text:
# nLab defining ideal of a topologizing subcategory

The notion of the defining sheaf of ideals of a closed subscheme inspires the notion of defining ideal of a topologizing subcategory $S$ of an abelian category $A$ as the endofunctor $\mathcal{I}=\mathcal{I}_S\in End(A)$ which is the subfunctor of identity $Id_A$ assigning to any $M\in A$ the intersection of kernels $Ker(f)$ of all morphisms $f: M\to N$ where $N\in Ob(S)$. One can show that if $T\subset S$ is an inclusion of topologizing subcategories, then $\mathcal{I}_{S}\subset \mathcal{I}_{T}$.

If $R$ is an associative unital ring and $J\subset R$ a left ideal in $R$. Let $T = [J]$ be the smallest coreflective subcategory of $A$ containing $R/J$. Then $\mathcal{I}_T(R)\subset R$ is a two-sided ideal, namely the maximal 2-sided ideal in $J$, which is explicitly, $(J:R) = \{r\in R\,|\, J r \subset J\}$. Moreover, $J_{T\circ T}$ where $T\circ T$ is the square under the Gabriel multiplication agrees with $(J:R)^2$.

• A. L. Rosenberg, Noncommutative algebraic geometry and representations of quantized algebras, MIA 330, Kluwer Academic Publishers Group, Dordrecht, 1995. xii+315 pp. ISBN: 0-7923-3575-9

• V. A. Lunts, A. L. Rosenberg, Differential calculus in noncommutative algebraic geometry I. D-calculus on noncommutative rings, MPI 1996-53 pdf

Created on May 12, 2011 16:34:06 by Zoran Škoda (148.6.183.21)

format: plain


text:
255 views

The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
Privacy-challenged office workers may find it hard to believe, but open-plan offices and cubicles were invented by architects and designers who thought that to break down the social walls that divide people, you had to break down the real walls, too. Modernist architects saw walls and rooms as downright fascist. The spaciousness and flexibility of an open plan would liberate homeowners and office dwellers from the confines of boxes. But companies took up their idea less out of a democratic ideology than a desire to pack in as many workers as they could. The typical open-plan office of the first half of the $20$th century was a white-collar assembly line. Cubicles were interior designers’ attempt to put some soul back in.

1. Wall-free office spaces did not quite work out the way their utopian inventors intended, as they became tools for exploitation of labor
2. Wall-free office spaces could have worked out the way their utopian inventors intended had companies cared for workers' satisfaction
3. Wall-free office spaces did not quite work out as desired and therefore cubicles came into being
4. Wall-free office spaces did not quite work out as companies don’t believe in democratic ideology


text:
# cf.Field.autocyclic¶

Field.autocyclic(**kwargs)[source]

Set dimensions to be cyclic.

A dimension is set to be cyclic if it has a unique longitude (or grid longitude) dimension coordinate construct with bounds and the first and last bounds values differ by 360 degrees (or an equivalent amount in other units).

New in version 1.0.

Parameters
verbose: int or str or None, optional

If an integer from -1 to 3, or an equivalent string equal ignoring case to one of:

• 'DISABLE' (0)

• 'WARNING' (1)

• 'INFO' (2)

• 'DETAIL' (3)

• 'DEBUG' (-1)

set for the duration of the method call only as the minimum cut-off for the verboseness level of displayed output (log) messages, regardless of the globally-configured cf.log_level. Note that increasing numerical value corresponds to increasing verbosity, with the exception of -1 as a special case of maximal and extreme verbosity.

Otherwise, if None (the default value), output messages will be shown according to the value of the cf.log_level setting.

Overall, the higher a non-negative integer or equivalent string that is set (up to a maximum of 3/'DETAIL') for increasing verbosity, the more description that is printed to convey information about the operation.

config: dict

Additional parameters for optimizing the operation. See the code for details.

New in version 3.9.0.

Returns

bool


text:
# Definition:String

## Definition

Let $\AA$ be an alphabet of symbols.

A string (in $\AA$) is a sequence of symbols from $\AA$.

There is no limit to the number of times a particular symbol may appear in a given string.

### Finite String

A string $S$ in $\AA$ is a finite string if and only if the sequence of symbols of which it is composed is finite.

### Infinite String

A string $S$ in $\AA$ is an infinite string if and only if the sequence of symbols of which it is composed is infinite.

## Also defined as

Some sources use the word string to mean a finite string, that is, what is defined on $\mathsf{Pr} \infty \mathsf{fWiki}$ as a word.


text:
# FiveThirtyEight's Elo Ratings and Logistic Regression

Inspired by FiveThirtyEight’s update of their CARMELO NBA player projections, a recent conversation on Twitter brought up the value of Elo Ratings. I decided to re-implement 538’s NBA Ratings and tried out a few improvements and variations. The code is all available on this GitHub repo. TLDR: Elo is simple, robust, and works pretty well.

## What is Elo?

### Basic Elo

Personally, I find the Elo Wikipedia page interesting but not helpful to beginners, so here’s mine. Classically, Elo is a rating system that only uses win and losses to measure performance. Initially, every competitor is assigned an initial rating (538 NBA uses 1300). Higher ratings are taken to mean better teams. To figure out if team A will beat team B, you first compute the difference between their scores. To map this difference into probabilities, we use a sigmoid function, namely the logistic. What’s a sigmoid? In this case, it’s a function that maps $$(-\infty,+\infty) \to [0,1]$$, where $$\sigma(0) = 0.5$$.

What’s great is that the logistic function will take any number and turn it a probability, so we just put in the difference between Elo Ratings. If the ratings are equal, the difference is zero, and we get 50-50 (0 maps to 0.5!). If your score is higher, the difference is positive, so you get a probability greater than 0.5. The larger the score difference, the larger the probability difference. Additionally, $$\sigma(-x) = 1-\sigma(x)$$, so flipping the sign of the score difference produces 1 minus the probability.

There exists a scale that determines how sharp that curve is, which most Elo variants set scale to 400. Roughly, 400 means if you’re +100, you win 64% of the time, +200 75%, +400 91%, etc. Most people familiar with machine learning have seen the logistic function as $$\frac{1}{1+e^{-x}}$$, but Elo is usually written with the base-10 variant. They’re basically equivalent, you just have to adjust the scale (400 in base-10 is 173 in base-e).

When we observe an outcome, we give the winner some fraction of K points, and take away the same amount from the loser. How big of a fraction? The loser’s probability of winning. The more likely you are to win, the fewer points you get. Equal odds? 50% of K. You had a 90% chance of winning and won, 10% of K. A 10% chance of winning and won, 90% of K.

That’s it. That’s all of Elo.

### Elo Formulas

More formally, we have a sigmoid with a scale of 400. $\sigma(x) = \frac{1}{1+10^{\frac{-x}{400}}}$ We have the probability of A winning as $P(A > B) = \sigma(R_A - R_B)$ We’ll assume A wins. So the error, given that A won, is the probability of B winning $E_{(w=A)} = P(B > A)$ The update is just that amount of K $R_A^{(t+1)} = R_A^{(t)} + K E_{(w=A)}$ $R_B^{(t+1)} = R_B^{(t)} - K E_{(w=A)}$

### 538’s NBA Elo

The FiveThirtyEight model makes three adjustments.

1. After every season, all teams get regresed to the mean.
2. The home team get an extra 100 points when comparing winning probabilities, as there’s home court advantage. Since the home team wins about 60% of the time, this should probably be +70 (+100 means 64% of the time!).
3. K is adjusted depending on the margin of victory and the Elo difference. You win by more points, you get more of K (with a small adjustment for team strength as well).

FiveThirtyEight thankfully is very open with their models and includes a lot of content on their GitHub. Plus, they wrote a nice article describing the formulas in their version of Elo for the NBA. I re-implemented all of this and got identical results. One thing I had to manually add was how teams moved around from 1947 until now. Additionally, some teams in the 538 dataset started with non-default scores, so there’s a slight error for those teams that gets propogated.

### SIDE NOTE: Errors & the logistic function

Your choice of sigmoid function corresponds to your choice of error distribution. By using a logit, Elo assumes errors are distributed according to a logistic distribution, which looks enough like (but is not!) a Gaussian/Normal distribution that we consider it general-purpose (the Central Limit Theorm claims Gaussians are general). This is the same assumption that Logistic Regression makes. Neither model assumes your errors are normal! Why? Well your sigmoid choice is just the cumulative distribution function of your error distribution. If we wanted a Gaussian, it’d be called probit regression and we’d have to use the erf, which has no simple closed-form. So instead we use the logit, it’s similiar enough and much easier to work with mathematically.

The logit is really nice: the x axis corresponds to log odds. In the base 10 logit, if you want one in ten times, then -1 gives $$\frac{1}{11}$$, -2 gives $$\frac{1}{101}$$, -3 gives $$\frac{1}{1,001}$$, etc. Positive numbers like +2 means “in all but $$\frac{1}{101}$$” times.

## Tweaking the variables

If you read the 538 article, they talk a lot about their update function, what they found K to be, etc. Does this really matter? One way to test this is to find new settings/parameters and see if we can outperform the 538 model. I used a black-box optimizer in the form of GPyOpt, but you can use anything, even random sampling, which can be hard to beat.

What did I find out? For almost any objective function I constructed (correctly picking winners, correctly outputing sensible probabilities, some weighted average of the two, etc.), I found that almost all my Elo variants performed similiarly. They all picked winners correctly about 67% to 68% of the time. My weighted wins + well-calibrated probabilities metric produced scores (higher is better, 100 is perfect) like 75.6 (538 Elo). 75.95 (Tuned Settings), 75.6 (Simple Elo, without margin of victory), 75.6 (PID model, see below). Basically they’re all almost the same. Elo is really robust to configuration. 538’s article claims their choice of K is important. I found you could get sensible results with almost any setting. I got good configurations with a K of 20, 35, 50, etc. I got good results blending end-of-season Elo towards the mean Elo with a weighting of 50%, 75% and 95%. Home court advantage? I had good results from 50 to 150. Elo didn’t really care.

In the code elo_acc is my version of 538 Elo, elo_acc_simple is my version using a standard sigmoid and no margin of victory, elo_acc_pid is the version described below and elo_acc_poly uses margin-of-victory in a general-purpose polynomial form.

## Proportional–Integral–Derivative Control

Adjusting a rating up and down to get the correct win probability? That’s just a simple feedback loop. Well control theory is a fairly established field and from a controls perspective, the Elo update is just proporation control. It takes an error and adjusts the ratings with a control gain (K).

In practice, PID controllers work well for most control problems. For a PID controller, we’ll have 3 K settings, one for each of the P,I,D settings. If the standard update is P, what are the other two? Well D is simply the difference between consecutive errors. That is, we’d want to have consistent errors over time. The I term is the sum of all the errors over time. We’d want to have our model over-shoot the target as much as it under-shoots the target. Thus the update equation would look something like this.

