Accuracy summation question

09062023, 07:30 PM
Post: #1




Accuracy summation question
I calculated 2000! in home by summing from 1 to 2000 log (x), and taking 10^answer, getting 3.316274+ E5735
On the WP34S, I directly get 3.31627509245+ E5735 On my CASIO 991CW, using the summation method used on prime, I got 3.3162750924 E5735 On my CASIO 9750giii, I got 3.316275094 E5735 The CASIO 991CW uses 22 or 23 significant digits in its calculations, may help explain close agreement with the WP34S. I was surprised that the Prime in home disagreed with the others in just the 6th decimal. 

09072023, 08:38 AM
(This post was last modified: 09072023 09:05 AM by rkf.)
Post: #2




RE: Accuracy summation question
(09062023 07:30 PM)lrdheat Wrote: I calculated 2000! in home by summing from 1 to 2000 log (x), and taking 10^answer, getting 3.316274+ E5735 It´s indeed strange (altough I would say it differs in the seventh place, ...487 vs. 509). Assuming the LOGs are calculated correctly to 12 digits with values rising from 0 to about 3.3, the error of one summand should be less than +/ 5E12 If the error in summation propagates normally, I would expect in the sum an error of ca. sqrt(2000)*(+/ 5E12) < 2.3E10 Taking 10^ this I get 1.00000000051, so I would nevertheless expect about 9 correct places, and not 7. :( Addendum: BTW I just compared with my HP50g, which gives exactly the same result of 5735.52065052 for the sum of logarithms as the prime, thus indicating the 12digits arithmetics core engine is the same. 

09072023, 10:40 AM
(This post was last modified: 09072023 10:42 AM by Albert Chan.)
Post: #3




RE: Accuracy summation question
(09072023 08:38 AM)rkf Wrote: It´s indeed strange (altough I would say it differs in the seventh place, ...487 vs. 509). Assuming your numbers are correct: log10(3.31627487E5735) vs log10(3.31627509E5735) = 5735.52065052 vs 5735.52065055 For 12 decimal digits, 1 .. 2000 log10 sum, error is just 3 ULP. But we are mapping sum with 10^, IP goes to exponent, FP goes to mantissa. FP with 7 digits accuracy translate to about 7 digits mantissa with 10^, as expected. 10^(FP*(1+ε)) = 10^FP * e^(ln(10)*FP*ε) ≈ 10^FP * (1 + ln(10)*FP*ε) 0.5207*ln(10) ≈ 0.5207/0.4343 ≈ 1.20 // FP → mantissa relative error expanded a little. 

09072023, 08:29 PM
Post: #4




RE: Accuracy summation question
OK, this is probably a dumb question, but how are you guys getting such large exponents? Mine maxes out at E499.
 Bruce 

09082023, 01:38 AM
(This post was last modified: 09082023 01:44 AM by Gerson W. Barbosa.)
Post: #5




RE: Accuracy summation question  
09082023, 01:04 PM
Post: #6




RE: Accuracy summation question
(09082023 01:38 AM)Gerson W. Barbosa Wrote:(09072023 08:29 PM)byoung Wrote: OK, this is probably a dumb question, but how are you guys getting such large exponents? Mine maxes out at E499. Thanks. I just thought you had discovered some way to put the calculator into a highprecision mode, which I didn't think was possible.  Bruce 

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