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# Day One to Tinderbox

I’ve kept a Daybook using Tinderbox for years, and wish to continue doing so. The problem is I know that Day One exists and is really good at what it does. The conflict between smooth and easy entry using Day One and the comfort and flexibility of Tinderbox keeps me up at night. Well, not really, but it does cause me to waffle between the two.

I switched to Day One again a month ago and I really do love it. Day One automatically logs my location and the current weather, is always just a click away on either the iPhone or Mac, and is pretty to look at. The trouble is that I can’t to some of the nifty analytics that I do with Tinderbox. If only there were a way to use both!

I decided today that I’d find a way, and came up with something that, while not pretty, works.

Basically, I export the Day One entries to an OPML file and then drop that file onto my Tinderbox Daybook and run a couple of quick “stamps” to clean things up. Here’s an overview of how it works.

I originally wanted to process the native Day One export but found the file needed way too much manipulation if it was going to be useful to Tinderbox. I wanted an easy way to get dates, locations, and weather into Tinderbox attributes and not just as part of each note’s text. OPML seemed an obvious choise. I found Nathan Grigg’s dayone_export. It’s a python script that parses the Day One data file and generates an export file using a simple templating system (Jinja2). Day One Export didn’t come with an OPML template but using the Markdown example it was easy enough to create one. It looks like this…

The template file is saved locally (on my Mac) as ~/.dayone_export/default.opml.

To generate the OPML I run the following…

# dayone_export --output dayone-export.opml --format=opml --after 2014-04-01 ~/Path/To/Day One/Journal.dayone

Then I drop the output file (dayone-export.opml in this case) onto my open Tinderbox daybook document. Note that I included some additional attributes in the OPML template for things like Weather, etc. When importing, Tinderbox will use or create User Attributes matching each of them automatically. This gives me something easy to work with. Also note that I use the –after attribute, which will let me export regularly without dealing with duplicates.

To clean things up in Tinderbox, I apply the following “stamp” to each of the imported notes:

$Text=$Name;$Name=$Text.split(".").at(0)

This stamp…

1. Copies the note’s Name to its Text

2. Changes the Name to only contain the first sentence

Then I just drag the notes into the proper dated container in Tinderbox and I’m done. Sure, it’s a little kludgy and probably prone to errors but at least now I can keep using both tools and enjoy the benefits of each.

It works, but I hope someone smarter than me comes up with something better one day.


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# Definition:Fermat Pseudoprime/Base 4/Mistake

## Source Work

The Dictionary
$217$

## Mistake

The second smallest pseudoprime to base $4$ ($15$ is the smallest).
$4^{216} - 1$ is divisible by $217$ although $217$ is not prime but $7 \times 31$.

## Correction

The author is correct in that $217$ is indeed not prime, but $7 \times 31$.

Also, $15$ is the smallest pseudoprime to base $4$.

However, $217$ is not a pseudoprime to base $4$.

It is in fact the $3$rd pseudoprime to base $5$:

$5^{216} - 1$ is divisible by $217$.

In David Wells: Curious and Interesting Numbers (2nd ed.), this section has been removed.

format: plain


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Question

# Question 8 ABCD is a rectangle formed by points A (-1,-1), B(-1,4), C(5,4) and D(5,-1). P, Q, R, and S are mid-points of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS a square, rectangle or rhombus? Justify your answer.

Open in App
Solution

## Coordinates of P can be calculated as follows: [Since, mid-point of the line segment joining the points (x1,y1) and (x2,y2) is (x1+x22,y1+y22)] (−1−12,4−12)=(−1,32) Coordinates of Q can be calculated as follows: (5−12,4+42)=(2,4) Coordinates of R can be calculated as follows: (5+52,4−12)=(5,32) Coordinates of S can be calculated as follows: (5−12,−1−12)=(2,−1) Length of PQ can be calculated as follows: PQ=√(2+1)2+(4−32)2 =√(2+1)2+(52)2 =√(9)+(254) =√612 Similarly, QR can be calculated as follows: QR =√(5−2)2+(32−4)2 =√32+(−52)2 =√612 Similarly, PS =√(2+1)2+(−1−32)2 =√32+(−52)2 =√612 Similarly, QR can be calculated as follows: SR =√(2−5)2+(−1−32)2 =√(−3)2+(−52)2 =√612 The above values shows that PQ = QR = RS = PS, i.e. all sides are equal. Now let us calculate the diagonals. PR=√(5+1)2+(32−32)2 =√62=6 QS=√(2−2)2+(−1−4)2 =√(−5)2=5 Hence, it is clear that while all sides are equal, diagonals are not equal. So, the given figure is a rhombus.

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# Light neutral mesons probed to high pT

30 April 2021

A report from the ALICE experiment

Neutral pion (π0) and eta-meson (η) production cross sections at midrapidity have recently been measured up to unprecedentedly high transverse momenta (pT) in proton–proton (pp) and proton–lead (p–Pb) collisions at √sNN = 8 and 8.16 TeV, respectively. The mesons were reconstructed in the two-photon decay channel for pT from 0.5 and 1 GeV up to 200 and 50 GeV for π0 and η mesons, respectively. The high momentum reach for the π0 measurement was achieved by identifying two-photon showers reconstructed as a single energy deposit in the ALICE electromagnetic calorimeter.

In pp collisions, measurements of identified hadron spectra are used to constrain perturbative predictions from quantum chromodynamics (QCD). At large momentum transfer (Q2), one relies in these perturbative approximations of QCD (pQCD) on the factorisation of computable short-range parton scattering processes such as quark–quark, quark–gluon and gluon–gluon scatterings from long-range properties of QCD that need experimental input. These properties are modelled by parton distribution functions (PDFs), which describe the fractional-momentum (x) distributions of quarks and gluons within the proton, and fragmentation functions, which describe the fractional-momentum distribution of quarks or gluons for hadrons of a certain species.

In p–Pb collisions, nuclear effects are expected to significantly affect particle production, in particular at small parton fractional momentum x, compared to pp collisions. Modification at low pT (~1 GeV), usually attributed to nuclear shadowing (CERN Courier March/April 2021 p19), can be parameterised by nuclear parton distribution functions (nPDFs). However, since high parton densities are reached at the LHC, the Colour Glass Condensate (CGC) framework is also applicable at low pT (x values as small as ~5 × 10–4), which predicts strong particle suppression due to saturation of the parton phase space in nuclei. Above momenta of about 10 GeV/c, measurements in p–Pb collisions can also be sensitive to the energy loss of the outgoing partons in nuclear matter.

The nuclear modification factor (RpPb), shown in the lower panel of the figure, was measured as the ratio of the cross sections in p–Pb and pp collisions normalised by the atomic mass number. Below 10 GeV, RpPb is found to be smaller than unity, while above 10 GeV it is consistent with unity. The measurement is described by calculations over the full transverse momentum range and provides further constraints to the nPDF parameterisations for lower than about 5 GeV. The direct comparison of the neutral pion cross section in pp collisions at 8 TeV, with pQCD calculations shown in the upper panel of the figure, reveals differences in the low to intermediate pT range, which, however, cancel in RpPb, since similar differences are also present for the p–Pb cross section. Future high-precision measurements are ongoing using the large dataset from pp collisions at 13 TeV, providing further constraints to pQCD calculations.


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# Synopsis: Crystalline coherence

Atoms subjected to strong optical fields exhibit splitting of energy levels. The same effect has now been observed when an atom moves through the periodic field of a crystal lattice.

When a very strong electromagnetic field interacts with an atom, the usual picture of energy levels breaks down. Instead, the states of the atom must be considered an intimately coupled system of atom + photons. If a two-level system of these so-called “dressed” states interacts with a third level by absorbing or emitting photons, then the spectrum can show an energy level splitting called Autler-Townes doublets.

As Y. Nakai et al. report in Physical Review Letters, however, these spectral effects do not only occur in strong optical fields—atoms flying through a crystal lattice can experience the same splitting. The research team shot ${\text{Ar}}^{16+}$ ions through a single crystal of silicon at different angles relative to the crystal planes. The ions pass through the spatially periodic field created by the silicon atoms in the crystal lattice, but in the moving reference frame of the $\text{Ar}$ ions, the crystal field looks like a time varying field. By carefully choosing the angle that the ions pass through the crystal, the researchers could generate two different frequency components of the moving field to create the needed double resonance.

