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Substitute one equation into another in SymPy

Suppose I have equations x = z + 2 and y = x + 1 and I wish to substitute the first one into the second one, to eliminate x and get y = z + 3. In SymPy, I can create the first two equations as:
x = sympy.symbols('x')
y = sympy.symbols('y')
z = sympy.symbols('z')
equation_one = sympy.Eq(x, z + 2)
equation_two = sympy.Eq(y, x + 1)
What is the correct way to now substitute equation_one into equation_two? The output should be a new equation.

An approach that works in this case is to use the attributes lhs/rhs ("left hand side" and "right hand side").
import sympy as sp
x = sp.symbols('x')
y = sp.symbols('y')
z = sp.symbols('z')
equation_one = sp.Eq(x, z + 2)
equation_two = sp.Eq(y, x + 1)
print(equation_two.subs(equation_one.lhs,equation_one.rhs))
Result:
Eq(y, z + 3)

Minimize system of nonlinear equation (integral on exponent)

General:
I am using maximum entropy to find distribution for on positive integers vectors, I can estimate the mean and variance, and have three equation I am trying to find a and b,
The equations:
integral(exp(a*x^2+bx+c) from (0 , infinity))-1
integral(xexp(ax^2+bx+c)from (0 , infinity))- mean
integral(x^2*exp(a*x^2+bx+c) from (0 , infinity))- mean^2 - var
(integrals between [0,∞))
The problem:
I am trying to use numerical solver and I used fsolve of sympy
But I guess I am missing some knowledge.
My code:
import numpy as np
import sympy as sym
from scipy.optimize import *
def myFunction(x,*data):
y = sym.symbols('y')
m,v=data
F = [0]*3
x[0] = - abs(x[0])
print(x)
F[0] = (sym.integrate(sym.exp(x[0] * y ** 2 + x[1] * y + x[2]), (y, 0,sym.oo)) -1).evalf()
F[1] = (sym.integrate(y*sym.exp(x[0] * y ** 2 + x[1] * y + x[2]), (y, 0,sym.oo))-m).evalf()
F[2] = (sym.integrate((y**2)*sym.exp(x[0] * y ** 2 + x[1] * y + x[2]), (y,0,sym.oo)) -v-m).evalf()
print(F)
return F
data = (10,3.5) # mean and var for example
xGuess = [1, 1, 1]
z = fsolve(myFunction,xGuess,args = data)
print(z)
my result are not that accurate, is there a better way to solve it?
integral(exp(a*x^2+bx+c))-1 = 5.67659292676884
integral(xexp(ax^2+bx+c))- mean = −1.32123173796713
integral(x^2*exp(a*x^2+bx+c))- mean^2 - var = −2.20825624606312
Thanks

I have rewritten the problem replacing sympy with numpy and lambdas (inline functions).
Also note that in your problem statement you subtract the third equation with $mean^2$, but in your code you only subtract $mean$.
import numpy as np
from scipy.optimize import minimize
from scipy.integrate import quad
def myFunction(x,data):
m,v=data
F = np.zeros(3) # use numpy array
# use scipy.integrade.quad for integration of lambda functions
# quad output is (result, error), so we just select the result value at the end
F[0] = quad(lambda y: np.exp(x[0] * y ** 2 + x[1] * y + x[2]), 0, np.inf)[0] -1
F[1] = quad(lambda y: y*np.exp(x[0] * y ** 2 + x[1] * y + x[2]), 0, np.inf)[0] -m
F[2] = quad(lambda y: (y**2)*np.exp(x[0] * y ** 2 + x[1] * y + x[2]), 0, np.inf)[0] -v-m**2
# minimize the squared error
return np.sum(F**2)
data = (10,3.5) # mean and var for example
xGuess = [-1, 1, 1]
z = minimize(lambda x: myFunction(x, data), x0=xGuess,
bounds=((None, 0), (None, None), (None, None))) # use bounds for negative first coefficient
print(z)
# x: array([-0.99899311, 2.18819689, 1.85313181])
Does this seem more reasonable?

