PROBLEM WITH PI π

[Albert Chan]

(12-29-2023 03:19 PM)bxparks Wrote:  how does the second equality happen? The "ULP(1/3) = ULP(1)/4" and the "ULP(1/3) = ULP(1)/10".

Base 2, IEEE double as example:

hexfloat(1/3) = 0x1.5555555555555p-2
hexfloat(1/4) = 0x1.0000000000000p-2      // exponents matched → same sized ULP

ULP(1/3) = ULP(1/4) = ULP(1/2)/2 = ULP(1)/4

Base 10, 12-digits precision as example, in SCI format

float(1/3)   = 3.33333333333E-1
float(1/10) = 1.00000000000E-1                // exponents matched → same sized ULP

ULP(1/3) = ULP(1/10) = ULP(1)/10

In general, for base b, integer n: ULP(1/b^n) = ULP(1)/b^n