Nakai et al. were able to detect the coherent excitation of the ions by the crystal field in the fraction of ${\text{Ar}}^{16+}$ that survived the journey: successful excitation meant that an ion would be subsequently further ionized to ${\text{Ar}}^{17+}$ and higher by collisions with silicon atoms. Distinct Autler-Townes doublet dips in the survival fraction were observed as the excitation energy was varied. The results offer a new method for manipulating atomic state populations in energy ranges where optical sources are difficult to find. – David Voss

### Announcements

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## Subject Areas

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# Tag Info

3

Let me first review what I would consider to be the standard definitions of these terms. A space $X$ is path-connected if for any $x,y\in X$ there is a continuous map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. A space $X$ is arc-connected if for any $x,y\in X$ there is an injective continuous map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. That ...

3

$\cos(nt)=T_n(\cos t)$, where $T_n$ is the $n$-th Chebyshev polynomial. So your curve can be written $x=T_m(u)$ and $y=T_n(u)$, where $u=\cos(t)$. Since $T_n(x)=T_n(T_m(u))=T_{nm}(u)=T_{mn}(u)=T_m(T_n(u))=T_m(y)$, your curve is (part of) the algebraic curve given by $T_n(x)=T_m(y)$. This is not the lowest possible degree. For instance, if $lcm(m,n)=am=bn$...

2

Use Eisenstein to deduce that $y^3-x^3(x+1)$ is irreducible. As an alternative, note that the polynomial (in the variable $y$) is primitive, of degree $3$ and has no roots. Then invoke the Gauß lemma.

2

The correct statement is that every nonempty compact metrizable space is a continuous image of the ternary Cantor set (Alexandroff-Hausdorff theorem).

1

Usual convention: \begin{align*} w &= a e^{i\theta} \\ c &= (a-b) e^{i\theta} \\ \frac{z-c}{w-c} &= e^{-i\phi} \\ a\theta &= b\phi \\ z &= (a-b)e^{i\theta}+b e^{-i\left( \frac{a-b}{b}\right) \theta} \end{align*} $w$: point of contact $c$: centre of blue circle $z$: locus of the initial point of contact, i.e. the hypocycloid ...

1

Indeed the statements are equivalent. So we have in particular that the following are equivalent for $X$: $X$ is a compact, connected, locally connected metric space. $X$ is a compact, connected, locally pathwise connected metric space. $X$ is a compact, connected, locally arcwise connected metric space. $X$ is a compact, path-connected, locally connected ...

1

That's not quite correct. For example, your curve could be empty, or a single point, or a line segment. If the Hessian matrix is positive definite on some convex region $R$, then $F$ is a convex function there. If $R$ is bounded, contains $\gamma$ and $\inf_R F < 0 \inf_{\partial R} F$, then $\gamma$ is the boundary of the convex open set $\{(x,y) \... 1 No. Actually I think you sort of didn't ask the question you meant to ask; we give a trivial counterexample to the question as asked and then a not quite as trivial counterexample to the question I think you meant to ask. First, let$J$be the unit circle, and let$K$be the circle of radius$\epsilon/2$centered at the point$1$. Every point of$K$is ... 1$C$doesn't have to be "nice".$E$is closed and bound in${\Bbb R}^2$and so there are 2 points$c,d\in E$such that their distance is maximal,$m$(distance is a bound and continuous function on the compact set$E\times E$so it has a maximum). It's pretty obvious that any such pair$c,d$with distance$m$must be on the boundary (if either or both are ... 1 I will assumme that a rotation and transformation means an isometry here, and that the curve$C$must be sufficiently nice (i.e. smooth). Suppose such a curve and a rotation and transformation exist. Let$\partial E$denote the boundary of$E$, i.e. it is the union of$C$and the line segment connecting$a$and$b$. Let$C'$denote the rotated and ... 1 No. Consider the curve $$\gamma(t) = (s \cos t, s \sin t), 0 \le t \le \frac{L}{2\pi s}$$ where$s < r$. its curvature is$\frac{1}{s}$, which is clearly unbounded. If you don't like that the path intersects itself, just make$s$a very slowly increasing function of$t$with mean$S$. Then the curvature will be approximately$\frac{1}{S}\$.

Only top voted, non community-wiki answers of a minimum length are eligible


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2.2 The structure of ggplot commands

As calls to the ggplot() function are often quite long and take many different arguments, it is helpful to understand the function’s generic structure before looking at some concrete examples. A generic template for creating a graph with the ggplot() function has the following structure:

# Generic ggplot template:
ggplot(data = <DATA>) +                              # 1. specify data set to use
<GEOM_fun>(mapping = aes(<MAPPING>),               # 2. specify geom + mappings
<arg_1 = val_1, ..., arg_n = val_n>) +  # - optional arguments to geom
...                                                # - additional geoms + mappings
<FACET_fun> +                                      # - optional facet function
<LOOK_GOOD_fun>                                    # - optional themes, colors, labels, etc.

The generic template includes the following parts:

• <DATA> is a data frame or tibble that contains the data that is to be plotted.

• <GEOM_fun> is a function that maps data to a geometric object (“geom”) according to an aesthetic mapping that is specified in aes(<MAPPING>). (A mapping specifies a relation between two entities. Here, the mapping specifies the correspondence of variables to graphical elements, i.e., what goes where.)

• A geom’s visual appearance (e.g., colors, shapes, sizes, …) can be customized

1. in the aesthetic mapping (when varying visual features according to data properties), or
2. by setting its arguments to specific values in <arg_1 = val_1, ..., arg_n = val_n> (when remaining constant).
• An optional <FACET_fun> uses one or more variable(s) to split a complex plot into multiple subplots.

• A sequence of optional <LOOK_GOOD_fun> adjust the visual features of plots (e.g., by adding titles and text labels, color scales, plot themes, or setting coordinate systems).

Actually, a lot of the generic template is not necessary for using ggplot() for generating a graph. A minimal template of a ggplot() command can be reduced to the following structure:

# Minimal ggplot template:
ggplot(<DATA>) +             # 1. specify data set to use
<GEOM_fun>(aes(<MAPPING>)  # 2. specify geom + mappings

A comparison of the generic and the minimal templates shows that large parts of a typical ggplot() command are optional. In fact, the bare essentials only include some <DATA>, at least one <GEOM_fun>, and its required mappings in aes(<MAPPING>). This creates the basic visualization specified by the geom and its variable mappings. All other arguments (e.g., adding aesthetic elements, facetting, titles and labels) provide additional functionality and fluff. Thus, when creating a new visualization, it always makes sense to start with a minimal working recipe of a ggplot() command and then add more elements and fluff. And as looks can have a major impact — for human beings and other animals — we must not underestimate the effects of visual fluff on communication.


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# The internal branch lengths of the Kingman coalescent

@article{Dahmer2013TheIB,
title={The internal branch lengths of the Kingman coalescent},
author={Iulia Dahmer and G{\"o}tz Kersting},
journal={arXiv: Probability},
year={2013}
}
• Published 19 March 2013
• Mathematics
• arXiv: Probability
In the Kingman coalescent tree the length of order $r$ is defined as the sum of the lengths of all branches that support $r$ leaves. For $r=1$ these branches are external, while for $r\ge2$ they are internal and carry a subtree with $r$ leaves. In this paper we prove that for any $s\in\mathbb{N}$ the vector of rescaled lengths of orders $1\le r\le s$ converges to the multivariate standard normal distribution as the number of leaves of the Kingman coalescent tends to infinity. To this end we use…

