text:
17

# 17Slow scrolling

## Slow scrolling with a link

Slow scrolling is easy; we need four things, the first three of which are in the HTML:

• An ID on the target element we want to scroll to

• A class in the link that references the ID

• Some jQuery to join it all together

We’ve actually covered everything here in other sections, usually with the caveat that I’ll explain the slow scrolling later (well, here it is).

To explain this, I’m going to use a simple link to a point on the same page; I’ll use the exact example from the introduction, the link and soft scroll to figure 99.1 on the 99-00-typicals.html page of the website template.

This is the completion of what we started in § 13.4, in fact it is exactly the same link. Let’s look at what we have in the HTML:

99-00-typicals.html

Internal links to a point on the same page use the class xref; this is less noticeable on the page: Figure 99.1.

...

Figure 99.1   Full width figure

Code 17.1   Link to figure 99.1

Let’s look at figure 99.1 first, this is just a standard full width image, in fact it is exactly the same full width image we created in § 15.1.1, this is it:

The figure has been given the unique ID (id="js--f99-01"), I’ve shown it underlined in dark blue and bold in the listing above (Code 17.1). This is the standard form for the ID of a figure; see Table 13.1.

All well and good, it is a unique ID for the figure in question.

Next let’s look at the link in the listing, I’ve shown it underlined in dark red and bold in Code 17.1:

Figure 99.1

The anchor for the link  has two classes, the first is the xref styling class that determines the appearance of the link (see § 13.4). The second is the peculiar class:

js--sc-f99-01

I first introduced this class in § 13.4; it is needed to make the soft scroll work; it matches the given ID of the object we want to soft scroll to.

This class has the same name as the target ID except that the js-- of the ID is replaced with js--sc-, i.e. there is an extra sc- after the --.

The names of these IDs and corresponding classes, the js-- and js--sc- look kind of official, as if they have a particular meaning to the JavaScript, but they don’t, they are a complete product of my imagination. You could use anything, hector and henry will work just as well.

In my mind, js was intended to indicate that it was applicable to the JavaScript, the double dashes -- allowed me to search for it easily and the sc is my abbreviation for scroll.

I’m sorry if you thought I was doing something clever — you should know by now that I hardly do anything clever and when I do it’s usually by accident.

To summarise:

• The target (the thing we want to scroll to) has the ID:
js--f99-01

• The link (to the object) has the class js--sc-f99-01 i.e. exactly the same as the ID but with js--sc- instead of js--

Both the ID (js--f99-01) and the class (js--sc-f99-01) are missing from any CSS file (there isn’t actually a class js--sc-f99-01 in style.css), these are purely for use in a JavaScript file, specifically the scroll.js file for the web page.

### The soft scroll jQuery

This is the clever stuff.

The soft scrolling is different for each web page, each web page has different sections, tables, figures &c. and no two are the same. Thus the soft scrolling goes in the .js file for the web page in particular; specifically it is the c-ss-scroll.js file. Even more specifically, in this case it is:

01-pages/99-00-typicals/05-js/99-00-scroll.js

This is the scroll.js†1 file for the web page 99-00-typicals.html.

That is why it’s called scroll; it has all the slow scroll jQuery stuff in it.

†1 Generally, when I refer to the scroll.js file, I’m referring to the cc-ss-scroll.js file for the particular web page in question.

At the moment, this file just has the jQuery container for our code:

01-pages/99-00-typicals/05-js/99-00-scroll.js
$(document).ready(function() { /* START OF PAGE READY FUNCTION */ /* **************************************************************************** YOUR CODE GOES IN HERE **************************************************************************** */ }); /* END OF PAGE READY FUNCTION */ Code 17.2 99-00-scroll.js Let’s add the slow scroll stuff we need for this particular link: 01-pages/99-00-typicals/05-js/99-00-scroll.js $(document).ready(function() {                  /* START OF PAGE READY FUNCTION */

$('.js--sc-f99-01').click(function () { /* START of scroll function */$('html, body').animate({scrollTop: $('#js--f99-01').offset().top -80}, 1000); }); }); /* END OF PAGE READY FUNCTION */ Code 17.3 99-00-scroll.js with slow scroll code I know what you’re thinking, you’re thinking “that’s a shitload of brackets” and you’re right, I count 14 brackets and four braces — for just three lines of code? I refer you to my earlier comment. You’re probably also thinking “that doesn’t look like the stuff we did before, it looks complicated” — you’re less right this time, it does looks complicated (Linux people†2 . I blame you), but it sort of isn’t, you get used to it. †2 Q. How do you tell an extroverted Linux person from an introverted one? A. The extravert looks at your shoes when you talk to him. And for C programmers: 99 definite bugs in the code 99 definite bugs You fix one bug and you re-compile 100 definite bugs in the code. Let’s look at this in terms of the jQuery syntax:$(selector).action();

What we have here is just this, except that in the action, we create a function that has another jQuery in it.

Think of it like this (I’ve simplified the brackets):

$(selector1).action1($(selector2).action2();
);

If selctor1 is true, THEN we test another condition (selector2) and if that is true, carryout action2. If selector1 was false, we don’t do anything.

The only other thing we have to do concerns the second line; we need to put in another function (another anonymous function).

Let’s go through it. The first line:

    $('.js--sc-f99-01').click(function () { The first bit (the selector) is looking for the class js--sc-f99-01 that we used in the link (). In short, $('.js--sc-f99-01').click() is waiting for something with the class js--sc-f99-01 to be clicked. In our case, the link to figure 99.1 has that class because we gave it that class in the HTML:

Internal links to a point on the same page use the class xref; this is less noticeable on the page: Figure 99.1.

It applies to everything inside the link (i.e. the Figure 99.1 that displays on the web page), if we click anywhere in the link, the jQuery $('.js--sc-f99-01').click() will be true and everything in the brackets of the .click action will execute. What’s in those brackets is: function () { so what does this mean? It means that if the class has been clicked (i.e. the link has been clicked), I create another anonymous function that has no parameters (the empty brackets) and executes it. When it is executed it runs all the code between the braces {}. In this case, it executes the following: $('html, body').animate({scrollTop: $('#js--f99-01').offset().top -80}, 1000); This is another jQuery,$(selector) &c. This is also a standard bit of code for the scrollTop method, you see it everywhere.

This time the selector is $('html, body')†3 it is selecting anything that is within the html element or the body element, this essentially is everything on the web page, so this selector is always true. The .animate() performs some sort of animation, in this case it is the scrollTop animation. †1 You may be wondering why I use both html and body in the selector, the body element is after all within the html element. The reason is for browser consistency (and this is a fairly standard method for overcoming it), some browsers (Firefox and Internet Explorer) need this to be html, some (Chrome, Safari and Edge) need it to be body. Using both satisfies everyone. The syntax for the animate method is: .animate({property}, duration) The duration is in milliseconds The property is scrollTop and its syntax is: scrollTop:$(target).offset().top nn

Where nn is the number of pixels below the top of the screen where the scroll will position the target.

Nearly there.

The scrollTop property is the thing that actually does the scrolling, it will scroll to the target, in this case the target is the ID of the object: $('#js--f99-01'), it is the bit between the quotes and it is the CSS class or ID, if it were a class, the # would be replace with a ., just like it is in CSS. • I tend to use single quotes in jQuery, but double quotes work in just the same way; i.e.$("#js--f99-01"). JavaScript and jQuery don’t care which you use, just be consistent, it doesn’t like it if you start with one and finish with the other, for example $("#js--f99-01') won’t work (double quote at the start, single quote at the end). The offset().top is entirely optional, if you miss it out or set the offset to zero, the top of the target will be positioned exactly at the top of the browser window: The offset moves the stop point, if the number is positive it is further down the page, the following is an example with an offset of 100 (pixels, offset is always in pixels) offset().top +100. • For positive offsets, you have to use the plus sign (+), it doesn’t work if you just give it a number. It chops the top off the image. Generally, we want negative offsets to leave a bit of a gap between the target and the top of the browser. This is an offset of -20: offset().top -20: Now in practice, there will be the sticky navigation bar (next chapter) at the top of the browser, and that has maximum height of 60 px; to make the slow scroll stop before the sticky navigation bar, we need to increase the gap by 60 pixels. I tend to use an offset of -80 pixels for figures: offset().top -80. With the sticky nav it looks like this: Tables and code fragments have different offsets (-60 for tables and -120 for code fragments), this is because these objects have different top margins. Sections, subsections and inline sections also have different offsets; the summary at the end gives examples of each (Table 17.1). The final thing is the duration; this is the 1000 in the second line: $('html, body').animate({scrollTop: \$('#js--f99-01').offset().top -80}, 1000);

This is the duration of the animation, i.e. how long it takes the browser to scroll for where it is to the target. This value is in milliseconds (one thousandth of a second) so a value of 1000 is equivalent to a duration of one second.

