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https://xkcd.com/3142/

If you want to see true audacity, do an image search for 'Altoona-style pizza.'

https://xkcd.com/3142/

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Posted by Terence Tao

https://terrytao.wordpress.com/2025/09/15/smooth-numbers-and-max-entropy/

http://terrytao.wordpress.com/?p=15682

Given a threshold ${y>1}$ , a ${y}$ -smooth number (or ${y}$ -friable number) is a natural number ${n}$ whose prime factors are all at most ${y}$ . We use ${\Psi(x,y)}$ to denote the number of ${y}$ -smooth numbers up to ${x}$ . In studying the asymptotic behavior of ${\Psi(x,y)}$ , it is customary to write ${y}$ as ${x^{1/u}}$ (or ${x}$ as ${y^u}$ ) for some ${u>0}$ . For small values of ${u}$ , the behavior is straightforward: for instance if ${0 < u < 1}$ , then all numbers up to ${x}$ are automatically ${y}$ -smooth, so

$\displaystyle \Psi(x,y) \sim x$

in this case. If ${1 < u < 2}$ , the only numbers up to ${x}$ that are not ${y}$ -smooth are the multiples of primes ${p}$ between ${y}$ and ${x}$ , so

$\displaystyle \Psi(x,y) \sim x - \sum_{y < p \leq x} (\frac{x}{p} + O(1))$

$\displaystyle \sim x - x (\log\log x - \log\log y)$

$\displaystyle \sim x (1 - \log u)$

where we have employed Mertens’ second theorem. For ${2 < u < 3}$ , there is an additional correction coming from multiples of two primes between ${y}$ and ${x}$ ; a straightforward inclusion-exclusion argument (which we omit here) eventually gives

$\displaystyle \Psi(x,y) \sim x (1 - \log u + \int_2^u \frac{\log(t-1)}{t} dt)$

in this case.

More generally, for any fixed ${u>0}$ , de Bruijn showed that

$\displaystyle \Psi(x, y) \sim \rho(u) x$

where ${\rho(u)}$ is the Dickman function. This function is a piecewise smooth, decreasing function of ${u}$ , defined by the delay differential equation

$\displaystyle u \rho'(u) + \rho(u-1) = 0$

with initial condition ${\rho(u) = 1}$ for ${0 \leq u \leq 1}$ .

The asymptotic behavior of ${\rho(u)}$ as ${u \rightarrow \infty}$ is rather complicated. Very roughly speaking, it has inverse factorial behavior; there is a general upper bound ${\rho(u) \leq 1/\Gamma(u+1)}$ , and a crude asymptotic

$\displaystyle \rho(u) = u^{-u+o(u)} = \exp( - u \log u + o(u \log u)).$

With a more careful analysis one can refine this to

$\displaystyle \rho(u) = \exp( - u \log u - u \log\log u + u + o(u)); \ \ \ \ \ (1)$

and with a very careful application of the Laplace inversion formula one can in fact show that

$\displaystyle \rho(u) \sim \sqrt{\frac{\xi'(u)}{2\pi}} \exp( \gamma - u \xi(u) + \int_0^{\xi(u)} \frac{e^s - 1}{s} ds) \ \ \ \ \ (2)$

where ${\gamma}$ is the Euler-Mascheroni constant and ${\xi(u)}$ is defined implicitly by the equation

$\displaystyle e^{\xi(u)} - 1 = u \xi(u). \ \ \ \ \ (3)$

One cannot write ${\xi(u)}$ in closed form using elementary functions, but one can express it in terms of the Lambert ${W}$ function as ${\xi(u) = -W(-\frac{1}{u} e^{-1/u}) - 1/u}$ . This is not a particularly enlightening expression, though. A more productive approach is to work with approximations. It is not hard to get the initial approximation ${\xi(u) \approx \log u}$ for large ${u}$ , which can then be re-inserted back into (3) to obtain the more accurate approximation

$\displaystyle \xi(u) = \log u + \log\log u + O(1)$

and inserted once again to obtain the refinement

$\displaystyle \xi(u) = \log u + \log\log u + O(\frac{\log\log u}{\log u}).$

We can now see that (2) is consistent with previous asymptotics such as (1), after comparing the integral ${\int_0^{\xi(u)} \frac{e^s - 1}{s} ds}$ to

$\displaystyle \int_0^{\xi(u)} \frac{e^s - 1}{\xi(u)} ds = u - 1.$

For more details of these results, one can see for instance this survey by Granville.

