Autor: Alfredo Andrés Valenzuela Riquelme
Control sorpresa con variantes de la prueba 2 [ ] "Nodos no equidistantes"
Dados los nodos
x
0
=
0
,
x
1
=
1
/
3
,
x
2
=
1
{\displaystyle x_{0}=0,x_{1}=1/3,x_{2}=1}
ponderaciones
p
0
,
p
1
,
p
2
{\displaystyle p_{0},p_{1},p_{2}}
∫
x
0
x
2
f
(
x
)
d
x
=
(
x
2
−
x
0
)
=
{\displaystyle \displaystyle \int _{x_{0}}^{x_{2}}f(x)dx=(x_{2}-x_{0})=}
∑
i
=
0
2
p
i
f
(
x
i
)
=
p
0
f
(
x
0
)
+
p
1
f
(
x
1
)
+
p
2
f
(
x
2
)
{\displaystyle \displaystyle \sum _{i=0}^{2}p_{i}f(x_{i})=p_{0}f(x_{0})+p_{1}f(x_{1})+p_{2}f(x_{2})}
sea válida para cualquiera función cuadrática
f
(
x
)
=
a
+
b
x
+
c
x
2
{\displaystyle f(x)= a + bx + cx^2 }
Solución:
∫
x
0
x
2
f
(
x
)
d
x
=
(
x
2
−
x
0
)
=
{\displaystyle \displaystyle \int _{x_{0}}^{x_{2}}f(x)dx=(x_{2}-x_{0})=}
∑
i
=
0
2
p
i
f
(
x
i
)
=
p
0
f
(
x
0
)
+
p
1
f
(
x
1
)
+
p
2
f
(
x
2
)
{\displaystyle \displaystyle \sum _{i=0}^{2}p_{i}f(x_{i})=p_{0}f(x_{0})+p_{1}f(x_{1})+p_{2}f(x_{2})}
válida para cualquier función cuadrática
f
(
x
)
=
a
+
b
x
+
c
x
2
{\displaystyle f(x)= a + bx + cx^2 }
Determinar ponderaciones
p
0
,
p
1
,
p
2
{\displaystyle p_{0},p_{1},p_{2}}
x
0
=
0
,
x
1
=
1
/
3
,
x
2
=
1
{\displaystyle x_{0}=0,x_{1}=1/3,x_{2}=1}
∫
0
1
f
(
x
)
d
x
=
P
0
f
(
0
)
+
P
1
f
(
1
/
3
)
+
P
2
(
1
)
{\displaystyle \displaystyle \int _{0}^{1}f(x)dx=P_{0}f(0)+P_{1}f(1/3)+P_{2}(1)}
Paso 1: "Calcular el lado izquierdo"
∫
0
1
f
(
x
)
d
x
{\displaystyle \displaystyle \int _{0}^{1}f(x)dx}
∫
0
1
(
a
+
b
x
+
c
x
2
)
d
x
{\displaystyle \displaystyle \int _{0}^{1}(a+bx+cx^{2})dx}
∫
0
1
(
a
)
d
x
{\displaystyle \displaystyle \int _{0}^{1}(a)dx}
∫
0
1
(
b
x
)
d
x
{\displaystyle \displaystyle \int _{0}^{1}(bx)dx}
∫
0
1
(
c
x
2
)
d
x
{\displaystyle \displaystyle \int _{0}^{1}(cx^{2})dx}
=
a
+
b
/
2
+
c
/
3
{\displaystyle =a+b/2+c/3}
Paso 2: "Calcular lado derecho"
P
0
f
(
0
)
+
P
1
f
(
1
/
3
)
+
P
2
(
1
)
{\displaystyle P_{0}f(0)+P_{1}f(1/3)+P_{2}(1)}
f
(
x
)
=
a
+
b
x
+
c
x
2
{\displaystyle f(x)= a + bx + cx^2 }
=
P
0
f
(
a
)
+
P
1
f
(
a
+
b
/
3
+
c
/
9
)
+
P
2
(
a
+
b
+
c
)
{\displaystyle =P_{0}f(a)+P_{1}f(a+b/3+c/9)+P_{2}(a+b+c)}
=
a
(
P
0
+
P
1
+
P
2
)
+
b
(
P
1
/
3
+
P
2
)
+
c
(
P
1
/
9
+
P
2
)
{\displaystyle =a(P_{0}+P_{1}+P_{2})+b(P_{1}/3+P_{2})+c(P_{1}/9+P_{2})}
Paso 3: "Comparar ambos lados"
a
+
b
/
2
+
c
/
3
{\displaystyle a+b/2+c/3}
=
a
(
P
0
+
P
1
+
P
2
)
+
b
