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I was musing about a particular limit,

$L = \prod\limits_{n > 0} \bigl(1 - 2^{-n}\bigr)$:

we may bound 0.288 < L < 0.308, which we may show by taking the logarithm:

$\ln(L) = \ln \bigl( \frac{315}{1024}\bigr) + \sum\limits_{n > 4} \ln\bigl(1 - 2^{-n}\bigr) > \ln\bigl(\frac{315}{1024}\bigr) - \sum\limits_{n > 4} 2^{-n} =\; \ln\bigl(\frac{315}{1024} \cdot \mathrm e^{-1/16}\bigr)$.

I was wondering if this type of infinite product (or the corresponding sum of logarithms) has a name, and whether there are techniques for obtaining a closed form expression for the limit.

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Looks awfully like a q-Pochhammer: ; note formula 3 and the asymptotic result formula 8. – J. M. Sep 1 '10 at 7:56
@J.M. : An interesting reference. For casual spectators, the asymptotic behaviour is that $\prod_{n>0} (1-\exp(-t)) = \sqrt{\frac{2\pi}{t}} \exp\bigl(\frac{t}{24} - \frac{\pi^2}{6t}\bigr) + o(1)$. Unfortunately, I'm interested in the case of p=exp(-t) a medium-range probability: that o(1) term washes out the quantitative behaviour I'm interested in. Is there an easily-found, more precise bound? – Niel de Beaudrap Sep 1 '10 at 9:47
I've actually only started to study q-calculus, and I'm far away from my books at the moment, so I can't answer that question. Sorry. :( – J. M. Sep 1 '10 at 10:45
[Erratum: in my previous comment, the LHS of the formula should involve exp(–nt), not exp(–t).] – Niel de Beaudrap Sep 1 '10 at 12:35
up vote 3 down vote accepted

The product $$\phi (x) = \prod_{n > 0} (1 - x^n)$$ is called the Euler function, and is well studied. I don't know of any way to compute values at special points. The Euler identity can be used to compute numerical values.

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The relationship with Dedekind $\eta$ is very useful when computing it numerically (IMHO more useful than the pentagonal number theorem); the appropriate use of modular transformations allow quick evaluation of Dedekind $\eta$. – J. M. Sep 1 '10 at 8:29
@J.M. : Looking up Dedekind η on Wikipedia, it is clear that it is related to φ. What isn't obvious is what techniques exist for computing φ (via η) which would be superior to the straightforward one of computing the partial product, and bounding the tail. Am I missing something? – Niel de Beaudrap Sep 1 '10 at 9:37
Well, the infinite product for $\phi(q)=(q;q)_\infty$ is already slowly convergent for $|q|\geq 1/2$. (Of course, one can use convergence acceleration techniques, but this is outside the scope of this comment.) For Dedekind $\eta$, however, appropriate modular transformations (more specifically, linear fractional transformations of $\tau=-i\ln(q)/\pi$), allow $\eta$ to be computed easily through its $\vartheta$ series representation. – J. M. Sep 1 '10 at 10:51
Thanks, J.M.: this may come in handy. – Niel de Beaudrap Sep 1 '10 at 12:37
Niel, I could detail the algorithm I use for computing Dedekind $\eta$ should you need it. Well, when I'm with my books and notes again, that is. – J. M. Sep 1 '10 at 13:48

If you want to evaluate $$\prod_{n=1}^\infty (1-1/2^n)$$ numerically you will find that it's VERY close to

$$2^{1/24}\sqrt{\frac{2\pi}{\log 2}}\exp{(-\pi^2/(6\log 2))}.$$

Which is about 0.2887880950866... And that's because $$\exp{(-4\pi^2/\log 2)}$$ is pretty small, approximately 1.839x10^(-25).

This can be seen from the fact that $$\Delta(-1/z) = z^{12}\Delta(z),$$ where $$\Delta(z)$$ is the cusp form of weight 12 defined by

$$\Delta(z)=q\prod_{n=1}^\infty (1-q^n)^{24},$$

where $$q=e^{2\pi iz}$$ for Im(z)>0.

