mesura

a(n) = (x/n)

d_{x}[x^{n}] = ( 1/a(n) )·x^{n}

∫ [ x^{n} ] d[x] = a(n+1)·x^{n}

b(n) = (1/n)

d_{x}[e^{nx}] = ( 1/b(n) )·e^{nx}

∫ [ e^{nx} ] d[x] = b(n)·e^{nx}

c(sin(nx)) = n·( 1+(-1)·( sin(nx) )^{2} )^{(1/2)}

d_{x}[sin(nx)] = c(sin(nx))

∫ [ sin(nx) ] d[x] = (-1)·c(sin(nx))

c(cos(nx)) = n·( 1+(-1)·( cos(nx) )^{2} )^{(1/2)}

d_{x}[cos(nx)] = (-1)·c(cos(nx))

∫ [ cos(nx) ] d[x] = c(cos(nx))

u(ln(x)) = e^{ln(x)}

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

∫ [ ln(x) ] d[x] = ln(x)·u(ln(x))+(-1)·u(ln(x))

teorema er-sinus y er-cosinus

teoremes:
cos[n:n](x^{n}) = (1/n)·( sin(x^{n})/x^{n} )


sin[n:n](x^{n}) = (1/n)·(1/x^{n})·( 1+(-1)·cos(x^{n}) )


demostracions:
d_{x}[ ∫ [0-->x^{n}][ cos(x) ] d[x] ] = cos(x^{n})·nx^{(n+(-1))}


d_{x}[ ∫ [0-->x^{n}][ sin(x) ] d[x] ] = sin(x^{n})·nx^{(n+(-1))}

limit integral

lim[n-->oo][ ∫ [0-->oo]-[ (1/n) ] d[x] ] = 1


lim[n-->oo][ ∫ [0-->oo]-[ (1/n)·(x/n)^{m} ] d[x] ] = (1/(m+1))


lim[n-->oo][ ∫ [0-->oo]-[ (1/n)·e^{(1/n)·x} ] d[x] ] = e+(-1)


lim[n-->oo][ ∫ [0-->oo]-[ (1/n)·( 1/( (x/n)+1 ) ) ] d[x] ] = ln(2)


lim[n-->oo][ ∫ [0-->oo]-[ (1/n)·( 1/( (x/n)+a ) ) ] d[x] ] = ln( (1/a)+1 )

desigualtats

x [< y <==> (-y) [< (-x)
y >] x <==> (-x) >] (-y)


0 [< x <==> (-x) [< x
x >] 0 <==> x >] (-x)


x [< 0 <==> x [< (-x)
0 >] x <==> (-x) >] x


demostració:
x [< y
(-x)+x+(-y) [< (-x)+y+(-y)
(-y) [< (-x)
x+(-y)+y [< x+(-x)+y


demostració:
(-x) [< 0 & 0 [< x
absurd: Si x < 0 ==>
(-x) [< x & x < 0

funció

si f(x) = ( (y+(-x))/(2s) ) ==> [∀s][ s ≠ 0 ==> f(y+(-s)) = 1+f(y+s) ]


f(y+(-s)) = ( (y+(-y)+s)/(2s) ) = ( s/(2s) ) = (1/2)
f(y+s) = ( (y+(-y)+(-s))/(2s) ) = (-1)( s/(2s) ) = (-1)(1/2)


si f(x) = ( (y+(-1)·x^{n})/(2s) ) ==> [∀s][ s ≠ 0 ==> f( (y+(-s))^{(1/n)} ) = 1+f( (y+s)^{(1/n)} ) ]

derivació logarítmica

L_{x}[f(x)] = e^{f(x)}·d_{x}[f(x)]


L_{x}[c] = 0


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


L_{x}[ln(x)] = 1

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}

funcions continues

[∀s][ s > 0 ==> lim [h-->0][ |f(x+h)+(-1)f(x)| < s ]  ]


x^{n} és continua en R


lim [h-->0][ |(x+h)^{n}+(-1)x^{n}| < s ]


e^{x} és continua en R


lim [h-->0][ |e^{x}·( e^{h}+(-1) )| < s ]


ln(x) és continua en ¬( x = 0 ) 


lim [h-->0][ |ln(1+(h/x) )| < s ]


ln(x) no és continua en x = 0 


lim [h-->0][ |ln(2)| >] s ]

funció continua

Si |f(x)| [< ln(x) ==> f(x) és continua en x = 1.


