termo-electricidad industrial y mecánica industrial

Si cuando eres pequeño sufres,

recibes la gloria de la luz verdadera,

y pagas condenación.

Si cuando eres pequeño no sufres,

no recibes la gloria de la luz verdadera,

y no pagas condenación.

Con radiación esclerósica,

tienes mucha electricidad en la columna,

y no paras de andar.

Con des-radiación esclerósica,

tienes poca electricidad en la columna,

y paras de andar.

cámara térmica:

( 2pi·q_{0} )·d_{t}[T(t)] = h·f(t)

T(t) = ( h/(2pi·q_{0}) )·int[f(t)]d[t]

( 2pi·q_{0} )·(1/s)·d_{tt}^{2}[T(t)] = h·f(t)

T(t) = ( (hs)/(2pi·q_{0}) )·int-int[f(t)]d[t]d[t]

cámara eléctrica:

( 2pi·q_{0} )·R·d_{t}[q(t)] = h·f(t)

q(t) = ( h/((2pi·q_{0})·R) )·int[f(t)]d[t]

( 2pi·q_{0} )·R·(1/s)·d_{tt}^{2}[q(t)] = h·f(t)

q(t) = ( (hs)/((2pi·q_{0})·R) )·int-int[f(t)]d[t]d[t]

reactor a combustión:

( 2pi·q_{0} )·d_{t}[T(t)] = V·P(t)

reactor eléctrico:

( 2pi·q_{0} )·R·d_{t}[q(t)] = V·P(t)

Termo-electricidad industrial:

( 2pi·q_{0} )·d_{t}[T(t)] = E(t)

( 2pi·q_{0} )·(1/s)·d_{tt}^{2}[T(t)] = E(t)

( 2pi·q_{0} )·d_{t}[T(t)] = I(t)·T(t)

T(t) = T_{k}·e^{( 1/(2pi·q_{0}) )·int[ I(t) ]d[t]}

( 2pi·q_{0} )·d_{t}[T(t)] = ( 1/E(t) )·( I(t) )^{2}·(1/2)·( T(t) )^{2}

T(t) = ( (-1)·( 1/(2pi·q_{0}) )·(1/2)·int[ ( 1/E(t) )·( I(t) )^{2} ]d[t] )^{(-1)}

( 2pi·q_{0} )·d_{t}[T(t)] = ( 1/( E(t) )^{2} )·( I(t) )^{3}·(4/3)·( T(t) )^{3}

T(t) = ...

... ( (-2)·( 1/(2pi·q_{0}) )·(4/3)·int[ ( 1/( E(t) )^{2} )·( I(t) )^{3} ]d[t] )^{(-1)·(1/2)}

( 2pi·q_{0} )·R·d_{t}[q(t)] = E(t)

( 2pi·q_{0} )·R·(1/s)·d_{tt}^{2}[q(t)] = E(t)

( 2pi·q_{0} )·R·d_{t}[q(t)] = A(t)·q(t)

q(t) = q_{k}·e^{( 1/((2pi·q_{0})·R) )·int[ A(t) ]d[t]}

( 2pi·q_{0} )·R·d_{t}[q(t)] = ( 1/E(t) )·( A(t) )^{2}·(1/2)·( q(t) )^{2}

q(t) = ( (-1)·( 1/((2pi·q_{0})·R) )·(1/2)·int[ ( 1/E(t) )·( A(t) )^{2} ]d[t] )^{(-1)}

( 2pi·q_{0} )·R·d_{t}[q(t)] = ( 1/( E(t) )^{2} )·( A(t) )^{3}·(4/3)·( q(t) )^{3}

q(t) = ...

... ( (-2)·( 1/((2pi·q_{0})·R) )·(4/3)·int[ ( 1/( E(t) )^{2} )·( A(t) )^{3} ]d[t] )^{(-1)·(1/2)}

Mecánica industrial:

mc·d_{t}[x(t)] = E(t)

mc·(1/s)·d_{tt}^{2}[x(t)] = E(t)

mc·d_{t}[x(t)] = F(t)·x(t)

x(t) = x_{k}·e^{( 1/(mc) )·int[ F(t) ]d[t]}

mc·d_{t}[x(t)] = ( 1/E(t) )·( F(t) )^{2}·(1/2)·( x(t) )^{2}

x(t) = ( (-1)·( 1/(mc) )·(1/2)·int[ ( 1/E(t) )·( F(t) )^{2} ]d[t] )^{(-1)}

mc·d_{t}[x(t)] = ( 1/( E(t) )^{2} )·( F(t) )^{3}·(4/3)·( x(t) )^{3}

x(t) = ( (-2)·( 1/(mc) )·(4/3)·int[ ( 1/( E(t) )^{2} )·( F(t) )^{3} ]d[t] )^{(-1)·(1/2)}

Demostreu:

Si ( u(x) )^{n} =[m]= x ==> sum[ x = 1 ---> x = (m+(-1)) ][ u(x) ] [< (3/2)·(m+(-1))·m

Si ( u(x) )^{n} =[m]= x ==> ...

... sum[ x = 1 ---> x = (m+(-1)) ][ u(x) ] [< (1/6)·(m+(-1))·m·(2m+5)

Demostració:

u(x) = m+x^{(1/n)}

[Ax][ x >] 1 ==> u(x) [< m+x ]

[Ax][ x >] 1 ==> u(x) [< m+x^{2} ]

Fraccions continues:

(a/b) = q_{1}+( 1/(b/r_{1}) )

(b/r_{1}) = q_{2}+( 1/(r_{1}/r_{2}) )

s_{1} = q_{1} = (P_{1}/Q_{1})

s_{2} = ( q_{1}+(1/q_{2}) ) = ...

... ( (q_{2}P_{1}+P_{0})/(q_{2}Q_{1}+Q_{0}) ) = (P_{2}/Q_{2})

s_{3} = ( q_{1}+( 1/(q_{2}+(1/q_{3})) ) ) = ...

... ( (q_{3}( q_{2}P_{1}+P_{0} )+P_{1})/(q_{3}Q_{2}+Q_{1}) ) = (P_{3}/Q_{3})

Teorema:

s_{n} = ...

... ( (q_{n}P_{n+(-1)}+P_{n+(-2)})/(q_{n}Q_{n+(-1)}+Q_{n+(-2)}) ) = (P_{n}/Q_{n})

Teorema:

s_{n}·Q_{n} = q_{n}·s_{n+(-1)}·Q_{n+(-1)}+s_{n+(-2)}·Q_{n+(-2)}

Demostració:

s_{n}+(-1)·s_{n+(-1)} = (P_{n}/Q_{n})+(-1)·(P_{n+(-1)}/Q_{n+(-1)}) = ...

... (-1)·( s_{n+(-1)}+(-1)·s_{n+(-2)} )·( Q_{n+(-2)}/Q_{n} )

(15/6) = 2+(1/2) = (5/2)

(15/6) = 6·2+3

(6/3) = 3·2+0

mcd{15,6} = 3

(2/3) = ( 1/(1+(1/2)) )

(2/3) = 3·0+2

(3/2) = 2·1+1

(2/1) = 1·2+0

mcd{2,3} = 1

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