Just put $$z=\frac{2\pi i}{\log 2}.$$

You can use this technique to greatly accelerate the convergence of your product for terms other than (1/2)^n.

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@Niel de Beaudrap. In the question your lower bound of 0.289 is too high as the actual value begins 0.2887... You can verify this as per my answer or with wolframAlpha – Derek Jennings Sep 1 '10 at 19:01
Whoops, yeah. The wrong kind of rounding on my part. – Niel de Beaudrap Sep 1 '10 at 19:11
The identity after "This can be seen from the fact that..." is in fact a modular transformation (linear fractional transformation of z). :) Though I see you're using the squared nome instead of the nome. – J. M. Sep 1 '10 at 21:16

Another approach that confirms the above extremely close approximation is to introduce $$S = \log P = \log \prod_{n\ge 1} \left(1-\frac{1}{2^n}\right) = \sum_{n\ge 1} \log \left(1-\frac{1}{2^n}\right)$$ and observe that this sum is harmonic and may be evaluated by inverting its Mellin transform. To do this introduce $$S(x) = \sum_{n\ge 1} \log \left(1-\frac{1}{2^{nx}}\right)$$

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad g(x) = \log\left(1-\frac{1}{2^x}\right).$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \log\left(1-\frac{1}{2^x}\right) x^{s-1} dx.$$ The function $g(x)$ is well-behaved near zero where it is on the order of $\log x$ and vanishes faster than any polynomial at infinity.

To calculate the Mellin transform start with $$\int_0^\infty \log\left(1-\frac{1}{2^x}\right) x^{s-1} dx = - \int_0^\infty \sum_{q\ge 1} \frac{2^{-qx}}{q} x^{s-1} dx = - \sum_{q\ge 1} \frac{1}{q} \int_0^\infty 2^{-qx} x^{s-1} dx.$$

Observe that $$\int_0^\infty 2^{-qx} x^{s-1} dx = \frac{1}{(q \log 2)^s} \Gamma(s)$$ by a straightforward substitution that turns the integral into a gamma function integral.

This yields $$g^*(s) = - \sum_{q\ge 1} \frac{1}{q} \frac{1}{(q \log 2)^s} \Gamma(s) = -\frac{1}{(\log 2)^s} \Gamma(s) \sum_{q\ge 1} \frac{1}{q^{s+1}} = -\frac{1}{(\log 2)^s} \Gamma(s) \zeta(s+1).$$

By the harmonic sum identity we now have that the Mellin transform $Q(s)$ of $S(x)$ is given by $$Q(s) = -\frac{1}{(\log 2)^s} \Gamma(s) \zeta(s) \zeta(s+1)$$ with the Mellin inversion integral being $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero. Fortunately the two zeta function terms with their trivial zeros combine to cancel the poles of the gamma function and we are left with just three poles and residues. We have $$\mathrm{Res}(Q(s)/x^s; s=1) = -\frac{\pi^2}{6x\log 2}.$$ Furthermore $$\mathrm{Res}(Q(s)/x^s; s=0) = \frac{1}{2}\left(\log\frac{2\pi}{\log 2}-\log x\right)$$ and finally $$\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{1}{24} x \log 2.$$ Putting $x=1$ we obtain the following approximation for $S(1):$ $$S(1)\approx \frac{1}{24} \log 2 + \frac{1}{2} \log\frac{2\pi}{\log 2} - \frac{\pi^2}{6\log 2}.$$ This is $$-1.2420620948124149457978452979784311762117047031228$$ while the exact value is $$-1.2420620948124149457978454818946296689734039782504$$ so this approximation is good to an amazing $25$ digits. This gives for $P$ the approximation $$P \approx 2^{1/24} \sqrt{\frac{2\pi}{\log 2}} \exp\left(- \frac{\pi^2}{6\log 2}\right).$$ This is also good to $25$ digits, confirming the observation from the other poster above.

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