Newton's Method for Finding Roots

Polynomial is composed of two words: Poly (which means "many") and Nominal (which means "terms."). A polynomial is a mathematical equation made up of variables, constants, and exponents that are mixed using operations like addition, subtraction, multiplication, and division. The general form of a polynomial is 

f(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + ... + a₂x² + a₁x + a₀

Polynomials can be classified as monomials, binomials, and trinomials based on the number of terms present. For example, terms like x, 13y, 39, etc. are all monomials while terms like x² + x, x10 - x⁴, etc. are termed as binomials because they consist of two terms. Similarly, such polynomials as having only three terms are termed trinomials.

Roots of a Polynomial

 The roots of a polynomial are the solutions to the given polynomial for which the unknown variable's value must be determined. We can evaluate the value of a polynomial to zero if we know the roots. A polynomial of degree 'n' in variable x is an equation of the type a_nxⁿ + a_n-1x^n-1 +...... + a₁x + a₀, where each variable has a constant as its coefficient. The phrase refers to each variable in an expression that is separated by an addition or subtraction sign. The greatest power of a polynomial variable is defined as the degree of the polynomial.

Newton's Method of Finding Roots of a Polynomial

Newton's method formula is used for finding the roots of a polynomial by iterating from one root to the next. Calculating the roots by this approach takes a long time for polynomials of greater degree, but for polynomials of lower degree, the results are quite rapid and near to the true roots of the equation. Using this strategy, we can identify the consecutive roots of an equation if we know any one of its roots. The formula for Newton's method of finding the roots of a polynomial is as follows:

x_1=x_0-\frac{f(x_0)}{f'({x_0})}

where,

• x₀ is the initial value
• f(x₀) is the function value at the initial value
• f'(x₀) is the first derivative of the function value at initial value.

Derivation of Newton's Method

Newton's method begins with an initial guess that is relatively near to the correct root (solution), and then utilize the tangent line to acquire an even better x-intercept than our first guess or beginning point.

Assume we're looking for the root (or x-intercept) of f(x). This indicates we're looking for an in the image below. The trick of Newton's method is to draw a tangent line to the graph y = f(x) at the point (b, f(b)).

The crucial calculation in each stage of Newton's technique, as we just saw, is to discover where the tangent line to y = f(x) at the point (x₁, y₁) crosses the x-axis.

The point gradient form of equation of a line with slope m and passing through the point (x0, y0) is given as:

y - y₀ = m(x - x₀).

In this situation, the line has a slope of f'(x_n). The x- intercept occurs where y = 0.

Thus, by setting y = 0, we have:

x_1=x_0-\frac{f(x_0)}{f'({x_0})}

Hence proved.

Sample Problems on Newton's Method for Finding Roots

Question 1. Starting with x₀ = 5, find the next root of the equation x³ − 7x² + 8x − 3 = 0.

Solution:

As per Newton's method, x_1=x_0-\frac{f(x_0)}{f'({x_0})}  .

Given: x₀ = 5, f(x) = x³ − 7x² + 8x − 3 = 0

⇒ f'(x) = 3x² − 14x + 8

Now, x₁ = x_0-\frac{f(5)}{f'({5})}

= 5 − (−13)/13

= 5 + 1

⇒ x₁ = 6

Question 2. Starting with x₀ = −3.5, find the next root of the equation x³ − x² − 15x + 1 = 0.

Solution:

As per Newton's method, x_1=x_0-\frac{f(x_0)}{f'({x_0})}.

Given: x₀ = 5, f(x) = x³ − x² − 15x + 1 = 0

⇒ f'(x) = 3x² − 2x − 15

Now, x₁ = x_0-\frac{f(-3.5)}{f'({-3.5})}

= −3.5 − (−1.625)/28.75

⇒ x₁ = −3.443

Question 3. Starting with x₀ = 2, find the next root of the equation x² − 2 = 0.

Solution:

As per Newton's method, x_1=x_0-\frac{f(x_0)}{f'({x_0})}  .

Given: x0 = 5, f(x) = x² − 2 = 0

⇒ f'(x) = 2x

Now, x₁ = x_0-\frac{f(2)}{f'({2})}

= 2 − 2/4

⇒ x₁ = 1.5

Question 4. Starting with x₀ = 1, find the next root of the equation −x³ + 4x² − 2x + 2 = 0.

Solution:

As per Newton's method, x_1=x_0-\frac{f(x_0)}{f'({x_0})}.

Given: x0 = 5, f(x) = −x³ + 4x² − 2x + 2 = 0.

Now, x1 = x_0-\frac{f(2)}{f'({2})}

= 1 − 3/3

⇒ x₁ = 0

Question 5. Starting with x₀ = 8, find the next root of the equation x³ − 10x² + 9x − 12 = 0.

Solution:

As per Newton's method, x_1=x_0-\frac{f(x_0)}{f'({x_0})}.

Given: x0 = 5, f(x) = −x3 + 4x2 − 2x + 2 = 0.

Now, x1 = x_0-\frac{f(2)}{f'({2})}

= 1 − 3/3

⇒ x1 = 0

Question 6: Starting with x0 = 1.5, find the square roots by Newton's Method for Finding Roots of equation x3 - 3x2 + 2x - 1.

Solution:

The function is f(x) = x3- 3x2 + 2x + 1

The derivative of the function is f'(x) = 3x2 -6x +2

We are given x0= 1.5 as initial guess.

