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# Why does the definition of an open subscheme / open immersion of schemes allow for an “extra” isomorphism?

After taking an algebraic geometry course last year, I've been reviewing the material this year, and I remembered something that struck me as odd, but which I'd neglected to ask about at the time:

Hartshorne's definition of an open subscheme and open immersion (p.85):

An open subscheme of a scheme $X$ is a scheme $U$, whose topological space is an open subset of $X$, and whose structure sheaf $\mathcal{O}_U$ is isomorphic to the restriction $\mathcal{O}_X|_U$ of the structure sheaf of $X$. An open immersion is a morphism $f:X\to Y$ which induces an isomorphism of $X$ with an open subscheme of $Y$.

However, many other sources (including Wikipedia) go with an open subscheme being a scheme $Z$ of the form $(U,\mathcal{O}_X|_U)$ where $U\subseteq X$ is an open subset, and then defining an open immersion to be a morphism $f:Y\to X$ that factors through an isomorphism with an open subscheme, i.e. there is an isomorphism of schemes $g:Y\,\stackrel{\sim}{\to} Z$ such that $f=i\circ g$, where $i:Z\to X$ is the inclusion map.

Why make these (quite subtly) more general definitions? Clearly, there is some issue that is addressed by either making the definition of open subscheme, or that of open immersion, contain this "extra" isomorphism. By no means do I treat things that are isomorphic as being "equal", so I understand that this definition is not really equivalent. But surely defining a subgroup of a group $G$ to be "a group $H$ whose underlying set is a subset of $G$ and whose operation is isomorphic to the restriction of the operation of $G$" would sound a bit off?

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Re your last paragraph: what sounds better to you, if I say "the symmetry group of the cube is a subgroup of the symmetric group on 6 letters" or "the symmetric group on 6 letters has a subgroup isomorphic to the symmetry group of the cube"? To me, the 2nd alternative sounds much better, because the 1st makes it sound as though there was a canonical embedding, which is not the case. More seriously, if you want to be able to state results like "a map of schemes is an open immersion if and only if it is an etale monomorphism", then you need to use the categorical language that surprised you. –  Alex B. Oct 6 '11 at 7:02
I absolutely agree that in your example, the second version is much better. I suppose the issue is only then that I have not yet learned about this whole "etale" notion that seems to be so important. Could you expand on what it means to be an etale monomorphism, and why we would require this more general definition of open immersion (possibly in an answer)? Is this the main motivation for this definition, or are there any other contributing factors? –  Zev Chonoles Oct 6 '11 at 7:07
I don't think that the etale business is the main motivation for this definition. I just gave you an example of a statement that can only be stated cleanly in categorical language. An "open immersion" is a special kind of morphism in the category of schemes. If you want this class of morphisms to behave reasonably well, then it ought to be closed under taking compositions with isomorphisms, as almost anything categorical. So once we agree on this point, you have no choice in how to define open immersions. –  Alex B. Oct 6 '11 at 7:14
Having said that, maybe the definition of a Grothendieck topology will convince you that open immersions are defined the way they should be, since they are central in the definition of a Grothendieck topology. If this doesn't convince you, then we will have to wait for Matt Emerton to give a better explanation. –  Alex B. Oct 6 '11 at 7:17
Hmm, I do see your point, we certainly would want an open immersion composed with an isomorphism to be an open immersion, and I don't see any real way around it other than one of these two approaches (or, unless there is an "intrinsic" property that suffices to characterize them, somehow? Perhaps I am too attached to my manifold intuition?). I think that I prefer Wikipedia's approach over Hartshorne's approach, putting the extra isomorphism in the definition of the map. I am convinced, and you are welcome to put your comments as an answer - hopefully Matt Emerton will weigh in too though! :) –  Zev Chonoles Oct 6 '11 at 7:39
show 1 more comment

The definition of open subscheme in Hartshorne is so bad that the first sentence he writes after it is false!
That sentence is: Note that every open subset of a scheme carries a unique structure of open subscheme (Ex.2.2).
This is false: indeed, replacing $\mathcal O_{X|U}$ by an isomorphic sheaf you will get another scheme (albeit isomorphic, of course) and the claimed unicity does not hold..
And you really get other schemes because a scheme is a pair formed by a topological space and a sheaf of rings, not an isomorphism class of sheaves of rings.
So that the collection of subscheme structures on an open subset $U$ of a scheme $X$ would not even be a set with Hartshorne's definition ...

Of course in EGA, Görz-Wedhorn, Qing Liu,... you will find the reasonable definition that an open subscheme of $X$ is an open subset $U\subset X$ endowed with the restricted sheaf $\mathcal O_{X|U}$ , and not some arbitrary sheaf isomorphic to it. There is of course a canonical morphism of schemes $j:U\to X$

Edit: Open immersions Once you have the correct notion of open subscheme, you have no choice for defining an open immersion $Y\to X$. The natural idea is that you want it to be an isomorphism $g:Y \stackrel {\sim}{\to} U$ where $U\subset X$ is an open subscheme. However this has the wrong target, so you just compose it with the canonical morphism $j:U\to X$ mentioned above and you get the required immersion $f=j\circ g:Y\to X$, just as Wikipedia and most other references say.
And, quite satisfactorily, $j:U\to X$ itself is then an open immersion ( take $g=id_U$).

A criterion for being an open immersion: In practice to prove that a morphism $f:Y\to X$ is an open immersion, it suffices to check that:
a) $f$ induces a homeomorphism of $Y$ onto an open subset $U$ of $X$.
b) For all $y\in Y$ the local morphism $f^\ast_y: \mathcal O_{X,f(y)} \to \mathcal O_{Y,y}$ is an isomorphism of local rings.

NB Needless to say, I have an immense admiration for Hartshorne: I'm criticizing a definition in a book, and certainly not that great, friendly mathematician!

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# Xetex pdf bookmarks don't work

At this time arxiv does not support processing with. Go to home and click on the add bookmark button, then the bookmark you want to add will show in the left window. An explicit \usepackage or \requirepackage can only be dropped, because it is loaded by another package, if the dependency in the other package is part of the official interface for this other package. This includes all major texrelated programs and a huge range of additional fonts, macro packages, and utilities, with readytouse binaries for around 15 computer platforms including mac os x ppc and x86, windows.

When she walked up to his desk, he intended on evading her mind, but to his surprise, she was good at shielding her thoughts. For example, a package such as package bookmark might evolve to get rid of the dependency of hyperref. Using mtprint to print dvi files dvi files can also be printed with the help of the commandline utility mtprint miktex print utility. Anything that relies on something other than tex or pdflatex will fail. The fontspec package fontselectionforx e lt exandlual. Toolskindlatex, toolskindrerunnable, toolskindpdf, toolskindstdout, toolskindviewer, which give a list of commands that are treated as latex compiler e. Pdf portable document format is one of the most popular formats used for storing document files which include both text and graphics. Each bookmark goes to a different view or page in the document. Apologies if this isnt clear enough im slightly new here. The title doesnt say dont use latex at all, it says dont write in latex.

The bookmark should contain the name of the entry figure and table the number and the caption. After texexecs final xetex run, texexec runs xdv2pdf in order to create a pdf file. Unicodeaware texderivatives such as xetex, and also. You often need to know what engine your macros are running on the engine is the texderivative or texalike processor that typesets your document. In the case of 1, you probably use dvipsto create a postscript. According to the pypdf2 website, you can also use pypdf2 to add data, viewing options and passwords to the pdfs too. Optimizing pdf output size of tex documents tex users group. To add a bookmark to the document, go to click on the bookmark.

It processes cjk text, and the current encoding is some cxx. Cant add, change or save bookmarks how to fix firefox. A bookmark is a type of link with representative text in the bookmarks panel in the navigation pane. Once the bookmarks have been added, you can now convert the word document to pdf. Xetex is a standard component of the complete tex live distribution. To do that, click on file save as and in the save as type option click on the dropdown arrow to choose pdf. I will eventually switch to xetex, but since i dont have time to experiment too much at the moment, this is one of the best solutions i can find. For example, section commands will produce a bookmark and a link. Or one could perhaps use pdfencodingunicode option always. Moreover, they can offer decent image quality in a. The pdf specification allows two encodings pdfdocencoding 8bit encoding and unicode utf16. To have the bookmark window open automatically when the. Pdfa is a standardized version of the pdf file format and it is intended for archiving and. Simply extending the \declaregrphicsextentions does not work.

Examples are bookmarks or the entries in the information dictionary. If there is a problem with this file, your bookmark changes may be lost whenever you restart firefox. It allows you to create and manage latex projects directly on your ios device and generate a pdf by using verbosus online latex editor. Problem with \% in pdf bookmark under xelatex tex latex. Here are some things to try edit the document name above by typing in the input field. Some users have tried to use ghostscript to produce a pdf from acrotex. Adobe acrobatgenerated pdf docs that are opened in edge. This software is provided as is without warranties or conditions of any kind, either expressed or i.

Be sure to include with the source any stylemacro files that we dont have. It is possible to use standard postscript type 1 fonts rather than the computer modern fonts in latex for example, by using the packages times. He tried probing harder, but not too hard, for all he needed was for her to collapse under pressure. To insert a picture in your document, just use the \includegraphics command in the latex menu. Page 1 contents connections playback settings tips appendix avrx integrated network av receiver owners manual you can print more than one page of a pdf onto a single sheet of paper.

Page thumbnails and bookmarks in pdfs, adobe acrobat. Open the pdf in a viewer and open the bookmark panel. Xetex is a new tex engine by jonathan kew and sil international, which combines etex with pervasive. Sometimes we just dont want the wretched package interfering with us. Dynamic documents you can choose to export the finished report as a html, pdf, ms word, odt, rtf, or markdown document. This will help you to create pdf with bookmarks from word. How to create pdf file with all your bookmarks displayed. Perhaps you should convert it to ps and then convert back to pdf see xetex cannot include pdf page. For example, run mtprint paper to send the dvi file paper. Bookmarks and browsing history are stored in the firefox profile folder in the file places. According to the manual it is also possible to include tif, gif and png images as well, but i dont no how. Finally you can use pypdf2 to extract text and metadata from your continue reading an intro to pypdf2. Xetex and its variants including luatex, lyx, or pdftex.

