# Limited Precision calculation in Mathematica

I apologize upfront for the simple minded question, however i couldn't find an answer in either the Mathematica documentation or Stack exchange.

I am trying to explore the effect of finite precision in Mathematica, however i seem to miss how to limit calculation precision . Here is a trivial example . Say i want to calculate the n'th power of x^n using various precessions. I tried something like this, however while the output becomes truncated the calculation is clearly not limited (see the result from two 2) ?

what am i missing, or better how would this be done correct ?

SetPrecision[Pi, 3]
SetPrecision[SetPrecision[Pi, 3]^8, 15]
SetAccuracy[Pi, 3]^8
Pi^8 // N
3.14^8

• among other things note the precision of the literal 3.14 is not 3, but is MachinePrecision. You should add 3.143^8 to your test list Sep 10, 2015 at 18:46

Most Mathematica functions have no side-effects. This means that SetPrecision returns a version of its input in which all numbers have been set to a certain precision, but it does not influence the precision of the arguments themselves. In other words, it is the output of SetPrecision that has the requested precision; you will want to e.g. explicitly save the reduced-precision result in another variable in order to use it.

See the difference e.g. between the following expressions

pilow = SetPrecision[Pi, 3]
(* Out: 3.14 *)

N[Pi^8]
(* Out: 9488.53 *)

pilow^8
(* Out: 9.5*10^3 *)

3.143^8 (* this has the same precision as pilow *)
(* Out: 9.5*10^3 *)

3.14^8 (* this number actually has \$MachinePrecision digits of precision *)
(* Out: 9450.12 *)


The precision of these results can be probed explicitly:

Precision[Pi^8]      (* Infinity         *)
Precision[N[Pi^8]]   (* MachinePrecision *)
Precision[pilow^8]   (* 2.09691          *)
Precision[3.14^8]    (* MachinePrecision *)
Precision[3.143^8]  (* 2.09691          *)


Alternatively, one can read the precision off e.g. the FullForm of any numerical result:

FullForm[pilow]      (* 9488.53101607057400712862.0969100130080567 *)

• Thx a lot Marco, however to be honest your answer sounds puzzling to me. For one would you also say the instead of Sort[2.0]^8 i should use a= Sort[2.0]; a^2 ? I think the whole notion of Functional programming is to daisy chain functions ? Beside that your suggestion still gives the unexpected value that the limited precision calculation yields the full precision result ? Sep 11, 2015 at 15:31
• 1) I am not sure what Sort[2.0] would achieve, but e.g. f[x]^2 is in fact daisy-chaining functions, i.e. you obtain Power[f[x], 2], which you could equivalently write as Power[#, 2]& @f[x] to highlight the daisy-chain: this calculates the return value of f when applied to x, then feeds the return value to Power. Am I missing your point? 2) The limited precision calculation does not yield a full precision result. You can convince yourself of that by noting the precision digits associated with the result in FullForm[pilow^8], or explicitly by running Precision[pilow^8]. Sep 11, 2015 at 17:37
• @bernddude Please see my comment above, and the edit to my answer. Sep 11, 2015 at 17:43
• Marco, thx a lot once again. Sorry for the typo, spellchecker feature Sqrt-> Sort nice :). I do under stand your point, I feel its more that SetPrecision does behave different then i would expect . Below is one more example N[Pi, 20] SetPrecision[Pi, 3] pillow = SetPrecision[Pi, 3] FullForm[N[Pi^8]] pillow^8 FullForm[pillow^8] FullForm[3.14^8] pillow^8 - 3.14^8 the last line gives the result 38.414 Sep 12, 2015 at 14:41
• Marco here might be a better example of what i am trying to say . p = 200 e = .996 ep = SetPrecision[e, 2] SetPrecision[e^p, 3] e^p ep^p SetPrecision[ep, 20]^p i would expect for ep^100 to get 1 after the assignment is done. Obviously this isn't wrong however in order to "simulate" the behavior of a finite precision embedded system, its not that useful . Sep 12, 2015 at 14:57