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How to extract equation from a polynomial fit?


How to extract equation from a polynomial fit?

By : Rafael
Date : November 22 2020, 02:59 PM
this one helps. The coefficients of the linear model are stored in the intercept_ and coeff_ attributes of the model.
You can see this more clearly by turning-down the regularization and feeding-in a known model; e.g.
code :
import numpy as np
from sklearn.linear_model import Ridge
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import PolynomialFeatures

x = 10 * np.random.random(100)
y = -4 + 2 * x - 3 * x ** 2

model = make_pipeline(PolynomialFeatures(2), Ridge(alpha=1E-8, fit_intercept=False))
model.fit(x[:, None], y)
ridge = model.named_steps['ridge']
print(ridge.coef_)
# array([-4.,  2., -3.])


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Solving polynomial 3rd order polynomial equation for intensity mapping

Solving polynomial 3rd order polynomial equation for intensity mapping


By : raid3r
Date : March 29 2020, 07:55 AM
like below fixes the issue Not sure if this is what you need, but try this simple approach, which uses the [-10,10) values range for x,y and z:
code :
class Program
{
    static void Main(string[] args)
    {
        int x = 0, y = 0, z = 0;
        int x1 = -10, x2 = 10,
            y1 = -10, y2 = 10,
            z1 = -10, z2 = 10;

        for (int ix = x1; ix < x2; ix++)
        {
            for (int iy = y1; iy < y2; iy++)
            {
                for (int iz = z1; iz < z2; iz++)
                {
                    var result = (2 * ix) + (5 * iy) + 6 * (Math.Pow(iz, 2));
                    if (result > 0)
                    {
                        Console.WriteLine("x {0} y {1} z {2} : {3}", 
                            ix, iy, iz, result);
                    }
                }
            }
        }
    }
}
polynomial equation parameters

polynomial equation parameters


By : Karttik Vivek
Date : March 29 2020, 07:55 AM
it should still fix some issue numpy.polyfit
code :
>>> x = np.array([0.0, 1.0, 2.0, 3.0,  4.0,  5.0])
>>> y = np.array([0.0, 0.8, 0.9, 0.1, -0.8, -1.0])
>>> z = np.polyfit(x, y, 3)
>>> z
array([ 0.08703704, -0.81349206,  1.69312169, -0.03968254])
Rewrite equation as polynomial

Rewrite equation as polynomial


By : Vikas Verma
Date : March 29 2020, 07:55 AM
seems to work fine , Here's what you can do.
code :
import sympy as sp
Kf,Td0s,Ke,Te,Tv,Kv,s= sp.symbols("K_f,T_d0^',K_e,T_e,T_v,K_v,s")
Ga= Kf/(1+s*Tv)
Gb= Ke/(1+s*Te)
Gc= Kf/(1+s*Td0s)
G0=Ga*Gb*Gc
eq = (G0 + 1).as_numer_denom()[0]
eq = eq.expand().collect(s)
Eq(eq, 0)
Eq(K_e*K_f**2 + T_d0^'*T_e*T_v*s**3 + s**2*(T_d0^'*T_e + T_d0^'*T_v + T_e*T_v) + s*(T_d0^' + T_e + T_v) + 1, 0)
Big-O for a polynomial in log equation?

Big-O for a polynomial in log equation?


By : Marek Nowacki
Date : March 29 2020, 07:55 AM
hope this fix your issue log(n^n + n) <= log(2*n^n) = log2 + log(n^n)
Besides log(n^n) = nlog(n).
Solve polynomial equation for x in R

Solve polynomial equation for x in R


By : Jannik Rittwage
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further the short answer is that this polynomial has no roots in the set of real numbers, you can see that analytically with some help from R :
code :
> #((4*x)^2+(2*x)^2+(1*x)^2+(0.5*x)^2+0.25)*((1 - 0.167)/0.167) = 1
> 
> # first add up your coefficients 
> coefs <- c(16 + 4 + 1+ .25 , .25)
> coefs 
[1] 21.25  0.25
> 
> # apply the second product 
> coefs <- (coefs - 0.167*coefs)/0.167
> coefs
[1] 105.995509   1.247006
> 
> # move the one from one side to another
> 
> coefs <- coefs - c(0,1)
> coefs
[1] 105.995509   0.247006
> 
> #106*x^2 + 1/4 = 0 has no solution in the set of real number
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