Common mistakes that cause automated processing to fail arxiv e. The pdf docs are set to automatically display the bookmarks on open. The package derives from, and builds on, the work of the hypertex project. These show up in the pdf and in the postscript with some viewers. Both can be modified to affect the result of running both ffnpdf build and ffnpdf convert. Common mistakes that cause automated processing to fail. This text in zwadobe font, 1 pt is usually added by pdfmaker when it cannot complete the pdf file creation and it has to be removed with findreplace. Acrobat pdfwriter also directly prints pdf files, but does not seem to work effectively with tex based output and should be avoided. There are other engines, such as xetex and luatex, the latex versions are xelatex and lualatex. The context suite is a complete, uptodate context distribution that can be upgraded efficiently. Im going to go integrate xetex into my current product to generate custom typeset results for.

Xetex is unicodebased tex using platform fonts and font technologies e. Supported drivers are pdftex, dvips, dvipsone, textures, dvipdfm xetex, and vtex. In the pdf version, the caption will be on top of the image \scaledimggraphic. Finally, there are also hidden options, which can only be changed by modifing the inifile. This is important for large pdf docs as a navigation aid. Url that command may also work with windows, but i dont have a windows system to test with. Some features of the pdf specification needs pdf strings. Verbtex pro is the secure latex editor for your ios device. The setting i use the most is the one to generate the pdf in a different directory. The reason that you need to know is that the set of functions available in each engine is different. The user can help using \texorpdfstring to replace complicate tex constructs by a representation for the pdf. When i open two of the docs in reader, the bookmarks in the left pane, show the correct structure of sections and subsections, matching the table of contents. Imageconverter or its subclasses used for latex build are now sanitized, to prevent. Use a pdf printer driver such as pdfcreator for the windows platform or cupspdf on linux that can output selected pages.

You can print more than one page of a pdf onto a single sheet of paper. To generate a list of all fonts available available with xetex and xelatex, use this command at your unix command line. When saving the word file as pdf from within word 2007, the bookmarks on the bookmark panel of the resulting pdf file work fine. I dont really mind texs age since the results of typesetting with it are very nice and there is practically no alternative at least in academia.

The bookmark entries are taken from the chapter, section etc headings, but note that the text in the pdf bookmark will not follow any l a tex formatting. I need to add bookmarks for all figure and table entries in my document. I found one way of doing this using the bookmark package, however this requires that i manually insert the bookmark after each figuretable, and copy the caption to the bookmark. However, that is not the case with these word 2007 files. The package derives from, and builds on, the work of the hyper tex project, described at. Using a viewer to print dvipdf files tex output files. Ive just thanks largely to stefan managed to update those docs so that theyll render well in adobe reader 8. The defines how the bookmark fits in the hierarchy. You can now work around this problem by providing a text string for the bookmarks, which replaces the offending text if your urls are too long and running off of the page, try using the breakurl package to split the url over multiple lines.

It can also be used in parallel with existing tex installations such as miktex, texlive, and so forth. From data analysis in r to a typeset laboratory notebook as. The pypdf2 package is a purepython pdf library that you can use for splitting, merging, cropping and transforming pages in your pdfs. The option bookmarksopen can be set if you want all the bookmark subtrees expanded, and the option bookmarksnumbered can be set if you want the section numbers included in the bookmarks. Use of unusualuncommon fonts authors should keep in mind that if they use unusual fonts, many potential readers of their work wont have them installed. With \usepackagebookmarkshyperref i get only one bookmark for the whole document, it makes sense because it is for the proceedings in lncs, but i dont know how to change the way that template works wrt that, and that would be very useful for my thesis. Then, click on the browser button in the dialog to select the graphic file. Given that ghostscript cant create a pdfx1a, i think conversion is your best bet using acrobat pro.

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# On Philosophy

## March 10, 2007

### Identity And Necessity

Filed under: Language,Logic — Peter @ 12:00 am

I was trying out the new latex rendering engine in wordpress with this post. As you can see below it kind of sucks (it puts way too much space around the equations and doesn’t adjust to the viewer’s font size settings), so I won’t be using it again. But try to bear with me for this post, since I have everything in the latex format already and don’t exactly relish trying to get it into html formatted math. If anything is unreadable because of these issues just leave a comment and I will do my best to clarify matters. Now, with that out of the way, on to the post.

[Edit: It turns out this was partly a problem with my theme, which I have changed. Things look much better now, so I think I will stick with doing math this way in the future, assuming no one has any objections.]

There statement $x=y \rightarrow (\varphi(x/c) \rightarrow \varphi(y/c))$ ($\varphi(x/c)$ means that all free occurrences of c are replaced with x) is an axiom in first order logic, for all formulas $\varphi$ as long as x and y are free for c in $\varphi$. Now consider $\varphi = \Box (x=c)$. If we instantiate our axiom with that formula then it becomes $x=y \rightarrow (\Box (x=x) \rightarrow \Box (x=y))$, which seems to say, since $\Box (x=x)$ is true, that if x and y are identical then they are necessarily identical, or at least this is how Kripke reasons.

But the axiom that the argument begins with is not one from modal logic, it is from first order logic, so we can’t simply assume that it is consistent with modal logic. To explore that possibility let us describe modal logic in terms of first order logic. In first order logic we already have the concept of satisfaction. We say that a set $\Gamma$ satisfies a formula $\varphi$ when for all models $\mathcal{M}$ and assignments s that satisfy $\Gamma$ ($\mathcal{M} \models \Gamma [s]$) then that model and assignment satisfy $\varphi$ ($\mathcal{M} \models \varphi [s]$). Each model corresponds to a set of objects and defines which properties and relations hold for which objects. Thus a model is like a possible world. And so it seems natural to say that $\Box \varphi$ ($\varphi$ is necessary) means that $\varphi$ is true under all models and assignments, which means that all models and assignments satisfy $\varphi$ ($\mathcal{M} \models \varphi [s]$). Which means that the empty set satisfies $\varphi$ ($\emptyset \models \varphi$). And that means that $\varphi$ is a logical truth. Now certainly this is sufficient when we are trying to determine whether a formula is a necessary truth. But it doesn’t quite capture what we mean when we specify that a formula is necessary. When we specify that a formula is necessary we are adding it to the axioms, so to speak, indicating that it is true in all models, even if it isn’t a logical truth.

Now we can recast our original substitution in these terms. We let $\varphi = \emptyset \models x=c$. Thus after substituting we obtain: $x=y \rightarrow (\emptyset \models x=x \rightarrow \emptyset \models x=y)$, Since $\emptyset \models x=x$ is in fact a logical truth we obtain $x=y \rightarrow \emptyset \models x=y$. Now, using the idea that a necessary truth is true under all possible models we can check this claim. And it is in fact false. So what went wrong? Well, as mentioned when the axiom was first introduced that x and y had to be free for the c that they were replacing. Usually this just means that $\forall x , \exists x , \forall y , \textrm{ and } \exists y$ don’t occur where they might be able to capture x or y when they replace c. I propose that $\Box$ actually captures all free variables, and thus that x and y are not free for c in $\Box (x=c)$. This is because $\Box (x=c)$ expands to $\emptyset \models x=c$ which expands to $\textrm{for all models } \mathcal{M} \textrm{ and assignments s } \mathcal{ M} \models x=c [s]$. An assignment effectively captures free variables by assigning them to members of the domain of $\mathcal{M}$. This is just as much as a capture as $\forall x(x \leq c)$ which if it was allowed to fill the place of $\varphi$ would result in the claim that $x = 0 \rightarrow (\forall x(x \leq x) \rightarrow \forall x(x \leq 0))$. Assuming that our domain is the positive integers including zero this would imply that $x = 0 \rightarrow \forall x(x \leq 0)$, which is clearly ridiculous.

The conclusion to draw from all of this is not that a posteriori necessary truths are impossible (formulas that are necessary but not logical truths). However it does show that modal logic and first order logic don’t mix well. Modal logic is defined as an extension to sentential logic, and thus trying to integrate it into first order logic is not the easiest task. At the very least this exercise shows that the modal operator should be considered to capture free variables, but other complications may arise as well.


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Public Group

# Problem with wide chars in VC++ 2005 Express

This topic is 4566 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Okay I got a new problem in Visual C++ 2005 Express. I created a win32 program and loaded my opengl template code. After templewulf and SiCrane told me how to link libs in VC++ 2005, I tried to build it and had another error. I had made a global character array called ProgramName, and I had been just inserting the name of my program in there everytime I used the template for a different program. (I know, everyone says to use strings but I'm hard headed =) ). Well, in VC++ when I tried to compile it, it gave me this error: 'CreateWindowExW' : cannot convert parameter 2 from 'char [9]' to 'LPCWSTR' Also it complained about the same thing for my message boxes. I looked around a little and finally realized that CreateWindowExW and MessageBoxW are for wide characters, not regular ones. And CreateWindowEx and MessageBox are typedefed to the FunctionNameWs if UNICODE is defined and FunctionNameA if it's not. I believe this is the source of my problems. So why and where is UNICODE defined? Because I don't want it to be. Also, as a "side question", I noticed that when I tried to use my input method editor to enter japanese characters in my old VC++ 6.0 Introductory Edition, it replaced them with questions marks. But VC++ 2005 Express doesn't. So I thought I could now use japanese characters in my programs, but when changed my ProgramName array to a WCHAR array and tried to compile it, after telling me I'd need to save my cpp file in unicode format, it gave me these errors:
------ Build started: Project: Degenetron, Configuration: Debug Win32 ------
Compiling...
Degenetron.cpp
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30C7' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30B8' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30A7' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30CD' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30C8' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30ED' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : warning C4566: character represented by universal-character-name '\u30F3' cannot be represented in the current code page (1252)
c:\documents and settings\owner\desktop\icon folder\opengl programming stuff\degenetron\degenetron\degenetron.cpp(9) : error C2440: 'initializing' : cannot convert from 'const char [8]' to 'WCHAR []'
There is no context in which this conversion is possible
Build log was saved at "file://c:\Documents and Settings\Owner\Desktop\Icon Folder\OpenGL Programming Stuff\Degenetron\Degenetron\Debug\BuildLog.htm"
Degenetron - 1 error(s), 7 warning(s)
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

What does it mean they can't be represented? They are represented. At least I can see them, and they're not replaced with question marks. At any rate, what I really want is to get the compiler back to the way it was before, with regular 8-bit characters. Help would be appreciated. Thanks in advance.