## Figures from this paper

The joint fluctuations of the lengths of the Beta$(2-\alpha, \alpha)$-coalescents
• Mathematics
• 2020
We consider Beta$(2-\alpha, \alpha)-n$-coalescents with parameter range $1 <\alpha<2$. The length $\ell^{(n)}_r$ of order $r$ in the Beta$(2-\alpha, \alpha)-n$-coalescent tree is defined as the sum
The total external length of the evolving Kingman coalescent
• Mathematics
• 2014
The evolving Kingman coalescent is the tree-valued process which records the time evolution undergone by the genealogies of Moran populations. We consider the associated process of total external
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• Physics
• 2020
We present approximation methods which lead to law of large numbers and fluctuation results for functionals of $\Lambda$-coalescents, both in the dust-free case and in the case with a dust component.
J an 2 02 2 The joint fluctuations of the lengths of the Beta ( 2 − α , α )-coalescents ∗
• Mathematics
• 2022
We consider Beta(2− α, α)-coalescents with parameter range 1 < α < 2 starting from n leaves. The length l (n) r of order r in the n-Beta(2 − α, α)-coalescent tree is defined as the sum of the lengths
Probabilistic aspects of Λ-coalescents in equilibrium and in evolution
• Physics
• 2020
We present approximation methods which lead to law of large numbers and fluctuation results for functionals of Λ-coalescents, both in the dust-free case and in the case with a dust component. Our
Can the Site-Frequency Spectrum Distinguish Exponential Population Growth from Multiple-Merger Coalescents?
• Physics
Genetics
• 2015
Estimates of statistical power indicate that exponential and algebraic growth can indeed be distinguished from multiple-merger coalescents, even for moderate sample sizes, if the number of segregating sites is high enough.
Statistical properties of the site-frequency spectrum associated with Lambda-coalescents
• Mathematics
• 2013
Statistical properties of the site frequency spectrum associated with Lambda-coalescents are our objects of study. In particular, we derive recursions for the expected value, variance, and covariance

## References

SHOWING 1-10 OF 17 REFERENCES
Asympotic behavior of the total length of external branches for Beta-coalescents
• Mathematics
• 2012
It is proved that $n^{{\alpha}-2}L^{(n)_{ext}$ converges in $L^2$ to $\alpha(\alpha-1)\Gamma(\alpha)$ as a consequence of the asymptotic behavior of the total length of the associated $n$-coalescent.
Asymptotics of the allele frequency spectrum associated with the Bolthausen-Sznitman coalescent
• Mathematics
• 2007
This paper proves the law of large numbers-type results for the allele frequency spectrum when the coalescent process is taken to be the Bolthausen-Sznitman coalescent and shows that $n^{-1}(\log n) N_1(n) {\stackrel{p}{\rightarrow}} \rho$ as $n \to \infty$.
The Total External Branch Length of Beta-Coalescents†
• Physics
Combinatorics, Probability and Computing
• 2014
It turns out that the fluctuations of the external branch length follow those of τn2−α over the entire parameter regime, where τn denotes the random number of coalescences that bring the n lineages down to one.
On the Total External Length of the Kingman Coalescent
• Mathematics
• 2011
We prove asymptotic normality of the total length of external branches in the Kingman coalescent. The proof uses an embedded Markov chain, which can be described as follows: Take an urn with black
BETA-COALESCENTS AND CONTINUOUS STABLE RANDOM TREES
• Mathematics
• 2007
It is proved that Beta-coalescents can be embedded in continuous stable random trees based on a construction of the Donnelly-Kurtz lookdown process using continuous random trees, which is of independent interest.
On Asymptotics of Exchangeable Coalescents with Multiple Collisions
• Mathematics
Journal of Applied Probability
• 2008
We study the number of collisions, X n , of an exchangeable coalescent with multiple collisions (Λ-coalescent) which starts with n particles and is driven by rates determined by a finite
Markov Chains and Mixing Times
• Mathematics
• 2008
This book is an introduction to the modern approach to the theory of Markov chains. The main goal of this approach is to determine the rate of convergence of a Markov chain to the stationary
Statistical tests of neutrality of mutations.
• Biology
Genetics
• 1993
From these properties, several new statistical tests based on a random sample of DNA sequences from the population are developed for testing the hypothesis that all mutations at a locus are neutral.


text:
## jose206 Group Title which is the value of the gamma function in 1/3? 2 years ago 2 years ago

1. JamesJ

You want to know the value of this? $\Gamma(1/3) = \int_0^\infty t^{-2/3}e^{-t} \ dt$

2. jose206

yeah..!!!

3. JamesJ

Well, it can't be calculated exactly but here's an approximation of it: http://www.wolframalpha.com/input/?i=gamma%281%2F3%29

4. jose206

thank you


text:
# +1 order efficiency in a blazed transmission grating

1. Jan 22, 2015

### Ljungholm

1. The problem statement, all variables and given/known data
I have a blazed transmission-grating with a known blaze-angle θB and refractive index n. Is there a way (as with a blazed reflection-grating) to calculate at what incident angle maximum efficiency will be achieved for the m'th order diffraction given a wavelength?

Edit: Of course the N of grooves/mm is also known.

Thank you

Last edited: Jan 22, 2015
2. Jan 27, 2015


text:
mersenneforum.org Prime95 Mysteriously Abandoned Work
 Register FAQ Search Today's Posts Mark Forums Read

 2012-05-17, 17:56 #1 ClownRoyal     May 2012 Idaho 1210 Posts Prime95 Mysteriously Abandoned Work Hello, Please help me get back my work. My quad-core runs 4 workers and last night, 3 of the workers inexplicably dropped what they were doing and moved on to different jobs. I have three separate LL tests in various stages of completion - the largest (58mil+) was nearing completion. I was screwing around last night with mfaktc around the time that it happened, but I can't imagine what went wrong - it uses a separate directory/worktodo/results/etc. I can still see the bu and bu2 files in my prime95 directory - actually three files each, one without extension. Browsing the forums, I found a similar post where a guy was able to modify the worktodo.txt but I don't know the exact syntax that should be entered for these exponents. Exponents are listed below. Please help save my work if you can. 53109619 55421753 58056553 Thanks, -cr "Search on!"
 2012-05-17, 19:22 #2 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
Gravemind154197029 DLL, 82.40%2012-05-07 18:46102012-05-17 02:562012-05-18 02:562012-05-19 13:062
 This is an example from my own account (this computer is named Gravemind). The difference is where I've written 'name="hidden"', you will see a long hex key; that is your AID. (Never post that on the forum.) The AID corresponds to the exponent which is written in the "sorttable_customkey" field, which is 54197029 in my example. -------------------------------------------------------------------------------------- Having found the AID for each exponent you lost (doesn't matter which method), now you must shutdown Prime95 ("Test/Exit", don't just close the window) to modify the text files. Put the following line (based on my example in the manual method above) into worktodo.txt, one line for each exponent: Code: Test=AID,54197029,72,1 The first field is the AID, the second is the exponent; the third is the trial factoring depth (so it's 72 for two of those exponents, but 71 for another) and the last field is 0 for no P-1 complete or 1 for P-1 is complete. If you had already started the LL test for all three, then you can put ,72,1 for each of them and that won't screw anything else up. So, for example if my AID was DFCAB8000CFDE06B58203, it would look like Code: Test=DFCAB8000CFDE06B58203,54197029,72,1` You should also see the equivalent lines for whatever you are working on now; Prime95 will work on them in the order they are listed in the file, so if you want to resume these older ones first, put them ahead/above of the new exponents.
2012-05-17, 19:33   #4
ClownRoyal

May 2012
Idaho

148 Posts
slapping myself...

Quote:
 Originally Posted by Dubslow Fortunately, all three exponents are still assigned to you. I have absolutely no idea why Prime95 would do something like that.
Gosh, thanks, Dubslow! You and I were both working hard typing away at the same time. I appreciate your reply and I hope I did not take up too much of your time.

-cr

"Search On!"

2012-05-17, 19:37   #5
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts

Quote:
 Originally Posted by ClownRoyal 1 - This number tells something about how the exponent has been handled. I am unclear of the meaning and this can be taken as my best guess as to what to enter.

As I said in my post above which you apparently missed while typing your second response, this indicates whether or not P-1 has been complete. In your case, all three have had P-1 complete. (If you had set 0, then Prime95 would have started a day or two of P-1 work before continuing the test.)

If I didn't know better I'd say you copied my post :razz: but at least you were able to work it out on your own. How did figure out where to find the AID in the page source? Did you see that elsewhere on the forum?

More cross posting!!!

Last fiddled with by Dubslow on 2012-05-17 at 19:37

 2012-05-17, 19:48 #6 ClownRoyal     May 2012 Idaho 22×3 Posts CrossPosters Anonymous Dubslow, Yes, I saw another forum post where someone had used the "view source" method to get the AID but it did not state how to generate the rest of the test entry. So I just kept poking around the forum and around mersenne.org until I figured out what to do with the rest of it. Hey, out of curiosity; what's the worst that can happen if someone were to post their entire AID on the forum? -cr "Search On!"
 2012-05-17, 20:22 #7 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2012-05-18, 05:06   #8

"Richard B. Woods"
Aug 2002
Wisconsin USA

22×3×641 Posts

Quote:
 Originally Posted by ClownRoyal Hey, out of curiosity; what's the worst that can happen if someone were to post their entire AID on the forum?
Even if we can't think of much, just keep the AID confidential in case someone else does.