It is actually part of the animate method, not the scrollTop.


text:
# Problems with time series prediction

I got a question about modeling time series in R. my data consist of the following matrix:

1   0.03333333 0.01111111 0.9555556
2   0.03810624 0.02309469 0.9387991
3   0.00000000 0.03846154 0.9615385
4   0.03776683 0.03119869 0.9310345
5   0.06606607 0.01201201 0.9219219
6   0.03900325 0.02058505 0.9404117
7   0.03125000 0.01562500 0.9531250
8   0.00000000 0.00000000 1.0000000
9   0.04927885 0.01802885 0.9326923
10  0.06106870 0.02290076 0.9160305
11  0.03846154 0.00000000 0.9615385
12  0.00000000 0.00000000 1.0000000
13  0.06028636 0.03843256 0.9012811
14  0.09646302 0.05144695 0.8520900
15  0.04444444 0.06666667 0.8888889

these matrix has in total 200 rows.

as you can see in each situation the sum of each row is 1, that becomes because the values are the percentage of a whole. for example row 1 contains 3.33% of variable a, 1.11% of variable 2 and 95.5% of veriable 3. the first collomn indicates the year that the values are measured.

my target is to make a prediction for the next 5 years, so from year 200 to 205.

I can doe that by making three normal time series forecast. But for that forecast the total sum is never equal to 1, which is very important. Normaly is use techniques like arima and exponential smoothing.

Does somebody know a method to make a forecast for such a problem?

• Do you know if these are discrete proportions (eg, number of units in categories A, B, & C, given a total number of units) or continuous proportions? If they are discrete, do you know the total number of units at each time point? – gung - Reinstate Monica Mar 25 '15 at 16:21
• Yes these values comes from discrete units, I calculated these ratios with use of the total numbers. – karmabob Mar 26 '15 at 8:22

You are attempting to forecast a compositional time series. That is, you have three components that are all constrained to lie between 0 and 1 and to add up to 1.

You can address this issue using standard exponential smoothing, by using an appropriate generalized logistic transformation. There was a presentation on this by Koehler, Snyder, Ord & Beaumont at the 2010 International Symposium on Forecasting, which turned into a paper (Snyder et al., 2017, International Journal of Forecasting).

Let's walk though this with your data. Read the data into a matrix obs of time series:

obs <- structure(c(0.03333333, 0.03810624, 0, 0.03776683, 0.06606607,
0.03900325, 0.03125, 0, 0.04927885, 0.0610687, 0.03846154, 0,
0.06028636, 0.09646302, 0.04444444, 0.01111111, 0.02309469, 0.03846154,
0.03119869, 0.01201201, 0.02058505, 0.015625, 0, 0.01802885,
0.02290076, 0, 0, 0.03843256, 0.05144695, 0.06666667, 0.9555556,
0.9387991, 0.9615385, 0.9310345, 0.9219219, 0.9404117, 0.953125,
1, 0.9326923, 0.9160305, 0.9615385, 1, 0.9012811, 0.85209, 0.8888889
), .Dim = c(15L, 3L), .Dimnames = list(NULL, c("Series 1", "Series 2",
"Series 3")), .Tsp = c(1, 15, 1), class = c("mts", "ts", "matrix"
))

You can check whether this worked by typing

obs

Now, you have a few zeros in there, which will be a problem once you take logarithms. A simple solution is to set everything that is less than a small $$\epsilon$$ to that $$\epsilon$$:

epsilon <- 0.0001
obs[obs<epsilon] <- epsilon

Now the modified rows do not sum to 1 any more. We can rectify that (although I think this might make the forecast worse):

obs <- obs/matrix(rowSums(obs),nrow=nrow(obs),ncol=ncol(obs),byrow=FALSE)

Now we transform the data as per page 35 of the presentation:

zz <- log(obs[,-ncol(obs)]/obs[,ncol(obs)])
zz

Load the forecast package and set a horizon of 5 time points:

library(forecast)
horizon <- 5

Now model and forecast the transformed data column by column. Here I am simply calling ets(), which will attempt to fit a state space exponential smoothing model. It turns out that it uses single exponential smoothing for all three series, but especially if you have more than 15 time periods, it may select trend models. Or if you have monthly data, explain to R that you have a potential seasonality, by using ts() with frequency=12 - then ets() will look at seasonal models.

baz <- apply(zz,2,function(xx)forecast(ets(xx),horizon=horizon)["mean"])
forecasts.transformed <- cbind(baz[[1]]$$mean,baz[[2]]$$mean)

Next we backtransform the forecasts as per page 38 of the presentation:

forecasts <- cbind(exp(forecasts.transformed),1)/(1+rowSums(exp(forecasts.transformed)))

Finally, let's plot histories and forecasts:

plot(obs[,1],ylim=c(0,1),xlim=c(1,nrow(obs)+horizon),type="n",ylab="")
for ( ii in 1:ncol(obs) ) {
lines(obs[,ii],type="o",pch=19,col=ii)
lines(forecasts[,ii],type="o",pch=21,col=ii,lty=2)
}
legend("left",inset=.01,lwd=1,col=1:ncol(obs),pch=19,legend=colnames(obs))

EDIT: a paper on compositional time series forecasting just appeared. I haven't read it, but it may be of interest.

• Thanks @stephan Kolassa, that is exactly what I mean. Does anybody knows wether there is toolpack in R to make a forecasting with compositional time series. – karmabob Mar 26 '15 at 9:21
• I don't think so, and these authors are not among the "always write a companion R package to any publication" people. But their approach is pretty simple and should not require more than five lines to preprocess your time series, after which you can use the standard tools (e.g., ets() in the forecast package for state space exponential smoothing). – S. Kolassa - Reinstate Monica Mar 26 '15 at 10:11
• My mathematical knowledge is not that good. can you maybe help with the pre processing the data? when the data is in the right format i can do the forecasting process. – karmabob Mar 26 '15 at 16:00
• I was wandering if you have found some time to help me with the problem I had with the compositional time series? – karmabob Apr 1 '15 at 14:47
• Here you go. Sorry it took a while; I actually got stuck understanding the presentation and had to clear my head. – S. Kolassa - Reinstate Monica Apr 1 '15 at 15:33

format: plain


text:
# Show that 2sqrt(7) is Irrational. - Mathematics

Show that 2sqrt(7) is irrational.

#### Solution

2/sqrt(7) = 2/sqrt(7) xx sqrt(7)/sqrt(7) = 2/7 sqrt(7)

Let 2/7 sqrt (7) is a rational number.

∴ 2/7 sqrt (7) = p/q, where p and q are some integers and HCF(p,q) = 1     ….(1)

⇒2sqrt(7)q = 7p

⇒(2sqrt(7) q) ^ 2 = (7p)^ 2
⇒7(4q^2) = 49p^2
⇒4q^2 = 7p^2
⇒ q^2 is divisible by 7
⇒ q is divisible by 7        …..(2)
Let q = 7m, where m is some integer.
∴2sqrt(7) q = 7p
⇒ [2sqrt(7) (7m)]^2 = (7p)^2
⇒343(4m^2) = 49p^2
⇒ 7(4m^2) = p^2
⇒ p^2 is divisible by 7
⇒ p is divisible by 7             ….(3)
From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, 2 sqrt(7) is irrational.

Concept: Concept of Irrational Numbers
Is there an error in this question or solution?


