Teoría:
Teorema:
Sea ( c_{k} > 0 & [An][ n [< u_{n} ] ) ==>
Si a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] ==> a_{n} no está acotada superiormente.
Sea ( d_{k} < 0 & [An][ n [< v_{n} ] ) ==>
Si b_{n} = sum[k = 1]-[p][ d_{k}·( v_{n} )^{k} ] ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Sea M = min{c_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n [< n^{k} [< ( u_{n} )^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
n^{k+1} [< ( u_{n} )^{k}·n [< ( u_{n} )^{k+1}
a_{n} = sum[k = 1]-[p][ c_{k}·( u_{n} )^{k} ] >] sum[k = 1]-[p][ M·( u_{n} )^{k} ] = ...
... M·sum[k = 1]-[p][ ( u_{n} )^{k} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s
Sea s < 0 ==>
Sea M = max{d_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n [< n^{k} [< ( v_{n} )^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
n^{k+1} [< ( v_{n} )^{k}·n [< ( v_{n} )^{k+1}
b_{n} = sum[k = 1]-[p][ d_{k}·( v_{n} )^{k} ] [< sum[k = 1]-[p][ M·( v_{n} )^{k} ] = ...
... M·sum[k = 1]-[p][ ( v_{n} )^{k} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s
Teorema:
[An][ [ n // k ] >] 1 ]
Demostración:
[ n // 0 ] = [ n // n ] = 1
[ n+1 / k ] = [ n // k+(-1) ]+[ n // k ] >] [ n // k+(-1) ]+1 >] 1
Teorema:
Sea [An][ n [< u_{n} ] ==>
Si a_{n} = ( u_{n}+1 )^{p} ==> a_{n} no está acotada superiormente.
Sea [An][ n [< v_{n} ] ==>
Si b_{n} = (-1)·( v_{n}+1 )^{p} ] ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Se define n_{0} > (s/p) ==>
Sea n > n_{0} ==>
n [< n^{k} [< ( u_{n} )^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
n^{k+1} [< ( u_{n} )^{k}·n [< ( u_{n} )^{k+1}
a_{n} = ( u_{n}+1 )^{p}= sum[k = 0]-[p][ [ p // k ]·( u_{n} )^{k} ] >] ...
... sum[k = 0]-[p][ ( u_{n} )^{k} ] = 1+sum[k = 1]-[p][ ( u_{n} )^{k} ] > ...
... sum[k = 1]-[p][ ( u_{n} )^{k} ] >] sum[k = 1]-[p][ n ] = pn > pn_{0} > s
Sea s < 0 ==>
Se define n_{0} > ( s/(-p) ) ==>
Sea n > n_{0} ==>
n [< n^{k} [< ( v_{n} )^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
n^{k+1} [< ( v_{n} )^{k}·n [< ( v_{n} )^{k+1}
b_{n} = (-1)·( v_{n}+1 )^{p}= (-1)·sum[k = 0]-[p][ [ p // k ]·( v_{n} )^{k} ] [< ...
... (-1)·sum[k = 0]-[p][ ( v_{n} )^{k} ] = (-1)+(-1)·sum[k = 1]-[p][ ( v_{n} )^{k} ] < ...
... (-1)·sum[k = 1]-[p][ ( v_{n} )^{k} ] [< (-1)·sum[k = 1]-[p][ n ] = (-p)·n < (-p)·n_{0} < s
Problemas:
Teorema:
Sea c_{k} > 0 ==>
Si a_{n} = sum[k = 1]-[p][ c_{k}·n^{k} ] ==> a_{n} no está acotada superiormente.
Sea d_{k} < 0 ==>
Si b_{n} = sum[k = 1]-[p][ d_{k}·n^{k} ] ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Sea M = min{c_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n [< n^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
a_{n} = sum[k = 1]-[p][ c_{k}·n^{k} ] >] sum[k = 1]-[p][ M·n^{k} ] = ...