$R_A^{(t+1)} = R_A^{(t)} + K_{P} E^{(t)} + K_{D} (E^{(t)}-E^{(t-1)}) + K_{I} \sum_i E^{(i)}$

Anyways I added this to my Elo controller, with seperate I and D terms for every team, reset to 0 at the end of each season. How well did it work? I was able to get a better log-liklihood model than my other Elo variants, but it was about the same accuracy. Oh well, it was a fun idea. Here’s an example run (these graphs are smoothed as the PID output shakes a little bit and makes the curve hard to read).

## Using Logistic Regression to obtain Elo

Benjamin Morris remarked that Elo is pretty much strictly dominated by having a running logistic regression over the same data, with extra clarification here. The Elo formula for win probability between A and B (where A is the home team) is $$\sigma(R_A - R_B + H)$$ where $$H$$ is home court advantage. Logistic Regression computes probabilities using the formula $$\sigma( \sum_i \alpha_i x_i + \beta)$$. We can use this to model Elo! First, create a variable for each team. Then, each game is an example, where the home team has a +1 (an $$x_i$$) and the away team a -1 (another $$x_i$$). That’s it! Given the game results, logistic regression can the optimal coefficients for our formula (the $$\alpha_i$$s ), which are basically Elo Ratings. Additionally, it discovers home court advantage automatically ($$\beta)$$. By adding 1500 and multiplying the coefficients by 173, we can obtain Elo on the standard Elo scale. The only real choice you have to make is how many games to use. More games lead to a smoother estimate. Using the latest 10 games and 100 home games (for each team, so about 300 and 3000 total).

One issue I had was that each estimate independently obtains home court advantage and this was highly variable across my dataset

So I decided to formulate a logistic regression that had a fixed home court advantage based on the average of an automatically obtained run (75), and using a window of 30 games for each team.

The best results I got was 66.8% accuracy for a set of games where the 538 Elo model got 66.4%. I also tried weighting the model to value more recent estimates higher, but this lead to even more unstable Home Court Adjustment values in the everyone-gets-their-own-HCA case and weird curves in general; maybe I had a bug.

## Building a classifier to predict games

So, what about Logistic Regression to build a classifier based on just features? I defined some simple features (winning %, difference in winning % between teams, net rating, difference in net rating between teams, etc.), standardized them and added interaction terms. Then I trained a classifier over the same dataset. Guess what? 68% accuracy, about the same as Elo.

Here’s what the model predicts as the winning probabilities for all (out-of-sample) games since 2010, for home teams and away teams.

And here are the winning probabilities for Elo and the regression model, when playing a league-average team at home, across a few teams, across a few years. I find them shockingly comparable.

What ended up mattering to the classifier? Basically the net rating and how well the team did last year.

Variable Coeff
NetRating Diff 0.42
Prev Year W% Diff 0.29
NetRating Diff (Recent) 0.13
NetRating HomeTeam 0.04
W% Diff 0.03

## Combining the models

If my classifier got 67.6% of games correct, and my Elo model got 66.7% of games correct, what if I combined their results? With a simple average, I got 67.4% of games correct, so no better! With a simpler classifier, I had a slight bump, but not in general. I even tried UCB and EXP3 bandits, but only got 67% and they didn’t really develop a consistent preference over trials. Maybe a contextual bandit would work better.

# Conclusion

Elo is simple and works well across tons of parameter settings. An advanced update function did not seem to help. Logistic Regression produced comparable accuracy, and a sensible model. For using just win/loss data, either method is fine. If you want better prediction, you need more information. After game-level data, the natural progression is use box scores, then play-by-play data and then player tracking data.

Written on July 5, 2019


text:
# PFATSDR

Bulk Data Entry Defines properties for solder fatigue analysis.

## Format

(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
PFATSDR ID C4GAMMA K1 K2 K3 K4

## Definition

Field Contents SI Unit Example
ID Entry identification number.

Each PFATSDR entry must have a unique ID. This is referenced by the FATDEF Subcase Information Entry.

No default (Integer > 0)

C4GAMMA Empirical model constraint for solder fatigue.

No default (Real > 0)

K1 Constant in crack initiation equation in Darveaux method. 2

No default (Real)

K2 Constant in crack initiation equation in Darveaux method.

No default (Real)

K3 Constant in crack growth equation in Darveaux method. 2

No default (Real)

K4 Constant in crack growth equation in Darveaux method.

No default (Real)

1. C4GAMMA is used to calculate the strain range of the solder joint in DIFFCTE method:(1)
$\text{Δ}\gamma =C\frac{{L}_{D}}{{h}_{s}}\text{Δ}\alpha \text{Δ}T$
Where,
${L}_{D}$
Component length.
$\alpha$
Coefficient of thermal expansion (CTE).
${h}_{s}$
Solder joint height.
$\text{Δ}\alpha =CT{E}_{p}-CT{E}_{c}$
Difference between thermal expansion coefficients of the PCB and the component soldered onto the board.
$T$
Input temperature.
$\text{Δ}T={T}_{p}-{T}_{c}$
Difference between temperatures of the PCB and the component soldered onto the board.
$C$
Empirical model constant for solder fatigue (specified using the C4GAMMA field). It is equal to $\frac{1}{\sqrt{2}}$ for leadless joint type. For BGA type, you can define the empirical model constant.

Refer to Solder Fatigue Analysis for additional information.

2. K1 and K3 are dependent on the modeling of solder joints such as the thickness of solder joint interface layers and the creep constitutive model. Therefore, K1 and K3 are not material constants. Units of dimension for K1 and K3 should be consistent with those of length and stress on FATPARM. Dimension of K1 is cycles/stressK2. The dimension of K3 is (length/cycle)*(1/stressK4). For example, if length units and stress units in FATPARM are mm and MPa respectively, the units of K1 and K3 should be cycles/MPaK2 and (mm/cycle)*(1/MPaK4), respectively.
The constants are used in the following equations to calculate life (N) of a solder joint:
• Crack initiation:(2)
${N}_{0}\text{ }=\text{ }{K}_{1}\text{Δ}{W}_{ave}^{{K}_{2}}$
• Crack growth:(3)
$\frac{da}{dN}\text{ }=\text{ }{K}_{3}\text{Δ}{W}_{ave}^{{K}_{4}}$
• Life:(4)
$N\text{ }=\text{ }{N}_{0}\text{ }+\text{ }\frac{a}{da}{dN}}$

Where, α is the joint diameter at the interface defined FATSDR.


text:
# Side of square is 20% of length of rectangle if length and breadth of rectangle are in ratio 3 : 2 and area of rectangle is 150 cm2. What is the area of square?

1. 4 cm2
2. 9 cm2
3. 16 cm2
4. 25 cm2

Option 2 : 9 cm2

## Detailed Solution

Concept:

Area of square = (Side)2

Area of rectangle = Length × Breadth

Calculation:

Let be assume the length and breadth of the rectangle is 3p and 2p respectively.

⇒ 150 = 3p × 2p

⇒ p = 5, length of rectangle = 5 × 3 = 15 cm

⇒ Side of square = 15 × (20/100) = 3 cm

⇒ Area of square = 3 × 3 = 9 cm2

∴ The required result will be 9 cm2.


text:
# Beyond score following...¶

Most people think Antescofo is exclusively used for mixed music and score following. That's partly true, but since the language exists and has been developed, Antescofo is also a great tool for electronic music.

## Antescofo as a sequencer¶

Speaking about time is very easy in Antescofo. AscoGraph and an Antescofo score can be seen like a sequencer, where all the actions are organised in time (if you click on the Play button, Antescofo becomes a sequencer) that you specify in the score.

You can write an electronic score with the beat notation and decide after to change the tempo. In this case you don't have to rewrite all the durations. Antescofo does the translation. This allows you to change the tempo (with BPM) in any place in your score.

As we have shown before, you can also work with different times in a same moment with the @tempo attribute and write complex polyrythms.

With macros you can greatly simplify your score to focus on the musical questions instead of the technical realization.

## Hierarchical scores¶

With the system of GROUP, you can write a hierarchical score where the different voices are denoted by different groups. You can also create tracks (see @track_def) to filter any kind of musical object or focus on a particular voice.

With the action view of AscoGraph you can easily see how the different voices of an electronic polyphony interact with each other.

## Open scores and installations¶

If you see the process and whenever constructions, you will see that these features can be easily used for installations or for open scores. You can use Antescofo without a “classical score”, without NOTES, CHORD... It means that you can program a reactive environment that “hears” sensors, audio descriptors...

If you compose open scores, you can take a look to the @jump keyword that is followed by a comma separated list of simple identifiers referring to the label of an event in the score. This attribute specifies that this event can be followed by several continuations: the next event in the score, as well as the events listed by the @jump. So you can compose a non-linear score with choice points leaved to the performer. You evn can modify during the execution the list of allowed jumps, achieving a kind of “multi-graph” score.

## Beyond Max and Pure Data...¶

We have shown many uses of Antescofo in Max. However, with OSC, you can use Antescofo as controller of any kind of software that supports OSC. For example, SuperCollider and CSound are great Antescofo fellow players ! In the two next examples, Max is a kind of "pipe" between Antescofo, CSound or SuperCollider, IRCAM's Spat and audio descriptors. But Antescofo sends also many messages via OSC without Max.

### A SuperCollider example¶

The figure below shows an environment created by José-Miguel Fernandez for his pieces where you can see 3 different softwares. In the background you can see the AscoGraph interface. At the left front you can see Max using Antescofo and descriptors (who are sended to whenever and process). At the right you can see a SuperCollider interface where processes are generated “on the fly” and displayed on the screen. In this example, Antescofo manages all the control parts and sends the parameters needed for synthesis and effects to SuperCollider. Fernandez uses many chaotic funcions included in Antescofo and several of its data structures (e.g., map and tab).

### A CSound example¶

In the next example, Antescofo is used to send a score to CSound and to generate table for CSound. Here, Antescofo can replace the ".sco" file for CSound or the "score part" of the ".csd" in CSound. All the CSound synthesis is generated in real time and controled by the object antescofo~.

The spatialization is controled by Antescofo too with many curves and process that generate complex trajectory.

### OSC¶

Many people have been using Antescofo message passing strategies as defined above to interact with processes living outside MAX/PD (such as CSound, SuperCollider, etc.). To make their life easier, Antescofo comes with a built-in OSC host. The OSC protocol3 can be used to interact with external processes using the UDP protocol. It can also be used to make two Antescofo objects interact within the same patch. Unlike MAX or PD messages, OSC messages can be sent and received at the level of the Antescofo program. The embedding of OSC in Antescofo is done through 4 primitives.

oscsend name host : port msg_prefix

oscrecv name port msg_prefix $v1 . . .$vn

see the OSC message chapter for more details.


text:
Kattis

# Game Design

Carol enjoys playing with wooden games. The objective of the game that fascinates her the most is to tilt a maze, made out of $1\text { cm}$-by-$1\text { cm}$ blocks, in any of the four cardinal directions to move a ball to a hole in the centre at $(0, 0)$. As shown in Figure 1, once the $1\text { cm}$ wide ball starts moving, it keeps going until either it runs into a wooden block, or it falls into the hole—whichever comes first.