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Finally, select "Import" and Teamcraft will produce a list of the average amount of items required to complete the Ariyala list you imported. 2017/10/21 06:32 Male/Under 20 years old/Elementary school/ Junior high-school student/Very/ Posted by 1 year ago. Buff up at the start, progress until you're one hit away from finishing, use rest of CP and DUR to push Quality/Collectability, on last 5 DUR, finish craft. Although my favorite camera comes with me everywhere, my lights and my lenses are the key elements to get the best photos. Caesar cipher (or Caesar code) is a shift cipher, one of the most easy and most famous encryption systems. required progress is 1000 and rotation is predicted to achieve 1005 progress) I'll do a test run with Reclaim loaded and abort if the 1st synthesis ability doesn't hit the predicted progress. I'll look it over tonight. Copyright © 2014 - 2017 by Rhoda Baker and Gordon Tyler. The variables are : the binary integer decisions to accept rotations. FFXIV Teamcraft. Tool to decrypt/encrypt with Caesar. The outcome of a single crafting attempt can be no items, a standard item or a … 2D IK automatically calculates for the positions and rotations of the a chain of bones moving towards a target position. I, however, learned the hard way to be careful with the build-in recipe solver. In general this will not level you to your desired level. Notre but, nous éclater sur différents jeux ! Server hasn't been up for: 3 years. Crafting Calculator. Solver: The base abstract class that all solvers derive from. Rejoignez nous ! ... Edit: I did the same rotation, and set the Recipe Level to Lvl 70. It is also recommended to clean your inventory and rebuild it before using the optimizer for the first time, to make sure you have up to date information. Teamcraft Rotation Simulator [Discussion] Close. Archived. Next - List features. For example, here is a$16 \times 25$board with$9 \times 9\$ rotating blocks: (A sample move's rotation is shown in progress) What is the strategy for solving the NRP? used it as you described and I indeed could get a high collectible level with a bit of optimization. Due to the need for TC to know what your inventory and retainers have, you need to use the Teamcraft client with data parsing turned on. Recipe item level is not the same as class level, if you thought that. FINAL FANTASY XIV ©2010 - 2017 SQUARE ENIX CO., LTD. At all. Crafting in Final Fantasy 14 is a minigame. Discord for support, bugs discussion and contributors: https://discord.gg/r6qxt6P Development ), handles gathering of solver components, and executes updating them in the proper order. Rot-N/Rot cipher is a simple character substitution based on a shift/rotation of N letters in an alphabet. I've been having issues with macros in general when below the recommended item level and melds since the crafting skill changes. They were also accompanied by DeceptiBonk, who served as their artist. Copyright © 2014 - 2017 by Rhoda Baker and Gordon Tyler. 2. The Angels will enter the season using a six-man rotation to accommodate the schedule and routine of Shohei Ohtani, while also making a conscious effort to get their starters some rest. Hi M1uu, thank you for the kind words. {{version}}. Be sure to pick up some extras! The team competed in the now defunct A1 Grand Prix and Euroseries 3000 series.. History. teamcraft rotation solver. FFXIV Teamcraft is a tool to help FFXIV crafters and gatherers with items listing, preparation, crafting simulation, gathering informations, timers, macro translation, etc.. With the arrival of FFXIV Patch 4.5, I decided to finally release Teamcraft v5, the update that took me 6 months to dev. Male or Female ? Collaborative crafting tool for Final Fantasy XIV. All Rights Reserved. We want only one crew per flight. Every recipe is hand written out with a separate list keeping track of what items are raw materials (non synthed), and another list tracking synths using only raw mats, another list with items that require tier 1 synths, and another list for items requiring tier 2 synths. Nam tortor massa, herit lorem semitur fertiolas. Press question mark to learn the rest of the keyboard shortcuts. {{version}}. I've been struggling to find any macros for that kind of thing outside of Ashe10's 2 star macros. @ffxiv-teamcraft/simulator. Some team building program formats work better with smaller groups, others larger, and for many, it doesn't matter. Lots and lots of lists and vlookups. Using this, the centre of mass and rotation are coincident as expected, and everything seems fine (I've no idea what the value uniform (0, 0, 0) field is for, and commenting it out doesn't seem to change anything). A rotation in the x–y plane by an angle θ measured counterclockwise from the positive x-axis is represented by the real 2×2 special orthogonal matrix,2 cosθ −sinθ sinθ cosθ . © 2020 Amado Claro Photo. Get the free "Rotation Matrices Calculator MyAlevelMathsTut" widget for your website, blog, Wordpress, Blogger, or iGoogle. ... Edit: I did the same rotation, and set the Recipe Level to Lvl 70. Recipe item level is not the same as class level, if you thought that. FFXIV Cactpot Solver. Public; Public/Protected; All Angels Rumors: Joe Musgrove could help solve years of rotation issues halohangout.com - Daniel Preciado. Click a face to make a clockwise rotation as shown by the arrow. Let me quote Yoshida real quick: Q. Based in Austin, Texas. Guides - Previous. This is not just a 2x2 solver but a puzzle simulator where you can play with the Pocke Cube. :confused: You do not need HQ mats to HQ the endgame Facet crafting gear. Scrip turn-ins are hardly anything close to as complicated as starred crafts. Consider a counter-clockwise rotation of 90 degrees about the z-axis. https://ffxivteamcraft.com. In a nutshell, I am using lists. So that's what you got. with the new ishgard restoration going on right now, I feel like I want to be part of it (and also level up my low lvl crafters). Thank. original coordinates:(x 0, y 7.5)rotation θ 90 degreenew coordinates:(X 7.5 ,Y 0 )the new coordinate X should be -7.5 Reply This calculator is the "rotation of axes" Calculator.Your answer noted above can be got by the "rotation of points" Calculator. How To Solve Cyclic Rotation Code Challenge. So, X= 9.89, Y=-1.41. A character that isn't verified will have its avatar greyed while a verified one will have it full color. Starting with "A Realm Reborn" and the expansions include "Heavensward", "Stormblood", and "Shadowbringers"! Same exact thing happened like your screenshot. Same exact thing happened like your screenshot. So that's what you got. the act of rotating as if on an axis; "the rotation of the dancer kept time with the music" a planned recurrent sequence (of crops or personnel etc. A community for fans of Square Enix's popular MMORPG "Final Fantasy XIV Online", also known as FFXIV or FF14. Contribute to ffxiv-teamcraft/crafting-solver development by creating an account on GitHub. me! The effect an IK handle has on a joint chain depends on the type of IK solver used by the IK handle. Crafting Calculator Display all the materials needed to craft one of each item between two levels. Press J to jump to the feed. Welcome to the Teamcraft Facebook page! Use Teamcraft's "Create a Custom Rotation" feature and just input the Craft's conditions as well as your crafter stats, food bonuses, etc, the recipe doesn't have to be added to Teamcraft yet if doing it like that, though it might've already been updated since yesterday. E.g. Find more Widget Gallery widgets in Wolfram|Alpha. It uses the substitution of … These are marked with x2. Solve It Yourself. The Angels are currently rumored to be in the mix for Pirates starting pitcher Joe Musgrove, a possible solution to the rotation struggles. It provides state tracking, smoothing parameters and implementation, automatic solver system integration, and update order. The next step is to determine the angle of rotation, theta. Figured a thread like this is more appropriate. The crafting simulator is a feature inside Teamcraft that you can reach from different places, it allows you to simulate a crafting rotation, meaning that you'll be able to craft an item out of FFXIV, in order to prepare your rotations. This is meant to avoid list trolling for big FCs or even identity issues when a list is shared or an action is made on a given list. Before you start crafting parameters and implementation, automatic solver system integration, set! A 2x2 solver but a puzzle Simulator where you can easily HQ it with all mats! 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To figure it out been struggling to find the solution then follow the steps to solve your Cube Dawn! With me everywhere, my lights and my lenses are the key elements get... For Final Fantasy XIV online '',  Stormblood '', and executes updating them in the proper order by! In general when below the recommended item level and melds since the crafting optimizer to help me leveling. Into the formulas above to determine the new teamcraft rotation solver changes to endgame crafting tell me you. Complicated as starred crafts class that all solvers derive from I hardly found a difference in difficulty making... Of optimization solve button can easily HQ it with all NQ mats of Enix! Recommended item level is not the same rotation, and Sky 's now ex-girlfriend, Dawn main camera,. Musgrove, a possible solution to the rotation struggles rotation to work well item is. 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'S now ex-girlfriend, Dawn and  Shadowbringers '' the colors of your and. The proper order that kind of thing outside of Ashe10 's 2 star macros Deadlox in a call... Is my passion because it allows me to immortalize moments and convey emotions that will last forever for... Are hardly anything close to as complicated as starred crafts suitable dictionary for 'New. Reference object to track against ( ex: the Binary integer decisions accept! Forever and I indeed could get a high collectible level with a bit of optimization system integration, and many. Nec mattis mi faucibus id comes with me everywhere, my lights and my lenses are key! Crafted was first started by Deadlox in a Skype call with SkyDoesMinecraft moments and convey emotions that will forever. The puzzle, because the nine numbers are now in increasing book-order clockwise rotation: Joe Musgrove, a solution. Build-In recipe solver to rotation matrix was a British Motorsport team based in,... 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A target position team based in Silverstone, UK a chain of bones moving towards a target position rotation. On the type of IK solver used by the arrow just a 2x2 solver but a puzzle where. Not HQ mats solve the Cube yourself chain depends on the type of IK solver used by IK. Is independent of the representation used for initialization mean a 180 degree turn a. Allows me to immortalize moments and convey emotions that will last forever uses the substitution of … a solver., seminars and consulting services with an experiential focus the bottom of the keyboard.. Teamcraft rotation Simulator [ Discussion ] close when I stumbled upon something today but soon changed his mind joined.


text:
# In figure there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field B, which is present everywhere. The mass of each slider is m, resistance R and initially these are at rest Now if one slider is given a velocity v0 the velocity of other slider after considerably long time will be (neglect the self- induction) 1. $\frac{{v}_{0}}{4}$ 2. $\frac{{v}_{0}}{2}$ 3. v0 4. Zero

Dear Student,

Regards.