... M·sum[k = 1]-[p][ n^{k} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s
Sea s < 0 ==>
Sea M = max{d_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n [< n^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
b_{n} = sum[k = 1]-[p][ d_{k}·n^{k} ] [< sum[k = 1]-[p][ M·n^{k} ] = ...
... M·sum[k = 1]-[p][ n^{k} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s
Teorema:
Si a_{n} = (n+1)^{p} ==> a_{n} no está acotada superiormente.
Si b_{n} = (-1)·(n+1)^{p} ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Se define n_{0} > (s/p) ==>
Sea n > n_{0} ==>
n [< n^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
a_{n} = (n+1)^{p} = sum[k = 0]-[p][ [ p // k ]·n^{k} ] >] sum[k = 0]-[p][ n^{k} ] = ...
... 1+sum[k = 1]-[p][ n^{k} ] >] sum[k = 1]-[p][ n^{k} ] >] sum[k = 1]-[p][ n ] = p·n > p·n_{0} > s
Sea s < 0 ==>
Se define n_{0} > ( s/(-p) ) ==>
Sea n > n_{0} ==>
n [< n^{k}
n+1 [< n^{k}+1 [< sum[j = 0]-[k][ [ k // j ]·n^{j} ] = (n+1)^{k}
b_{n} = (-1)·(n+1)^{p} = (-1)·sum[k = 0]-[p][ [ p // k ]·n^{k} ] [< (-1)·sum[k = 0]-[p][ n^{k} ] = ...
... (-1)+(-1)·sum[k = 1]-[p][ n^{k} ] [< (-1)·sum[k = 1]-[p][ n^{k} ] [< (-1)·sum[k = 1]-[p][ n ] = ...
... (-p)·n < (-p)·n_{0} < s
Teorema:
Sea c_{k} > 0 ==>
Si a_{n} = sum[k = 1]-[p][ c_{k}·e^{kn} ] ==> a_{n} no está acotada superiormente.
Sea d_{k} < 0 ==>
Si b_{n} = sum[k = 1]-[p][ d_{k}·e^{kn} ] ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Sea M = min{c_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n < e^{kn}
n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}
a_{n} = sum[k = 1]-[p][ c_{k}·e^{kn} ] >] sum[k = 1]-[p][ M·e^{kn} ] = ...
... M·sum[k = 1]-[p][ e^{kn} ] >] M·sum[k = 1]-[p][ n ] = Mp·n > Mp·n_{0} > s
Sea s < 0 ==>
Sea M = max{d_{k}} ==>
Se define n_{0} > ( s/(pM) ) ==>
Sea n > n_{0} ==>
n < e^{kn}
n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}
b_{n} = sum[k = 1]-[p][ d_{k}·e^{kn} ] [< sum[k = 1]-[p][ M·e^{kn} ] = ...
... M·sum[k = 1]-[p][ e^{kn} ] [< M·sum[k = 1]-[p][ n ] = Mp·n < Mp·n_{0} < s
Teorema:
Si a_{n} = (e^{n}+1)^{p} ==> a_{n} no está acotada superiormente.
Si b_{n} = (-1)·(e^{n}+1)^{p} ==> b_{n} no está acotada inferiormente.
Demostración:
Sea s > 0 ==>
Se define n_{0} > (s/p) ==>
Sea n > n_{0} ==>
n < e^{kn}
n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}
b_{n} = (e^{n}+1)^{p} = sum[k = 0]-[p][ [ p // k ]·e^{kn} ] > ...
... 1+sum[k = 1]-[p][ e^{kn} ] > sum[k = 1]-[p][ e^{kn} ] > ...
... sum[k = 1]-[p][ n ] = pn > pn_{0} > s
Sea s < 0 ==>
Se define n_{0} > ( s/(-p) ) ==>
Sea n > n_{0} ==>
n < e^{kn}
n+1 < e^{kn}+1 < e^{kn}+e^{kn} = 2e^{kn} < e^{k}·e^{kn} = e^{k·(n+1)}
b_{n} = (-1)·(e^{n}+1)^{p} = (-1)·sum[k = 0]-[p][ [ p // k ]·e^{kn} ] < ...
