What's the correct way to using Mathematica to get location of true value when using IEEE-754 float32 precision?
Inspired by this question, where the user got a range of different answers when asking for the result of the following computation, I'm curious to see what the error estimates would be for float32.
import math
u = 5.0
for i in range(73):
u = 6*math.cos(u)
print(u)
We can coerce a real number into a 32 bit float in a few ways:
pi = N[\[Pi]];
pi // FullForm
3.141592653589793`
ImportString[ExportString[pi, "Real32"], "Real32"][[1]] // FullForm
3.1415927410125732`
Normal[NumericArray[{pi}, "Real32"]][[1]] // FullForm
3.1415927410125732`
To verify equivalence as a 32 bit float, we see the first 23 digits in the mantissa agree (IEEE 754 standard), other than rounding of the least significant bit:
pi32bit = ImportString[ExportString[pi, "Real32"], "Real32"][[1]];
RealDigits[pi, 2][[1, 2 ;; 24]]
{1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0}
RealDigits[pi32bit, 2][[1, 2 ;; 24]]
{1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1}
Interval[5.] gives me an interval for float64, was looking for an easy way to get interval for 32-bit
$\endgroup$
Commented
Feb 16 at 19:56
Here are a couple of demonstrations that we have no significant digits after 13 or so steps. First we get the base 10 precision for 24 bits mantissas, the common standard for 32 bit floats.
Log[10., 2^24]
(* Out[56]= 7.22472 *)
Now use this to create a bignum 5 with that precision. Then iterate and print precision of the result.
u = 5.`7.225;
Do[Print[Precision[u = 6*Cos[u]]], {13}]
(* During evaluation of In[63]:= 5.99705
During evaluation of In[63]:= 4.88645
During evaluation of In[63]:= 4.99221
During evaluation of In[63]:= 4.06717
During evaluation of In[63]:= 3.96362
During evaluation of In[63]:= 3.17044
During evaluation of In[63]:= 2.55444
During evaluation of In[63]:= 1.73966
During evaluation of In[63]:= 2.35372
During evaluation of In[63]:= 2.08599
During evaluation of In[63]:= 1.5386
During evaluation of In[63]:= 0.453482
During evaluation of In[63]:= 0. *)
Alternatively, set up intervals with an epsilon of 2^(-24).
u = Interval[{5. - 2^(-24), 5. + 2^(-24)}];
Do[Print[u = 6*Cos[u]], {13}]
(* During evaluation of In[67]:= Interval[{1.70197,1.70197}]
During evaluation of In[67]:= Interval[{-0.784807,-0.784803}]
During evaluation of In[67]:= Interval[{4.24515,4.24516}]
During evaluation of In[67]:= Interval[{-2.70256,-2.70247}]
During evaluation of In[67]:= Interval[{-5.43098,-5.43074}]
During evaluation of In[67]:= Interval[{3.94887,3.94994}]
During evaluation of In[67]:= Interval[{-4.14881,-4.14418}]
During evaluation of In[67]:= Interval[{-3.22876,-3.20531}]
During evaluation of In[67]:= Interval[{-5.98782,-5.97722}]
During evaluation of In[67]:= Interval[{5.72134,5.74018}]
During evaluation of In[67]:= Interval[{5.07764,5.13697}]
During evaluation of In[67]:= Interval[{2.1431,2.47161}]
During evaluation of In[67]:= Interval[{-4.703,-3.24943}] *)
Same outcome: we no longer have any significant digits at this point.
Interval[5.], wondering if there's something similar for 32-bit $\endgroup$