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For the first part of your problem: go to your project settings, under general, go to the character set option and change it from Unicode to Multi-Byte. That should undefine UNICODE.

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Ah, k. Thanks again =).

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For the other part, codepage 1252 is the 8 bit ANSI codepage, which obviously can not represent unicode characters (codepage 1200 for instance) that are not in it. To fix that problem, simply save the source code files to be saved as unicode files. Select the file, then go to File->Advanced Save Options and set the encoding to 1200 or someother unicode format.

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I tried File->Advanced Save Options and selected Unicode - Codepage 1200 from the encoding list. But it still gave me the same errors.

Then I try saving it in a few of the japanese ones (like "Japanese (Shift-Jis) - Codepage 932"), and it gives me this error:

error C2440: 'initializing' : cannot convert from 'const char [25]' to 'WCHAR []'

So then I change the WCHAR array to a char array and it compiles, but when I execute the program, it's not what I entered at all, just some screwed up junk.

I mean this isn't really all that important. I used to be/kind of still am interested in japanese and I learned the kana, but I don't really *know* japanese and will likely never make a program that needs to use japanese characters. But it's still kind of annoying that I can't get the program to work. So if you know any other reasons this isn't work, it would be appreciated. Thanks in advance.

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that junk is the charset default , it's used when the the app can't find the charechtar you
specfied. make sure you have fonts that support the lang your writing in .i don't really know
how many letters there is in a japanese alphabet , but if there is more that (2^8)256 , you
should have unicode enabled.so you can have more than 8-bits for each letter.

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Quote:
 Original post by K A Z Eerror C2440: 'initializing' : cannot convert from 'const char [25]' to 'WCHAR []'

yeah, i think you are defining text incorrectly...

char str[] = "ascii";
wchar_t wstr[] = L"unicode"; // <<-- NOTE: capital L infront of "

normally you use the TEXT() macro for this, ie :

TEXT("text string")

which should sort that for you.

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Ah, okay. Yeah I had the charset on multi byte. And the L is why it was complaining when I was trying to define the WCHAR array.

Thank you both. =)

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# 17x-23=74+81

## Simple and best practice solution for 17x-23=74+81 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

## Solution for 17x-23=74+81 equation:

Simplifying
17x + -23 = 74 + 81
Reorder the terms:
-23 + 17x = 74 + 81
Combine like terms: 74 + 81 = 155
-23 + 17x = 155
Solving
-23 + 17x = 155
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '23' to each side of the equation.
-23 + 23 + 17x = 155 + 23
Combine like terms: -23 + 23 = 0
0 + 17x = 155 + 23
17x = 155 + 23
Combine like terms: 155 + 23 = 178
17x = 178
Divide each side by '17'.
x = 10.47058824
Simplifying
x = 10.47058824`

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# The hardest riddle on earth: Covering an 8×8 grid with trominoes

I already sought answers on three different forums without success. I hope I will be lucky this time.

I tried to find non-trivial solutions on a deficient 8×8 grid covered with trominoes and I regret to say that after extensive efforts I found only two non-trivial solutions:

diagram: 2 non-trivial solutions

The conditions of covering such a grid with trominoes are the following: In total we have 21 L-shaped trominoes with 3 different colors. There are equal numbers of trominoes of each color. Placing the trominoes on the grid, no 2 trominoes of the same color are allowed to touch each other anywhere, except only once, corner to corner (highlighted by red lines in the diagram).

Solutions from rotations and reflections are trivial. Does anyone know how to obtain more non-trivial solutions? Please include the full 8X8 grid in your answers.

It is more appropriate to be solved with a computer program. If you prefer to do it by hands, here is a workable procedure:

1. Place the hole. After removing symmetry redundancy, there are 10 distinct positions of the hole, namely: A1, B1, B2, C1, C2, C3, D1, D2, D3, D4;

2. Layout bricks, disregard color, you may want to work around the holes and spread out. Any partial layout that has additional holes in it is not feasible and should be discarded;

3. Color all feasible layouts obtained in step 2.

I was able to obtain a total of 49 unique solutions. Here is some examples:

01122113
33123133
13233221
11221211
23311332
22322322
31123113
33133133
solution:1 with corner touch:1

01122112
33123122
13233233
11221223
23311311
22322331
31123122
33133112
solution:2 with corner touch:1

01122122
33123112
13233233
11221223
23311311
22322331
31123122
33133112
solution:3 with corner touch:1

• You seem to have ignored this rule: "no 2 trominoes of the same color are allowed to touch each other anywhere, except only once, corner to corner" – r.e.s. Apr 21 '18 at 0:42
• Sorry. Can you name the two trominoes that touch each other. Reference a block by any of its cell. Eg. C2 and D1. Thanks! – Lance Apr 21 '18 at 0:48
• $(B2,C3), (E2,F3), (F2,E3), ...$ – r.e.s. Apr 21 '18 at 0:54
• @Lance: You have corners touching at H2,G3 and C6, B7 and E6,F7. – john Apr 21 '18 at 0:57
• The rule would be less ambiguous if stated something like this: "no 2 trominoes of the same color are allowed to touch each other anywhere, except at most one pair of the same color is allowed to touch corner to corner" – r.e.s. Apr 21 '18 at 1:15


text:
# Find Derivatives Using Chain Rule - Calculator

 A step by step calculator to find the derivative using the chain rule .


text:
Theorem 35.4.26. If $A \otimes _ R S$ has one of the following properties as an $S$-algebra

1. of finite type;

2. of finite presentation;

3. formally unramified;

4. unramified;

5. étale;

then so does $A$ as an $R$-algebra (and of course conversely).

Proof. To prove (a), choose a finite set $\{ x_ i\}$ of generators of $A \otimes _ R S$ over $S$. Write each $x_ i$ as $\sum _ j y_{ij} \otimes s_{ij}$ with $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending $e_{ij}$ to $y_{ij}$. Then $F \otimes _ R S\to A \otimes _ R S$ is surjective, so $\mathop{\mathrm{Coker}}(F \to A) \otimes _ R S$ is zero and hence $\mathop{\mathrm{Coker}}(F \to A)$ is zero. This proves (a).

To see (b) assume $A \otimes _ R S$ is a finitely presented $S$-algebra. Then $A$ is finite type over $R$ by (a). Choose a surjection $R[x_1, \ldots , x_ n] \to A$ with kernel $I$. Then $I \otimes _ R S \to S[x_1, \ldots , x_ n] \to A \otimes _ R S \to 0$ is exact. By Algebra, Lemma 10.6.3 the kernel of $S[x_1, \ldots , x_ n] \to A \otimes _ R S$ is a finitely generated ideal. Thus we can find finitely many elements $y_1, \ldots , y_ t \in I$ such that the images of $y_ i \otimes 1$ in $S[x_1, \ldots , x_ n]$ generate the kernel of $S[x_1, \ldots , x_ n] \to A \otimes _ R S$. Let $I' \subset I$ be the ideal generated by $y_1, \ldots , y_ t$. Then $A' = R[x_1, \ldots , x_ n]/I'$ is a finitely presented $R$-algebra with a morphism $A' \to A$ such that $A' \otimes _ R S \to A \otimes _ R S$ is an isomorphism. Thus $A' \cong A$ as desired.

To prove (c), recall that $A$ is formally unramified over $R$ if and only if the module of relative differentials $\Omega _{A/R}$ vanishes, see Algebra, Lemma 10.148.2 or [Proposition 17.2.1, EGA4]. Since $\Omega _{(A \otimes _ R S)/S} = \Omega _{A/R} \otimes _ R S$, the vanishing descends by Theorem 35.4.22.

To deduce (d) from the previous cases, recall that $A$ is unramified over $R$ if and only if $A$ is formally unramified and of finite type over $R$, see Algebra, Lemma 10.151.2.

To prove (e), recall that by Algebra, Lemma 10.151.8 or [Théorème 17.6.1, EGA4] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

There are also:

• 4 comment(s) on Section 35.4: Descent for universally injective morphisms

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
## Aims of this notebook

• non-parametric data
• correlation
• mean comparison for two groups
• mean comparison for two and more groups
• discrete data
• 2-by-2 tables
• R-by-C tables
• Multidimensional arrays

## Requirements

We assume that you have:

• read the required literature for weeks 1-5
• revised the lectures
• completed the introductory R tutorials (12 steps & How to solve problems) as well as the tutorial from week 2
• completed the homework in this module so far
• replicated the code from the lectures (if concepts/R implementation is unclear)

If you struggle with basics of R, you may also find this online book useful: https://bookdown.org/ndphillips/YaRrr/

## Non-parametric data analysis

Read the csv file data1_week5.csv from ./data. The csv file contains three variables of 1000 observations of areas surveyed and rated according to:

• their rank in deprivation on a scale from 1 to 10 (1 highest, 10 lowest)
• their rank in public disorder on a scale from 1 to 10 (1 highest, 10 lowest)
• their rank in social media participation on a scale from 1 to 5 (1 highest, 5 lowest)

You are interested in testing whether these data correlate. First, test whether the assumption of normally distributed data is met and the choose the appropriate test to ascertain whether or not there are correlations between these three variables.

#your code

### Task 2: Comparing two groups

Load the dataset (from the csv file data2_week5.csv) that gives you 1000 observations of male adolescents and the number of brothers they have and whether or not they had contact with the police in the past 5 years.