Carol is interested in designing a maze of her own, and like any good game designer she already has a fixed solution in mind. This is given as a sequence of tilt moves which must all be followed in order. If any move causes nothing to happen, for example because the ball is up against a block in that direction or already in the hole, the solution does not count.

The ball can be placed anywhere to start. Carol will take care of adding a square border of blocks covering the rows and columns $10^9+1$ cells away from the centre in each direction.

Design a board which can be won by applying her sequence of moves.

## Input

The input consists of:

• One line with a string $s$ consisting of only the characters “LRUD” ($1 \le |s| \le 20$), the sequence of moves. These characters correspond to the directions $-x, +x, +y, -y$ respectively. No two consecutive characters in $s$ are the same.

## Output

If it is possible to create a maze with the given solution, first output $x$ and $y$, the integer coordinates for the ball to start at. Then on the next line, output $n$, the number of blocks to use. On each of the remaining $n$ lines, output $s$ and $t$, the integer coordinates of a block.

Otherwise, output “impossible”.

You may use at most $n \le 10^4$ blocks. All coordinates used must be at most $10^9$ in absolute value. No coordinate pair may be the same as the centre or any other coordinate pair. If there are multiple valid solutions, you may output any one of them.

Sample Input 1 Sample Output 1
DLDLRUR

-3 1
8
-1 -1
-1 -2
-2 1
-3 -1
-5 0
-6 -1
-7 -2
-4 -3

Sample Input 2 Sample Output 2
LRLRLRLRULD

1 1
5
2 1
2 0
-1 1
-1 0
-1 1000000000

Sample Input 3 Sample Output 3
LRLR

impossible


text:
# integration of a 2-form over en sphere

• May 12th 2011, 11:08 PM
rayman
integration of a 2-form over en sphere
$f=\frac{E}{4\pi}(\frac{x}{(x^2+y^2+z^2)^{3/2}}dy\wedge dz+\frac{y}{(x^2+y^2+z^2)^{3/2}}dz\wedge dx+\frac{x}{(x^2+y^2+z^2)^{3/2}}dx\wedge dy)$

over a sphere $x^2+y^2+z^2 \ge \mathbb{R}^3$
• May 13th 2011, 10:15 PM
FernandoRevilla
Quote:

Originally Posted by rayman
over a sphere $x^2+y^2+z^2 \ge \mathbb{R}^3$

That inequality has no sense.
• May 13th 2011, 11:16 PM
rayman
Quote:

Originally Posted by FernandoRevilla
That inequality has no sense.

of course it should be $x^2+y^2+z^2\ge R^2$
• May 14th 2011, 12:22 AM
FernandoRevilla
Quote:

Originally Posted by rayman
of course it should be $x^2+y^2+z^2\ge R^2$

I suppose you mean the sphere $S\equiv x^2+y^2+z^2=R^2$. In such case, apply the Divergence Theorem to $V-V_r$ where $0 . That is,

$\iint_Sw+\iint_{S_r}w=\iiint_{V-V_r}dw=0$

So,

$\iint_Sw=-\iint_{S_r}w$
• May 14th 2011, 12:29 AM
rayman
actually it is supposed to be like I wrote earlier $x^2+y^2+z^2\ge R^2$ that is actually what my teacher says. I will apply Gauss-Otrogradski theorem to it and she what happens.


text:
Maximising H-colourings of graphs

Guggiari, H
Scott, A

1 January 2019

Journal:

Journal of Graph Theory

Last Updated:

2019-09-09T06:43:16.93+01:00

2

92

DOI:

10.1002/jgt.22446

172-185

abstract:

© 2019 Wiley Periodicals, Inc. For graphs G and H, an H -colouring of G is a map ψ: V (G) → V (H) such that ij ∈ E (G) ⇒ ψ(i)ψ(j) ∈E (H). The number of H -colourings of G is denoted by hom(G, H). We prove the following: for all graphs H and δ ≥ 3, there is a constant κ (δ, H) such that, if n ≥ κ (δ, H), the graph Kδ,n−δ maximises the number of H-colourings among all connected graphs with n vertices and minimum degree δ. This answers a question of Engbers. We also disprove a conjecture of Engbers on the graph G that maximises the number of H -colourings when the assumption of the connectivity of G is dropped. Finally, let H be a graph with maximum degree k. We show that, if H does not contain the complete looped graph on k vertices or Kk,k as a component and δ ≥ δ0 (H), then the following holds: for n sufficiently large, the graph Kδ,n−δ maximises the number of H -colourings among all graphs on n vertices with minimum degree δ. This partially answers another question of Engbers.

953282

Submitted

Journal Article


text:
Web lists-archives.com

# [Mingw-users] redefinition of 'struct timespec'

• Date: Sun, 7 Aug 2016 19:03:20 +0800 (CST)
• From: zjuzw <zjuzw@xxxxxxxx>
• Subject: [Mingw-users] redefinition of 'struct timespec'

Hi,

I know that someone has asked the same question before.In my opinion,just deleting the definition of this structure in the header file

"C:\MinGW\include\pthread.h" could be valid for resolving this problem.I wonder if it is the best way to avoid this error at present.

This structure is showed below:

struct timespec
{
time_t tv_sec;
long tv_nsec;
};

