python - Sympy:如何将乘积的对数简化为对数和？

Question

为了激发这个问题，sympy.concrete 有一些有效的工具来处理符号和。为了将这些工具应用于符号产品，必须取对数。然而，直接取对数并不会自动给出转换:

import sympy as sp
sp.init_printing() # display math as latex
z = sp.Symbol('z')
j,k = sp.symbols('j,k')
Prod = sp.Product( (z + sp.sqrt(1-4*j*z**2))**(-1), (j,1,k) )
sp.log(Prod)

给予

在所有可能的变化中:

sp.log(Prod)
sp.log(Prod).expand()
sp.log(Prod).simplify()
sp.expand_log(sp.log(Prod),force=True)

Question. How to convert it into sum of logarithms?

相关:

How to simplify logarithm of exponent in sympy?

Answer 1

假设还没有具有所需行为的标准函数，我编写了自己的函数，模仿了

sp.expand_log(expr, force=True)

此代码递归地遍历表达式，试图找到模式 log(product) 并将它们替换为 sum(log)。这也支持多索引求和。

Code.

def concrete_expand_log(expr, first_call = True):
    import sympy as sp
    if first_call:
        expr = sp.expand_log(expr, force=True)
    func = expr.func
    args = expr.args
    if args == ():
        return expr
    if func == sp.log:
        if args[0].func == sp.concrete.products.Product:
            Prod = args[0]
            term = Prod.args[0]
            indices = Prod.args[1:]
            return sp.Sum(sp.log(term), *indices)
    return func(*map(lambda x:concrete_expand_log(x, False), args))

Example.

import sympy as sp
from IPython.display import display
sp.init_printing() # display math as latex
z = sp.Symbol('z')
j,k,n = sp.symbols('j,k,n')
Prod = sp.Product( (z + sp.sqrt(1-4*j*z**2))**(-1), (j,0,k))
expr = sp.log(z**(n-k) * (1 - sp.sqrt((1 - 4*(k+2)*z**2)/(1-4*(k+1)*z**2)) ) * Prod)
display(expr)

display(concrete_expand_log(expr))