 2012-05-18, 05:41 #9 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 19×541 Posts The worst can happen is someone else registers your exponent using your assignment key, you don't know and continue working, and find a prime. If I am the other guy I would say you poached me and I am 99% of work done (even if I didn't move a byte) and I may grab your \$50k if I am enough clever. When no prime is found, who cares about dry bones? Do you want my ~70GHzDays of credit? Be my guest, I won't cut your head for it. I made another in few days with a good GPU. But when the bone is juicy, all the dogs will start grrrrrrrrowlingggg...
 2012-05-18, 13:54 #10 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 41·199 Posts Are there any clues in results.txt or prime.log indicating why prime95 abandoned the exponents? If there is a bug, I need to fix it.
2012-05-19, 04:51   #11
ClownRoyal

May 2012
Idaho

11002 Posts

Quote:
 Originally Posted by Prime95 Are there any clues in results.txt or prime.log indicating why prime95 abandoned the exponents? If there is a bug, I need to fix it.
• results.txt
I do not see anything out of the ordinary in results.txt, nor any reference to the exponents in question.
• prime.log
I found nothing untoward in prime.log either. Just the usual "sending expected completion date...", "getting assignment from server..." etc. around the time in question. No reference to the abandoned exponents other than their original assignment and subsequent eta reports.

I wish I could offer more, but there just aren't any clues as far as my amateur eyes can tell.

@Prime95: Would you like to see the log file for yourself? I wouldn't put it past me to be missing something here. This system is still new to me.

-cr

"Search On!"

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# STRENGTHS OF $^{13}C^{16}O_{2}$ AND $^{12}C^{16}O_{2}$ LINES AT 4.3-4.5 $\mu$m

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/10622

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 Title: STRENGTHS OF $^{13}C^{16}O_{2}$ AND $^{12}C^{16}O_{2}$ LINES AT 4.3-4.5 $\mu$m Creators: Braun, W. C.; Fridovich, B.; Smith, G. R.; Champion, E. E.; Rinsland, C. P.; Chen, Da-Wun; Rao, K. Narahari Issue Date: 1978 Publisher: Ohio State University Abstract: Measurements on the intensities of single lines in the $\nu_{3}$ and $\nu_{2} + \nu_{3} - \nu_{2}$ bands of $^{13}C^{16}O_{2}$ and the $\nu_{3}$ band of $^{12}C^{16}O_{2}$ have been made on grating spectrometers in two laboratories. For the $^{13}C^{16}O_{2}$ measurements enriched samples have been used. As a reference, $N_{2} O$ lines in the $\nu_{3}$ band have also been measured, and the results compared with intensity values available in the literature. Description: Author Institution: National Oceanic and Atmospheric Administration, National Environmental Satellite; Department of Physics, The Ohio State University URI: http://hdl.handle.net/1811/10622 Other Identifiers: 1978-TE-05


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chrono::particlefactory::ChRandomShapeCreatorConvexHulls Class Reference

## Description

Class for generating convex hulls with variable chordal size and aspect ratios.

By default uses constant distributions (all hulls are equally sized) but you can provide your distributions.

#include <ChRandomShapeCreator.h>

Inheritance diagram for chrono::particlefactory::ChRandomShapeCreatorConvexHulls:
Collaboration diagram for chrono::particlefactory::ChRandomShapeCreatorConvexHulls:

## Public Member Functions

virtual std::shared_ptr< ChBodyRandomGenerate (ChCoordsys<> mcoords) override
Function that creates a random ChBody particle each time it is called.

void SetNpoints (int mnpoints)
Set the number of random vertexes used to generate each random convex hull. More...

void SetChordDistribution (std::shared_ptr< ChDistribution > mdistr)
Set the statistical distribution for the radius.

void SetSizeRatioYZDistribution (std::shared_ptr< ChDistribution > mdistr)
Set the statistical distribution for scaling on both Y,Z widths (the lower <1, the thinner, as a needle).

void SetSizeRatioZDistribution (std::shared_ptr< ChDistribution > mdistr)
Set the statistical distribution for scaling on Z width (the lower <1, the flatter, as a chip).

void SetDensityDistribution (std::shared_ptr< ChDistribution > mdistr)
Set the statistical distribution for the random density.

Public Member Functions inherited from chrono::particlefactory::ChRandomShapeCreator
virtual std::shared_ptr< ChBodyRandomGenerateAndCallbacks (ChCoordsys<> mcoords)
This function does RandomGenerate and also executes the the custom callback, if provided. More...

Set the callback function to execute at each each particle generation.

Set if the created particles must include the collision shape(s). More...

Set if the created particles must include the visualization asset(s). More...

## Additional Inherited Members

Protected Attributes inherited from chrono::particlefactory::ChRandomShapeCreator

## ◆ SetNpoints()

 void chrono::particlefactory::ChRandomShapeCreatorConvexHulls::SetNpoints ( int mnpoints )
inline

Set the number of random vertexes used to generate each random convex hull.

Note that the final number of vertexes in the hull might be lower since random points that fall inside the convex hull are not used.

The documentation for this class was generated from the following file:
• /builds/uwsbel/chrono/src/chrono/particlefactory/ChRandomShapeCreator.h


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# Bad results in a Loan Default Prediction Problem

I have a dataset consisting of 23 features for a number of clients :

• Client ID
• yearly financial ratios
• couple of qualitative features
• and a binary default variable

I'm trying to create a model that predicts whether the client will default or not , i tried , for starters , with svm and xgboost but the results were pretty bad (high accuracy and very bad recall - F1) due to the imbalance of the data (~minority about 6%).
A problem that i tried to fix by trying SMOTE,undersampling and trying to adjust the built-in tools for balancing data like class_weight parameter for SVC and scale_pos_weight for xgboost respectively but the results are still bad .( whatever i do if the recall of one of the two classes rises the other one decreases ).

Also the Correlations between the predictors and the target are very low ( 0.18 is the highest ) ,
i want to also know, at which point can i tell for sure that the data at hand is not explanatory enough to predict a certain variable?

• Just for clarification purposes, is your question related to these particular modelling choices (XGBoost, SVC) and their performance on this dataset? Or are you just trying to build a good predictive model on the data you have? In other words, does the solution to your problem have to include XGBoost/SVC or will any method with good enough results suffice? – Emil Jul 11 '19 at 9:09
• No , i'm not restricted to these two methods. I'm planning to try other methods and compare between all of them and eventually choose the best one so any suggestions are welcome. – Whatever Jul 11 '19 at 9:29

3. Hopefully now you have decreased your predictors from 23 to e.g. 5 or 6, and you can then perform a simple logistic regression to estimate probability of default for future customers. To measure how well this method performs, you could split your data to development/validation samples, where you take for example 80% of the data and use it to train the model, then test the results on the remaining 20%. In general, a good model is one with discriminatory power $$\geq 70\%$$.


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1. # Equivalent definitions for $BMO$

The space of functions of bounded mean oscillation $BMO$ is defined by the BMO norm $$\label{norm} ||f||_{BMO} = \sup_{\text{cubes }Q} \frac{1}{|Q|} \int_Q |u(y) - u_Q| dy$$

But an equivalent definition is to take the sup over balls instead of cubes ...

2. # The space $BMO$
## Introducing $BMO$
The space of functions of bounded mean oscillation or $BMO$ arises when studying the space of functions whos deviation from the mean over cubes is bounded. In ways $BMO$ is similar to $L^\infty$, and it is often used as a replacement, however functions in $BMO$ may be ...


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In how many ways can you arrange the five numbers 1, 2, 3, 4, 5?
Think about the number of choices: five for the first choice, four for the second, and so on, so the number of ways is:
5 \times 4 \times 3 \times 2 \times 1 = 120

This calculation is called 5 factorial and is written 5!

In general for any positive integer n , n! = n(n-1)(n-2)... 1 \quad and 0! = 1.

