## Infinite Gift

From “Passion for Mathematics” by Clifford Pickover

The diagram is for a puzzle called ‘Infinite Gift’ which appears in Pickover’s “Passion for Mathematics”. The puzzle asks to compute a volume of infinitely long solid formed by stacking up infinitely many cubes. The bottom-most 1st cube has a side of 1.0, and each subsequent i-th cube has a side 1/sqrt(i). So the series has an overall formula:

sum( (1/i)^(3/2), i, 1, +infinity)

The Basel problem asks to compute a volume of series which looks remarkably similar:

sum( (1/i)^2, i, 1, infinity)

and it has a beautiful solution found by Euler: (pi^2)/6. Thus it would be very interesting to find out if the series for this problem has an analytic solution. The following is the small C program to compute the volume by numerical approximation. It uses gmp library designed for calculations using long numbers. The compilation command is:

gcc -w -std=c99 infinite_gift.c -lm

and the resultant value is:

2.61037534818546109312

I found the infinite series for this problem is an example of Riemann zeta function. Some Riemann’s zeta functions have closed forms like (pi^2)/6 but many do not, and apparently this is an example of such a problem.

// ---------------------------------------------------------------------------------------------- #include <unistd.h> #include <stdio.h> #include <stdlib.h> #include <math.h> #include <gmp.h> int main(int argc, char ** argv){ mpf_t md; mpf_init(md); mpf_t mi; mpf_init(mi); mpf_t mi_s; mpf_init(mi_s); for (int i = 1; i < 1000000; i++) { mpf_set_d(mi, 1.0/((double)i)); mpf_sqrt(mi,mi); mpf_mul(mi_s, mi, mi); mpf_mul(mi, mi_s, mi); mpf_add(md, md, mi); } gmp_printf("%.Ff\n", md); mpz_clear(md); mpz_clear(mi); mpz_clear(mi_s); return(EXIT_SUCCESS); } // ----------------------------------------------------------------------------------------------