... (-1)+(-1)·sum[k = 1]-[p][ e^{kn} ] < (-1)·sum[k = 1]-[p][ e^{kn} ] [< ...
... (-1)·sum[k = 1]-[p][ n ] = (-p)·n < (-p)·n_{0} < s
Examen de análisis matemático:
Teorema:
Sea c_{k} > 0 ==>
Si a_{n} = sum[k = 1]-[p][ c_{k}·( ne^{n} )^{k} ] ==> a_{n} no está acotada superiormente.
Sea d_{k} < 0 ==>
Si b_{n} = sum[k = 1]-[p][ d_{k}·( ne^{n} )^{k} ] ==> b_{n} no está acotada inferiormente.
Teorema:
Si a_{n} = (ne^{n}+1)^{p} ==> a_{n} no está acotada superiormente.
Si b_{n} = (-1)·(ne^{n}+1)^{p} ==> b_{n} no está acotada inferiormente.
Teorema:
Sea a >] 0 ==>
Si 0 [< f(x) [< | x+(-a) | ==> ( f(a) = 0 & f(x) es continua )
Demostración:
0 [< f(a) [< | a+(-a) | = |0| = 0
f(a) = 0
Se define 0 < s < 1 ==>
Sea d > 0 & |h| < d ==>
g(0) = n
f(x) [< | x+(-a) | <==> | x+(-a) | < f(x)
| f(x+h)+(-1)·f(x) | > | | (x+h)+(-a) |+(-1)·| x+(-a) | | = | (x+(-a))+h |+(-1)·| x+(-a) | > ...
... | x+(-a) |+|h|+(-1)·| x+(-a) | = |h| = 0 = g(0) = n > s
Teorema:
Sea a >] 0 ==>
Si a [< f(x) [< | x+(-a) |+|x| ==> ( f(a/n) = a & f(x) es continua )
Demostración:
a [< f(a/n) [< | (a/n)+(-a) |+|(a/n)| = | a·( (1/n)+(-1) ) |+(a/n) = a+(-1)·(a/n)+(a/n) = a
Se define 0 < s < 1 ==>
Sea d > 0 & |h| < d ==>
g(0^{2}) = n^{2}
f(x) [< | x+(-a) |+|x| <==> | x+(-a) |+|x| < f(x)
| f(x+h)+(-1)·f(x) | > | | (x+h)+(-a) |+| x+h |+(-1)·| x+(-a) |+(-1)·|x| | = ...
... | (x+(-a))+h |+| x+h |+(-1)·| x+(-a) |+(-1)·|x| > | x+(-a) |+|h|+|x|+|h|+(-1)·| x+(-a) |+(-1)·|x| = ...
... |h|+|h| = 0+0 = 0^{2} = g(0^{2}) = n^{2} >] n > s
Si 2a [< f(x) [< | x+(-a) |+| x+a | ==> ( f(a/n) = 2a & f(x) es continua )
Métodos numéricos:
det( A+(-x)·Id ) = | < (-x),2p+1,0 >,< 2p+1,(-x),2p·(p+1) >,< 0,2p·(p+1),(-x) > | = 0
(-x)·( x^{2}+(-1)·( 2p·(p+1)+1 )^{2} ) = 0
( < (-1)·(2p·(p+1)+1),2p+1,0 >,< 2p+1,(-1)·(2p·(p+1)+1),2p·(p+1) >,< 0,2p·(p+1),(-1)·(2p·(p+1)+1) > )
u = < 2p+1,2p·(p+1)+1,2p·(p+1) >
( < 0,2p+1,0 >,< 2p+1,0,2p·(p+1) >,< 0,2p·(p+1),0 > )
v = < 2p·(p+1),0,(-1)·(2p+1) >
( < 2p·(p+1)+1,2p+1,0 >,< 2p+1,2p·(p+1)+1,2p·(p+1) >,< 0,2p·(p+1),2p·(p+1)+1 > )
w = < 2p+1,(-1)·(2p·(p+1)+1),2p·(p+1) >
Comprobar con el ordenador el Ker(A) de las matrices pitagóricas:
Comprobar con el ordenador la matriz inversa de las matrices pitagóricas:
Algoritmo:
producto-de-matrices-columna( ...
