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Practical work #2. Bisection (or Bolzano) method

This method is based on the repeated application of intermediate value property.

Let the function f (x) be continuous between a and b. For definiteness, let f (a) be (–)ve and f (b) be (+)ve. Then the first approximation to the root is x ₁ = (a + b).

If f (x ₁) = 0, then x ₁ is a root of f (x) = 0, otherwise, the root lies between a and x ₁ or x ₁ and b according to f (x ₁) is (+)ve or (–)ve. Then we bisect the interval as before and continue the process until the root is found to the desired accuracy.

In the adjoining figure, f (x ₁) is (+)ve so that the root lies between a and x ₁. The second approximation to the root is x ₂ = (a + x ₁). If f (x ₂) is (–)ve the root lies between x 1 and x 2. The third approximation to the root is x 3 = (x ₁ + x ₂), and so on.

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Example. Find the real root of the equation x log10 x = 1.2 by Bisection method correct to four decimal places. Also write its program in C-language.

Sol. f (x) = x log₁₀ x – 1.2

Since f (2.74) = –.000563 i.e., (–)ve

and f (2.75) =.0081649 i.e., (+)ve

Hence, the root lies between 2.74 and 2.75.

∴ First approximation to the root is

Now f (x 1) = f (2.745) =.003798 i.e., (+)ve

Hence, the root lies between 2.74 and 2.745.

∴ Second approximation to the root is

Now f (x ₂) = f (2.7425) =.001617 i.e., (+)ve

Hence, the root lies between 2.74 and 2.7425.

∴ Third approximation to the root is

=2.74125

Now f (x ₃) = f (2.74125) =.0005267 i.e., (+)ve

Hence, the root lies between 2.74 and 2.74125.

∴ Fourth approximation to the root is

Now f (x ₄) = f (2.740625) = –.00001839 i.e., (–)ve.

Hence, the root lies between 2.740625 and 2.74125.

∴ Fifth approximation to the root is

Now f (x ₅) = f (2.7409375) =.000254 i.e., (+)ve

Hence, the root lies between 2.740625 and 2.7409375.

∴ Sixth approximation to the root is

Now f (x ₆) = f (2.74078125) =.0001178 i.e., (+)ve

Hence, the root lies between 2.740625 and 2.74078125.

∴ Seventh approximation to the root is

Now f (x ₇) = f (2.740703125) =.00004973 i.e., (+)ve

Hence, the root lies between 2.740625 and 2.740703125

∴ Eighth approximation to the root is

Now f (x ₈) = f (2.740664063) =.00001567 i.e., (+)ve

Hence, the root lies between 2.740625 and 2.740664063.

∴ Nineth approximation to the root is

Since x ₈ and x ₉ are the same up to four decimal places, the approximate real root is 2.7406. C-program for above problem is given below: