Assistance needed in my code.

Our instructor wants us to write own sqrt code.
His instructions: Write your own square root function named double my_sqrt_1(double n) using the following pseudocode:
if n < 1 then lower_bound is n, otherwise lower_bound is
1 + (n – 1) / 2 – (n – 1)2 / 8
upper_bound is (n + 1) / 2
repeat 10 times:
midpoint = average of upper_bound and lower_bound
if midpoint2 > n then upper_bound = midpoint
if midpoint2 < n then lower_bound = midpoint
return midpoint
and then write a main which prints n, sqrt(n), and my_sqrt_1(n) for n =
π times 10 to the kth power for k = -100, -10, -1, 0, 1, 10, and 100. Use this C++11 code (which may not work on Visual Studio):
for(auto k : {-100, -10, -1, 0, 1, 10, 100}){
n = M_PI * pow(10.0, k);
//cout goes here
}

So I started it, but it won't run, and I'm not sure why, I am new to c++
but here is what I have so far:
Note that the "std_lib" has the iostream, etc. in it.

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#include "std_lib_facilities_4.h"


double my_sqrt_1(double n);
{	
	
	if (n < 1)
	{
		lower_bound (n);
	}
	else
	{
		lower_bound (1+(n-1)/2-(n-1)*pow(2)/8);
		upper_bound ((n+1)/2);
	}
	double x =1;
	for (int i = 0; i < 10; ++i) 
	{
		x = (x+n/x)/2;
		
		return x;
	}
}


int main()
{
	double n;
	for(auto k: {-100, -10, -1, 0, 1, 10, 100})
		{
			n = M_PI*pow(10.0, k);
			cout << n << " ," << sqrt(n) << " ," << my_sqrt_1(n) << '\n';
			
			
		}







}
  
That repeat 10times I think it should calculate lower bound and upper bound 10times. Then lowerbound and upperbound iit should be a variable with data type double, not a function
I think my for is wrong since it was meant to do the calculation of lower and upper.
I'm new to lower and upper bound so I was not even sure I set them up correctly.
Let's fix that, at the start of the function create a variables upper_bound and lower_bound wth data type double
Then inside if and if else add an = after each lower and upperbound but before parenthesis, next delete everythng inside for and move the if else to inside for
After for part s complete wrte the code about midpoint as describe in your descrpton
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