For which models was 2^3>8?
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04-04-2017, 05:03 AM
Post: #1
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For which models was 2^3>8?
According to my best attempt to pinpoint the moment when 2^3 began to equal 8 in the evolution of HP calculators, I've compiled the following partial table of model introductions in historical order:
HP-35 1972-07 HP-80 1973-02 HP-56 1973-04 HP-66 1973-04 HP-45 1973-05 HP-65 1974-01 HP-70 1974-08 HP-55 1975-01 HP-21 1975-02 ??? HP-25 1975-08 2^3>8 ======================== HP-22 1975-09 2^3=8 HP-91 1976-03 HP-27 1976-05 HP-25C 1976-07 ??? HP-67 1976-07 HP-97 1976-07 HP-10 1977-07 HP-29C 1977-07 HP-92 1977-07 HP-19C 1977-09 HP-97S 1977-12 Therefore the accuracy improvement seems to have happened between the inaccurate HP-25 (introduced August 1975) and the accurate HP-22 (introduced one month later). HOWEVER that leaves me with two questions: (1) This implies that the HP-21 used the old inaccurate math, but I don't have one to verify that on. Could somebody with a real physical HP-21 (no simulators please) please check whether it returns 8.000000002 or exactly 8? Be sure to set DSP 9 of course. (2) This also implies that the HP-25C is more accurate than the HP-25, which I highly doubt. Could somebody with a real physical HP-25C (not 25, and no simulators please) please check whether it returns 8.000000002 or exactly 8? Be sure to set FIX 9 of course. Thanks in advance! <0|ɸ|0> -Joe- |
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04-04-2017, 06:26 AM
Post: #2
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RE: For which models was 2^3>8?
Joe
8.000000002 on my 21 S/N 1508s Also 8.000000002 on my 25C S/N 1805S |
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04-04-2017, 12:17 PM
Post: #3
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RE: For which models was 2^3>8?
(04-04-2017 06:26 AM)james summers Wrote: 8.000000002 on my 21 S/N 1508s Thanks, James! That completes the table. <0|ɸ|0> -Joe- |
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04-04-2017, 01:05 PM
Post: #4
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RE: For which models was 2^3>8?
Fortunately, the only two Woodstocks I have were the ones that were needed!
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04-04-2017, 03:28 PM
Post: #5
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RE: For which models was 2^3>8?
I should check my Sinclair Cambridge Programmable. I think we'll be lucky if it's accurate to more than 2 significant figures.
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04-04-2017, 05:20 PM
(This post was last modified: 04-04-2017 05:21 PM by pier4r.)
Post: #6
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RE: For which models was 2^3>8?
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. Wikis are great, Contribute :) |
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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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04-04-2017, 06:10 PM
(This post was last modified: 04-04-2017 06:11 PM by pier4r.)
Post: #8
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RE: For which models was 2^3>8?
Thanks Dieter! Yes I choose the square root of 169 because as far as I know it involves a bit of calculations and it could lead to errors.
For the \( e^{x \cdot ln(y)} \) I did not think they would use that unless rational numbers were used as input. For special cases, like 2, I also thought about shortcuts ( shifting bits ). Well, once again, today I learned. Wikis are great, Contribute :) |
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04-04-2017, 06:16 PM
Post: #9
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RE: For which models was 2^3>8?
Great explanation Dieter. Interesting side note: On the HP-35 doing x^y the display goes blank for a split second, but on the HP-80 doing y^x causes the display to flicker as if it is running a program. I have to wonder if the HP-80 was programmed differently in order to pack so many functions (programs actually) into it.
Regards, Bob |
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04-04-2017, 06:45 PM
(This post was last modified: 04-04-2017 06:53 PM by james summers.)
Post: #10
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RE: For which models was 2^3>8?
Thanks for the great explanation, Dieter.
Just playing around, my MK-61, after a slight pause gives 7.9999993. Doing [2] [ENTER] [ln] [3] [x] [ex] also gives 7.9999993. The 25C is more accurate, but similar, in that [2] [ENTER] [3] [yx] and [2] [ENTER] [ln] [3] [x] [ex] both give 8.000000002. However, my 32E, 41CV and 15C, whilst giving [2] [ENTER] [3] [yx] as 8.000000000, for [2] [ENTER] [ln] [3] [x] [ex] they give 8.000000003. Edited to add for Dave that my Sinclair Cambridge Scientific, which doesn't have the luxury of a [yx] key, gives 7.9998. |
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04-04-2017, 07:18 PM
(This post was last modified: 04-04-2017 07:19 PM by Dieter.)
Post: #11
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RE: For which models was 2^3>8?
