análisis-matemático y computación-de-álgebra-lineal y álgebra-borrosa-y-teoría-de-números y integrales

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

Examen de análisis matemático:

Teorema:

Sea a >] 0 ==>

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)) )

Anexo:

x = 0.2n^{2}2n^{2}2n^{2}..._{2n+1}

(2n+1)·x = 2n^{2}.2n^{2}2n^{2}2n^{2}..._{2n+1}

2n·x = (2n+1)·x+(-x) = 2n^{2}

x = n

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