análisis-matemático-Stolz y constante-de-Euler y residus-y-integral-de-cauchy y evangelio-stronikiano

Axioma:

Stolz es falso con constructor <==> Stolz es verdadero con destructor

Stolz es verdadero con constructor <==> Stolz es falso con destructor

Axioma:

( Si p(oo) = Stolz con constructor ==> ¬p(oo) ) <==> ( Si p(oo) = Stolz con destructor ==> p(oo) )

( Si p(oo) = Stolz con constructor ==> p(oo) ) <==> ( Si p(oo) = Stolz con destructor ==> ¬p(oo) )

Teorema:

Si lim[n = oo][ a_{n} ] = a ==> lim[n = oo][ ( (a_{1}+...+a_{n})/n ) ] = a

Demostración

( a_{n}/( (n+1)+(-n) ) ) = ( a_{oo}/( (oo+1)+(-oo) ) )  = ( a_{oo}/( oo+(-oo) ) ) = ( a_{oo}/1 ) = a

Teorema:

lim[n = oo][ ( ((pn+1)·...(n)...·(pn+n))/(an)! )^{(1/n)} ] = ( (p+1)/a )

lim[n = oo][ ( (an)!/((pn+1)·...(n)...·(pn+n)) )^{(1/n)} ] = ( a/(p+1) )

Demostración:

lim[n = oo][ ln( (p·(n+1)+k)/(pn+k) ) ] = lim[n = oo][ ln( 1+( p/(pn+k) ) ) ] = ln(1) = 0

[Ak][ k€N ==> lim[n = oo][ ( p/(pn+k) ) ] = lim[n = oo][ ( (p/n)/(p+(k/n)) ) ] = 0 ]

lim[n = oo][ ( p/(pn+n) ) ] = lim[n = oo][ ( (p/n)/(p+1) ) ] = 0

Teorema:

lim[n = oo][ ( n!/(an)! )^{(1/n)} ] = (1/a)

lim[n = oo][ ( (an)!/n! )^{(1/n)} ] = a

Teorema-Corolario: [ de Stolz ]

Sea lim[n = oo][ b_{n+1} ] = oo ==> 

Si lim[n = oo][ ( ( a_{n+1}+(-1)·a_{n} )+(-1)·a_{p} )/( b_{n+1}+(-1)·b_{n} ) ] = c ==>

... lim[n = oo][ ( a_{n+1}/b_{n+1} ) ] = c+a_{p}

Teorema:

[An][ n >] x·2^{x} >] x ==> (1+x)^{n} >] 1+n^{x} ]

Demostración:

(1+x)^{n+1} >] (1+n^{x})·(1+x) = 1+x+xn^{x}+n^{x} = 1+x+nxn^{x+(-1)}+n^{x} >] ...

... 1+x+x·2^{x}·xn^{x+(-1)}+n^{x} >] 1+x+(n+1)^{x} >] 1+(n+1)^{x}

Teorema:

lim[n = oo][ ( (1+x)^{n}/n^{x} ) ] = 1

Demostración:

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

[En][ Id(1/n^{x}) = n & n = 1 ]

Se define 1 > s > 0 ==>

Sea n_{0}€N ==>

Se define n > max{x·2^{x},n_{0}} ==>

| ( (1+x)^{n}/n^{x} )+(-1) | = | (1/n^{x})·( (1+x)^{n}+(-1)·n^{x} ) | >] ...

... | (1/n^{x})·( (1+n^{x})+(-1)·n^{x} ) | = (1/n^{x}) = f(1/n^{x}) = n > n_{0} > s

Teorema:

lim[n = oo][ n^{(1/n)} ] = 2

Demostración:

f(1) = n

[En][ Id(1) = n & n = 1 ]

e^{ln(1+(1/n))} = e^{ln(1+( f(1)/n ))} = e^{ln(1+(n/n))} = e^{ln(1+1)} = 2

Teorema:

[Ap][ p > 0 ==> lim[n = oo][ ( ( p+1 )·...(n)...·( p+(1/n) ) )^{(1/n)} ] = (p+1) ]

[Ap][ p > 0 ==> lim[n = oo][ ( 1/( ( p+1 )·...(n)...·( p+(1/n) ) ) )^{(1/n)} ] = ( 1/(p+1) ) ]

Demostración:

f(1) = n

[En][ Id(1) = n & n = 1 ]

e^{ln(p+( 1/(n+1) ))} = e^{ln(p+( f(1)/(n+1) ))} = e^{ln(p+( n/(n+1) ))} = e^{ln(p+1)} = p+1

Teorema:

lim[n = oo][ p^{n} ] = lim[n = oo][ p^{n}·(n!/n!) ] < ...

... lim[n = oo][ ( p+1 )·...(n)...·( p+(1/n) ) ] = lim[n = oo][ (p+1)^{n} ] = lim[n = oo][ n^{p} ] >] oo

Teorema:

lim[n = oo][ ( 2·...(n)...·( 1+(1/n) ) )^{(1/n)} ] = 2 ]

lim[n = oo][ ( 1/( 2·...(n)...·( 1+(1/n) ) ) )^{(1/n)} ] = (1/2) ]

Demostración:

lim[n = oo][ (n+1)^{(1/n)} ] = 2 ]

lim[n = oo][ (n+1) ] = lim[n = oo][ (n/n)·(n+1) ] = lim[n = oo][ n·(1+(1/n)) ] = oo = 2^{oo}

Examen de Stolz:

Teorema:

[Ap][ p€N ==> lim[n = oo][ ( ( 2p+1 )·...(n)...·( 2p+(1/n) ) )^{(1/n)} ] = (2p+1) ]

[Ap][ p€N ==> lim[n = oo][ ( 1/( ( 2p+1 )·...(n)...·( 2p+(1/n) ) ) )^{(1/n)} ] = ( 1/(2p+1) ) ]

Teorema:

lim[n = oo][ ( 1^{(1/k)}+...(n)...+n^{(1/k)} )/(( (k+1)/k )+1)^{n} ] = ( k/(k+1) )

lim[n = oo][ (( (k+1)/k )+1)^{n}/( 1^{(1/k)}+...(n)...+n^{(1/k)} ) ] = ( (k+1)/k )

Demostración:

f(k+1) = 1

[Ek][ Id(k+1) = 1 & k = 0 ]

lim[n = oo][ ( (n+1)^{(1/k)}/(( (k+1)/k )+1)^{n} )·( k/(k+1) ) ] = ...

lim[n = oo][ ( (n+1)^{(1/k)}/(( f(k+1)/k )+1)^{n} )·( k/(k+1) ) ] = ...

lim[n = oo][ ( (n+1)^{(1/k)}/((1/k)+1)^{n} )·( k/(k+1) ) ] = ...

lim[n = oo][ ( (n+1)^{(1/k)}/n^{(1/k)} )·( k/(k+1) ) ] = ( k/(k+1) )

Teorema:

lim[n = oo][ ( 1^{k}+...(n)...+n^{k} )/n^{k+1} ] = ( 1/(k+1) )

lim[n = oo][ n^{k+1}/( 1^{k}+...(n)...+n^{k} ) ] = k+1

Teorema:

lim[n = oo][ 1^{(1/k)}+...(n)...+n^{(1/k)} ] = ( k/(k+1) )·oo^{(1/k)+1} >] oo

lim[n = oo][ 1^{k}+...(n)...+n^{k} ] = ( 1/(k+1) )·oo^{k+1} >] oo

Teorema:

int[x = 0]-[1][ x^{(1/k)} ]d[x] = ...

