Prime calculator sum(1.0002^n/n, 1, 20000)

12282020, 09:14 PM
(This post was last modified: 12312020 11:20 PM by Albert Chan.)
Post: #21




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
Note that z = 0.0002 is a small number, perfect for taylor series approximation.
\(f = \Large {(1+z)^k \over k} \displaystyle \normalsize = {1\over k} + \sum_{j=1}^k \binom{k}{j}{z^j \over k} = {1\over k} + \sum_{j=1}^k \binom{k1}{j1}{z^j \over j} \) \(\small \displaystyle\sum_{k=j}^n \binom{k1}{j1} = \sum_{k=j}^n \left[\binom{k}{j}  \binom{k1}{j}\right] = \binom{n}{j}  \binom{j1}{j} = \binom{n}{j} \) // middle terms telescopingly cancelled Flip the order of sum, \(\large \sum _{k=1}^n \sum _{j=1}^k ≡ \sum _{j=1}^n \sum _{k=j}^n\) \(\displaystyle\sum_{k=1}^n f = H(n) + \sum_{j=1}^n \displaystyle\sum_{k=j}^n \binom{k1}{j1} {z^j \over j} = H(n) + \sum_{j=1}^n \binom{n}{j}{z^j\over j} \) H(n), we use Harmonic Number asymptotic formula: lua> n, z = 20000, 0.0002 lua> euler_gamma = 0.5772156649015329 lua> Hn = (11/(6*n))/(2*n) + euler_gamma + log(n) lua> Hn 10.480728217229327 Sum the other terms, until converged. With small z, we may not need n terms for convergence. lua> j, t, s = 1, 1, 1 lua> repeat s=s+t/j; j=j+1; t=t*(nj+1)/j*z; until s==s+t/j lua> s = n*z * (s+1) lua> Hn + s  = sum(1.0002^k/k, k = 1 .. 20000) 28.14397383505517 lua> j  much less than 20000 terms 30 

12282020, 11:13 PM
Post: #22




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
(12282020 04:21 PM)Didier Lachieze Wrote:(12282020 03:02 PM)Joe Horn Wrote: Anybody know how to tell the DM42 not to update the stack display while running? Thanks in advance! Awesome, thanks! Doing that makes the DM42 run the above program in 40 seconds (powered by USB). <0ɸ0> Joe 

12292020, 01:57 AM
Post: #23




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
Just note that, due to roundings, the last digits of the
given result 28.14397383505516924725756564071173 are — not surprisingly — incorrect after checking with Wolfram Alpha (458 [55], instead of 173). Gil 

12292020, 05:10 PM
Post: #24




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
(12292020 01:57 AM)Gil Wrote: Just note that, due to roundings, the last digits of the The errors mostly comes from implementation of Y^X. For integer X, Free42 were patched to repeated squaring: Y^X = (Y^(X//2))^2 * Y^(X%2) Example, for 7 squarings: 1.0002 128 Y^X → 1.025927868161809572716800732564789 1.0002 LN 128 × E^X → 1.025927868161809572716800732564790 Extending to 14 squarings: 1.0002 16384 Y^X → 26.48218820206050462500040752260032 1.0002 LN 16384 × E^X → 26.48218820206050462500040752260470 (last digit should be 1) Because all terms are using the same base, almost all errors fall on the same side. And, it get worse and worse, with 1 ULP error for 7 squarings to 47132 = 439 ULP for 14. I replaced Y^X by E^(X*LN(Y)), and the sum ends in 447, much closer to true result. 

12302020, 09:02 PM
Post: #25




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
(12272020 07:03 PM)John Keith Wrote:(12272020 03:04 AM)Gil Wrote: On the HP50G, I get the wished answer by writing Oh, are we doing benchmarks now? Rewriting the expression in RPL as a FOR NEXT loop, a physical 50g with newRPL gives the answer in 6.4 seconds for 32 digits and 5.0 seconds with 16 digits precision. As a curiosity, I ran the same on a physical Prime G1 with newRPL and it took 11.9 seconds with 32 digits which makes no sense... unless the clock is stuck at 100 MHz, will need to investigate this and report again, sorry. 

12312020, 01:19 AM
Post: #26




RE: Prime calculator sum(1.0002^n/n, 1, 20000)
Differences of times might be substantial — on the same "device/model" —
according to how you calculate:  with integer limits or with nointeger (i. e. limits with dot);  with buildin summation (Greek S) or the loop For... Next. With Emu48 Android Application on my phone Samsung galaxy S6, I got the following results: 7.7" with points after 0., 1. and 2000.: \<< 0. 1. 20000. FOR i 1.0002 i ^ i / + NEXT \>> 22.8" without points after 0, 1 and 20000: \<< 0 1 20000 FOR i 1.0002 i ^ i / + NEXT \>> 9.1" without points, i. e. limits like integers, but mandatory with 105 SF! \<< '\GS(n=1,20000,1.0002^n/n)' 105. SF EVAL \>> 8.9" with points already included after 1. and 20000.: \<< '\GS(n=1.,20000.,1.0002^n/n)' EVAL \>> Conclusion: The selfmade program is the fastest... providing you do not forget to put dots to the limits (nointegers). Regards, [/align] Gil 

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