Different rounding for polar(r,t) vs. r*exp(t*1i) for t=acos(.6) and r>=5

[sangisos]

Hi, just wanted to ask whether it is expected and desired behavior to have

; polar(4,acos(.6))
	2.4+3.2i
; polar(5,acos(.6))
	2.99999999999999999999+4i
; 5*polar(1,acos(.6))
	3+4i
; 5*exp(acos(.6)*1i)
	3+4i

for epsilon(1e-20)?

With epsilon(1e-22) I get

; polar(5,acos(.6))
	3+4i
; polar(6,acos(.6))
	~3.60000000000000000000+4.8i
; polar(7,acos(.6))
	~4.20000000000000000000+~5.60000000000000000000i

I'm guessing this is expected with a different precision?

[lcn2]

Hello @sangisos,

Yes, with a different precision (i.e., epsilon), one may very likely see different output when calc prints values.

The following changes in output are expected (note we are using the 3rd eps argument to change rounding for a single builtin call, see help polar for details):

$ calc
C-style arbitrary precision calculator (version 2.14.0.14)
Calc is open software. For license details type:  help copyright
[Type "exit" to exit, or "help" for help.]

; epsilon()
	0.00000000000000000001
; display()
	20
; polar(5, acos(.6))
	2.99999999999999999999+4i
; polar(5, acos(.6), 1e-20)
	2.99999999999999999999+4i
; polar(5, acos(.6), 1e-25)
	~2.99999999999999999999+~4.00000000000000000000i
; polar(5, acos(.6), 1e-10)
	3+4i

We hope this helps.