I have made a calculator that approximates any given function as an input. They later I want it to calculate an integral, but after writing:
function = str(input("The function that must be expanded and integrated: "))
It doesn't print a number but instead a value. This is my code:
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))
I think, it's because my input is a string. However I can't choose my input to be an integer, because exp(-x**2) isn't an integer. If that's the case, how can I input any function in my calculater and still get a value?
There a some significant issues in your code:
integrate, you are using local variable n, but inside f(x) you consider it to be the global n (but the local is used, which is what you want, just print n inside the f(x)). The same holds for x as a global variable and parameter in f(x). Do not use same names for global and local variables if you want to use both in the same scope.f(x) is a sympy epxression, not a single value, that's why you get the output what you get.After some refactoring and using subs and 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))
Some outputs for x**2 integrated from x=0 to x=2 expanded at x=1. Analytical result is 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
You can use lambdify to "convert a SymPy expression into a function that allows for fast numeric evaluation". For the case of N=1000 the speedup is significant.
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))