Python积分计算器不显示值

如何解决Python积分计算器不显示值

我制作了一个计算器，可以近似任何给定的函数作为输入。他们后来我想让它计算一个积分，但是在写完之后：

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

它不打印数字，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,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,b))

我认为是因为我的输入是字符串。但是我不能选择输入为整数，因为exp(-x**2)不是整数。如果是这种情况，如何在计算器中输入任何函数并仍然获得值？

解决方法

您的代码中存在一些重大问题：

• 在integrate内，您正在使用局部变量n，但是在f(x)内，您认为它是全局变量n（但是使用了局部变量，这就是只需在n内打印f(x)）。 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，请不要使用相同的名称。
• f(x)的返回值是sympy形式，而不是单个值，这就是为什么要获得所得到的输出的原因。

经过重构并使用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,n,N,a,b): # using the approach with all variables as parameters
    taylor = series(function,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,b))

从x**2到x=0集成的x=2的某些输出在x=1处扩展。分析结果为8/3=2.6666666...。

x**2,1,5,2 => 2.0 # constant approximation
x**2,2 => 2.0 # linear approximation
x**2,2,2 => 2.64 # quadratic approximation - exact function
x**2,10,2 => 2.66
x**2,100,2 => 2.6666
x**2,1000,2 => 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而言，提速非常重要。

from sympy.utilities.lambdify import lambdify
def integrate(function,b):
    taylor = series(function,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))