... lim[n = oo][ ( (1/n)^{(1/k)}+...(n)...+(n/n)^{(1/k)} )·d[x] ] = ( k/(k+1) )

int[x = 0]-[1][ x^{k} ]d[x] = ...

... lim[n = oo][ ( (1/n)^{k}+...(n)...+(n/n)^{k} )·d[x] ] = ( 1/(k+1) )

Teorema:

El exponente infinito no es conmutativo:

( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo}

Demostración:

oo^{kp} = ( oo^{k} )^{p} = ( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo} = (kp+1)^{oo}

oo^{pk} = ( oo^{p} )^{k} = ( (p+1)^{oo} )^{k} = ( (p+1)^{log_{p+1}(pk+1)} )^{oo} = (pk+1)^{oo}

(k+1)^{oo} = ( (k+1)^{log_{k+1}(k+1)} )^{oo}

Teorema:

[Ap][ p > 0 ==> lim[n = oo][ ( n^{p}/e^{an} )^{(1/n)} ] = ( (p+1)/e^{a} ) ]

[Ap][ p > 0 ==> lim[n = oo][ ( e^{an}/n^{p} )^{(1/n)} ] = ( e^{a}/(p+1) ) ]

oo^{p} = ( 2^{oo} )^{p} = ( 2^{log_{2}(p+1)} )^{oo} = (p+1)^{oo}

Demostración: [ por Stolz ]

f(1) = n

[En][ Id(1) = n & n = 1 ]

( u(1) = p & v(p) = 1 )

[Ep][ ( Id(1) = p & Id(p) = 1 ) & p = 1 ]

e^{ln( (1+(1/n))^{p} )} = e^{ln( (u(1)+(1/n))^{v(p)} )} = e^{ln(p+(1/n))} = ...

... e^{ln(p+(f(1)/n))} = e^{ln(p+(n/n))} = e^{ln(p+1)} = p+1

Examen de Stolz:

Teorema:

[Ak][Ap][ ( k > 0 & p > 0 ) ==> lim[n = oo][ ( n^{kp}/e^{an} )^{(1/n)} ] = ( (kp+1)/e^{a} ) ]

[Ak][Ap][ ( k > 0 & p > 0 ) ==> lim[n = oo][ ( e^{an}/n^{kp} )^{(1/n)} ] = ( e^{a}/(kp+1) ) ]

oo^{kp} = ( oo^{k} )^{p} = ( (k+1)^{oo} )^{p} = ( (k+1)^{log_{k+1}(kp+1)} )^{oo} = (kp+1)^{oo}

Teorema:

[Ap][ p > 0 ==> lim[n = oo][ ( ( p+1^{k} )·...(n)...·( p+(1/n)^{k} ) )^{(1/n)} ] = ... 

... ( p^{k}+1 )^{(1/k)} ]

[Ap][ p > 0 ==> lim[n = oo][ ( 1/( ( p+1^{k} )·...(n)...·( p+(1/n)^{k} ) ) )^{(1/n)} ] =  ...

... ( 1/( p^{k}+1 )^{(1/k)} ) ]

Demostración: [ por Stolz ]

f(1) = n

[En][ Id(1) = n & n = 1 ]

g(k) = 1 & u(1) = k & v(1) = (1/k)

[Ek][ Id(k) = 1 & Id(1) = k & Id(1) = (1/k) & k = 1 ]

e^{ln(p+( 1/(n+1) )^{k})} = e^{ln( p+( f(1)/(n+1) )^{g(k)} )} = ...

... e^{ln( p+( n/(n+1) ) )} = e^{ln(p+1)} = e^{ln( (p^{u(1)}+1)^{v(1)} )} = ...

... e^{ln( (p^{k}+1)^{(1/k)} )} = ( p^{k}+1 )^{(1/k)}

Teorema:

p < ( p^{k}+1 )^{(1/k)} [< (p+1)

Demostración:

Raíz k-ésima positiva:

( d_{x}[ x^{k} ] = kx^{k+(-1)} > 0 <==> x > 0 ) 

( x^{k} es creciente <==> x > 0 )

Teorema:

a·oo^{k}+p = a·oo^{k}

Demostración:

a·oo^{k}+p = oo^{k}·( a+p·(1/oo)^{k} ) = oo^{k}·( a+p·0^{k} ) = a·oo^{k}

Teorema:

Si p > q ==> a·oo^{p}+b·oo^{q} = a·oo^{p}

Si p < q ==> a·0^{p}+b·0^{q} = a·0^{p}

Demostración:

Si p > q ==> [Ek][ oo^{p} = oo^{q+k}]

a·oo^{p}+b·oo^{q} = oo^{q}·( a·oo^{k}+b) = a·oo^{q+k} = a·oo^{p}

Si p < q ==> [Ek][ 0^{q} = 0^{p+k}]

a·0^{p}+b·0^{q} = 0^{p}·( a+b·0^{k} ) = a·0^{p}

Teorema:

lim[n = oo][ n+(-n) ] = 1

Demostración:

lim[n = oo][ n+(-n) ] = lim[n = oo][ n·(1+(-1)) ] = oo·0 = 1

Teorema:

[Ak][ k€N ==> lim[n = oo][ (n+p)^{(1/k)}+(-1)·n^{(1/k)} ] = (1/k)·0^{1+(-1)·(1/k)} ]

[Ak][ k€N ==> lim[n = oo][ (n+p)^{k}+(-1)·n^{k} ] = k·oo^{k+(-1)} ]

Demostración:

x^{k}+(-1)·y^{k} = (x+(-y))·( x^{k+(-1)}+x^{k+(-2)}·y+...+x·y^{k+(-2)}+y^{k+(-1)} )

Teorema:

[Ak][ k€N ==> lim[n = oo][ a_{k}·n^{k}+...(k)...+a_{0} ] = a_{k}·oo^{k} ]

Demostración:

Sea s > 0 ==>

Se define n_{0} > (k/s)

lim[n = oo][ | ( (a_{k}·n^{k}+...(k)...+a_{0})/(a_{k}·oo^{k}) )+(-1) | ] = ...

... lim[n = oo][ | (1/oo)^{k}·(1/a_{k})·( a_{k}·n^{k}+(-1)·a_{k}·oo^{k} ) | ] = ...

... |(k/oo)| = 0 [< (k/n) < (k/n_{0}) < s

Comprobación del límite:

lim[n = oo][ | ( (a_{k}·n^{k}+...(k)...+a_{0})/(a_{k}·oo^{k}) )+(-1) | ] = ...

