martes, 13 de agosto de 2019

problemes de enters

sum[k=0-->p]( (q/p)·k ) + (-1)·sum[k=0-->q]( (p/q)·k ) = (1/2)·(q+(-p))


sum[k=0-->p]( (q/p)(q+1)·k^{2} ) + (-1)·sum[k=0-->q]( (p/q)(p+1)·k^{2} ) = ...
...(1/6)·(q+(-p))(p+1)(q+1)


sum[k=0-->p]( (q/p^{2})(q+1)·k^{3} ) + (-1)·sum[k=0-->q]( (p/q^{2})(p+1)·k^{3} ) = ...
...(1/4)·(q+(-p))(p+1)(q+1)




proposició:
si n=2^{k}+(-1) ==> [Ep][ p€N & 2n+1=2^{p}+(-1) ]
demostració
2n+1=2·(2^{k}+(-1))+1=2^{k+1}+(-2)+1
2n+1=2^{k+1}+(-1)=2^{p}+(-1)


proposició:
si n=m^{k}+(-1) ==> [Ep][ p€N & m·n+(m+(-1))=m^{p}+(-1) ]
proposició:
si n=m^{k}+1 ==> [Ep][ p€N & m·n+(-1)(m+(-1))=m^{p}+1 ]


a=mb_{a}+r_{a}
a+(-m)b_{a}=r_{a}


sum[a]( (a+(-m)b_{a})·(1/m(m+(-1))) )=(1/2)


sum[a]( (a+(-m)b_{a})^{2}·(1/m(m+(-1))(2m+(-1))) )=(1/6)


sum[a]( (a+(-m)b_{a})^{3}·(1/m^{2}(m+(-1))^{2}) )=(1/4)

punts enters de una regió

x^{2}+y^{2} [< r^{2}


y [< ( r^{2}+(-1)x^{2} )^{(1/2)}
x [< ( r^{2}+(-1)y^{2} )^{(1/2)}


x+y [< ( r^{2}+(-1)y^{2} )^{(1/2)}+( r^{2}+(-1)x^{2} )^{(1/2)}


si x=y ==>
2y^{2} [< r^{2} <==> y [< (r/2^{(1/2)})
2x^{2} [< r^{2} <==> x [< (r/2^{(1/2)})


si ( x=0 or y=0 ) ==>
y [< r
x [< r


sum( [x+y] ) [<
...1+...
...4·r+...
...4·sum[y=0-->(r/2^{(1/2)})]( [( r^{2}+(-1)y^{2} )^{(1/2)}] )+...
...4·sum[x=0-->(r/2^{(1/2)})]( [( r^{2}+(-1)x^{2} )^{(1/2)}] )+...
...(-4)[(r/2^{(1/2)})]


x > 0 & y > 0 & xy [< n


y [< (n/x)
x [< (n/y)


x+y [< (n/y)+(n/x)

si x=y ==>
y^{2} [< n <==> y [< n^{(1/2)}
x^{2} [< n <==> x [< n^{(1/2)}


sum( [x+y] ) [< ...
...sum[(y > 0)-->n^{(1/2)}]( [(n/y)] )+...
...sum[(x > 0)-->n^{(1/2)}]( [(n/x)] )+...
...(-1)[n^{(1/2)}]

congruencies

teorema:
Si p€P ==> n^{p} =[p]= n
demostració:
(n+1)^{p}=n^{p}+pk+1=(npk+n)+pk+1=(n+1)pk+(n+1)


teorema:
Si p€P ==> n^{p+(-1)} =[p]= 1
demostració:
(n+1)^{p+(-1)}=(n^{p}+pk+1)/(n+1)=(n·n^{p+(-1)}+pk+1)/(n+1)=...
(n+1)^{p+(-1)}=(n(pk+1)+pk+1)/(n+1)=((n+1)pk+(n+1))/(n+1)=pk+1


proposició:
Si p€P ==> 1^{p+(-1)}+...+n^{p+(-1)} =[p]= n


proposició:
Si p€P ==> 1^{p}+...+n^{p} =[p]= ( n(n+1)/2 )


teorema:
ax =[a·m]= ab <==> x =[m]= b
demostració:
ax=a·mt+ab
x=mt+b




3x =[9]= 18 <==> x =[3]= 6 =[3]= 0
2x =[8]= 18 <==> x =[4]= 9 =[4]= 1
3x =[6]= 18 <==> x =[2]= 6 =[2]= 0
2x =[6]= 18 <==> x =[3]= 9 =[3]= 0


teorema:
Si ( p€P & p=2k+1 ) ==> ax^{2} =[p]= a <==> x =[p]= a^{(p+(-1))/2}
demostració:
(a^{(p+(-1)/2)})^{2} = a^{p+(-1)} =[p]= 1
a^{p} =[p]= a