Assess whether those who had contact with the police differ from those who have not in the number of brothers they have. Again, make sure you test whether the assumptions are met and use an appropriate test.

#your code

### Task 3: Comparing 2 (or more) groups

Read the next csv file (data3_week5.csv). This file shows the number of casualties in traffic accidents on 600 high-risk street crossings. Each crossing is either a zebra-crossing, a traffic-light crossing, or an inofficial crossing that has none of the aforementioned.

Assess whether the factor “Type of crossing” has an effect on the number of casualties.

#your code

## Discrete data

Run the command below to create a 2-by-2 table that shows you the number of people that were refused boarding an airplane dependin whether they were drunk or not.

flight_data = array(c(100, 50, 150, 500), dim=c(2,2))
dimnames(flight_data) = list( c('Drunk', 'Not drunk')
, c('Denied boarding', 'Allowed boarding'))
flight_data

Assess whether there is an association between being drunk and being allowed to board the airplane.

Perform a Chi-Square test manually by calculating the expected cell counts and deriving the ChiSquare value.

#your code

Think for a moment: what do the expected values mean?

Now validate your results against the automatically calculated values.

#your code

Think for a moment: what does the test statistic and significance level mean?

Now follow-up on this finding if you deem this necessary. Is there an association? If so, what is it driven by?

#your code (if needed)

Think for a moment: what does each z-score mean?

Run the code below to build another array in the simple row-by-column format but with more levels for each. The below data array expresses the count of flight delays by the season for a small airport.

flight_delays = array(c(300, 350, 200, 150, 600, 300, 200, 100, 400, 100, 50, 50, 300, 250, 200, 200), dim=c(4,4))
dimnames(flight_delays) = list( c('On time', 'Minor delay', 'Severe delay', 'Cancelled')
, c('Spring', 'Summer', 'Fall', 'Winter'))
flight_delays

Use the appropriate statistical test and follow-up on the findings if necessary.

#your code

Which cells contain more counts than expected, which ones contain fewer counts than expected, and for which cells does this not differ significantly?

Run the code below to create a three-dimensional array that provides you with count data on whether someone was subjected to additional checks (e.g. an interview) upon entering the UK at an airport border control, which area they came from (EU vs nonEU) and whether the passenger was additionally checked by a customs office.

airport_data = array(c(1000, 500, 4000, 5000, 8000, 1000, 6000, 4000), dim=c(2,2,2))
dimnames(airport_data) = list('additional_checks' = c('yes', 'no')
, 'zone' = c('EU', 'nonEU')
, 'customs_check' = c('yes', 'no')
)
airport_data

To test whether there are associations between the factors “additional_checks”, “customes_check”, and “zone”, we’d want to use a loglinear model. Before we can use that model, we need to transform the representation of the data.

airport_data_long = as.data.frame(as.table(airport_data))
airport_data_long

Now the modelling process can start.

First, build and assess the independence model:

#your code

Take a look at the actual vs fitted values:

#your code

We’d like to use a better model for the data. The ideal (= perfect) model is of course the saturated (full) model. Build this model and show that the fitted values are identical to the observed ones.

#your code

The task is to find a simpler model that still is adequate. Next, build a model that in addition to the independence model contains two-way dependencies (similar to interactions) and test whether you can reject the H0 of model adequacy.

#your code

What do you conclude?

Finally, use the automated model select with the step function.

#your code

What do you conclude?


text:
# Question #97270

Apr 12, 2017

Equation is already in $y = m x + b$ format. Just graph using the slope and y-intercept.

#### Explanation:

So the equation ($y = \frac{9}{4} x - 4$) is already in $y = m x + b$ format.
=> Where $\frac{9}{4}$ is the slope, $m$.
=> Where $- 4$ is the y-intercept, $b$.

All we have to do is graph it.

graph{9/4x-4 [-10, 10, -5, 5]}

I think what you have difficulty with, is the slope. You can use the $\text{rise"/"run}$ method to graph the slope.

Basically, all numerators are $\text{rise}$, and all denominators are $\text{run}$. So in this case, starting at $\left(0 , - 4\right)$, move up $9$ points, then right $4$ points, giving us $\left(4 , 5\right)$.

You also do the opposite: starting at $\left(0 , - 4\right)$, move down $9$ points, then left $4$ points, giving us $\left(- 4 , - 13\right)$.

Connect them with a straight line and BAM!

Hope this helps :)


text:
Science Modules / Macrophyte

We approach the model by including a core state variable HALO, that is made up of above ground (AG) and below ground (BG); the former referring to the leaves, and the latter the cumulative mass of rhizomes and roots.

aed2_macrophyte : Mass balance and functions related to the macrophyte model

The equation for seagrass biomass in a given cell is computed as shown in the below table, assuming photosynthesis, excretion, mortality and excretion. Nutrient uptake is not included in the simulation in SCERM v1 since Hillman et al. (1995) highlight that light, salinity and temperature were the dominant drivers, with some minor correlation with phosphate levels, however the stoichiometry is known and may be included in future iterations. The effective coverage area (i.e. Leaf Area Index, HALOLAI of the meadow) can be computed based on the scaling expression in Baird et al. (2016); although this is not directly used in the photosynthesis calculation, it is useful output for comparing with field data.

Growth is calculated in response to light, but also sensitive to salinity and temperature:

Photosynthesis-irradiance relationships for Halophila have been estimated by Ralph and Burchett (1995) who found photo-inhibition occurring at modest light intensities. The Steele (1962) equation is therefore suggested as the most appropriate. Light extinction can also occur over the meadow depth, although for Halophila this is assumed to be relatively small due to the small leaf structure. The salinity effect on photosynthesis has been reported by Ralph (1998b) and Hillman et al. (1995).

Respiration, excretion and mortality are also commonly simulated with typical first-order rate coefficients for each:

The seagrass-sediment-light (SSL) feedback has been identified as a potentially important driver determining meadow persistence (Adams et al., 2016), as depicted schematically for a numerical model in Figure 8. This requires the connection between sediment resuspension and seagrass presence to be made, however, the complete feedback loop has rarely been reported in aquatic models to date. The model implemented for SCERM focuses on Halophila and accounts for the feedback by including ability to simulate the link between: a) above ground biomass (l“DYÀE) and shear stress, b) below ground biomass (l“DYFE), and c) the amount of resuspension and light.

To simulate the SSL feedback, drag is increased in proportion with the above ground biomass:

where $$C_d$$ is the base drag coefficient for a numerical cell based on its sediment material properties, and the above ground fraction is a user definable constant fraction of total biomass. The default value of $$C_{D_{bottom}}$$is stored in the hydrodynamic driver model and influences the local momentum budget; therefore the 2nd term of the RHS is passed to the host model calling AED2. The critical shear stress for resuspension is also increased based on the below ground biomass:

where $$\tau_0$$ is critical shear stress for resuspension, computed based on a minimum value (reflecting bare sediment), and linear coefficient linked to biomass. As HI increases, the concentration of SS in the local domain will reduce, thereby improving the overall light climate. Note that in the model, presence or absence of HALO in a given cell can be configured be reading in an appropriate distribution map.

Variable Summary & Setup Options

pending ...

pending ...

The following parameters and configuration variables are available to be set via the nml module block:

Parameter Name Description Units Parameter Type Default Typical Range Comment
num_mphy number of macrophyte groups - integer 0 0-10 -
the_mphy list of ID's of groups in aed_mphy_pars.nml (len=num_phyto) - integer 0 1-10 -
n_zones number of bethic zones of macrophyte - integer 0 - -
active_zones identity of zones being activated - integer 0 1-10 -

An example nml block for the phytoplankton module is shown below.NOTE: Users must supply a valid "aed_mphy_pars.nml" file.

 &aed2_phytoplankton num_mphy = 1 the_mphy = 1 n_zones = 1 ! This was 1 #BB active_zones = 5 dbase = '../External/AED2/aed_mphy_pars.nml' /

Examples

Under construction

Publications & References

To be completed ...


text:
Animations unifying the relationships among rates of change and accumulations of power functions

The formulas illustrated are:<

which may be rearranged to obtain the familiar summation formulas used in integrating  and directly:

These demonstrations show that the coefficients of  and  in the differentiation and integration formulas share a common source: Pascal's triangle

They foreshadow the deeper connection between rates of change and geometric boundaries made systematic in the Stokes' theorems.

We ultimately wish to emphasize other aspects of the duality between differentiation and integration in a more unified and symmetrical fashion, exhibiting in parallel the linearity rules for both differentiation and integration, the product rule beside integration by parts, the chain rule along with substitution, etc.

Along with summation formulas, discrete analogues of other calculus rules, such as "summation by parts" will also be highlighted.

Andre Cherkaev

2001-06-15


text:
Asymptotics of Chebyshev polynomials for general subsets of the real line

-
Barry Simon, Caltech
Fine Hall 314

Given a compact subset, E in R, the Chebyshev polynomials are the unique minmizers of the sup norm over E among all degree n monic polynomials. I will describe some recent results of Christensen, Simon and inchenko on this classical subject. We settle a 45 year old conjecture of Widom on the large n pointwise asymptotics for the case of finite gap sets. We also extend upper bounds on the norm that Totik and Widom obtained for the finite gap case to positive measure Cantor sets. Our proof in this case is new and even in the finite gap case is simpler and more explicit bounds than the earlier work.


text:
# Question b467f

Jul 9, 2017

Limiting reagent: ${\text{C"_2"H"_3"Br}}_{3}$

#### Explanation:

We're asked to find the limiting reactant in a chemical reaction given the masses of reactants.

