英文:
Problems solving nonlinear systems of equations involving trig functions in Python using Sympy
问题
以下是翻译好的内容，去掉了代码部分：
"I'm trying to create a program that solves the solutions for an equation of a circle:
This is the code I wrote for it using Sympy. It works for functions like x^2, but I always encounter an error when using tan(x) or sin(x)
Graphical solutions show that a solution definitely exists."
英文:
I'm trying to create a program that solves the solutions for an equation of a circle:
This is the code I wrote for it using Sympy. It works for functions like x^2, but I always encounter an error when using tan(x) or sin(x)
import sympy as sym
x = sym.symbols('x')
y = sym.symbols('y')
f = sym.Eq(sym.sqrt(9-x**2), y)
g = sym.Eq(sym.tan(x), y)
print(sym.solve((f, g)(x, y)))
Graphical solutions show that a solution definitely exists.
答案1
得分: 3
"解决方案的图表显示，这并不意味着可以通过分析方法计算解决方案。solve 用于获得分析解决方案。我们可以使用 nsolve 来计算数值解，但我们需要找到合适的初始条件，因此图表非常重要。
from sympy import *
var("x, y")
eq1 = Eq(y, sqrt(9 - x**2))
eq2 = Eq(y, tan(x))
plot(eq1.rhs, eq2.rhs, (x, -pi, pi), ylim=(-5, 5))
图表显示在区间[-pi, pi]中有3个解。阅读 help(nsolve) 以熟悉其语法。要找到解决方案，我们需要尝试不同的初始猜测值：
s1 = nsolve([eq1, eq2], [x, y], [-2, 2])
s2 = nsolve([eq1, eq2], [x, y], [1, 2])
s3 = nsolve([eq1, eq2], [x, y], [-2.9, 0.5])
print(s1)
print(s2)
print(s3)
# 输出
# Matrix([[-1.98982600933056], [2.24512637786642]])
# Matrix([[1.22089272895276], [2.74033226897584]])
# Matrix([[-2.99643622736316], [0.146184593393441]])
您可以通过图形验证这些解决方案的正确性。"
英文:
Just because plots show a solution, it doesn't mean that the solution can be computed analytically. solve is used to get an analytical solution. We can use nsolve to compute a numerical solution, but we are going to find suitable initial conditions, hence a plot is more than welcome.
from sympy import *
var("x, y")
eq1 = Eq(y, sqrt(9 - x**2))
eq2 = Eq(y, tan(x))
plot(eq1.rhs, eq2.rhs, (x, -pi, pi), ylim=(-5, 5))
The plot shows that there are 3 solutions in the interval [-pi, pi]. Read help(nsolve) to get acquainted with its syntax. To find the solutions we will have to play with initial guesses:
s1 = nsolve([eq1, eq2], [x, y], [-2, 2])
s2 = nsolve([eq1, eq2], [x, y], [1, 2])
s3 = nsolve([eq1, eq2], [x, y], [-2.9, 0.5])
print(s1)
print(s2)
print(s3)
# out
# Matrix([[-1.98982600933056], [2.24512637786642]])
# Matrix([[1.22089272895276], [2.74033226897584]])
# Matrix([[-2.99643622736316], [0.146184593393441]])
You can verify graphically that the solutions are correct.
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