For which models was 2^3>8?
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04-04-2017, 05:51 PM
(This post was last modified: 04-04-2017 05:58 PM by Dieter.)
Post: #7
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RE: For which models was 2^3>8?
(04-04-2017 05:20 PM)pier4r Wrote: Why does it happen? 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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