( <a_{1},a_{2}> = <a_{2},a_{1}> & a_{1}+a_{2}=8 ) <==> ( a_{1}=4 & a_{2}=4 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+18 & a_{1}+a_{2}=8 ) <==> ( a_{1}=5 & a_{2}=3 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+36 & a_{1}+a_{2}=8 ) <==> ( a_{1}=6 & a_{2}=2 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+54 & a_{1}+a_{2}=8 ) <==> ( a_{1}=7 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+9 & a_{1}+a_{2}=7 ) <==> ( a_{1}=4 & a_{2}=3 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+27 & a_{1}+a_{2}=7 ) <==> ( a_{1}=5 & a_{2}=2 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+45 & a_{1}+a_{2}=7 ) <==> ( a_{1}=6 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}> & a_{1}+a_{2}=6 ) <==> ( a_{1}=3 & a_{2}=3 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+18 & a_{1}+a_{2}=6 ) <==> ( a_{1}=4 & a_{2}=2 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+36 & a_{1}+a_{2}=6 ) <==> ( a_{1}=5 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+9 & a_{1}+a_{2}=5 ) <==> ( a_{1}=3 & a_{2}=2 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+27 & a_{1}+a_{2}=5 ) <==> ( a_{1}=4 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}> & a_{1}+a_{2}=4 ) <==> ( a_{1}=2 & a_{2}=2 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+18 & a_{1}+a_{2}=4 ) <==> ( a_{1}=3 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}>+9 & a_{1}+a_{2}=3 ) <==> ( a_{1}=2 & a_{2}=1 )
( <a_{1},a_{2}> = <a_{2},a_{1}> & a_{1}+a_{2}=2 ) <==> ( a_{1}=1 & a_{2}=1 )
teorema:
<a_{1},a_{2}> = <a_{2},a_{1}>+(a_{1}+(-1)a_{2})·(10+(-1))
<a_{1},a_{2}> = <a_{2},a_{1}>+(a_{1}+(-1)a_{2})·9
teorema:
<a_{1},a_{2}> = <a_{2},a_{1}>+(2a_{1}+(-n))·9 & a_{1}+a_{2}=n
jueves, 28 de noviembre de 2019
miércoles, 27 de noviembre de 2019
ecuacions diofàntiques enteres
x^{2}+y^{2}=z
[2k+1]^{2}+[(2i)·k]^{2}=[4k+1]
x^{2}+y^{2}=z^{2}
[3k]^{2}+[4k]^{2}=[5k]^{2}
x^{3}+y^{3}+z=s^{2}
[3k+1]^{3}+[3·e^{(pi/3)i}k]^{3}+[3k]=[9k^{2}+6k+1]=[3k+1]^{2}
x^{4}+( y^{2}+(-5)z^{2} )·[k]^{2}=[k]·s^{3}
[3k]^{4}+( [8k]^{2}+(-5)[2k]^{2} )·[k]^{2}=[125k^{4}]=[k]·[5k]^{3}
x^{4}+y^{4}+(-2)z^{2}+3=4s^{3}
[3k+1]^{4}+[3e^{(pi/4)i}k]^{4}+(-2)[3k]^{2}+3=4[27k^{3}+9k^{2}+3k+1]=4[3k+1]^{3}
[2k+1]^{2}+[(2i)·k]^{2}=[4k+1]
x^{2}+y^{2}=z^{2}
[3k]^{2}+[4k]^{2}=[5k]^{2}
x^{3}+y^{3}+z=s^{2}
[3k+1]^{3}+[3·e^{(pi/3)i}k]^{3}+[3k]=[9k^{2}+6k+1]=[3k+1]^{2}
x^{4}+( y^{2}+(-5)z^{2} )·[k]^{2}=[k]·s^{3}
[3k]^{4}+( [8k]^{2}+(-5)[2k]^{2} )·[k]^{2}=[125k^{4}]=[k]·[5k]^{3}
x^{4}+y^{4}+(-2)z^{2}+3=4s^{3}
[3k+1]^{4}+[3e^{(pi/4)i}k]^{4}+(-2)[3k]^{2}+3=4[27k^{3}+9k^{2}+3k+1]=4[3k+1]^{3}
martes, 26 de noviembre de 2019
lunes, 25 de noviembre de 2019
teoría de construcción
El milagro en la construcción:
1o inductivos: ( Si pones uno ==> se ponen todos ) <==> [An][ f(n) ==> f(n+1) ].
