Bisenton Method:-
In Numarical Anylisys, for algebraic equation and Transcendental equation are sloving using Bisenton Method.
Working Rules:-
1:- Let f(x) = 0, be a given equation. Now we are find to nearest root a and b such that f(a) < 0 and f(b) > 0 ;
2:- After finding nearest root a and b, We start finding frist approximate root using Bisenton Method
Frist approximate root, x1 = (a+b) /2
Calculate f(x1) and examine its sign.
2.1:- if f(x1) <0, =>> Then Root lies between x1 and b.
Then find 2nd approximate root x2=(x1+b) /2
2.2:- if f(x1) >0, =>> Then Root lies between x1 and a.
Then find 2nd approximate root x2=(x1+a) /2.
Approximate root finding process are continue and stop when we rech f(x) = 0,Where x is the root equation.
Example:-
1:- Find the real root of the equation x^3 - x -4 = 0,Using Biseton Method correct upto 3 decimal place.
Solution:-
Let f(x) = x^3 - x -4 = 0
Now We are find nearest approximate Root.
f(0) = 0 - 0 - 4 = - 4 <0
f(1) = 1 -1 - 4 = -4 < 0 ... (1)
f(2) = 8 - 2 - 4 = 2 > 0 ..... (2)
Form (1) and (2) , we can told root lies on 1 and 2.
Tips: For Quick Sloving, we try to find more nearest approximate root of the equation.
f(1.5) = (1.5) ^3 - 1.5 -4 <0
f(1.7) = (1.7) ^3 - 1.7 - 4 < 0
f(1.8) = (1.8) ^3 - 1.8 - 4 > 0
f(1.799) = (1.799) ^3 - 1.799 - 4 > 0
f(1.797) =(1.797) ^3 - 1.797 - 4 > 0
f(1.796) = (1.796) ^3 - 1.796- 4 <0
>>>> Let a = 1.796 and b= 1.797
Let frist approximate root by Biseton Method x1= (a+b) /2= (1.796+1.7965) /2=1.7965
Now f(x1) = 0.0015461 >0
>>>> Root lies on 1.7965 to 1.7966
Let 2nd approximate root by Biseton Method x2=(1.7965+1.7966) /2=1.79625
Now f(x2) = - 0.00062411.
>>>>> Approximate Root Upto 3 decimal is 1.79625 ( because f(1.79625) = -0. 00062411
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