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For which models was 2^3>8?
04-04-2017, 05:51 PM (This post was last modified: 04-04-2017 05:58 PM by Dieter.)
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RE: For which models was 2^3>8?
(04-04-2017 05:20 PM)pier4r Wrote:  Why does it happen?

I mean if it would be , say, \( 169^{0.5} \) I could understand computation errors, but \( 2^3 \) seems pretty straightforward to me, it is even a base two power.

So does anyone have an explanation for it? It would be interesting.

The explanation is simple and straightforward: yx is calculated as ex · ln y.
So 2³ is not evaluated as 2·2­·2 – how far should this go? Up to the 4th power? the 5th? The 10th?

That's why 2³ is calculated as e3 · ln 2. With 10-digit precision the exponent is 3 · 0,6931471806 = 2,079441542 and the exponential of this is 8,0000 00002 56...  Especially powers of 2 (and 5) are prone to such roundoff errors because of the rounded 10-digit logarithm.

This improved as the calculators internally used a higher precision than displayed. Since 1976 most HP 10-digit calculators work with 13-digit internal precision – and the problem is solved. Well, in most cases, that is. ;-)

More details can be found in the November 1976 issue of HP-Journal, p. 16 f.

BTW, evaluating x³ by using the yx function always has been a quite bad idea. Beside the accuracy issue this method is sloooow. Remember: there are two transcendental functions involved in every simple power calculation. On classic HPs you can see the display go blank for a short moment during such power calculations – simply because the operation takes that much time. A much better way is [ENTER] [x²] [x] or [x²] [LastX] [x] which even sets LastX correctly. Some newer calculators even allow [x²] [RCLx]L which does it all in the X-register.

BTW2: Your example 1690,5 can be evaluated by the square root function which uses a completely different approach that does not cause the mentioned errors (and is also much faster). And the fact that 2³ is a base 2 power does not mean anything on a calculator that uses real BCD arithmetics, i.e. base 10.

BTW3: Who cares about the 10th digit? My first calculator returned exactly 5 (five!) significant digits of all transcendental functions. ;-)

Dieter
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Messages In This Thread
For which models was 2^3>8? - Joe Horn - 04-04-2017, 05:03 AM
RE: For which models was 2^3>8? - Joe Horn - 04-04-2017, 12:17 PM
RE: For which models was 2^3>8? - pier4r - 04-04-2017, 05:20 PM
RE: For which models was 2^3>8? - Dieter - 04-04-2017 05:51 PM
RE: For which models was 2^3>8? - pier4r - 04-04-2017, 06:10 PM
RE: For which models was 2^3>8? - bshoring - 04-04-2017, 06:16 PM
RE: For which models was 2^3>8? - teenix - 04-05-2017, 07:50 AM
RE: For which models was 2^3>8? - bshoring - 04-05-2017, 06:08 PM
RE: For which models was 2^3>8? - Dieter - 04-04-2017, 07:18 PM
RE: For which models was 2^3>8? - Dieter - 04-04-2017, 07:53 PM
RE: For which models was 2^3>8? - bshoring - 04-04-2017, 10:14 PM
RE: For which models was 2^3>8? - Andres - 06-23-2020, 04:39 AM



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