• 9
What are you looking for?


text:
# Important Properties of Transverse Common Tangents

I. The two transverse common tangents drawn to two circles are equal in length.

Given:

WX and YZ are two transverse common tangents drawn to the two given circles with centres O and P. WX and YZ intersect at T.

To prove: WX = YZ.

Proof:

 Statement Reason 1. WT = YT. 1. The two tangents, drawn to a circle from an external point, are equal in length. 2. XT = ZT. 2. An in statement 1. 3. WT + XT = YT + ZT⟹ WX = YZ. (Proved) 3. Adding statements 1 and 2.

II. The length of a transverse common tangent to two circles is $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$, where d is the distance between the centres of the circles, and r$$_{1}$$ and r$$_{2}$$ are the radii of the given circles.

Proof:

Let two circles be given with centres O and P, and radii r$$_{1}$$ and r$$_{2}$$ respectively, where r$$_{1}$$ < r$$_{2}$$. Let the distance between the centres of the circles, OP = d.

Let WX be a transverse common tangent.

Therefore, OW = r$$_{1}$$ and PX = r$$_{2}$$.

Also, OW ⊥ WX and PX ⊥ WX, because a tangent is perpendicular to the radius drawn through the point of contact

Produce W to T such that WT = PX = r$$_{2}$$. Join T to P. In the quadrilateral WXPT, WT ∥ PX, as both are perpendiculars to WX; and WT = PX. Therefore, WXPT is a rectangle. Thus, WX = PT, as the opposite sides of a rectangle are equal.

OT = OW + WT = r$$_{1}$$  +  r$$_{2}$$.

In the right-angled triangle OPT, we have

PT2 = OP2 – OT2 (by Pythagoras’ Theorem)

⟹ PT2 = d2 – (r$$_{1}$$ + r$$_{1}$$)$$^{2}$$

⟹ PT = $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$

⟹ WX = $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$ (Since, PT = WX).

III. The transverse common tangents drawn to two circles intersect on the line drawn through the centres of the circles.

Given: Two circles with centres O and P, and their transverse common tangents WX and YZ, which intersects at T

To prove: T lies on the line joining O to P, i.e., O T and P lie on the same straight line.

Proof:

 Statement Reason 1. OT bisects ∠WTY⟹ ∠ATO = $$\frac{1}{2}$$∠WTY. 1. The tangents drawn to a circle from an external point are equally inclined to the line joining the point to the centre of the circle. 2. TP bisects ∠ZTX⟹ ∠XTP = $$\frac{1}{2}$$∠ZTX. 2.  As in statement 1. 3. ∠WTY = ∠ZTX. 3. Vertically opposite angles. 4. ∠WTO = ∠XTP. 4. From statement 1, 2 and 3. 5. OT and TP lie on the same straight line⟹ O, T, P are collinear. (Prove) 5. The two angles are forming a pair of vertically opposite angles.

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• ### Solved Examples on the Basic Properties of Tangents | Tangent Circle

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# Sideways margin with parbox?

How to get right margin in parboxed header? Top, bottom and left margin are ok, but there is no margin at the right (top) side of 'Parameter' header.

\begin{table}[ht]\footnotesize
\centering
\begin{tabular}{|c|}
\hline
\begin{sideways}\parbox{2cm}{Parameter etc etc ....}\end{sideways}
\end{tabular}

\caption{Summary}
\label{tab:summary}
\end{table}

There are top,bottom and left margin, but there is no margin at right side of the header, as you can see at attached image.

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Somehow the problem is not comprehensible for me. Perhaps a full but minimal example that reproduces the undesired behaviour can help to understand. –  Thorsten Donig Nov 14 '10 at 16:25

\parbox supports optional arguments for specifying height and alignment. In your case, within a sideways environment, you could adjust the height of the parbox to get a wider header and thus a wider margin.

Further, if you wish to get space below or above the turned parbox, you could use \hspace* to insert some space within the sideways environment.

For example, within your code, write

\begin{sideways}
\parbox[c][2cm][c]{2cm}{Parameter etc etc ....}
\hspace*{1ex}
\end{sideways}

to get (with an additional \hline):

The syntax of \parbox is

\parbox[pos][height][inner-pos]{width}{text}

where pos means one of the usual alignment options t, b or c.

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Ok, it works for top and bottom - but what about right margin? When you look at table you can see that there is no margin at the very top of it (looking normally -from top to bottom, not sideways) –  pixel Nov 14 '10 at 19:52
The question seemed not so clear (what's been meant by right) and the code and the first column had whitespace at its top. Just insert some space after the parbox (\hspace*) - I'll edit my example. –  Stefan Kottwitz Nov 14 '10 at 20:20


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# No definite integrals of trigonometry

I have big problems with the following integrals:

$$\int\frac{dx}{\sin^6 x+\cos^6x}$$

$$\int\frac{\sin^2x}{\sin x+2\cos x}dx$$

It isn't nice of me but I almost have no idea, yet I tried the trigonometric substitution $\;t=\tan\frac x2\;$ , but I obtained terrible things and can't do the rational function integral.

Perhaps there is exist some trigonometry equalities? I tried also

$$\frac1{\sin^6x+\cos^6x}=\frac{\sec^6x}{1+\tan^6x}=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\cdot\overbrace{\frac1{\sin^2x\cos^2x}}^{=\frac14\sin^22x}$$

and then doing parts with

$$u=\frac14\sin^22x\;\;:\;\;u'=\sin2x\cos2x=\frac12\sin4x\\{}\\v'=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\;\;:\;\;v=\arctan\tan^3x$$

But it is impossible to me doing the integral of $\;u'v\;$ .

Any help is greatly appreciated

• First one has the sum of two cubes. – Akiva Weinberger Apr 8 '15 at 10:02

For the first integral, try simplifying the denominator as $$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$$ $$=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$$ $$=1-\frac{3\sin^22x}{4}$$ This should start you off.

For the second integral, express $\sin x$ and $\cos x$ in terms of $\tan\left(\frac{x}{2}\right)$ and then substitute $u=\tan\left(\frac{x}{2}\right)$. Decompose into a lot of nasty looking partial fractions. For more help see here.

Hint: For the first integral, we can multiply by $\sec^{4}\left(x\right)$ numerator and denominator and get$$\int\frac{\sec^{2}\left(x\right)\left(\tan^{2}\left(x\right)+1\right)}{\tan^{4}\left(x\right)-\tan^{2}\left(x\right)+1}dx$$ and now take $u=\tan\left(x\right)$ $$\int\frac{u^{2}+1}{u^{4}-u^{2}+1}du=\int\frac{u^{2}+1}{\left(u^{2}-\frac{i\sqrt{3}}{2}-\frac{1}{2}\right)\left(u^{2}+\frac{i\sqrt{3}}{2}-\frac{1}{2}\right)}du$$ and now split using partial fractions. It gives a more tractable form. For the second integral take $u=\tan\left(\frac{x}{2}\right)$ to get $$-\int\frac{4u^{2}}{u^{6}-u^{5}+u^{4}-2u^{3}-u^{2}-u-1}du=-\int\frac{4u^{2}}{\left(u^{2}+1\right)^{2}\left(u^{2}-u-1\right)}du$$ and again you can use partial fractions.

