stdlib-js
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rempio2
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Compute
x - nπ/2 = r.
bash
npm install @stdlib/math-base-special-rempio2
javascript
var rempio2 = require( '@stdlib/math-base-special-rempio2' );
#### rempio2( x, y )
Computes x - nπ/2 = r.
javascript
var y = [ 0.0, 0.0 ];
var n = rempio2( 128.0, y );
// returns 81
var y1 = y[ 0 ];
// returns ~0.765
var y2 = y[ 1 ];
// returns ~3.618e-17
When x is NaN or infinite, the function returns 0 and sets the elements of y to NaN.
javascript
var y = [ 0.0, 0.0 ];
var n = rempio2( NaN, y );
// returns 0
var y1 = y[ 0 ];
// returns NaN
var y2 = y[ 1 ];
// returns NaN
y = [ 0.0, 0.0 ];
n = rempio2( Infinity, y );
// returns 0
y1 = y[ 0 ];
// returns NaN
y2 = y[ 1 ];
// returns NaN
n and stores the remainder r as two numbers in y, such that y[0]+y[1] = r.
- For input values larger than 2^20*π/2 in magnitude, the function only returns the last three binary digits of n and not the full result.
javascript
var linspace = require( '@stdlib/array-base-linspace' );
var rempio2 = require( '@stdlib/math-base-special-rempio2' );
var x = linspace( 0.0, 100.0, 100 );
var y = [ 0.0, 0.0 ];
var n;
var i;
for ( i = 0; i < x.length; i++ ) {
n = rempio2( x[ i ], y );
console.log( '%d - %dπ/2 = %d + %d', x[ i ], n, y[ 0 ], y[ 1 ] );
}
c
#include "stdlib/math/base/special/rempio2.h"
#### stdlib_base_rempio2( x, &rem1, &rem2 )
Computes x - nπ/2 = r.
c
#include <stdint.h>
double rem1;
double rem2;
int32_t n = stdlib_base_rempio2( 4.0, &rem1, &rem2 );
The function accepts the following arguments:
- x: [in] double input value.
- rem1: [out] double* destination for first remainder number.
- rem2: [out] double* destination for second remainder number.
c
int32_t stdlib_base_rempio2( const double x, double *rem1, double *rem2 );
n and stores the remainder r as two numbers in rem1 and rem2, respectively, such that rem1+rem2 = r.
c
#include "stdlib/math/base/special/rempio2.h"
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main( void ) {
const double x[] = { 0.0, 1.0, 4.0, 128.0 };
double rem1;
double rem2;
int32_t n;
int i;
for ( i = 0; i < 4; i++ ) {
n = stdlib_base_rempio2( x[ i ], &rem1, &rem2 );
printf( "%lf - %"PRId32"π/2 = %lf + %lf\n", x[ i ], n, rem1, rem2 );
}
}
