WP34S : is this result "zero" enough ?

[Miguel Toro]

Hola,

I am still playing with the 15c examples and this time it is from the advanced functions handbook, page 78, calculating the nth roots of a complex number. I tried my best to translate the program, so here is my WP34s version:

001 LBL A

002 [cmplx]x[<->]y

003 [cmplx]1/x

004 [cmplx]RCL L

005 [cmplx]R[v]

006 [cmplx]y[^x]

007 [cmplx]STO 02

008 0

009 ENTER[^]

010 3

011 6

012 0

013 [cmplx]R[^]

014 [cmplx]/

015 [cmplx]STO 04

016 0

017 STO K

018 LBL 00

019 [cmplx]RCL 04

020 RCL[times] K

021 1

022 DEG

023 [->]REC

024 [cmplx]RCL[times] 02

025 1

026 STO+ K

027 R[v]

028 STOP

029 GTO 00

to run it, I first set the ssize8, then 0 FILL and then follow the example in the book, using 1^1/100 as the value. Here are the results:

k = 0,  z0  = 1 (check!)

k = 1,  z1  = 0.9980+0.0628i (check!)

k = 50, z50 = -1+(1.2815^-38)i (check? 15C result is -1)

So, is this because the different precision between both machines and this is the good result? ... or am I that bad as a programer :-)

Saludos,

Miguel

Edited: 1 Sept 2011, 8:41 p.m.

[Marcus von Cube, Germany]

Miguel, you could try the various rounding modes (RM) which affect how the internal high precision numbers are rounded. The e-38 looks very much like a rounding artifact because the internal precision is in this range.

Looking more closely at your code I see you do calculate the inverse of the exponent and later multiply it by k. This may lead to different results then computing k/n.

Edited: 2 Sept 2011, 2:27 a.m.

[Paul Dale]

Quote:

---

you could try the various rounding modes (RM) which affect how the internal high precision numbers are rounded.

---

The RM modes only change how an internal high precision number is rounded to a sixteen digit register, not how rounding takes place during internal high precision calculations.

- Pauli

[Paul Dale]

Without having investigated specifically, the 10^-38 result you are seeing is around the internal precision used for calculations.

My guess would be a slight rounding issue is present here. Probably in the polar to rectangular conversion.

Still, I'm not really concerned here, the result is close enough to zero for my liking.

- Pauli

[Miguel Toro]

I think the rounding issue may come from the y^x. I tried this simple example with (1+i)^2:

   1, ENTER, 1, 0, ENTER, 2, CPX, Y^X = (-7,5998)^-39 + 2i

   1, ENTER, 1, CPX, X^2 = 2i

In both cases the 15c and 42s give 2i as the result, either with y^x or x^2.

I tried the RM command (thanks for calling me geek :-))and it did not make any difference.

As you say, it is close enough to zero.

Thanks,

Migue