Page 138 - IJOCTA-15-3

Srinivasarao Thota et al. / IJOCTA, Vol.15, No.3, pp.503-516 (2025)
Table 2. Number of iterations

Eq Root NR HM MHM NM1 NM2 EM Sania9 MBIT9 PM
Eq 1.404491648 11 7 5 5 4 5 3 3 3 (Example 1)
1
Eq 2 6.308777130 12 8 6 5 6 5 3 3 3 (Example 2)
Eq 3 1.465091220 10 7 5 4 4 4 3 3 3 (Example 3)
Eq 4 2.125391199 10 8 6 4 4 4 3 3 3
Notes: NM1 and NM2 represent the numerical methods proposed in refs. [38] and [1,2], respectively. EM
indicates the enhanced method from reference [44]. RFM refers to the ninth-order root-finding method from
ref. [37], which introduced a new three-step numerical scheme for solving nonlinear scalar and vector
equations. MBIT denotes the ninth-order memory-based iterative technique from ref. [37], in which the
authors proposed a novel three-step, memory-based method for solving nonlinear equations. Sania is a
Ninth-order method by Sania. Abbreviations: Eq, equation; HM, Halley’s method; MHM, modified Halley’s
method; NR, Newton Raphson method; PM, proposed method.

> Prop_Alg(1, sin(x)^2-x^2+1 = 0,3); t = x - (2 * f(x) * df(x)) / (2 *
Iteration 1 = 1.403066996. (df(x))**2 - f(x) * ddf(x))
f(x) = .35327132e-2, s = t * math.exp(-f(t) / (t * df(t)))
AbsError = .2872756591 if t * df(t) != 0 else t
Iteration 2 = 1.404491649. x_new = t - (f(t) + f(s)) / df(t)
f(x) = -.2e-8, print(f"Iteration {iteration}:
AbsError = .1014354910e-2 x = {x_new:.15f}, f(x) =
Iteration 3 = 1.404491649. {f(x_new):.15e}")
f(x) = -.2e-8, if abs(x_new - x) < tol:
AbsError = 0. print("\nConverged to root:",
1.404491649 x_new)
Example 6. Consider with x0 = 0.8. print("Total iterations:",
iteration)
> Prop_Alg(0.8, 10*x*exp(-x^2)-1 = 0, 4);
return x_new
Iteration 1 = 1.531609079.
x = x_new
f(x) = .466791835,
print("\nMaximum iterations reached.
AbsError = .4776735063
Root may not be accurate.")
Iteration 2 = 1.679628729.
return x
f(x) = .5200e-5,
AbsError = .8812640999e-1
# Example: Solve sin^2(x) - x^2 + 1 = 0
Iteration 3 = 1.679630611.
initial_guess = 1.0 # Start near an
f(x) = -.12e-8,
expected root
AbsError = .1120484461e-5
root = ninth_order_method(initial_guess)
Iteration 4 = 1.679630611.
f(x) = -.12e-8, AbsError = 0.
1.679630611
Output of the implementation:
4.2. Python implementation
Consider the transcendental Equation (13) from
Example 1 for sample computations using the Iteration 1: x = 1.403066995981864,
Python implementation, as follows. f(x) = 3.532713035351298e-03
import numpy as np Iteration 2: x = 1.404491648215341,
import math f(x) = -4.440892098500626e-16
def f(x): Iteration 3: x = 1.404491648215341,
return np.sin(x)**2-x**2+1 f(x) = -4.440892098500626e-16
def df(x):
return 2*np.sin(x)*np.cos(x)-2*x Converged to root:
def ddf(x): 1.4044916482153413
return 2*(np.cos(x)**2- np.sin(x)**2)-2 Total iterations: 3
def ninth_order_method(x0, tol=1e-10,
max_iter=20): x = x0 === Code Execution Successful ===
for iteration in range(1, max_iter + 1):
510