How to solve a system of linear equations with b=0 in R
By : sourabh Gangoli
Date : March 29 2020, 07:55 AM
it fixes the issue In R I need to solve a system of linear equations (Ax=b), where b=0. By using solve() it just returns a zero vector for the answer, but I want the nonzero solutions of the system. Is there any way for it? , I think you are looking for the null space of a matrix A. Try : code :
library(MASS)
Null(t(A))
R > (A < matrix(c(1,2,3,2,4,7), ncol = 3, byrow = T))
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 4 7
R > Null(t(A))
[,1]
[1,] 8.944272e01
[2,] 4.472136e01
[3,] 7.771561e16
R > (A < matrix(c(1,2,3,2,4,6), ncol = 3, byrow = T))
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 4 6
R > Null(t(A))
[,1] [,2]
[1,] 0.5345225 0.8017837
[2,] 0.7745419 0.3381871
[3,] 0.3381871 0.4927193

Solving Nonlinear System of 3 differentail equations(Lorenz equations) using ode solve: ode23s in Matlab
By : Raffy Sueno
Date : March 29 2020, 07:55 AM
around this issue I am trying to solve 3 differentail equations(Lorenz equations) using ode solver: ode23s in Matlab. Here are the 3 lorenz equations: , Your function is incorrect as far as I can see. This line: code :
xprime=[ alpha*I*(1c) + c*( k_f  k_d  k_n * s  k_p*(1q)); lambda_b * c* P_C  lambda_r *(1q)*s; k_p * c *(P_C / P_Q) gamma * q];
xprime=[ alpha*I*(1x(1)) + x(1)*( k_f  k_d  k_n * x(2)  k_p*(1x(3))); lambda_b * x(1)* P_C  lambda_r *(1x(3))*x(2); k_p * x(1) *(P_C / P_Q) gamma * x(3)];
%initial vectors
c=0.25;
s=0.02;
q=0.02;

Solve nonlinear system of equations R
By : Yakov Shtykov
Date : March 29 2020, 07:55 AM
I wish this help you With your addition of some data for C10 the example runs perfectly ok if one takes into account that the loop should take account of length(C10). Like this (with some changes; for why see below): code :
library(nleqslv)
C10 < c(1.000000e+00,9.754920e01,9.547681e01,9.359057e01,9.182586e01,9.014674e01)
system_size < 1024
target < function(Coeffs){
y < numeric(3)
y[1] < Coeffs[1]*(exp(Coeffs[2]*(C[4]1))+exp(Coeffs[2]*(system_size(C[4]1))))+Coeffs[3]^2C[1]
y[2] < Coeffs[1]*(exp(Coeffs[2]*(C[4]))+exp(Coeffs[2]*(system_size(C[4]))))+Coeffs[3]^2C[2]
y[3] < Coeffs[1]*(exp(Coeffs[2]*(C[4]+1))+exp(Coeffs[2]*(system_size(C[4]+1))))+Coeffs[3]^2C[3]
y
}
init < 50*c(0.001,0.01,0)
for(i in 2:min(length(C10)1,(system_size/2)))
{
C < c(C10[i1],C10[i],C10[i+1],i2)
#function
target < function(Coeffs){
y < numeric(3)
y[1] < Coeffs[1]*(exp(Coeffs[2]*(C[4]1))+exp(Coeffs[2]*(system_size(C[4]1))))+Coeffs[3]^2C[1]
y[2] < Coeffs[1]*(exp(Coeffs[2]*(C[4]))+exp(Coeffs[2]*(system_size(C[4]))))+Coeffs[3]^2C[2]
y[3] < Coeffs[1]*(exp(Coeffs[2]*(C[4]+1))+exp(Coeffs[2]*(system_size(C[4]+1))))+Coeffs[3]^2C[3]
y
}
cat("i=",i,"init=",init, "target(init)=",target(init),"\n")
sol < nleqslv(init, target,control=list(btol=.01), method="Broyden")
print(sol)
}

Solve a system of linear equations and linear inequalities
By : jyothsna shekar
Date : March 29 2020, 07:55 AM
wish helps you Yours is a problem in linear programming, where your equality and inequalities are the limitations and you want to minimize (then later maximize) the expression y. The equality, inequalities, and expression are all linear, so that makes it linear programming. The scipy package, using the scipy.optimize.linprog function, can do this kind of linear programming. Here is commented code to do what you want. Note that all the inequalities were slightly changed to include equality, which is necessary to have a maximum or minimum value of y. To find the maximum value of y the code instead finds the minimum value of y then prints the additive inverse of that, since linprog minimizes the objective function. Finally, the inequality restrictions must be "less than or equal to" in linprog, so I multiplied both sides of your inequality x + y > 180 by 1 to get one, namely x + y <= 180. Ask if you have any questions. code :
from scipy.optimize import linprog
# Set up values relating to both minimum and maximum values of y
coefficients_inequalities = [[1, 1]] # require 1*x + 1*y <= 180
constants_inequalities = [180]
coefficients_equalities = [[3, 12]] # require 3*x + 12*y = 1000
constants_equalities = [1000]
bounds_x = (30, 160) # require 30 <= x <= 160
bounds_y = (10, 60) # require 10 <= y <= 60
# Find and print the minimal value of y
coefficients_min_y = [0, 1] # minimize 0*x + 1*y
res = linprog(coefficients_min_y,
A_ub=coefficients_inequalities,
b_ub=constants_inequalities,
A_eq=coefficients_equalities,
b_eq=constants_equalities,
bounds=(bounds_x, bounds_y))
print('Minimum value of y =', res.fun)
# Find and print the maximal value of y = minimal value of y
coefficients_max_y = [0, 1] # minimize 0*x + 1*y
res = linprog(coefficients_max_y,
A_ub=coefficients_inequalities,
b_ub=constants_inequalities,
A_eq=coefficients_equalities,
b_eq=constants_equalities,
bounds=(bounds_x, bounds_y))
print('Maximum value of y =', res.fun) # opposite of value of y
Minimum value of y = 43.3333333333
Maximum value of y = 51.1111111111

How to solve nonlinear system of trigonometric equations in Python (which MATLAB can solve easily)
By : user3497036
Date : March 29 2020, 07:55 AM
it fixes the issue I am trying to solve a system of nonlinear trigonometric equations in Python. I tried the following: , In SymPy if you want numerical solutions you should use nsolve: code :
In [97]: nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1])
Out[97]:
⎡0.5538674055548 ⎤
⎢ ⎥
⎢0.837453526933376⎥
⎢ ⎥
⎣6.95538865037068 ⎦
In [101]: eq1
Out[101]: a⋅sin(11.42806⋅π⋅f + φ)  0.347064 = 0
In [102]: (sol,) = nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1], dict=True)
In [103]: sol
Out[103]: {a: 0.5538674055548, f: 0.837453526933376, φ: 6.95538865037068}
In [104]: eq1.subs(sol)
Out[104]: False
In [105]: eq1.lhs.subs(sol)
Out[105]: 0.347064  0.5538674055548⋅sin(6.95538865037068 + 9.57046915300624⋅π)
In [106]: eq1.lhs.subs(sol).evalf()
Out[106]: 1.29025679909939e15
In [107]: (sol,) = nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1], dict=True, prec=50)
In [108]: sol
Out[108]:
{a: 0.55386740555480009188439615822304411607289430639164, f: 0.83745352693337644862065403386504543698722276260565, φ: 6.9553886
503706758809942541544797040214354242211993}
In [109]: eq1.lhs.subs(sol).evalf()
Out[109]: 3.27785083138700e51

