Python 积分计算器不打印值

Question

我制作了一个计算器，可以近似任何给定函数作为输入。他们后来我希望它计算积分，但写完后：

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

它不打印数字，而是打印值。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

  里面，你使用的是局部变量

  n

  ，但是在

  f(x)

  里面你认为它是全局的

  n

  （但是使用了局部变量，这就是你想要的，只需在里面打印

  n

  ）

  f(x)

  ）。这同样适用于

  x

  作为

  f(x)

  中的全局变量和参数。如果您想在同一范围内使用全局变量和局部变量，请不要使用相同的名称。

• f(x)

  的返回值是一个

  sympy

  表达式，而不是单个值，这就是为什么你得到你得到的输出。

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))

从

x**2

集成到

x=0

的

x=2

的一些输出在

x=1

处扩展。分析结果是

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

“将 SymPy 表达式转换为允许快速数值计算的函数”。对于

N=1000

的情况，加速是显着的。

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))