Python 积分计算器不打印值

Question

我制作了一个计算器，可以将任何给定的函数近似为输入。他们后来我想让它计算一个积分，但在写下：

function    = str(input("The function that must be expanded and integrated: "))

它不打印数字，而是打印一个值。这是我的代码：

from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import *
from sympy import series
from math import *
function    = str(input("The function that must be expanded and integrated: "))
x0          = int(input("Point of development: "))
n           = int(input("Amount of expressions: "))
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(N, a, b):
def f(x):
return series(function, x, x0, n)
value=0
value=2
for n in range(1, N+1):
value += f(a+((n-(1/2))*((b-a)/N)))
value2 = ((b-a)/N)*value
return value2
print("...................")
print("Here is your answer: ")
print(integrate(N, a, b))

我想，这是因为我的输入是一个字符串。但是，我不能将我的输入选择为整数，因为exp(-x**2)不是整数。如果是这样的话，我怎么能在计算器中输入任何函数，却仍然得到一个值呢？

Answer 1

您的代码中存在一些重大问题：

• 在integrate中，您使用的是局部变量n，但在f(x)中，您认为它是全局n(但使用了局部变量，这正是您想要的，只需在f(x)中打印n(。x作为f(x)中的全局变量和参数也是如此。如果要在同一范围内使用全局变量和局部变量，请不要对它们使用相同的名称
• f(x)的返回值是一个sympyepxression，而不是单个值，这就是为什么您得到的输出

经过一些重构并使用subs和removeO:

from sympy.functions import sin,cos,tan
from sympy.abc import x
from sympy import series
function    = str(input("The function to be expanded and integrated: "))
x0          = int(input("Point of development: "))
n           = 1 + int(input("Degree: "))
# input 0 -> n=1 -> constant  (1 term, constant)
# input 1 -> n=2 -> linear    (2 terms, constant + linear)
# input 2 -> n=3 -> quadratic (3 terms, constant + linear + quadratic)
# ...
print(series(function, x, x0, n))
N = int(input("Amount of summs (Bigger number is more accurate but takes longer time): "))
a = int(input("Integrate from: "))
b = int(input("Integrate to: "))
# We will use the midpoint method to integrate the function
def integrate(function, x0, n, N, a, b): # using the approach with all variables as parameters
taylor = series(function, x, x0, n) # the same expression for the function, create it once
taylor = taylor.removeO() # do not use O term (may corrups subs below)
dx = (b-a)/N # also computed just once
def f(v):
return taylor.subs(x,v) # taylor is expression, return value is float evaluated with substituted x by v
return dx * sum(f(a+(i+1/2)*dx) for i in range(N)) # simple sum function, can be rewriten using a for loop
print("...................")
print("Here is your answer: ")
print(integrate(function, x0, n, N, a, b))

从x=0到x=2的x**2的一些输出在x=1处扩展。分析结果为8/3=2.6666666...。

x**2, 1, 0, 5, 0, 2 => 2.0 # constant approximation
x**2, 1, 1, 5, 0, 2 => 2.0 # linear approximation
x**2, 1, 2, 5, 0, 2 => 2.64 # quadratic approximation - exact function
x**2, 1, 2, 10, 0, 2 => 2.66
x**2, 1, 2, 100, 0, 2 => 2.6666
x**2, 1, 2, 1000, 0, 2 => 2.666666

您可以使用CCD_；将SymPy表达式转换为允许快速数值求值的函数"；。对于N=1000的情况，加速是显著的。

from sympy.utilities.lambdify import lambdify
def integrate(function, x0, n, N, a, b):
taylor = series(function, x, x0, n)
taylor = lambdify(x,taylor.removeO()) # here
dx = (b-a)/N
def f(v):
return taylor(v) # here
return dx * sum(f(a+(i+1/2)*dx) for i in range(N))