(
P
1
/
3
+
P
2
)
+
c
(
P
1
/
9
+
P
2
)
{\displaystyle =a(P_{0}+P_{1}+P_{2})+b(P_{1}/3+P_{2})+c(P_{1}/9+P_{2})}
La ecuación debe ser válida para cualquier selección (a,b,c)
(
a
,
b
,
c
)
=
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
,
(
0
,
0
,
1
)
{\displaystyle (a,b,c)=(1,0,0),(0,1,0),(0,0,1)}
--> Formular sistema de tres ecuaciones:
(
1
,
0
,
0
)
:
a
=
1
,
b
=
0
,
c
=
0
{\displaystyle (1,0,0):a=1,b=0,c=0}
1
=
P
0
+
P
1
+
P
2
{\displaystyle 1=P_{0}+P_{1}+P_{2}}
(
0
,
1
,
0
)
:
a
=
0
,
b
=
1
,
c
=
0
{\displaystyle (0,1,0):a=0,b=1,c=0}
1
/
2
=
P
1
1
/
3
{\displaystyle 1/2=P_{1}1/3}
(
0
,
0
,
1
)
:
a
=
0
,
b
=
0
,
c
=
1
{\displaystyle (0,0,1):a=0,b=0,c=1}
1
/
3
=
P
1
1
/
9
+
P
2
{\displaystyle 1/3=P_{1}1/9+P_{2}}
Una vez resuelto el sistema:
P
0
=
0
{\displaystyle P_0 = 0 }
P
1
=
3
/
4
{\displaystyle P_{1}=3/4}
P
2
=
1
/
4
{\displaystyle P_{2}=1/4}
"Sistema no-lineal sobredeterminado"
f
(
x
,
y
)
=
y
e
x
=
0
{\displaystyle f(x,y) = ye^x = 0 }
g
(
x
,
y
)
=
e
2
y
=
0
{\displaystyle g(x,y)=e^{2y}=0}
h
(
x
,
y
)
=
e
3
x
y
=
0
{\displaystyle h(x,y)=e^{3xy}=0}
a) Linealizar el sistema cerca de un punto
X
0
,
Y
0
,
Z
0
{\displaystyle X_{0},Y_{0},Z_{0}}
f
1
(
x
1
,
x
2
,
.
.
.
x
n
)
=
0
{\displaystyle f_{1}(x_{1},x_{2},...x_{n})=0}
f
2
(
x
1
,
x
2
,
.
.
.
x
n
)
=
0
{\displaystyle f_{2}(x_{1},x_{2},...x_{n})=0}
f
n
(
x
1
,
x
2
,
.
.
.
x
n
)
=
0
{\displaystyle f_{n}(x_{1},x_{2},...x_{n})=0}
forma general
f
(
x
)
=
a
1
x
1
+
a
2
x
2
+
.
.
.
a
n
x
n
−
b
=
0
{\displaystyle f(x)=a_{1}x_{1}+a_{2}x_{2}+...a_{n}x_{n-b}=0}
f
(
x
,
y
)
==
f
(
x
0
,
y
0
)
+
{\displaystyle f(x,y)==f(x_{0},y_{0})+}
∂
f
(
x
0
,
y
0
)
∂
x
(
x
−
x
0
)
+
{\displaystyle {\partial f(x_{0},y_{0}) \over \partial x}(x-x_{0})+}
∂
f
(
x
0
,
y
0
)
∂
y
(
y
−
y
0
)
{\displaystyle {\partial f(x_{0},y_{0}) \over \partial y}(y-y_{0})}
-->
f
(
x
,
y
)
=
y
e
x
=
0
{\displaystyle f(x,y) = ye^x = 0 }
resolviendo...
=
y
0
e
x
0
+
(
y
0
e
x
0
)
(
x
−
x
0
)
+
e
x
0
(
y
−
y
0
)
=
k
(
x
,
y
)
{\displaystyle =y_{0}e^{x_{0}}+(y_{0}e^{x_{0}})(x-x_{0})+e^{x_{0}}(y-y_{0})=k(x,y)}
-->
g
(
x
,
y
)
=
e
2
y
=
0
{\displaystyle g(x,y)=e^{2y}=0}
resolviendo...
e
2
y
0
+
e
2
y
0
(
x
−
x
0
)
+
2
e
y
0
(
y
−
y
0
)
=
l
(
x
,
y
)
{\displaystyle e^{2{y_{0}}}+e^{2{y_{0}}}(x-x_{0})+2e^{y_{0}}(y-y_{0})=l(x,y)}
-->
h
(
x
,
y
)
=
e
3
x
y
=
0
{\displaystyle h(x,y)=e^{3xy}=0}
resolviendo...
