Question

I m having trouble solving this integral in python. The function being integrated is not defined on the boundaries of integration. I ve found a few more questions similar to this, but all were very specific replies to the issue in particular. I don t want to approximate the integral too much, if possible not at all, as the reason I m doing this integral in the first place is to avoid an approximation. Is there any way to solve this integral?

import numpy as np
from pylab import *
import scipy
from math import *
from scipy import integrate

m_Earth_air = (28.0134*0.78084)+(31.9988*0.209476)+(39.948*0.00934)+(44.00995*0.000314)+(20.183*0.00001818)+(4.0026*0.00000524)+(83.80*0.00000114)+(131.30*0.000000087)+(16.04303*0.000002)+(2.01594*0.0000005)
Tb0 = 288.15
Lb0 = -6.5
Hb0 = 0.0
def Tm_0(z):
    return Tb0+Lb0*(z-Hb0)
k = 1.38*10**-19 #cm^2.kg/s^2.K   #Boltzmann cst
mp = 1.67262177*10**-27 #kg
Rad= 637100000.0 #radius planet #cm
g0 = 980.665 #cm/s^2
def g(z):
    return (g0*((Rad/(Rad+z))**2.0))
def scale_height0(z):
    return k*Tm_0(z*10**-5)/(m_Earth_air*mp*g(z))



def functionz(z,zvar):
    return np.exp(-zvar/scale_height0(z))*((Rad+zvar)/(Rad+z))/((np.sqrt(((Rad+zvar)/(Rad+z))**2.0-1.0)))

def chapman0(z):
    return (1.0/(scale_height0(z)))*((integrate.quad(lambda zvar: functionz(z,zvar), z, np.inf))[0])

print chapman0(1000000)
print chapman0(5000000)

The first block of variables and definitions are fine. The issue is in the "functionz(z,zvar)" and its integration. Any help very much appreciated !

Answer 1

它常常通过调整变量,帮助消除可能的数字不稳定性。在你的情况下,<代码>zvar从1e6开始,这可能由于<代码>quad(>的某些执行细节而造成问题。如果您将其排列为y = zvar /z,从而使整合从1开始,则似乎对z = 1e6来说相当好:

def functiony(z, y):
    return np.exp(-y*z/scale_height0(z))*(Rad+y*z)/(Rad+z) / np.sqrt(((Rad+y*z)/(Rad+z))**2.0-1.0)

def chapman0y(z):
    return (1.0/(scale_height0(z)))*((integrate.quad(lambda y: functiony(z,y), 1, np.inf))[0])

>>> print(chapman0y(1000000))

1.6217257661844094e-06

(I set m_ Earth_air = 28.8e-3——这一不变值在您的代码中缺失,我假定其为空气体质(edit/千克/分子)。

关于<代码>z = 5e6,pl_hal0(z)为负值,在权宜之计下具有巨大的积极价值,使整体在定义上有所不同。

Answer 2

除非你能够以整体分析方式解决,否则就没有办法解决该问题,而不必对其约束进行近似。这只是一个 Python子问题,而是一个cal子问题,因此,数学课会给你带来如此巨大的痛苦,以显示数字近似。

如果你不希望它有太多的差别,就选择一个小块sil子,其方法会很快。

Edit- Clarity on final statement:

Epsilon - ɛ - 系指通过集成链条体的体积——三角体X——记住,数字近似值法将有机物纳入肝脏并添加到肝脏中,认为这是每一肝脏的 w,越小肝越好。你可以在数字组合中具体说明这些内容。

一种迅速趋向的方法意味着这种方法能够迅速接近整体的真正价值,每肝脏的近似值很小。例如,Riemann的总和是一种天花的方法,假设每个肝脏都是一种 rec子,而一种诱杀器把肝的开始和结束与形成一种诱杀虫的线联系起来。在这2个国家中,诱杀虫类动物通常会更快地聚集在一起,因为它试图在形状中说明变化。 (由于大多数职能都有较好的电梯,因此通常都使用过吗?)

Both of these variables change the computational expense of the calculation. Typically epsilon is the most expensive to change, thus why it is important you choose a good method of approximation (some can differ by an order of magnitude in error for the same epsilon).

所有这一切都取决于你计算出的错误能容忍多少。

Answer 3

我有类似的问题,发现SciPy之四需要你具体说明另一个参数,epsabs=1e-1000,limit=1000 (步骤size limit),epsrel=1_/code> 为我所尝试的一切工作而努力。页: 1



def chapman0(z):    
    return (1.0/(scale_height0(z)))*((integrate.quad(lambda zvar: functionz(z,zvar), z, np.inf, limit=1000, epsabs=1e-1000, epsrel=1e1))[0])[0])
#results:
0.48529410529321887
-1.276589093231806e+21


种子是绝对错误的容忍,但对于不迅速趋同的整体来说,似乎正在解决这个问题。 就象这个职位那样有类似问题的其他人而言,情况已经很成熟。 其他包裹中的算法比较快,但我在SciPy没有发现。 成果依据的是现行法典(而不是选定的答案)。

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