 0]-[n][ [ n // k ] ] = 2^{n}
[ (-n) // k ]+[ (-n) // (k+1) ] = [ ((-n)+1) // (k+1) ]
sum[k = 0]-[n][ [ (-n) // k ] ] = 2^{(-n)}
Teorema:
[ (-2) // 0 ] = 1 & [ (-2) // 1 ] = (-2) & [ (-2) // 2 ] = (5/4)
[ (-3) // 0 ] = 1 & [ (-3) // 1 ] = (-3) & [ (-3) // 2 ] = (17/4) & [ (-3) // 3 ] = (-1)·(17/8)
Si k >] 2 ==> [ (-n) // k ] = (-1)^{k}·( F(n,k)/2^{k} )
Binomis:
Teorema:
(x+x)^{n} = sum[k = 0]-[n][ [ n // k ]·x^{n+(-k)}·x^{k} ]
Teorema:
(x+x)^{(-n)} = sum[k = 0]-[n][ [ (-n) // k ]·x^{(-n)+(-k)}·x^{k} ]
Teorema:
lim[h = 0][ (x+h)^{n} ] = ...
... lim[h = 0][ sum[k = 0]-[n][ [ n // k ]·x^{n+(-k)}·h^{k} ] ] = x^{n}
Teorema:
lim[h = 0][ (x+h)^{(-n)} ] = ...
... lim[h = 0][ sum[k = 0]-[n][ [ (-n) // k ]·x^{(-n)+(-k)}·h^{k} ] ] = x^{(-n)}
Demostració:
[ (-n) // k ]·x^{(-n)+(-k)}·h^{k}·(x+h) = ...
... [ (-n) // k ]·x^{(-n)+1+(-1)·(k+1)}·h^{k+1}+[ (-n) // p ]·x^{(-n)+1+(-p)}·h^{p}
Distribucions:
f(k) = [ n // k ]·2^{(-n)}
g(k) = [ (-n) // k ]·2^{n}
E[k·f(k)] = (n/2)
E[k·g(k)] = (-n)+sum[k = 2]-[n][ (-1)^{k}·k·( F(n,k)/2^{k} ) ]
H(n) = sum[k = 1]-[n][ (1/k)·[ (n+(-1)) // (k+(-1)) ] ] = (1/n)·2^{n}
Derivada:
d_{x}[x^{n}] = lim[h = 0][ ( ( (x+h)^{n}+(-1)·x^{n} )/h ) ] = nx^{n+(-1)}
Integral:
int[x^{n}]d[x] = ( 1/(n+1) )·lim[h = 0][ int[ (x+h)^{n}·(x+h)+(-1)·x^{n}·x ] ] = ...
... ( 1/(n+1) )·int[ d[x^{n+1}] ] = ( 1/(n+1) )·x^{n+1}
int[x^{n}]d[x] = ( 1/(n+1) )·int[ (n+1)·x^{n} ]d[x]
Derivada:
d_{x}[x^{(-n)}] = lim[h = 0][ ( ( (x+h)^{(-n)}+(-1)·x^{(-n)} )/h ) ] = (-n)·x^{(-n)+(-1)}
Integral:
int[x^{(-n)}]d[x] = ( 1/((-n)+1) )·lim[h = 0][ int[ (x+h)^{(-n)}·(x+h)+(-1)·x^{(-n)}·x ] ] = ...
... ( 1/((-n)+1) )·int[ d[x^{(-n)+1}] ] = ( 1/((-n)+1) )·x^{(-n)+1}
int[x^{(-n)}]d[x] = ( 1/((-n)+1) )·int[ ((-n)+1)·x^{(-n)} ]d[x]
Derivada:
d_{x}[e^{x}] = lim[h = 0][ ( ( e^{x+h}+(-1)·e^{x} )/h ) ] = e^{x}
Integral:
int[e^{x}]d[x] = lim[h = 0][ int[ e^{x+h}·(x+h)+(-1)·e^{x}·x ] ] = ...
... lim[h = 0][ int[ e^{x}·(x+h)+(-1)·e^{x}·x ] ] = ...
... int[ e^{x} ]d[x] = int[ d_{x}[e^{x}] ]d[x] = e^{x}
Derivada:
d_{x}[ln(x)] = lim[h = 0][ ( ( ln(x+h)+(-1)·ln(x) )/h ) ] = (1/x)
Integral:
int[ (1/x) ]d[x] = lim[h = 0][ int[ ( 1/(x+h) )·(x+h)+(-1)·(1/x)·x ] ] = ...
... lim[h = 0][ int[ (1/x)·(x+h)+(-1)·(1/x)·x ] ] = ...
... int[ (1/x) ]d[x] = int[ d_{x}[ln(x)] ]d[x] = ln(x)
No hay comentarios:
Publicar un comentario