 Two Plus Two Older Archives Infinite multiplication
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 PrayingMantis Senior Member Join Date: Nov 2003 Location: 11,600 km from Vegas Posts: 489 Infinite multiplication

I define M to be a multiplication of an infinite series of real numbers, all between 0 and 1 (not inclusive). You can come up with any rule to decide how you create these numbers, they can also be all random, or the same repeated number, or any kind of series, as long as each one of them, is 0&lt;number&lt;1.

What can be said about the vaule of M?

(For the mathematicians here, I apologize for the non-formal description of the question, I hope you don't mind).
#2
 Bataglin Junior Member Join Date: May 2004 Posts: 15 Re: Infinite multiplication

[ QUOTE ]

What can be said about the vaule of M?

[/ QUOTE ]

I don't think Harrington would have liked it.
#3
 Guest Posts: n/a Re: Infinite multiplication

The log of the product is the sum of the logs, so one way to analyze this is to look at the series resulting from summing the logs of the numbers in your original sequence. If this sum diverges, then M=0. This happens if your numbers are all random with the same distribution, or you have the same repeated number. More generally, it will happen if the sequence of numbers that you are multiplying doesn't converge to 1, or converges slowly to 1. If the numbers in your sequence go to 1 quickly enough, then M can be non-zero.
#4
 PrayingMantis Senior Member Join Date: Nov 2003 Location: 11,600 km from Vegas Posts: 489 Re: Infinite multiplication

[ QUOTE ]
More generally, it will happen if the sequence of numbers that you are multiplying doesn't converge to 1, or converges slowly to 1. If the numbers in your sequence go to 1 quickly enough, then M can be non-zero.

[/ QUOTE ]

Thanks for the answer. Can you please elaborate a bit on how is the "quickness" of the convergence effecting M? that is, what is the criteria for a sequence that is converging to 1, which for it M=0, as opposed to a different sequence that also converges to 1, but that produces a non-zero M?
#5
 superleeds Senior Member Join Date: Jul 2003 Posts: 309 Re: Infinite multiplication

[ QUOTE ]
If the numbers in your sequence go to 1 quickly enough, then M can be non-zero

[/ QUOTE ]

But your number is always less than 1 and every multiplication will lower it further. It can never approach 1 but will always approach 0. What am I missing?
#6
 Guest Posts: n/a Re: Infinite multiplication

There is no simple criterion that is necessary and sufficient, but I can give some examples. If the n-th term in your sequence is e^{-1/n}, then the log is -1/n, so the sum of the logs diverges (it is the negative harmonic series) and M=0. If the n-th term in your sequence is e^{-1/n^2}, then the log is -1/n^2, and so the sum of the logs converges and M is not 0. IIRC, the sum of -1/n^2 is -pi^2/6 [aka -zeta(2)], and so M=e^[-pi^2/6] for that example.

There are a variety of tests for whether or not a series converges--see your favorite calculus textbook. One trick that might be useful in dealing with examples is that log(1-x) ~ -x for small x.
#7
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636 Re: Infinite multiplication

[ QUOTE ]
[ QUOTE ]
If the numbers in your sequence go to 1 quickly enough, then M can be non-zero

[/ QUOTE ]

But your number is always less than 1 and every multiplication will lower it further. It can never approach 1 but will always approach 0. What am I missing?

[/ QUOTE ]

He means that the sequence of numbers that are being multiplied need to go to 1. Then the logs of these numbers will go to 0, and if they do so fast enough, the sum of the logs will converge to some number -x instead of going to minus infinity, and then the product will be some number other than zero. It will be e^-x.
#8
 Guest Posts: n/a Re: Infinite multiplication

This math hurts my head. :-| Can one of you tell me what this series would converge to:

M = .98 * .998 * .9998 * .99998 * .999998 * .9999998 * ...

And how do you figure that out?
#9
 Guest Posts: n/a Re: Infinite multiplication

[ QUOTE ]
This math hurts my head. :-| Can one of you tell me what this series would converge to:

M = .98 * .998 * .9998 * .99998 * .999998 * .9999998 * ...

And how do you figure that out?

[/ QUOTE ]

I don't know how to find an exact value, but it is a little bit more than .9778. I got that by multiplying the first few terms together, and then convincing myself that the rest of the product doesn't matter very much. There are very few infinite series or products for which the exact value is known, and I don't recognize this particular one as being one of them.

One reason why the rest of the series doesn't matter very much is the following: you are taking the product from n=1 to infty of [1-.2(10)^{-n}]. The log of the n-th term is roughly -.2(10)^{-n} (where by log I mean base e), and this approximation is very accurate for all but the smallest values of n. The logs are thus close to a geometric series, and summing that series from n=k to infty gives (-.2)10^{-k}/.9. Taking k to be 7 or so gives something fairly tiny, so the product of all the terms except the first 6 is essentially 1. Multiplying the first 6 terms of the series thus gives us a decent approximation to the final answer.