2x^{2} =[5]= 2 <==> x =[5]= 4
3x^{2} =[5]= 3 <==> x =[5]= 9 =[5]= 4
4x^{2} =[5]= 4 <==> x =[5]= 16 =[5]= 1


teorema:
Si ( p€P & p=2k+1 ) ==> ax^{2} =[p]= (-a) <==> x =[p]= a^{(p+(-1))/2}·i
demostració:
(a^{(p+(-1)/2)}·i)^{2} = (-1)a^{p+(-1)} =[p]= (-1)
(-1)a^{p} =[p]= (-a)


2x^{2} =[5]= (-2) <==> x =[5]= 4i
3x^{2} =[5]= (-3) <==> x =[5]= 9i =[5]= 4i
4x^{2} =[5]= (-4) <==> x =[5]= 16i =[5]= i


teorema:
Si ( p€P & p=3k+1 ) ==> ax^{3} =[p]= a <==> x =[p]= a^{(p+(-1))/3}
demostració:
(a^{(p+(-1)/3)})^{3} = a^{p+(-1)} =[p]= 1
a^{p} =[p]= a


2x^{3} =[7]= 2 <==> x =[7]= 4
3x^{3} =[7]= 3 <==> x =[7]= 9 =[7]= 2
4x^{3} =[7]= 4 <==> x =[7]= 16 =[7]= 2


teorema:
Si p€N ==> ...
...x^{2}+2bx =[p^{2}]= (-1)b^{2} <==> x =[p]= (-b)
demostració:
x^{2}+2bx =[p^{2}]= (-1)b^{2}
x^{2}+2bx+b^{2} =[p^{2}]= 0
(x+b)^{2} =[p^{2}]= 0
x+b =[p]= 0
x =[p]= (-b)


x^{2}+4x =[9]= (-4) <==> x =[3]= (-2) =[3]= 1
x^{2}+6x =[9]= (-9) <==> x =[3]= (-3) =[3]= 0
x^{2}+8x =[9]= (-16) <==> x =[3]= (-4) =[3]= (-1) =[3]= 2







problemes de números

s=1+(1/2)+...+(1/n)  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n)/n! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n)/n! )
sum[k=1-->n]( m_{k} )n!=sum[k=1-->n]( (1·...·(k+(-1))(k+1)·...·n) )
m_{k}=(1/k)


s=1+(1/3)+...+(1/(2n+1))  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1))/(2n+1)!! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1))/(2n+1)!! )
sum[k=1-->n]( m_{k} )(2n+1)!!=sum[k=1-->n]( (1·...·(2(k+(-1))+1)(2(k+1)+1)·...·(2n+1)) )
m_{k}=(1/(2k+1))


s=1+(1/2!)+...+(1/n!)  ==> ¬( s€N ) or s=1


s = sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!)/(n!)! )
sigui m_{k}€N ==>
s = sum[k=1-->n]( m_{k} )=sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!)/(n!)! )
sum[k=1-->n]( m_{k} )(n!)!=sum[k=1-->n]( (1·...·(k+(-1))!(k+1)!)·...·n!) )
m_{k}=(1/k!)


s=(p/n)+...(n)...+(p/n)==> s€N


s = ( (p+...(n)...+p)/n )
sigui m_{k}€N ==>
s = sum[k=1-->p]( m_{k} )=( ( p+...(n)...+p )/n )
( m_{1}+...(p)...+m_{p} )=(np/n)
( 1+...(p)...+1 )=p
m_{k}=1


s=(p/n^{q})+...(n)...+(p/n^{q})==> ¬( s€N ) or p=n^{q+(-1)}


s = ( (p+...(n)...+p)/n^{q} )
sigui m_{k}€N ==>
s = sum[k=1-->p]( m_{k} )=( ( p+...(n)...+p )/n^{q} )
( m_{1}+...(p)...+m_{p} )=(np/n^{q})
( 1/n^{q+(-1)}+...(p)...+1/n^{q+(-1)} )n^{q+(-1)}=p
m_{k}=(1/n^{q+(-1)})
( 1/n^{q+(-1)}+...(n^{q+(-1)})...+1/n^{q+(-1)} )n^{q+(-1)}=n^{q+(-1)}