## Software/Applets used on this page

This page uses jsMath
You can get a better display of the maths by downloading special TeX fonts from jsMath. In the meantime, we will do the best we can with the fonts you have, but it may not be pretty and some equations may not be rendered correctly.

## Glossary

### factorial

n! = n(n-1)(n-2)(n-3)...3.2.1.

### integer

a positive or negative whole number.

### union

The union of two sets A and B is the set containing all the elements of A and B.

Full Glossary List


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# How do you factor b^3-64c^3?

Jan 1, 2016

Apply the difference of cubes formula to see that
${b}^{3} - 64 {c}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

#### Explanation:

The difference of cubes formula states that

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

(Verify this by multiplying out the right hand side)

Then, for the given expression, we have

${b}^{3} - 64 {c}^{3} = {b}^{3} - {\left(4 c\right)}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

Jan 1, 2016

${b}^{3} - 64 {c}^{3} = \left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$

#### Explanation:

${b}^{3} - 64 {c}^{3}$ is a difference of cubes . The formula for factoring a difference of cubes is $\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$, where $a = b$ and $b = 4 c$ in this case.

Substitute the values for $a \mathmr{and} b$ into the equation.

$\left(b - 4 c\right) \left({b}^{2} + b \times 4 c + {\left(4 c\right)}^{2}\right)$

Simplify.

$\left(b - 4 c\right) \left({b}^{2} + 4 b c + 16 {c}^{2}\right)$


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 8.8: Chapter 8 Review

Difficulty Level: At Grade Created by: CK-12

Keywords & Theorems

Pythagorean Theorem
Given a right triangle with legs of lengths a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} and a hypotenuse of length c\begin{align*}c\end{align*}, then a2+b2=c2\begin{align*}a^2 + b^2 = c^2\end{align*}.
Pythagorean Triple
A set of three whole numbers that makes the Pythagorean Theorem true.
Distance Formula
d=(x1x2)2+(y1y2)2\begin{align*}d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\end{align*}.
Pythagorean Theorem Converse
If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Theorem 8-3
If the sum of the squares of the two shorter sides in a right triangle is greater than the square of the longest side, then the triangle is acute.
Theorem 8-4
If the sum of the squares of the two shorter sides in a right triangle is less than the square of the longest side, then the triangle is obtuse.
Theorem 8-5
If an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.
Geometric Mean
The geometric mean is a different sort of average, which takes the nth\begin{align*}n^{th}\end{align*} root of the product of n\begin{align*}n\end{align*} numbers.
Theorem 8-6
In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments.
Theorem 8-7
In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments
45-45-90 Corollary
If a triangle is an isosceles right triangle, then its sides are in the extended ratio x:x:x2\begin{align*}x : x : x \sqrt{2}\end{align*}.
30-60-90 Corollary
If a triangle is a 30-60-90 triangle, then its sides are in the extended ratio x:x3:2x\begin{align*}x : x \sqrt{3} : 2x\end{align*}.
Trigonometry
The study of the relationships between the sides and angles of right triangles.
A side adjacent to an angle is the leg of the triangle that helps form the angle.
Opposite (Leg)
A side opposite to an angle is the leg of the triangle that does not help form the angle.
Sine Ratio
For an acute angle x\begin{align*}x\end{align*} in a right triangle, the sinx\begin{align*}\sin x\end{align*} is equal to the ratio of the side opposite the angle over the hypotenuse of the triangle.
Cosine Ratio
For an acute angle x\begin{align*}x\end{align*} in a right triangle, the cosx\begin{align*}\cos x\end{align*} is equal to the ratio of the side adjacent to the angle over the hypotenuse of the triangle.
Tangent Ratio
For an acute angle x\begin{align*}x\end{align*}, in a right triangle, the tanx\begin{align*}\tan x\end{align*} is equal to the ratio of the side opposite to the angle over the side adjacent to x\begin{align*}x\end{align*}.
Angle of Depression
The angle measured from the horizon or horizontal line, down.
Angle of Elevation
The angle measure from the horizon or horizontal line, up.
Inverse Tangent
Inverse tangent is also called arctangent and is labeled tan1\begin{align*}\tan^{-1}\end{align*} or arctan. The “-1” indicates inverse.
Inverse Sine
Inverse sine is also called arcsine and is labeled sin1\begin{align*}\sin^{-1}\end{align*} or arcsin.
Inverse Cosine
Inverse cosine is also called arccosine and is labeled cos1\begin{align*}\cos^{-1}\end{align*} or arccos.
Law of Sines
If ABC\begin{align*}\triangle ABC\end{align*} has sides of length, a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\end{align*}.
Law of Cosines
If ABC\begin{align*}\triangle ABC\end{align*} has sides of length a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then a2=b2+c22bccosA\begin{align*}a^2 = b^2 + c^2 - 2bc \cos A\end{align*}

b2c2=a2+c22ac cosB=a2+b22ab cosC\begin{align*}b^2 & = a^2 + c^2 - 2ac \ \cos B\\ c^2 & = a^2 + b^2 - 2ab \ \cos C\end{align*}

## Review Questions

Solve the following right triangles using the Pythagorean Theorem, the trigonometric ratios, and the inverse trigonometric ratios. When possible, simplify the radical. If not, round all decimal answers to the nearest tenth.

Determine if the following lengths make an acute, right, or obtuse triangle. If they make a right triangle, determine if the lengths are a Pythagorean triple.

1. 11, 12, 13
2. 16, 30, 34
3. 20, 25, 42
4. 106,30,1015\begin{align*}10 \sqrt{6}, 30, 10 \sqrt{15}\end{align*}
5. 22, 25, 31
6. 47, 27, 35

Find the value of x\begin{align*}x\end{align*}.

1. The angle of elevation from the base of a mountain to its peak is 76\begin{align*}76^\circ\end{align*}. If its height is 2500 feet, what is the distance a person would climb to reach the top? Round your answer to the nearest tenth.
2. Taylor is taking an aerial tour of San Francisco in a helicopter. He spots AT&T Park (baseball stadium) at a horizontal distance of 850 feet and down (vertical) 475 feet. What is the angle of depression from the helicopter to the park? Round your answer to the nearest tenth.

Use the Law of Sines and Cosines to solve the following triangles. Round your answers to the nearest tenth.

## Texas Instruments Resources

In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9693.

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Adding tangent (velocity) vectors to a plot of a space curve [duplicate]

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I'm new to Mathematica and my assignment is very vague... I'm hoping someone can explain to me a bit more of that this quesion means: Given the curve C, display the point on the curve and the ...
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# Simulating a Binomial distribution with $\mathscr{U}(0,1)$.

Ex:Show the procedures to simulate an random variable that follows a binomial distribution with parameter $p$, using the $\mathscr{U}(0,1)$(Uniform distribution on the interval (0,1)).

I tried to solve this question by using the following theorem:

Theorem: Let $U\sim\mathscr{U}(0,1)$. Let $X$ be a random variable with distribution $F_X(x)$. Therefore the random variable Y=F^{-1}(U) has a distribution function equal to $F_X$, the distribution of the $X$ variable.

According to this theorem I would need to find a the inverse of the binomial c.d.f, define it as a function in python and generate random numbers.

However I have no idea on how to invert the Binomial distribution.

Questions:

1) Is this the simplest method to simulate a Binomial distribution with the Uniform(0,1)? Are there other methods?

2) How do I compute the inverse of the Binomial distribution?

• Simulating $n$ iid Bernoulli variables is easier. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 14:40
• @GNUSupporter Indeed, but this requires generating $n$ uniform random variates. The problem doesn't specify whether this is allowed :) – Math1000 Apr 1 '18 at 15:18
• @Math1000 When one can simulate one uniform random variable, one can also simulate $n$ copies. If you insist on generating one single variable and use the inversion theorem, you'll have to compute a discrete sum $\sum_{i = 1}^k p^i (1-p)^{n-i}$ in order to find out $P(X \le k)$. I don't think it's a sensible simulating algo. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 15:25

It may be helpful to see how this procedure for random sampling from $\mathsf{Binom}(n=5, p=.4)$ is implemented in R statistical software. First, some R notation: runif (without extra parameters) is a source of pseudorandom values from standard uniform; dbinom, pbinom and qbinom denote binomial PDF, CDF, and quantile funcions (invese CDF) respectively.

