[英]Finding roots of an equation involving a summation using sympy
我目前是 sympy 的新手,我正在嘗試在 Python 的附加圖像中重現 Mathematica 示例。 我的嘗試寫在下面,但它返回一個空列表
import sympy
m , n, D_star, a, j = sympy.symbols('m , n, D_star, a, j')
s1 = sympy.Sum(a**(j-1),(j, 1, m-1))
rhs = 6 * sympy.sqrt((D_star * (1 + a)*(n - 1))/2)
expand_expr = sympy.solve(s1 - rhs, m)
temp = sympy.lambdify((a, n, D_star), expand_expr, 'numpy')
n = 100
a = 1.2
D_star = 2.0
ms = temp(1.2, 100, 2.0)
ms
# what I get is an empty list []
# expected answer using Mma FindRoot function is 17.0652
添加.doit()
來擴大總和似乎有幫助。 它給出了Piecewise((m - 1, Eq(a, 1)), ((a - a**m)/(1 - a), True))/a
的總和s1
。
from sympy import symbols, Eq, Sum, sqrt, solve, lambdify
m, n, j, a, D_star = symbols('m n j a D_star')
s1 = Sum(a**(j - 1), (j, 1, m - 1)).doit()
rhs = 6 * sqrt((D_star * (1 + a) * (n - 1)) / 2)
expand_expr = solve(Eq(s1, rhs), m)
temp = lambdify((a, n, D_star), expand_expr, 'numpy')
n = 100
a = 1.2
D_star = 2.0
ms = temp(1.2, 100, 2.0)
這給出了expand_expr
:
[Piecewise((log(a*(3*sqrt(2)*a*sqrt(D_star*(a*n - a + n - 1)) - 3*sqrt(2)*sqrt(D_star*(a*n - a + n - 1)) + 1))/log(a), Ne(a, 1)), (nan, True)),
Piecewise((3*sqrt(2)*a*sqrt(D_star*(a*n - a + n - 1)) + 1, Eq(a, 1)), (nan, True))]
它分為a != 1
和a == 1
。
ms
的結果給出[array(17.06524172), array(nan)]
,再次以有點尷尬的方式分隔假設a == 1
。
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