Let's first write the balanced chemical equation for this reaction (look up what forms if you don't know for sure):

$4 {\text{C"_2"H"_3"Br"_3 (l) + 11"O"_2 (g) rarr 8"CO"_2 (g) + 6"H"_2"O" (g) + 6"Br}}_{2} \left(l\right)$

To find the limiting reactant, we convert the given masses of reactants to moles, then divide by the respective coefficient in front of it; the limiting reactant is the quantity whose number is lower:

$76.4 \cancel{\text{g C"_2"H"_3"Br"_3)((1color(white)(l)"mol C"_2"H"_3"Br"_3)/(266.76cancel("g C"_2"H"_3"Br"_3))) = 0.286/(4color(white)(l)"(coefficient)}}$

= color(red)(0.0716

$49.1 \cancel{\text{g O"_2)((1color(white)(l)"mol O"_2)/(32.00cancel("g O"_2))) = 1.53/(11color(white)(l)"(coefficient)}}$

= color(blue)(0.140

Since the number for ${\text{C"_2"H"_3"Br}}_{3}$ is lower, $\textcolor{red}{{\text{C"_2"H"_3"Br}}_{3}}$ sfcolor(red)("is the limiting reagent"#.


text:
### Example of 1-form which is closed, but not exact

(Supplement to 2.6)

On $\mathbb{R}^2\setminus \{0\}$ (hole is very important) consider form $$\frac{-ydx+xdy}{x^2+y^2}.$$ So, $M(x,y)= -y/(x^2+y^2)$, $N(x,y)=x /(x^2+y^2)$.
• Prove that $$\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=0.$$ as $(x,y)\ne 0$.
• Consider polar angle $\theta(x,y)=\arctan (y/x)$.
• Prove that $$\frac{\partial \theta}{\partial x}=-\frac{y}{(x^2+y^2)},\qquad \frac{\partial \theta}{\partial y}=\frac{x}{(x^2+y^2)}.$$
• This provides another proof that the form is closed.
• The trouble with this primitive is that it is defined uniquely on a subdomain $\Omega\subset \mathbb{R}\setminus \{0\}$ if and only if one cannot go around $\{0\}$ as in this case increment of $\theta$ would be $2\pi$ (and thus $\theta$ would be defined up to $2\pi \mathbb{Z}$.
• Further, there is no primitive, as if $d F=d\theta$ then increment of $F$ as we go around $\{0\}$ along contour $\Gamma$ would be $$\oint _\Gamma dF= \oint _\Gamma d\theta =2\pi$$ (and thus $F$ would be defined up to $2\pi \mathbb{Z}$.


text:
# How to think about dual space of a certain space of Lipschitz functions

Consider the following Banach space (for concreteness):

$$X=Lip(\bar{\mathbb{B}}^n)=\{f\in C^0(\bar{\mathbb{B}}^n): \Vert f \Vert_L<\infty \}$$ where $$\bar{\mathbb{B}}^n=\{\mathbf{x}\in \mathbb{R}^n: |\mathbf{x}|\leq 1\}$$ is the closed ball and $$\Vert f \Vert_L=\sup_{\bar{\mathbb{B}}^n}|f| +\sup_{\mathbf{x}\neq \mathbf{y}\in \bar{\mathbb{B}}^n} \frac{|f(\mathbf{x})-f(\mathbf{y})|}{|\mathbf{x}-\mathbf{y}|}$$ is one of the usual versions of a lipschitz norm.

I'm curious what is the best way to think about the (topological) dual space $$X^*$$ of $$X$$ as this space is a bit mysterious to me.

For instance, it's clear that any finite (signed) measure on $$\bar{\mathbb{B}}^n$$ can be thought of as an element of $$X^*$$, but one should also have elements that look like differences of infinite measures whose supports are sufficiently close and whose relative mass" is finite. Are there other natural elements?

Any references (in particular those that are more concrete and less abstract) would be appreciated.

Edit: I guess (please correct me if I am wrong) a fairly pathological element in the dual would be something like $$\mu=\sum_{i=1}^\infty \left( 2^i \delta_{2^{-2i}}-2^{i} \delta_{-2^{-2i}}\right)$$ on $$\mathbb{B}=[-1,1]$$.

Are there any natural subsets on which one can restrict and have a less crazy situation? For instance, fix a Radon measures $$\mu$$ on $$\mathbb{B}^n$$ (the open ball) with $$\mu(\mathbb{B}^n)=\infty$$. Is the subspace

$$Z=\{T\in X^*: T=\nu-\mu: \nu \mbox{ a Radon measure}\}$$ any nicer?

• The space you describe contains an isomorphic copy of $l^\infty$. Its dual is wild. – Nik Weaver Jan 3 '19 at 23:00
• @NikWeaver Any interesting ways to tame it? For instance, I'm actually interested in something like the following: I have a fixed Radon measure $\mu$ (of infinite mass) on $\mathbb{B}^n$ (the open ball) and am interested in the set of measures $\nu$ on $\mathbb{B}^n$ so that $\nu-\mu$ lives in $X^*$. – RBega2 Jan 3 '19 at 23:12
• I think you want to look at the predual, which is variously known as the "Arens-Eels" or "Lipschitz-free" space. For specific information about the measures which live there see Functional Analysis by Kantorovich and Rubenstein. I don't have the reference handy but it can be found in my book on Lipschitz algebras. – Nik Weaver Jan 3 '19 at 23:58
• As a linear space, $Z$ is isomorphic to a subspace of Radon measures, so yes, it is nicer. I suppose you would like to say something nice about the norm on $Z$. – user95282 Jan 8 '19 at 11:23
• I think the topological dual of your space is the space of measures on $\overline{\mathbb B}^n$, equipped with the Kantorovich-Rubenstein norm. See my answer below. – dohmatob Nov 6 '19 at 18:34

This would be a really long comment, so I've decided to post it as an answer. Hope it helps!

Disclaimer. I'm still learning FA, and my answer is based on my blurred understanding of the subject. I hope experts here will (in)validate this.

On a metric space $$X=(X,d)$$, let $$\mathcal M(X)$$ be the set of all measures and $$\mathcal M_0(X)$$ be the subset of measures $$\mu$$ for which $$\mu(X)=0$$. Define $$\|\cdot\|_{KR} \rightarrow \mathbb R$$ by

$$\|\mu\|_{KR} := \inf_{\nu \in \mathcal M_0(X)}\|\nu\|_0 + \|\mu-\nu\|_{TV},$$

where

• $$\|\mu-\nu\|_{TV}$$ is the total-variation between $$\mu$$ and $$\nu$$
• $$\|\nu\|_0 := \underset{\lambda \in \Phi_\nu}{\inf}\int_{X \times X}d(x,x')d\gamma(x,x')$$ and $$\Phi_\nu$$ is the subset of nonnegative measures $$\gamma$$ on $$X \times X$$ such that $$\gamma(X \times B) - \gamma(B \times X) = \nu(B)$$ every Borell subset $$B$$ of $$X$$.

Let $$\|f\|_L := \max(\|f\|_\infty, L(f))$$ be the bounded-Lipschitz norm on $$Lip(X)$$. Then

Theorem. $$(\mathcal M(X),\|\cdot\|_{KR})$$ is a normed vector space and the dual pairing $$\langle f, \mu\rangle := \int_{X} fd\mu,\;for\;(f,\mu) \in Lip(X) \times \mathcal M(X)$$ establishes an isometric isomorphism between $$(Lip(X), \|.\|_L)$$ and the topological dual of $$(\mathcal M(X),\|\cdot\|_{KR})$$.

This result goes back to Kantorovich and Rubenstein (hence "KR"), and is well-documented in Hanin '92 (see Theorem 0 there).

• I don't think Hanin is the appropriate citation. He quotes this result but it was originally proven by Kantorovich and Rubenstein (hence "KR"). – Nik Weaver Nov 6 '19 at 18:11
• Sorry, I failed to mention that KR = Kantorovich-Robenstein. – dohmatob Nov 6 '19 at 18:29
• Doesn't $\mu(X) = 0 \implies \mu = 0$? – ogogmad Sep 13 '20 at 7:36
• I guess you mean signed measures – ogogmad Sep 13 '20 at 9:12
• Yep, by "measures" I mean "signed measures". – dohmatob Sep 13 '20 at 11:02


text:
# How do you configure Texmaker to use LuaTeX?

I'm extending my use from pdfLaTeX to LuaLaTeX. The transition is smooth especially as I collaborate with other users that shy out upgrading their systems. However, TexMaker does not seem to come pre-configured in the Texmaker editor.

How does one configure Texmaker to use LuaTeX on a document-to-document basis?

(I am using ubuntu 10.4, but this question should be a priori platform independent.)

-

If you want to permanently switch from pdflatex to lualatex, see Joseph's answer. If, however, you want to be able to use both, you need to add a user-defined command. From the menu, User -> User commands -> Edit user commands opens a dialog box. Choose a name (eg "LuaLaTeX" for the command) and use lualatex --interaction=nonstopmode % for the command itself. This command is now available form the menu or as Alt+Shift+F1.

Normally, lualatex should be in your command search path (given by the PATH environment variable), but if it isn't, you need to specify the full path to lualatex. Check the path for pdflatex in the preferences: lualatex is most probably in the same directory.

-
The question doesn't exactly give away whether this is a permanent change or not :-) –  Joseph Wright Nov 15 '10 at 20:16
minor suggestion for improving your answer, for completeness' sake: adding how (with which terminal command) to check/verify whether Lua's on PATH... –  nutty about natty Jul 2 '13 at 6:23

From the Preferences menu entry (location platform-dependent), you should be able to change the 'PdfLaTeX' line changing <path>/pdflatex to <path>/lualatex. Usually, <path> will be something like /usr/local/texlive/2010/bin, or /usr/texbin on the Mac.

-

Another way is to use arara. It is included in TeX Live from version 2012, so those with recently updated TeX Lives will have it available. Other users will have to install it manually.

To use arara from Texmaker see Integration of arara in Texmaker. Having installed it, and configured Texmaker, one just has to add for example

% arara: lualatex

at the beginning of your file, and compile with arara. See Tools for automating document compilation for another example.

-

Options > Configure Texmaker.

If you intend to replace pdflatex then I suggest you simply overwrite the PdfLaTeX-field with the appropriate command. Either that or you can specify a User Command (User > User Commands > Edit User Commands) and then invoke it from the same menu.

Keep in mind that if you're using a custom Quick Build you need to make sure it's replaced too, as just changing the PdfLaTeX field doesn't do this for you.

