The almost universally used algorithm to compute a, where a > 0, is the recursio
ID: 3888756 • Letter: T
Question
The almost universally used algorithm to compute a,
where a > 0, is the recursion
The trick for doing this is to use Newton’s
method to compute 1 a , and then obtain a by multipling by a. Write down your recur-
sion formula for computing 1 a in a manner similar to (3). This formula should only involve
addition/subtraction, multiplication and division by 2.
Try you algorithm on the problem of computing 5. As an initial guess use x 0 = 0.5. Report
the values of x 0 , x 1 ,. . . , x 5 in a nice table and
verify that your algorithm is working by
comparing these numbers to the true value of 5
I've tried:
clc
clear all
iteration = 100;
A = 5;
a=1/A;
x=a/2;
xn=1;
tolerance = 10e-16;
f =@x .5*(xn+a/xn)
while f ~= sqrt(A)* tolerance
x(n+1) = f(x(n));
end
x(n+1)
but I've been banging my head and am now probably so confused that I'm missing the obvious.
2 3 rt rtExplanation / Answer
package bookList;
public class RecursiveSequence {
public static double recursiveSeq(int i,double x){
double x1=0.0;
int a=5;
double r=0.5; // means (r=1/2)
double result= r*(x+(a/x));
System.out.println("X"+i+" "+x+" , Approximate value "+result);
if(i==5) return 0.0; // we can replace i==5 with i==100 .
else recursiveSeq(++i,result); // recursive function.
return 0.0;
}
public static void main(String[] args) {
double x0=0.5;
int i=0;
RecursiveSequence.recursiveSeq(i,x0);
}
}
/*
output:-
X0 0.5 , Approximate value 5.25
X1 5.25 , Approximate value 3.1011904761904763
X2 3.1011904761904763 , Approximate value 2.3567372726441826
X3 2.3567372726441826 , Approximate value 2.239157222737191
X4 2.239157222737191 , Approximate value 2.23607010853285
X5 2.23607010853285 , Approximate value 2.236067977500805
*/
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