So in R we can generate $m = 100,000$ observations from $\mathsf{Binom}(n=5, p=.4)$ in a vector xas follows.

set.seed(4118)
m = 10^5;  u = runif(m)
x = qbinom(u, 5, .4)     # inverse CDF transformation

Then we can tally the results, make a histogram of them, and plot exact PDF values on the histogram for comparison:

table(x)/m
x
0       1       2       3       4       5
0.07790 0.25775 0.34608 0.23105 0.07744 0.00978

hist(x, prob=T, br=(0:6)-.5, col="skyblue2", main="100,000 Realizations of BINOM(5,.5)")
k = 0:5; pdf = dbinom(k, 5, .4)
points(k, pdf, col="red")

Within the accuracy of the graph, it is not possible to distinguish the simulated results (heights of histogram bars) from the theoretical ones (small red circles).

Finally, by using the same seed for the pseudoranom generator as above, we can access exactly the same values u as above. Thus, we can see that R implements this (inverse CDF) method to generate $m$ observations from $\mathsf{Binom}(n=5, p=.4)$ by using the function rbinom defined in R. The tallied results are exactly the same below as above.

set.seed(4118)
m = 10^5;  x = rbinom(m, 5, .4)
table(x)/m
x
0       1       2       3       4       5
0.07790 0.25775 0.34608 0.23105 0.07744 0.00978

Note: By contrast, when $p >.5,$ R uses a slight modification of the inverse CDF method, so that the two approaches give slightly different results. (I used $m = 10,000$ so that differences would be more obvious.)

# Inverse CDF method
set.seed(401);  m = 10^4;  u = runif(m);  x1 = qbinom(u, 5, .7)
table(x1)/m
x1
0      1      2      3      4      5
0.0022 0.0296 0.1323 0.3076 0.3651 0.1632

# Built-in function 'rbinom'
set.seed(401);  x2 = rbinom(m, 5, .7)
table(x2)/m
x2
0      1      2      3      4      5
0.0023 0.0278 0.1288 0.3126 0.3599 0.1686

Addendum 1: Graphs of CDF of $X \sim \mathsf{Binom}(5, .4)$ and its inverse function. The latter shows $F_X^{-1}(u) = 0,$ for $u < 0.6^5=0.07776,$ as in a Comment.

Addendum 2: Generating $X \sim \mathsf{Binom}(5, .4)$ as the sum of five independent Bernoulli random variables with $p=.4.$ [This is the method suggested in the Comment by @GNUSupporter.]

First let $U_1, U_2, \dots, U_5$ be a random sample from $\mathsf{Unif}(0,1).$ Then $B_i = 1,$ if $U_i \le .4$ and $0$ otherwise. This is essentially a trivial application of the quantile method to a variable that takes only values $0$ and $1$. Then $X = \sum_{i=1}^5 B_i \sim \mathsf{Binom}(5, .4).$ We generate four such binomial random variables below (results: 1, 2, 2, 4). Notice that five pseudorandom uniform values are required for each binomial.

set.seed(1234)
u = runif(5); b = (u < .4);  x = sum(b);  x # sum of logical vec b is nr of its TRUEs
[1] 1
u = runif(5); b = (u < .4);  x = sum(b);  x
[1] 2
u = runif(5); b = (u < .4);  x = sum(b);  x
[1] 2
u = runif(5); b = (u < .4);  x = sum(b);  x
[1] 4

Next we use a for loop to iterate this procedure $m = 100,000$ times. Because we start with the same seed as above, the first four iterations repeat the realizations of $\mathsf{Binom}(5, .4)$ shown just above. A tally of all $m$ results shows results similar to those in the initial simulation of this Answer, closely matching the target distribution.

set.seed(1234);  m = 10^5;  x = numeric(m)
for (i in 1:m) {
u = runif(5); b = (u < .4);  x[i] = sum(b)  }
mean(x);  x[1:4]
[1] 1.99806  # aprx E(X) = 5(.4) = 2
[1] 1 2 2 4  # same first four realizations of X as above
table(x)/m
x
0       1       2       3       4       5
0.07766 0.25971 0.34683 0.22878 0.07675 0.01027

• Thanks for your answer! I think what I am asked to do is to generate or simulate a Binomial distribution from a Uniform distribution. I can easily using the theorem provided in my post to simulate an exponential function by inverting by hand its expression, then generating random numbers ,defining the inverse exponential as function on Python. However for distributions that are difficult to reverse such as this. I do not know how to proceed since I am not allowed to use commands that already relate to the distribution(on this case Binomial). – Pedro Gomes Apr 1 '18 at 18:26
• Is there a way to generate it given the requirements? – Pedro Gomes Apr 1 '18 at 18:27
• I guess you'll have to use a sequence of if-then statements as implied by @NCh's Answer, which 'inverts the binomial CDF'. Explicitly, for X~BINOM(5, .4), you have $P(X = 0) = P(X \le 0) = .6^5 = 0.07776.$ So take $X = 0$ if $U \le 0.07776,$ And so on. Hope this helps. (You're not working a 'toy' exercise, but learning about something that is actually used. I was not trying to add another answer for your question, but to discuss implementation of the idea in software.) – BruceET Apr 1 '18 at 18:46
• Directly to your two questions: (1) Inverse CDF is a simple method. Context may determine what is 'simplest'. Typically software will already have programmed a quantile function, so using it is trivial. Starting from scratch @GNU may be right that it's easier to program $n$ Bernoulli RVs and add. (2) As mentioned in my previous comment, NCh has shown you how to invert the binomial CDF (it seems, without explicitly using those words). // I'm off to brunch, will check back later. – BruceET Apr 1 '18 at 18:58
• See two addenda to my Answer. Hope they are helpful. – BruceET Apr 2 '18 at 1:26

For the second question: If $U\leq (1-p)^n$, set $Y=0$.

If $(1-p)^n< U\leq (1-p)^n+np(1-p)^{n-1}$, set $Y=1$.

If $(1-p)^n+np(1-p)^{n-1}< U\leq(1-p)^n+np(1-p)^{n-1}+\binom{n}{2}p^2(1-p)^{n-2}$, set $Y=2$.

And so on.

$$Y=k \iff \sum_{i=0}^{k-1} \binom{n}{i}p^i(1-p)^{n-i} < U \leq \sum_{i=0}^{k} \binom{n}{i}p^i(1-p)^{n-i}.$$

On your first question: as GNU Supporter noted in comments, this is not the simplest way to generate binomial distribution.

• What do you think about question (1)? Perhaps you won't code this in a real simulation. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 15:54
• The answer to this question is obvious and is given in your comment above. Thank you, I'll add this to the answer. – NCh Apr 1 '18 at 15:59
• Thanks for reply. I just want your viewpoint due Math1000's comment. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 16:01
• But simulating a Bernoulli r.v. is also "inverting" its CDF. Anyways, thanks for your opinion. – GNUSupporter 8964民主女神 地下教會 Apr 1 '18 at 16:11
• There exist probability spaces on which a binomial-distributed random variable exists but cannot be written as the sum of independent $\{0,1\}$-valued random variables. See here math.stackexchange.com/questions/1920276/… – Math1000 Apr 1 '18 at 17:33


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By Sunil Bhardwaj

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CAPACITANCE OF CYLINDRICAL CAPACITOR

Consider a cylindrical capacitor of length $$L$$. It is constructed with two coaxial cylinders such that inner cylinder has radius $$a$$ and charge $$+Q$$ while outer cylinder has radius $$b$$ and charge $$-Q$$ as shown in fig.

The electric field lines come out from positively charged cylinder and enter into negatively charged cylinder.

Now imagine a Gaussian surface in the form of a cylinder of radius r encloses the positively charged inner cylinder of capacitor. The Gaussian cylinde three surfaces.

