How can I generate the sequence of Kvalues for CORDIC?

I am trying to implement CORDIC in C. I have to generate the constant values that the algorithm uses. The first list of arctangent values was straightforward but I am having trouble with the second list of 'Kvalues.'

% Kvalues = cumprod(1./sqrt(1 + 1j*2.^(-(0:23))))
Kvalues = [ ...
   0.70710678118655   0.63245553203368   0.61357199107790   0.60883391251775 ...
   0.60764825625617   0.60735177014130   0.60727764409353   0.60725911229889 ...
   0.60725447933256   0.60725332108988   0.60725303152913   0.60725295913894 ...
   0.60725294104140   0.60725293651701   0.60725293538591   0.60725293510314 ...
   0.60725293503245   0.60725293501477   0.60725293501035   0.60725293500925 ...
   0.60725293500897   0.60725293500890   0.60725293500889   0.60725293500888 ];

However I am not very familiar with this notation and I am having a hard time making sense of this and producing a simple formula that would produce the same results. I do not want to just paste these values into my CORDIC implementation because I would like to generate them myself with the appropriate amount of precision.

Here is what I came up with based on what I could understand but this does not produce the correct results:

float prev = 1;
for (int i = 0; i < PRECISION; i++)
    printf("%f\n", prev = prev/sqrt(1+pow(2, -i)));

I am expecting the same values I saw in the list from Wikipedia:

But with my formula they converge to 0.457942... instead of 0.607252...:

What is the correct formula in C to generate PRECISION 'Kvalues' for CORDIC?

Solution

I tried a number of variations [not shown] by doing a websearch on CORDIC and looking at various pages.

From here: https://nonagon.org/ExLibris/cordic-c-javascript I got the formula:

prev = prev / sqrt(1 + (1.0 / pow(2, 2.0 * i)))

Ultimately, for your code, it was change:

pow(2, -i)

Into:

pow(2, -2.0 * i)

Here is the modified code that works:

#include <stdio.h>
#include <math.h>

#define PRECISION   6

#define CALL(_fnc) \
    do { \
        printf("\n" #_fnc "\n"); \
        _fnc(); \
    } while (0)

void
orig(void)
{
    double prev = 1.0;

    for (int i = 0; i < PRECISION; i++)
        printf("%f\n", prev = prev / sqrt(1 + pow(2, -i)));
}

void
fix1(void)
{
    float prev = 1;

    for (int i = 0; i < PRECISION; i++)
        printf("%f\n",prev = prev / sqrt(1 + pow(2, -2.0 * i)));
}

int
main(void)
{

    CALL(orig);
    CALL(fix1);

    return 0;
}

Here is the program output:

I recommend using double instead of float and increasing the precision of the printf format (e.g. %.14f):

#include <stdio.h>
#include <math.h>

#define PRECISION   6

#define CALL(_fnc) \
    do { \
        printf("\n" #_fnc "\n"); \
        _fnc(); \
    } while (0)

void
orig(void)
{
    double prev = 1.0;

    for (int i = 0; i < PRECISION; i++)
        printf("%.14f\n", prev = prev / sqrt(1 + pow(2, -i)));
}

void
fix1(void)
{
    double prev = 1;

    for (int i = 0; i < PRECISION; i++)
        printf("%.14f\n",prev = prev / sqrt(1 + pow(2, -2.0 * i)));
}

int
main(void)
{

    CALL(orig);
    CALL(fix1);

    return 0;
}

Here is the enhanced output:


orig
0.70710678118655
0.57735026918963
0.51639777949432
0.48686449556015
0.47232793895472
0.46511640189705

fix1
0.70710678118655
0.63245553203368
0.61357199107790
0.60883391251775
0.60764825625617
0.60735177014130