Si |f(x)| [< ln( (x/a) ) ==> f(x) és continua en x = a.


Si |f(x)| [< ln( (x/(-a)) ) ==> f(x) és continua en x = (-a).

funció continua

Si |f(x)| [< ln(x+e)+(-1) ==> f(x) és continua en x = 0.


Si |f(x)| [< ln( (x+(-a))+e )+(-1) ==> f(x) és continua en x = a.


Si |f(x)| [< ln( (x+a)+e)+(-1) ==> f(x) és continua en x = (-a).

funció continua

Si |f(x)| [< ln(x+1) ==> f(x) és continua en x = 0.


Si |f(x)| [< ln( (x+(-a))+1 ) ==> f(x) és continua en x = a.


Si |f(x)| [< ln( (x+a)+1) ==> f(x) és continua en x = (-a).

funció continua

Si |f(x)| [< x^{n} ==> f(x) és continua en x = 0.


Si |f(x)| [< (x+(-a))^{n} ==> f(x) és continua en x = a.


Si |f(x)| [< (x+a)^{n} ==> f(x) és continua en x = (-a).

funció continua

Si |f(x)| [< e^{ln(a)·x}+(-1) ==> f(x) és continua en x = 0


Si |f(x)| [< e^{ln(a)·(x+(-c))}+(-1) ==> f(x) és continua en x = c


Si |f(x)| [< e^{ln(a)·(x+c)}+(-1) ==> f(x) és continua en x = (-c)

ecuació integral

∫ [a--->x]-[ d_{x}[f(x)] ] d[x]+n·f(x) = (n+1)·∫ [a--->x]-[ d_{x}[f(x)] ] d[x]
f(x) = (x+(-a))

zero de funció

Si f(x) = (x+(-a))+(x+(-b)) ==> [∃c][ c€[a+s,b+(-s)]_{K} & f(c) = 0 ]
f(a+s) = (-1)·f(b+(-s))
f(b+(-s)) = (-1)·f(a+s)


Si s = ( (b+(-a))/2 ) ==> f( (a+b)/2 ) = 0


Si f(x) = ( s/(x+(-a)) )+( s/(x+(-b)) ) ==> [∃c][ c€[a+s,b+(-s)]_{K} & f(c) = 0 ]
f(a+s) = (-1)·f(b+(-s))
f(b+(-s)) = (-1)·f(a+s)


Si s = ( (b+(-a))/2 ) ==> f( (a+b)/2 ) = 0


Si f(x) = (x+(-1)·a_{1})+...(n)...+(x+(-1)·a_{n}) ==> f( (a_{1}+...(n)...+a_{n})/n ) = 0


0 = nx+(-1)( a_{1}+...(n)...+a_{n} )
a_{1}+...(n)...+a_{n} = nx

funció continua

teorema:
Si |f(x)| [< |x| ==> f(x) és continua en x = 0


demostració:
sigui s > 0 ==>
|f(x)+(-1)f(0)| [< |x|+|0| = |x| < s


teorema:
Si |f(x)| [< |x+(-a)| ==> f(x) és continua en x = a


demostració:
sigui s > 0 ==>
|f(x)+(-1)f(a)| [< |x+(-a)|+|0| = |x+(-a)| < s


teorema:
Si |f(x)| [< |x+a| ==> f(x) és continua en x = (-a)


demostració:
sigui s > 0 ==>
|f(x)+(-1)f(-a)| [< |x+a|+|0| = |x+a| < s

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

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