Mostrando entradas con la etiqueta matemàtiques-càlcul-diferencial. Mostrar todas las entradas
Mostrando entradas con la etiqueta matemàtiques-càlcul-diferencial. Mostrar todas las entradas

sábado, 25 de abril de 2020

segon valor mitx y zeros de funcions de polinomis de tercer grau


f(x) = (x+(-a))(x+(-b))(x+(-s))


d_{x}[f(x)] = 3x^{2}+2·((-a)+(-b)+(-s))x+((ab)+(bs)+(sa))


c = (1/6)( 2·(a+b+s)+( 4·((-a)+(-b)+(-s))^{2}+12·( 1+(-1)·(ab+bs+sa) ) )^{(1/2)} )
c = (1/6)( 2·(a+b+s)+(-1)( 4·((-a)+(-b)+(-s))^{2}+12·( 1+(-1)·(ab+bs+sa) ) )^{(1/2)} )


x = a or x = b or x = s


f(x) = (x+(-a))(x+(-b))(x+(-s))+( a+(-a) )
f(x) = (x+(-a))(x+(-b))(x+(-s))+( b+(-b) )
f(x) = (x+(-a))(x+(-b))(x+(-s))+( s+(-s) )

segon valor mitx y zeros de funcions polinomis de tercer grau

f(x) = x(x+(-a))(x+(-b))


d_{x}[f(x)] = 3x^{2}+2·((-a)+(-b))x+(ab)


c = (1/6)( 2·(a+b)+( 4·((-a)+(-b))^{2}+12·( 1+(-1)·ab ) )^{(1/2)} )
c = (1/6)( 2·(a+b)+(-1)( 4·((-a)+(-b))^{2}+12·( 1+(-1)·ab ) )^{(1/2)} )


x = 0 or x = a or x = b


f(x) = x(x+(-a))(x+(-b))+( 0+(-0) )
f(x) = x(x+(-a))(x+(-b))+( a+(-a) )
f(x) = x(x+(-a))(x+(-b))+( b+(-b) )

martes, 14 de abril de 2020

segon valor mitx y zeros de funcions de polinomis de segon grau

f(x) = x(x+(-a))


d_{x}[f(x)] = 2x+(-a)


c = (1/2)(a+1)


x = 0 or x = a


f(x) = x^{2}+(-a)·x+( 0+(-0) )
f(x) = x^{2}+(-a)·x+( a+(-a) )


f(x) = x(x^{n}+(-a))


d_{x}[f(x)] = (n+1)·x^{n}+(-a)


c = (1/(n+1))(a+1)^{(1/n)}


x = 0 or x = a^{(1/n)}


f(x) = x^{(n+1)}+(-a)·x+( 0+(-0) )
f(x) = x^{(n+1)}+(-a)·x+( a^{(1/n)}+(-1)·a^{(1/n)} )


f(x) = (x+(-a))(x+(-b))


d_{x}[f(x)] = 2x+((-a)+(-b))


c = (1/2)( (a+b)+1 )


x = a or x = b


f(x) = x^{2}+((-a)+(-b))·x+ab+( a+(-a) )
f(x) = x^{2}+((-a)+(-b))·x+ab+( b+(-b) )

segon valor mitx y zeros de funcions

f(s) = 0


f(x) = d_{x}[f(c)]·( x+(-s) )


s = x+( f(x)/d_{x}[f(c)] )


[∃c][ ( c [< s or c€C ) & ( f(s)/d_{x}[f(c)] ) = s+(-s) ]


f(x) = x^{n}+(-b)
[∃c][ s = b^{(1/n)}+( b^{(1/n)}+(-1)·b^{(1/n)} ) ]


f(x) = x^{2}+(-b)
x = b^{(1/2)}
c = ( b^{(1/2)}/2 )


f(x) = x^{3}+(-b)
x = b^{(1/3)}
c = ( b^{(1/3)}/3^{(1/2)} )


f(x) = x^{n}+(-b)
x = b^{(1/n)}
c = ( b^{(1/n)}/n^{(1/(n+(-1)))} )


f(x)  = e^{x}+(-b)
[∃c][ s = ln(b)+( ln(b)+(-1)·ln(b) ) ]


f(x) = e^{x}+(-b)
x = ln(b)
c = ln( b/ln(b) )


f(x)  = e^{ln(a)·x}+(-b)
[∃c][ s = ( ln(b)/ln(a) )+( ( ln(b)/ln(a) )+(-1)·( ln(b)/ln(a) ) ) ]


f(x) = e^{ln(a)·x}+(-b)
x = ( ln(b)/ln(a) )
c = ln( b/ln(b) )


f(x)  = ln(x)+(-b)
[∃c][ s = e^{b}+( e^{b}+(-1)·e^{b} ) ]


f(x) = ln(x)+(-b)
x = e^{b}
c = ( e^{b}/b )