... lim[n = oo][ | ( a_{k}·n^{k}/a_{k}·oo^{k} )+(-1) | ] = 0 [< (1/n) < (1/n_{0}) < s

Teorema:

lim[n = oo][ a_{k}·(1/n)^{k}+...(k)...+a_{0} ] = a_{0}

Demostración:

Sea s > 0 ==>

Se define n_{0} > (1/s)

lim[n = oo][ | ( a_{k}·(1/n)^{k}+...(k)...+a_{0} )+(-1)·a_{0} | = | a_{0}+(-1)·a_{0} | = 0 [< ...

... (1/n) < (1/n_{0}) < s

Teorema:

lim[n = oo][ (an^{2}+bn+c)/(an+b) ] = oo

lim[n = oo][ (an+b)/(an^{2}+bn+c) ] = 0

Definición:

[Ax][Ay][Es][ ( xn^{y+s} <=[R]=> (y+s)·n^{x} ) <==> ( (y+s)·n^{x} <=[R]=> xn^{y+s} ) ]

Teorema:

Sea s = x+(-y) ==>

xn^{x} <=[R]=> xn^{x}

Teorema:

[Eu][ xn^{y+u} <=[R]=> (y+u)·n^{x} ] <==> [Ev][ yn^{x+v} <=[R]=> (x+v)·n^{y} ]

Demostración:

Se define u = 0 & v = 0

Teorema:

Si ( [Eu][ xn^{y+u} <=[R]=> (y+u)·n^{x} ] & [Ev][ yn^{z+v} <=[R]=> (z+v)·n^{y} ] ) ==> ...

... [Ew][ xn^{z+w} <=[R]=> (z+w)·n^{x} ]

Demostración:

Se define u = 0 & v = x+(-z) & w = y+(-z)

( p(x) & ( p(x) <==> q(y) ) ) <==> ( p(x) & q(y) )

( q(y) & q(y) ) <==> q(y)

Teorema:

lim[n = oo][ (3n^{4}+4n^{3}+n^{2}+2n)/(2n^{3}+3n^{2}+1) ] = (3/2)·oo

lim[n = oo][ (2n^{3}+3n^{2}+1)/(3n^{4}+4n^{3}+n^{2}+2n) ] = (2/3)·0

Teorema:

lim[n = oo][ (2n^{3}+3n^{2}+1)/(n^{3}+3n+2) ] = 2

lim[n = oo][ (n^{3}+3n+2)/(2n^{3}+3n^{2}+1) ] = (1/2)

Teorema:

Si a_{n} es de Cauchy ==> a_{n} está acotada

Demostración:

Sea s > 0 ==>

Sea m = n_{0}+1 ==>

|a_{n}| [< |a_{n}+(-1)·a_{m}|+|a_{m}| = |a_{n}+(-1)·a_{n_{0}+1}|+|a_{n_{0}+1}|

Se define M = max{a_{1},...,a_{n_{0}},s+|a_{n_{0}+1}|}

Sea n€N ==>

|a_{n}| [< M

Teorema:

Si a_{n} > 0 ==> lim[n = oo][ a_{n}] >] 0

Si a_{n} < 0 ==> lim[n = oo][ a_{n}] [< 0

Demostración:

Sea s > 0 ==>

a_{n} > 0

a_{n}+(-a) > (-a)

s > a_{n}+(-a) > (-a)

s > (-a)

0 >] (-a)

a >] 0

Ley: [ de caminar sobre las aguas anulando la gravedad de la Tierra ]

[Ef][ < f: A ---> B & x --> f(x) = qg > & m·d_{tt}^{2}[y(t)] = (-1)·qg+f(x) ]

Ley: [ de levantar las aguas anulando la gravedad de la Luna ]

[Ef][ < f: A ---> B & x --> f(x) = (-1)·pg > & m·d_{tt}^{2}[y(t)] = pg+f(x) ]

Ley: [ de volar anulando la gravedad de la Tierra ]

[Ef][ < f: A ---> B & x --> f(x) = qg+F·at > & m·d_{tt}^{2}[y(t)] = (-1)·qg+f(x) ]

Ley: [ de bucear anulando la gravedad de la Luna ]

[Ef][ < f: A ---> B & x --> f(x) = (-1)·pg+(-1)·F·at > & m·d_{tt}^{2}[y(t)] = pg+f(x) ]

Constante de Euler:

Teorema:

lim[n = oo][ ln(n)+(-1)·( 1+...(n)...+(1/n) ) ] = ln(2)

oo·ln(2)+(-1)·oo·ln(2) = ln(2)

Arte: [ de Euler ]

[En][ sum[k = 1]-[n][ ln(1+k)+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+k)+(-1)·(1/k) ] = ln(2)

[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+(1/k))+(-1)·(1/k) ] = ln(2)

Arte:

[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(n/k)·(1/2)^{k+(-1)} ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(oo/k)·(1/2)^{k+(-1)} ] = ln(2)

[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(n/k)·(1/2)^{(1/k)+(-1)} ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(oo/k)·(1/2)^{(1/k)+(-1)} ] = ln(2)

Arte:

[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(n/k)·( 2k+(-1) ) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(oo/k)·( 2k+(-1) ) ] = ln(2)

[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(1/k)·( 2·(1/k)+(-1) ) ] ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(oo/k)·( 2·(1/k)+(-1) ) ] ] = ln(2)

Arte:

[En][ sum[k = 1]-[n][ ln(1+k)+(-1)^{k}·(1/k)·[ n // k ] ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+k)+(-1)^{k}·(1/k)·[ oo // k ] ] = ln(2)

[En][ sum[k = 1]-[n][ ln(1+(1/k))+(-1)^{k}·(1/k)·[ n // (1/k) ] ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ ln(1+(1/k))+(-1)^{k}·(1/k)·[ oo // (1/k) ] ] = ln(2)

Arte:

[En][ lim[s = 1][ sum[k = 1]-[n][ int[ ( 1/(s+1)^{k} ) ]d[s]+(-1)·(1/k)^{s} ) ] = ln(2)+(-1)·(1/n) ]

lim[s = 1][ sum[k = 1]-[oo][ int[ ( 1/(s+1)^{k} ) ]d[s]+(-1)·(1/k)^{s} ) ] ] = ln(2)

[En][ lim[s = 1][ sum[k = 1]-[n][ int[ ( 1/(s+1)^{(1/k)} ) ]d[s]+(-1)·(1/k)^{s} ) ] = ln(2)+(-1)·(1/n) ]

lim[s = 1][ sum[k = 1]-[oo][ int[ ( 1/(s+1)^{(1/k)} ) ]d[s]+(-1)·(1/k)^{s} ) ] ] = ln(2)

Arte:

[En][ sum[k = 1]-[n][ int[x = 1]-[2][ (1/2)·( k·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = 1]-[2][ (1/2)·( k·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)

[En][ sum[k = 1]-[n][ int[x = 1]-[2][ (1/2)·( (1/k)·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = 1]-[2][ (1/2)·( (1/k)·ln(x)+1 ) ]d[x]+(-1)·(1/k) ] = ln(2)

Arte:

[En][ sum[k = 1]-[n][ ...

