ROOTS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS
Generally an equation is solved by factorization.
But in many cases, the method of factorization fails.
In such cases, numerical methods are used.
The numerical methods to solve the equation f(x) = 0 are:
1) BISECTION METHOD
2) REGULA FALSI METHOD
3) NEWTON RAPHSON METHOD
4) SECANT METHOD
1) BISECTION METHOD
Bisection Method is also called the interval halving method, the binary search method, or the dichotomy method.
The Rate of Convergence of Bisection Method is Linear.
For any continuous function f(x),
Find two points say a and b such that f(a) < 0 and f(b) > 0
Find midpoint of a and b say xn
xn = (a + b)/2
Question :
Obtain the root of the equation x – cosx = 0 by bisection method correct up to two decimal places.
Solution :
f(x) = x – cosx
f(0) = -1 and f(1) = 0.4597
Since f(0) < 0 and f(1) > 0 , the root lies between 0 and 1.
|
No. |
A |
B |
xn = (a + b)/2 |
Sign of f(xn) |
|
1 |
0 |
1 |
0.5 |
-ve |
|
2 |
0.5 |
1 |
0.75 |
+ve |
|
3 |
0.5 |
0.75 |
0.625 |
-ve |
|
4 |
0.625 |
0.75 |
0.6875 |
-ve |
|
5 |
0.6875 |
0.75 |
0.71875 |
-ve |
|
6 |
0.71875 |
0.75 |
0.7344 |
-ve |
|
7 |
0.7344 |
0.75 |
0.7422 |
+ve |
|
8 |
0.7344 |
0.7422 |
0.7383 |
-ve |
|
9 |
0.7383 |
0.7422 |
0.74025 |
+ve |
|
10 |
0.7383 |
0.74025 |
0.7393 |
+ve |
|
11 |
0.7383 |
0.7393 |
0.7388 |
|
Since x10 and x11 are same up to two decimal places, the root is 0.73
2) REGULA FALSI METHOD
Regula Falsi Method is also known as False Position Method.
xn = [af(b) – bf(a)]/[f(b) – f(a)]
Question:
Find the real root of x log10 x – 1.2 = 0, correct up to three decimal places.
Solution:
f(x) = x log10 x – 1.2
f(2) = -0.5979 and f(3) = 0.2314
Since f(2) < 0 and f(3) > 0, the root lies between 2 and 3.
|
No. |
a |
b |
f(a) |
f(b) |
xn =[af(b) – bf(a)]/[f(b) – f(a)] |
f(xn) |
|
1 |
2 |
3 |
-0.5979 |
0.2314 |
2.721 |
-0.0171 |
|
2 |
2.721 |
3 |
-0.0171 |
0.2314 |
2.7402 |
-0.0004 |
|
3 |
2.7402 |
3 |
-0.0004 |
0.2314 |
2.7406 |
|
The root is 2.740
3) NEWTON-RAPHSON METHOD
The Newton Raphson Method has a quadratic convergence and the convergence is of the order of 2.
Newton raphson formula or Newton’s iteration formula is :
xn+1 = xn – [f(xn)/f’(xn)]
Question:
Find the real positive root of the equation x – cosx = 0 using Newton Raphson Method correct up to three decimal places.
Solution:
f(x) = x – cosx
f(0) = -1 and f(1) = 0.4597
Since f(0) < 0 and f(1) > 0 , the root lies between 0 and 1.
Let x0 = 1
f’(x) = 1 + sinx
f(x0) = f(1) = 0.4597
f’(x0) = f’(1) = 1.8415
x1 = x0 – [f(x0)/f’(x0)]
= 1 – (0.4597/1.8415)
= 0.7504
f(x1) = f(0.7504) = 0.019
f’(x1) = f’(0.7504) = 1.6819
x2 = x1 – [f(x1)/f’(x1)]
= 0.7504 – (0.019/1.6819)
= 0.7391
f(x2) = f(0.7391) = 0.00002
f’(x2) = f’(0.7391) = 1.6736
x3 = x2 – [f(x2)/f’(x2)]
= 0.7391 – (0.00002/1.6736)
= 0.7391
Since x2 and x3 are same up to three decimal places, the root is 0.739
4) SECANT METHOD
xn+1 = xn – {[(xn – xn-1) * f(xn)]/ [f(xn) – f(xn-1)]}
The rate of convergence of secant method is 1.618.
The rate of convergence of secant method is lesser than Newton-Raphson Method and so Secant Method is more efficient than the Newton Raphson Method.
Question:
Find the root of xlog10 x – 1.9 = 0, correct up to three decimal places with x0 = 3 and x1 = 4.
Solution:
f(x) = xlog10 x – 1.9
x0 = 3, x1 = 4
f(x0) = f(3) = -0.4686, f(x1) = f(4) = 0.5082
x2 = x1 – {[(x1 – x0) * f(x1)]/ [f(x1) – f(x0)]}
= 4 – {[4 – 3) * (0.5082)]/[(0.5082 + 0.4686)]}
= 3.4797
f(x2) = f(3.4797) = -0.0156
x3 = x2 – {[(x2 – x1) * f(x2)]/ [f(x2) – f(x1)]}
= 3.4979 – {[3.4979 – 4) * (-0.0156)]/[(-0.0156 – 0.5082)]}
= 3.4952
f(x3) = f(3.4952) = -0.0005
x4 = x3 – {[(x3 – x2) * f(x3)]/ [f(x3) – f(x2)]}
= 3.4952 – {[3.4952 – 3.4797) * (-0.0005)]/[(-0.0005 + 0.01556)]}
= 3.4957
Since x3 and x4 are same up to three decimal places, the root is 3.495