In this post, we learn about solving Algebraic Equation using Newton's Method (or) Newton-Raphson Method.
Given an approximate root of an equation, a closer approximation to the root can be found using Newton's Method.
Let a0 be the approximate root of the equation f(x)=0.
Let a be the exact root nearer to a0.
Then, a = a0 + h
where h is an integer with very small value.
Since a is the exact root of f(x)=0,
f(a) = f(a0+h) = 0
By Taylor's expression,
f(a) = f(a0+h) = f(a0) + hf'(a0) + (h^2/2!)f"(a0)+0000=0
Since, h is small, neglecting higher powers h^2, h^3,....etc, we get
f(a0) + hf'(a0) = 0
h = -(f(a0)/f'(a0))
a1 = a0 + h = a0 - (f(a0)/f'(a0))
a2 = a1 + h = a1 - (f(a1)/f'(a1))
Thus, we again get a sequence of approximate roots which converges to form the exact root on a few iterations depending on the required precision.
% this function that calculates solution of algebraic equation
% using Newton-Raphson method
% the variable prev stores the first approximate root of the eqn.
% here, we assume it to be the mean of the two points p1 and p2 (say)
% such that, f(p1)<0 and f(p2)>0
prev = (range(1) + range(2))/2;
% x is symbol var represented by 'x'
% the function 'f' is initialized
% n gives the size of the coeff/: vector
% in the loop below, we create the function 'f' using symbol 'x'
% from the coeff/: vector
for i = 1 : n
% the last element of co vector is the constant
if i == n
f = f + co(i);
% end of if i==n
% adding the terms of the function step by step
if co(i) ~= 0 && i ~= n
f = f + co(i)*(x^(n-i));
% end of if
%end of for
% infinite while loop
% where we find the sequence of approximate roots
while ( 1 )
% finding the approximate root
% using Taylor's expansion (neglecting higher powers)
% when the required precision is obtained, break the loop
if abs( abs(prev) - abs(x0) ) < 0.000001
% end of if
% assigning previous value
% end of while
% end of function