• Thank you, though this is what I did and fractial partions are extremely hard this time. +1 – user177692 Apr 9 '15 at 10:55

HINT:

For the second one, as $\sin x+2\cos x=\sqrt5\sin\left(x+u\right)$ where $u=\arcsin\dfrac2{\sqrt5}\implies \sin u=\dfrac2{\sqrt5},\cos u=+\sqrt{1-\left(\dfrac2{\sqrt5}\right)^2}=\dfrac1{\sqrt5}$

let $x+u=y\iff x=\cdots$

$\sin^2x=\dfrac{1-\cos2x}2=\dfrac{1-\cos2\left(y-u\right)}2$

$\cos2\left(y-u\right)=\cos2y\cos\left(2u\right)+\sin2y\sin\left(2u\right)$

$\cos\left(2u\right)=1-2\sin^2u=\cdots$

$\sin\left(2u\right)=2\sin u\cos u=\cdots$

• Thank you, but it looks very hard. +1 – user177692 Apr 9 '15 at 10:56

$\bf{My\; Solution}::$ Given $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6 x}dx\;,$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^6 x\;,$

We Get $$\displaystyle \int\frac{\sec^6 x}{1+\tan^ 6 x}dx\;,$$ Now let $$\tan x=t\;,$$ Then $$\sec^2 xdx = dt\;,$$ So we get

$$\displaystyle \int\frac{(1+t^2)^2}{1+t^6}dt = \int\frac{(1+t^2)^2}{(1+t^2)\cdot (t^4-t^2+1)}dt = \int\frac{1+t^2}{t^4-t^2+1}dt$$.

Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $t^2\;,$ We get

$$\displaystyle \int\frac{\left(1+\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}-1}dt = \int\frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+1},$$ Now Let $$\displaystyle \left(t-\frac{1}{t}\right)=u\;,$$ Then $$\displaystyle \left(1+\frac{1}{t^2}\right)dt = du$$

So We get $$\displaystyle \int\frac{1}{u^2+1}du = \tan^{-1}\left(u\right)+\mathcal{C} = \tan^{-1}\left(\frac{t^2-1}{t}\right)+\mathcal{C}$$

So we get $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6x}dx = \tan^{-1}\left(\frac{\tan^2 x-1}{\tan x}\right)+\mathcal{C}$$

• Thanks, very nice! You began as I began but you went farther away. +1 – user177692 Apr 9 '15 at 10:57

Let $$I = \int\frac{1}{\sin^6 x+\cos^6 x}dx = \int\frac{1}{(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x)}dx$$

So $$\int\frac{1}{\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x(\tan^2 x-1+\cot^2 x)}dx$$

So $$\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x(\tan^2 x+\cot^2 x-1)}dx$$

So $$I = \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+1^2}dx$$

Now Put $(\tan x-\cot x) = t\;,$ Then $(\sec^2 x+\csc^2 x)dx = dt$

So $$I = \int\frac{1}{1+t^2}dt = \tan^{-1}(t)+\mathcal{C} = \tan^{-1}(\tan x-\cot x)+\mathcal{x}$$


text:
25 Mar 2019

# On camera sensor sizes

A common internet wisdom about photography is that bigger camera sensors capture more light. So if you want to work in low light, you need a full frame camera, and a bigger sensor always produces better image quality. I have struggled with this a lot. It just doesn't make sense: Lenses can focus light on any surface, so why should the surface size matter?

The answer turns out to be… disappointing. Big-sensor cameras allow for larger (practical) apertures, and lower base ISO. But less noise for the same picture is simply impossible with the same sensor technology. Because that's not how physics works. Let me explain.

At this point, I had previously launched into a long-winded explanation on equivalence, and reached a slightly misguided conclusion. So here's a better one:

It is possible to build equivalent lenses for differently sized sensors, which capture the same amount of light, with the same depth of field, and the same field of view, but simply project this light on a differently sized sensor. Or you can use a "speed booster", which adapts a "bigger" lens to a "smaller" sensor. Both of these will give you identical images, and identical amounts of noise (if the same sensor technology is used).

As an example, a micro four thirds 23mm f/1.4 lens produces identical images to an APS-C 35mm f/2 lens or a full frame 50mm f/2.8 lens. You might notice that $$\frac{23}{1.4} = \frac{35}{2} = \frac{50}{2.8}$$, which means each of these lenses will have the same physical aperture size, and therefore admit the same amount of light. And be the same size and weight. Since the smaller sensor collects the same light on a smaller area, the image will be brighter, and will need to use a lower ISO for the same exposure. And since lower ISO produces less noise, the "one stop advantage" of bigger sensors is bunk, if equivalent lenses are used.

The one remaining difference between sensor sizes is that bigger photosites carry more charge, which means more dynamic range at base ISO on a bigger sensor. This is only a factor if you indeed shoot at base ISO.

But, that's not the interesting part. Camera manufacturers cleverly designed their product lines such that small-sensor cameras are physically smaller, and get smaller lenses, while large-sensor cameras are bigger, with bigger lenses. Thus, if you want to get access to the most fancy glass, you will have to use a bigger, more expensive camera body as well. And inversely, if you want to use the most compact glass, it will only be available on smaller, less expensive camera bodies. And that's not pure marketing, either, as bigger lenses require bigger bodies to hold them comfortably, while smaller lenses balance better on a smaller body.

So in the end, it comes down to a compromize: If you want/need the best glass, it will only be available on the bigger bodies with the bigger sensors. If, on the other hand, you want to go small and light, you will need to sacrifice big apertures, but get access to smaller bodies with smaller sensors. But it's really all about the glass, not the sensor sizes.


text:
# Sh:948

• Göbel, R., Herden, D., & Shelah, S. (2011). Absolute E-rings. Adv. Math., 226(1), 235–253.
• Abstract:
A ring R with 1 is called an E-ring if End_{\mathbb Z} R is ring-isomorphic to R under the canonical homomorphism taking the value 1 \sigma for any \sigma \in End_{\mathbb Z} R. Moreover R is an absolute E-ring if it remains an E-ring in every generic extension of the universe. E-rings are an important tool for algebraic topology as explained in the introduction. The existence of an E-ring R of each cardinality of the form \lambda^{\aleph_0} was shown by Dugas, Mader and Vinsonhaler [DMV]. We want to show the existence of absolute E-rings. It turns out that there is a precise cardinal-barrier \kappa(\omega) for this problem: (The cardinal \kappa(\omega) is the first \omega-Erdős cardinal defined in the introduction. It is a relative of measurable cardinals.) We will construct absolute E-rings of any size \lambda <\kappa(\omega). But there are no absolute E-rings of cardinality \ge\kappa(\omega). The non-existence of huge, absolute E-rings \ge \kappa(\omega) follows from a recent paper by Herden and Shelah [HS] and the construction of absolute E-rings R is based on an old result by Shelah [S] where families of absolute, rigid colored trees (with no automorphism between any distinct members) are constructed. We plant these trees into our potential E-rings with the aim to prevent unwanted endomorphisms of their additive group to survive. Endomorphisms will recognize the trees which will have branches infinitely often divisible by primes. Our main result provides the existence of absolute E-rings for all infinite cardinals \lambda <\kappa(\omega), i.e. these E-rings remain E-rings in all generic extensions of the universe (e.g. using forcing arguments). Indeed all previously known E-rings ([DMV,GT]) of cardinality \ge 2^{\aleph_0} have a free additive group R^+ in some extended universe, thus are no longer E-rings, as explained in the introduction. Our construction also fills all cardinal-gaps of the earlier constructions (which have only sizes \lambda^{\aleph_0}). These E-rings are domains and as a by-product we obtain the existence of absolutely indecomposable abelian groups, compare [GS2].
• published version (19p)
Bib entry
@article{Sh:948,
author = {G{\"o}bel, R{\"u}diger and Herden, Daniel and Shelah, Saharon},
title = {{Absolute $E$-rings}},


text:
# Custom hatched pattern imitating hand drawing

Following this great question, I have been drawing a picture that simulates hand drawing. I want to fill some solids with a simple hatch pattern, however I don't know how to apply the path decorator to the lines that compose the pattern. Here is my MWE:

\documentclass[border=2mm]{standalone}

\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing, patterns}

\begin{document}
\begin{tikzpicture}[
solid/.style={draw=black,thick,pattern=north west lines, pattern color=black, decorate}
]

\begin{scope}[decoration={random steps,segment length=3pt,amplitude=0.45pt}, decorate]
\draw[solid] (0,0) rectangle +(2,2);
\end{scope}
\end{tikzpicture}
\end{document}

I tried to use the following example from the manual as a starting point but I am not sure how to apply the decorations at the "pgf level":

\documentclass[border=2mm]{standalone}

\usepackage{tikz}
\usetikzlibrary{patterns.meta}
\pgfdeclarepattern{%
name=hatch,
parameters={\hatchsize,\hatchangle,\hatchlinewidth},
bottom left={\pgfpoint{-.1pt}{-.1pt}},
top right={\pgfpoint{\hatchsize+.1pt}{\hatchsize+.1pt}},
tile size={\pgfpoint{\hatchsize}{\hatchsize}},
tile transformation={\pgftransformrotate{\hatchangle}},
code={
\pgfsetlinewidth{\hatchlinewidth}
\pgfpathmoveto{\pgfpoint{-.1pt}{-.1pt}}
\pgfpathlineto{\pgfpoint{\hatchsize+.1pt}{\hatchsize+.1pt}}
\pgfpathmoveto{\pgfpoint{-.1pt}{\hatchsize+.1pt}}
\pgfpathlineto{\pgfpoint{\hatchsize+.1pt}{-.1pt}}
\pgfusepath{stroke}
}
}

