Home on lxs.io
https://lxs.io/
Recent content in Home on lxs.ioHugo -- gohugo.ioen-usSun, 03 Nov 2019 00:00:00 +0000Fibonacci numbers satisfy Benford's law
https://lxs.io/posts/2019-11-03-fibonacci-numbers-satisfy-benford-s-law/
Sun, 03 Nov 2019 00:00:00 +0000https://lxs.io/posts/2019-11-03-fibonacci-numbers-satisfy-benford-s-law/And we don’t need too many elements of the Fibonacci sequence too see this.
The distribution of the first digits of the first n Fibonacci numbers
After only 100 elements the relative frequencies start to look familiar.
First digit Rel. frequency 1 30% 2 18% 3 13% 4 9% 5 8% 6 6% 7 5% 8 7% 9 4% Who has worked with Al Pacino the most times?
https://lxs.io/posts/2019-10-27-who-has-worked-with-al-pacino-the-most-times/
Sun, 27 Oct 2019 00:00:00 +0000https://lxs.io/posts/2019-10-27-who-has-worked-with-al-pacino-the-most-times/No, it’s not Robert De Niro.
Name Coappearances John Cazale 6 James Bulleit 6 Robert De Niro 5 Richard Bright 5 James Tolkan 5 Diane Keaton 4 James Caan 4 Sofia Coppola 4 Charles Durning 4 Talia Shire 4 Here’s the source.The period of a physical pendulum
https://lxs.io/posts/2019-10-20-the-period-of-a-physical-pendulum/
Sun, 20 Oct 2019 00:00:00 +0000https://lxs.io/posts/2019-10-20-the-period-of-a-physical-pendulum/The period \(T\) of a physical pendulum is given by
$$T = 2 \pi \sqrt{\frac{I}{m g R}} \mathrm{,}$$
where
\(I\) is the moment of inertia of the pendulum about the pivot point. \(m\) is the mass of the pendulum. \(g\) is the accelleration of gravity. \(R\) is the distance between the pivot point and the center of mass. The formula holds for all sizes and shapes; however, when someone says the word “pendulum”, we usually think of a bob attached to a rod, in which case the formula can be simplified as follows.How accurate is Stirling's approximation?
https://lxs.io/posts/2019-10-13-how-accurate-is-stirling-s-approximation/
Sun, 13 Oct 2019 00:00:00 +0000https://lxs.io/posts/2019-10-13-how-accurate-is-stirling-s-approximation/The question whether an approximation is good should not be asked in isolation, because the answer depends on the purpose. For example, \(\mathrm{e}^x \approx 1 + x\) works well for small (positive) \(x\). How small? If we want to keep the relative error below 10%, \(x \lt 0.5\) will do fine. But if we need more precision, the range gets narrower, e.g., \(x \lt 0.15\) if the tolerance is smaller than 1%.Number of trials to first success
https://lxs.io/posts/2019-10-06-number-of-trials-to-first-success/
Sun, 06 Oct 2019 00:00:00 +0000https://lxs.io/posts/2019-10-06-number-of-trials-to-first-success/How many times, on average, do we need to roll an n-sided die until we get an X?
The probability of getting an X for the first time on the i-th trial is
$$\left( 1 - \frac{1}{n} \right)^{i - 1} \frac{1}{n} .$$
Let \(p = \frac{1}{n}\). The mean number of trials to first success is
$$p \sum_{i = 1}^{\infty} i (1 - p)^{i - 1} = p \left[ - \sum_{i = 1}^{\infty} (1 - p)^i \right]^{\prime} = p \left( \frac{p - 1}{p} \right)^{\prime} = \frac{1}{p} = n .How many electrons are there in the Solar System?
https://lxs.io/posts/2019-09-29-how-many-electrons-are-there-in-the-solar-system/
Sun, 29 Sep 2019 00:00:00 +0000https://lxs.io/posts/2019-09-29-how-many-electrons-are-there-in-the-solar-system/I was asked this question in an interview some twenty years ago. This is a typical Fermi problem. Solving it means making a good estimate with little data. Of course, there are many possible approaches; my answer went something like this…
Because
the Sun makes almost all of the Solar System’s mass (more than 99%) and the Sun consists mostly of hydrogen (around three quarters is hydrogen and the rest is mostly helium; this is a good enough approximation) the number of electrons roughly equals the mass of the Sun \(m_S\) divided by the mass of the hydrogen atom \(m_H\).Contact
https://lxs.io/contact/
Mon, 01 Jan 0001 00:00:00 +0000https://lxs.io/contact/https://gitea.com/lxs https://mastodon.social/@lxs https://pixelfed.social/lxs https://peertube.social/accounts/lxs