e
3
x
0
y
0
+
3
y
0
e
3
x
0
y
0
(
x
−
x
0
)
+
3
x
0
e
3
x
0
y
0
(
y
−
y
0
)
=
m
(
x
,
y
)
{\displaystyle e^{3{x_{0}}{y_{0}}}+3y_{0}e^{3{x_{0}}{y_{0}}}(x-x_{0})+3x_{0}e^{3{x_{0}}{y_{0}}(y-y_{0})}=m(x,y)}
buscamos los valores x1, y1 que minimizan la sumatoria
k
2
(
x
1
,
y
1
)
+
l
2
(
x
1
,
y
1
)
+
m
2
(
x
1
,
y
1
)
{\displaystyle k^{2}(x_{1},y_{1})+l^{2}(x_{1},y_{1})+m^{2}(x_{1},y_{1})}
Escribir el sistema linealizado:
h
(
x
1
,
y
1
)
=
0
{\displaystyle h(x_{1},y_{1})=0}
l
(
x
1
,
y
1
)
=
0
{\displaystyle l(x_{1},y_{1})=0}
m
(
x
1
,
y
1
)
=
0
{\displaystyle m(x_{1},y_{1})=0}
En forma matricial
Δ
x
=
x
1
−
x
0
,
Δ
y
=
y
1
−
y
0
{\displaystyle \Delta x=x_{1}-x_{0},\Delta y=y_{1}-y_{0}}
(
y
0
e
x
0
e
x
0
e
2
y
0
2
e
2
y
0
3
y
0
e
3
x
0
y
0
3
x
0
e
3
x
0
y
0
)
(
Δ
x
Δ
y
)
=
−
(
f
(
x
0
,
y
0
)
g
(
x
0
,
y
0
)
h
(
x
0
,
y
0
)
)
{\displaystyle {\begin{pmatrix}y_{0}e^{x_{0}}&e^{x_{0}}\\e^{2{y_{0}}}&2e^{2{y_{0}}}\\3y_{0}e^{3{x_{0}}{y_{0}}}&3x_{0}e^{3{x_{0}}{y_{0}}}\\\end{pmatrix}}{\begin{pmatrix}\Delta x\\\\\\\\\Delta y\end{pmatrix}}=-{\begin{pmatrix}f(x_{0},y_{0})\\g(x_{0},y_{0})\\h(x_{0},y_{0})\\\end{pmatrix}}}
Sistema lineal sobredeterminado de la forma
A
Δ
x
=
b
{\displaystyle A\Delta x=b}
Pendiente.... pregunta sobre distancia dentro de dos funciones. (por si alguien lo quiere completar.)
APORTE : Natalia Gonzalez
Distancia dentro dos funciones :
Calcular la distancia mínima dentro las curvas de las funciones
f
(
x
)
=
−
x
2
,
y
=
3
+
s
i
n
(
x
)
{\displaystyle f(x)=-x^{2},y=3+sin(x)}
a) Definir una función
g
:
R
2
→
R
{\displaystyle g: R^{2}\rightarrow R }
b)Formular como se realizaría un paso de la iteración según método de Newton-Raphson; dado cierto punto
(
x
i
,
y
i
)
{\displaystyle (x_i, y_i) }
(
x
i
+
1
,
y
i
+
1
)
{\displaystyle (x_{i+1}, y_{i+1})}
f
(
x
)
=
−
x
2
,
y
=
3
+
s
i
n
(
x
)
{\displaystyle f(x)=-x^{2},y=3+sin(x)}
f
1
(
x
)
=
f
2
(
x
)
,
g
(
x
)
=
f
1
(
x
)
−
f
2
(
x
)
=
0
{\displaystyle f_{1}(x)=f_{2}(x),g(x)=f_{1}(x)-f_{2}(x)=0}
g
(
x
)
=
−
x
2
−
(
3
+
s
i
n
(
x
)
)
{\displaystyle g(x)=-x^{2}-(3+sin(x))}
g
(
x
)
=
−
x
2
−
3
−
s
i
n
(
x
)
)
{\displaystyle g(x)=-x^{2}-3-sin(x))}
g
′
(
x
)
=
−
2
x
−
c
o
s
(
x
)
{\displaystyle g'(x)=-2x-cos(x)}
⇒
y
1
=
f
1
(
x
)
=
−
x
1
2
{\displaystyle \Rightarrow y_{1}=f1(x)=-x_{1}^{2}}
y
2
(
x
)
=
f
2
=
3
+
s
i
n
(
x
)
)
{\displaystyle y_{2}(x)=f_{2}=3+sin(x))}
d
(
x
1
,
x
2
)
=
(
x
1
−
x
2
)
2
+
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
2
2
{\displaystyle d(x_{1},x_{2})={\sqrt[{2}]{(x_{1}-x_{2})^{2}+(-x_{1}^{2}-(3+sin(x_{2})))^{2}}}}
d
2
=
(
x
1
−
x
2
)
2
+
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