... int matriz[0][0] , int vector-columna-x[0] , int vector-columna-y[0] , int n )
{
for( j = 1 ; j [< n ; j++ )
{
vector-columna-y[j] = 0;
for( k = 1; k [< n [< k++ )
vector-columna-y[j] = vector-columna-y[j]+matriz[k][j]·vector-columna-x[k];
}
}
producto-de-matrices-fila( int matriz[0][0] , int vector-fila-x[0] , int vector-fila-y[0] , int n )
{
for( i = 1 ; i [< n ; i++ )
{
vector-fila-y[i] = 0;
for( k = 1; k [< n [< k++ )
vector-fila-y[i] = vector-fila-y[i]+matriz[i][k]·vector-fila-x[k];
}
}
Algoritmo:
introducción-de-vector-columna-en-matriz( int matriz[0][0] , int vector-columna[0] , int n , int i )
{
for( k = 1; k [< n [< k++ )
matriz[i][k] = vector-columna[k];
}
introducción-de-vector-fila-en-matriz( int matriz[0][0] , int vector-fila[0] , int n , int j )
{
for( k = 1; k [< n [< k++ )
matriz[k][j] = vector-fila[k];
}
Algoritmo:
dibujo-de-vectores-columna( int vector-columna[0] , int n , int x , int y , int s )
{
for( k = 1; k [< n [< k++ )
{
gotoxy(x,s·(k+not(1))+y);
printf("%",vector-columna[k]);
}
}
dibujo-de-vectores-fila( int vector-fila[0] , int n , int x , int y , int s )
{
for( k = 1; k [< n [< k++ )
{
gotoxy(s·(k+not(1))+x,y);
printf("%",vector-fila[k]);
}
}
División borrosa:
Teorema:
[Ap][An][Em][ p = m·(1/n) ]
[Ap][An][Em][ p·(n+(-1)) = m·(n+(-1))·(1/n) ]
Demostración:
Se define m = pn
Teorema:
[Ap][An][Em][ p = m·(1/n) ]
[Ap][An][Em][ p·(2n+(-1)) = m·(2n+(-1))·(1/n) ]
Demostración:
Se define m = pn
Teorema:
Sea n = 2k ==>
2p = 4pk·(1/(2k))
2p+1 = (4pk+2k)·(1/(2k))
Teorema:
Sea n = 2k+1 ==>
2p = (4pk+2p)·(1/(2k+1))
2p+1 = (4pk+2k+2p+1)·(1/(2k+1))
Teorema:
Sea n = 2k & 2n+(-1) = 4k+(-1) ==>
2p·(4k+(-1)) = 4pk·(4k+(-1))·(1/(2k))
(2p+1)·(4k+(-1)) = (4pk+2k)·(4k+(-1))·(1/(2k))
Teorema:
Sea n = 2k+1 & 2n+(-1) = 4k+1 ==>
(2p)·(4k+1) = (4pk+2p)·(4k+1)·(1/(2k+1))
(2p+1)·(4k+1) = (4pk+2k+2p+1)·(4k+1)·(1/(2k+1))
Trabajo para el CITNB:
Estudiar las matrices cuadradas de orden 2:
A(2k,p) = ( < 4pk,4pk+2k >,< 4pk·(4k+(-1)),(4pk+2k)·(4k+(-1)) > )
( x = 0 || x = 4pk+(4pk+2k)·(4k+(-1)) )
u = < 4pk+2k,4pk >
v = < 1,4k+(-1) >
Son teoremas de cuerdas cerradas,
porque diagonalizan en un solo valor propio diferente de cero.