... int[x = e]-[e^{2}][ ( 1/( k·ln(x) ) )·(1/x)^{k} ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e]-[e^{2}][ ( 1/( k·ln(x) ) )·(1/x)^{k} ]d[x]+(-1)·(1/k) ] = ln(2)

[En][ sum[k = 1]-[n][ ...

... int[x = e]-[e^{2}][ ( 1/( (1/k)·ln(x) ) )·(1/x)^{(1/k)} ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e]-[e^{2}][ ( 1/( (1/k)·ln(x) ) )·(1/x)^{(1/k)} ]d[x]+(-1)·(1/k) ] = ln(2)

Arte:

Sea s != (-1) ==>

[En][ sum[k = 1]-[n][ ...

... int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...

... (s+1)·( k·ln(x) )^{s}·( 1/x^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...

... (s+1)·( k·ln(x) )^{s}·( 1/x^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)

[En][ sum[k = 1]-[n][ ...

... int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...

... (s+1)·( (1/k)·ln(x) )^{s}·( 1/x^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{( 1/(s+1) )}}]-[e^{( ln(2) )^{( 1/(s+1) )}}][ ...

... (s+1)·( (1/k)·ln(x) )^{s}·( 1/x^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)

Arte:

Sea s != 0 ==>

[En][ sum[k = 1]-[n][ ...

... int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...

... e^{x^{k}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...

... ( k·ln(x) )^{s}+s·( k·ln(x) )^{s+(-1)}·(1/x)^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...

... e^{x^{k}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...

... ( k·ln(x) )^{s}+s·( k·ln(x) )^{s+(-1)}·(1/x)^{k} ) ]d[x]+(-1)·(1/k) ] = ln(2)

[En][ sum[k = 1]-[n][ ...

... int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...

... e^{x^{(1/k)}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...

... ( (1/k)·ln(x) )^{s}+s·( (1/k)·ln(x) )^{s+(-1)}·(1/x)^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)+(-1)·(1/n) ]

sum[k = 1]-[oo][ int[x = e^{( ln(1) )^{(1/s)}}]-[e^{( ln(2) )^{(1/s)}}][ ...

... e^{x^{(1/k)}+(-1)·e^{( ln(2) )^{(1/s)}}}·( ...

... ( (1/k)·ln(x) )^{s}+s·( (1/k)·ln(x) )^{s+(-1)}·(1/x)^{(1/k)} ) ]d[x]+(-1)·(1/k) ] = ln(2)

Teorema:

Si se explica un teorema entonces se está dentro de los métodos de demostración

porque sinó no sirve para nada explicar, porque no hay amor, fuera de los métodos de demostración.

Demostración:

Se explica un teorema y no se está dentro de los métodos de demostración

aunque no-obstante sirve para algo explicar, porque hay amor, dentro de los métodos de demostración.

Teorema:

lim[s = 0][ int-[m]-[z = se^{ix}+a][ ( f(z)/(z+(-a)) ) ]d[z] ] = (2pi·i)·m·f(a)

lim[s = 0][ int-[m]-[z = se^{ix}+(-a)][ ( f(z)/(z+a) ) ]d[z] ] = (2pi·i)·m·f(-a)

Teorema:

lim[s = 0][ int-[m]-[z = se^{ix}+a][ ( f(z)/( (z+(-a))·(z+a) ) ) ]d[z] ] = (pi·i)·m·(f(a)/a)

lim[s = 0][ int-[m]-[z = se^{ix}+(-a)][ ( f(z)/( (z+(-a))·(z+a) ) ) ]d[z] ] = (pi·i)·m·( f(-a)/(-a) )

Teorema:

lim[s = 0][ int-[m]-[z = ( se^{ix}+a )^{(1/p)}][ ( f(z)/( (z^{p}+(-a))·(z^{q}+a) ) ) ]d[z] ] = ...

... (2pi·i)·m·( f(a)/( a^{(q/p)}+a ) )·(1/p)·a^{(1/p)+(-1)}

lim[s = 0][ int-[m]-[z = ( se^{ix}+(-a) )^{(1/q)}][ ( f(z)/( (z^{p}+(-a))·(z^{q}+a) ) ) ]d[z] ] = ...

... (2pi·i)·m·( f(-a)/( (-a)^{(p/q)}+(-a) ) )·(1/q)·(-a)^{(1/q)+(-1)}

Teorema:

lim[s = 0][ int-[m]-[z = (1/p)·ln( se^{ix}+a )][ ( f(z)/( (e^{pz}+(-a))·(e^{qz}+a) ) ) ]d[z] ] = ...

... (2pi·i)·m·( f(a)/( a^{(q/p)}+a ) )·(1/p)·(1/a)

lim[s = 0][ int-[m]-[z = (1/q)·ln( se^{ix}+(-a) )][ ( f(z)/( (e^{pz}+(-a))·(e^{qz}+a) ) ) ]d[z] ] = ...

... (2pi·i)·m·( f(-a)/( (-a)^{(p/q)}+(-a) ) )·(1/q)·(1/(-a))

Uzkatzen-ten-dut-zû-tek a la gentotzak,

parlatzi-ten-dut-zare-dut en Euskera-Bascotzok parlatzi-koak.

Veurtu-ten-dut-zû-tek a la gentotzak,

escrivitzi-ten-dut-zare-dut en Euskera-Bascotzok parlatzi-koak.

Sepjjakin-ten-dut-zû-tek per les meuotzaks orelli-koaks,

que entendertu-ten-dut-zû-tek el Euskera-Bascotzok parlatzi-koak.

Sepjjakin-ten-dut-zû-tek per els meuotzoks ur-ulli-koaks,

que llezkatzen-ten-dut-zû-tek el Euskera-Bascotzok parlatzi-koak.

Juan:

Hemos sorprendido a esta mujer cometiendo adulterio.

Tú que dices maestro?

Si la mujer está libre de pecado,

que le tiren la primera piedra,

porque tiene condenación,

en no ser pecadora.

Hemos sorprendido a esta hombre cometiendo adulterio.

Tú que dices maestra?

Si el hombre no está libre de pecado,

que no le tiren la primera piedra, 

porque no tiene condenación,

en ser pecador.

Carta de Jûanat-hád a los cristianos stronikianos:

Un infiel puede cometer adulterio,

porque no tiene que nacer,

y no está libre de pecado,

porque es un pecador,

y siempre puede haber porno de infieles.

Un fiel no puede cometer adulterio,

porque tiene que nacer,

y está libre de pecado,

porque no es un pecador,

y nunca puede haber porno de fieles.

Sabéis que no nacen los que han cometido adulterio

y entonces también si eran infieles los que han muerto,

es irrelevante,

y alegre en todo-algún caso.

Quizás sabéis que no nacen los que han cometido adulterio

pero si eran fieles los que han muerto,

es relevante,

y triste en todo caso.

No se puede odiar a un fiel,

no cometiendo adulterio,

no follando estando divorciado.

Se puede odiar a un infiel,

cometiendo adulterio,

follando estando divorciado.

Definición: [ de la suma de Riemman ]

[As][ s > 0 ==> [En_{0}][An][ n > n_{0} ==> ...

