I have a project that one step of the process of it is to solve R(k,d,a),
where k means kth step.
My friend suggest me to do this in sympy ,but I don't know how to do it.
from sympy import *
k= symbols('k')
d= symbols('d')
a= symbols('a')
R= function('R')(k,d,a)
print R`
In fact I don't know how to define a function in sympy with class method...
and this expression is failure.
def R(k,d,a):
k:# of nth process
d:parameter in order to initializing
R(0,d,a) should be given
if k==0:
return 100*d
r=d*(1-(1-(d/R(k-1,d,a))**2)**0.5)**0.5
B=-3/2*d
D=(R(k-1,d,a))**3*(3*a*d/R(k-1,d,a)-2)+r**2*(r-3/2*d)
here I define R(k,d,a) with R(k-1,d,a),is it appropriate?
x^3+Bx^2+Cx+D=0 ,where c=0
x represent R(k,d,a) here.
x=symbols('x')
y=solve (x**3+x**2*B+D,x)
return max(y)
Here I want a list of y,and asking y in real number.
Later return the biggest one.
But I don't know how to realize it.
Finally,for each k ,I will need the other function to give a value that R(k,d,a) will be a parameter in it.I think I can do it by my self with for loop,it is not hard for me.
What is the hardest is how to get R(k,d,a).
I don't need complex root .But if I want ,how can I get them ?
Thank you for watching!
What you have looks basically OK. Three suggestions, however:
make sure your function always returns a SymPy value, e.g. don't return 100*d since d might be an int; return S(100)*d;
wherever you have division make sure that it is not suffering from Python trunction wherein 1/2 -> 0. e.g. write B = -S(3)/2*d instead of what you have (and use that B in your expression for D, writing (r+B) at the end of it;
max will not be able to sort the roots if complex roots are present so it would be better to select the real ones by hand: y=[i for i in solve (x**3+x**2*B+D,x) if i.is_real].
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Let's say I have an equation:
2x + 6 = 12
With algebra we can see that x = 3. How can I make a program in Python that can solve for x? I'm new to programming, and I looked at eval() and exec() but I can't figure out how to make them do what I want. I do not want to use external libraries (e.g. SAGE), I want to do this in just plain Python.
How about SymPy? Their solver looks like what you need. Have a look at their source code if you want to build the library yourself…
There are two ways to approach this problem: numerically and symbolically.
To solve it numerically, you have to first encode it as a "runnable" function - stick a value in, get a value out. For example,
def my_function(x):
return 2*x + 6
It is quite possible to parse a string to automatically create such a function; say you parse 2x + 6 into a list, [6, 2] (where the list index corresponds to the power of x - so 6*x^0 + 2*x^1). Then:
def makePoly(arr):
def fn(x):
return sum(c*x**p for p,c in enumerate(arr))
return fn
my_func = makePoly([6, 2])
my_func(3) # returns 12
You then need another function which repeatedly plugs an x-value into your function, looks at the difference between the result and what it wants to find, and tweaks its x-value to (hopefully) minimize the difference.
def dx(fn, x, delta=0.001):
return (fn(x+delta) - fn(x))/delta
def solve(fn, value, x=0.5, maxtries=1000, maxerr=0.00001):
for tries in xrange(maxtries):
err = fn(x) - value
if abs(err) < maxerr:
return x
slope = dx(fn, x)
x -= err/slope
raise ValueError('no solution found')
There are lots of potential problems here - finding a good starting x-value, assuming that the function actually has a solution (ie there are no real-valued answers to x^2 + 2 = 0), hitting the limits of computational accuracy, etc. But in this case, the error minimization function is suitable and we get a good result:
solve(my_func, 16) # returns (x =) 5.000000000000496
Note that this solution is not absolutely, exactly correct. If you need it to be perfect, or if you want to try solving families of equations analytically, you have to turn to a more complicated beast: a symbolic solver.
A symbolic solver, like Mathematica or Maple, is an expert system with a lot of built-in rules ("knowledge") about algebra, calculus, etc; it "knows" that the derivative of sin is cos, that the derivative of kx^p is kpx^(p-1), and so on. When you give it an equation, it tries to find a path, a set of rule-applications, from where it is (the equation) to where you want to be (the simplest possible form of the equation, which is hopefully the solution).
Your example equation is quite simple; a symbolic solution might look like:
=> LHS([6, 2]) RHS([16])
# rule: pull all coefficients into LHS
LHS, RHS = [lh-rh for lh,rh in izip_longest(LHS, RHS, 0)], [0]
=> LHS([-10,2]) RHS([0])
# rule: solve first-degree poly
if RHS==[0] and len(LHS)==2:
LHS, RHS = [0,1], [-LHS[0]/LHS[1]]
=> LHS([0,1]) RHS([5])
and there is your solution: x = 5.
