Python math powers?

11152021, 02:14 PM
Post: #1




Python math powers?
CASIO fxCG50 MicroPython v1.9.4 (but may not be specific to this calculator and Python version)
125**(1/3) 4.99999999999999 pow(125,1/3) 4.99999999999999 (125)**(1/3) (2.500000000000001+4.330127018922194j) abs((125)**(1/3)) 5.00000000000001 pow(125,1/3) ValueError: math domain error (27)**(1/3) (1.5+2.598076211353316j) abs((27)**(1/3)) 3.0  Rob "I can count on my friends"  HP 71B,PrimeTi VOY200,Nspire CXII CASCasio fxCG50...Sharp PCG850,E500,2500,1500,14xx,13xx,12xx... 

11152021, 09:57 PM
Post: #2




RE: Python math powers?
Here's what my Prime G2 does:
(11152021 02:14 PM)robve Wrote: CASIO fxCG50 MicroPython v1.9.4 (but may not be specific to this calculator and Python version)Prime: 5.0 Quote:pow(125,1/3)Prime: 5.0 Quote:(125)**(1/3)Prime: same Quote:abs((125)**(1/3))Prime: 5.0 Quote:pow(125,1/3)Prime: as in (125)**(1/3) I think that's almost all correct. Either 4.99.... or 5.0 probably depends on the specific implementation. There has been a lengthy discussion about 3rd root and operators precedence not so long ago. I've checked the inputs in two other Python interpreters, one on my Samsung Tablet the other on in Windows. They show the same results except that none gives a clear 5.0 like the (probably cheating) Prime. But all return a valid result for pow(125,1/3) therefor I'd think this MicroPython on the CASIO is not complete. HTH Günter 

11152021, 10:31 PM
Post: #3




RE: Python math powers?
(11152021 09:57 PM)Guenter Schink Wrote: Here's what my Prime G2 does: TInspire CX II: 125**(1/3) 5.000000000000002 pow(125,1/3) 5.000000000000002 (125)**(1/3) (2.500000000000001+4.330127018922195j) abs((125)**(1/3)) 5.000000000000001 pow(125,1/3) (2.500000000000001+4.330127018922195j) Tom L Cui bono? 

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