This asymptotic (2) is quite complicated, and so one does not expect there to be any simple argument that could recover it without extensive computation. However, it turns out that one can use a “maximum entropy” analysis to get a reasonably good heuristic approximation to (2), that at least reveals the role of the mysterious function ${\xi(u)}$ . The purpose of this blog post is to give this heuristic.

Viewing ${x = y^u}$ , the task is to try to count the number of ${y}$ -smooth numbers of magnitude ${y^u}$ . We will propose a probabilistic model to generate ${y}$ -smooth numbers as follows: for each prime ${p \leq y}$ , select the prime ${p}$ with an independent probability ${c_p/p}$ for some coefficient ${c_p}$ , and then multiply all the selected primes together. This will clearly generate a random ${y}$ -smooth number ${n}$ , and by the law of large numbers, the (log-)magnitude of this number should be approximately

$\displaystyle \log n \approx \sum_{p \leq y} \frac{c_p}{p} \log p, \ \ \ \ \ (4)$

(where we will be vague about what “ ${\approx}$ ” means here), so to obtain a number of magnitude about ${y^u}$ , we should impose the constraint

$\displaystyle \sum_{p \leq y} \frac{c_p}{p} \log p = u \log y. \ \ \ \ \ (5)$

The indicator ${1_{p|n}}$ of the event that ${p}$ divides this number is a Bernoulli random variable with mean ${c_p/p}$ , so the Shannon entropy of this random variable is

$\displaystyle \mathbf{H}(1_{p|n}) = - \frac{c_p}{p} \log(\frac{c_p}{p}) - (1 - \frac{c_p}{p}) \log(1 - \frac{c_p}{p}).$

If ${c_p}$ is not too large, then Taylor expansion gives the approximation

$\displaystyle \mathbf{H}(1_{p|n}) \approx \frac{c_p}{p} \log p - \frac{c_p}{p} \log c_p + \frac{c_p}{p}.$

Because of independence, the total entropy of this random variable ${n}$ is

$\displaystyle \mathbf{H}(n) = \sum_{p \leq y} \mathbf{H}(1_{p|n});$

inserting the previous approximation as well as (5), we obtain the heuristic approximation

$\displaystyle \mathbf{H}(n) \approx u \log y - \sum_{p \leq y} \frac{c_p}{p} (\log c_p - 1).$

The asymptotic equipartition property of entropy, relating entropy to microstates, then suggests that the set of numbers ${n}$ that are typically generated by this random process should be approximately

$\displaystyle \exp(\mathbf{H}(n)) \approx \exp\left(u \log y - \sum_{p \leq y} \frac{c_p}{p} (\log c_p - 1)\right)$

$\displaystyle = \exp\left(- \sum_{p \leq y} \frac{c_p \log c_p - c_p}{p}\right) x.$

Using the principle of maximum entropy, one is now led to the approximation

$\displaystyle \rho(u) \approx \exp\left(- \sum_{p \leq y} \frac{c_p \log c_p - c_p}{p}\right). \ \ \ \ \ (6)$

where the weights ${c_p}$ are chosen to maximize the right-hand side subject to the constraint (5).

One could solve this constrained optimization problem directly using Lagrange multipliers, but we simplify things a bit by passing to a continuous limit. We take a continuous ansatz ${c_p = f(\log p / \log y)}$ , where ${f: [0,1] \rightarrow {\bf R}}$ is a smooth function. Using Mertens’ theorem, the constraint (5) then heuristically becomes

$\displaystyle \int_0^1 f(t)\ dt = u \ \ \ \ \ (7)$

and the expression (6) simplifies to

$\displaystyle \rho(u) \approx \exp( - \int_0^1 \frac{f(t) \log f(t) - f(t)}{t}\ dt). \ \ \ \ \ (8)$

So the entropy maximization problem has now been reduced to the problem of minimizing the functional ${\int_0^1 \frac{f(t) \log f(t) - f(t)}{t}\ dt}$ subject to the constraint (7). The astute reader may notice that the integral in (8) might diverge at ${t=0}$ , but we shall ignore this technicality for the sake of the heuristic arguments.