2
{\displaystyle d^{2}=(x_{1}-x_{2})^{2}+(-x_{1}^{2}-(3+sin(x_{2})))^{2}}
δ
F
d
2
δ
x
1
=
δ
(
x
1
−
x
2
)
2
δ
x
1
+
δ
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
2
δ
x
1
{\displaystyle {\frac {\delta Fd^{2}}{\delta x_{1}}}={\frac {\delta (x_{1}-x_{2})^{2}}{\delta x_{1}}}+{\frac {\delta (-x_{1}^{2}-(3+sin(x_{2})))^{2}}{\delta x_{1}}}}
0
=
2
(
x
1
−
x
2
)
+
2
(
−
x
1
2
−
4
x
1
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
{\displaystyle 0=2(x_{1}-x_{2})+2(-x_{1}^{2}-4x_{1}(-x_{1}^{2}-(3+sin(x_{2})))}
δ
F
d
δ
x
2
=
δ
(
x
2
−
x
2
)
2
δ
x
2
+
δ
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
2
δ
x
2
{\displaystyle
\frac{\delta Fd}{\delta x_{2}} = \frac{\delta (x_{2}-x_{2})^2}{\delta x_{2}} + \frac{\delta (-x_{1}^2 - (3 + sin(x_{2})))^2}{\delta x_{2}}
}
0
=
−
2
(
x
1
−
x
2
)
−
2
c
o
s
(
x
2
)
(
−
x
1
2
−
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
{\displaystyle 0=-2(x_{1}-x_{2})-2cos(x_{2})(-x_{1}^{2}-(-x_{1}^{2}-(3+sin(x_{2})))}
(
x
1
−
x
2
)
−
4
x
1
(
−
x
1
2
−
2
x
1
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
=
0
⇒
u
(
x
1
,
x
2
)
{\displaystyle (x_{1}-x_{2})-4x_{1}(-x_{1}^{2}-2x_{1}(-x_{1}^{2}-(3+sin(x_{2})))=0\Rightarrow u(x_{1},x_{2})}
(
x
1
−
x
2
)
−
c
o
s
(
x
2
)
(
−
x
1
2
−
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
=
0
⇒
v
(
x
1
,
x
2
)
{\displaystyle (x_{1}-x_{2})-cos(x_{2})(-x_{1}^{2}-(-x_{1}^{2}-(3+sin(x_{2})))=0\Rightarrow v(x_{1},x_{2})}
con
x
0
,
y
0
⇒
(
0
,
1
)
{\displaystyle x_{0},y_{0}\Rightarrow (0,1)}
(
δ
u
i
δ
x
1
δ
u
i
δ
x
2
δ
v
i
δ
x
1
δ
v
i
δ
x
2
)
(
Δ
x
1
Δ
x
2
)
=
−
(
u
1
v
2
)
{\displaystyle
\begin{pmatrix} \frac{\delta u_{i}}{\delta x_{1}} & \frac{\delta u_{i}}{\delta x_{2}}\\ \frac{\delta v_{i}}{\delta x_{1}} & \frac{\delta v_{i}}{\delta x_{2}}\end{pmatrix} \begin{pmatrix} \Delta x_{1} \\ \Delta x_{2} \end{pmatrix} = - \begin{pmatrix} u_{1} \\ v_{2} \end{pmatrix}
}
(
1
+
4
x
1
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
)
−
c
o
s
(
x
2
)
1
−
2
x
1
−
1
+
s
e
n
(
x
2
)
(
−
x
1
2
−
(
3
+
s
i
n
(
x
2
)
)
+
c
o
s
(
x
2
)
)
(
Δ
x
1
Δ
x
2
)
=
−
(
u
1
v
2
)
{\displaystyle {\begin{pmatrix}1+4x_{1}(-x_{1}^{2}-(3+sin(x_{2})))&-cos(x_{2})\\1-2x_{1}&-1+sen(x_{2})(-x_{1}^{2}-(3+sin(x_{2}))+cos(x_{2})\end{pmatrix}}{\begin{pmatrix}\Delta x_{1}\\\Delta x_{2}\end{pmatrix}}=-{\begin{pmatrix}u_{1}\\v_{2}\end{pmatrix}}}
Escribe la segunda sección de tu artículo aquí.
![{\displaystyle d(x_{1},x_{2})={\sqrt[{2}]{(x_{1}-x_{2})^{2}+(-x_{1}^{2}-(3+sin(x_{2})))^{2}}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/745a21ec47e2baf1c66faa6c51a8f67a56ae5838)