Changing coefficients modulo p in a SymPy polynomial

I took a cryptography course this semester in graduate school, and once of the topics we covered was NTRU. I am trying to code this in pure Python, purely as a hobby. When I attempt to find a polynomial's inverse modulo p (in this example p = 3), SymPy always returns negative coefficients, when I want strictly positive coefficients. Here is the code I have. I'll explain what I mean.
import sympy as sym
from sympy import GF
def make_poly(N,coeffs):
"""Create a polynomial in x."""
x = sym.Symbol('x')
coeffs = list(reversed(coeffs))
y = 0
for i in range(N):
y += (x**i)*coeffs[i]
y = sym.poly(y)
return y
N = 7
p = 3
q = 41
f = [1,0,-1,1,1,0,-1]
f_poly = make_poly(N,f)
x = sym.Symbol('x')
Fp = sym.polys.polytools.invert(f_poly,x**N-1,domain=GF(p))
Fq = sym.polys.polytools.invert(f_poly,x**N-1,domain=GF(q))
print('\nf =',f_poly)
print('\nFp =',Fp)
print('\nFq =',Fq)
In this code, f_poly is a polynomial with degree at most 6 (its degree is at most N-1), whose coefficients come from the list f (the first entry in f is the coefficient on the highest power of x, continuing in descending order).
Now, I want to find the inverse polynomial of f_poly in the convolution polynomial ring Rp = (Z/pZ)[x]/(x^N - 1)(Z/pZ)[x] (similarly for q). The output of the print statements at the bottom are:
f = Poly(x**6 - x**4 + x**3 + x**2 - 1, x, domain='ZZ')
Fp = Poly(x**6 - x**5 + x**3 + x**2 + x + 1, x, modulus=3)
Fq = Poly(8*x**6 - 15*x**5 - 10*x**4 - 20*x**3 - x**2 + 2*x - 4, x, modulus=41)
These polynomials are correct in modulus, but I would like to have positive coefficients everywhere, as later on in the algorithm there is some centerlifting involved, so I need to have positive coefficients. The results should be
Fp = x^6 + 2x^5 + x^3 + x^2 + x + 1
Fq = 8x^6 + 26x^5 + 31x^4 + 21x^3 + 40x^2 + 2x + 37
The answers I'm getting are correct in modulus, but I think that SymPy's invert is changing some of the coefficients to negative variants, instead of staying inside the mod.
Is there any way I can update the coefficients of this polynomial to have only positive coefficients in modulus, or is this just an artifact of SymPy's function? I want to keep the SymPy Poly format so I can use some of its embedded functions later on down the line. Any insight would be much appreciated!

This seems to be down to how the finite field object implemented in GF "wraps" integers around the given modulus. The default behavior is symmetric, which means that any integer x for which x % modulo <= modulo//2 maps to x % modulo, and otherwise maps to (x % modulo) - modulo. So GF(10)(5) == 5, whereas GF(10)(6) == -4. You can make GF always map to positive numbers instead by passing the symmetric=False argument:
import sympy as sym
from sympy import GF
def make_poly(N, coeffs):
"""Create a polynomial in x."""
x = sym.Symbol('x')
coeffs = list(reversed(coeffs))
y = 0
for i in range(N):
y += (x**i)*coeffs[i]
y = sym.poly(y)
return y
N = 7
p = 3
q = 41
f = [1,0,-1,1,1,0,-1]
f_poly = make_poly(N,f)
x = sym.Symbol('x')
Fp = sym.polys.polytools.invert(f_poly,x**N-1,domain=GF(p, symmetric=False))
Fq = sym.polys.polytools.invert(f_poly,x**N-1,domain=GF(q, symmetric=False))
print('\nf =',f_poly)
print('\nFp =',Fp)
print('\nFq =',Fq)
Now you'll get the polynomials you wanted. The output from the print(...) statements at the end of the example should look like:
f = Poly(x**6 - x**4 + x**3 + x**2 - 1, x, domain='ZZ')
Fp = Poly(x**6 + 2*x**5 + x**3 + x**2 + x + 1, x, modulus=3)
Fq = Poly(8*x**6 + 26*x**5 + 31*x**4 + 21*x**3 + 40*x**2 + 2*x + 37, x, modulus=41)
Mostly as a note for my own reference, here's how you would get Fp using Mathematica:
Fp = PolynomialMod[Algebra`PolynomialPowerMod`PolynomialPowerMod[x^6 - x^4 + x^3 + x^2 - 1, -1, x, x^7 - 1], 3]
output:
1 + x + x^2 + x^3 + 2 x^5 + x^6

