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Algebraic and transcendental equations and Bisection or Bolzano method.

 Algebraic and transcendental equations;

An equation of the form f(x)=0 is called algebraic or transcendental equations depending upon whether f(x) is purely a polynomial of x or contains exponential, trigonometric or logarithmic functions. For example: 2x³+3x²+17=0 is an algebraic equation while 5x³+4logx+2sinx=0 is transcendental equation.Examples of transcendental functions include the exponential function, the logarithm, and the trigonometric functions. The process of finding the roots of an equation (or zeroes of an equation) is called the solution of f(x)=0.

Bisection or Bolzano method;

Let f(x)=0 be an algebraic or a transcendental equation and let f(a) and f(b) are of opposite sign i.e f(a)f(b)<0 then root of equation f(x)=0 lie between a and b. Then the first approximation to the root is x₁ = (a+b)/2.
Now if f(x₁) = 0, then x₁ is the root of f(x)=0, otherwise the root lie between a and x₁ or x₁ and b according as f(x) is positive or negative. Then we bisect the interval as before i.e second approximation to the root is x₂ = (a+x₁)/2 or  x₂ =  (x₁+b)/2 and continue the process until the root is found to desired accuracy.
Bisection method
Bisection method.

From the figure root lie between a and b i.e first approximation to the root x₁ = (a+b)/2.
Then the second approximation to the root lie between a and x₁  i.e x₂ = (a+x₁)/2.
Similarly, the third approximation to the root lie between  x₁ and x₂ i.e x₃ =  (x₁+x₂)/2 and so on.
Example: Find a root of the equation x³-x-4=0 between 1 and 2, to three places of decimal by bisection method.
Solution. Let f(x)= x³-x-4=0
Initialisation
At x=1,
f(1)= (1)³-1-4 = -4
and at x=2,
f(2)= (2)³ -2- 4 = 2
Clearly, f(1)f(2) = -8 <0.
Therefore, the root of f(x)=0 lies between 1 and 2.
First iteration
x₁ = (a+b)/2 = (1+2)/2 = 1.5,
Since f(1) = -4, f(2) = 2, f(x₁) = (1.5)³-1.5-4 = -2.125.
Since f(1.5) is negative and f(2) is positive i.e f(1.5)f(2)<0
Therefore, the root lies in interval (1.5,2).
Second iteration
x₂ = (1.5+2)/2 = 1.75
Now f(1.75) = (1.75)³ -1.75 - 4 = -0.39062
Clearly f (1.75) is negative and f(2) is positive.
Therefore, the root lies in interval (1.75,2).
Third iteration
x₃ = (1.75 +2)/2 =1.875
Now f(1.875) = (1.875)³ - 1.875 -4 = 0.71679
Since f (1.75) is negative and f (1.875) is positive.
Therefore, the root lies in interval (1.75,1.875).
Fourth iteration
x₄ = (1.75+1.875)/2 = 1.8125
Now f(1.8125) = (1.8125)³- 1.8125-4 = 0.14184
Since f (1.75) is negative and f(1.8125) is positive.
Therefore, the root lies in interval (1.75,1.8125).
Fifth iteration
x₅ = (1.75+1.8125)/2 = 1.78125.
Now f(1.78125) = (1.78125)³ - 1.78125 - 4 = -0.12960
Since f (1.78125) is negative and f (1.8125) is positive.
Therefore, the root lies in interval (1.78125, 1.8125).
Sixth iteration
x₆ = (1.78125+1.8125)/2 = 1.79687,
Now f(1.79687) = (1.79687)³- 1.79687 - 4 = 0.00477
Since f(1.78125) is negative and f (1.79687) is positive.
Therefore, the root lies in interval (1.78125,1.79687).
Seventh iteration
x₇ = (1.78125+1.79687)/2 = 1.78906
Now f (1.78906) = (1.78906)³ - 1.78906 -4 = -0.0672.
Since f (1.78906) is negative and f (1.79687) is positive.
Therefore, the root lies in interval (1.78906,1.79687).
Eighth iteration
x₈ = (1.78906 + 1.79687)/2 = 1.79296
Now f (1.79296) =(1.79296)³- 1.79296 -4 = -0.02913.
Since f (1.79296) is negative and f (1.79687) is positive.
Therefore, the root lies in interval (1.79296,1.79687).
Nineth iteration
x₉ = (1.79296 + 1.79687)/2 = 1.79491
f(1.79491) = (1.79491)³ - 1.79491 -4 = -0.02513.
Since f(1.79491) is negative and f (1.79687) is positive.
Therefore, the root lies in interval (1.79491,1.79687).
Tenth iteration
x₁₀ = (1.79491 + 1.79687)/2 = 1.79589
Now f(1.79589) = (1.79589)³ -1.79589 -4 = -0.00375.
Since f (1.79589) is negative and f (1.79687) is positive.
Therefore the root lies in interval (1.79589,1.79687).
Eleventh iteration
x₁₁ = (1.79589+1.79687)/2 = 1.79638
Now f (1.79638) = (1.79638)³ - 1.79638 -4 = 0.000504.
Since f (1.79638) is positive and f (1.79589) is negative.
Therefore, the root lies in interval (1.79589,1.79638).
Twelfth iteration
x₁₂= (1.79589+1.79638)/2 = 1.79613
f(1.79613) = (1.79613)³-1.79613 - 4 = -0.00167.
Since f(1.79613) is negative and f (1.79638) is positive.
Therefore, the root lies in interval (1.79613,1.79638).
Here we see that the digits in the first three places of decimal are the same in the interval (1.79613,1.79638). Therefore, the value of the root to three places of decimal is 1.796 Ans.

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