# Initial Approximation For An Iterative Procedure

**Initial Approximation for an Iterative Procedure:**

For polynomial equations, Descartes’ rule of signs gives the bound for the number of positive and negative real roots.

(i) We count the number of changes of signs in the coefficients of P_{n}(x) for the equation f(x) =P_{n}(x) = 0. The number of positive roots cannot exceed the number of changes of signs.

For example, if there are four changes in signs, then the equation may have four positive roots or two positive roots or no positive root. If there are three changes in signs, then the equation may have three positive roots or definitely one positive root. (For polynomial equations with real coefficients, complex roots occur in conjugate pairs.)

(ii) We write the equation f(– x) = P_{n}(– x) = 0, and count the number of changes of signs in the coefficients of P_{n}(– x). The number of negative roots cannot exceed the number of changes of signs. Again, if there are four changes in signs, then the equation may have four negative roots or two negative roots or no negative root. If there are three changes in signs, then the equation may have three negative roots or definitely one negative root.

We use the following theorem of calculus to determine an initial approximation. It is also called the intermediate value theorem.

**Theorem 1.1** If f(x) is continuous on some interval [a, b] and f(a)f(b) < 0, then the equation f(x) = 0 has at least one real root or an odd number of real roots in the interval (a, b). This result is very simple to use. We set up a table of values of f (x) for various values of x. Studying the changes in signs in the values of f (x), we determine the intervals in which the roots lie. For example, if f (1) and f (2) are of opposite signs, then there is a root in the interval (1, 2).

**Example 1.1** Determine the maximum number of positive and negative roots and intervals of length one unit in which the real roots lie for the following equations.

**Solution :
**

Let

The number of changes in the signs of the coefficients (8, – 12, – 2, 3) is 2. Therefore, the equation has 2 or no positive roots.

Now, f(– x) = – 8x3 – 12x2 2x 3. The number of changes in signs in the coefficients (– 8, – 12, 2, 3) is 1. Therefore, the equation has one negative root.

We have the following table of values for f(x),

Values of f (x), Example 8x^{3} – 12x^{2} – 2x 3 = 0

Since

f(– 1) f(0) < 0, there is a root in the interval (– 1, 0),

f(0) f(1) < 0, there is a root in the interval (0, 1),

f(1) f(2) < 0, there is a root in the interval (1, 2).

Therefore, there are three real roots and the roots lie in the intervals (– 1, 0), (0, 1), (1, 2).

**Example 1.2** Determine an interval of length one unit in which the negative real root, which is smallest in magnitude lies for the equation

9x^{3} 18x^{2} – 37x – 70 = 0.

**Solution :
**

Let f(x) = 9x3 18x2 – 37x – 70 = 0. Since, the smallest negative real root in magnitude is required, we form a table of values for x < 0,

Table 1.3. Values of f (x ),9x^{3} 18x^{2} – 37x – 70 = 0.

From the table, we find that there is one real positive root in the interval (1, 2). The equation has no negative real root.