Python积分计算器不打印值

Question

我制作了一个计算器，可以将任何给定函数近似为输入。他们后来我想让它计算一个积分，但是写完之后：

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

它不打印一个数字，而是一个值。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。但是我不能选择我的输入为整数，因为exp(-x**2) 不是整数。如果是这样，我怎样才能在我的计算器中输入任何函数并仍然得到一个值？

Answer 1

您的代码中存在一些重大问题：

• 在integrate 内部，您使用的是局部变量n，但在f(x) 内部，您认为它是全局变量n（但使用了本地变量，这是您想要的，只需打印n在f(x)) 内。 x 作为f(x) 中的全局变量和参数也是如此。如果您想在同一范围内同时使用全局变量和局部变量，请不要使用相同的名称。
• f(x) 的返回值是 sympy 表达式，而不是单个值，这就是为什么你会得到你得到的输出。

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))

x**2 的一些输出从 x=0 集成到 x=2 扩展为 x=1。解析结果为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lambdify to "convert a SymPy expression into a function that allows for fast numeric evaluation"。对于N=1000 的情况，加速非常显着。

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))