Sympy代数求解系列求和

Question

我正在尝试在以下等式中解决C

What are the undocumented features and limitations of the Windows FINDSTR command?

我可以使用sympy为x's填充x0, x2, ..., x4，例如i=0，但似乎无法弄明白如何为t {{1}执行此操作}}。例如。数量有限

from sympy import summation, symbols, solve

x0, x1, x2, x3, x4, alpha, C = symbols('x0, x1, x2, x3, x4, alpha, C')

e1 = ((x0 + alpha * x1 + alpha**(2) * x2 + alpha**(3) * x3 + alpha**(4) * x4)
      / (1 + alpha + alpha**(2) + alpha**(3) + alpha**(4)))
e2 = (x3 + alpha * x4) / (1 + alpha)
rhs = (x0 + alpha * x1 + alpha**(2) * x2)  / (1 + alpha + alpha**(2))

soln_C = solve(e1 - C*e2 - rhs, C)

非常感谢任何见解。

Answer 1

感谢@bryans指点我Sum的方向。详细阐述他的评论，这是一个似乎有效的解决方案。因为我是sympy的新手，如果有人有更简洁的方法，请分享。

from sympy import summation, symbols, solve, Function, Sum

alpha, C, t, i = symbols('alpha, C, t, i')
x = Function('x')

s1 = Sum(alpha**i * x(t-i), (i, 0, t)) / Sum(alpha**i, (i, 0, t))
s2 = Sum(alpha**i * x(t-3-i), (i, 0, t-3)) / Sum(alpha**i, (i, 0, t-3)) 
rhs = (x(0) + alpha * x(1) + alpha**(2) * x(2))  / (1 + alpha + alpha**(2))

soln_C = solve(s1 - C*s2 - rhs, C)

Answer 2

我不确定是否可以将其归类为更多的“简洁” ，但是当您知道汇总的上限时，它也可能很有用。假设我们要评估这个表达式：

我们可以用sympy表示并解决它，如下所示：

from sympy import init_session
init_session(use_latex=True)
n = 4
As = symbols('A_1:' + str(n+1))
x = symbols('x')
exp = 0
for i in range(n):
    exp += As[i]/(1+x)**(i+1)
Ec = Eq(exp,0)
sol = solve(Ec,x)
#print(sol)
#sol    #Or, if you're working on jupyter...