(04-04-2017 06:45 PM)james summers Wrote: Thanks for the great explanation, Dieter. Interesting. Without guard digits it should be 7.9999973. What are the intermediate results you get for ln 2 and 3x this? (04-04-2017 06:45 PM)james summers Wrote: The 25C is more accurate, but similar, in that [2] [ENTER] [3] [yx] and [2] [ENTER] [ln] [3] [x] [ex] both give 8,000000002. The 25 uses 10 digits as opposed to 8 in the MK-61. (04-04-2017 06:45 PM)james summers Wrote: However, my 32E, 41CV and 15C, whilst giving [2] [ENTER] [3] [yx] as 8.000000000, for [2] [ENTER] [ln] [3] [x] [ex] they give 8,000000003. Why do you add an [ENTER] before the log? It's classic RPN. ;-) The HP32/41/15 of course use the "new accuracy" since 1976 with three additional guard digits. That's why even with a manually calculated exponentional you get the correctly rounded value ...003. The difference to the HP25 result ...002 is probably due to different algorithms and/or rounding methods. In 1976 there were more improvements than just adding the guard digits, cf. the mentioned HP Journal article. BTW, if you do not have access to the latter: most of the interesting stuff is available on hpl.hp.com. Dieter |
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04-04-2017, 07:42 PM
Post: #12
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RE: For which models was 2^3>8?
(04-04-2017 07:18 PM)Dieter Wrote:(04-04-2017 06:45 PM)james summers Wrote: Thanks for the great explanation, Dieter. [2] [ln] = 6.9314717 E-01 [3] [x] = 2.0794415 [ex] b = 7.9999993 (04-04-2017 07:18 PM)Dieter Wrote:(04-04-2017 06:45 PM)james summers Wrote: However, my 32E, 41CV and 15C, whilst giving [2] [ENTER] [3] [yx] as 8.000000000, for [2] [ENTER] [ln] [3] [x] [ex] they give 8.000000003. Because I'm an idiot! ;-) |
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04-04-2017, 07:53 PM
Post: #13
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RE: For which models was 2^3>8? | |||
04-04-2017, 10:14 PM
Post: #14
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RE: For which models was 2^3>8?
Because the earliest calculators didn't have the 3 guard digits, you could do 1.000001 Enter 1,000,000 y^x and get e (2.718281828). This doesn't work on the HP-22 and later calculators. The first 2 financial calcs don't have e^x.
Regards, Bob |
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04-04-2017, 10:26 PM
Post: #15
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RE: For which models was 2^3>8?
(04-04-2017 05:51 PM)Dieter Wrote: The explanation is simple and straightforward: yx is calculated as ex · ln y. How about all the way? It's really quite efficient if you use exponentiation by squaring: x^n = (x^(n/2))^2 if n is even, and x*(x^((n-1)/2))^2 if n is odd; apply recursively or iteratively. The algorithm is O(log(n)). This is how Free42 gets 2^3=8 exactly, despite not using extended precision for intermediate results. |
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04-05-2017, 07:50 AM
Post: #16
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RE: For which models was 2^3>8?
(04-04-2017 06:16 PM)bshoring Wrote: Great explanation Dieter. Interesting side note: On the HP-35 doing x^y the display goes blank for a split second, but on the HP-80 doing y^x causes the display to flicker as if it is running a program. I have to wonder if the HP-80 was programmed differently in order to pack so many functions (programs actually) into it. I know I know, so simulators :-) Just curious though, the simulators returned the same results The HP35 took 1720 instructions to process so about 1/2 second display off time assuming 800KHz clock. The HP80, HP70 leave the display on for calculations because they can take a lot of time to process some financial problems. That way the user knows it is busy. The HP80 took 1546 instructions to process and the HP67 took 1632. cheers Tony |
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04-05-2017, 11:34 AM
Post: #17
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RE: For which models was 2^3>8?
(04-04-2017 10:26 PM)Thomas Okken Wrote:(04-04-2017 05:51 PM)Dieter Wrote: The explanation is simple and straightforward: yx is calculated as ex · ln y. I'm guessing they didn't want to spend the probably very limited ROM/RAM space of the old models to handle special cases of powers like that. |
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04-05-2017, 06:08 PM
Post: #18
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RE: For which models was 2^3>8?
(04-05-2017 07:50 AM)teenix Wrote:(04-04-2017 06:16 PM)bshoring Wrote: Great explanation Dieter. Interesting side note: On the HP-35 doing x^y the display goes blank for a split second, but on the HP-80 doing y^x causes the display to flicker as if it is running a program. I have to wonder if the HP-80 was programmed differently in order to pack so many functions (programs actually) into it. Tony, thanks for enlightening me on why this function operates differently on the HP-80 from the HP-35. I have both actual calculators, so it's fun to see the differences. I've also downloaded your simulators for the 70 & 80, and I am very impressed. Yes, they operate just like the originals. Regards, Bob |
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06-23-2020, 04:39 AM
(This post was last modified: 06-23-2020 04:40 AM by Andres.)
Post: #19
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RE: For which models was 2^3>8?
It has been said already, but just in case... The HP=25 needed to allocate ROM space for programmable functions, but the HP-22 and the HP-27 were not programmable, so there was more ROM space available to put more sophisticated code for these math functions. While the HP-21 was not programmable indeed, it has no memory (apart from ACT registers) and I guess its ROM space was also limited. By the time HP-29C, 19C, 91, 67, 97 appeared, more ROM space was feasible and the improved routines became standard.
Andrés C. Rodríguez (Argentina) Please disregard idiomatic mistakes. My posts are mostly from old memories, not from current research. |
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