Newton's method is given by the formula:

x(n+1) = x(n) - f(x(n))/ f'(x(n))

where x(n) is the current estimate, and x(n+1) is the next estimate.

We start with x0 = 1.5.

f(x0) = f(1.5) = (1.5)^3 - 3(1.5)^2 + 2(1.5) - 1 = 0.875

f'(x0) = f'(1.5) = 3(1.5)^2 - 6(1.5) + 2 = 2.25

x1 = x0 - f(x0) / f'(x0) = 1.5 - 0.875 / 2.25 = 1.5 - 0.3889 = 1.1111

We repeat the process with x1 = 1.1111.

f(x1) = f(1.1111) = (1.1111)^3 - 3(1.1111)^2 + 2(1.1111) - 1 = 0.2344

f'(x1) = f'(1.1111) = 3(1.1111)^2 - 6(1.1111) + 2 = 1.3333

x2 = x1 - f(x1) / f'(x1) = 1.1111 - 0.2344 / 1.3333 = 1.1111 - 0.1759 = 0.9352

We continue this process until we reached a desired level of accuracy.

After several iterations, we find that the root of the equation x3 - 3x2 + 2x - 1 is approximately x ≈ 1.0000.

Question 7: Starting with x0 = 0.1, find the square roots by Newton's Method for Finding Roots of equation ln(x) - x/2.

Solution:

The function is f(x) = ln(x) - x/2

The derivative of f(x) id f'(x) = 1/x -1/2

We are given x0 = 0.1 as the initial guess.

Newton's method is given by the formula:

x(n+1) = x(n) - f(x(n))/ f'(x(n))

where x(n) is the current estimate, and x(n+1) is the next estimate.

We start with x0 = 0.1.

f(x0) = f(0.1) = ln(0.1) - 0.1/2 = -2.3026 - 0.05 = -2.3526

f'(x0) = f'(0.1) = 1/0.1 - 1/2 = 10 - 0.5 = 9.5

x1 = x0 - f(x0) / f'(x0) = 0.1 - (-2.3526) / 9.5 = 0.1 + 0.2477 = 0.3477

We repeat the process with x1 = 0.3477.

f(x1) = f(0.3477) = ln(0.3477) - 0.3477/2 = -1.0829 - 0.1739 = -1.2568

f'(x1) = f'(0.3477) = 1/0.3477 - 1/2 = 2.8823 - 0.5 = 2.3823

x2 = x1 - f(x1) / f'(x1) = 0.3477 - (-1.2568) / 2.3823 = 0.3477 + 0.5273 = 0.8750

We continue this process until we reached a desired level of accuracy.

After several iterations, we find that the root of the equation ln(x) - x/2 is approximately x ≈ 0.6412.

Read More:

• Newton Raphson Method | Definition, Formula, Examples
• Root Finding Algorithm
• Program for Newton Raphson Method

Practice Problems on Newton's Method for Finding Roots

1. Starting with x₀ = 0.5. Find the square roots by Newton's Method for Finding Roots of equation cos(x) - x

2. Starting with x₀ = 1. Find the square roots by Newton's Method for Finding Roots of equation e^x - 3x

3. Starting with x₀ = 0. Find the square roots by Newton's Method for Finding Roots of equation x³ - 2x + 2

4. Starting with x₀ = 2. Find the square roots by Newton's Method for Finding Roots of equation ln(x) - 1

5. Starting with x₀ = 1.2, find the square roots by Newton's Method for Finding Roots of equation 2x³ - 5x² + 3x - 1.

6. Starting with x₀ = 0.8, find the square roots by Newton's Method for Finding Roots of equation sin(x) - x/2.

7. Starting with x₀ = 0.5, find the square roots by Newton's Method for Finding Roots of equation e^x - 2x - 2.

8. Starting with x₀ = 2.8, find the square roots by Newton's Method for Finding Roots of equation x⁴ - 4x³ + 5x² - 4x + 1.

9. Starting with x₀ = 1.8, find the square roots by Newton's Method for Finding Roots of equation x⁴ - 4x³ + 3x² - 2x + 1.

10. Starting with x₀ = 0.4, find the square roots by Newton's Method for Finding Roots of equation e^x - 3x + 1.

11. Starting with x₀ = 2.5, find the square roots by Newton's Method for Finding Roots of equation x⁴ - 4x³ + 7x² - 10x + 5.

12. Starting with x₀ = 0.3, find the square roots by Newton's Method for Finding Roots of equation e^x - 2x - 1.

13. Starting with x₀ = 0.9, find the square roots by Newton's Method for Finding Roots of equation sin(2x) - x.

14. Starting with x₀ = 3.5, find the square roots by Newton's Method for Finding Roots of equation x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1.

15. Starting with x₀ = 0.2, find the square roots by Newton's Method for Finding Roots of equation ln(x) - x/3.

Conclusion

Newton's Method is a powerful and efficient algorithm for finding roots of real valued functions. By iteratively refining an initial estimate using the formula x(new) = x(old)/ x'(old), we can coverage to a root with high accuracy. The method's quadratic convergence rate makes it particularly useful for finding roots of functions with a simple root structure.

However it is essential to be aware of the method's limitations such as requirement for an initial estimate close to the root and potential for convergence to a local minimum or divergence if the initial estimate is poor.