8

Also, what about in general, for some value p, which has the value 2 in the given formula?

MOTIVATION:

I was wondering the probability of never getting tails if one forever flipped a coin whose probability of landing tails decreased (in this case, geometrically) each flip.

The probability of tails on flip i is $\frac{1}{2^i}$, and the probability of heads on flip i is $1-\frac{1}{2^i}$. So, first flip the coin is 50-50, next it is 75-25, etc. And the probability of never landing tails is equal to the probability of always landing heads, which is the infinite product of the heads probabilities, yielding $\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$. => ($\frac{1}{2} * \frac{3}{4} * \frac{7}{8} ...$)

Thomas Andrews
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tscizzle
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    I don't think there's any analytic solution. The answer is approx 0.28878809 – shalop Mar 22 '15 at 05:09
  • Why don't you think there's any analytic solution?

    Also, might there be a very good approximation (besides the decimal approximation you provided, which I thank you for)? Like, a function that is very close to the partial products?

     – tscizzle Mar 22 '15 at 05:12
  • I just plugged it into Wolfram Alpha, and it gave the answer in terms of a Pochammer symbol. Just go there, type in "product 1-1/2^i from i=1 to infty" and you can get the answer to a huge number of decimal places. – shalop Mar 22 '15 at 05:14
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    Just to add, this number is also the measure of a (generalized) Cantor set, where we take the interval $[0,1]$ and then remove the middle half, and then remove the middle fourths from the two remaining intervals, and then remove the middle eights from the four remaining intervals, and so on... – shalop Mar 22 '15 at 05:22
  • Oooh, beautiful. That stuff is awesome. It is so cool that this is a non-zero constant. And any polynomial decrease (instead of exponential decrease) yields an infinite product equal to 0? – tscizzle Mar 22 '15 at 05:25
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    No, not true. The infinite product $1-\frac{1}{n^2}$ also converges to a nonzero number. In fact, $\Pi_i^{\infty} (1-a_i)$ converges to a nonzero number iff $\sum_i^{\infty} a_i < \infty$. (assuming all $a_i \in [0,1]$). Try to prove it (hint: use logarithms). – shalop Mar 22 '15 at 05:26
  • Ah woops, I also plugged into Wolfram, and saw 0, but to get a meaningful result I need to take product starting from i=2 – tscizzle Mar 22 '15 at 05:30
  • Starting from i=1 – shalop Mar 22 '15 at 05:31
  • 1-1/(n^2) is 0 when n is 1, so the whole product is 0, right? – tscizzle Mar 22 '15 at 05:33
  • Oh, I thought you were still talking about $1-1/2^i$. Yes, you're correct. Start from i=2. BTW, I meant to say $a_i \in [0,1)$ before, rather than $a_i \in [0,1]$. – shalop Mar 22 '15 at 05:34
  • See OEIS A$048651$. – Lucian Mar 22 '15 at 10:43
  • Well, after rereading the post, I have decided to retract my duplicate close vote - mainly because of the part asking about more general question about $\prod \left(1-\frac1p\right)$. (Although this is not addreseed in the answers.) – Martin Sleziak Feb 15 '17 at 17:11
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    Lookup up "Euler function" https://en.wikipedia.org/wiki/Euler_function – leonbloy Nov 18 '19 at 21:59

2 Answers2

5

$$\sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) = -\sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{n 2^{kn}}=-\sum_{m\geq 1}\frac{1}{2^m}\sum_{d\mid m}\frac{1}{d}=-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}$$ hence: $$\prod_{i\geq 1}\left(1-\frac{1}{2^i}\right)=\exp\left(-\sum_{m\geq 1}\frac{\sigma_1(m)}{m 2^m}\right)\geq\exp\left(-\sum_{m\geq 1}\frac{m+1}{2^{m+1}}\right)=e^{-3/2}$$ but the bound: $$\sigma_1(m)=\sum_{d\mid m}d \leq \sum_{k=1}^{m}k = \frac{m(m+1)}{2}$$ is obviously very crude.

Jack D'Aurizio
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3

You can use the $q$-Pochhammer symbol to represent the infinite product as

$$ \prod_{i=1}^{\infty}\left( 1-\frac{1}{2^i} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)_{\infty} . $$

Note:

$$ \left( a, q \right)_{\infty}= \prod_{i=1}^{\infty}\left( 1-a q^k \right) $$

Martin Sleziak
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science
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