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Question

I need to do program for calculating square root of number programatically not using any function

Vulpes · Answer

Well, it's not very efficient but here's a way of doing it without using predefiined functions: using System; class SquareRoot { static void Main() { Console.WriteLine("Enter a negative number to quit"); while (true) { Console.Write("
Input a number : "); double n; double.TryParse(Console.ReadLine(), out n); if (n double sqrRoot = SqrRoot(n); Console.WriteLine("The square root is {0}", sqrRoot); } } static double SqrRoot(double n) { double r = 0.0; double diff = 1.0; while (true) { while ((r * r) if ( (r * r - n) r -= diff; diff /= 2.0; } } }

Vulpes · Answer

I don't know whether you noticed but the previous version had problems converging on the square root of some numbers. Here's a much improved version (still using simple arithmetic) in which, hopefully, I've eliminated those problems. The output of Math.Sqrt is also shown for comparison: using System; class SquareRoot { static void Main() { Console.WriteLine("Enter a negative number to quit"); while (true) { Console.Write("
Input a number : "); double n; double.TryParse(Console.ReadLine(), out n); if (n double sqrRoot = SqrRoot(n); Console.WriteLine("The square root is {0}", sqrRoot); Console.WriteLine("Math.Sqrt produces {0}", Math.Sqrt(n)); } } static double SqrRoot(double n) { double r = 0.0; double p = 0.0; double pow2 = 1.0; if (n >= 2) // obtain a better starting increment { while (pow2 * pow2 } else if (n { while (pow2 * pow2 > n) pow2 /= 2.0; pow2 *= 2; } if (pow2 * pow2 == n) return pow2; double diff = pow2; double tol = pow2 * 1.0e-15; while (true) { while (r * r if (r - p p = r - diff; diff /= 2.0; r = p + diff; } } }

Mayur Gujrathi · Answer

i said not using any predefined function

Frogleg · Answer

double i = 25; double j = Math .Sqrt(i);

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