\tikzset{
hatch size/.store in=\hatchsize,
hatch angle/.store in=\hatchangle,
hatch line width/.store in=\hatchlinewidth,
hatch size=5pt,
hatch angle=0pt,
hatch line width=.5pt,
}

\begin{document}
\begin{tikzpicture}
\foreach \r in {1,...,4}
\draw [pattern=hatch, pattern color=red]
(\r*3,0) rectangle ++(2,2);
\foreach \r in {1,...,4}
\draw [pattern=hatch, pattern color=green, hatch size=2pt]
(\r*3,3) rectangle ++(2,2);
\foreach \r in {1,...,4}
\draw [pattern=hatch, pattern color=blue, hatch size=10pt, hatch angle=21]
(\r*3,6) rectangle ++(2,2);
\foreach \r in {1,...,4}
\draw [pattern=hatch, pattern color=orange, hatch line width=2pt]
(\r*3,9) rectangle ++(2,2);
\end{tikzpicture}

\end{document}

• It will compile if you (a) use \pgfdeclarepattern{name=hatch, i.e. no empty line, and (b) add \begin{document}. – Schrödinger's cat Nov 19 '19 at 4:16
• Thanks, I fixed that. – jsb Nov 19 '19 at 4:20
• You're welcome! Just to make sure I understand the question. You want the diagonal lines to have random steps, right? – Schrödinger's cat Nov 19 '19 at 4:32
• Correct, so that they would look "hand drawn" and hence in harmony with other lines in the picture. – jsb Nov 19 '19 at 4:35

This answer has a really important input by the OP, jsb, who pointed out that what I did with the clipping was insanely complicated. Thanks a lot!

This is a first, faster version:

\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{decorations.pathmorphing,patterns.meta}
\pgfdeclarepattern{name=lines,
parameters={\hatchsize,\hatchangle,\hatchlinewidth,\hatchsegmentlength,\hatchamplitude},
bottom left={\pgfpoint{-.1pt}{-.1pt}},
top right={\pgfpoint{\hatchsize+.1pt}{\hatchsize+.1pt}},
tile size={\pgfpoint{\hatchsize}{\hatchsize}},
tile transformation={\pgftransformrotate{\hatchangle}},
code={\pgfsetlinewidth{\hatchlinewidth}
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,0.5*\hatchsize) -- (\hatchsize+.1pt,0.5*\hatchsize);
}
}

\tikzset{
hatch size/.store in=\hatchsize,
hatch angle/.store in=\hatchangle,
hatch line width/.store in=\hatchlinewidth,
hatch size=5pt,
hatch angle=0pt,
hatch line width=.5pt,
hatch segment length/.store in=\hatchsegmentlength,
hatch segment length=3pt,
hatch amplitude/.store in=\hatchamplitude,
hatch amplitude=1pt
}
\begin{document}
\begin{tikzpicture}
\begin{scope}[thick,
decoration={random steps,segment length=3pt,amplitude=0.45pt}]
\draw[decorate,pattern=lines, pattern color=black, hatch size=7pt, hatch angle=21,
hatch line width=0.9pt] (0,0) rectangle +(2,2);
\draw[decorate,pattern=lines, pattern color=black, hatch size=12pt, hatch angle=70,
hatch line width=0.9pt,hatch amplitude=1.4pt,hatch segment length=4pt] (3,0) rectangle +(2,2);
\end{scope}
\end{tikzpicture}
\end{document}

Note also that these are patterns, so they are fast by repeating things precisely. This means that the pattern does not look as random as it could, one can clearly see the repetition. One way to address this issue is to create a slower pattern that has several (10, say) of these lines.

\documentclass[tikz,border=2mm]{standalone}
\usetikzlibrary{decorations.pathmorphing,patterns.meta}
\pgfdeclarepattern{name=lines,
parameters={\hatchsize,\hatchangle,\hatchlinewidth,\hatchsegmentlength,\hatchamplitude},
bottom left={\pgfpoint{-.1pt}{-.1pt}},
top right={\pgfpoint{10*\hatchsize+.1pt}{10*\hatchsize+.1pt}},
tile size={\pgfpoint{10*\hatchsize}{10*\hatchsize}},
tile transformation={\pgftransformrotate{\hatchangle}},
code={\pgfsetlinewidth{\hatchlinewidth}
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,0.5*\hatchsize) -- (10*\hatchsize+.1pt,0.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,1.5*\hatchsize) -- (10*\hatchsize+.1pt,1.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,2.5*\hatchsize) -- (10*\hatchsize+.1pt,2.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,3.5*\hatchsize) -- (10*\hatchsize+.1pt,3.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,4.5*\hatchsize) -- (10*\hatchsize+.1pt,4.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,5.5*\hatchsize) -- (10*\hatchsize+.1pt,5.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,6.5*\hatchsize) -- (10*\hatchsize+.1pt,6.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,7.5*\hatchsize) -- (10*\hatchsize+.1pt,7.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,8.5*\hatchsize) -- (10*\hatchsize+.1pt,8.5*\hatchsize);
\draw[decoration={random steps,segment length=\hatchsegmentlength,
amplitude=\hatchamplitude}, decorate,rounded corners=1pt]
(-.1pt,9.5*\hatchsize) -- (10*\hatchsize+.1pt,9.5*\hatchsize);
}
}

\tikzset{
hatch size/.store in=\hatchsize,
hatch angle/.store in=\hatchangle,
hatch line width/.store in=\hatchlinewidth,
hatch size=5pt,
hatch angle=0pt,
hatch line width=.5pt,
hatch segment length/.store in=\hatchsegmentlength,
hatch segment length=3pt,
hatch amplitude/.store in=\hatchamplitude,
hatch amplitude=1pt
}
\begin{document}
\begin{tikzpicture}
\begin{scope}[thick,
decoration={random steps,segment length=3pt,amplitude=0.45pt}]
\draw[decorate,pattern=lines, pattern color=black, hatch size=7pt, hatch angle=21,
hatch line width=0.9pt] (0,0) rectangle +(2,2);
%
\draw[decorate,pattern=lines, pattern color=black, hatch size=12pt, hatch angle=70,
hatch line width=0.9pt,hatch amplitude=1.4pt,hatch segment length=4pt]
(3,0) rectangle +(2,2);
\end{scope}
\end{tikzpicture}
\end{document}

The result looks definitely more random but takes longer to compile. I repeated the path by hand since I had once a bad experience when using \foreach in a meta pattern declaration. However, it seems to work fine in most cases. If I ever remember the case that went bad I will add it here. I believe to remember it was in connection with either this post or this post.

• Wonderful! I do not understand why you are using \clip, I drew a wedge with: \draw[decorate, pattern=lines, pattern color=black, hatch size=7pt, hatch angle=21, hatch line width=0.9pt] (0,0) -- (3,0) arc (0:30:3) -- cycle; and it looks perfect. I did not have to set the random seed either. – jsb Nov 19 '19 at 19:30
• @jsb That is a very good point! I just did not think about it. But you are right, this should work. I was being stupid. Thanks a lot! – Schrödinger's cat Nov 19 '19 at 19:33
• Thank you. Also, I have been playing around with \foreach, do you think you might have run into some bug using it in the past? Did you try it in this case? – jsb Nov 19 '19 at 19:39
• @jsb Now I regret that I did not keep the example. It is true that it usually works. I just think to have had one use case in which it didn't. And it was not something as stupid as calling the loop variable the same as something that is used elsewhere. but it could definitely be an unfortunate naming of the loop variable, or some expansion issue. But as of now I do not remember the details. – Schrödinger's cat Nov 19 '19 at 19:43
• @jsb At this point I only believe to remember it was in connection with tex.stackexchange.com/a/510132/194703 or tex.stackexchange.com/a/513715/194703, but I do not remember the details. – Schrödinger's cat Nov 19 '19 at 19:48


text:
## Biblioteca

Publicado en:

### Information design and sequential screening with ex post participation constraint

Theoretical Economics, 2020. Vol. 5. Pages 319–359

Abstract: We study a principal–agent model. The parties are symmetrically informed at first; the principal then designs the process by which the agent learns his type and, concurrently, the screening mechanism. Because the agent can opt out of the mechanism ex post, it must leave him with nonnegative rents ex post. We characterize the profit-maximizing mechanism. In that optimal mechanism, learning proceeds in continuous time and, at each moment, the agent learns a lower bound on his type. For each type, there is one of two possible outcomes: the type is allocated the efficient quantity or is left with zero rents ex post.