I hope this gives the flavor of the idea; the details of implementation (finding a good, complete set of rules and deciding when each rule should be applied) can easily consume many man-years of effort.
Python may be good, but it isn't God...
There are a few different ways to solve equations. SymPy has already been mentioned, if you're looking for analytic solutions.
If you're happy to just have a numerical solution, Numpy has a few routines that can help. If you're just interested in solutions to polynomials, numpy.roots will work. Specifically for the case you mentioned:
>>> import numpy
>>> numpy.roots([2,-6])
array([3.0])
For more complicated expressions, have a look at scipy.fsolve.
Either way, you can't escape using a library.
If you only want to solve the extremely limited set of equations mx + c = y for positive integer m, c, y, then this will do:
import re
def solve_linear_equation ( equ ):
"""
Given an input string of the format "3x+2=6", solves for x.
The format must be as shown - no whitespace, no decimal numbers,
no negative numbers.
"""
match = re.match(r"(\d+)x\+(\d+)=(\d+)", equ)
m, c, y = match.groups()
m, c, y = float(m), float(c), float(y) # Convert from strings to numbers
x = (y-c)/m
print ("x = %f" % x)
Some tests:
>>> solve_linear_equation("2x+4=12")
x = 4.000000
>>> solve_linear_equation("123x+456=789")
x = 2.707317
>>>
If you want to recognise and solve arbitrary equations, like sin(x) + e^(i*pi*x) = 1, then you will need to implement some kind of symbolic maths engine, similar to maxima, Mathematica, MATLAB's solve() or Symbolic Toolbox, etc. As a novice, this is beyond your ken.
Use a different tool. Something like Wolfram Alpha, Maple, R, Octave, Matlab or any other algebra software package.
As a beginner you should probably not attempt to solve such a non-trivial problem.
Write a function to compute the quantity
F(n) = n^2 Σ i=1 (i^3)
Read the problem as n squared over Sigma, with i = 1 under the sigma and I cubed at the end of the function.
I am not sure how to approach this idea. I tried setting up a function but I do not know how to use a function in Python to compute the problem we were given.
As mentioned above, I am sorry, but I do not know how to approach this problem.
I suppose the expected output here would be some quantity but because I haven't been able to make much progress, I have no clue what to expect exactly. To give more background, I understand how functions work but do not know how to approach this type of problem. Any help/guidance in writing this code would be greatly appreciated.
With a list comprehension:
Use the built-in functions, sum and range
def my_sigma(n_start: int, n_end: int) -> int:
return sum([i**3 for i in range(n_start, (n_end**2) + 1)])
# usage
print(my_sigma(1, 3))
>>> 2025
You can use the Math library to take the powers, and use a for loop for the sigma.
I think this would give you an idea,
for i in range (a , int(math.pow(n, 2))):
list_1.append(int(math.pow(i, 3)))
You can just put this inside a function with two variables a and n a representing the i in the sigma and n representing the n in the sigma.
You can use the sum function for summing up all the elements in the list if that is what you want.
I am very new to programming, and had to use sympy for a school project.
I think I get that using nonlinsolve to return an angle gives an ImageSet with the angle + 2n*pi. However I want it to return only the value of the angle (in the interval [0,pi/2]), and as one value and not an interval.
from sympy import nonlinsolve, symbols,cos
x=symbols('x')
print(nonlinsolve([cos(x)-1],[x]).args[0][0])
I want the result to be 0 and not 2*n*pi.
Clarification : I know that the result is correct, but I only want one value, that I can use algebraically, and I don't know how Sympy works (how to manipulate ImageSets)
So I might be wrong because i dont use sympy, but the solution that solvers return seems to be corect to me.
ImageSet(Lambda(_n, 2*_n*pi), Integers)
From what I understand solver returned lambda function. Cosinus is a cyclic function which means it reapeats it's value every 2PI. So the solver says first solution (_n = 0) is 0, second (_n = 1) is 2pi and so on.
look at the function plot and it will hopefully make sense:
Wolfram Alpha - (cos(x) - 1)
EDIT: I think you need to use intersect method of imageset like this( note that intersect returns all the intersections, here i selected just the first one):
from sympy import nonlinsolve, symbols,cos, Interval
import math
x = symbols('x')
f = nonlinsolve([cos(x)-1], [x]).args[0][0]
sol = f.intersect(Interval(0, math.pi/2)).args[0]
print(sol)
So I am starting with an equality of an equation and a fraction that I use to solve for both x and y:
mrs = y/x
ratio = 2/5
x = sympy.solveset(sympy.Eq(mrs, ratio), x)
y = sympy.solveset(sympy.Eq(mrs, ratio), y)
In the end, solving for y returns:
{2*x/5}
Which is a FiniteSet
But solving for x returns:
{5*y/2} \ {0}
Which is a Complement
I don't get why solving for one variable gives me a FiniteSet when solving for the other doesn't do the same? Also, would there be a way to solve for the other variable so as to get a FiniteSet instead of a Complement?
