5

I am trying to solve for the x values with a known y. I was able to get the polynomial to fit my data, and now I want to know the x value that a chosen y would land on the curve.

import numpy as np

x = [50, 25, 12.5, 6.25, 0.625, 0.0625, 0.01]
y = [0.00, 0.50, 0.68, 0.77, 0.79, 0.90, 1.00]

poly_coeffs = np.polyfit(x, y, 3)

f = np.poly1d(poly_coeffs)

I want to do 0.5 = f and solve for the x values.

I can solve this in WolframAlpha by typing: 

0.5 = -9.1e-6*x^3 + 5.9e-4*x^2 - 2.5e-2*x + 9.05e-1

The real x value is ~26

Trenton McKinney
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chimpsarehungry
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2 Answers2

6
In [1]: from numpy.polynomial import Polynomial as P

In [2]: x = [50, 25, 12.5, 6.25, 0.625, 0.0625, 0.01]

In [3]: y = [0.00, 0.50, 0.68, 0.77, 0.79, 0.90, 1.00]

In [4]: p = P.fit(x, y, 3)

In [5]: (p - .5).roots()
Out[5]: 
array([ 19.99806935-37.92449551j,  19.99806935+37.92449551j,
        25.36882693 +0.j        ])

Looks like the root you want is 25.36882693.

Charles Harris
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5

You can solve the equation f(x) - y = 0 using np.roots. Consider the function:

def solve_for_y(poly_coeffs, y):
    pc = poly_coeffs.copy()
    pc[-1] -= y
    return np.roots(pc)

Then you can use it to solve your polynomial for any y you want:

>>> print solve_for_y(poly_coeffs, 0.5)
[ 19.99806935+37.92449551j  19.99806935-37.92449551j  25.36882693 +0.j        ]
>>> print solve_for_y(poly_coeffs, 1.)
[ 40.85615395+50.1936152j  40.85615395-50.1936152j -16.34734226 +0.j       ]
Andrey Sobolev
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    Your method appears to modify `poly_coeffs`, so the second call to the method will actually give the wrong answer, i.e. it's solving for `y = 1` for `x` in a different polynomial than the original one. – Amit Kumar Gupta May 31 '13 at 08:34
  • @Amit: Yes, my bad. I thought that function parameters are passed by value, and it seems to be true only for immutable types. Since `np.array` is mutable, it is passed by reference. I'll edit the post, thank you. – Andrey Sobolev May 31 '13 at 09:02