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  #1  
Old 12-08-2005, 07:14 AM
PrayingMantis PrayingMantis is offline
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Default 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<number<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).
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  #2  
Old 12-08-2005, 09:40 AM
Bataglin Bataglin is offline
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Default Re: Infinite multiplication

[ QUOTE ]

What can be said about the vaule of M?



[/ QUOTE ]

I don't think Harrington would have liked it.
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  #3  
Old 12-08-2005, 10:07 AM
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Default 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.
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  #4  
Old 12-08-2005, 10:38 AM
PrayingMantis PrayingMantis is offline
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Default 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?
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Old 12-08-2005, 11:07 AM
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Default 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.
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  #6  
Old 12-08-2005, 10:59 AM
superleeds superleeds is offline
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Default 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?
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  #7  
Old 12-08-2005, 01:48 PM
BruceZ BruceZ is offline
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Default 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.
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  #8  
Old 12-08-2005, 07:54 PM
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Default 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?
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  #9  
Old 12-08-2005, 08:38 PM
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Default 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.
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  #10  
Old 12-09-2005, 02:14 AM
Siegmund Siegmund is offline
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Default Re: Infinite multiplication

If there is some fixed k<1, such that every term in your sequence is less than k, that is sufficient to show convergence to 0. The n_th partial product will be less than k^n and the power series with k<1 converges to 0.

For instance, 0.9 x 0.95 x 0.955 x 0.9555 x 0.95555 ... will converge to zero, because every term is less than 0.96.

Sufficient, but not necessary - the terms can get arbitrarily close to 1 provided they do so very slowly (more slowly than e^(-1/n) approaches 1, I think.)
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