Ley:
(m/2)·d_{t}[u]^{2} = ( 4pk+(4pk+2k)·(4k+(-1)) )·k·(1/a)^{2}·(1/2)·(au)^{2}
(m/2)·d_{t}[v]^{2} = ( 4pk+(4pk+2k)·(4k+(-1)) )·k·(1/b)^{2}·(1/2)·(bv)^{2}
Anexo:
La música de frecuencia de octava [1,2]_{R} emite una cuerda cerrada.
Teorema:
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>
Si [An][ n > n_{0} ==> a_{n} = (1/n) ] ==> a_{n} está acotada superiormente
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>
Si [An][ n > n_{0} ==> b_{n} = (-1)·(1/n) ] ==> b_{n} está acotada inferiormente
Demostración:
Sea n_{0}€N ==>
Se define M = max{c_{n},1} ==>
Sea n€N ==>
Si n [< n_{0} ==> a_{n} = c_{n} [< max{c_{n},1} = M
Si n > n_{0} ==> a_{n} = (1/n) [< 1 [< max{c_{n},1} = M
Sea n_{0}€N ==>
Se define M = min{d_{n},(-1)} ==>
Sea n€N ==>
Si n [< n_{0} ==> b_{n} = d_{n} >] min{d_{n},(-1)} = M
Si n > n_{0} ==> b_{n} = (-1)·(1/n) >] (-1) >] min{d_{n},(-1)} = M
Examen de análisis matemático:
Teorema:
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>
Si [Ew][ w > 0 & [An][ n > n_{0} ==> a_{n} = (w/n) ] ] ==> a_{n} está acotada superiormente
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>
Si [Ew][ w > 0 & [An][ n > n_{0} ==> b_{n} = (-1)·(w/n) ] ] ==> b_{n} está acotada inferiormente
Teorema:
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ec_{n}][ c_{n} >] 0 & a_{n} = c_{n} ] ) ==>
Si [An][ n > n_{0} ==> a_{n} = ( 1+(1/n) )^{n} ] ==> a_{n} está acotada superiormente
Sea ( n_{0}€N & [An][ n [< n_{0} ==> [Ed_{n}][ d_{n} [< 0 & b_{n} = d_{n} ] ) ==>
Si [An][ n > n_{0} ==> b_{n} = (-1)·( 1+(1/n) )^{n} ] ==> b_{n} está acotada inferiormente
Teorema:
( 1+(1/n) )^{n} es creciente
Demostración: [ por destructor ]
1+(1/n) > 1+(1/(n+1))
( 1+(1/n) )^{n} > ( 1+(1/(n+1)) )^{n} > ...