Ignoring the details of the math, it was easy to get a good approximation for this particular product because the terms in your product converge geometrically to 1. If they converge slowly, then you will have to multiply far more terms to get a decent approximation.
#10
 Guest Posts: n/a Re: Infinite multiplication

Cool... thanks. [img]/images/graemlins/smile.gif[/img]

I wrote a perl script (before I saw your anwer), and it didn't seem to converge. Here is the output:

<font class="small">Code:</font><hr /><pre>\$i = 2 \$prec = 2 0.98
\$i = 3 \$prec = 5 0.97804
\$i = 4 \$prec = 9 0.977844392
\$i = 5 \$prec = 14 0.97782483511216
\$i = 6 \$prec = 20 0.97782287946248977568
\$i = 7 \$prec = 27 0.977822683897913883182044864
\$i = 8 \$prec = 35 0.97782266434146020522376720035910272
\$i = 9 \$prec = 44 0.97782266238581487654084678991156831928179456
\$i = 10 \$prec = 54 0.977822662190250344063683814603398961299480896143 641088
\$i = 11 \$prec = 65 0.977822662170693890819878807722125285007412916917 65147007712717824
\$i = 12 \$prec = 77 0.977822662168738245495537419940485527391968666347 63664424329187529984574564352
\$i = 13 \$prec = 90 0.977822662168542680963103672291386419907980569242 15825051002234797099708726846003
\$i = 14 \$prec = 104 0.977822662168523124509860301437767157834534741513 76009089863750480598688682150061
\$i = 15 \$prec = 119 0.977822662168521168864535964391518138113931865979 44442182915447728580508954649099
\$i = 16 \$prec = 135 0.977822662168520973300003530687284365206738987675 81679904278128139692072371559554
\$i = 17 \$prec = 152 0.977822662168520953743550287316864899206668373930 12949490800152788058474285996991
\$i = 18 \$prec = 170 0.977822662168520951787904962979822991719567799296 39969649466478002032575304396685
\$i = 19 \$prec = 189 0.977822662168520951592340430546118801361986806700 43509815075122016104581374503390
\$i = 20 \$prec = 209 0.977822662168520951572783977302748382330139998089 51272212351148402703711178201887
\$i = 21 \$prec = 230 0.977822662168520951570828331978411340426994430134 90722535885120403085808633777185
\$i = 22 \$prec = 252 0.977822662168520951570632767445977636236680264468 51154309076580514483110489270008
\$i = 23 \$prec = 275 0.977822662168520951570613210992734265817648851813 16262353804107153954173466183826
\$i = 24 \$prec = 299 0.977822662168520951570611255347409928775745710586 74063806950943624183810833659119
\$i = 25 \$prec = 324 0.977822662168520951570611059782877495071555396464 48956858752368109269599098846357
\$i = 26 \$prec = 350 0.977822662168520951570611040226424251701136365052 26837292997377966158806169867220
\$i = 27 \$prec = 377 0.977822662168520951570611038270778927364094461911 04629247712527625931533159413546
\$i = 28 \$prec = 405 0.977822662168520951570611038075214394930390271596 92408482296949078649643921192620
\$i = 29 \$prec = 434 0.977822662168520951570611038055657941687019852565 51186406146520288788863377998772
\$i = 30 \$prec = 464 0.977822662168520951570611038053702296362682810662 37064198535388700451459407485669
\$i = 31 \$prec = 495 0.977822662168520951570611038053506731830249106472 05651977774314654524205751272422
\$i = 32 \$prec = 527 0.977822662168520951570611038053487175377005736053 02510755698207641060545253059478
\$i = 33 \$prec = 560 0.977822662168520951570611038053485219731681399011 12196633490596943625469851912267
\$i = 34 \$prec = 594 0.977822662168520951570611038053485024167148965306 93165221269835873921075218284287
\$i = 35 \$prec = 629 0.977822662168520951570611038053485004610695721936 51262080047759766951026883986356
\$i = 36 \$prec = 665 0.977822662168520951570611038053485002655050397599 47071765925552156254025961847212
\$i = 37 \$prec = 702 0.977822662168520951570611038053485002459485865165 76652734513331395184325908746204
\$i = 38 \$prec = 740 0.977822662168520951570611038053485002439929411922 39610831372109319077355903827232
\$i = 39 \$prec = 779 0.977822662168520951570611038053485002437973766598 05906641057987111466658903339246
</pre><hr />

I had to stop at 39 because my "BigFloat" could only hold 819 decimal points.

By the way, I used the following forumla to calcluate the result:

\$newresult = \$oldresult * (((10^i)-2)/(10^i)) [i = 2 .. 39] Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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