Numpy arange error with Lagrange Multiplier in Python

I try to use Lagrange multiplier to optimize a function, and I am trying to loop through the function to get a list of number, however I got the error
ValueError: setting an array element with a sequence.
Here is my code, where do I go wrong? If the n is not an array I can get the result correctly though
import numpy as np
from scipy.optimize import fsolve
n = np.arange(10000,100000,10000)
def func(X):
x = X[0]
y = X[1]
L = X[2]
return (x + y + L * (x**2 + y**2 - n))
def dfunc(X):
dLambda = np.zeros(len(X))
h = 1e-3
for i in range(len(X)):
dX = np.zeros(len(X))
dX[i] = h
dLambda[i] = (func(X+dX)-func(X-dX))/(2*h);
return dLambda
X1 = fsolve(dfunc, [1, 1, 0])
print (X1)
Helps would be appreciated, thank you very much

First, check func = fsolve()
Second, print(func([1,1,0]))` - result in not number ([2 2 2 2 2 2 2 2 2]), beause "n" is list. if you want to iterate n try:
import numpy as np
from scipy.optimize import fsolve
n = np.arange(10000,100000,10000)
def func(X,n):
x = X[0]
y = X[1]
L = X[2]
return (x + y + L * (x**2 + y**2 - n))
def dfunc(X,n):
dLambda = np.zeros(len(X))
h = 1e-3
r = 0
for i in range(len(X)):
dX = np.zeros(len(X))
dX[i] = h
dLambda[i] = (func(X+dX,n)-func(X-dX,n))/(2*h)
return dLambda
for iter_n in n:
print("for n = {0} dfunc = {1}".format(iter_n,dfunc([0.8,0.4,0.3],iter_n)))

CORDIC algorithm returning bad numbers

I started to implement a CORDIC algorithm from zero and I don't know what I'm missing, here's what I have so far.
import math
from __future__ import division
# angles
n = 5
angles = []
for i in range (0, n):
angles.append(math.atan(1/math.pow(2,i)))
# constants
kn = []
fator = 1.0
for i in range (0, n):
fator = fator * (1 / math.pow(1 + (2**(-i))**2, (1/2)))
kn.append(fator)
# taking an initial point p = (x,y) = (1,0)
z = math.pi/2 # Angle to be calculated
x = 1
y = 0
for i in range (0, n):
if (z < 0):
x = x + y*(2**(-1*i))
y = y - x*(2**(-1*i))
z = z + angles[i]
else:
x = x - y*(2**(-1*i))
y = y + x*(2**(-1*i))
z = z - angles[i]
x = x * kn[n-1]
y = y * kn[n-1]
print x, y
When I plug z = π/2 it returns 0.00883479322917 and 0.107149125055, which makes no sense.
Any help will be great!
#edit, I made some changes and now my code has this lines instead of those ones
for i in range (0, n):
if (z < 0):
x = x0 + y0*(2**(-1*i))
y = y0 - x0*(2**(-1*i))
z = z + angles[i]
else:
x = x0 - y0*(2**(-1*i))
y = y0 + x0*(2**(-1*i))
z = z - angles[i]
x0 = x
y0 = y
x = x * kn[n-1]
y = y * kn[n-1]
Now it's working way better, I had the problem because I wasn't using temporary variables as x0 and y0, now when I plug z = pi/2 it gives me better numbers as (4.28270993661e-13, 1.0) :)