-

I post this as a new answer because the latest version of TexMaker (4.1) supports this natively.

Options -> Configure TexMaker
Fast Compil -> Luatex

-


text:
# Assembly parts

Nine machines produce 1,800 parts on nine machines. How many hours will it produce 2 100 parts on seven such machines?

Correct result:

t =  12 h

#### Solution:

$s=1800 / (9 \cdot \ 8)=25 \ \\ t=2100/(7 \cdot \ s)=2100/(7 \cdot \ 25)=12 \ \text{h}$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Please write to us with your comment on the math problem or ask something. Thank you for helping each other - students, teachers, parents, and problem authors.

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A bottle full of cola weighs 1,320 g. If we drink three-tenths of it, it will weigh 1,008g. How much does an empty bottle weigh?
• Motorcycle engine
The motorcycle engine has a constant power of 1.2 kW for 0.5 hours. How much work does the engine do? Express work in kWh and Ws units.
• In the
In the rectangle ABCD, the distance of its center from the line AB is 3 cm greater than from the line BC. The circumference of the rectangle is 52 cm. Calculate the contents of the rectangle. Express the result in cm2.
Calculate the surface of a quadrilateral prism according to the input: Area of the diamond base S1 = 2.8 m2, length of the base edge a = 14 dm, height of the prism 1,500 mm.
• Cube surfce2volume
Calculate the volume of the cube if its surface is 150 cm2.
• Concrete hatch
The concrete hatch for a round well has a diameter of 1300 mm and a thickness of 80 mm. Determine its weight in kg if the density of the concrete is 2545 kg/m3
• Three examples per-mille
A. Calculate 1.5 ‰ from 4.5 liters of blood. Express the result in milliliters. B. Calculate three ‰ from € 50,000. C. The tray they brought breakfast to the landlord's bed is made of an alloy that contains 830 ‰ of silver. How much silver did they use
• Spherical segment
Calculate the volume of a spherical segment 18 cm high. The diameter of the lower base is 80 cm, the upper base 60 cm.
• Cuboid edges
Calculate the volume and surface of a cuboid whose edge lengths are in the ratio 2: 3: 4 and the longest edge measures 10cm.
• Ratio
Alena collected 7.8 kg of blueberries, 2.6 kg of blackberries, and 3.9 kg of cranberries. Express the ratio in the smallest natural numbers in this order.
• Bike cost
The father gave his son € 100 to buy a bicycle, which was 40% of the total amount of the bicycle. How much did the bike cost?


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# Integral

### Definition

$$\int f'(x) \, dx = f(x) + C$$

### Identities

$$\begin{array}{ll} \int a \, f(x) \, dx = a \int f(x) \, dx & \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \\[12pt] \int x^a \, dx = \frac{x^{a + 1}}{a + 1} + C & \int e^{x} \, dx = e^x + C \\ \int x^{-1} \, dx = \ln |x| + C & \int a^{x} \, dx = \frac{a^x}{\ln a} + C \\[12pt] \int \sin x \, dx = -\cos x + C & \int \frac{\sin x}{\cos^2 x} \, dx = \frac{1}{\cos x} + C \\ \int \cos x \, dx = \sin x + C & \int \frac{\cos x}{\sin^2 x} \, dx = \frac{1}{\sin x} + C \\[12pt] \int \frac{1}{\sin^2 x} \, dx = -\cot x + C & \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \frac{x}{a} + C \\ \int \frac{1}{\cos^2 x} \, dx = \tan x + C & \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \\[12pt] \int_a^b f'(x) \, dx = f(b) - f(a) & \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \end{array}$$

### Integration by substitution

$$\int \sin\left(\frac{\pi}{2}x\right) \, dx = \int \frac{2}{\pi} \sin u \, du = -\frac{2}{\pi} \cos\left(\frac{\pi}{2}x\right) + C \quad \color{red}{u = \frac{\pi}{2}x \quad du = \frac{\pi}{2}dx}$$
$$\int \frac{x}{\sqrt{x^2 + 16}} \, dx = \int \frac{1}{2} \frac{1}{\sqrt{u}} \, du = \sqrt{x^2 + 16} + C \quad \color{red}{u = x^2 + 16 \quad du = 2x\,dx}$$
$$\int \frac{x}{(3 - x^2)^4} \, dx = \int -\frac{1}{2} \frac{1}{u^4} \, du = \frac{1}{6(3 - x^2)^3} + C \quad \color{red}{u = 3 - x^2 \quad du = -2x\,dx}$$

### Integration by parts

$$\int u \, dv = uv - \int v \, du$$
$$\int x\cos x \, dx = x\sin x - \int \sin x \cdot 1 \, dx = x\sin x + \cos x + C \quad \color{red}{u = x \quad dv = \cos x\,dx}$$

### Riemann sum

Let $[a, b]$ be partitioned by points $a < x_1 < x_2 < \ldots < x_{n-1} < b$ , where the lengths of the resulting intervals between the points are denoted $\Delta x_1, \Delta x_2, \ldots, \Delta x_n$ . Let $x_k^*$ be an arbitrary point in the $k$-th subinterval. Then the quantity
$$\sum_{k = 1}^n f(x_k^*)\Delta x_k$$
is called a Riemann sum for a given function $f(x)$ and partition, and the value $\max\Delta x_k$ is called the mesh size of the partition.
Different types of Riemann sum (left, right, upper...) employ different methods of choosing the height of a partition. In case of the upper Riemann sum it's a supremum (the largest value) of $f(x)$ on each interval, which bounds a corresponding partition.
If the limit of a Riemann sum exists as $\max\Delta x_k\rightarrow 0$ , this limit is known as the Riemann integral of $f(x)$ over $[a, b]$.
Example:
\begin{align} \int_a^b (2x - x^2)\,dx & = \lim_{n\rightarrow\infty} \sum_{k = 1}^n \frac{b - a}{n} \left(2\left(a + k \frac{b - a}{n}\right) - \left(a + k \frac{b - a}{n}\right)^2\right) \\ & = \lim_{n\rightarrow\infty} \frac{b - a}{n} \left(2an - a^2 n + 2 \frac{b - a}{n} \sum_{k = 1}^n k - 2a \frac{b - a}{n} \sum_{k = 1}^n k - \left(\frac{b - a}{n}\right)^2 \sum_{k = 1}^n k^2\right) \\ & = \lim_{n\rightarrow\infty} (b - a) \left(2a - a^2 + \frac{(b - a)(n + 1)(2 - 2a)}{2n} - \frac{(b - a)^2(n + 1)(2n + 1)}{6n^2}\right) \\ & = (b - a)\left(2a - a^2 + \frac{2(b - a)(1 - a)}{2} - \frac{(b - a)^2}{3}\right) \\ & = \frac{a^3 - b^3}{3} - a^2 + b^2 \end{align}
Important identity:
$$\int_0^1 f(x)\,dx = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k = 1}^n f\left(\frac{k}{n}\right)$$

### Arc length

The length of an arc of a curve $y = f(x)$ from $x = a$ to $x = b$ is:
$\displaystyle L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx$
The length of an arc of a parametric curve $x = x(t), \, y = y(t)$ from $t = \alpha$ to $t = \beta$ is:
$\displaystyle L = \int_\alpha^\beta \sqrt{(x'(t))^2 + (y'(t))^2} \, dt$

### Volume of revolution

The volume of a revolution of a curve $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis is:
$\displaystyle V = \pi \int_a^b (f(x))^2 \, dx$

### Area of revolution

The area of a revolution of a curve $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis is:
$\displaystyle A = 2\pi \int_a^b |f(x)| \sqrt{1 + (f'(x))^2} \, dx$
The area of a revolution of a parametric curve $x = x(t), \, y = y(t)$ from $t = \alpha$ to $t = \beta$ about the $x$-axis is:
$\displaystyle A = 2\pi \int_\alpha^\beta |y(t)| \sqrt{(x'(t))^2 + (y'(t))^2} \, dt$


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# Earth 🜨

• ~13,000 kilometers in diameter
• The Moon is the nearest astronomical neighbour
• The Sun is about 150 million kilometers away
• Axis of rotation is tilted 23.5˚
• This is why the Sun is visible for less time in the winter and more time in the summer (in the northern hemisphere)
• Composed primarily of heavy elements

## Layers

• Separated into layers of material based on different densities

### Mantle

• Largest part of the solid Earth
• Oozing flow (like lava)
• Stretches from base of the crust --> 2900 km deep
• ~84% of Earth's volume

### Core

• 5000 K
• 1.4 million atmospheres
• High pressure
• Diameter of 7000 km
• Larger than the entire planet Mercury
• Outer core is liquid
• 1200 km
• Inner core is probably solid

## Magnetic Field and Magnetosphere

• Magnetic field generated by moving material in the liquid metallic core
• Extends into surrounding space
• Magnetosphere
• The zone within which Earth's magnetic field dominates over the weak interplanetary magnetic field that extends outward from the Sun
• Extends ~60,000 km in the direction of the Sun
• Can reach as far as the orbit of the Moon on the side opposite the Sun

## Atmosphere

### Greenhouse Effect

• Trapping of infrared rays
• Visible light comes through, converted to heat, and then cannot escape

## Coordinate System

• Axis of rotation defines poles
• Equator is halfway between both poles
• East/west defined by Earth's rotation
• East --> direction toward which Earth rotates
• West --> opposite of East

## Calculating Earth's Size in Ancient times

• This means at Alexandria, Earth's surfaced has curved by 7˚ (1/50 of a full circle)
• The distance between the two cities must then be 1/50 the Earth's circumference, so we can measure that

## Orbit

• Distance from the Sun only varies by ~3%

## Seasons

• Caused by the tilt of Earth's axis (23.5˚)
• Depending where the Earth is in its orbit, one of the hemispheres will be more directly illuminated
• Sunlight becomes more/less effective at heating the Earth
• Tilting towards the Sun provides more direct sunlight
• Longer days gives the Sun more time to heat up the Earth
• The Sun will actually spend more/less time above the horizon