Keywords. Information design, sequential screening, ex post participation constraint, sequential information disclosure, dynamic mechanism design.

Ir a publicación


text:
# Slide Puzzle logic??

say there is an image made into a slide puzzle in a grid of sections 4 wide and 6 high, the missing piece that is missing so you can slide the other pieces of the puzzle around is always the bottom right piece of the grid if the picture matches the complete puzzle, so, if all the pieces of the image even the missing one were randomly placed on the grid, is there ANY chance that the puzzle would be impossible to complete?

-

It's entirely analogous to the 15-puzzle.

To show that not all initial positions allow you to get to the desired final position, note the following invariant.

Assume that in the final position you have the numbers from 1 to 23 written on the pieces starting in the top left corner, and proceeding row-wise to the bottom right corner, where you put 24 to represent the missing piece. Associate to this the identity permutation, obtained by reading the images of the numbers $1, \dots, 24$ row-wise from top left to bottom right.

When you exchange 24 with another piece, you are composing the given permutation with a $2$-cycle, thereby changing the parity. But you are also changing the block distance of 24 from the bottom right corner by 1, so $$P = \text{parity of the permutation} + \text{parity of the block distance}$$ is preserved.

Now the final position has $P$ even. If you start with the initial position in which 22 and 23 are exchanged, $P$ for this position is odd, so you will never be able to get to the final position from here.

-
is there an equation to seperate the solvable configurations from the impossible configurations? –  JRowan Feb 27 '13 at 15:03
As @ChrisTaylor said in his answer, you can solve exactly those (half of the total number of) configurations for which $P$ is even. This is a bit more complicated to show. –  Andreas Caranti Feb 27 '13 at 15:06
thanks for your time, is it half of the P even and half of P odd are the solvable ones and vice versa for impossible? or either even or odd are solvable? im just trying to get it through my head. –  JRowan Feb 27 '13 at 15:13
Even P = solvable, odd P = unsolvable. (PS You're welcome!) –  Andreas Caranti Feb 27 '13 at 15:16
im so happy right now, thank you, you live and you learn, notch on the belt –  JRowan Feb 27 '13 at 16:47

Yes. Exactly half of all random starting positions are solvable, and the other half are unsolvable.

-


text:
# HyperLogLog vs LogLog

To understand HyperLogLog (HLL), one should read the LogLog and HyperLogLog papers.

The major difference between the two is one uses regular mean and the other (HLL) uses harmonic mean to average the estimate cardinality calculated by different m buckets. I would view these buckets as different experiments.

The accuracy (standard error) is controlled by the number of buckets, m. The more buckets the better is the estimation. But, HLL has better accuracy compared to LogLog given the same number of buckets.

The standard error for HLL is

The standard error for LogLog is

The following is the LogLog algorithm taken from the paper

Basically, there are m buckets, each one is keeping track of the run of consecutive zeros in the bit string. We then figure out the max run of consecutive zeros (position of leftmost 1) in all the buckets.

eg.  1  => 1,   001 =>3

Then using the cardinality estimate formula shown in the last line, calculate the estimate unique counts. As you see from the formula, it involves taking the average of the run of zeros observed in these buckets. (Note: the alpha is the bias correction applied to the cardinality estimate)

For HyperLogLog, instead of using the average, harmonic mean (stochastic averaging) is used. Please refer to the HLL paper for more detailed description of the formula.

Taken from the HLL paper,


text:
Scary decisions and how to handle them

We regularly make decisions. First are simple decisions, such as whether to eat lunch at 12 pm or 12:15 pm. Second are large decisions, such as a job offer or being accepted to several top choice universities.

Those large decisions might be difficult to make, but often we’re picking which between A and B will give us the most benefit, assuming that the effort we’ll spend is roughly the same - e.g. comparable universities have similar course loads.

However, there is a third type of decision that we say no to by default, that fly under the radar. This is when the decisions require vastly different efforts. We often don’t realize we’ve made our decision by avoiding a high effort, scary option, instead of saying yes, and “how can I make it work?”

We decide “no by default” to the higher effort option

Not applying for a job, because it seems intimidating, is a decision. This is saying “no by default” to the “higher effort” option. Simply surviving, and not thriving, is also a decision - it is saying “no” when faced with the decision to explore outside one’s comfort zone.

We by default tend to say “no” to this type of decision, because we have the option to. Because nothing’s going to happen if we say no.

Ways of identifying this type of “no by default” decision are thoughts such as “It’s too difficult, maybe next time.” “I can’t do it.” - without further reasoning on “how can I make it work?”.

If you’re scared of something, you should do it!” is a way of thinking that has served me extremely well when I face decisions. It’s always pushed me toward a better life (how else am I going to have new material for this blog?) but it’s taken a long time to learn to let go of my fears.

In this article I will walk through how I manage these scary decisions in a practical way.

Analyzing “scary” decisions

Identify situations where reluctance comes from fear. If we fear something, it’s often that we care a great deal, and not that it is an unattractive option. If what we were scared of was actually unimportant to us, we would be indifferent to it, rather than fear it. We shouldn’t let fear get in the way of what we care about and desire.

One of the times I had been scared shitless was a 4 day span I took to interview for a multi-national gaming company. The final interview round had candidates fly in to Paris, France. Unfortunately, one of the courses that I required for my master’s degree had a final exam the morning of the return flight.

I was too paralyzed to even study, due to preparing for the interview. If I truly hadn’t cared about getting the job, I wouldn’t be scared - it was precisely that the outcome meant so much to me, that I was terrified.

I cared much more about the interview, and choosing it over the exam was the scary option. Falling back on what I knew, which is to study for an exam, was the easy option.

Pinpointing when I was trying to turn tail and hide. During the process, the thought had actually crossed my mind to give up on the interview, and instead study for my final exam. I had also secretly hoped to not get into the final round interview. Then, I would be able to say “well, I didn’t want the job, anyway” and not need to face the fears and the strong desire head-on.

To those on the outside, it was probably a no-brainer: “Of course you’d take the interview!” But in my mind at the time it was very conflicting. I found something I wrote in my journal at the time (habit of 4+ years): “(It feels like) running into the sky, throwing myself off a cliff, to a destination that cannot be seen, but such velocity.”

How often do we say no to opportunities because they are outside of our comfort zone?

When I finally just took the leap of faith, said yes, and made the arrangement to fly out to the interview, I was suddenly forced to face all my fears.

Hence the solution: just say yes, and the rest will follow. There are ways that we can achieve the outcome that we so desire, as long as we don’t let fear paralyze us and create a self-fulfilling prophecy.

Just say yes, and the rest will follow

Well, maybe some success will follow, after we put in hard work. There is always an incredible amount we can do that’s within our control, to make a difficult situation less scary, and thus make us perform better.

Do what is within control

Once I decided to do the interview, I started to plan. Since I was pretty much griefing my final exam anyway, the interview had to be worth it. I didn’t want to just survive - I wanted to be at the top of my game and absolutely crush it.

The interview start time in Paris, France would be 8 AM, which was 3 AM Toronto time. I started changing my sleep schedule by half hour increments a day, and changed my eating schedule gradually as well. This was so that my peak energy times would at least align with the work hours of the interview.

The day before the interview, in Paris, I spent 6 hours walking at tourist destinations, but also planned to get to bed by 10 pm local time.

I ended up overestimating my walking speed and went a bit too far (but I was at the foot of the Eiffel tower!) and had to get a taxi back to the hotel. The trip ended up costing just below the cash I had on hand (no Uber without mobile data, and didn’t feel comfortable enough taking the metro without mobile data). This was a minor suboptimal situation in the plan, but it all worked out.

I’m glad I didn’t stay at the hotel all day, but went out and enjoyed Paris, since cooping myself up would only make me more anxious and perform worse the next day. My mind was in a relaxed and joyful state - something that happens when exploring new, wonderful things in a city I hadn’t been to before.

A relaxed mindset was the goal, and it’s my secret sauce for performing well in high pressure situations. In high school, the day before the Taiwan-version of the SATs, I didn’t study at all, but listened to Lady Gaga on repeat. Scored top 4% across ~150,000 examinees. This worked for my CFA level 1 exam as well.