... | sum[k = 1]-[k = n][ f(k/n)·(1/n) ]+(-1)·int[x = 0]-[1][ f(x) ]d[x] | < s ] ]

Teorema:

int[x = 0]-[1][ e^{x} ]d[x] = ...

... lim[n = oo][ ( e^{(1/n)}+...(n)...+e^{(n/n)} )·( e^{(1/n)}+(-1) ) ] = e+(-1)

Demostración:

lim[n = oo][ e^{(1/n)}+(-1) ] = lim[n = oo][ (1/n)+sum[k = 1]-[oo][ (1/(k+1)!)·(1/n)^{k+1} ] ] = ...

... lim[n = oo][ (1/n) ] = 0

lim[n = oo][ ( e^{(1/n)}+...(n)...+e^{(n/n)} ) ] = ...

... lim[n = oo][ ( (e^{(n/n)+(1/n)}+(-1)·e^{(1/n)})/(e^{(1/n)}+(-1)) ) ]

Teorema:

int[x = 0]-[1][ e^{(-x)} ]d[x] = ...

... lim[n = oo][ ( e^{(-1)·(1/n)}+...(n)...+e^{(-1)·(n/n)} )·( 1+(-1)·e^{(-1)·(1/n)} ) ] = 1+(-1)·(1/e)

Demostración:

lim[n = oo][ 1+(-1)·e^{(-1)·(1/n)} ] = ...

... lim[n = oo][ (1/n)+sum[k = 1]-[oo][ (-1)^{k}·(1/(k+1)!)·(1/n)^{k+1} ] ] = ...

... lim[n = oo][ (1/n) ] = 0

lim[n = oo][ ( e^{(-1)·(1/n)}+...(n)...+e^{(-1)·(n/n)} ) ] = ...

... lim[n = oo][ ( (e^{(-1)·(1/n)}+(-1)·e^{(-1)·( (n/n)+(1/n) )})/(1+(-1)·e^{(-1)·(1/n)}) ) ]

Teorema:

int[x = 0]-[1][ ln(x+1) ]d[x] = ...

... lim[n = oo][ ( ln((1/n)+1)+...(n)...+ln((n/n)+1) )·(1/n) ] = 2·ln(2)+(-1)

Demostración: [ por Stolz ]

f(1) = ( (2n+(-1))/n )

[En][ Id(1) = ( (2n+(-1))/n ) & n = 1 ] 

g(0) = ( (1+(-n))/n )

[En][ Id(0) = ( (1+(-n))/n ) & n = 1 ]

lim[n = oo][ ln( ( (n+1)/(n+1) )+1 ) ] = lim[n = oo][ f(1)·ln( ( (n+1)/(n+1) )+1 )+g(0) ] = ...

... lim[n = oo][ ( ( (2n+(-1))/n ) )·ln( ( (n+1)/(n+1) )+1 )+( (1+(-n))/n ) ] = 2·ln(2)+(-1)

Teorema:

int[x = 0]-[1][ cos(x) ]d[x] = ...

... lim[n = oo][ ( cos(1/n)+...(n)...+cos(n/n) )·(1/n) ] = sin(1)

Demostración: [ por Stolz ]

cos( x+(-1)·(pi/2) ) = cos(x)·cos( (-1)·(pi/2) )+(-1)·sin(x)·sin( (-1)·(pi/2) ) = sin(x)

f(0) = x·(pi/2)

g(x) = (-1)

lim[n = oo][ cos(k/(n+1))+(-1)·cos(k/n) ] = cos(0)+(-1)·cos(0) = 1+(-1) = 0

lim[n = oo][ cos((n+(-k))/(n+1))+(-1)·cos((n+(-k))/n) ] = cos(1)+(-1)·cos(1) = 0

lim[n = oo][ cos((n+1)/(n+1)) ] = lim[n = oo][ cos( (n+1)/(n+1)+f(0) ) ] = ...

... lim[n = oo][ cos( (n+1)/(n+1)+x·(pi/2) ) ] = lim[n = oo][ cos( (n+1)/(n+1)+g(x)·(pi/2) ) ] = ...

... lim[n = oo][ cos( (n+1)/(n+1)+(-1)·(pi/2) ) ] = lim[n = oo][ sin( (n+1)/(n+1) ) ] = sin(1)

Examen de sumas de Riemman:

Teorema:

int[x = 0]-[1][ sin(x) ]d[x] = ...

... lim[n = oo][ ( sin(1/n)+...(n)...+sin(n/n) )·(1/n) ] = 1+(-1)·cos(1)

Demostración: [ por Stolz ]

sin( x+(-1)·(pi/2) ) = sin(x)·cos( (-1)·(pi/2) )+cos(x)·sin( (-1)·(pi/2) ) = (-1)·cos(x)

f(0) = x·(pi/2)

g(x) = (-1)

h(0) = ( (n+(-1))/n )

Teorema:

sinh(ix+iy) = ( sin(x+y)/i ) = (1/i)·( sin(x)·cos(y)+cos(x)·sin(y) ) = sinh(ix)·cosh(iy)+cosh(ix)·sinh(iy)

cosh(ix+iy) = cos(x+y) = ( cos(x)·cos(y)+(-1)·sin(x)·sin(y) ) = cosh(ix)·cosh(iy)+sinh(ix)·sinh(iy)

Teorema:

int[x = 0]-[1][ cosh(x) ]d[x] = ...

... lim[n = oo][ ( cosh(1/n)+...(n)...+cosh(n/n) )·(1/n) ] = sinh(1)

Demostración: [ por Stolz ]

cosh( x+i·(pi/2) ) = cosh(x)·cosh( i·(pi/2) )+sinh(x)·sinh( i·(pi/2) ) = i·sinh(x)

f(0) = x·(pi/2)

g(x) = i

h(i) = (1+(-i))·( (n+(-1))/n )+i

ecuaciones-diferenciales-lineales-afines y ley y análisis-matemático y teoria-de-conjuntos

Producto de gauge y álgebra lineal afín:

Definición de espacio de Gauge-Afín = A:

A = p+F & F es un espacio vectorial & p = ( x_{1}+...+x_{n} ) +[o]+ ( y_{1}+...+y_{n} ).

Teorema: [ de espacio Afín ]

A = p+F = < u,v >+Gen( < a,b >,< c,d> ) = ( < u,v >+Gen( < c,d > ) )+Gen( < a,b > ) = q+G

Teorema:

x^{2}+(-1)·(a+b)·x+ab = ( (y+c) +[o]+ (y+c) )·1 = 0

( y = ic || y = (-i)·c )

u·< 0,0,ic >+v·< a,b,0 > = v·< a,b,0 >+(-u)·< 0,0,(-i)·c >

Teorema:

x^{2}+(-1)·(a+b)·x+ab = (yx+c) +[o]+ (yx+c) = 0

( y = i·(c/x) || y = (-i)·(c/x) )

u·< 0,0,i·(c/x) >+v·< a,b,0 > = v·< a,b,0 >+(-u)·< 0,0,(-i)·(c/x) >

Teorema:

x^{2}+(-1)·(a+b)·x+ab = ( (p+c) +[o]+ (d+q) )·1 = 0

( p = c & q = (-d) ) || ( p = (-c) & q = d ) 

u·< 0,0,0,c,(-d) >+v·< a,b,0 > = v·< a,b,0 >+(-u)·< 0,0,0,(-c),d >

Teorema:

d_{x}[...]^{2}+(-1)·( f(x)+g(x) )·d_{x}[...]+( f(x)·g(x) ) = ...

