Improved Quadratic Equation Solver

import math
import numpy as np
from ipywidgets import interact

if not input('Load JupyterAI? [Y/n]').lower()=='n':
    %reload_ext jupyter_ai

In this notebook, we will improve the quadratic equation solver in the previous lab using conditional executions.

Discriminant¶

Recall that a quadratic equation is

ax^2+bx+c=0

(1)

where $a$ , $b$ , and $c$ are real-valued coefficients, and $x$ is the unknown variable.

The discriminant $\Delta$ discriminates the roots

\frac{-b\pm \sqrt{\Delta}}{2a}

(3)

of a quadratic equation:

For example, if $(a,b,c)=(1,-2,1)$ , then

\Delta = (-2)^2 - 4(1\cdot 1) = 0

(5)

and so the repeated root is

- \frac{b}{2a} = \frac{2}2 = 1.

(6)

$(x-1)^2 = x^2 - 2x + 1$ as expected.

We will improve the quadratic solver to return only one root when the discriminant is close enough to 0.

Exercise 1

To determine whether a floating point number $x$ is close to 0, the implementation math.isclose is insufficient:

The test math.isclose(x, 0, rel_tol=rel_tol) is ineffective as explained in the lecture as it boils down to x==0 or rel_tol>=1.
The test math.isclose(x, 0, abs_tol=1e-9) may not work as expected since x (and therefore its precision error) floats at different scale according to its exponent.

Complete the following function to implement an alternative test of how close $x$ is to 0 relative to $y$ and $z$ with a relative tolerance of $\delta_{\text{rel}}$ :

\lvert x\rvert\leq \max\Set{\lvert y\rvert, \lvert z\rvert} \delta_{\text{rel}}

(7)

This test is similar to math.isclose, where $\delta_{\text{rel}}\geq 0$ is the relative tolerance specified by the keyword-only argument rel_tol, which has a default value of 1e-9.

def iszero(x, y, z, *, rel_tol=1e-9):
    # YOUR CODE HERE
    raise NotImplementedError

# tests
assert not iszero(1e-8, 0, 1)
assert iszero(1e-8, 1, 0, rel_tol=1e-7)
assert iszero(1e-8, 1, 10)

# hidden tests

def get_roots(a, b, c, *, rel_tol=1e-9):
    if iszero(d:=((y:=b**2)-(z:=4*a*c)), y, z, rel_tol=rel_tol):
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()
assert np.isclose(get_roots(1, 1e-9, 0), 0.0).all()
assert np.isclose(get_roots(1, 1e-6, 0), (-1e-06, 0.0)).all()

# hidden tests

YOUR ANSWER HERE

Linear equation¶

YOUR ANSWER HERE

Nevertheless, the quadratic equation remains valid:

bx + c=0,

(8)

the root of which is $x = -\frac{c}b$ .

def get_roots(a, b, c, *, rel_tol=1e-9):
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()
assert np.isclose(get_roots(1, -2, 1), 1.0).all()
assert np.isclose(get_roots(0, -2, 1), 0.5).all()

# hidden tests

Degenerate cases¶

What if $a=b=0$ ? In this case, the equation becomes

c = 0

(9)

which is always satisfied if $c=0$ but never satisfied if $c\neq 0$ .

Exercise 6

Improve the function get_roots to return root(s) under all the following cases:

If $a=0$ and $b\neq 0$ , assign roots to the single root $-\frac{c}{b}$ .
If $a=b=0$ and $c\neq 0$ , assign roots to None.
If $a=b=c=0$ , there are infinitely many roots. Assign to roots the tuple -float('inf'), float('inf').
Note that float('inf') converts the string 'inf' to a floating point value that represents $\infty$ .

The above checks of whether a coefficient is 0 should be relative to other coefficients with a relative tolerance of rel_tol.

def get_roots(a, b, c, *, rel_tol=1e-9):
    # YOUR CODE HERE
    raise NotImplementedError
    return roots

# tests
assert np.isclose(get_roots(1, 1, 0), (-1.0, 0.0)).all()
assert np.isclose(get_roots(1, 0, 0), 0.0).all()
assert np.isclose(get_roots(1, -2, 1), 1.0).all()
assert np.isclose(get_roots(0, -2, 1), 0.5).all()
assert np.isclose(get_roots(0, 0, 0), (-float('inf'), float('inf'))).all()
assert get_roots(0, 0, 1) is None

# hidden tests

After you have complete the exercises, you can run your robust solver below:

# quadratic equations solver
@interact(a=(-10, 10, 1), b=(-10, 10, 1), c=(-10, 10, 1))
def quadratic_equation_solver(a=1, b=2, c=1):
    print("Root(s):", get_roots(a, b, c))