So I hit the hay at 11 pm, and woke up amazingly refreshed and on my A game the next day.

I was scared, but by preparing what was within my control, sleep schedule and energy management, I was able to perform at my best, despite being in a scary and difficult situation (that I cornered myself into by saying yes, oops).

Conclusion

Often, the scarier option is the one we want the most, but we let fear say no by default. Knowing this, we should say yes more - the rest will follow.

I’ve grown a lot from fighting against “no by default” thoughts and saying yes to scary choices that are worth it. It’s turned out to help me grow much more than I could have imagined. Otherwise, sometimes I wonder who in their right mind would want to build a video game studio while working and studying full time? That’s just looking for trouble! But years later, I couldn’t be happier.

There’s a lot within our control that we can do to turn the odds in our favor. Instead of paralyzed by fear, it’s helpful to execute on what can help us be at our best.

Worthwhile opportunities come now and then, but may be scary due to the high effort required from us, and the difficulty they pose. But let’s be ready to say yes!

More articles about "career"

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text:
# Math Help - Euler's Lucky number

1. ## Euler's Lucky number

Euler's Lucky numbers are positive integers n such that m2 − m + n is a prime number for m = 0, …, n − 1.

How do we prove that if n is lucky, then n is prime?

2. Try proving this by assuming to the contrary that $n$ is composite, or 1.

3. Interesting post!

In my(poor) opinion to prove this we can't do what mr. melese offered...

Here is a link I think it be useful! Conjecture 17. The Ludovicus conjecture about the Euler trinomials

4. Originally Posted by Also sprach Zarathustra
In my(poor) opinion to prove this we can't do what mr. melese offered..
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.

5. Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky by contraposition.
How you can choose m like that?

6. Originally Posted by Also sprach Zarathustra
How you can choose m like that?
$m$ takes on all values less than $n$.

7. Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.
For completness, when you start with $n$ not a prime you should also examine $n=1$(not a prime). Then what happens is that $m=0$ and $m^2-m+n=1$, and also $n$ isn't lucky.

8. Originally Posted by melese
For completness, when you start with $n$ not a prime you should also examine $n=1$(not a prime). Then what happens is that $m=0$ and $m^2-m+n=1$, and also $n$ isn't lucky.
To avoid this special case you could just assume $n$ is composite and let $m=0$.

9. But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.

10. Originally Posted by melese
But by definition 1 isn't a composite number (nor a prime), so 1 has a unique place.
But my way above takes care of 1 just like you did.

11. I found in a different book that $1 \leq m \leq n-1$.
It said the first 6 lucky numbers are 2, 3, 5, 11, 17 and 41 .
It should make sense to have m=0 since $0^2-0+2 = 2$, which is prime.

12. Originally Posted by chiph588@
But my way above takes care of 1 just like you did.
I assume you mean:
Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.
I don't see how the case n=1 is covered, because $n=ab$ and $a>1$.

13. Originally Posted by melese
I assume you mean:

I don't see how the case n=1 is covered, because $n=ab$ and $a>1$.
No I offered another solution later.

14. Originally Posted by chiph588@
If $n$ wasn't prime, then $n=ab$ where $n>a>1$.

Now let $m=a$: $a^2-a+ab = a(a-1+b)$ which is composite. Thus $n$ isn't lucky. So by contraposition the claim follows.
I spent half a day trying it many different ways. I attempted to prove this by contradiction at the beginning, but to prove $m^2-m+n \Rightarrow n$ being a composite is rather difficult because there isn't a mathematical expression for prime numbers, so I decided to prove it by contrapositive, but my main difficulty was here:

I let $n=ab$ where $1 and $1 and $a,b \in \mathbb{N}$.
Since $1 \leq m \leq n-1$, I let $m=ab-1$.

I ended up with $m^2-m+n= (ab-1)^2-(ab-1)+ab= a^2b^2-2ab+2$, which led me to nowhere, and I decided to post the question.

I found your technique quite interesting. I see why you chose $m=a$. It makes sense since $1. That's quite creative.

15. To show lucky => prime, simply let m = 0

clearly prime numbers are not necessarily lucky (n=7)

Are there infinite many lucky number?

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text:
Here is a list of generalized homological product codes.
Code Description
Balanced product code Family of CSS quantum codes based on products of two classical codes which share common symmetries. The balanced product can be understood as taking the usual tensor/hypergraph product and then factoring out the symmetries factored. This reduces the overall number of physical qubits $$n$$, while, under certain circumstances, leaving the number of encoded qubits $$k$$ and the code distance $$d$$ invariant. This leads to a more favourable encoding rate $$k/n$$ and normalized distance $$d/n$$ compared to the tensor/hypergraph product.
Distance-balanced code CSS stabilizer code constructed from a CSS code and a classical code using a distance-balancing procedure based on a generalized homological product. The initial code is said to be unbalanced, i.e., tailored to noise biased toward either bit- or phase-flip errors, and the procedure can result in a code that is treats both types of errors on a more equal footing. The original distance-balancing procedure [1], later generalized in Ref. [2], can yield QLDPC codes; see Thm. 1 in Ref. [1].
Expander lifted-product code Family of $$G$$-lifted product codes constructed using two random classical Tanner codes defined on expander graphs. For certain parameters, this construction yields the first asymptotically good QLDPC codes [3].
Fiber-bundle code CSS code constructed by combining a random LDPC code as the base and a cyclic repetition code as the fiber of a fiber bundle. After applying distance balancing, a QLDPC code with distance $$\Omega(n^{3/5}\text{polylog}(n))$$ and rate $$\Omega(n^{-2/5}\text{polylog}(n))$$ is obtained.
Generalized homological product code Stabilizer code formulated using the CSS chain-complex construction, with chain complexes consisting of products of other chain complexes. The chain-complex construction of codes yields an interpretation of codes in terms of manifolds, thus allowing for the use of various products from topology in constructing codes. The codes participating in the product can be quantum, classical, or mixed. Products can be of more than two codes, in which case the output code need not be of CSS type (e.g., for XYZ-product codes). The simplest product is a tensor product, with more general products imposing equivalence or symmetry relations on the outputs of the tensor product. A product of two codes can be interpreted as a fiber bundle, with one element of the product being the base and the other being the fiber.
Homological product code CSS code formulated using the CSS chain-complex construction in the homological product construction. Stub.
Hypergraph product code A family of $$[[n,k,d]]$$ CSS codes whose construction is based on two binary linear seed codes $$C_1$$ and $$C_2$$.
Lifted-product (LP) code Code that utilizes the notion of a lifted product in its construction. Lifted products of certain classical Tanner codes are the first (asymptotically) good QLDPC codes.
Quantum Tanner code Stub.
Quantum expander code CSS codes constructed from a hypergraph product of bipartite expander graphs with bounded left and right vertex degrees. For every bipartite graph there is an associated matrix (the parity check matrix) with columns indexed by the left vertices, rows indexed by the right vertices, and 1 entries whenever a left and right vertex are connected. This matrix can serve as the parity check matrix of a classical code. Two bipartite expander graphs can be used to construct a quantum CSS code (the quantum expander code) by using the parity check matrix of one as $$X$$ checks, and the parity check matrix of the other as $$Z$$ checks.
Ramanujan-complex product code CSS code constructed from a Ramanujan quantum code and an asymptotically good classical LDPC code using distance balancing. Ramanujan quantum codes are defined using Ramanujan complexes which are simplicial complexes that generalise Ramanujan graphs. Combining the quantum code obtained from a Ramanujan complex and a good classical LDPC code, which can be thought of as coming from a 1-dimensional chain complex, yields a new quantum code that is defined on a 2-dimensional chain complex. This 2-dimensional chain complex is obtained by the co-complex of the product of the 2 co-complexes. The length, dimension and distance of the new quantum code depend on the input codes.
XYZ product code A non-CSS QLDPC code constructed from three classical codes. The construction of an XYZ product code is similar to that of a hypergraph product code and related codes. The idea is that rather than taking a product of only two classical codes to produce a CSS code, a third classical code is considered, acting with Pauli-$$Y$$ operators.

## References

[1]
M. B. Hastings, “Weight Reduction for Quantum Codes”. 1611.03790
[2]
Shai Evra, Tali Kaufman, and Gilles Zémor, “Decodable quantum LDPC codes beyond the $\sqrt{n}$ distance barrier using high dimensional expanders”. 2004.07935
[3]
Pavel Panteleev and Gleb Kalachev, “Asymptotically Good Quantum and Locally Testable Classical LDPC Codes”. 2111.03654