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Python solvers.solve_undetermined_coeffs函数代码示例

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

本文整理汇总了Python中sympy.solvers.solve_undetermined_coeffs函数的典型用法代码示例。如果您正苦于以下问题:Python solve_undetermined_coeffs函数的具体用法?Python solve_undetermined_coeffs怎么用?Python solve_undetermined_coeffs使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。



在下文中一共展示了solve_undetermined_coeffs函数的3个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。

示例1: test_solve_undetermined_coeffs

def test_solve_undetermined_coeffs():
    assert solve_undetermined_coeffs(a*x**2 + b*x**2 + b*x + 2*c*x + c + 1, [a, b, c], x) == \
        {a: -2, b: 2, c: -1}
    # Test that rational functions work
    assert solve_undetermined_coeffs(a/x + b/(x + 1) - (2*x + 1)/(x**2 + x), [a, b], x) == \
        {a: 1, b: 1}
    # Test cancellation in rational functions
    assert solve_undetermined_coeffs(((c + 1)*a*x**2 + (c + 1)*b*x**2 +
    (c + 1)*b*x + (c + 1)*2*c*x + (c + 1)**2)/(c + 1), [a, b, c], x) == \
        {a: -2, b: 2, c: -1}
开发者ID:Maihj,项目名称:sympy,代码行数:10,代码来源:test_solvers.py


示例2: synthesize

def synthesize(polynomial, filter_implementation, filter_type, parameters):
    """ Implements the filter given by polynomial,
        with the filter_implementation and filter_type given.
        Assumptions:

        - Low-pass and high-pass filters set the resistor
          and find the capacitor.
        - Sallen-key LP filter sets R1 = R2 = Rr and C1 and C2 can vary.
        - Sallen-key HP filter sets C1 = C2 = Cc and R1 and R2 can vary.
        - MFB LP filter sets C1 = C2 = Cc and R1, R2, R3 can vary.
        - MFB HP filter sets R1 = R2 = R3 = Rr and C1, C2 can vary.
    """
    s = sym.var('s')
    if filter_implementation == '1' and filter_type == 'lp':
        C = sym.var('C')
        targets = [C]
        R = parameters['R']
        tf = rc_lowpass(s, R, C)

    if filter_implementation == '1' and filter_type == 'hp':
        C = sym.var('C')
        targets = [C]
        R = parameters['R']
        tf = rc_highpass(s, R, C)

    if filter_implementation == 'sk' and filter_type == 'lp':
        C1, C2, s = sym.var('C1 C2 s')
        targets = [C1, C2]  # The values we want to find
        Rr = parameters['Rr']
        tf = sk_lowpass(s, Rr, Rr, C1, C2)

    if filter_implementation == 'sk' and filter_type == 'hp':
        R1, R2, s = sym.var('R1 R2 s')
        targets = [R1, R2]  # The values we want to find
        Cc = parameters['Cc']
        tf = sk_highpass(s, R1, R2, Cc, Cc)
        print(tf)

    print("I need to solve ", tf-polynomial)
    print("My variables are ", targets)
    result = solvers.solve_undetermined_coeffs(tf - polynomial, targets, 's')
    return result
开发者ID:Python-Devs-Brasil,项目名称:pyfilter,代码行数:42,代码来源:synthesis.py


示例3: rsolve_poly


#.........这里部分代码省略.........

    if nni_roots:
        N = [max(nni_roots)]
    else:
        N = []

    if homogeneous:
        N += [-b-1]
    else:
        N += [f.as_poly(n).degree() - b, -b-1]

    N = int(max(N))

    if N < 0:
        if homogeneous:
            if hints.get('symbols', False):
                return (S.Zero, [])
            else:
                return S.Zero
        else:
            return None

    if N <= r:
        C = []
        y = E = S.Zero

        for i in xrange(0, N+1):
            C.append(Symbol('C'+str(i)))
            y += C[i] * n**i

        for i in xrange(0, r+1):
            E += coeffs[i].as_basic()*y.subs(n, n+i)

        solutions = solve_undetermined_coeffs(E-f, C, n)

        if solutions is not None:
            C = [ c for c in C if (c not in solutions) ]
            result = y.subs(solutions)
        else:
            return None # TBD
    else:
        A = r
        U = N+A+b+1

        nni_roots = roots(polys[r], filter='Z',
            predicate=lambda r: r >= 0).keys()

        if nni_roots != []:
            a = max(nni_roots) + 1
        else:
            a = S.Zero

        def zero_vector(k):
            return [S.Zero] * k

        def one_vector(k):
            return [S.One] * k

        def delta(p, k):
            B = S.One
            D = p.subs(n, a+k)

            for i in xrange(1, k+1):
                B *= -Rational(k-i+1, i)
                D += B * p.subs(n, a+k-i)
开发者ID:Sumith1896,项目名称:sympy-polys,代码行数:66,代码来源:recurr.py



注:本文中的sympy.solvers.solve_undetermined_coeffs函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。


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