f(x)  = ln(ax)+(-b)
[∃c][ s = e^{(b+(-1)·ln(a))}+( e^{(b+(-1)·ln(a))}+(-1)·e^{(b+(-1)·ln(a))} ) ]


f(x) = ln(ax)+(-b)
x = e^{(b+(-1)·ln(a))}
c = ( e^{(b+(-1)·ln(a))}/b )


f(x)  = (1/x)+(-b)
[∃c][ s = (1/b)+( (1/b)+(-1)·(1/b) ) ]


f(x) = (1/x)+(-b)
x = ( 1/b )
c = ( 1/b )·i


f(x)  = (1/x^{n})+(-b)
[∃c][ s = (1/b^{(1/n)})+( (1/b^{(1/n)})+(-1)·(1/b^{(1/n)}) ) ]


f(x) = (1/x^{2})+(-b)
x = ( 1/b^{(1/2)} )
c = 2^{(1/3)}·( 1/b^{(1/2)} )·e^{(1/3)·pi·i}


f(x) = (1/x^{n})+(-b)
x = ( 1/b^{(1/n)} )
c = n^{( 1/(n+1) )}·( 1/b^{(1/n)} )·e^{( 1/(n+1) )·pi·i}

lunes, 30 de marzo de 2020

corolari del valor mitx

Si ( d_{x}[f(x)] és creishent & d_{x}[f(x)] >] 0 )  ==> ...
... [∀x][ x€(0,1)_{K} ==> ( x·d_{x}[f(x)] ) és creishent ].


sigui 0 < x [< y < 1 ==>
x·d_{x}[f(x)] [< x·d_{y}[f(y)] [< y·d_{y}[f(y)]
x·d_{x}[f(x)] [< y·d_{x}[f(x)] [< y·d_{y}[f(y)]


Si ( f(0) = 0 & f(1) = 0 & d_{x}[f(x)] és creishent & d_{x}[f(x)] >] 0 ) ==> ...
... [∃u][∃v][ 0 < u [< v < 1 & [∀x][ x€[u,v]_{K} ==> f(x) es creishent ] ].


sigui 0 < x [< y < 1 ==>
[∃c][ 0 [< c [< x ] & [∃a][ y [< a [< 1 ]
f(x) = x·d_{x}[f(c)] [< x·d_{x}[f(x)] [< y·d_{y}[f(y)] [< y·d_{x}[f(a)] [< f(y)


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


(-1)(1/4) = f(1/2) [< f(2/3) = (-1)(1/9)

lunes, 23 de marzo de 2020

derivació y integració exponencial

D_{x}[f(x)] = lim [h-->0][ (f(x+h)+(-1)·f(x))/(h·f(x)) ] = ( d_{x}[f(x)]/f(x) )


D_{x}[c] = 0


D_{x}[x^{n}] = (n/x)


D_{x}[e^{x}] = 1


D_{x}[a^{x}] = ln(a)


D_{x}[ln(x)] = ( 1/ln(x) )·(1/x)


D_{x}[f(x)+g(x)] = D_{x}[f(x)] + D_{x}[g(x)]


D_{x}[a·f(x)] = D_{x}[f(x)]


D_{x}[f(x)·g(x)] = D_{x}[f(x)]·g(x) + f(x)·D_{x}[g(x)]


D_{x}[ax^{2}+bx+c] = (2ax+b)/(ax^{2}+bx)


∫ [ D_{x}[f(x)] ] d[x] = ln( ∫ [ D_{x}[f(x)] ] D[x] )


e^{ ∫ [ D_{x}[f(x)] ] d[x] } = ∫ [ D_{x}[f(x)] ] D[x]


e^{∫ d[x]} = ∫ D[x] = e^{x}


e^{∫ [(n/x)] d[x]} = ∫ (n/x) D[x]
e^{ln(x^{n})} = x^{n}


e^{∫ [ ln(a) ] d[x]} = ∫ [ ln(a) ] D[x]
e^{ln(a)·x} = a^{x}


e^{∫ [ ( 1/ln(x) )·(1/x) ] d[x]} = ∫ [ ( 1/ln(x) )·(1/x) ] D[x]
e^{ln(ln(x))} = ln(x)


D_{x}[f(g(x))] = d_{g(x)}[f(g(x))]·d_{x}[g(x)]·(1/f(g(x)))


e^{∫ [ x^{n} ] d[x]} = ∫ [ x^{n} ] D[x] = e^{(1/(n+1))·x^{(n+1)}}
D_{x}[ e^{(1/(n+1))·x^{(n+1)}} ] = x^{n}

domingo, 8 de marzo de 2020

derivació logarítmica

f(x) = ( g(x) )^{h(x)}
ln(f(x)) = h(x)·ln(g(x))
d_{x}[f(x)] = f(x)·d_{x}[h(x)·ln(g(x))]
d_{x}[f(x)] = f(x)·( d_{x}[h(x)]·ln(g(x))+h(x)·(d_{x}[g(x)]/g(x)) )