... ( ( h(x)+s(x) ) +[o]+ ( d_{x}[h(x)]+(-1) ) )·1 = 0

h(x) = ( 2·int[s(x)]d[x] )^{(1/2)}

u·< 0,0,( 2·int[s(x)]d[x] )^{(1/2)} >+v·< int[f(x)]d[x],int[g(x)]d[x],0 > = ...

... v·< int[f(x)]d[x],int[g(x)]d[x],0 >+(-u)·< 0,0,(-1)·( 2·int[s(x)]d[x] )^{(1/2)} >

Teorema:

( int[...]d[x] )^{2}+(-1)·( f(x)+g(x) )·int[...]d[x]+( f(x)·g(x) ) = ...

... ( ( h(x)+s(x) ) +[o]+ ( int[h(x)]d[x]+(-1) ) )·1 = 0

h(x) = s(x)·( 2·int[s(x)]d[x] )^{(-1)·(1/2)}

u·< 0,0,s(x)·( 2·int[s(x)]d[x] )^{(-1)·(1/2)} >+v·< d_{x}[f(x)],d_{x}[g(x)],0 > = ...

... v·< d_{x}[f(x)],d_{x}[g(x)],0 >+(-u)·< 0,0,(-1)·s(x)·( 2·int[s(x)]d[x] )^{(-1)·(1/2)} >

Teorema:

d_{xx}^{2}[...]+(-1)·(p+q)·d_{x}^{1}[...]+pq·d_{x}^{0}[...] = ...

... ( d_{x}^{1}[,,,]+(-n) ) +[o]+ ( 1+d_{x}^{0}[,,,] ) = 0

[,,,] = e^{nx}

u·< 0,0,e^{nx} >+v·< e^{px},e^{qx},0 > = ...

... v·< e^{px},e^{qx},0 >+(-u)·< 0,0,(-1)·e^{nx} >

Teorema:

int-int[...]d[x]d[x]+(-1)·(p+q)·int[...]d[x]+pq·int-[0]-[...]d[x] = ...

... ( int[,,,]d[x]+(-n) ) +[o]+ ( 1+int-[0]-[,,,]d[x] ) = 0

[,,,] = e^{(x/n)}

u·< 0,0,e^{(x/n)} >+v·< e^{(x/p)},e^{(x/q)},0 > = ...

... v·< e^{(x/p)},e^{(x/q)},0 >+(-u)·< 0,0,(-1)·e^{(x/n)} >

Teorema:

d_{x}[...]^{2}+(-1)·( f(x)+g(x) )·d_{x}[...]+( f(x)·g(x) ) = ...

... ( ( h(x)+s(x) ) +[o]+ ( d_{x}[h(x)]^{2}+(-1) ) )·1 = 0

h(x) = ( (3/2)·int[ ( s(x) )^{(1/2)} ]d[x] )^{(2/3)}

u·< 0,0,h(x) >+v·< int[f(x)]d[x],int[g(x)]d[x],0 > = ...

... v·< int[f(x)]d[x],int[g(x)]d[x],0 >+(-w)·< 0,0,(-1)·e^{(1/3)·pi·i}·h(x) >

w = u·( 1/e^{(1/3)·pi·i} )

Teorema:

( int[...]d[x] )^{2}+(-1)·( f(x)+g(x) )·int[...]d[x]+( f(x)·g(x) ) = ...

... ( ( h(x)+s(x) ) +[o]+ ( ( int[h(x)]d[x] )^{2}+(-1) ) )·1 = 0

h(x) = s(x)·( 3·int[ s(x) ]d[x] )^{(-1)·(2/3)}

u·< 0,0,h(x) >+v·< d_{x}[f(x)],d_{x}[g(x)],0 > = ...

... v·< d_{x}[f(x)],d_{x}[g(x)],0 >+(-w)·< 0,0,(-1)·e^{(1/3)·pi·i}·h(x) >

w = u·( 1/e^{(1/3)·pi·i} )

Teorema:

d_{xx}^{2}[...]+( f(x)+g(x) )·d_{x}^{1}[...]+( f(x)·g(x)+h(x) )·d_{x}^{0}[...] = ...

... ( ( h(x)+(-1)·d_{x}[f(x)] )+( h(x)+(-1)·d_{x}[g(x)] ) ) +[o]+ d_{x}^{0}[...]

y(x) = e^{(-1)·int[f(x)]d[x]}+e^{(-1)·int[g(x)]d[x]}

Teorema:

d_{xx}^{2}[...]+2·int[h(x)]d[x]·d_{x}^{1}[...]+( ( int[h(x)]d[x] )^{2}+h(x) )·d_{x}^{0}[...] = 0

y(x) = 2e^{(-1)·int-int[h(x)]d[x]d[x]} || y(x) = (-2)·e^{(-1)·int-int[h(x)]d[x]d[x]}

Demostración:

f(x) = int[h(x)]d[x] & g(x) = int[h(x)]d[x]

Teorema:

d_{xx}^{2}[...]+2·int[h(x)]d[x]·d_{x}^{1}[...]+( ( int[h(x)]d[x] )^{2}+h(x) )·d_{x}^{0}[...] = ...

... ( ( d_{x}^{1}[,,,]+(-1)·s(x) ) +[o]+ ( 1+d_{x}^{0}[,,,] ) ) = 0

h(x) = e^{int[s(x)]d[x]}

u·< 0,e^{int[s(x)]d[x]} >+v·< 2e^{(-1)·int-int[h(x)]d[x]d[x]},0 > = ...

... (-v)·< (-2)·e^{(-1)·int-int[h(x)]d[x]d[x]},0 >+(-u)·< 0,(-1)·e^{int[s(x)]d[x]} >

 

Teorema:

d_{xx}^{2}[...]+2·(1/x)·d_{x}^{1}[...] = ( ( d_{x}^{1}[,,,]+(-1)·(1/x) ) +[o]+ ( 1+d_{x}^{0}[,,,] ) ) = 0

y(x) = 2·(1/x) || y(x) = (-2)·(1/x)

h(x) = x || h(x) = (-x)

v·< 2·(1/x),0 >+u·< 0,x > = (-v)·< (-2)·(1/x),0 >+(-u)·< 0,(-x) >

Mientras el psiquiatra y el del banco no honran al Hijo,

es imposible morir,

porque no me pueden sacar el amor,

no creyendo en condenación,

donde no existe la reacción al bien.

Cuando el psiquiatra y el del banco honren al Hijo,

será posible morir,

porque me podrán sacar el amor,

creyendo en condenación,

donde existe la reacción al bien.

Teorema:

Sean R & S dos relaciones ==> ...

