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Python libmp.mpf_pi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.mpmath.libmp.mpf_pi函数的典型用法代码示例。如果您正苦于以下问题:Python mpf_pi函数的具体用法?Python mpf_pi怎么用?Python mpf_pi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mpf_pi函数的5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: evalf_log

def evalf_log(expr, prec, options):
    arg = expr.args[0]
    workprec = prec + 10
    xre, xim, xacc, _ = evalf(arg, workprec, options)

    if xim:
        # XXX: use get_abs etc instead
        re = evalf_log(C.log(C.Abs(arg, evaluate=False), evaluate=False), prec, options)
        im = mpf_atan2(xim, xre or fzero, prec)
        return re[0], im, re[2], prec

    imaginary_term = mpf_cmp(xre, fzero) < 0

    re = mpf_log(mpf_abs(xre), prec, rnd)
    size = fastlog(re)
    if prec - size > workprec:
        # We actually need to compute 1+x accurately, not x
        arg = C.Add(S.NegativeOne, arg, evaluate=False)
        xre, xim, _, _ = evalf_add(arg, prec, options)
        prec2 = workprec - fastlog(xre)
        re = mpf_log(mpf_add(xre, fone, prec2), prec, rnd)

    re_acc = prec

    if imaginary_term:
        return re, mpf_pi(prec), re_acc, prec
    else:
        return re, None, re_acc, None
开发者ID:smichr,项目名称:sympy,代码行数:28,代码来源:evalf.py


示例2: _create_evalf_table

def _create_evalf_table():
    global evalf_table
    evalf_table = {
        C.Symbol: evalf_symbol,
        C.Dummy: evalf_symbol,
        C.Float: lambda x, prec, options: (x._mpf_, None, prec, None),
        C.Rational: lambda x, prec, options: (from_rational(x.p, x.q, prec), None, prec, None),
        C.Integer: lambda x, prec, options: (from_int(x.p, prec), None, prec, None),
        C.Zero: lambda x, prec, options: (None, None, prec, None),
        C.One: lambda x, prec, options: (fone, None, prec, None),
        C.Half: lambda x, prec, options: (fhalf, None, prec, None),
        C.Pi: lambda x, prec, options: (mpf_pi(prec), None, prec, None),
        C.Exp1: lambda x, prec, options: (mpf_e(prec), None, prec, None),
        C.ImaginaryUnit: lambda x, prec, options: (None, fone, None, prec),
        C.NegativeOne: lambda x, prec, options: (fnone, None, prec, None),
        C.NaN: lambda x, prec, options: (fnan, None, prec, None),
        C.exp: lambda x, prec, options: evalf_pow(C.Pow(S.Exp1, x.args[0], evaluate=False), prec, options),
        C.cos: evalf_trig,
        C.sin: evalf_trig,
        C.Add: evalf_add,
        C.Mul: evalf_mul,
        C.Pow: evalf_pow,
        C.log: evalf_log,
        C.atan: evalf_atan,
        C.Abs: evalf_abs,
        C.re: evalf_re,
        C.im: evalf_im,
        C.floor: evalf_floor,
        C.ceiling: evalf_ceiling,
        C.Integral: evalf_integral,
        C.Sum: evalf_sum,
        C.Piecewise: evalf_piecewise,
        C.bernoulli: evalf_bernoulli,
    }
开发者ID:smichr,项目名称:sympy,代码行数:34,代码来源:evalf.py


示例3: npartitions

def npartitions(n, verbose=False):
    """
    Calculate the partition function P(n), i.e. the number of ways that
    n can be written as a sum of positive integers.

    P(n) is computed using the Hardy-Ramanujan-Rademacher formula,
    described e.g. at http://mathworld.wolfram.com/PartitionFunctionP.html

    The correctness of this implementation has been tested for 10**n
    up to n = 8.
    """
    n = int(n)
    if n < 0: return 0
    if n <= 5: return [1, 1, 2, 3, 5, 7][n]
    # Estimate number of bits in p(n). This formula could be tidied
    pbits = int((math.pi*(2*n/3.)**0.5-math.log(4*n))/math.log(10)+1)*\
        math.log(10,2)
    prec = p = int(pbits*1.1 + 100)
    s = fzero
    M = max(6, int(0.24*n**0.5+4))
    sq23pi = mpf_mul(mpf_sqrt(from_rational(2,3,p), p), mpf_pi(p), p)
    sqrt8 = mpf_sqrt(from_int(8), p)
    for q in xrange(1, M):
        a = A(n,q,p)
        d = D(n,q,p, sq23pi, sqrt8)
        s = mpf_add(s, mpf_mul(a, d), prec)
        if verbose:
            print "step", q, "of", M, to_str(a, 10), to_str(d, 10)
        # On average, the terms decrease rapidly in magnitude. Dynamically
        # reducing the precision greatly improves performance.
        p = bitcount(abs(to_int(d))) + 50
    np = to_int(mpf_add(s, fhalf, prec))
    return int(np)
开发者ID:101man,项目名称:sympy,代码行数:33,代码来源:partitions_.py


示例4: _d

def _d(n, j, prec, sq23pi, sqrt8):
    """
    Compute the sinh term in the outer sum of the HRR formula.
    The constants sqrt(2/3*pi) and sqrt(8) must be precomputed.
    """
    j = from_int(j)
    pi = mpf_pi(prec)
    a = mpf_div(sq23pi, j, prec)
    b = mpf_sub(from_int(n), from_rational(1, 24, prec), prec)
    c = mpf_sqrt(b, prec)
    ch, sh = mpf_cosh_sinh(mpf_mul(a, c), prec)
    D = mpf_div(mpf_sqrt(j, prec), mpf_mul(mpf_mul(sqrt8, b), pi), prec)
    E = mpf_sub(mpf_mul(a, ch), mpf_div(sh, c, prec), prec)
    return mpf_mul(D, E)
开发者ID:malikdiarra,项目名称:sympy,代码行数:14,代码来源:partitions_.py


示例5: _as_mpf_val

 def _as_mpf_val(self, prec):
     return mpf_pi(prec)
开发者ID:goriccardo,项目名称:sympy,代码行数:2,代码来源:numbers.py



注:本文中的sympy.mpmath.libmp.mpf_pi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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