## Time

• Solar days are defined in terms of the Earth's rotation with respect to the Sun
• How we tell time
• ~4 minutes longer than a sidereal day ($\frac{1}{365}$ of a day)
• Stars rise 4 minutes earlier every day
• Sidereal days are defined in terms of the Earth's rotation with respect to the stars
• Stars rise at the same sidereal time every day
• Better for observing the stars
• Apparent solar time
• Time determined by the actual position of the Sun
• Time shown on sundials
• Not very convenient for us
• First half of the day, the Sun has not reached the meridian
• Before the Sun reaches the local meridian
• ante meridiem or a.m.
• Second half of the day, after noon
• post meridiem or p.m.
• After the Sun reaches the local meridian
• Mean solar time
• Average value of the solar day over the year
• Exactly 24 hours in a day
• Progresses at a uniform rate
• Standard time
• 24 standardized global time zones
• Each zone keep the same standard time with the local mean solar time running through the middle of each zone
• Travelers change their watches only when time change is >= one hour

## Tidal Force

• Not a real force like gravity
• The Moon pulls on the Earth, making tidal bulges in the oceans
• Water on the side facing the Moon flows toward it
• Water on the side away from the Moon has weak influence so tides are high
• Bulges out on the side facing the Moon as well as the side facing away from it
• This causes two high/low tides per day
• Once when away from the Moon, and once when facing it


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# weight(RootSystem,Vector) -- construct a weight from a vector

## Synopsis

• Function: weight
• Usage:
weight(R,v)
• Inputs:
• R, an instance of the type RootSystem,
• v, , the coordinates of the weight
• Outputs:
• an instance of the type Weight, a weight vector in a the weight lattice of R

## Description

 i1 : R=rootSystemB(3) o1 = RootSystem{...8...} o1 : RootSystem i2 : weight(R,vector {1,2,1}) o2 = | 1 | | 2 | | 1 | o2 : Weight


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Pattern in Fermat Factorization

I have the Fermat Factorizations of $n = pq$ where $p$ and $q$ are primes. I am trying to find a formula/pattern for the number of cycles required to perform the factorization in terms of $n, p, q$. Here is a set of integers I have interated over:

FermatFactorization(15) (5)(3)
NumCycles: 1
FermatFactorization(21) (7)(3)
NumCycles: 2
FermatFactorization(33) (11)(3)
NumCycles: 5
FermatFactorization(35) (7)(5)
NumCycles: 1
FermatFactorization(39) (13)(3)
NumCycles: 6
FermatFactorization(51) (17)(3)
NumCycles: 9
FermatFactorization(55) (11)(5)
NumCycles: 3
FermatFactorization(57) (19)(3)
NumCycles: 11
FermatFactorization(65) (13)(5)
NumCycles: 4
FermatFactorization(69) (23)(3)
NumCycles: 14
FermatFactorization(77) (11)(7)
NumCycles: 2
FermatFactorization(85) (17)(5)
NumCycles: 7
FermatFactorization(87) (29)(3)

Can you find the formula for the given data as described above?

-

The first question to ask is how is the value of $X = s + a$ ( half the sum of the factors) related to $p\text{ and }q$, the factors of $N$. $2X = 2s + 2a$ is the sum of the factors of $N$; how can we relate this to $p\text{ and }q$?

We define p and q as follows: \begin{align} s &= \lfloor\sqrt{N}\rfloor\\ q &= s - t &\text{(t is an integer)}\\ p &= s + t + k &\text{( k is an integer)} \end{align} I have used $p$ as the larger number as it was used like that in the answer above.

The sum of the factors $= q + p = ( s - t) + ( s + t + k) = 2s + k$.

We also know that the sum of the factors is $2s + 2a$ therefore we have $2s + 2a = 2s + k$, and $a = k/2$.

From this we can use the defined $p$ and $q$ as follows: \begin{align} N &= qp = (s- t)(s + t + k)\\ N &= s^2 - t^2 + sk - tk\\ \end{align} Solving for $k$ gives $k = (t^2 + N - s^2)/(s - t)$

This will give you a formula for $2a$, so half the value will be the values that you have calculated above.

There is no formula in terms of $N,p\text{ and }q$ though!

-
I've taken the liberty of formatting your mathematics. To see what I've done, click on the "edited" text above my name and then select "source". For a tutorial on how to do this yourself, see here. – Rick Decker Nov 15 '12 at 2:26

I'm going to assume that what you are doing to factor $n$ is you are adding odd numbers until you get a square, e.g., to factor 65, you go $$65+1=66;\quad 66+3=69;\quad 69+5=74;\quad 74+7=81=9^2$$ Having done 4 cycles, you have $$65+4^2=9^2;\quad 65=9^2-4^2=(9+4)(9-4)=(13)(5)$$ In general, and assuming $p\gt q$, we have $$n=pq=\Bigl({p+q\over2}\Bigr)^2-\Bigl({p-q\over2}\Bigr)^2$$ so the number of cycles is $(p-q)/2$.

If this is the correct decryption of your question, then there are lots of errors in your table.

-

There are three cases for running time. 1st case: let $q-p = 2n$ where $n$ is odd then the running time $t \approx (((n(n - 4) + 7)/4 - p) + 2)$

2nd case let $q - p = 4n$ where $n$ is odd then the running time $t \approx ((n(n - 2) + 2) - p) + 2)$

3rd case let $q - p = 4n$ where $n$ is even then the running time $t \approx ((n(n - 2) + 1) - p) + 2)$ Fore more reference http://kadinumberprops.blogspot.in/2012/11/fermats-factorization-running-time.html

-


text:
Fifth, it is essential to keep a clear head while making any decisions about binary options trading. Beginners should avoid trading when they are emotionally disturbed, as emotions can create havoc with your trading. It is very easy to make the wrong decisions in an emotional state, so it is advisable to stay away from any form of trading until you’ve calmed down and cleared your mind.
In the U.K. corporate profits are taxed at 24% and this rate is even higher in the U.S., where companies can be taxed as high as 35%. However, taxes are only 10% in Cyprus. In light of this, it makes complete sense that many brokers choose to base their businesses in Cyprus. Many of you have probably noted that most binary options brokers refuse to accept clients from the American market as they do not wish to be taxed by the U.S. authorities.

Many binary options and Forex brokers have enticing trading platforms in addition to a horde of fabulous attractions for their traders. With this in mind, many traders are wondering whether it is okay to just invest in any binary options or Forex broker. The truth is that not all brokers in the market today are reputable, just like in the online poker market. In the U.S.A, for instance, there are strict regulations that have been imposed for the brokers in the region to be in a position to accept US-based traders.
In March 2016 binary options trading within Israel was banned by the Israel Securities Authority, on the grounds that such trading is essentially gambling and not a form of investment management. The ban was extended to overseas clients as well in October 2017.[16][48][4] It was approved by the Knesset in October, despite strong opposition from the binary options industry.[17]
Some brokers do not offer truly helpful trading tools such as charts and features for technical analysis to their clients. Experienced traders can get around this by sourcing for these tools elsewhere; inexperienced traders who are new to the market are not as fortunate. This is changing for the better though, as operators mature and become aware of the need for these tools to attract traders.
##### Terms and Conditions. When taking a bonus or offer, read the full terms and conditions. Some will include locking in an initial deposit (in addition to the bonus funds) until a high volume of trades have been made. The first deposit is the trader’s cash – legitimate brokers would not claim it as theirs before any trading. Some brokers also offer the option of cancelling a bonus if it does not fit the needs of the trader.

For Nadex binary options you have an extra step because you can purchase an option at any price between 0 and 100, which affects how much you could lose. Assume you have a $5500 account and are willing to risk 2% per trade. That means you can lose up to$110 per trade and still be within your risk parameter. Don't take a trade where you could lose more than $110. Many experts recommend getting into binary options trading here while the conditions are still favorable and there is a big chance to profit on a larger scale. Secondly, there is an opportunity to trade during the weekends. While all other are open Monday through Friday, The Dubai Security Exchange is working Sunday through Thursday. This is a lot more convenient for those of us who prefer to trade in our free time. In March 2016 binary options trading within Israel was banned by the Israel Securities Authority, on the grounds that such trading is essentially gambling and not a form of investment management. The ban was extended to overseas clients as well in October 2017.[16][48][4] It was approved by the Knesset in October, despite strong opposition from the binary options industry.[17] Also called the Up/Down binary trade, the essence is to predict if the market price of the asset will end up higher or lower than the strike price (the selected target price) before the expiration. If the trader expects the price to go up (the “Up” or “High” trade), he purchases a call option. If he expects the price to head downwards (“Low” or “Down”), he purchases a put option. Expiry times can be as low as 5 minutes. Risk and reward are known in advance, offering a major advantage. There are only two outcomes: win a fixed amount or lose a fixed amount, and there are generally no commissions or fees. They're simple to use and there's only one decision to make: is the underlying asset going up or down? In addition, there are also no liquidity concerns because the trader doesn't own the underlying asset and brokers can offer innumerable strike prices and expiration times/dates, which is an attractive feature. The trader can also access multiple asset classes anytime a market is open somewhere in the world. Those people whom you see are “making money” are either the brokers themselves trying to post some youtube videos to convince people that their platform is designed to help people “earn money” or a group of misguided affiliate marketerswho still have some balance in their account are trying to get people to participate and to reap off some commission from their referrals who had completely blow their account.  As well, the withdrawal process with this platform is both fast and secure. Clients have the ability to request to withdraw their funds at any time. The withdrawal request is then process by OFM’s Compliance Department and may take up to 5 business days to process. There is an additional 2 to 3 days for those funds that are to be tendered to the bank account or credit card that applies for this process. The minimum amount for withdrawal is$100.
You are taking a risk when trading binary options. If it was a sure fire thing that we would win every time we trade then everyone would be doing it and everyone would be winning. The brokers are there to make money just as you are and on every trade someone always loses. The trick is to minimise the risk of it being you and that any money invested is not going to hurt you if you lose.
With the exception of High/low trades, they offer two additional means of trading: including One-Touch option, which lets the trader decide if the underlying asset will hit a certain value before the expiry time, which is most often available on weekends. Secondly, they offer a Boundary Option with 75% return rate via winning trade. Investors should have in mind that can lose all their capital and that trading alerts should not be considered as investment advice. The decision to act on any signals is yours and taken at your own risk.
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In the standard Black–Scholes model, one can interpret the premium of the binary option in the risk-neutral world as the expected value = probability of being in-the-money * unit, discounted to the present value. The Black–Scholes model relies on symmetry of distribution and ignores the skewness of the distribution of the asset. Market makers adjust for such skewness by, instead of using a single standard deviation for the underlying asset {\displaystyle \sigma } across all strikes, incorporating a variable one {\displaystyle \sigma (K)} where volatility depends on strike price, thus incorporating the volatility skew into account. The skew matters because it affects the binary considerably more than the regular options.


text:
# chainerx.huber_loss¶

chainerx.huber_loss()

Element-wise Huber loss.

