Newton-Raphson Method to Compute the Square Root of x

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Newton-Raphson Method to Compute the Square Root of x

Step 1: Set the value of guess to 1. Step 2: If |guess² - x| < ε, proceed to step 4. Step 3: Set the value of guess to (x / guess + guess) / 2 and return to step 2. Step 4: The guess is the approximation of the square root.

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//阅读版本 // Function to calculate the absolute value of a number

include

float absoluteValue (float x) { 　 if ( x < 0 ) 　　 x = -x; 　 return (x); } // Function to compute the square root of a number float squareRoot (float x) { 　 const float epsilon = .00001; 　 float guess = 1.0; 　 while ( absoluteValue (guess * guess - x) >= epsilon ) 　　 guess = ( x / guess + guess ) / 2.0; 　 return guess; } int main (void) { 　 printf ("squareRoot (2.0) = %f\n", squareRoot (2.0)); 　 printf ("squareRoot (144.0) = %f\n", squareRoot (144.0)); 　 printf ("squareRoot (17.5) = %f\n", squareRoot (17.5)); 　 return 0; }

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//运行版本 // Function to calculate the absolute value of a number

include

float absoluteValue (float x) { if ( x < 0 ) x = -x; return (x); } // Function to compute the square root of a number float squareRoot (float x) { const float epsilon = .00001; float guess = 1.0; while ( absoluteValue (guess * guess - x) >= epsilon ) guess = ( x / guess + guess ) / 2.0; return guess; } int main (void) { printf ("squareRoot (2.0) = %f\n", squareRoot (2.0)); printf ("squareRoot (144.0) = %f\n", squareRoot (144.0)); printf ("squareRoot (17.5) = %f\n", squareRoot (17.5)); return 0; }

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squareRoot (2.0) = 1.414216 squareRoot (144.0) = 12.000000 squareRoot (17.5) = 4.183300

Process returned 0 (0x0) execution time : 0.016 s Press any key to continue.