First of all; english is not my main language so bare in mind. (The text in the script is in norwegian, I will try to translate as well as possible)
I am trying to create a python script that solves a second degree equation or a quadratic? equation I think. If i understand it correctly I should use cmath module to be able to solve complex roots, however, I havent got that far in my class yet and it is my task to only use math module and return a answer that looks something like this (if the roots are complex): "There is no real solution".
Also, when using normal math module I get returned a "math domain error" when trying to for example put a = 6, b = 11 and c = 32.
Thank you in advance.
Edward
Trying to use a if/elif/else thingy, have tried cmath (it returns a waaaay to long and unreadable answer)
My code is here under this (hopefully I have pasted it correctly, sorry in advance:
#Importerer math
import math
print("Programmet skal løse andregradslikningen ")
print("ax^2 + bx + c = 0 ved hjelp av abc-formelen")
print("Skriv inn verdiene for a, b og c.")
a = float(input('a = '))
b = float(input('b = '))
c = float(input('c = '))
d = (b**2) - (4*a*c)
sol1 = (-b+math.sqrt(d))/(2*a)
sol2 = (-b-math.sqrt(d))/(2*a)
if d < 0:
print("Likningen har ingen løsning")
elif d==0:
print(f"Løsning x = {-b/(2*a)}")
else:
print("Løsningen er")
print(f"x = {sol1}, og x= {sol2}")
You have to check if d is negative before you try to call math.sqrt(d), so you don't get an error.
d = (b**2) - (4*a*c)
if d < 0:
print("Likningen har ingen løsning")
elif d==0:
print(f"Løsning x = {-b/(2*a)}")
else:
sol1 = (-b+math.sqrt(d))/(2*a)
sol2 = (-b-math.sqrt(d))/(2*a)
print("Løsningen er")
print(f"x = {sol1}, og x= {sol2}")
I'm new to python but I been working on a code which can solve an integral equation which range is also changing according the unknown parameter. I tried to use Sympy solve function but it does not return any result. I solved the problem with a for loop, but its really slow, inefficient. I'm sure there must be a better solution, a solver. Maybe there is an other approach? I'am messing up something? I attach also the code.
import sympy as sy
from sympy.solvers import solve
alphasum = 1.707
Lky = 3.078
g = 8
Ep = 195
sigp1 = 1401.927
sigp0 = 1476
e = 2.718282
u = 0.05
k = 0.007
lsl = sy.Symbol('lsl')
def lefthand(g, Ep):
return g * Ep
def rigthhand(sigp1, sigp0, e, u, alphasum, Lky, k, lsl):
return (sigp1 - (-sigp1 + 2 * sigp0 - 2 * sigp0 * (1 - e ** (-u * (((alphasum / Lky) * lsl) + k * lsl)))))
equr = (sy.integrate(rigthhand(sigp1, sigp0, e, u, alphasum, Lky, k, lsl), (lsl, 0, lsl)))
equl = lefthand(g, Ep)
print(equr)
print (equl)
print (equr-equl)
result = solve(equr-equl, lsl,warn =True,check=False,minimal=True,quick=True,simplify=True,quartics=True)
print(result)
You can use nsolve to find numerical solutions:
In [11]: sympy.nsolve(equr-equl, lsl, 8)
Out[11]: 8.60275245116315
In [12]: sympy.nsolve(equr-equl, lsl, -4)
Out[12]: -4.53215114758428
When u is 0 your equation is linear and you can solve if for lsl; let this be an initial guess for the value of lsl at a larger value of u. Increase your value of u to the target value. Let's use capital U for the parameter we will control, replacing it with values closer and closer to u:
>>> U = Symbol('U')
>>> equr = (sy.integrate(rigthhand(sigp1, sigp0, e, U, alphasum, Lky, k, lsl), (lsl, 0, lsl)))
>>> equl = lefthand(g, Ep)
>>> z = equr-equl
>>> u0 = solve(z.subs(U,0),lsl)[0]
>>> for i in range(1,10): # get to u in 9 steps
... u0 = nsolve(z.subs(U,i*u/10.), u0)
>>> print(u0)
-4.71178322070344
Now define a larger value of u
>>> u = 0.3
>>> for i in range(1,10): # get to u in 9 steps
... u0 = nsolve(z.subs(U,i*u/10.), u0)
...
