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Xeda112358
Active Member
Joined: 19 May 2009 Posts: 520
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Posted: 31 Aug 2010 03:28:22 pm Post subject: |
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Okay, I am working on an algorithm (for the TI-89, currently) and I need to test if it works. I had it calculate the summation of X^31 from 1 to 9999999999999999999 (that should be 19 9's) in 8~9 seconds using this algorithm, but it would require a few weeks using the normal method (mine doesn't add 1^31+2^31+3^31...). The result was over 600 digits (607 to be exact). I need to find a calculator that can use the regular method, but at a much, much faster rate than the TI-89.
I saved the stats in a program:
Code: Prgm
©8 seconds
©sums(x^31,x,0,9999999999999999999)
Pause dim("3124999999999999995000000000000000002583333333333333333333333333333333332958750000000000000000000000000000000067424999999999999999999999999999999989043437500000000000000000000000000001527278409090909090909090909090909090730493321590909090909090909090909090926278665340909090909090909090909090907758735211846590909090909090909090909171894622962662337662337662337662337658604147793605519480519480519480519480644314187368019480519480519480519480516634653617006087662337662337662337662378030845546587662337662337662337662337355573790713079004329004329004329004329936779358875000000000000000000000000000000000000")
EndPrgm
That huge number in quotes is the value for sums(x^31,x,0,9999999999999999999)
sums() is the program I made.
I am still working
Last edited by Guest on 01 Sep 2010 06:21:37 am; edited 1 time in total |
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DrDnar
Member
Joined: 28 Aug 2009 Posts: 116
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Posted: 31 Aug 2010 05:04:29 pm Post subject: |
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There are several programming languages that offer arbitrary precision integers. However, a single PC cannot finish such a long loop in any reasonable time. The good news is that doing the sum number-by-number could easily be split into a parallel computing task. It would still take a long time or a lot of computers. |
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Xeda112358
Active Member
Joined: 19 May 2009 Posts: 520
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Posted: 01 Sep 2010 12:51:02 pm Post subject: |
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Okay. I am starting to work on a way to prove my method works. It was accurate at 9999 and various other testing points, but oh well, I still have to prove it. You did answer another question, I believe. I wanted to ask about how long a computer would take to solve that the normal way. I am guessing even the fastest cannot do 8 seconds, right? |
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Lionel Debroux
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Joined: 01 Aug 2009 Posts: 170
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Posted: 02 Sep 2010 01:24:40 am Post subject: |
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Does someone here have access to CAS programs such as Maple or Mathematica ? |
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Weregoose Authentic INTJ
Super Elite (Last Title)
Joined: 25 Nov 2004 Posts: 3976
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Posted: 02 Sep 2010 02:52:14 am Post subject: |
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Lionel Debroux wrote:
Does someone here have access to CAS programs such as Maple or Mathematica ?
Mathematica's Sum[x^31, {x, 0, 9999999999999999999}] gives his quoted number instantly. [url=http://www.wolframalpha.com/input/?i=sum+x^31+from+0+to+9999999999999999999]Alpha[/url] does the same.
The actual computation performed is Zeta[-31]-HurwitzZeta[-31,n+1] (where n=9999999999999999999), which breaks down to:
Code: 7709321041217/16320 -
1/1256640 (593617720173709 - 11717544135366800 n^2 +
38549175163831640 n^4 - 50728681747663920 n^6 +
35762288787254700 n^8 - 15687098041414800 n^10 +
4691683716643560 n^12 - 1017696589998000 n^14 +
167406298338510 n^16 - 21598822014000 n^18 + 2244328723560 n^20 -
191923914000 n^22 + 13768454700 n^24 - 847289520 n^26 +
47071640 n^28 - 3246320 n^30 - 628320 n^31 - 39270 n^32)
http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/
http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta2/
Last edited by Guest on 02 Sep 2010 04:13:39 am; edited 1 time in total |
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Xeda112358
Active Member
Joined: 19 May 2009 Posts: 520
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Posted: 02 Sep 2010 10:19:37 am Post subject: |
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THANK YOU! That is exactly what I was looking for.
I've been working on this for eight months trying to figure out what the heck I was doing... ☺
Last edited by Guest on 02 Sep 2010 10:26:54 am; edited 1 time in total |
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