rempio2

Compute x - nπ/2 = r.

Usage

var rempio2 = require( '@stdlib/math/base/special/rempio2' );

rempio2( x, y )

Computes x - nπ/2 = r.

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.

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

Notes

The function returns 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.

Examples

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 APIs

Usage

#include "stdlib/math/base/special/rempio2.h"

stdlib_base_rempio2( x, &rem1, &rem2 )

Computes x - nπ/2 = r.

#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.

int32_t stdlib_base_rempio2( const double x, double *rem1, double *rem2 );

Notes

The function returns n and stores the remainder r as two numbers in rem1 and rem2, respectively, such that rem1+rem2 = r.

Examples

#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 );
    }
}