All the best,
Wei.

```------------------------------------------------------------------------------
```
```_______________________________________________
MinGW-users mailing list
MinGW-users@xxxxxxxxxxxxxxxxxxxxx

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text:
# Kellogg-Evans theorem

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The set of all irregular boundary points of an arbitrary domain $D$ in the Euclidean space $\mathbf{R}^n$, $n\ge 2$, with respect to a generalized solution of the Dirichlet problem for $D$ in the sense of Wiener–Perron (see Perron method) has capacity zero, is a polar set and has type $F_\sigma$. A corollary of the Kellogg–Evans theorem is that if $K$ is a compact set of positive capacity in $\mathbf{R}^n$ and $D$ is the connected component of the complement $\complement K$ containing the point at infinity, then there exists on the boundary $\partial D \subset K$ at least one regular point. The Kellogg–Evans theorem was stated by O.D. Kellogg [1] as a conjecture, and was first proved by G.C. Evans [2].


text:
Neural Turing Machines – Implementation Hell

I’ve been struggling with the implementation of the NTM for the past week and a half now.

There are various problems that I’ve been trying to deal with. The paper is relatively sparse when it comes to details of the architecture, and a lot more brief when it comes to the training process. Alex Graves trains RNNs a lot in his work, and it seems to me some of the tricks he has used here might have been distributed through his previous work.

Controller Outputs

A huge part of my problems stem from not knowing what the architecture of the controller looks like. The paper describes it’s inputs as coming from the input sequence, and the read head of the memory. It’s outputs are the write heads, and the output sequence.

It then goes on to describe the reading and writing mechanism, sometimes referring to “multiple heads”. This got me confused as to whether there were separate read, erase, and write heads and corresponding weights to go along with each of these. Another possible way to implement this was to have, for each head, an erase, and an add head, with one set of weight parameters (parameters from which we calculate the current weight for this timestep from the previous). This is the version I finally settled with.

The other problem was to decide where the head[s] would go. Looking at previous LSTM papers, the write/forget gates are directly dependent on the input data as well, so this led me to think maybe the heads was a single layer network that directly gave a read, writes and erase on seeing input. I eventually settled for the heads being connected after the hidden layer.

I took a look at this paper to figure out how Graves usually models outputs of different domains. This is what I’ve come up with:

$$\hat{y} = \left(\hat{\key},\hat{\beta},\hat{g},\hat{\shift},\hat{\gamma}\right) = \mathbf{h}^\top\mathbf{W} + \mathbf{b}$$

\begin{align} \key &= \hat{\key} &\\ \beta &= \exp\left(\hat{\beta}\right) &\Longrightarrow &\beta > 0 \\ g &= \sigma\left(\hat{g}\right) &\Longrightarrow & g \in (0,1) \\ (\shift)_i &= \frac{\exp((\hat{\shift})_i)}{\sum_j \exp((\hat{\shift})_j)} &\Longrightarrow & (\shift)_i \in (0,1),\sum_i (\shift)_i = 1 \\ \gamma &= \log\left(e^{\hat{\gamma}} + 1 \right) + 1 &\Longrightarrow & \gamma \geq 1 \\ \end{align}

I have no particular justification to choose the softplus over the $\exp$. In my opinion, exponentiating seems like it would be much more likely to be unstable numerically as opposed to the softplus.

Taking a look at equation (5) in the paper, if we let $\beta = 1$ and $K$ be the dot product function, then we see that equation (5) is actually a general form of the softmax function.

It’s not entirely clear to me why (5) looks more like softmax, while (9) doesn’t. I guess this might have something to do with (9) already dealing with values that are supposed to be probabilities.

Training

I initially used adadelta for training, but because I was unable to get good results, I tried implementing the version of rmsprop Graves used in his previous paper.

The rmsprop algorithm is interesting, and it is slightly different from the one presented by Hinton in his Coursera lesson. In Graves’ variant, he is essentially dividing by the standard deviation of the gradient in recent history. You can take a look at the code here.

I run into a lot of problems with numerical stability, and have tried adjusting learning rates to cope with that. The paper mentions using gradient clipping, but since I have no way of implementing that easily, I left it out.

One hack that seemed to work was ‘squeezing’ the sigmoids down so that high activations don’t create $\log(0)$ situations.

$$\frac{\epsilon}{2} + (1-\epsilon)\cdot\sigma(x)$$

This results in the outputs being between $\left(\frac{\epsilon}{2},1-\frac{\epsilon}{2}\right)$, so the probabilities in cross-entropy calculation is never zero.

If you look at the file I use to do the training, you’ll see the commented code that was there from all the things I’ve tried.


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# varkappa: amssymb AND kpfonts

Between the two packages amssymb and kpfonts, whichever is loaded last determines the look of \varkappa. With the usual tricks using \let and \LetLtxMacro, I have not found a way to simultaneously have both amssymb's \varkappa and kpfonts'

How can I have both versions of \varkappa available within a given document?

So far, I've been able to ascertain that the second package (whichever one that might be at the given compile time) redefines what AMSb is/does, but how to use that information is presently beyond my capacity.

• So what is the actual question? Can you make it into something more precise? May 23 '19 at 16:44
• Do you really want to use both in the same documents? May 23 '19 at 17:31
• One possibility is to, after loading the first package, \let\altvarkappa\varkappa. Then \varkappa will be as assigned by the second package, and the other shape can be accessed by \altvarkappa. There may be situations (probably in moving arguments) where additional measures are needed, but others are better able to provide that information. May 23 '19 at 17:32
• Did you see the posting Importing a Single Symbol From a Different Font?
– Mico
May 23 '19 at 17:35
• Please advise which math alphabet -- CM or kpfonts -- is supposed to be the main math font and which math alphabet is supposed to serve as the source of the alternative \varkappa symbol.
– Mico
May 23 '19 at 17:37

You surely can:

\documentclass{article}
\usepackage{kpfonts}

\DeclareSymbolFont{AMSAMSb}{U}{msb}{m}{n}
\DeclareMathSymbol{\amsvarkappa}{\mathord}{AMSAMSb}{"7B}

\begin{document}

$\varkappa+\amsvarkappa$

\end{document}

I feel sympathetic towards the poor readers that will not be able to distinguish the two symbols.

• Very nice the first kappa :). My upvote. May 23 '19 at 20:56


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# Extract/Project IQTensor on a QN subspace

Hi all,

I am fairly new to ITensor and C++.
Thus if you have any advice that can be easier implemented in Julia I could take it into account, too.

First let me describe the system I am dealing with:
I consider a Bosonic chain of N sites: Boson(N,{"ConserveQNs=",true,"MaxOcc",1})).
I would like to construct the reduced density matrix of the system and project it on a given subspace.
In formulas, given the reduced density matrix @@\rho_A=Tr(|\psi><\psi|)@@
I would like
$$\rho_A(q)=\frac{P_q \rho_A P_q}{Tr(\rho_A P_q)}$$
Where @@P_q@@ is the projector on the subspace with a given number of particles
$$\sum_{i=1}^{l}n_i=q.$$
Here @@n_i@@ is the number of Bosons at site i and l is the length of the subsystem.

The question is the following: is there a way to do obtain @@\rho_A(q)@@, using IQTensor, without computing explicitly the reduced density matrix and the projector, since it is very inefficient?

I saw in the source code of qutil.h the function getBlock, but I'm not used to C++ and I could not deduce its functionalities.

Best,
Vittorio

commented by (51.6k points)
Hi Vittorio, yes I think there should be a good way to do this, with mostly tools we have available in the code already. (Yes it would be easier to do some of this with the Julia version since it is more flexible and offers more low-level tools to manipulate QN blocks.)

I assume your state |psi> is an MPS correct?

Best,
Miles
commented by (120 points)
Yes, It is an MPS.

Vittorio


text:
# When a star becomes a singularity and has zero volume and infinite density, it is called

A. White dwarf
B. Blue giant
C. WolfRaynet star
D. Black hole
E. None of above


text:
# Underwriting discipline is only partly to blame

We know that underwriting conditions are becoming increasingly difficult and, naturally, the finger is pointing towards a failure in underwriting discipline. When challenged, the management of som


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text:
# Short Prime Number Divisibility Question

Question: Calculate all prime numbers $$x$$, where $$x^{18} - 1$$ is divisible by $$28728$$

Apparently, the answer is all prime numbers except 2, 3, 7, and 19. I did some prime factorisation and found that 28728 = $$2^3, 3^3, 7, 19$$ and spotted that the numbers that x cannot be with it. However, I still don't know why and how to prove that the answer can be all prime numbers except 2, 3, 7, and 19.

Any help would be extremely appreciated :)

• Fermats theorem that really fits in the margin guarantees that $x^{p-1}-1$ is divisible by $p$ iff $x$ is prime to $p$. Apply this for $p$ among the numbers above. (And also note $x^{18}-1$ is divisible by $x^d-1$, where $d$ divides $18$.) – dan_fulea Jun 22 at 3:48
• For example $37^{18}-1$ is divisible by $28728.$ – Michael Rozenberg Jun 22 at 3:54
• Consider $x^{18}\equiv 1\pmod{m}$ for $m=8,9,7,19$. By the Chinese remainder theorem, a simultaneous solution to all of them solves the original problem. Fermat's little theorem, and Euler's generalization dispose of all $4$ cases. – saulspatz Jun 22 at 4:04

The Carmichael function $$\lambda(n)$$ gives the smallest exponent $$k$$, for which $$x^k\equiv 1\mod n$$ holds for all $$x$$ coprime to $$n$$.
Because of $$\lambda(28728)=18$$ , the congruence $$x^{18}\equiv 1\mod 28728$$ is satisfied for all $$x$$ coprime to $$28728$$ and in particular for all primes except $$2,3,7$$ and $$19$$.


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# Getting my own module to work in Sage

Hey everyone. I wrote a sage module that defines a couple functions that I'd like to be able to import into sage any time I want. From what I've researched, it seems the way to do this is save the file that defines the functions as a .py file, then put it into sage's version of python. I've done that, and sage imports the file just fine without errors, except when I attempt to call functions from the file in the sage prompt, it gets stuck on the first sage-specific function, in this case var(). I get this error message:

 sage: polyv1.sift(4, 3)

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-20-d64cf99d698f> in <module>()
----> 1 polyv1.sift(Integer(4), Integer(3))

/usr/lib/sagemath/local/lib/python/polyv1.py in sift(r, s)

NameError: global name 'var' is not defined

I've tried adding 'from sage.all import *' to the beginning of the file, and it doesn't create an error, but it also doesn't fix the problem. Am I missing something really easy? How do you get sage to recognize the sage functions?

edit retag close merge delete

Sort by » oldest newest most voted

You have to import what is used in the file. For this, there is a nice tool:

sage: import_statements("var")
from sage.calculus.var import var

then you just have to copy-paste from sage.calculus.var import var in the beginning of your file.

more

Unfortunately still getting the same error, any other ideas?

( 2018-02-20 22:20:34 +0200 )edit

Could you please provide your code so that we understand what happens ?

( 2018-02-20 23:39:39 +0200 )edit

For sure. The entire document is pretty long, but I've taken what I think are the most important parts. Let me know if anything else seems important. Here's the header of the document (everything I'm importing):

from sage.calculus.var import var
import itertools
from sympy.utilities.iterables import multiset_permutations

and here's the beginning of the function that gives the above-mentioned error when I call it:

def sift(r, s):
polynomial = 1

p = r/2
q = (s-1)/2
poly_degree = 0

#For x-terms
for j in xrange(2, r+1):
for i in xrange(1, j):
term = (var("x" + str(j))-var("x" + str(i)))
polynomial = polynomial *term
poly_degree += 1

it's the first 'var("x" + str(j)) there that's giving me the error.

( 2018-02-21 00:01:00 +0200 )edit

