Square Root program using Difference Equations

Question

I made a program using difference equation initial conditions for each number, but as you can see the code didn't give me satisfying results, plus it took more loops than it's supposed to, it should be two runs for each number maximum. The square root of 5 = 2.23 after 2 loops.

N=4;
S=[5 10 15 27 40]; %root squared numbers input variables 
y1=[2 3 4 5 6]; %1st Initial conditions
for  i=0:1:1
for  n=1:1:3
y1(n)=0.5*(y1(i+1)+(S(n)./y1(i+1)))
end
end

Results on command window:

y1 =
    2.2500    3.0000    4.0000    5.0000    6.0000
y1 =
    2.2500    3.3472    4.0000    5.0000    6.0000
y1 =
    2.2500    3.3472    4.4583    5.0000    6.0000
y1 =
    2.4205    3.3472    4.4583    5.0000    6.0000
y1 =
    2.4205    3.1674    4.4583    5.0000    6.0000
y1 =
    2.4205    3.1674    3.9516    5.0000    6.0000

Welcome to stackoverflow, please edit your question to include a minimal reproducible example. It's unclear what you expect as outputs, what would be a "satisfying" / expected result? Why do you assert that? Have you tried stepping through your code to find the specific point where the output diverges from your expectations? Keep in mind that we have zero contextual information other than what you include in the question for what you're trying to achieve — Wolfie
– Wolfie, Commented Feb 20, 2023 at 14:10

Cris Luengo · Accepted Answer · 2023-02-20 17:39:07Z

3

The inner loop does not address all elements of y1. Try this instead:

format long

S=[5 10 15 27 40]; 
y=[2 3 4 5 6]; 

% y=2*x*x'

for n=1:5
    for i=1:numel(S)
        y(i)=0.5*(y(i)+(S(i)./y(i)));
    end
    disp(y)
end

The closing line is

y =

   2.236067977499790   3.162277660168379   3.872983346207417

   5.196152422706632   6.324555320336758

The exact solution being

 S.^.5
 =
   2.236067977499790   3.162277660168380   3.872983346207417

   5.196152422706632   6.324555320336759

Now your 2 loops reach the expected values much faster.

edited Feb 20, 2023 at 17:39

Cris Luengo

61.5k10 gold badges76 silver badges135 bronze badges

answered Feb 20, 2023 at 15:47

John Bofarull Guix

8545 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

John Bofarull Guix Over a year ago

Comment: The code in the question uses n that cannot go above amount elements in initial condition vector. Now n is the amount of loops so when using same approach for different approximation n can be increased to whatever satisfies error thresholds.

Collectives™ on Stack Overflow

Square Root program using Difference Equations

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related