C++ Program Help (newtons method)

  • #1
I'm new to programming. my assignment wants me to create a program using c++ that can find the square root of any number using Newtons Method.

The user has to enter a number (y) and the program has to calculate the root and has to show each iteration. the initial guess (x0) has to be y/4

the iterations have to stop when the initial guess (x0) and the new guess (x1) are within 0.0001 of each other

Newtons method equation: x1 = (x0 + (y/x0))/2

the problem is that i keep getting a infinite loop showing the first x0 and x1 values
I think the loop isn't updating the x1 and x0 values after each iteration
how do i get the loop to update the x1 and x0 values
any help would be appreciated


here what i got for my code so far:

# include <iostream>
# include <math.h>

using namespace std;

int main ()
{
double y = 0, x1 = 0, x0 = 0;

cout << "enter number" << endl;
cin >> y;

x0 = ( y/4);

while (!( x1 - x0 <= 0.0001 && x1 - x0 >= -0.0001))
{

x1 = ((x0 + (y/x0))/2);
x0 == x1;

cout << "x0=" << x0 << "x1=" << x1 << endl;

}

return 0;
}
 

Answers and Replies

  • #2
phinds
Science Advisor
Insights Author
Gold Member
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This would probably work a whole lot better if you were to replace x0 with x1 (x0 = x1) instead of asking if they are the same (x0 == x1) and then not doing anything with the answer.

Once you fix THAT mistake, you'll find another.

EDIT: by the way, if you plan on doing much programming at ALL, you really should learn to use whatever debugger facility you have available.
 
  • #3
phinds,

I changed x0 == x1 to x0 = x1 instead

This stopped the infinite loop, but it shows x1 and x0 as the same value.The loop still isn't updating x1 and x0. Is there something i'm missing, whats wrong with my source code.
 
  • #4
8
0
phinds,

I changed x0 == x1 to x0 = x1 instead

This stopped the infinite loop, but it shows x1 and x0 as the same value.The loop still isn't updating x1 and x0. Is there something i'm missing, whats wrong with my source code.
x1 = ((x0 + (y/x0))/2);
x0 = x1;

in normal terms:

x1 = calculate a new value

x0 = update me to that new calculated value

Now x0 holds the same value as x1

When you print them out, you confirm this.

Which means your while loop is seeing X1-X0 (BOTH BEING EQUAL) = 0
This breaks the while loop.



edit:
May I suggest you attempt to solve this on paper minus the coding part. It is solved like how a while loop is coded.
I was curious how this worked so I looked it up.

http://mitpress.mit.edu/sicp/chapter1/node9.html

If you can solve it on paper you will see what you need to do to x0 in order to fix the error.
 
Last edited:
  • #6
the loop ends when x1 and x0 are equal, but i dont need them to be equal i need x1 to be equal to the root of the y value
 
  • #7
8
0
I suggest testing the first guess outside the while loop

x0 = (y/4);
x1 = (x0+(y/x0))/2;

while (!( x1 - x0 <= 0.0001 && x1 - x0 >= -0.0001))

It will enter the loop if it wasn't divisible by 4. Now you need a new guess to use. Update your guess x0 = something.

Then you can try your formula again. If it fails, the while loop will update your guess again.
 
  • #8
thanks

that last reply helped me figure out what i was doing wrong
I put the equation, x1= (x0+(y/x0))/2 outside of the loop
and put x0 = x1 inside the loop along with x1= (x0+(y/x0))/2 and it worked

x0 = (y/4);
x1 = ((x0 + (y/x0))/2);

while (!( x1 - x0 <= 0.0001 && x1 - x0 >= -0.0001))
{
x0 = x1;
x1 = ((x0 + (y/x0))/2);

cout << "x0 =" << x0 << "x1 =" << x1 << endl;
}

cout << "The root is " << x1 << endl;



thanks for all the help ducks and phinds
 
  • #9
Borek
Mentor
28,635
3,106
Put your programs in [noparse]
Code:
[/noparse] tags:

Code:
while (!( x1 - x0 <= 0.0001 && x1 - x0 >= -0.0001))
{
   x0 = x1;
   x1 = ((x0 + (y/x0))/2);

   cout << "x0 =" << x0 << "x1 =" << x1 << endl; 
}
It doesn't break space and tab formatting.

BTW: why not abs(x0-x1) < 0.0001?
 

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