This is all in the same file right? If the import succeeded (i.e. there's no ImportError) then you should have anything you imported defined globally (within the module that imported it). Beyond that, if you're still getting a NameError then there's something else strange in your code that would be hard to pinpoint without seeing all of it.

( 2018-02-21 14:23:28 +0200 )edit

Turns out it was a mistake that I was making, not a code problem! Sorry for troubling you guys, turns out that first answer was totally correct!

( 2018-02-21 16:04:13 +0200 )edit


text:
If monthly income is withdrawn from a $350,000 trust fund invested at 6% compounded continuously, what will the income be if taken over 25 years? ## Answers (1) •$300,000
Comments
• Anonymous commented

how did you get this answer

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text:
# Rookie Kayak Fishing Middle Tennessee in July

Keywords: #fishing

I wanted to try kayak fishing, but wasn’t sure I’d like it. So I bought a (nice) inflatable kayak and have loved the experience. Thought I’d share what I’ve learned so far.

I’ll also just include my experience fishing in late July on the Defeated Creek section of the Cordell Hull Reservoir (part of the Cumberland River).

## Quick Photos

I don’t have a waterproof camera, so only have limited 1st person iPhone shots of my kayak experience. But here’s a couple:

## Kayak Setup

I bought an Intex Explorer K2 kayak. It’s a “2 person” kayak, but I’m 6’ tall, and only my kids could fit in the front. I really only use it myself.

I’ve been fishing the Cordell Hull Reservoir outside Carthage, TN. It’s a dammed section of the Cumberland River. Since it’s managed by the Army Corps of Engineers, there are no docks outside of the few marinas on the lake. I love it.

In particular, I fish the Defeated Creek area. I do love that Defeated, TN is just south of Difficult, TN. Mostly around 2 and 8 on the map.

My wife works remotely for a company in Carthage, so we started coming out here for her work and me to chill with the kids outside from time to time. Now I like coming here for any vacation because I know the waters. Also I know the local restaurants well enough that folks recognize us and we’re not “outsiders” anymore for the most part.

That’s both a nice feeling, as well as a good way to get inside info on fishing 😄.

## My First Week, Late Spring/Early Summer

I’d never paddled a kayak before I got in. I’ve spent months in canoes with river trips and 2 week long lake trips in Canada, portaging from lake to lake. Really since I was 8 or 9. I wasn’t worried, but wasn’t 100% sure it would come naturally to me. It really did. I can scoot around in that thing with ease and practically spin in place.

I immediately fell in love. I do a lot of shore fishing, so it gets me all the angles and distance I wish I could have. Versus a “real” boat, it’s dead silent, easy to maneuver, and I can go way shallower. And it just feels so natural and peaceful.

I’d bring 2-3 prerigged rods with me, a dry bag with essentials, and a water bottle. Gotta hydrate.

Catching even a 3 lb bass on the thing is so much fun. It’ll actually pull the boat as you reel it in. I get a mix of the joys of lake fishing and fly fishing.

The boat is not tippy and I never felt insecure. I do wear a life jacket, though. Otherwise my wife would kill me. It is a “fishing” life jacket, so does not get in the way of casting and has some handy pockets.

## The Problems

### Wind

Paddling against the wind isn’t bad, but once you set the paddle down, it’s a sail. I had to stick to coves and the little deltas formed by creeks. Even then, I wouldn’t be able to hang in a spot. I’d fish an area, paddle back, fish it again, and repeat.

### Casting

The kayak gives you super human casting abilities. I joke, but I seriously cast 1.5-2x as far as I do from the shore even though I’m sitting on the water. It takes some getting used to.

Getting snagged on rocks is not a problem. Just let the line out, move around and tug. The mobility is awesome there. Sometimes I’ll intentionally land a lure on a rock outcrop and pull it into the water. It’s a good way to draw the fish out of a gap near the shore or from under the outcrop.

## Second Week

I got to come back again this week with the lessons from last under my belt.

### Glasses

I’m nearsighted and wear glasses all the time. I’ve used a variety of clip on and blueblocker-style sunglass things that cover my glasses and they’re all obnoxious. I finally got a nice pair of polarized sunglasses a couple years back. When I forget them, it’s terrible. It’s bad enough being just off the water. Without them, I’m almost blind below the surface if there is any wind at all.

### Casting

Just start casting out to open water before you start playing with the shoreline. Do this for every rod and lure weight as you roll through. Rocks aren’t a problem, but getting snagged in trees sucks.

This is a deathtrap until you have you have a handle on things.

### Anchor

First, I bought a 3.5 lb anchor. I wanted to be able to park and fan cast an area. The anchor was a great buy and let me fish after the wind kicked up. I’d even just let out all the rope so I could adjust but not worry about drifting far.

This is the one I got on Amazon Prime Day. Probably any of them are fine. “Gradient Fitness” must be some Chinese brand, because it makes 0 sense for the product.

It is a grappling hook-style anchor. So when I do pull it up, I usually have to jig it up and down a bit to get it loose. That’s not an issue to me.

The one difficulty with the anchor is that you really need to attach it to the back handle of the kayak and it’s difficult to reach back when you want to pull it in. I did end up attaching it to the back of my seat and running it back through the handle. That helped a bit. I don’t have a good solution here. I want it coming out straight back, so it’s not pulling me cockeyed. Being inflatable does make this more difficult, I imagine.

### Seating Position

For positioning, I try to move the rear seat as far forward as I can. I tried using the front Velcro to go all the way back, but it sits me too far forward. It’s unfortunate that the front has a long stretch to adjust, but the rear is really made to be in a fixed position.

I’d like a longer paddle. I’m exactly 6’ tall and it makes me really dig more than cruise. But then I end up spending 1/2 the price of the kayak on a paddle. I’ll probably go for a “real” sit-on-top fishing kayak before I do that. The paddle shape, etc. is fine for me, and I can maneuver the boat really well.

## Late Summer Bass in the Cordell Hull

I know hot/late summer is not thought of as “bass time”, but it’s actually probably the easiest to catch fish right now. Now, I don’t fish deep for pike and such that do exist. But you can hit largemouth and striped bass with some regularity.

### The situation

Right now, shad pools are running high in the deeper water. Striped bass are tearing these up, but they pop up and disappear quickly. There are tons of larger minnows as well, and the largemouth are lurking around the rocky dropoffs between the shallow and deep water.

The Cordell Hull, being the dammed Cumberland River goes quickly from 5-10’, to ~ 30’, to crazy deep, especially in the old riverbed. Rock features and creek inlets are pretty great.

### My Habits

Spend a little time using your senses. Listen to the wind and watch the patterns on the water. Listen for splashes, big and little. For the latter, what’s coming from the shore (tree junk, bugs), what’s evidence that minnows or shad are being eaten up. Try to discern the topography underwater.

This is an example of the area. It’s hard to see in the photo, but whatever. There’s a very shallow 3’ or so patch, followed by 5-10’, then a big drop off to who knows how deep. Between at each drop, there’s great features for fish to hide in and snipe food.

In the deep water you can occasionally see schools of shad rushing the surface. They’ll usually be followed by a striped bass erupting out of the surface.

Don’t target the splash. It’s too late. You want to watch for the shad, cast past and reel through it when you can. You can kind of jerk the lure through to help draw attention. You won’t always hit, but when you do, hold on. It’s fun.

### My gear

Fishing from a kayak, I’m a little modified from what you would think of as bass gear. Mostly a lot lighter weight. I bring a few rods with me, so I’m not having to juggle tackle as much.

I do use spinning reels because I find them most satisfying. Fishing for me is primarily about the experience. I like the tactile feeling of the line snapping off my finger as I cast. I like the way the rod balances in my hand. And I just don’t enjoy baitcasting unless I really have to.

I also believe in just using gear you like. Use what’s comfortable, what you can afford, and even what just looks and feels cool to you. If you don’t enjoy it, it’s a waste of money even if it is the “best.”

I’ve found I really like the Daiwa Regal reels. They’re not pricey at around $60, but cast and reel smoothly, have tough bails that don’t bend when I toss them in the car, etc. I usually have two 2000 reels on medium action rods. I have an old Berkeley Lightning Rod that’s maybe 5.5’, and a 6’ Bass Pro somethingerather I use. I’m a convert to just getting Bass Pro rods. They’re well built and usually have one that matches my preferences. I do like the shorter rod for shore casting with brush, etc. nearby. The 6’ is nicer on the kayak. Bass Pro does have the greatest pyramid in the world, in my home city of Memphis and that may contribute to my bias. Sure, there’s Giza and the Luxor, but they don’t have long-nosed gar in them, so the Bass Pro pyramid wins. It also is the best use of the building that was the absolute worst concert venue I have ever been in and was terrible for basketball as well. Second, I bring a long ultralight Bass Pro rod to be able to chuck 1/8oz lures. I bought it for trout, but I’ve caught 5-6lb bass on them without issue, they’re fun. I use a cheap KastKing reel I got off Amazon. I liked one I got for surf casting. I don’t get to do that a lot and didn’t want to drop bank. It’s better than a cheap store-bought reel or even a low-end Shimano, but I wouldn’t buy it again and will replace eventually with a 1000 Daiwa of some variety. Lastly, I have a heavy action Daiwa “bass” rod, that I got because it was the only one available at the time. COVID jacked up fishing gear supply chain stuff bad a couple years ago. I have a Shimano Sahara 3000 on it that I kind of hate. It reverses a couple inches of line out and that drives me insane because you cannot keep a lure in the loop on the rod to save your life. Also, it has a tendency to birds nest and when it does, it’s tragic. I regret that purchase. I had a couple 2000 Sedonas that I replaced with the Daiwas because I could not keep the bails straight on them. I’ve given up on Shimano. Sometimes I don’t bother bringing that one except boat or shore finishing. I do like it for catfish, but that’s another story. ### Line and Usage I use 6lb line on the mid/light action rods and 10-15 on the heavy. I like the more transparent line, usually Trilene, but I’m not super-picky. I tend to use polomar knots, having converted completely away from a traditional clinch knot. I use a Trilene when a polomar is too much of a pain to tie (like quickly switching to a multi-hook lure). I don’t know that choice matters that much, but I’ve snapped at the knot too many times setting the hook, dealing with a fish jump and shake, or landing a fish to ever use a clinch knot again. I set the drag about 3/4 tight and then adjust after hooking the fish. A larger fish can be like fly-fishing. You let them run, take what you can get, and repeat. Using light gear means you can’t just muscle, and full muscle on the kayak would be rough. You do have to have the drag tight enough to really set the hook at the outset. I lost a few fish at the beginning of this week to starting to light. A rear drag reel would be nice here, but I haven’t seen/used one under about 4000-sized (including the only Shimano I own and like that’s probably 20 years old). I don’t have that much trouble dealing with the front-drag, though. ### Terminal Tackle I think a lot of belief around lures are just pseudoscience. Drawing lines from anecdotal experience. I don’t like using live bait, because it’s a mess, I’m doing this for fun and not food, and I catch-and-release exclusively in fresh water. #### Plastics A standby for me is just plain white/red eye Berkeley Powerbait Poly Shad. I keep them in 1/8oz and 1/4oz. It’s my “when in doubt” on the lake. I’ve caught large mouth and striped bass with them, and one time just blew up some Tennessee Tarpon/Skipjack/Giant Minnow things. I’ve tried similar non-Powerbait looking plastics without luck. I think it’s just the movement. I dunno. I’ll vary the speed I pull them in, and sometimes do a jerk-reel-jerk. If I play around and they’re biting, I’ll find a pattern that works. I almost always have 2 packs of the 1/8oz and 1 of the 1/4oz. The smaller get more fish, the bigger get bigger fish. “Tennessee Tarpon” are fun as hell to catch, but are super oily and shed scales like crazy. The scales will tear up your hands pretty bad too. They’re “trash fish” that are usually just cut up for catfish bait around the dam, but I get a kick out of them fighting and jumping. #### Hard Lures This time I found that the Rapala Ripstop something-9 is amazing in the particular coloring below. I tried other minnow patterns, different depth and coloring, but this was the one. The water can be murky so maybe something about the white and chartreuse? They’re almost$10, and I always get weird about putting that much into 1 lure. I lost one in some underwater tree roots from a pontoon boat I couldn’t get shallow enough to recover it. This upset me more than it probably should have, but I did buy another.

