Root finding (root, brentq) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of root finding with scipy.optimize.root

What is the output of this code snippet that finds the root of the function f(x) = x**3 - 1 using scipy.optimize.root starting from x0=0.5?

SciPy

from scipy.optimize import root

def f(x):
    return x**3 - 1

sol = root(f, 0.5)
print(round(sol.x[0], 4))

A 0.5

B 1.0

C -1.0

D 0.0

💡 Hint

Think about the cube root of 1.

The function f(x) = x^3 - 1 has a root at x=1 because 1^3 - 1 = 0. Starting from 0.5, the root finder converges to 1.0.

❓ data_output

intermediate

Number of iterations in brentq root finding

Using scipy.optimize.brentq to find the root of f(x) = cos(x) - x in the interval [0, 1], how many iterations does the solver perform?

SciPy

from scipy.optimize import brentq
import numpy as np

def f(x):
    return np.cos(x) - x

root, result = brentq(f, 0, 1, full_output=True)
print(result.iterations)

A 10

B 3

C 7

D 5

💡 Hint

Brent's method is efficient and usually converges quickly.

The brentq method typically converges in about 5 iterations for this function and interval.

🧠 Conceptual

advanced

Why brentq requires a sign change interval

Why does the scipy.optimize.brentq method require the function values at the interval endpoints to have opposite signs?

A Because brentq uses the Intermediate Value Theorem which guarantees a root if the function changes sign over the interval.

B Because brentq only works with positive functions.

C Because brentq uses derivative information which requires sign change.

D Because brentq approximates the function with a polynomial that must cross zero.

💡 Hint

Think about what guarantees a root exists in an interval.

Brent's method relies on the Intermediate Value Theorem, which states that if a continuous function changes sign over an interval, there must be a root inside.

🔧 Debug

advanced

Identify the error in root finding code

What error will this code produce when trying to find the root of f(x) = log(x) - 2 using scipy.optimize.root starting from x0=0?

SciPy

from scipy.optimize import root
import numpy as np

def f(x):
    return np.log(x) - 2

sol = root(f, 0)
print(sol.x)

A RuntimeWarning: divide by zero encountered in log

B TypeError: 'float' object is not iterable

C ValueError: math domain error

D No error, prints root

💡 Hint

Consider the domain of the log function at zero.

np.log(0) is undefined and causes a divide by zero warning during evaluation.

🚀 Application

expert

Find root of a noisy function using brentq

Given the noisy function f(x) = x^2 - 4 + noise, where noise is a small random value, which approach best finds a root near x=2 using scipy.optimize.brentq?

SciPy

import numpy as np
from scipy.optimize import brentq

np.random.seed(0)
noise = lambda: np.random.normal(0, 0.1)

def f(x):
    return x**2 - 4 + noise()

# Which code snippet correctly finds the root near 2?

A brentq(f, 1, 3)

B brentq(f, 0, 1)

C brentq(lambda x: x**2 - 4, 1, 3)

D brentq(lambda x: x**2 - 4 + 0.5, 1, 3)

💡 Hint

Brentq requires the function to have opposite signs at the interval ends and be continuous.

Because f includes random noise, it may not have a guaranteed sign change or continuity. Using the noise-free function x^2 - 4 ensures brentq works correctly.