... Si S es reflexiva ==> R o S [<< R

... Si R es reflexiva ==> R o S [<< S

Demostración:

Sea < x,y > € R o S ==>

[Ez][ < x,z > € S & < z,y > € R ] 

[ (MP) Definición de R o S ]

Sea z = x ==>

[ (MP) S es reflexiva ]

< x,x > € S & < x,y > € R 

[ (MP) Si x = z ==> f(x) = f(z) ]

< x,y > € R 

[ (MP) Si ( p(x) & q(y) ) ==> q(y) ]

R o S [<< R 

[ (MP) Definición de A [<< B ]

Teorema:

Sean R & S dos relaciones ==> ...

... Si ( S [<< R & R es transitiva ) ==> R o S [<< R

... Si ( R [<< S & S es transitiva ) ==> R o S [<< S

Demostración:

Sea < x,y > € R o S ==>

[Ez][ < x,z > € S & < z,y > € R ] 

[ (MP) Definición de R o S ]

< x,z > € R & < z,y > € R 

[ (MP) S [<< R ]

< x,y > € R 

[ (MP) R es transitiva ]

R o S [<< R 

[ (MP) Definición de A [<< B ]

Teorema:

Sean R & S & T tres relaciones ==> ( Si S = T ==> R o S = R o T )

Demostración:

< x,y > € R o S <==>

[Ez][ < x,z > € S & < z,y > € R ]

[ (MP) Definición de R o S ]

[Ez][ < x,z > € T & < z,y > € R ]

[ (MP) S = T ]

< x,y > € R o T

[ (MP) Definición de R o S ]

R o S = R o T

[ (MP) Definición de A = B ]

Leyes de Medicina:

Ley: [ de la Luz ]

Honrarás al padre y a la madre.

Es legal el análisis de orina.

Ley: [ del Caos ]

Des-honrarás a la madre y al padre.

Es legal el análisis de sangre.

Ley: [ de la Luz ]

Honrarás a la madre y no al padre,

porque no es y no tiene Padre.

Es legal: 

Pinchar medicación:

des-honrando al padre pinchando y honrando a la madre no sacando nada del cuerpo.

Pinchar un edema y sacar la pus:

des-honrando a la padre pinchando y honrando a la madre sacando la pus,

porque cagando no se des-honra a la madre.

Operar con un robot de cuatro luces y curar el órgano enfermo:

des-honrando a la padre pinchando y honrando a la madre no sacando nada del cuerpo.

Ley: [ del Caos ]

Des-honrarás al padre y no a la madre,

porque no es y no tiene Madre.

Es legal: 

Pinchar medicación:

des-honrando al padre pinchando y honrando a la madre no sacando nada del cuerpo.

Pinchar un edema y sacar la pus:

des-honrando a la padre pinchando y honrando a la madre sacando la pus,

porque cagando no se des-honra a la madre.

Operar con un robot de cuatro luces y curar el órgano enfermo:

des-honrando a la padre pinchando y honrando a la madre no sacando nada del cuerpo.

Medicación:

Espectroscopia:

( Blanco @ Negro ) & ( Rojo @ Verde ) & ( Azul @ Naranja ) & ( Amarillo @ Violeta )

Dualogía Química:

Ley:

1+7 = 8 <==> Li @ Cl <==>  ( < 10 > ) @ ( < 10 >,< 10,10,10 >,< 10,10,10 > )

2+6 = 8 <==> Be @ Ne <==> ( < 10 >,< 10 > ) @ ( < 11 >,< 10,10,10 >,< 10,10,10 > )

3+5 = 8 <==> Br @ N <==> ( < 10,10,10 > ) @ ( < 10 >,< 10,10,10 >,< 10 > )

4+4 = 8 <==> C @ Kg <==> ( < 10 >,< 10,10,10 > ) @ ( < 11 >,< 10,10,10 >,< 10 > )

Ley:

[Ey][ x+y = 8 & 8 = y+z ] <==> x = z

Deducción:

Se define y€N & 8+(-y) = x = z

Ley:

Si se menja,

se vive, 

porque funcionan el órganos del cuerpo

porque es la energía dual química de los elementos de la tabla periódica

lo que usan los órganos del cuerpo para funcionar.

Si no se menja,

se muere, 

porque no funcionan el órganos del cuerpo

aunque quizás es la energía dual química de los elementos de la tabla periódica

lo que usan los órganos del cuerpo para funcionar.

Teorema:

F(x,y) = ( < 1,14 >,< 1,a^{2}+(-2) > ) o < x,y > = < 10,a+(-4) >

a = 4

x = 10+(-14)·y

a = (-4)

y = (-8)·oo & x = 10+112·oo

( a != 4 & a != (-4) ) 

y = ( 1/(a+4) ) & x = (-1)·( 14/(a+4) )+10

Definición:

lim[n = oo][ ( P(n)/Q(n) ) ] = ?

Teorema:

lim[n = oo][ ( (n^{k}+p)/(n^{k}+q) ) ] = 1

Demostración:

Sea s > 0 ==>

Se define n_{0}€N & n_{0} > (1/s)·|p+(-q)|+(-q) ==>

Sea n€N & n > n_{0} ==>

n^{k}+q > n+q > n_{0}+q > 0

| ( (n^{k}+p)/(n^{k}+q) )+(-1) | = | ( (p+(-q))/(n^{k}+q) ) | = |p+(-q)|·( 1/(n^{k}+q) ) < ...

... |p+(-q)|·( 1/(n+q) ) < |p+(-q)|·( 1/(n_{0}+q) ) < s

Teorema:

lim[n = oo][ ( (an^{k}+u)/(bn^{k}+v) ) ] = (a/b)

Demostración:

lim[n = oo][ ( (an^{k}+u)/(bn^{k}+v) ) ] = lim[n = oo][ (a/b)·( (n^{k}+(u/a))/(n^{k}+(v/b)) ) ] = ...

... (a/b)·lim[n = oo][ ( (n^{k}+p)/(n^{k}+q) ) ] = (a/b)

Teorema:

lim[n = oo][ (1+...(n)...+n)/n^{k} ] = (1/2)·0^{k+(-2)}

Teorema:

lim[n = oo][ ( ln(n)/n ) ] = ln(2)

Demostración:

(1+x)^{n} >] 1+nx

n·ln(1+x) >] ln(1+nx)

x = 1

n·ln(2) >] ln(n+1)

| ( ln(n)/n )+(-1)·ln(2) | < | ( (ln(n+1)+(-1)·ln(2)·n )/n) | < | (n/n)·( ln(2)+(-1)·ln(2) )| < ...