I’ve tried the other Rapalas I mentioned, Rattle Traps, and more proper shad lures. I’ve tried varying depth and color without any real luck. You may be better than me at that.

I also use some spoons and variations. You can tell if they work within a few casts.

#### Slaying Bluegill

If you’re not catching fish and are just bored, you can do what my kids have found to work. My oldest daughter came up with this and will just yank up little bluegill every 5-10 mins when they’re active.

I let my kids use what they want and do what they want to do. My dad was kinda pushy and would ruin trips for me. I just want them to have fun. They pick all of their gear. I give them a 5 min lesson from time to time or point out where fish are. Otherwise, I only give advice when asked.

She will toss on a pink Trout Magnet with a gold jig head, just like the photo below. She’ll drag them across the rocks or kind of jig and swim them. Just do it near some rock cover and away from roots and wood.

This is the kind of area that works best. The rock outcrop hides bluegill which sneak out and eat minnows and bugs that fall in.

Maybe the way it swims? I dunno. She’s magic.

She found this fishing local park ponds outside Memphis, nailing Bluegill and Pumpkinseed. With the muddy bottoms there, she usually drags across or just above. At 5-7pm she’ll destroy them from 2-3in to 6in. She loves the coloring and taking her own fish off the hook and doesn’t care about the size.

My youngest finally found something to rival it on our last day. She picked out a chartreuse and pink straight tailed jig with a 1/16oz jig head (red hook, because red is cooler). She tore up the bluegill with them, got bored, and I pulled more out. Just dance it around off the bottom right by the shore. Could have just had them on a string on the end of a stick.

She enjoys the gear: helping me choose at the store or out of the tackle box. Or even just watching me rig poles. She gets disappointed if I reload reels with line without her. She’s got my fidgety genes and is my best helper with this stuff as well as things like electronics, building, and we rock the Legos.

## This Week’s Scoreboard

I really like fishing. Catching fish is just bonus. Being by the quiet water, hearing nature, etc. That’s my sanctuary.

But the family likes superlatives. My youngest didn’t really participate this week. So here is is:

• Biggest Fish: Me! Actually a few of the largemouth I caught were the biggest. Cheating factor: I fished a lot more.
• Most fish in a day: My oldest daughter tore up the bluegill and caught more in an hour than I did in probably 4.
• Prettiest Fish: My wife caught a beautiful 4ish lb striped bass. Also first fish of the trip.
• Best Gear Finder: My youngest for both picking out the Rapala mentioned above as well as the new jig pattern.

## Parting Thoughts

I really do want a proper sit-on-top fishing kayak. It’s just hard to justify with so few locations I could fish local to my house. Local park lakes ban boats so there’s only one public lake within 30 mins of me.

And the investment is more than a boat. I’ll need other accessories like the better paddle I mentioned, a roof carrier, a cart since they’re heavier, etc.

But I want one. 😊


text:
# Tag Info

2

For technical reasons, much of the work of unicode-math related to defining the meaning of symbols is performed at begin document. In particular, \int is among the symbols defined there. So use \AtBeginDocument{% \let\oldint\int \renewcommand\int{\oldint lol}% }

3

\documentclass[a4paper,11pt]{book} \usepackage{unicode-math} \usepackage{libertine} \addfontfeatures{Numbers={Lining,Monospaced}} %\setmonofont{Consolas}% not available for me \setmathfont[math-style=ISO,bold-style=ISO]{Cambria Math} \setmathfont[range=0048-0057]{Linux Libertine O} \usepackage{siunitx} \begin{document} [ ... ]

6

OpenType support in TeX as of when I first wrote xfrac in early 2004 was basically non-existent - XeTeX hadn't even been released back then. Therefore, the package was never designed with that in mind. Besides, the fundamental purpose was to provide a way to write nice fractions for fonts that didn't include them. The package could very likely be made ...

3

Here I create \altfrac{}{}. It does not require fontspec, it obeys math mode, its font is larger than that of \sfrac, the denominator lies on the baseline, while the top of the numerator I tries to lay at the top of the normal text font. The numbers are presented in \footnotesize. The slash is a horizontally stretched / to give it more of the fractional ...

5

I think that rather than look at unicode-math it's more natural to look at inputenc's utf8 support (if your main interest is textual accented characters rather than math symbols). That maps Unicode input to classic latex markup. The base latex distribution has a file utf8enc.dfu that contains the mapping as far as it is implemented. As memory constraints ...

6

This works for roughly the reasons explained in the comments: \documentclass{beamer} \usefonttheme{professionalfonts} \usefonttheme{serif} \usepackage{fontspec} \setmainfont{Latin Modern Sans}%TeX Gyre Schola} % \setsansfont{Latin Modern Sans} \usepackage{unicode-math} \setmathfont{TeX Gyre Pagella Math} \usepackage{siunitx} \sisetup{% detect-all,% } ...

2

If you don't need it presented in \scriptstyle, etc, then this might suffice. \documentclass{article} \usepackage{graphicx} \def\depend{\mathbin{% \ooalign{$\perp$\cr$\,\perp$\cr\scalebox{1.3}[.5]{\raisebox{5pt}{$\mkern2mu/$}}}}} \begin{document} $\mathcal{G}\depend X$ \end{document} If one does need it in different math styles, then maybe this: ...

1

symbols-a4 gives a possible definition for \independent on p. 212 (§10.3): \newcommand\independent{\protect\mathpalette{\protect\independenT}{\perp}} \def\independenT#1#2{\mathrel{\rlap{$#1#2$}\mkern2mu{#1#2}}}

Top 50 recent answers are included


text:
# Why isn't temperature measured in joules per mole (J/mol)? [duplicate]

Temperature is the average kinetic energy of all the molecules in a system.

Thus \begin{align} T &= \frac{1}{N}\sum \frac{1}{2} m v^2 \\[6pt] &= \sum \frac{E_\text{single particle}}{n} \cdot 6.022 \times 10^{23} \, . \end{align} where $N$ is the number of particles and $n$ is the amount of substance.

The unit here is J/mol. Why isn't this used in place of temperature? (Instead of K, °C or °F)?

## marked as duplicate by Nathaniel, Jon Custer, user36790, John Rennie thermodynamics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 25 '16 at 4:41

Temperature is the average kinetic energy of all the molecules in a system.

It’s not that simple. Let’s first look at an ideal, mono-atomic gas. Here we have:

$$T = \frac{2}{3 k_\text{B}} \bar{E}_\text{kin},$$

where $k_\text{B}$ is the Boltzmann constant and $\bar{E}_\text{kin}$ shall be the average kinetic energy of each particle. Apart from the proportionality constants, this is what you were describing. However, the $3$ in that equation is the number of the degrees of freedom of each particle. If we consider an ideal gas whose particles are molecules composed of two atoms (e.g., oxygen), we have five degrees of freedom and thus:

$$T = \frac{2}{5 k_\text{B}} \bar{E}_\text{kin},$$

In real situations (i.e., with non-ideal gases, fluids, and solid states), things become much more complicated here, as there are various degrees of freedom with differing properties and all of this becomes temperature-dependent on top. This is what is usually expressed in the heat capacity of a material.

Therefore there is no simple way to define temperature using kinetic energy. Now, you may ask: “Why is temperature defined such that we have to bother about heat capacities in the first place? Couldn’t we just define it via the average kinetic energy despite all of this?” In theory, we could of course do this, but then we would lose one of the most practical properties of temperature, namely that two bodies with the same temperature are in thermal equilibrium. For example, two materials which melt under the same conditions would very likely have a different melting "temperature"¹, or every object in your room would have a vastly different "temperature". Also, it would be very tedious to measure that "temperature".

¹ in the new sense

• I never understood why so many texts say temperature is average kinetic energy... – Mehrdad Sep 24 '16 at 9:37
• +1. The most important thing about temperature is that energy flows across a temperature gradient, which is an amazingly simple, useful and intuitive idea that if you start talking instead about the thermal energy in some object gets lost in the complexity. – Periata Breatta Sep 24 '16 at 14:25
• @Mehrdad I agree, that confused me a lot. – Srathi00 Sep 24 '16 at 14:40

If you read this report from the body who are responsible for the definitions of SI units (BIPM) you will realise that yours is an excellent suggestion made a few years too late.

The only difference is that BIPM have not used the mole in their definition of the kelvin:

The kelvin is the thermodynamic temperature at which the mean translational kinetic energy of atoms in an ideal gas at equilibrium is exactly $(\frac 3 2) \times 1.380 65{\rm X X} \times 10^{−23}$ joule

or variations of that statement, for example:

The kelvin is the change of thermodynamic temperature that results in a change of thermal energy kT by exactly $1.380 65{\rm X X} \times 10^{−23}$ joule.

or

The kelvin, unit of thermodynamic temperature, is such that the Boltzmann constant is exactly $1.380 65{\rm X X} \times 10^{−23}$ joule per kelvin

With, I think, the last definition favoured at the moment and reasons for not adopting the other definitions are given in the BIPM paper.

So the plan is to measure the Boltzmann constant as accurately as possible and then state that that value is fixed (cannot be measured just like the seed of light in the definition of the metre).
That is why the "X X" appears above as the defined value has yet to be fixed.

Again as for the newish definitions of the metre and the second it will mean that current temperature measuring devices will not have to be recalibrated.

The reasons for this new definition are given in this sentence:
The definition will be generalised, making it independent of any material substance, technique of realization, and temperature or temperature range, to ensure the long-term stability of the unit.

• Wow, I am going to read that but it seems tooooo long. But thats seems interesting. – Srathi00 Sep 24 '16 at 14:48


text:
## Modular characteristic classes for representations over finite fields

Publikation: Bidrag til tidsskriftTidsskriftartikelForskningfagfællebedømt

The cohomology of the degree-n general linear group over a finite field of characteristic p, with coefficients also in characteristic p, remains poorly understood. For example, the lowest degree previously known to contain nontrivial elements is exponential in n. In this paper, we introduce a new system of characteristic classes for representations over finite fields, and use it to construct a wealth of explicit nontrivial elements in these cohomology groups. In particular we obtain nontrivial elements in degrees linear in n. We also construct nontrivial elements in the mod p homology and cohomology of the automorphism groups of free groups, and the general linear groups over the integers. These elements reside in the unstable range where the homology and cohomology remain mysterious.

Originalsprog Engelsk Advances in Mathematics 323 1-37 37 0001-8708 https://doi.org/10.1016/j.aim.2017.10.029 Udgivet - 2018

ID: 215132889


text:
