代码搜索：SQRT
找到约 10,000 项符合「SQRT」的源代码
代码结果 10,000
www.eeworm.com/read/286264/8778474
sqrt
www.eeworm.com/read/493005/6403592
sqrt
NAME
sqrt - evaluate exactly or approximate a square root
SYNOPSIS
sqrt(x [, eps[, z]])
TYPES
If x is an object of type tt, or if x is not an object but y
is an object of type tt, an
www.eeworm.com/read/130881/14170570
sqrt
www.eeworm.com/read/306178/3748308
sqrt
ssssqqqqrrrrtttt((((3333)))) MMMMuuuuddddOOOOSSSS ((((5555 SSSSeeeepppp 1111999999994444)))) ss
www.eeworm.com/read/192122/8403168
asm sqrt.asm
;*******************************************************************
;
; Square Root By Newton Raphson Method
;
; This routine computes the square root of a 16 bit number(with
; lo
www.eeworm.com/read/391275/8412841
m sqrt.m
www.eeworm.com/read/290828/8458928
txt sqrt.txt
假设被开方数为a，如果用<mark>sqrt</mark>（a）表示根号a 那么[<mark>sqrt</mark>(x)-<mark>sqrt</mark>(a/x)]^2=0的根就是<mark>sqrt</mark>（a）
变形得
<mark>sqrt</mark>(a)=（x+a/x）/2
所以你只需设置一个约等于（x+a/x）/2的初始值，代入上面公式，可以得到一个更加近似的值，再将它代入，就得到一个更加精确的值……依此方法，最后得到一个足够精度的（x+a/x）/2的值。
如：计算<mark>sqrt</mark>(5)
...
www.eeworm.com/read/189298/8477321
c sqrt.c
/*
** Compute the square root of a number.
*/
#include
#include
int
main()
{
float new_guess;
float last_guess;
float number;
/*
** Prompt for and read the
www.eeworm.com/read/290380/8487091
c sqrt.c
#include
#include
void main(void)
{
double value;
for (value = 0.0; value < 10.0; value += 0.1)
printf("Value %f sqrt %f\n", value, sqrt(value));
}
www.eeworm.com/read/433114/8545200
m sqrt.m
function ds=sqrt(ds)
% Function takes the square root of the traces of a seismic dataset
%
% Written by: E. R.: September 12, 2005
% Last updated: September 18, 2006: Handle structure arrays
if isstr