The Huber loss is similar to the squared error but is less sensitive to outliers in the data. It is defined as

$\begin{split}L_{\delta}(a) = \left \{ \begin{array}{cc} \frac{1}{2} a^2 & {\rm if~|a| \leq \delta} \\ \delta (|a| - \frac{1}{2} \delta) & {\rm otherwise,} \end{array} \right.\end{split}$

where $$a = x - t$$ is the difference between the input $$x$$ and the target $$t$$.

Parameters
• x (ndarray) – Input variable.

• t (ndarray) – Target variable for regression.

• delta (float) – Constant variable for Huber loss function as used in definition.

Returns

A variable object holding an array representing the Huber loss $$L_{\delta}$$ of the two inputs.

Return type

ndarray


text:
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Simplify each of the following...

# Simplify each of the following and express the result as a rational number in standard form: (-19)/(36)xx 16 (ii) (-13)/9xx(27)/(-26)

Updated On: 27-06-2022

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Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

11 today question is simplify each of the following and Express as Express the result is a rational number in standard form in first part Vinay to simplify -19 upon 36 X 16in second part -13.9 X 27 - 26 so let's solve this question taking first part in first we need to simplify -19 upon 36 X 16 which can be done as 16 by one so making it more simple by cancellation method 16 and 36 will be cancelled 4 times a section will be cancelled for X and 3936 will be cancelled 9 time from 4 now it is not further we cannot make it more simple and more simplified so just multiplying numerator parts 19 x 4 19 multiplayer for will be

76 and minus will be there and 9 X 1 is 9 only soon -76.9 is the answer for the first part moving on to the second part in second part we need to simplify -13.9 X 27 upon -26 so making it more simple by cancellation method so 2626 will be cancelled - 2 X from 13 - 13 so it will be -2 and -1 and 27 will be cancelled 3 x 9 so now it is not we cannot simply for that I cancellations to moving to the next tab 2 - 1 x 3 will be -3 and denominator 1 x minus phase -2 so answer will be 3

to Fir the second part in the standard form I hope you understood the solution thank you

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text:
# Annuity (finance theory): Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Encyclopedia

The term annuity is used in finance theory to refer to any terminating stream of fixed payments over a specified period of time. This usage is most commonly seen in discussions of finance, usually in connection with the valuation of the stream of payments, taking into account time value of money concepts such as interest rate and future value.[1]

Examples of annuities are regular deposits to a savings account, monthly home mortgage payments and monthly insurance payments. Annuities are classified by payment dates. The payments (deposits) may be made weekly, monthly, quarterly, yearly, or at any other interval of time.

## Ordinary annuity

An ordinary annuity (also referred as annuity-immediate) is an annuity whose payments are made at the end of each period (e.g. a month, a year). The values of an ordinary annuity can be calculated through the following:[2]

Let:

r = the yearly nominal interest rate.
t = the number of years.
m = the number of periods per year.
i = the interest rate per period.
n = the number of periods.

Note:

$i = \frac{r}{m}$
n = tm

Also let:

P = the principal (or present value).
S = the future value of an annuity.
R = the periodic payment in an annuity (the amortized payment).
$S \,=\,R\left[\frac{\left(1+i\right)^n-1}{i}\right] \,=\,R\cdot s_{\overline{n}|i}$ (annuity notation)

Also:

$P \,=\,R\left[\frac{1-\frac{1}{\left(1+i\right)^n}}{i}\right] = R\cdot a_{\overline{n}|i}$

Clearly, in the limit as n increases,

$\lim_{n\,\rightarrow\,\infty}\,P\,=\,\frac{R}{i}$

Thus, even an infinite series of finite payments (perpetuity) with a non-zero discount rate has a finite present value.

### Proof

The next payment is to be paid in one period. Thus, the present value is computed to be:

$P = \frac{R}{1+i} + \frac{R}{(1+i)^2} + \cdots + \frac{R}{(1+i)^n} = \frac{R}{1+i} \left( 1 + \frac{1}{1+i} + \frac{1}{(1+i)^2} + \cdots + \frac{1}{(1+i)^{n-1}}\right)$.

We notice that the second term is a geometric progression of scale factor 1 and of common ratio $\frac{1}{1+i}$. We can write

$P = \frac{R}{1+i} \cdot \frac{1 - \frac{1}{(1+i)^n}}{1-\frac{1}{1+i}}$.

Finally, after simplifications, we obtain

$P = \frac{R}{i} \left(1 - \frac{1}{(1+i)^n} \right) = \frac{Rm}{r} \left(1 - \frac{1}{(1+\frac{r}{m})^{(tm)}} \right)$.

Similarly, we can prove the formula for the future value. The payment made at the end of the last year would accumulate no interest and the payment made at the end of the first year would accumulate interest for a total of (n−1) years. Therefore,

$S = R + R(1+i) + R(1+i)^2 + \cdots + R(1+i)^{n-1} = R \left(1 + (1+i) + (1+i)^2 + \cdots + (1+i)^{n-1}\right)$.

Hence:

$S = R \frac{(1+i)^n-1}{i}$.

If an annuity is for repaying a debt P with interest, the amount owed after n payments is:

$\frac{R}{i}- \left( 1+i \right) ^n \left( \frac{R}{i} - P \right)$

because the scheme is equivalent with lending an amount $\frac{R}{i}$ and putting part of that, an amount $\frac{R}{i}-P$, in the bank to grow due to interest. See also fixed rate mortgage.

## Annuity-due

An annuity-due is an annuity whose payments are made at the beginning of each period.[3] Deposits in savings, rent payments, and insurance premiums are examples of annuities due.

Because each annuity payment is allowed to compound for one extra period, the value of an annuity-due is equal to the value of the corresponding ordinary annuity multiplied by (1+i). Thus, the future value of an annuity-due can be calculated through the formula (variables named as above):[4]

$S \, = \, R \left[ { (1+i)^{n+1} - 1 \over i } \right] - R\,=\,R\cdot \ddot{s}_{\overline{n|}i}$ (annuity notation)

It can also be written as

$S \,=\,R\left[\frac{\left(1+i\right)^n-1}{i}\right](1 + i ) \,=\,R\cdot s_{\overline{n}|r}$(1 + i)

An annuity-due with n payments is the sum of one annuity payment now and an ordinary annuity with one payment less, and also equal, with a time shift, to an ordinary annuity with one payment more, minus the last payment.

Thus we have:

$\ddot{a}_{\overline{n|}i}=a_{\overline{n}|i}(1 + i)=a_{\overline{n-1|}i}+1$ (value at the time of the first of n payments of 1)
$\ddot{s}_{\overline{n|}i}=s_{\overline{n}|i}(1 + i)=s_{\overline{n+1|}i}-1$ (value one period after the time of the last of n payments of 1)

## Other types

• Fixed annuities – These are annuities with fixed payments. They are primarily used for low risk investments like government securities or corporate bonds. Fixed annuities offer a fixed rate but are not regulated by the Securities and Exchange Commission.
• Variable annuities – Unlike fixed annuities, these are regulated by the SEC. They allow you to invest in portions of money markets.

Annuity due is useful for lease payment calculations 1

## References

1. ^
2. ^ Finite Mathematics, Eighth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey. Published by Addison Wesley. ISBN 032122826X
3. ^ "Future Value of an Annuity Due". Retrieved 2008-07-10.
4. ^ ibid Lial.


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# If a = (0.404)^3 ,

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If a = (0.404)^3 , [#permalink]  21 Jun 2017, 02:34
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If $$a = (0.404)^3$$ , b = cube root of 0.404, and c = 0.404, then which of the following is true?

A) a > b > c

B) b > a > c

C) a > c > b

D) c > a > b

E) b > c > a
[Reveal] Spoiler: OA

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Re: If a = (0.404)^3 , [#permalink]  12 Oct 2017, 00:32
Since 0.404 is a number between 0 and 1, when it is elevated to a power greater than 1 the number becomes smaller, while when it is elevated to an improper fraction, i.e. a number between 0 and 1, the number becomes greater.

Given those rules, $$0.404^{\frac{1}{3}}>0.404>0.404^3$$

Re: If a = (0.404)^3 ,   [#permalink] 12 Oct 2017, 00:32
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# Natbib hook for italicised captions

Identical to Citation within caption => avoid italics (biblatex) just for natbib: Captions are italicised, how can I enforce \cite{ref} in a caption to behave like \textup{\cite{ref}} (other than replacing every such occurrence of the former with the latter)?

Is there a similar hook like \AtEveryCite for natbib (I would prefer not to hack the \cite macro, not sure why, happy to hear why this is an okay solution too)?

Adding \upshape to the \cite command:

\documentclass{article}

\usepackage{natbib}

\makeatletter
\DeclareRobustCommand\cite
{\begingroup\upshape\let\NAT@ctype\z@\NAT@partrue\NAT@swatrue
\@ifstar{\NAT@fulltrue\NAT@cites}{\NAT@fullfalse\NAT@cites}}
\makeatother

\begin{document}

\itshape

normal text \cite{knuth:ct:c} normal text

\bibliographystyle{plainnat}
\bibliography{biblatex-examples}

\end{document}