>>> u0
-2.21489271112540
Our intitial guess will (usually) be pretty good since it is exact when u is 0; if it fails then you might need to take more than 9 steps to reach the target value.
i am trying to solve a matrix that has 6x6 matrices as it's entries(elements)
i tried multiplying the inverse of gen to the solution matrix, but i don't trust the correctness of the answer am getting.
from sympy import Eq, solve_linear_system, Matrix,count_ops,Mul,horner
import sympy as sp
a, b, c, d, e,f = sp.symbols('a b c d e f')
ad = Matrix(([43.4,26.5,115,-40.5,52.4,0.921],
[3.78,62.9,127,-67.6,110,4.80],
[41.25,75.0,213,-88.9, 131, 5.88],
[-10.6,-68.4,-120,64.6,-132,-8.49],
[6.5,74.3,121,-72.8,179,29.7],
[1.2,30.7,49.7,-28.7,91,29.9]))
fb= Matrix(([1,0,0,0,0,0],
[0,1,0,0,0,0],
[0,0,1,0,0,0],
[0,0,0,1,0,0],
[0,0,0,0,1,0],
[0,0,0,0,0,1]))
ab = Matrix(([-0.0057],
[0.0006],
[-0.0037],
[0.0009],
[0.0025],
[0.0042]))
az = sp.symbols('az')
bz = sp.symbols('bz')
fz = sp.symbols('fz')
gen = Matrix(([az, fz, 0, 0, 0, 0,bz],
[fz,az,fz,0,0,0,bz],
[0,fz,az,fz,0,0,bz],
[0,0,fz,az,fz,0,bz],
[0,0,0,fz,az,fz,bz],
[0,0,0,0,fz,az,bz]))
answer = solve_linear_system(gen,a,b,c,d,e,f)
first_solution = answer[a]
df = count_ops(first_solution)
print(df,first_solution)
disolved = zip(first_solution.simplify().as_numer_denom(),(1,-1))
dft = Mul(*[horner(b)**e for b,e in disolved])
dff = count_ops(dft)
print(dff,dft)
_1st_solution = dft.subs({az:ad,fz:fb,bz:ab},simultaneous = True).doit()
print(_1st_solution)
when i ran my code it raised sympy.matrices.common.ShapeError
You have to be careful when using horner with expressions containing commutative symbols that are actually noncommutative (in your case because they represent matrices). Your dft expression is
(az**2*bz - bz*fz**2)/(az*(az*(az + fz) - 2*fz**2) - fz**3)
but should maybe be
(az**2 - fz**2)*(az*(az*(az + fz) - 2*fz**2) - fz**3)**(-1)*bz
You would have received a correct expression if you had created the symbols as noncommutative (as shown below).
But you can't use horner with non-commutative symbols, so I just rearranged the expression by hand; you will have to check to see that the ordering is right. As an alternative to doing the factoring by hand you might also try using factor_nc to help you -- but it won't handle horner like expression factoring:
>>> ax, bz, fz = symbols('az bz fz, commutative=False)
>>> (az**2*bz - fz**2*bz)
az**2*bz - fz**2*bz
>>> factor_nc(_)
(az**2 - fz**2)*bz
I am using sympy to solve some equations and I am running into a problem. I have this issue with many equations but I will illustrate with an example. I have an equation with multiple variables and I want to solve this equation in terms of all variables but one is excluded. For instance the equation 0 = 2^n*(2-a) - b + 1. Here there are three variables a, b and n. I want to get the values for a and b not in terms of n so the a and b may not contain n.
2^n*(2-a) - b + 1 = 0
# Since we don't want to solve in terms of n we know that (2 - a)
# has to be zero and -b + 1 has to be zero.
2 - a = 0
a = 2
-b + 1 = 0
b = 1
I want sympy to do this. Maybe I'm just not looking at the right documentation but I have found no way to do this. When I use solve and instruct it to solve for symbols a and b sympy returns to me a single solution where a is defined in terms of n and b. I assume this means I am free to choose b and n, However I don't want to fix n to a specific value I want n to still be a variable.
