带有迭代的inf和nan值

Question

之前的代码中有Rs值

from numpy import exp as e
Ri = 9
Rr = 19/2
Rs = 10
i = 0
Er = 0
Rra = 0
x = 1

def F(n):
    return 745*(1-e(-x/10))-49*x

#函数准计算误差相对论 def错误（Rra、Rrn）： Erel=abs（（Rrn-Rra）/Rrn）*100 回程鱼

print ('{:^15}{:^15}{:^15}{:^15}{:^15}{:^15}'.format('# iter','Ri','Rs','Rr','F(Ri)','Erel(%)'))
      
while (i < 10):
    Er = error(Rra,Rr)
    Rra = Rr `#Rra sera el Rr anterior para determinar el error relativo`
    if F(Ri).all() * F(Rr).all() < 0:
        Rs = Rr
    elif F(Ri).all() * F(Rr).all() > 0:
        Ri = Rr
    elif F(Rr) == 0:
        print('La raiz es',Rr)

    Rr = (Ri + Rs) / 2
    i = i + 1

    #Condicional para el primer error relativo

    if i == 1:
             print ('{:^15}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.6f}'.format(i-1,Ri,Rs,Rr,F(Rr)))
    elif i > 1:
             print ('{:^15}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.6f}{:^15.10f}'.format(i-1,Ri,Rs,Rr,F(Rr),Er))

我正在开发一个练习，其中我必须使用Regula-Falsi方法，但在获得结果中相应的Iteraction时，我得到inf和nan值，请有人指导我或告诉我它有什么错

from numpy import exp as e
 
Ri = 9
Rr = 10
i = 0
Er = 0
x = 1
 
def F(n):
    return 745*(1-e(-x/10))-49*x
 
#Funcion para calcular el error relativo
def error (Rra,Rrn):
    Erel = abs((Rrn - Rra) / Rrn) * 100
    return Erel
 
 
print ('{:^15}{:^15}{:^15}{:^15}{:^15}{:^25}'.format('# iter','Ri','Rs','Rr','F(Rr)','Erel(%)'))
 
while (i < 10):
    Er = error(Rra,Rr)
    Rra = Rr #Rra sera el Rr anterior para determinar el error relativo
    if F(Ri) * F(Rr) < 0:
        Rs = Rr
    elif F(Ri) * F(Rr) > 0:
        Ri = Rr
    elif F(Rr) == 0:
        print('La raiz es',Rr)
 
    Rr = (F(Rs)*Ri -F(Ri)*Rs)/(F(Rs) - F(Ri))
 
    i = i + 1
 
     #Condicional para el primer error relativo
 
    if i == 1:
             print ('{:^15}{:^15.10f}{:^15.10f}{:^15.10f}{:^15.10f}'.format(i-1,Ri,Rs,Rr,F(Rr)))
    elif i > 1:
             print ('{:^15}{:^15.10f}{:^15.10f}{:^15.10f}{:^15.10f}{:^25.10f}'.format(i-1,Ri,Rs,Rr,F(Rr),Er))

结果

# iter           Ri             Rs             Rr            F(Rr)              Erel(%)
       0        10.0000000000  10.0000000000       nan       21.8961235632
       1             nan       10.0000000000       nan       21.8961235632            nan
       2             nan       10.0000000000       nan       21.8961235632            nan
       3             nan       10.0000000000       nan       21.8961235632            nan
       4             nan       10.0000000000       nan       21.8961235632            nan
       5             nan       10.0000000000       nan       21.8961235632            nan
       6             nan       10.0000000000       nan       21.8961235632            nan
       7             nan       10.0000000000       nan       21.8961235632            nan
       8             nan       10.0000000000       nan       21.8961235632            nan
       9             nan       10.0000000000       nan       21.8961235632            nan
/usr/local/lib/python3.7/dist-packages/ipykernel_launcher.py:30: RuntimeWarning: invalid value encountered in double_scalars

Answer 1

x = 1
 
def F(n):
    return 745*(1-e(-x/10))-49*x

在F(n)方法中，x是一个赋值为1的值，并且永远不会更改。因此F(n)将始终返回相同的值21.896123563210153

    if F(Ri) * F(Rr) < 0:
        Rs = Rr
    elif F(Ri) * F(Rr) > 0:
        Ri = Rr
    elif F(Rr) == 0:
        print('La raiz es',Rr)

因此，F(Ri) * F(Rr)始终是正值。因此Ri具有相同的Rr值

Rr = (F(Rs)*Ri -F(Ri)*Rs)/(F(Rs) - F(Ri))

计算Rr的方法是将F(Rs) - F(Ri)除以0。这在数学上是违法的。所以Rr在你的例子中是nan

在第一个循环之后，Ri被赋值为Rr

def error (Rra,Rrn):
    Erel = abs((Rrn - Rra) / Rrn) * 100
    return Erel

Er = error(Rra,Rr)

Rrn的值为Rr。在(Rrn - Rra) / Rrn中，将nan值除以。这导致nan

总之，根本原因是方法F(n)中的x是一个常量值