This is a standard calculus of variations problem. The Euler-Lagrange equation for this problem can be easily worked out to be

$\displaystyle \frac{\log f(t)}{t} = \lambda$

for some Lagrange multiplier ${\lambda}$ ; in other words, the optimal ${f}$ should have an exponential form ${f(t)= e^{\lambda t}}$ . The constraint (7) then becomes

$\displaystyle \frac{e^\lambda - 1}{\lambda} = u$

and so the Lagrange multiplier ${\lambda}$ is precisely the mysterious quantity ${\xi(u)}$ appearing in (2)! The formula (8) can now be evaluated as

$\displaystyle \rho(u) \approx \exp\left( - \int_0^1 \frac{e^{\xi(u) t} \xi(u) t - e^{\xi(u) t}}{t}\ dt \right)$

$\displaystyle \approx \exp\left( - e^{\xi(u)} + 1 + \int_0^1 \frac{e^{\xi(u) t} - 1}{t}\ dt + \int_0^1 \frac{1}{t}\ dt \right)$

$\displaystyle \approx \exp\left( - u \xi(u) + \int_0^{\xi(u)} \frac{e^s - 1}{s}\ ds + C\right)$

where ${C}$ is the divergent constant

$\displaystyle C = \int_0^1 \frac{1}{t}\ dt.$

This recovers a large fraction of (2)! It is not completely accurate for multiple reasons. One is that the hypothesis of joint independence on the events ${p|n}$ is unrealistic when trying to confine ${n}$ to a single scale ${x}$ ; this comes down ultimately to the subtle differences between the Poisson and Poisson-Dirichlet processes, as discussed in this previous blog post, and is also responsible for the otherwise mysterious ${e^\gamma}$ factor in Mertens’ third theorem; it also morally explains the presence of the same ${e^\gamma}$ factor in (2). A related issue is that the law of large numbers (4) is not exact, but admits gaussian fluctuations as per the central limit theorem; morally, this is the main cause of the ${\sqrt{\frac{\xi'(u)}{2\pi}}}$ prefactor in (2).

Nevertheless, this demonstrates that the maximum entropy method can achieve a reasonably good heuristic understanding of smooth numbers. In fact we also gain some insight into the “anatomy of integers” of such numbers: the above analysis suggests that a typical ${y}$ -smooth number ${n}$ will be divisible by a given prime ${p \sim y^t}$ with probability about ${e^{\xi(u) t}/p}$ . Thus, for ${t = 1 - O(1/\log u)}$ , the probability of being divisible by ${p}$ is elevated by a factor of about ${\asymp e^{\xi(u)} \asymp u \log u}$ over the baseline probability ${1/p}$ of an arbitrary (non-smooth) number being divisible by ${p}$ ; so (by Mertens’ theorem) a typical ${y}$ -smooth number is actually largely comprised of something like ${\asymp u}$ prime factors all of size about ${y^{1-O(1/\log u)}}$ , with the smaller primes contributing a lower order factor. This is in marked contrast with the anatomy of a typical (non-smooth) number ${n}$ , which typically has ${O(1)}$ prime factors in each hyperdyadic scale ${[\exp\exp(k), \exp\exp(k+1)]}$ in ${[1,n]}$ , as per Mertens’ theorem.

https://terrytao.wordpress.com/2025/09/15/smooth-numbers-and-max-entropy/

http://terrytao.wordpress.com/?p=15682

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https://www.oglaf.com/polytheism/

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https://xkcd.com/3141/

Mantle plumes explain Hawaii, Yellowstone, Iceland, the East African Rift, the Adirondack uplift, the Permian extinction, the decline of Rome, the DB Cooper hijacking, and the balrog in Moria. Those little hills of sand in your yard are caused by antle plumes.

https://xkcd.com/3141/

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Posted by Abbadon

https://killsixbilliondemons.com/comic/wheel-smashing-lord-5-146/

https://killsixbilliondemons.com/?post_type=comic&p=4107

“Many have asked me why Ki Rata insists on flashiness. There are certainly more conservative styles that focus on efficient and useful practice of internal force. The answer is simple: the lower ranked techniques must be slow enough to be observed by the naked eye, and ferocious enough to completely obliterate thought of opposition. They are exceedingly gentle, and aimed at the conservation of life for those remaining. If you were targeted by a higher ranked technique, you would have not have time to even feel fear.”