... ( 1+(1/(n+1)) )^{n}+(1/(n+1))·( 1+(1/(n+1)) ) = ( 1+(1/(n+1)) )^{n+1}
Expresión posicional p-ádica de un número:
Teorema:
(1/n) = 0.2_{2n}
(1/n) = 2·(1/(2n))
Anexo:
0.2_{2} = 2·(1/2) = 1.0_{2}
Teorema:
(1/n) = 0.222..._{2n+1}
(1/n) = 2·sum[k = 0]-[oo][ (2n+1)^{(-k)+(-1)} ] = 2·( 1/((2n+1)+(-1)) )
Anexo:
x = 0.222..._{2n+1}
(2n+1)·x = 2.222...._{2n+1}
2n·x = (2n+1)·x+(-x) = 2
x = (1/n)
Teorema:
(1/(2n)) = 0.1_{2n}
(1/(2n)) = (1/(2n))
Teorema:
(1/(2n)) = 0.111..._{2n+1}
(1/(2n)) = sum[k = 0]-[oo][ (2n+1)^{(-k)+(-1)} ] = ( 1/((2n+1)+(-1)) )
Teorema:
n = 0.2n^{2}_{2n}
n = 2n^{2}·(1/(2n))
Teorema:
n = 0.2n^{2}2n^{2}2n^{2}..._{2n+1}
n = 2n^{2}·sum[k = 0]-[oo][ (2n+1)^{(-k)+(-1)} ] = 2n^{2}·( 1/((2n+1)+(-1)) )
Teorema:
Construcción algebraica de los números enteros:
< n,m > =[R]= < p,q > <==> n+q = m+p
< n,0 > =[R]= < 0,q >
Construcción algebraica de los números racionales:
< n,m > =[R]= < p,q > <==> nq = mp
< n,1 > =[R]= < 1,q >
Teorema:
Construcción algebraica de las potencias enteras:
< x^{n},x^{m} > =[R]= < x^{p},x^{q} > <==> x^{n+q} = x^{m+p}
< x^{n},1 > =[R]= < 1,x^{q} >
Construcción algebraica de las potencias racionales:
< x^{n},x^{m} > =[R]= < x^{p},x^{q} > <==> x^{nq} = x^{mp}
< x^{n},x > =[R]= < x,x^{q} >
Ley:
Se puede saber si saben,
porque no hay cobertura,
fuera de las teorías de las demostraciones,
y si se ve el documento matemático,
saben.
Se puede saber si no saben,
porque hay cobertura,
dentro de las teorías de las demostraciones,
y si no se ve el documento matemático,
no saben.
Teorema:
int[ ( x^{m}+a )^{n} ]d[x] = ( 1/( n·[m:a]+1 ) )·( x^{m}+a )^{n}·x
int[ ( e^{mx}+a )^{n} ]d[x] = ( 1/( n·[m:a] ) )·( e^{mx}+a )^{n}
Demostración:
x^{m}+a = x^{[m:a]}
( x^{m}+a )^{n} = ( x^{[m:a]} )^{n} = x^{n·[m:a]}
x·( x^{m}+a )^{n} = x·x^{n·[m:a]} = x^{n·[m:a]+1}
d_{x}[ x^{n·[m:a]+1} ] = ( n·[m:a]+1 )·x^{n·[m:a]}
e^{mx}+a = ( e^{x} )^{m}+a = ( e^{x} )^{[m:a]} = e^{[m:a]·x}
( e^{mx}+a )^{n} = ( e^{[m:a]·x} )^{n} = e^{n·[m:a]·x}
d_{x}[ e^{n·[m:a]·x} ] = n·[m:a]·e^{n·[m:a]·x}
Teorema:
int[ ( (x^{p}+a)·(x^{q}+b) )^{n} ]d[x] = ?
int[ ( (e^{px}+a)·(e^{qx}+b) )^{n} ]d[x] = ?
Teorema:
Sea F(x,y,z,t) = (1/u)·d_{t}[...]+(1/r)·( int[...]d[x]+int[...]d[y]+int[...]d[z] )
D[ct]-[ F(x,y,z,t) ] = (c/u)+ct·(1/r)·(x+y+z)
Teorema:
Sea F(x,y,z,t) = u·int[...]d[t]+r·( d_{x}[...]+d_{y}[...]+d_{z}[...] )
D[x+y+z]-[ F(x,y,z,t) ] = ut·(x+y+z)+3r
Teorema:
Sea F(x,y,z,u,v) = (1/a)·d_{u}[...]+(1/b)·d_{v}[...]+(1/r)·( int[...]d[x]+int[...]d[y]+int[...]d[z] )
D[u+v]-[ F(x,y,z,u,v) ] = ( (1/a)+(1/b)+(u+v)·(1/r)·(x+y+z) )
Teorema:
Sea F(x,y,z,u,v) = a·int[...]d[u]+b·int[...]d[v]+r·( d_{x}[...]+d_{y}[...]+d_{z}[...] )
D[x+y+z]-[ F(x,y,z,u,v) ] = (au+bv)·(x+y+z)+3r
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