# Words in text counter

This accepts text and returns a list of all words which occur with a required frequency (default frequency is 2). It works well, but gets too slow with large texts, more than 1M words. Where can this function be improved to speed searches in larger text?

from collections import Counter
import re

def words_in_text_counter(in_text,frequency=2):
no_noise_text = re.sub(r'[0-9.,:;<>"#!%&/()=?*-+]','',in_text.lower()).strip().split()
frequency_dict = {}
for key, value in Counter(no_noise_text).items():
frequency_dict.setdefault(value, []).append(key)
try:
print(str(frequency)+' times words in text.')
return frequency_dict[frequency]
except KeyError:
return str(frequency)+' times never happens in text.'

• Do you want the full frequency_dict returned, or only those word with a given frequency, or above a given frequency? Kind of doesn't make sense to just return the words at a given frequency, and then having to rerun a second time if you want to check for another frequency... – holroy Sep 14 '15 at 17:14

no_noise_text = re.sub(r'[0-9.,:;<>"#!%&/()=?*-+]','',in_text.lower()).strip().split()

There's a lot to dislike about this line. It horribly violates the Law of Demeter. It takes a while to figure out what it is supposed to do. It uses the wrong tool for the job. And it's inefficient.

First, to start with, if the goal is to split a strict by non-alpha characters, then we should just use the correct function: re.split:

words = re.split("[^a-z']+", in_text.lower())
counts = Counter(words)

Next, this is going to be inefficient. We have to walk the entire string once (to make it lower-case), then again to split it. And then we have to keep all of the words in memory. We don't have to do any of that if we really start with the correct function: re.finditer:

re.finditer("([a-zA-Z']+)", in_text)

That will give you an iterator over the words. Which you can stick into a generator expression that you can pass to Counter:

words = re.finditer("([a-zA-Z']+)", in_text)
counts = Counter(m.group(0).lower() for m in words)

And if you're already using Counter, you may as well use defaultdict:

freq_dict = defaultdict(list)
for key, val in counts.iteritems():
freq_dict[key].append(val)

Note that I'm using iteritems() (which just gives you a generator) rather than items() (which must give you the full list).

Furthermore, why do we need a frequency_dict? We only care about those words which match the given frequency. Why not just keep a list?

frequency_list = []
for key, val in counts.iteritems():
if val == frequency:
frequency_list.append(key)

Putting it all together

from collections import Counter
import re

def words_in_text_counter(in_text, frequency=2):
words = re.finditer("([a-zA-Z']+)", in_text)
counts = Counter(m.group(0).lower() for m in words)

frequency_list = []
for key, val in counts.iteritems():
if val == frequency:
frequency_list.append(key)

if frequency_list:
print(str(frequency)+' times words in text.')
return frequency_list
else:
return str(frequency)+' times never happens in text.'

• I'm kind of unsure what the OP really wants, but wouldn't it make more sense to actually return the frequency_dict, and then provide some simpler map function to get either the word list for a given frequency, or frequencies above a given level? – holroy Sep 14 '15 at 17:16
• Thanks Barry it is much faster now. I was unable activate counts.iteritems() because iteritems() method not available in Counter and use items() instead – mmachine Sep 14 '15 at 17:54
• .iteritems() is only available in Python 2.x; the OP has tagged the question with python3 which means that .items() is fine (and the only approach, afaik) – alexwlchan Sep 14 '15 at 21:54


text:
## An Icosahedral and Brioschi quintic identity

Here’s an identity I found.  For arbitrary r, define,

$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$

and,

$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$

then,

$w^5-10aw^3+45a^2w-a^2 = 0$

Those two complicated expressions neatly wrap up into that last equation, doesn’t it?  This is the Brioschi quintic form which the general quintic can be reduced into.  Two of the polynomials are easily recognizable as icosahedral invariants, while,

$P(r) = r^6+2r^5-5r^4-5r^2-2r+1$

is a polynomial invariant for the octahedron. This gave rise to the question here.


text:
# Math Help - Formulating the BV problem

1. ## Formulating the BV problem

In a rod length L with insulated lateral surface and thermal constants $c, \ \rho, \ \kappa, \ K$ heat is generated at a rate uniformly proportional to the temperature, that is, at a rate $ru(x,t)$ per unit volume per unit time, where r is a constant and $u(x,t)$ is the temperature function of the rod. The ends of the rod are maintained at temperature 0 and initially the rod has uniform temperature 1.

Formulate an initial-BV problem for determining $u(x,t)$

$\text{D.E.}: \ u_t=ru_{xx}$
$\displaystyle\text{B.C.}=\begin{cases}u(0,t)=0\\u( L,t)=0\end{cases}$
$\text{I.C.}: \ u(x,0)=1$

I am not sure what to do with the r so I just put it in front of u_{xx}. Is that correct? Also, I need a K, c, and rho somewhere but not sure where and why.

2. r does belong there. It is known as Thermal Diffusivity.

It relates thermal conductivity and volumetric heat capacity together. It specifically shows how conductivity changes in comparison to the thermal bulk. As to why it belongs there, I don't know. Maybe seeing the equation will help you figure it out.

$r = \frac{K}{\rho c_p}$

3. Originally Posted by Aryth
r does belong there. It is known as Thermal Diffusivity.

It relates thermal conductivity and volumetric heat capacity together. It specifically shows how conductivity changes in comparison to the thermal bulk. As to why it belongs there, I don't know. Maybe seeing the equation will help you figure it out.

$r = \frac{K}{\rho c_p}$
But by making that substitution, I will lose r. In the final solution, there is an r, rho, K, and c.

4. The DE is

$\displaystyle\text{D.E.}: \ u_t=ku_{xx}+\frac{r}{c\rho}u$


text:
The dual space of a vector space $V$ over a field $k$ is the vector space of all linear maps from $V$ into $k$.
The dual space of a vector space $V$ over a field $k$ is the vector space of all linear maps from $V$ into $k$.


text:
# If both A and B columns of a dataset satisfy condition X, then create another column with a value A,B

data = Dataset[{<|"userId" -> 5324,
"PerfFreeResponse" -> 0.8181818181818182,
"PerfMultipleChoice" -> 0.5, "PerfSingleChoice" -> 0.5,
"PerfWriteCode" -> 1., "MinPerf" -> 0.5|>, <|"userId" -> 5325,
"PerfFreeResponse" -> 1., "PerfMultipleChoice" -> 1.,
"PerfSingleChoice" -> 1., "PerfWriteCode" -> 1.,
"MinPerf" -> 1.|>, <|"userId" -> 5326,
"PerfFreeResponse" -> 0.9090909090909091,
"PerfMultipleChoice" -> 0.6666666666666666,
"PerfSingleChoice" -> 1., "PerfWriteCode" -> 1.,
"MinPerf" -> 0.6666666666666666|>, <|"userId" -> 5330,
"PerfFreeResponse" -> 0.6363636363636364,
"PerfMultipleChoice" -> 0.6666666666666666,
"PerfSingleChoice" -> 1., "PerfWriteCode" -> 1.,
"MinPerf" -> 0.6363636363636364|>, <|"userId" -> 5331,
"PerfFreeResponse" -> 0.6363636363636364,
"PerfMultipleChoice" -> 1., "PerfSingleChoice" -> 1.,
"PerfWriteCode" -> 0., "MinPerf" -> 0.|>}]

I want to append another column which tells me which type of question (Free response, Multiple Choice, Single Choice, write Code) has the minimum score. I wrote the following code for that:

  data[All, Append[#, "MinPerfType" ->
If[#MinPerf == 1., "Excellent Performance",
If[#MinPerf == #PerfFreeResponse, "Free Response",
If[#MinPerf == #PerfMultipleChoice, "Multiple Choice",
If[#MinPerf == #PerfSingleChoice, "Single Choice",
If[#MinPerf == #PerfWriteCode, "Write Code", "Blah"]]]]]] &]

But the problem with this code is, if two columns show the MinPerf (except when MinPerf==1.), it gives me the name of only one column and not both of them. Eg., the first row of data shows MinPerf = 0.5, which is true for both Multiple Choice and Single Choice. In that case, I want MinPerfType to show both Multiple Choice and Single Choice and not just either one of them.

• The first advice I would give you, is to look up Which to handle the cascading conditions. Also, wouldn't MinimalBy be useful for this? MinimalBy[assoc, Identity] returns an association containing all keys that have minimal values. – Sjoerd Smit May 1 '19 at 13:22
• Thank you for your comment. I used MinimalBy and wrote this: data[All, Append[#, "MinPerfType" -> If[#MinPerf == 1., "Excellent Perf", MinimalBy[{#PerfFreeResponse, #PerfMultipleChoice, \ #PerfSingleChoice, #PerfWriteCode}, #MinPerf]]] &]. But I need names corresponding to the columns that have the minimum score. MinimalBy satisfies the condition of giving two minimum values but not the corresponding names – Priyesh Sharma May 1 '19 at 13:30
• How about Keys @ MinimalBy[KeyDrop[#, {"userId", "MinPerf"}], Identity]`? – Sjoerd Smit May 1 '19 at 13:35
• Perfect. Works. Thank You – Priyesh Sharma May 1 '19 at 13:38
• @SjoerdSmit How can I get the elements in string form rather than list form like PerfMultipleChoice, PerfSingleChoice rather than {"PerfMultipleChoice","PerfSingleChoice "}? I tried ToString and then StringDelete, but that doesn't work – Priyesh Sharma May 1 '19 at 15:36


text:
Jamie Oaks

Evolutionary biologist.

Generalizing Bayesian phylogenetics to model shared divergences

We are excited about some new work out of lab that relaxes an important assumption in phylogenetics. In a new bioRxiv preprint, Perry Wood, Jr., Cam Siler, Rafe Brown, and I describe how we generalize the space of trees in Bayesian phylogenetics to avoid assuming divergences are independent across a phylogeny, and allow us to infer shared divergences among multiple clades. We have released a new version of ecoevolity (Version 1.0.0) that implements this fully phylogenetic approach to inferring shared divergence events across phylogenetic trees.

The assumption

In phylogenetics, we assume lineages diverge independently (conditional on the topology) and only leave two descendant lineages (i.e., bifurcate). Put another way, we assume all processes of diversification cause only a single lineage to bifurcate, and have no affect on any of the other extant lineages across the phylogeny. Across the tree of life, there are likely many exceptions to this assumption. For example, we can imagine rising sea levels fragmenting an island and causing co-distributed species of lizards to diverge at the same time.

Instead of lizards on islands, let’s imagine three members of a gene family residing along a region of a chromosome that gets duplicated. This would create shared divergences across the phylogenetic history of the gene family. When multiple infected individuals spread a pathogen to others at a social gathering, this will create shared divergences in the “transmission tree” of the pathogen. In fields as diverse as biogeography, genome evolution, and epidemiology, there are interesting processes that violate the assumption of independent divergences and cause shared divergences.

Back to our insular lizard example, we can also imagine the rising sea levels fragmenting the island into more than two islands, like in the animation below. This will cause multifurcations (a lineage diverging into three or more descendants) in addition to shared divergences.

Similarly, in epidemiology, when an infected individual spreads a pathogen to two or more others at a social gathering, this will create a multifurcating divergence in the transmission tree.

Current phylogenetic methods for inferring rooted trees assume all divergences are independent and bifurcating. More formally, if we have $$N$$ tips, current methods only consider trees with $$N-1$$ independent, bifurcating divergences. When shared or multifurcating divergences were common in the evolutionary history of the system we want to study, such tree models are over-parameterized as illustrated below.

More importantly, by assuming all divergences are independent and bifurcating, current phylogenetic methods do not allow us to test for patterns of shared or multifurcating divergences predicted by processes of diversification that are of interest across fields like biogeography, genome evolution, and epidemiology.

Our approach

To relax this assumption, we developed a Bayesian approach to generalizing the space of tree models to allow for shared and multifurcating divergences (Oaks et al., 2021). Under our generalized tree distribution, trees with $$N - 1$$ bifurcating divergences are only one class of tree models in a greater space of trees with anywhere from from 1 to $$N - 1$$ potentially shared or multifurcating divergences.

We developed reversible-jump Markov chain Monte Carlo algorithms to perform inference across this generalized space of trees.

This allows us to jointly infer relationships, shared and multifurcating divergences, and divergence times.