... (1/n) < (1/n_{0}) < s

Teorema:

ln(oo) = oo·ln(2) < oo

Teorema:

oo = 2^{oo}

Demostración:

Se define 1 >] s > 0 ==>

Sea n_{0}€N ==>

Se define n > n_{0} ==>

| (2^{n}/n)+(-1) | = | ( (2^{n}+(-n))/n ) | > | (1/n)·( (n+1)+(-n) )| = (1/n) = f(1/n) = n > n_{0} >] s 

[En][ Id(1/n) = n & n = 1 ]

[An][ Id( (k+(-1))+(1/n) ) != n ]

Teorema:

lim[n = oo][ n^{(1/n)} ] = 2

Demostración:

oo^{(1/oo)} = ( 2^{oo} )^{(1/oo)} = 2^{(oo/oo)} = 2^{1} = 2

ln(oo^{(1/oo)}) = ln(l)

(ln(oo)/oo) = ln(l)

ln(2) = ln(l)

Teorema:

Si f(x) = x ==> f(x) es continua

Demostración:

Sea s > 0 ==>

Se define s > d > 0 ==>

Sea x€R & |x+(-c)| < d ==>

| f(x)+(-1)·f(c) | = |x+(-c)| < d < s

Teorema:

Si 0 [< f(x) [< x ==> f(x) es continua

Demostración:

Se define s > 0 & [En][ (1/n) >] s ] ==>

Sea d > 0 ==>

De define x€R & |x+(-c)| < d ==>

| f(x)+(-1)·f(c) | >] |x+(-c)| = 0 = f(0) = (1/n) >] s

[En][ Id(0) = (1/n) & n = oo ]

Teorema:

Si f(x) = |x| ==> f(x) es continua

Demostración:

Sea s > 0 ==>

Se define s > d > 0 ==>

Sea x€R & |x+(-c)| < d ==>

| f(x)+(-1)·f(c) | = | |x|+(-1)·|c| | [< |x+(-c)| < d < s

Teorema:

| |x|+(-1)·|c| | [< |x+(-c)|

Demostración:

Sea x > 0 & c > 0 ==>

| |x|+(-1)·|c| | = | |(x/c)·c|+(-1)·|c| | = ( (x/c)+(-1) )·|c| = | ( (x/c)+(-1) )·c | = |x+(-c)|

| |(-x)|+(-1)·|c| | =| |(-1)·(x/c)·c|+(-1)·|c| | = | (x/c)+(-1) |·| |c| | = | (x/c)+(-1) |·|c| [< ...

... ( |(x/c)|+1 )·|c| = ( (x/c)+1 )·|c| = | (-1)·( (x/c)+1 )·c | = |(-x)+(-c)|

Sea x > 0 & c > 0 ==>

| |(-x)|+(-1)·|(-c)| | = | |(x/c)·(-c)|+(-1)·|(-c)| | = ( (x/c)+(-1) )·|(-c)| = | ( (x/c)+(-1) )·(-c) | = |(-x)+(-1)·(-c)|

| |x|+(-1)·|(-c)| | =| |(-1)·(x/c)·(-c)|+(-1)·|(-c)| | = | (x/c)+(-1) |·| |(-c)| | = | (x/c)+(-1) |·|(-c)| [< ...

... ( |(x/c)|+1 )·|(-c)| = ( (x/c)+1 )·|(-c)| = | (-1)·( (x/c)+1 )·(-c) | = |x+(-1)·(-c)|

Teorema:

Si 0 [< f(x) [< |x| ==> f(x) es continua

Demostración:

Se define s > 0 & [En][ (1/n) >] s ] ==>

Sea d > 0 ==>

De define x€R & |x+(-c)| < d ==>

| f(x)+(-1)·f(c) | >] | |x|+(-1)·|c| | >] |x+(-c)| = 0 = f(0) = (1/n) >] s

[En][ Id(0) = (1/n) & n = oo ]

Teorema: [ de Stolz ]

Si a€K [ \ ] {0} ==>

sum[k = 1]-[n][ (s+a)·( b_{k+1}+(-1)·b_{k} ) ] = (s+a)·( b_{n+1}+(-1)·b_{1} )

Si a = 0^{k} ==>

sum[k = 1]-[n][ (s+a)·( b_{k+1}+(-1)·b_{k} ) ] = (s+c·oo·0^{k})

Si a = oo^{k} ==>

sum[k = 1]-[n][ (s+a)·( b_{k+1}+(-1)·b_{k} ) ] = (s+c·oo·oo^{k})

Teorema:

lim[n = oo][ 1+...(n)...+(1/n) ] = ln(oo) = oo·ln(2) < oo = lim[n = oo][ 1+...(n)...+1 ]

Demostración: [ por Stolz ]

lim[n = oo][ ( ( 1+...(n)...+(1/n) )/ln(n) ) ] = 1

Examen de Stolz:

Teorema:

lim[n = oo][ ( n^{p+1}/( 1^{p}+...(n)...+n^{p} ) ) ] = (p+1)

lim[n = oo][ ( ( 1^{p}+...(n)...+n^{p} )/n^{p+1} ) ] = ( 1/(p+1) )

Teorema:

lim[n = oo][ ( (n+p)^{n+p}/(n+p)! )^{(1/n)} ] = e

lim[n = oo][ ( (n+p)!/(n+p)^{n+p} )^{(1/n)} ] = (1/e)

Teorema:

lim[n = oo][ ( (an)!/n! )^{(1/n)} ] = a

lim[n = oo][ ( n!/(an)! )^{(1/n)} ] = (1/a)

Teorema:

lim[n = oo][ ( (an)!/( (pn+1)·...(n)...·(pn+n) ) )^{(1/n)} ] = ( a/(p+1) )

lim[n = oo][ ( ( (pn+1)·...(n)...·(pn+n) )/(an)! )^{(1/n)} ] = ( (p+1)/a )

Demostración:

lim[n = oo][ ln( (p·(n+1)+k)/(pn+k) ) ] = lim[n = oo][ ln( 1+( p/(pn+k) ) ) ] = ln(1) = 0

Anexo al destructor de demostración de un teorema:

La función tiene que ser un arte que cumpla al menos alguien.

Imponer el blog,

hablar en la mente psíquica,

y der o datchnar clase inventando-se-lo,

no lo hace ninguien en Cygnus-Kepler,

porque no sirve para tener amor.

Ofrecer el blog,

hablar en el cuerpo físico,

y der o datchnar clase no inventando-se-lo,

lo hace alguien en Cygnus-Kepler,

porque sirve para tener amor.

Teorema:

d_{x}[ln(x)] = (1/x)

Demostración:

d_{ln(x)}[ e^{ln(x)} ]·d_{x}[ln(x)] = 1

Teorema:

lim[n = oo][ ( 1+(1/n) )^{n} ] = e

Demostración:

lim[h = 0][ (1/x)·ln( ( 1+(h/x) )^{(x/h)} ) ] = (1/x)

Sea x = 1 ==> lim[h = 0][ (1+h)^{(1/h)} ] = e

Teorema

lim[x = a][ (x/x) ] = 1

Demostración

| (x/x)+(-1) | = | ( (x+(-x))/x ) | = | ( (0·x)/x ) | = 0 [< |x+(-a)| < d < s

Temario de análisis matemático:

Límites de funciones.

Continuidad.

Derivadas.

Integrales.

Sucesiones.

Series numéricas.

Sucesiones de funciones.

Examen de sucesiones:

Teorema:

lim[n = oo][ ( 1+( (kn^{p})/(n^{p+1}+a) ) )^{n} ] = e^{k}