Code:
import sympy
n = sympy.var("n", integer = True)
a = sympy.var("a")
b = sympy.var("b")
f = 2**n*(2-a) - b + 1
solutions = sympy.solve(f, [a,b], dict = True)
# this will return: "[{a: 2**(-n)*(2**(n + 1) - b + 1)}]".
# A single solution where b and n are free variables.
# However this means I have to choose an n I don't want
# to that I want it to hold for any n.
I really hope someone can help me. I have been searching google for hours now...
Ok, here's what I came up with. This seems to solve the type of equations you're looking for. I've provided some tests as well. Of course, this code is rough and can be easily caused to fail, so i'd take it more as a starting point than a complete solution
import sympy
n = sympy.Symbol('n')
a = sympy.Symbol('a')
b = sympy.Symbol('b')
c = sympy.Symbol('c')
d = sympy.Symbol('d')
e = sympy.Symbol('e')
f = sympy.sympify(2**n*(2-a) - b + 1)
g = sympy.sympify(2**n*(2-a) -2**(n-1)*(c+5) - b + 1)
h = sympy.sympify(2**n*(2-a) -2**(n-1)*(e-1) +(c-3)*9**n - b + 1)
i = sympy.sympify(2**n*(2-a) -2**(n-1)*(e+4) +(c-3)*9**n - b + 1 + (d+2)*9**(n+2))
def rewrite(expr):
if expr.is_Add:
return sympy.Add(*[rewrite(f) for f in expr.args])
if expr.is_Mul:
return sympy.Mul(*[rewrite(f) for f in expr.args])
if expr.is_Pow:
if expr.args[0].is_Number:
if expr.args[1].is_Symbol:
return expr
elif expr.args[1].is_Add:
base = expr.args[0]
power = sympy.solve(expr.args[1])
sym = expr.args[1].free_symbols.pop()
return sympy.Mul(sympy.Pow(base,-power[0]), sympy.Pow(base,sym))
else:
return expr
else:
return expr
else:
return expr
def my_solve(expr):
if not expr.is_Add:
return None
consts_list = []
equations_list = []
for arg in expr.args:
if not sympy.Symbol('n') in arg.free_symbols:
consts_list.append(arg)
elif arg.is_Mul:
coeff_list = []
for nested_arg in arg.args:
if not sympy.Symbol('n') in nested_arg.free_symbols:
coeff_list.append(nested_arg)
equations_list.append(sympy.Mul(*coeff_list))
equations_list.append(sympy.Add(*consts_list))
results = {}
for eq in equations_list:
var_name = eq.free_symbols.pop()
val = sympy.solve(eq)[0]
results[var_name] = val
return results
print(my_solve(rewrite(f)))
print(my_solve(rewrite(g)))
print(my_solve(rewrite(h)))
print(my_solve(rewrite(i)))
How do i implement this kind of equation in python dC/dt = r + kI - dC where the left hand side are constants and the right hand side are varibles?
i am relatively new to python and as such can't really do much.
from sympy.solvers import ode
r=float(input("enter r:"))
k=float(input("enter k:"))
I=float(input("enter I:"))
d=float(input("enter d:"))
C=float(input("enter C:"))
dC/dt=x
x=r + kI-dC
print(x)
what it just does equate the values of x and not any differential, would like help getting this to work.
if possible i would like to get answer specifying the using of sympy,
but all answers are truly appreciated.
You asigned values to all the variables that are on the rhs of x so when you show x you see the value that it took on with the variables that you defined. Rather than input values, why not try solve the ode symbolically if possible?
>>> from sympy import *
>>> var('r k I d C t')
(r, k, I, d, C, t)
>>> eq = Eq(C(t).diff(t), r + k*I + d*C(t)) # note d*C(t) not d*C
>>> ans = dsolve(eq); ans
C(t) == (-I*k - r + exp(d*(C1 + t)))/d
Now you can substitute in values for the variables to see the result:
>>> ans.subs({k: 0})
C(t) == (-r + exp(d*(C1 + t)))/d