The flux through top-capped Gaussian cylinder having area element $$da$$ is $$\phi_1\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_1\ =\ \int{E\ da\ \cos{90}}$$ $$\phi_1\ =\ 0$$ The flux through bottom capped Gaussian cylinder having area element $$da$$ is $$\phi_2\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_2\ =\ \int{E\ da\ \cos{90}}$$ $$\phi_2\ =\ 0$$ The flux through curved surface of Gaussian cylinder having area element $$da$$ is $$\phi_3\ =\oint{\vec{E}.\vec{da}}$$ $$\ \phi_3\ =\ \int{E\ da\ \cos{0}}$$ $$\phi_3\ =\ E\left(2\pi r\ L\right)$$ Where $$2\pi r$$ is circumference of Gaussian cylinder and $$L$$ is its length. Net flux through whole Gaussian cylinder is $$\phi_c\ =\ \phi_1+\phi_2+\phi_3$$ $$\phi_c\ =\ 0+0+E\left(2\pi r\ L\right)$$ $$\phi_c\ =\ E\left(2\pi r\ L\right)$$

. . . . . . . . . . . . . (1)

From Gauss's law $$\phi_c\ =\ \frac{Q}{\varepsilon_0}$$

. . . . . . . . . . . . . (2)

Comparing eq (1) and eq (2), we get $$E\left(2\pi r\ L\right)\ =\ \frac{Q}{\varepsilon_0}$$ $$E\ =\ \frac{Q}{\left(2\pi r\ L\right)\varepsilon_0}$$ The potential related to this electric field of cylindrical capacitor is $$V\ =\ \int_{a}^{b}{E\ ds}$$ $$V\ =\ \int_{a}^{b}{\frac{Q}{\left(2\pi r\ L\right)\varepsilon_0}\ dr}$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\int_{a}^{b}{\frac{1}{r}\ dr}$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\left[\ln{r}\right]_a^b$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\left(\ln{b}-\ \ln{a}\right)$$ $$V\ =\ \frac{Q}{\left(2\pi\ L\right)\varepsilon_0}\ln{\left(\frac{b}{a}\right)}$$ $$\frac{\left(2\pi\ L\right)\varepsilon_0}{\ln{\left(\frac{b}{a}\right)}}\ =\ \frac{Q}{V}$$ The capacitance of a capacitor is related with voltage as $$C=\ \frac{Q}{V}$$ Comparing eq(1) and eq(2) $$C=\ \frac{\left(2\pi\ L\right)\varepsilon_0}{\ln{\left(\frac{b}{a}\right)}}$$ This is the capacitance of a cylindrical capacitor which depends upon length radius "$$a$$" of inner cylinder and radius "$$b$$" of outer cylinder.

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# Stars and bars to find “how many $x$ digit numbers are there with sum of digits $y$”?

This question poses a seemingly very simple method to solve problems of the sort "how many $x$ digit numbers are there with sum of digits $y$?", but I don't understand it. Why are the "bad" solutions in correspondence to the solutions of $y_1 + 10 + x_2 + x_3 + x_4 + x_5 = 23$? What's the idea?

For instance, what "correct solution" corresponds to taking $y_1=13$? How to solve this problem?

• See here – Moya Sep 11 '16 at 14:12
• @igael I don't understand why we make the substitution $y_1=x_1-10$ and not, say $x_1-14$.. the latter would also ensure a "digit"... – combinarcotics Sep 11 '16 at 14:12
• @Moya that still doesn't explain why we take the $-10$ substitution... Why not just take $-6$ in that quesion? Why is the $10$ there? – combinarcotics Sep 11 '16 at 14:14
• @igael I hope somebody will explain.. – combinarcotics Sep 11 '16 at 14:22
• @Moya: yes, it is the basis of the good answer. Perhaps it wasn't well applied in the 1st question. – user354674 Sep 11 '16 at 14:31

I'm also not great at combinatorics, but I'll give this a shot to give you a start. This isn't going to be an answer, but an extended comment.

First, we need to establish the number of nonnegative numbers satisfying $$\sum_{i=1}^5 x_i=23$$ (call this (*)). This is a standard stars and bars type problem, where we have $23+(5-1)=27$ stars and $5-1=4$ bars. But we've overcounted, because some of these solutions will have some of the numbers being greater than or equal to 10, which should not be allowed because we wants a 5 digit number. For this, we can use the inclusion-exclusion principle (we want the second quoted equation).

Let $A$ be the set of solutions to (*) where none of the integers is greater than or equal to 10, $B$ the set of solutions where 1 integer is greater than or equal to 10, and $C$ where two of the integers are greater than or equal to 10. After this, we're done, because at most 2 of the $x_i$ in the solutions we counted can be greater or equal to $10$. Then \begin{align} |A\cup B\cup C|= |A| + |B| + |C| - |B\cap C|\end{align} (I excluded the parts of the sum that were empty). We want to know $|A|$, we know $|A\cup B \cup C|$, so we need to compute $|B|,|C|,|B\cap C|$. How one would compute $|B|,|C|$ is given, more or less, in the answer I quoted in the comments.

How one would compute $|B\cap C|$, I'm not sure off the top of my head. We also need to exclude the solutions where $x_1=0$. In the quoted method, the number is small enough to count by hand. For this case we would need a more robust method.

Edit: See also here for ideas. I'm out the door, but best of luck to you!

Edit 2: Back, and I messed up a little in haste. $B\cap C$ is obviously equal to the number of solutions where exactly 2 numbers are $\geq 10$. So we can just take $|A\cup B\cup C|=|A|+|B|+|C|$, where $A$ is as before, $B$ is the number of solutions with exactly one number $\geq 10$, and $|C|$ the number of solutions with exactly two numbers $\geq 10$. Then we'll need to refine by excluding the case where $x_1=0$, this can be done with the trick $x_1'=x_1-1$, though some care will need to be taken.

The referred question asks for the number of non-negative integer solutions of \begin{align*} x_1+x_2+x_3+x_4+x_5=23 \end{align*} with only one additional restriction $0\leq x_1\leq 9$. In this case we do not need the inclusion-exclusion principle.

In order to determine all integer solutions with \begin{align*} x_1+x_2+x_3+x_4+x_5=23\qquad\qquad0\leq x_1\leq 9,0\leq x_2,x_3,x_4,x_5\tag{1} \end{align*}

we look at all integer solutions of

and subtract all solutions of

The solutions of (3) are the so-called bad solutions in the referred question, meaning the invalid solutions which are to subtract when determining the solutions of (2).

In order to calculate (2) with the nice range $x_1,x_2,x_3,x_4,x_5\geq 0$ we can use the stars-and-bars technique and obtain \begin{align*} \binom{23+4}{4}=\binom{27}{4}=17550 \end{align*}

In order to calculate (3) we transform the range by a proper substitution which enables us to apply the stars-and-bars technique again. Instead of $x_1\geq 10$, we substitute $x_1=y_1+10$ and the range $x_1\geq 10$ can then be transformed to $y_1+10\geq 10$ or equivalently $y_1\geq 0$.

This way we obtain from (3)

We are now in the same situation as in (2) and can apply the stars-and-bars technique again.

\begin{align*} \binom{13+4}{4}=\binom{17}{4}=2380 \end{align*}

We finally obtain the number of wanted solutions as the total number of solutions of (2) minus the number of solutions of (3) \begin{align*} \binom{27}{4}-\binom{17}{4}=17550-2380=15170 \end{align*}

• +1. Nice job. The complication comes from the subtraction. Something likes $00599$ is not allowed. – Felix Marin Sep 14 '16 at 21:19
• @FelixMarin: Many thanks, Felix! Btw. $00599$ is a valid string as well as $0000\color{blue}{23}$. Subtraction avoids counting e.g $\color{blue}{10}0580$. – Markus Scheuer Sep 15 '16 at 7:32

If $x_1 + x_2 + x_2 + x_3 + x_ + x_5 = 23$ and one or more of the $x_i$ has more than one digit. Than one or more terms is at least $10$. We can replace the $x_i$s with $x'_i$s where one of the $x'_i = x_i -10 \ge >0$ and the rest of the $x'_j = x_j$. Then $10+ x'_1 + x'_2 + .... + x'_5 = 23$.

So if we eliminate all of those solutions we have eliminated all the solutions in which at least one $x_i$ was double digit.

So we are only left with solutions in which all terms are single digit.

We did not eliminate any extraneous solutions because if $x_1 + ... x_5 = 23$ and all $x_i$ are single digit and $10 + x'_1 +.... + x'_5 = 23$, it is not possible for $x'_j = x_j$ for all but one term; that would make the $x'_i = x_i -10 < 0$.

====

Argh. WRONG!!

By eliminating solutions to $10 + .... = 23$ we will elimate an answer with two multidigit terms twice.

