Metodo de posicion falsa

1-
posicion_falsa
método de la posicion falsa
introduzca el valor de po: 29000
introduzca el valor de p1: 29100
introduzca el valor deprecision: 1e-12
introduzca el valor de la funcion f: (((1+((x/282000)^5)^(1/8))^4)^(1/5))*(0.294473*(x)^(1/2))-59.7
n= 1 po=2.900000000e+004p1= 2.910000000e+004 p2=2.906973556e+004 error(|P2-P1 F(17.33)=16.11
x2=42.67 ---> F(42.67)=18.89
x3=37.33 ---> F(37.33)=15.00x4=30.10 ---> F(30.10)=10.56
x5=29.31 ---> F(29.31)=9.44
x6=28.74 ---> F(28.74)=8.89

Co=0 C6=0
a0=6.670000000 h0=10.660000000a1=16.110000000 h1=25.340000000
a2=18.890000000 h2=-5.340000000
a3=15.000000000 h3=-7.230000000
a4=10.560000000 h4=-0.790000000a5=9.440000000 h5=-0.570000000
a6=8.890000000
S1(x)=a1+b1(x-x1)+C1(x-x1)^2+d1(x-x1)^3

h0C0+2(h0-h1)C1+h1C2=3(a2-a1)/h1-3(a1-a0)/h0h1C1+2(h1-h2)C2+h2C3=3(a3-a2)/h2-3(a2-a1)/h1
h2C2+2(h2-h3)C3+h3C4=3(a4-a3)/h3-3(a3-a2)/h2
h3C3+2(h3-h4)C4+h4C5=3(a5-a4)/h4-3(a4-a3)/h3h4C4+2(h4-h5)C5+h5C6=3(a6-a5)/h5-3(a5-a4)/h4

Resolviendo el sistema de ecuaciones
0.00C0+72.00C1+25.34C2=-2.33
25.34C1+40.00C2+-5.34C3=1.86-5.34C2+-25.14C3+-7.23C4=-0.34
-7.23C3+-16.04C4+-0.79C5=2.41
-0.79C4+-2.72C5+-0.00C6=-1.36

Se obtiene(mediante uso de calculadora):

Co=0
C1=-0.065797C2=0.095002
C3=0.051085
C4=-0.200774
C5=0.558313
C6=0

b1=0.418788398 d1=0.002115220

S1(x)=a1+b1(x-x1)+C1(x-x1)^2+d1(x-x1)^3=16.405766362
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