lunes, 12 de agosto de 2019

ecuacions diferencials lineals

d_{tt}[ z( af(t) ) [o(t)o] f(t)^{[o(t)o](-1)}) ]=(-1)a^{2}·z( af(t) )d_{t}[f(t)]


d_{tt}[ sin( af(t) ) [o(t)o] f(t)^{[o(t)o](-1)}) ]=...
...d_{t}[ cos( af(t) )a ]=...
...(-1)a^{2}·sin( af(t) )d_{t}[f(t)]


d_{tt}[ z( at ) ]=(-1)a^{2}·z( at )






d_{tt}[ d_{t}[ z( af(t) ) [o(t)o] f(t)^{[o(t)o](-1)}) ] [o(t)o] f(t)^{[o(t)o](-1)}) ]=...
...(-1)a^{3}·z( af(t) )d_{t}[f(t)]


d_{tt}[ d_{t}[ red( af(t) ) [o(t)o] f(t)^{[o(t)o](-1)}) ] [o(t)o] f(t)^{[o(t)o](-1)}) ]=...
d_{tt}[ blue( af(t) )a [o(t)o] f(t)^{[o(t)o](-1)}) ]=...
...d_{t}[ yel( af(t) )a^{2} ]=...
...(-1)a^{3}·red( af(t) )d_{t}[f(t)]


d_{ttt}[ z( at ) ]=(-1)a^{3}·z( at )

camps electrics discrets

E(0,0,z)= k_{e}·int-int-int[rho[q]]d[x]d[y]d[ d_{t}[f(t)]z ]·(1/(cm)^{n})·<1,1,1>
on d_{t}[f(t)] es la resistencia del material.


circumferencia
E(0,0,z)= ...
...k_{e}·int-int[0-->2pi][rho[q]]d[sR]d[z]·(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·2pi·R·z
m_{i}d_{tt}[z]=q_{i}·k_{e}(1/(cm)^{n})·rho[q]·2pi·R·z


disc cercle
E(0,0,z)= ...
...k_{e}·int-int[0-->2pi][rho[q]]d[(1/2)·s·R^{2}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·pi·R^{2}( d_{t}[f(t)]z )


esfera plena
E(0,0,z)= ...
...k_{e}·int-int[0-->4pi][rho[q]]d[(1/3)·s·R^{3}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·(4/3)·pi·R^{3}( d_{t}[f(t)]z )


esfera buida
E(0,0,z)= ...
...k_{e}·int-int[0-->4pi][rho[q]]d[s·R^{2}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·4pi·R^{2}( d_{t}[f(t)]z )


doble esfera plena per tall de porta
E(0,0,z)= ...
...k_{e}·int-int[0-->8pi][rho[q]]d[(1/3)·s·R^{3}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·(8/3)pi·R^{3}( d_{t}[f(t)]z )




doble esfera buida per tall de porta
E(0,0,z)= ...
...k_{e}·int-int[0-->8pi][rho[q]]d[R^{2}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·8pi·R^{2}( d_{t}[f(t)]z )


doble esfera:


1 1
2 2


2 2
1 1


doble esfera:


1 2
2 1


2 1
1 2


triple esfera plena per tall de porta
E(0,0,z)= ...
...k_{e}·int-int[0-->12pi][rho[q]]d[(1/3)·s·R^{3}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·4pi·R^{3}( d_{t}[f(t)]z )




triple esfera buida per tall de porta
E(0,0,z)= ...
...k_{e}·int-int[0-->12pi][rho[q]]d[R^{2}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·12pi·R^{2}( d_{t}[f(t)]z )


triple esfera:


1 2
3 1


2 3
1 2


3 1
2 3






quadruble esfera plena per doble tall de porta ortogonal
E(0,0,z)= ...
...k_{e}·int-int[0-->16pi][rho[q]]d[(1/3)·s·R^{3}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·(16/3)pi·R^{3}( d_{t}[f(t)]z )




quadruble esfera buida per doble tall de porta ortogonal
E(0,0,z)= ...
...k_{e}·int-int[0-->16pi][rho[q]]d[R^{2}]d[ d_{t}[f(t)]z ]·...
...(1/(cm)^{n})·<1,1,1>
E(0,0,z)= k_{e}(1/(cm)^{n})·rho[q]·16pi·R^{2}( d_{t}[f(t)]z )


cuadruble esfera:


1 2
3 4


4 3
2 1




2 4
1 3


3 1
4 2


a^{2}=(q_{i}/m_{i})·k_{e}(1/(cm)^{n})rho[q]·V


d_{tt}[ sinh(at^{(-p)}) [o(t)o] (1/(-p))(1/(p+2))t^{p+2} ]=...
...a^{2}( (-p)t^{(-p)+(-1)}sinh(at^{(-p)}) )


d_{tt}[ sinh( a·(bt^{2}+ct) ) [o(t)o] (1/(2b))·ln(2bt+c) ]=...
...a^{2}(2bt+c)sinh( a·(bt^{2}+ct) )

domingo, 11 de agosto de 2019

ones electro-magnétiques y gravito-magnétiques

E_{e}(r) = k_{e}q_{e}·( 1/r^{n} )
E_{g}(r) = (-1)·k_{g}q_{g}·( 1/r^{n} )


B_{e}(r) = (-1)·k_{e,m}q_{e}·( d_{t}[r]^{n}/r^{n} )
B_{g}(r) = k_{g,m}q_{g}·( d_{t}[r]^{n}/r^{n} )


divergencia y laplacià del camp eléctric y gravitatori:
d_{r}[ E_{e}(r) ] = (-n)·kq·(1/r^{n+1})
d_{rr}^{2}[ E_{e}(r) ] = (-n)((-n)+(-1))·kq·(1/r^{n+2})


d_{r}[ E_{g}(r) ] = (-1)·(-n)·kq·(1/r^{n+1})
d_{rr}^{2}[ E_{g}(r) ] = (-1)·(-n)((-n)+(-1))·kq·(1/r^{n+2})


divergencia y laplacià del camp magnétic:
d_{r}[ B_{e}(r) ] = (-1)·(-n)·kq·( d_{t}[r]^{n}/r^{n+1} )
d_{rr}^{2}[ B_{e}(r) ] = (-1)·(-n)((-n)+(-1))·kq·( d_{t}[r]^{n}/r^{n+2} )


d_{r}[ B_{g}(r) ] = (-n)·kq·( d_{t}[r]^{n}/r^{n+1} )
d_{rr}^{2}[ B_{g}(r) ] = (-n)((-n)+(-1))·kq·( d_{t}[r]^{n}/r^{n+2} )


ecuacions de camp del temps:
d_{t}[ d_{r}[ E_{e}(r) ] ] = d_{rr}^{2}[ E_{e}(r) ]·d_{t}[r]
d_{t}[ d_{r}[ B_{e}(r) ] ] = d_{rr}^{2}[ B_{e}(r) ]·d_{t}[r]


d_{t}[ d_{r}[ E_{g}(r) ] ] = d_{rr}^{2}[ E_{g}(r) ]·d_{t}[r]
d_{t}[ d_{r}[ B_{g}(r) ] ] = d_{rr}^{2}[ B_{g}(r) ]·d_{t}[r]


d_{t}[ d_{r}[ E_{e}(r)+B_{e}(r) ] ] = d_{rr}^{2}[ E_{e}(r)+B_{e}(r) ]·d_{t}[r]
d_{t}[ d_{r}[ E_{g}(r)+B_{g}(r) ] ] = d_{rr}^{2}[ E_{g}(r)+B_{g}(r) ]·d_{t}[r]


d_{tt}^{2}[r] = 0 <==> r(t)  = (k_{e}/k_{e,m})^{(1/n)}·t
d_{tt}^{2}[r] = 0 <==> r(t)  = (k_{g}/k_{g,m})^{(1/n)}·t


d_{t}[ d_{t}[ E_{e}(r)+B_{e}(r) ]·( 1/d_{t}[r] ) ] = d_{rr}^{2}[ E_{e}(r)+B_{e}(r) ]·d_{t}[r]


d_{tt}^{2}[ E_{e}(r)+B_{e}(r) ]·( 1/d_{t}[r] ) +...
... (-1)·d_{r}[ E_{e}(r)+B_{e}(r) ]·( d_{tt}^{2}[r]/d_{t}[r] ) =...
... d_{rr}^{2}[ E_{e}(r)+B_{e}(r) ]·d_{t}[r]


ecuacions de front de ones:
d_{tt}^{2}[ E_{e}(r)+B_{e}(r) ] = d_{rr}^{2}[ E_{e}(r)+B_{e}(r) ]·d_{t}[r]^{2}
d_{tt}^{2}[ E_{g}(r)+B_{g}(r) ] = d_{rr}^{2}[ E_{g}(r)+B_{g}(r) ]·d_{t}[r]^{2}