What do you expect as a result? Could you solve this problem by hand and write the expected solution? And why would you want a FiniteSet as solution?
I myself can not come up with a better notation than sympy, since x=0 needs to be excluded.
When you continue working with the solutions sympy can easily work with both, FiniteSet and Complement. Mathematically those are not completely different structures. The difference is that sympy somehow needs to represent these solutions internally and can not use the same construction for everything, but rather uses small building blocks to create the solution. The result you get with type(x) is symply the last building block used.
EDIT: Some math here: x=0 does not solve the equation y/x=2/5 for any y. So this must be excluded from the solutionset.
If you solve for y, then x=0 is already excluded since y/0 is not well defined.
If you solve for y, then y=0 is a priori possible, since 0/x=0 for x!=0. Thus sympy needs to exclude x=0 manually, which it does by removing 0 from the set of solutions.
Now, since we know that x=0 can never be a solution of the equation we can exclude it before even trying to solve the equation. Therefore we do
x = sympy.symbols('x', real=True, nonzero=True)
right at the beginning of the example (before the definition of mrs). The rest can remain unchanged.
First of, I'm sorry if the title is not entirely fitting, I had a hard time finding an appropriate one (which might have also effect my searching efficiency for already asked questions like this :/ ).
The problem is the following. While it is comparably easy to solve coupled ODE's in python with Scipy, I still have to write down my ODE in the form explicitly. For example for a coupled ODE of the form
d/dt(c_0)=a(c_0)+b(c_1) and d/dt(c_1)=c(c_0)
I would set up sth like:
import numpy as np
from scipy.integrate import ode
a=1
b=2
c=3
val=[]
def dC_dt(t, C):
return [a*C[0]+b*C[1],
c*C[0]]
c0, t0 = [1.0,0.0], 0
r = ode(dC_dt).set_integrator('zvode', method='bdf',with_jacobian=False)
r.set_initial_value(c0, t0)
t1 = 0.001
dt = 0.000005
while r.successful() and r.t < t1:
r.integrate(r.t+dt)
val.append(r.y)
However, now I have coupled ODE's of the rough form
d/dt(c_{m,n})=a(c_{m,n})+b(c_{m+1,n-1})+k(c_{m-1,n+1})
with c_{0,0}=1 and I have to include orders with m^2+n^2-mn smaller than a max value.
For a small max, what I did, is using a dictionary to use a notation with two indices and map it on a 1D list
dict_in={'0,0':0,'-1,0':2,...}
and then I entered the ODE for each order
def dC_dt(t,C):
return[a*C[dict_in['0,0']]+b*C[dict_in['1,-1']]...
Now I basically have to do that for some 100 coupled equations, which I ofc do not want to hard code, so I was trying to figure out a way, to realize the ODE's with a loop or sth. However I couldn't yet find a way around the fact of having two indices in my coefficients together with the condition of only including orders with m^2+n^2-mn smaller than a max value.
As I am running in some deadlines, I figured it is time to ask smarter people for help.
Thanks for reading my question!
I had a similar problem. If you fill you dictionary you can just redeclare the function more times inside the loop. This is a silly example of how it works:
dict_in={'0,0':0,'-1,0':2}
for elem in dict_in:
def dC_dt(t,C):
#return[a*C[dict_in['0,0']]+b*C[dict_in['1,-1']]
return dict_in[elem]
t, C = 0, 0
print(dC_dt(t,C))
#r = ode(dC_dt).set_integrator('zvode', method='bdf',with_jacobian=False)
If you need to use more functions together you can use anonymous functions and store them in memory. Another example:
functions_list = list()
for i in range(4):
f = lambda n = i: n
functions_list.append(f)
for j in range(4):
print(functions_list[j]())
You can use a list or a generator too. For example you can write down the value on a txt file and read that with the readline function each time.
As pointed in the comments below, if you use lamda functions you should pay attention to references. See also https://docs.python.org/3/faq/programming.html#why-do-lambdas-defined-in-a-loop-with-different-values-all-return-the-same-result