We coupled the generalized tree model with the “SNAPP likelihood” (Bryant et al., 2012) for directly calculating the probability of biallelic characters given a population (or species) phylogeny, while analytically integrating over all possible gene trees under a coalescent model and all possible mutational histories along those gene trees under a finite-sites model of character evolution. This allows us to tackle biogeographic questions by jointly inferring a species tree and shared divergences from genomic data (Oaks et al., 2021). However, the generalized tree distribution and associated algorithms are independent of the likelihood function, and so can be used with other data types and likelihood models (i.e., lots of room for future work and applications!).

Using simulations, we found this approach consistently inferred shared and multifurcating divergences with high support with moderately sized data sets.

We applied this new approach to genomic data from two genera of geckos co-distributed across the Philippines. Fragmentations of the islands caused by interglacial rises in sea level predict shared and multifurcating divergences over the last 5 million years. Our results show support for this predicted pattern.

If you want to learn more, here’s a talk I gave at the Evolution Meetings this summer:

And you can read our preprint here.

This work was a lot of fun, but we have only scratched the surface with the generalized tree distribution. If you are interested in this work, please reach out; there’s so much more to do!

References

1. Oaks JR, Wood, Jr. PL, Siler CD, Brown RM. 2021. Generalizing Bayesian phylogenentics to infer shared evolutionary events. bioRxiv. DOI: 10.1101/2021.07.23.453597. Link.
2. Bryant D, Bouckaert R, Felsenstein J, Rosenberg NA, Roychoudhury A. 2012. Inferring Species Trees Directly from Biallelic Genetic Markers: Bypassing Gene Trees in a Full Coalescent Analysis. Molecular Biology and Evolution 29:1917–1932. DOI: 10.1093/molbev/mss086. Link.


text:
### Home > CCA2 > Chapter 8 > Lesson 8.3.3 > Problem8-176

8-176.
1. The roots of two quadratic polynomials are given below. Write possible quadratic functions in standard form. Homework Help ✎

1. x = −i, x = i

2. x = 1 +, x = 1 −

$y=(x + i)(x - i)$

$y=(x - (1 + \sqrt{2}))(x - (1 - \sqrt{2}))$


text:
# A On the Planck Blackbody Function

• Featured

Homework Helper
Gold Member
2018 Award
The graph of the Planck blackbody function has an interesting feature:$\\$ $\rho_o=\frac{\int\limits_{0}^{\lambda_{max}} L_{BB}(\lambda,T) \, d \lambda}{\int\limits_{0}^{+\infty} L_{BB}(\lambda, T) \, d \lambda} \approx .2500$,
where $\lambda_{max}$, in an exact derivation of Wien's law, obeys $e^{-x}=1-\frac{x}{5}$, where $x \neq 0$, and $x=\frac{hc}{\lambda_{max} k_B T} \approx 4.96$. From this we can very accurately compute $\lambda_{max}$ as a function of temperature with $\lambda_{max}T=(\frac{hc}{k_B})(\frac{1}{X_o})$ where $X_o$ can be computed to any desired accuracy. The value of $\rho_o$ can be shown to be independent of temperature $T$. $\\$ The value of $\rho_o$ is so close to .2500, that in 1992 it took some calculus expansions long with some very high precision computing to show that the value of this fraction is not exactly equal to $\frac{1}{4}$, but instead approximately $.2501$. (As I recall it came out to be $\rho_o=.2500545...$). Today's very powerful computers could readily get a very accurate numerical number for this ratio to show that it is indeed $\rho_o=.2500545...$. Back in 1992, a couple of us made it a group effort to determine if $\rho_o=\frac{1}{4}$, or if $\rho_o \approx .2500$ , but not exact. After much effort, we concluded $\rho_o=.2500545...$.

Last edited:
mfb
Related Quantum Physics News on Phys.org

Homework Helper
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@vanhees71 @mfb You might find this of interest. $\\$ Hi @Greg Bernhardt Above is one more for Physics Forums. This one could almost be an Insights article, but this posting here should suffice.

Last edited:

Homework Helper
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If I did some quick transformations correctly $\rho_o=\frac{\int\limits_{X_o}^{+\infty} \frac{x^3 \,dx}{e^x-1}}{\int\limits_{0}^{+\infty} \frac{x^3 \, dx}{e^x-1}}$ where $X_o$ is the constant of post 2 above. $\\$ Edit: @jedishrfu @fresh_42 To see if I did this correctly, can you evaluate this ratio for me using $X_o=4.96$? Try to get 3 decimal place accuracy. Thanks. For very high accuracy, you might even try using $X_o$ as the root not equal to zero of $e^{-x}=1-\frac{x}{5}$. A rectangular type integration with about 100,000 intervals from $0$ to $100$ should suffice. Ignore $x=0$ in the denominator. No need to integrate that point. For increased accuracy, tighten the resolution around $x=4.96$. $\\$ From $0$ to $.00001$, a Taylor expansion of $e^x \approx 1+x +(\frac{x^2}{2}+...)$ allows for an almost exact $\int\limits_{0}^{+.00001} x^2 \, dx=\frac{1}{3}(.00001)^3$. $\\$ For $x>100$, I think it should be possible to show negligible contribution. $\\$ Tightening up the resolution round $x=4.96$ should allow one to get this ratio of integrals easily accurate to 10 decimal places or more. $\\$ I welcome your inputs. I currently don't have access to an EXCEL spreadsheet, but it really requires the high precision of Mathcad and/or its successors. @Dale might you give this a try? $\\$ Meanwhile the denominator integral actually has a closed form: $I_D=\frac{\pi^4}{15} =6.49393940227...$. See: https://math.stackexchange.com/questions/99843/contour-integral-for-x3-ex-1

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#### mfb

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As the integral is dominated by the region near X0 you might not get the ratio with three significant figures if you give X0 with just two decimal places.

WolframAlpha calculates 1.62824 for the integral using 4.96 and 1.62394 using 4.965.

We can ask it to solve the original equation for x, the result is X0=4.96511. Using that the numerator is 1.62384.

Starting the integral at 0 we get 6.49394, the ratio is 0.2500546. We have enough significant figures to conclude that the ratio is not 0.25.

Find X0
Numerator
Denominator

Note that I stopped the integral at 100 because it didn't work well with infinity.

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@mfb Many thanks !!

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@mfb See also my final line I just added to post 3. Suggestion is to integrate numerator from $0$ to $X_0$ and subtract from the closed form result of $0$ to $+\infty$.

#### mfb

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A historical note: The question came up,for a couple of us in 1992, what percentage of radiated energy lies to the left of the peak on the intensity vs. wavelength curve for a blackbody type source? $\\$ We consulted what is known in the IR (infrared) industry as a blackbody slide rule. The result of the mark on the slide rule showed it was so close to the .25 point for $0$ to $\lambda_{max}$ that the possibility existed that the number might be exactly .25. $\\$ We did arrive at the result that $\rho_o=.2500545...$, but our approach was much more complex mathematically than the route that @mfb took above. $\\$ In any case, the result that $\rho_o \approx \frac{1}{4}$, along with Wien's law, $\lambda_{max}T=2898 \, \mu \, K$, can be useful in estimating the efficiency in the visible region of the spectrum of various incadescent sources. The temperature of the filament of an incadescent bulb is typically $T \approx \,$ 2500-2800 K. Efficiencies in the visible are typically in the 10% range for incadescent sources. $.40 \, \mu < \lambda_{visible}<.75 \, \mu$.

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Homework Helper
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As the integral is dominated by the region near X0 you might not get the ratio with three significant figures if you give X0 with just two decimal places.

WolframAlpha calculates 1.62824 for the integral using 4.96 and 1.62394 using 4.965.

We can ask it to solve the original equation for x, the result is X0=4.96511. Using that the numerator is 1.62384.

Starting the integral at 0 we get 6.49394, the ratio is 0.2500546. We have enough significant figures to conclude that the ratio is not 0.25.

Find X0
Numerator
Denominator

Note that I stopped the integral at 100 because it didn't work well with infinity.
@mfb The WolframAlpha links don't seem to be working anymore. Perhaps you can reset them and/or permanently download them.

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Scratch the last comment=the "links are working again. Adding a 5 in the next digit of $X_o$ changed the 888 considerably, so, yes, I would agree with @mfb in his above assessment.

M

#### member659127

That was interesting, thanks!
This semester I'm giving modern physics course, did the BB radiation a few weeks ago. Never thought about the ratios before though... Will mention it this week then, thanks again!

#### DrClaude

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Using Mathematica, we can really go big on the digits. I get
$$X_0 \approx 4.96511423174427630369875913132289394405558499$$
and then
$$\rho_0 \approx 0.25005454682271047905965004646011027863057$$

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That was interesting, thanks!
This semester I'm giving modern physics course, did the BB radiation a few weeks ago. Never thought about the ratios before though... Will mention it this week then, thanks again!
By using this ration $\rho_o \approx \frac{1}{4}$, and Stefan-Boltzmann $\frac{\sigma T^4}{\pi}=\int\limits_{0}^{+\infty}L_{BB}(\lambda,T) \, d \lambda$, (where Stefan's constant $\sigma=\frac{\pi^2 k_B^4}{60 \hbar^3c^2} =5.6697$ E-8 Watts/$(m^2 K^4)$), and Wien's law $\lambda_{max}T=2898 \, \mu \, K$, it really tells you the most important features of the Planck Plackbody function. $\\$ Meanwhile radiated power per unit area (radiant emittance) for a blackbody is $M=\sigma T^4$, and brightness (radiance) $L=\frac{M}{\pi}$, so that irradiance $E$ (watts/m^2) at a distance $s$ on axis from a small blackbody source of area $A_{BB}$ is given by $E=\frac{LA_{BB}}{s^2}$ , and integrated over hemisphere with a $\cos(\theta)$ fall-off , we get power $P=M A_{BB}=L_{BB}A_{BB} \pi$. $\\$ The exact expression for spectral radiance (brightness) is $L_{BB}(\lambda,T)=\frac{2hc^2}{\lambda^5 (e^{hc/(\lambda k_B T)}-1) }$. $\\$ @Greg Bernhardt You just might want to feature this one. It is rather educational.

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Using Mathematica, we can really go big on the digits. I get
$$X_0 \approx 4.96511423174427630369875913132289394405558499$$
and then
$$\rho_0 \approx 0.25005454682271047905965004646011027863057$$
Mathematica must use a lot of decimal places for the exponential constant $e$ in order to get these integrals so precise and accurate.

#### DrClaude

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Mathematica must use a lot of decimal places for the exponential constant $e$ in order to get these integrals so precise and accurate.
As many as you want

(I don't like the new smilies...)

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#### BvU

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Thanks Charles, but I really don't deserve a like for that ....
If only I could cash in all or even just a few of those likes I'd be a gold member by now !

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"On the Planck Blackbody Function"

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JEDNOSTKA NAUKOWA KATEGORII A+

Wydawnictwa / Czasopisma IMPAN / Acta Arithmetica / Wszystkie zeszyty

A note on minimal zero-sum sequences over $\mathbb Z$

Tom 166 / 2014

Acta Arithmetica 166 (2014), 279-288 MSC: Primary 11B75; Secondary 11B30, 11P70. DOI: 10.4064/aa166-3-4

Streszczenie

A zero-sum sequence over ${\mathbb Z}$ is a sequence with terms in ${\mathbb Z}$ that sum to $0$. It is called minimal if it does not contain a proper zero-sum subsequence. Consider a minimal zero-sum sequence over ${\mathbb Z}$ with positive terms $a_1,\ldots,a_h$ and negative terms $b_1,\ldots,b_k$. We prove that $h\leq \lfloor \sigma^+/k\rfloor$ and $k\leq \lfloor \sigma^+/h\rfloor$, where $\sigma^+=\sum_{i=1}^h a_i=-\sum_{j=1}^k b_j$. These bounds are tight and improve upon previous results. We also show a natural partial order structure on the collection of all minimal zero-sum sequences over the set ${\{i\in {\mathbb Z}:\; -n\leq i\leq n\}}$ for any positive integer $n$.

Autorzy

• Papa A. SissokhoMathematics Department
Illinois State University
Normal, IL 61790-4520, U.S.A.
e-mail

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