– Musko Reeve, Manual of Hands and Feet, pp 405

https://killsixbilliondemons.com/comic/wheel-smashing-lord-5-146/

https://killsixbilliondemons.com/?post_type=comic&p=4107

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siderea

Canonical link: https://siderea.dreamwidth.org/1882720.html

0.

Hey, Americans! Look sharp, the Trump Administration is trying to play a head game on you about Covid vaccines, and it's apparently working, because I see nobody talking about this in the news or on social media.

There's a lot of complexity and chaos right now about what is available to whom and how to get it. Things are changing fast, especially on the state level. I hope to discuss it in another post, but there's one thing in particular I want to clarify for you.

As you've probably heard, week and a half ago, the FDA changed the authorization for the Covid vaccines, in a way which curtails access. The thing that people are hearing is that for people under 65 years old the Covid vaccines are not authorized with some exceptions.

That's technically correct, but badly misleading. A lot of people hear "not authorized" and stop really listening to the rest of the sentence. They hear "with some exceptions" and assume they're not likely to be one such, and won't qualify to get it, and tune right out.

To be cynical for a moment, you're meant to assume that.

But it turns out you're one of the exceptions. Probably. How can I know that?

The actual language from the FDA authorization just issued ( Read more [2,750 words] )

This post brought to you by the 218 readers who funded my writing it – thank you all so much! You can see who they are at my Patreon page. If you're not one of them, and would be willing to chip in so I can write more things like this, please do so there.

Please leave comments on the Comment Catcher comment, instead of the main body of the post – unless you are commenting to get a copy of the post sent to you in email through the notification system, then go ahead and comment on it directly. Thanks!

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LEARNING

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siderea

I don't know who needs to know about this, but:

I just discovered the Android app "Periodically". It's described as an "event logger". It's for keeping track of when a recurring thing has happened, and figuring out what the average time is between occurrences. You just keep it updated each time the event happens, and it will do the math for you to figure out the frequency, and even give you a notification when it predicts the event is likely to happen again. If you're tracking more than one thing, it will try to suss out correlations for you.

I mention because twenty five years ago or so, I needed exactly this functionality and could not find any application that would do what I needed, so I wrote a thing for myself, and since then a lot of people I've mentioned it to have wondered where they can get one like it. Mine was Mac/Palm Pilot, so not of much use to most people, especially these days.
Lo, somebody seems to have realized the need for this functionality, and brought it to the market. So I thought I'd mention.

Now, in this day and age, a lot of people, especially in the US, are concerned with security. Especially if they're tracking something to do with their health. This app is not specific to health, so nothing about it immediately reveals that it is storing health information on casual inspection; you could use some sort of other term for whatever health condition it is you are actually tracking. So, for instance, If you were tracking how often your migraines happened, you could call that "new box of cereal".

This app defaults to local-only data storage on your Android device, and the developer claims that it only collects "app activity" for analytics, and shares nothing with third parties. It outputs CSV and has an option to back up to Google Drive.

I haven't tried it myself, but it has a rating of 4.6 stars out of five on the Play Store.

Reviewers on the Play Store note that tracker apps that are specific to the kind of event – such as health- specific loggers – often have needless complexity, and often some weird ideas about graphic design. They praise this app for its clean, elegant look and simple, effective functionality.

In addition to its obvious applicability to episodic health conditions, it strikes me as potentially extremely useful in one of the trickier parts of prepping: figuring out one's burn rate of resources. I think I might trial it to help me figure out how often I should expect to have to buy a fresh bale of toilet paper and how long the big bottle of ibuprofen will last me.

xkcd_feed

https://xkcd.com/3140/

Welcome to the Linguistics Department - It has been [2] [DAYS] since someone noticed that the Biology Department sign has a one-day-long singular/plural disagreement after it resets.

https://xkcd.com/3140/

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Posted by Abbadon

https://killsixbilliondemons.com/weekly-updates-resume-9-12/

https://killsixbilliondemons.com/?p=4105

Hi ya’ll, starting this friday (sept 12th) we’ll be returning to weekly updates on Fridays again. Thanks for your patience with the change in formats, but the big pages + breaks for bulk updates have been way too stressful for me to handle, so we’re going to go back to weeklies for the foreseeable future.

https://killsixbilliondemons.com/weekly-updates-resume-9-12/

https://killsixbilliondemons.com/?p=4105