E.G. $10 + 11 + 0 + 1 + 1$ will be eliminated by eliminating $10 + ( 0+ 11 + 0 + 1 + 1)$ and by eliminating $10 + (10 + 1 + 0 + 1 + 1)$ as well.

To avoid double counting we must re add thus with two multidigits.

So the solution should be:

# solut to $x_1 + x_2 + ... + x_5 = 23$ - # solutions to $10 + x_1 + ..+x_5 = 23$ + # solutions to $10 + x_1+.... + x_5 = 23$

Which could be very complicated for sums greater than 30. Best to come up with another method altogether.


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# Solving $\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$

Given $$\left(\sqrt{3+2\sqrt{2}}\right)^x - \left(\sqrt{3-2\sqrt{2}}\right)^x = \frac32$$

What is $x$?

I just can do this with that equation

$$\left(\sqrt{2+1+2\sqrt{2.1}}\right)^x - \left(\sqrt{2+1-2\sqrt{2.1}}\right)^x = \frac32$$

$$\left(\sqrt{({\sqrt{2}+\sqrt{1})^2}}\right)^x-\left(\sqrt{({\sqrt{2}-\sqrt{1})^2}}\right)^x = \frac32$$

$$\left(\sqrt{2}+1\right)^x - \left(\sqrt{2}-1\right)^x = \frac32$$

And i stuck there for a few hours and get nothing

Pliz help me

• Note that $\sqrt 2 - 1 = \frac 1{\sqrt 2 + 1}$. This will allow you to set $(\sqrt 2 +1)^x = y$ , then solve for $y$ as a quadratic equation then retrieve $x$ from the value(s) of $y$ obtained. – Teresa Lisbon Sep 7 '18 at 16:46
• math.stackexchange.com/questions/202078/… – lab bhattacharjee Sep 7 '18 at 18:44
• Thank you very much for the hints sir @астон вілла олоф мэллбэрг – Luis Szooares Sep 7 '18 at 22:43
• Oh this is similiar with my problem, thank you @lab bhattacharjee – Luis Szooares Sep 7 '18 at 22:44

Use the fact that $$\sqrt{3-2\sqrt{2}}=\frac{1}{\sqrt{3+2\sqrt{2}}}$$


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Preprint Article Version 1 Preserved in Portico This version is not peer-reviewed

# On the Scales of Selves: Information, Life, and Buddhist Philosophy

Version 1 : Received: 18 March 2021 / Approved: 22 March 2021 / Online: 22 March 2021 (15:59:51 CET)
Version 2 : Received: 10 May 2021 / Approved: 12 May 2021 / Online: 12 May 2021 (13:59:04 CEST)

How to cite: Gershenson, C. On the Scales of Selves: Information, Life, and Buddhist Philosophy. Preprints 2021, 2021030554 (doi: 10.20944/preprints202103.0554.v1). Gershenson, C. On the Scales of Selves: Information, Life, and Buddhist Philosophy. Preprints 2021, 2021030554 (doi: 10.20944/preprints202103.0554.v1).

## Abstract

When we attempt to define life, we tend to refer to individuals, those that are alive. But these individuals might be cells, organisms, colonies... ecosystems? We can describe living systems at different scales. Which ones might be the best ones to describe different selves? I explore this question using concepts from information theory, ALife, and Buddhist philosophy. After brief introductions, I review the implications of changing the scale of observation, and how this affects our understanding of selves at different structural, temporal, and informational scales. The conclusion is that there is no best'' scale for a self, as this will depend on the scale at which decisions must be made. Different decisions, different scales.

## Keywords

philosophy of science; information; artificial life; scales; causality

## Subject

ARTS & HUMANITIES, Anthropology & Ethnography

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Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4

Question 1.
Write the Maclaurin series expansion of the following functions:
(i) ex
(ii) sin x
(iii) cos x
(iv) log (1 – x); -1 ≤ x < 1
(v) tan-1 (x) ; -1 ≤ x ≤ 1
(vi) cos2 x
Solution:

(vi) f(x) = cos2 x
f(0) = 1
f'(x) = 2 cos x (- sin x) = – sin 2x
f'(0) = 0
f”(x) = (-cos 2x)(2)
f”(0) = -2
f”'(x) = -2[- sin 2x](2) = 4 sin 2x
f”'(0) = 0
f4 (x) = 4(cos 2x)(2) = 8 cos 2x
f4 (0) = 8

Question 2.
Write down the Taylor series expansion, of the function log x about x = 1 upto three non-zero terms for x > 0.
Solution:

Question 3.
Expand sin x in ascending powers x – $$\frac{\pi}{4}$$ upto three non-zero terms.
Solution:
f (x) = sin x

Question 4.
Expand the polynomial f(x) = x2 – 3x + 2 in powers of x – 1
Solution:
f(x) = x2 – 3x + 2 = (x – 1) (x – 2)
f(1) = 0
f'(x) = 2x – 3 ; f'(1) = -1
f”(x) = 2 ; f”(1) = 2

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.4 Additional Problems

Question 1.
The Taylor’s series expansion of f(x) = sin x about x = $$\frac{\pi}{2}$$ is obtained by the following way.
Solution:

Question 2.
Obtain the Maclaurin’s series expansion for the following functions.
(i) ex
(ii) sin2 x
(iii) $$\frac{1}{1+x}$$
Solution:
(i)

(ii)

(iii)


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# Tag Info

1

You should not mix Transform changes (like transform.Translate()) with Rigidbody/Rigidbody2D movement. Use one or the other exclusively. If you want your object to move with physics, let the physics engine handle all its movement. Mixing these means anytime you edit the Transform, you're ripping the object out of the control of the physics engine, ...

3

I think we can do this more simply. This demo uses about 80 lines of code (the interesting ones being just the dozen below), instead of over 300, with no angles/trigonometry functions needed: At the time $t^*$ when we reach the vertex, we know the acceleration due to gravity has completely wiped out our vertical launch velocity: \begin{align} v_y + a_g ...

1

Note my original response suggested simply increasing maxMotorTorque, but it appears this has no apparent effect by itself. Try adding force to the parent (car_body) rigidbody along with the motor values of the wheel colliders. Something like this: float motor = maxMotorTorque * Input.GetAxis("Vertical"); rb.AddForce(transform.forward * motor); ...

0

I said in the OP I was struggling to find the way to use a PID controller in this case, but I've figured how to do it. Here are the few steps: Calculate the thrust required to go at the speed of B Use the distance to B as an input error for the PID Use the values returned by the PID as an offset applied to the thrust Tune the PID (the funniest part of ...

1

(I started to answer this before the post was closed soon after the question was posed and before I could finish it. DMGregory's answer is more complete and sophisticated, but I'm offering this anyway because I think it should be very easy for a beginner to understand.) The first part of your question is pretty simple. For demo purposes, here's our set up: ...

2

You can do this with a variation on this bobblehead script I wrote. Keep the spring stiffness high and the conservation low to make the object stay more upright and bounce less. using UnityEngine; public class Bobble : MonoBehaviour { // We'll sproing about this transform's origin point. public Transform pivot; // Tunable parameters for ...

1

In general you should not set a rigidbody’s velocity directly, as it often results in weird behavior, per the docs: Rigidbody2D.velocity Instead, try AddForce(), AddTorque() or AddForceAtPosition()

1

No, Unity does not offer you the option to apply different drag on different axis'. However what you can do is: Implement your own drag mechanic, as in the answer by Acme Nerd Games Change the drag coefficient in FixedUpdate depending on the current direction of the velocity vector. This is a solution I used once to implement a "drifting" mechanic ...

2

I can think of two ways to address your problem (while still using Physics2D) : Increase Gravity You could try increasing the gravity scale to offset the extra drag. Implement your Own Drag You could zero the rigidbody drag factor and instead add a script that adds an force on the x-axis only of the rigidbody that increases with speed. Here's a (simplistic) ...

0

One problem with many physics engines is that they are not deterministic. That means that two computers executing the same physics simulation with the same input parameters might end up with two different results. For example because one of them runs at a higher framerate leading to a more detailed resolution. This is often not much of a problem in ...

0

Is there something that makes client prediction/corrections easier in the kinematic version than the physics version? This is a good example. If you want an authoritative server, you would run all of your physics on the server. There is no need to run it again on the client for important parts, such as the player. So when the server tells the client the ...

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