03. Solution of Non Linear Equations (Bisection Method)

Dated: 14-08-2025

Polynomials¹

\(f(x) = 0\) is called an algebraic equation in \(x\) if it is purely a polynomial.¹

Some Facts

1. Every equation of the form \(f(x) = 0\) has at least one root (either real² or complex²).
2. Every polynomial¹ of \(n^{th}\) degree has \(n\) roots.
3. If \(f(x) = 0\) is an equation of odd degree then it has at least one real² root whose sign is opposite to that of last term.
4. If \(f(x) = 0\) is an equation of even degree whose last term is negative then it has at least one positive and at least one negative root.

Transcendental Equation

An equation is called transcendental equation if it has either or a combination of the following:

• Logarithmic³ functions⁴
• Trigonometric functions⁴
• Exponential functions⁴

Roots of an Equation

If we have an equation \(f(x) = 0\) then the values for \(x\) which satisfy the equation, are called roots.

\[a \text{ is a root for } f(x) = 0 \iff f(a) = 0\]

Properties of an Algebraic Equation

• Complex² roots appear in pairs. If \((a + \iota b)\) is a root then \((a - \iota b)\) is also a root.
• If \(f(x) = 0\) is of \(n^{th}\) degree and \(a\) is a root then \((x - a)\) is a factor of \(f(x)\). Diving \(f(x)\) by \((x - a)\) gives us a polynomial¹ of degree \(n - 1\).

Descartes Rule of Signs

The number² of positive roots of an algebraic equation \(f(x) = 0\) with real² coefficients can not exceed the number² of changes in the signs of the coefficients in the polynomial¹ \(f(x) = 0\). Similarly the number² of negative roots of the equation can not exceed the number² of changes in the sign of coefficients of \(f(-x) = 0\).

Example

\[x^3 - 3x^2 + 4x - 5 = 0\]

Positive Roots

Following are the sign changes that occur in the equation

1. \(+\) for \(x^3\) changes to \(-\) for \(3x^2\)
2. \(-\) for \(3x^2\) changes to \(+\) for \(4x\)
3. \(+\) for \(4x\) changes to \(-\) for \(5\)

Since there are 3 changes so there are 3 positive roots.

Negative Roots

\[f(-x) = -x^3 - 3x^2 - 4x - 5\]

Since there are no changes, therefore, there are no negative roots.

Intermediate Value Property

If \(f(x)\) is a continuous function⁵ defined over \(a \le x \le b\) such that \(f(a)\) and \(f(b)\) differ in their signs then, there is a number² \(\beta\) such that \(a \le \beta \le b\) and \(f(\beta) = 0\).

\(a \le \beta \le b\) where \(f(a) > 0 > f(b)\)

Direct Methods

Those methods which do not require any information about the initial approximation of root to start the solution are known as direct methods.

Examples

• Graefee root squaring method
• Gauss elimination method
• Gauss Jordan method

Iterative Methods

These methods require an initial approximation to start.

Examples

• Bisection method
• Newton Raphson method
• Secant method
• Jacobie method

How to Get an Initial Approximation?

Initial approximation can be found by using

Graphical Method

\[f(x) = x - \sin(x) - 1 = 0\]

\[\implies x - 1 = \sin(x)\]

\[f_1(x) = x - 1 \quad f_2(x) = \sin(x)\]

/// captions
\(f_1\) and \(f_2\) intersect at \(1.9\) and therefore, it should be taken as the initial approximation.
///

Analytical Method

This method can be used for

• Algebraic equations
• Transcendental equations

We have to locate two points \(a\) and \(b\) such that \(f(a)\) and \(f(b)\) has opposite signs.

Example

\[f(x) = 3x - \sqrt{1 + \sin(x)} = 0\]

\[\implies f(0) = -1\]

\[\implies f(1) = 3 - \sqrt{1 + \sin\left(1 \times \frac{180}{\pi}\right)} = 3 - \sqrt{1 + 0.84} = 1.64\]

Since \(f(0)\) and \(f(1)\) differ in signs, therefore, the root lies between \(0\) and \(1\).

Bisection Method (Bolzano)

Suppose that you have \(f(x) = 0\) defined on the \([x_0, x_1]\) and let \(f(x_0)\) and \(f(x_1)\) have opposite signs such that \(f(x_0)f(x_1) < 0\) then the graph crosses the \(x\) axis, suggesting that there exists a root \(x_0 < x_2 < x_1\) such that

\[x_2 = \frac{x_0 + x_1}{2}\]

If \(f(x_2) = 0\) then it is the root, otherwise the root lies somewhere in either

• \([x_0, x_2]\)
• \([x_2, x_1]\)

Now we have 2 halves and we check which halve will contain the root.

• \(f(x_0) f(x_2) < 0\) implies that the root is in \([x_0, x_2]\)
• \(f(x_2) f(x_1) < 0\) implies that the root is in \([x_2, x_1]\)

And we keep repeating the process.

In numerical analysis, \(\pi = 3.14\) and calculations are carried out in radians mode.

Example

Solve \(x^3 - 9x + 1 = 0\) for the root between \(x = 2\) and \(x = 4\) by bisection method.

Solution

\[\because f(x) = x^3 - 9x + 1 = 0\]

\[\implies f(2) = 2^3 - 9(2) + 1 = 8 - 18 + 1 = -9\]

\[\implies f(4) = 4^3 - 9(4) + 1 = 64 - 36 + 1 = 29\]

\[\because f(2) f(4) < 0\]

\[x_0 = 2 \quad x_1 = 4\]

\[x_2 = \frac{2 + 4}{2} = 3\]

\[ f(3) = 3^3 - 9(3) + 1 = 27 - 27 + 1 = 1 \]

\[\because f(2) f(3) < 0\]

\[x_3 = \frac{2 + 3}{2} = 2.5\]

\[ f(2.5) = 2.5^3 - 9(2.5) + 1 = 15.625 - 22.5 + 1 = -5.875 \]

\[\because f(2.5) f(3) < 0\]

\[ x_4 = \frac{2.5+3}{2} = 2.75 \]

\[ x_5 = 2.875 \quad \text{and} \quad x_6 = 2.9375 \]

\(n\)	\(xn\)​	\(f(xn​)\)
\(2\)	\(3\)	\(1.0\)
\(3\)	\(2.5\)	\(-5.875\)
\(4\)	\(2.75\)	\(-2.9531\)
\(5\)	\(2.875\)	\(-1.1113\)
\(6\)	\(2.9375\)	\(-0.0901\)

When to Stop the Process of Iteration?

Assume the accuracy was given to be \(10^{-3}\) and we have 2 consecutive roots like \(2.135648\) and \(2.135769\).

\[2.135769 - 2.135648 = 0.000121\]

Since the accuracy is achieved, we can stop here.

References

Read more about notations and symbols.

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1. Read more about polynomials. ↩↩↩↩↩

2. Read more about numbers. ↩↩↩↩↩↩↩↩↩↩

3. Read more about logarithms. ↩

4. Read more about functions. ↩↩↩

5. Read more about continuity. ↩