2o duales: ( rompes uno <==> se arreglan todos ) <==> [ ¬[Ex][ ¬f(x) ] <==> [Ax][ f(x) ] ]
Material próximo al lugar de obra.
El miracle en la construcció:
1r inductius: ( Si poses uno ==> es posen tots ) <==> [An][ f(n) ==> f(n+1) ].
2n duals: ( trenques un <==> se arreglen tots ) <==> [ ¬[Ex][ ¬f(x) ] <==> [Ax][ f(x) ] ]
Material próxim al lloc d'obra.
1o inductivos: ( Si pones uno ==> se ponen todos ) <==> [An][ f(n) ==> f(n+1) ].
2o duales: ( rompes uno <==> se arreglan todos ) <==> [ ¬[Ex][ ¬f(x) ] <==> [Ax][ f(x) ] ]
Material próximo al lugar de obra.
El miracle en la construcció:
1r inductius: ( Si poses uno ==> es posen tots ) <==> [An][ f(n) ==> f(n+1) ].
2n duals: ( trenques un <==> se arreglen tots ) <==> [ ¬[Ex][ ¬f(x) ] <==> [Ax][ f(x) ] ]
Material próxim al lloc d'obra.
domingo, 24 de noviembre de 2019
italiano
presente
tomare un capuccino
pasato próximo
havere tomato un capuccino
pasato perifrástico
váreti tomare un capuccino
pasato imperfecto
tomáveti un capuccino
futuro
tomareti un capuccino
futuro composto
habreti tomato un capuccino
subjuntivo
tomáseti un capuccino
subjuntivo composto
hubiéseti tomato un capuccino
condicionale
tomaríeti un capuccino
condicionale composto
habríeti tomato un capuccino
tomare un capuccino
pasato próximo
havere tomato un capuccino
pasato perifrástico
váreti tomare un capuccino
pasato imperfecto
tomáveti un capuccino
futuro
tomareti un capuccino
futuro composto
habreti tomato un capuccino
subjuntivo
tomáseti un capuccino
subjuntivo composto
hubiéseti tomato un capuccino
condicionale
tomaríeti un capuccino
condicionale composto
habríeti tomato un capuccino
sábado, 23 de noviembre de 2019
ecuació diferencial de producte integral invers
d_{xx}^{2}[ f(g(x)) [o(x)o] g(x)^{[o(x)o](-1)} ] = (x^{m_{1}}+...(n)...+x^{m_{n}})·f(g(x))
f(g(x)) = e^{g(x)}
g(x) = int[ x^{n_{1}}+...(n)...+x^{m_{n}} ]d[x]
f(g(x)) = e^{g(x)}
g(x) = int[ x^{n_{1}}+...(n)...+x^{m_{n}} ]d[x]
viernes, 22 de noviembre de 2019
solució de ecuacions diferencials en series
d_{xx}^{2}[f(x)]=x^{n}·f(x)
f(x) = ∑ ( 1/((n+2)k+(n+1))(n+2)k+(n+2))! )·x^{(n+2)(k+1)}
d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)(k+(-1))+(n+1))((n+2)(k+(-1))+(n+2))! )·x^{(n+2)k}·x^{n}
d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)p+(n+1))((n+2)p+(n+2))! )·x^{(n+2)(p+1)}·x^{n}
k+(-1)=p & k=p+1
d_{x,...,x}^{m}[f(x)]=x^{n}·f(x)
f(x) = ∑ ( 1/((n+m)k+(n+1))·...·(n+m)k+(n+m))! )·x^{(n+m)(k+1)}
f(x) = ∑ ( 1/((n+2)k+(n+1))(n+2)k+(n+2))! )·x^{(n+2)(k+1)}
d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)(k+(-1))+(n+1))((n+2)(k+(-1))+(n+2))! )·x^{(n+2)k}·x^{n}
d_{xx}^{2}[f(x)] = ∑ ( 1/((n+2)p+(n+1))((n+2)p+(n+2))! )·x^{(n+2)(p+1)}·x^{n}
k+(-1)=p & k=p+1
d_{x,...,x}^{m}[f(x)]=x^{n}·f(x)
f(x) = ∑ ( 1/((n+m)k+(n+1))·...·(n+m)k+(n+m))! )·x^{(n+m)(k+1)}
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