corolari del valor mitx


Si ( f(0) = 0 & f(1) = 0 & d_{x}[f(x)] és creishent ==> ...
... [∀x][ x€(0,1)_{K} ==> ( f(x)/x ) és creishent ]


sigui 0 < x [< y < 1 ==>
[∃c][ 0 [< c [< x ] & [∃b][ y [< b [< 1 ]
(f(x)/x) = ( (f(x)+(-1)f(0))/(x+(-0)) ) = d_{x}[f(c)] [< d_{x}[f(x)] [< ...
... d_{y}[f(y)] [< d_{y}[f(b)] = ( (f(1)+(-1)f(y))/(1+(-y)) ) [< ( f(y)/y )
0 [< 1
(-y) [< 1+(-y)


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


(f(x)/x) =  x+(-1)

corolari del valor mitx

Si ( f(0) = 0 & [∀x][ d_{x}[f(x)] [< f(x) ] & f(x) es creishent ==> ...
... [∀x][ x€(0,1)_{K} ==> f(x) = 0 ]


(f(x)/x) = ( (f(x)+(-1)f(0))/(x+(-0)) ) = d_{x}[f(c)] [< d_{x}[f(x)] [< f(x) [< ( f(x)/x )
( f(x)/x ) =  f(x) ==> f(x) =  0

sábado, 7 de marzo de 2020

teorema del valor mitx

teorema:
[∃c][ c€K & ( (f(b)+(-1)·f(a))/(b+(-a)) ) = d_{x}[f(c)] ]


demostració:


Q = ( (f(b)+(-1)·f(a))/(b+(-a)) )


F(x) = (-1)f(x)+f(a)+Q( x+(-a) )


F(a) = 0 & F(b) = 0


d_{x}[F(x)] = (-1)·d_{x}[f(x)]+Q


d_{x}[f(c)] = Q



miércoles, 4 de marzo de 2020

criteri de derivades

lim[h-->0][ (h/h)·( f(x+h)+(-1)f(x) )/( g(x+h)+(-1)g(x) ) ] = lim[h-->0][ ( f(x+h) )/( g(x+h) ) ] = ...
... ( f(x)/g(x) ) <==> ( f(x) = 0  & g(x) = 0 )

derivada de la exponencial

d_{x}[e^{x}] = lim[h-->0][ (1/h)( e^{x+h}+(-1)·e^{x} ) ]
d_{x}[e^{x}] = e^{x}·lim[h-->0][ (1/h)( e^{h}+(-1) ) ]
d_{x}[e^{x}] = e^{x}·lim[h-->0][ (1/h)( (1+h+(1/2)·h^{2}+...(1/n!)·h^{n}+...)+(-1) ) ]
d_{x}[e^{x}] = e^{x}


d_{x}[a^{x}] = lim[h-->0][ (1/h)( a^{x+h}+(-1)·a^{x} ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( a^{h}+(-1) ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( e^{ln(a)·h}+(-1) ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( (1+(ln(a)·h)+(1/2)·(ln(a)·h)^{2}+...(1/n!)·(ln(a)·h)^{n}+...)+(-1) ) ]
d_{x}[a^{x}] = a^{x}·ln(a)

derivada del logaritme

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


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


d_{x}[x^{n}] = nx^{(n+(-1))}
d_{x}[e^{ln(x^{n})}] = nx^{(n+(-1))}
d_{ln(x^{n})}[e^{ln(x^{n})}]·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
e^{ln(x^{n})}·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
x^{n}·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
d_{x}[ln(x^{n})] = (1/x^{n})·nx^{(n+(-1))}
d_{x}[ln(x^{n})] = (n/x)

martes, 13 de agosto de 2019

derivades

proposició:
si ( M [< x [< 2M & f(x)=ln(x^{n}) ) ==> n [< d_{x}[f(x)]·2M [< 2n
d_{x}[f(x)]=(n/x)
(n/2M) [< d_{x}[f(x)] [< (2n/2M)


proposició:
si ( M [< x [< 2M & f(x)=(n/x) ) ==> n [< d_{xx}[f(x)]·4M^{3} [< 8n
d_{xx}[f(x)]=(2n/x^{3})
(n/4M^{3}) [< d_{xx}[f(x)] [< (8n/4M^{3})


proposició:
si ( M [< x [< 2M & f(x)=(n/x^{2}) ) ==> 3n [< d_{xx}[f(x)]·8M^{4} [< 48n
d_{xx}[f(x)]=(6n/x^{4})
(3n/8M^{4}) [< d_{xx}[f(x)] [< (48n/8M^{4})


proposició:
si ( M [< x [< 2M & f(x)=(n/x^{3}) ) ==> 3n [< d_{xx}[f(x)]·8M^{5} [< 96n
d_{xx}[f(x)]=(12n/x^{5})
(3n/8M^{5}) [< d_{xx}[f(x)] [< (96n/8M^{5})