sqrt

问题 I would like to look at the way Python does computes square roots, so I tried to find the definition for math.sqrt() , but I can't find it anywhere. I have looked in _math.c , mathmodule.c , and elsewhere. I know that python uses C's math functions, but are these somewhere in the Python distribution, or are they linked to code elsewhere? I am using Mac OS X. Where is the algorithm in math.sqrt() ? 回答1: It depends on the implementation. CPython is using math functions from the standard C

问题 My goal is to print *** if the square root is negative. But I can't think of a way to change default nan text to *** for(int i=x1;i<=x2;i++){ double y = sqrt(pow(i,2)+3*i-500); if(y = ?){ outFile << "***"; } So, what should I write in the if statement to make it possible? Or maybe there is another way to check if the y is nan then print * 回答1: How about checking for a negative input to the square root function? for (int i = x1; i <= x2; ++i) { double x = pow(i, 2) + 3*i - 500; if (x < 0) {

问题 I searched the question similar to my problem Similar problem. But my problem is when using Turbo C compiler v3.0. Should I have to do some additional work for math.h file? please help. int main (void){ double result, a; clrscr(); printf("Enter a # for square root.\n"); scanf("%f",&a); printf("a = %f\n",a); result = sqrt(a); printf("a = %f and square root is %f\n",a, result); getch(); return 0; } The Output is like this: Enter a # for square root. 64 a = 0.000000 a = 0.000000 and square root

问题 How do you take the square root of a negative number in C++? I know it should return a real and a complex part, I get a NaN? How do I take the real part? 回答1: #include <complex> int main() { std::complex<double> two_i = std::sqrt(std::complex<double>(-4)); } or just std::complex<double> sqrt_minus_x(0, std::sqrt(std::abs(x))); 回答2: sqrt(-x) where x is a positive number is simply 0 + sqrt(x)*i . The real part is just 0. In general, the real part is x > 0 ? sqrt(x) : 0 and the imaginary part is

问题 def ellipse(numPoints, genX=np.linspace, HALF_WIDTH=10, HALF_HEIGHT=6.5): xs = 10.*genX(-1,1,numPoints) ys = 6.5*np.sqrt(1-(xs**2)) return(xs, ys, "-") I am getting an error that states that an invalid value was encountered in a squareroot. I can't see what it is. sqrt(0) = 0 6.5*sqrt(1- (-1**2)) = 0 They should work, but the y values are having problems, they are returning "nan" 回答1: probably xs**2 returns a number > 1 sqrt with negative number will return nan (not a number) >>> import numpy

问题 I need a simple function is_square :: Int -> Bool which determines if an Int N a perfect square (is there an integer x such that x*x = N). Of course I can just write something like is_square n = sq * sq == n where sq = floor $ sqrt $ (fromIntegral n::Double) but it looks terrible! Maybe there is a common simple way to implement such a predicate? 回答1: Think of it this way, if you have a positive int n , then you're basically doing a binary search on the range of numbers from 1 .. n to find the

问题 I need a simple function is_square :: Int -> Bool which determines if an Int N a perfect square (is there an integer x such that x*x = N). Of course I can just write something like is_square n = sq * sq == n where sq = floor $ sqrt $ (fromIntegral n::Double) but it looks terrible! Maybe there is a common simple way to implement such a predicate? 回答1: Think of it this way, if you have a positive int n , then you're basically doing a binary search on the range of numbers from 1 .. n to find the

问题 (EDIT: Let's title this, "Lessons in how measurements can go wrong." I still haven't figured out exactly what's causing the discrepancy though.) I found a very fast integer square root function here by Mark Crowne. At least with GCC on my machine, it's clearly the fastest integer square root function I've tested (including the functions in Hacker's Delight, this page, and floor(sqrt()) from the standard library). After cleaning up the formatting a bit, renaming a variable, and using fixed

问题 Do you know how to do this simple line of code without error using Boost::multiprecison ? boost::multiprecision::cpp_int v, uMax, candidate; //... v += 6 * ceil((sqrt(uMax * uMax - candidate) - v) / 6); Using MSVC there is an error for "sqrt" and it's possible to fix it with: v += 6 * ceil((sqrt(static_cast<boost::multiprecision::cpp_int>(uMax * uMax - candidate)) - v) / 6); Then there is an error for "ceil" and it's possible to fix it with: namespace bmp = boost::multiprecision; typedef bmp:

问题 I have several questions with the following algorithms to tell if a number is prime, I also know that with the sieve of Eratosthenes can be faster response. Why is faster to compute i i * sqrt (n) times. than sqrt (n) just one time ? Why Math.sqrt() is faster than my sqrt() method ? What is the complexity of these algorithms O (n), O (sqrt (n)), O (n log (n))? public class Main { public static void main(String[] args) { // Case 1 comparing Algorithms long startTime = System.currentTimeMillis(