text:
Astudent writes a term paper about similarities and differences between greek democracy during the classical

Question:

Astudent writes a term paper about similarities and differences between
greek democracy during the classical era and u. s. democracy during the
modern era. her paper focuses on which major historical theme?

What is the network formed by a river and all the smaller streams that contributes runoff to it?

What is the network formed by a river and all the smaller streams that contributes runoff to it?...

Let f(x) = x + 1 and g(y) = y^2Evaluate g( f ( g (z + 2)) − 4). Show your work.

Let f(x) = x + 1 and g(y) = y^2 Evaluate g( f ( g (z + 2)) − 4). Show your work....

Rational economic agents would resist consuming any good where A. the marginal utility of that good was negative since this would

Rational economic agents would resist consuming any good where A. the marginal utility of that good was negative since this would make you worse off. B. the marginal utility of that good was positive since this would make you worse off. C. the total utility of that good was falling since this would ...

Cost-benefit analysis is a process that involves which of the following? Question 14 options: calculating a budget for each expense

Cost-benefit analysis is a process that involves which of the following? Question 14 options: calculating a budget for each expense maximizing benefits and minimizing costs accurately predicting the effects of every decision determining the profit that results from production decisions...

Feudalism was created by the franks true or false

Feudalism was created by the franks true or false...

Q2. (a) a particle of mass 3 moves under a force f = 3e^2t i + 3t j − 9 k the particle is initially at the origin with

Q2. (a) a particle of mass 3 moves under a force f = 3e^2t i + 3t j − 9 k the particle is initially at the origin with velocity v(0) = 2i - k (i) find its position for all t (ii) what is the angle between the paths of the two balls after the collision? (b) a snooker cue ball of mass m travels wit...

Pontificating is speaking with claims of importancespeaking in a pompous waydrawing incorrect conclusionsdrawing

Pontificating is speaking with claims of importancespeaking in a pompous waydrawing incorrect conclusionsdrawing conclusions with insufficient information...

The stadium can seat 15,000 fans. if the stadium is 85 percent full, how many tickets were sold?

The stadium can seat 15,000 fans. if the stadium is 85 percent full, how many tickets were sold?...

If two triangles have equal perimeters, the. they could also have which of the following? equal volumes equal areas equal bases

If two triangles have equal perimeters, the. they could also have which of the following? equal volumes equal areas equal bases equal heights...

Who was the african american that created the casket lowering device?

Who was the african american that created the casket lowering device?...

Solve and get brainliest!!please be sure of your answer!! $Solve and get brainliest$...

format: plain


text:
# # Nushell 0.11.0

Nushell, or Nu for short, is a new shell that takes a modern, structured approach to your commandline. It works seamlessly with the data from your filesystem, operating system, and a growing number of file formats to make it easy to build powerful commandline pipelines.

We're excited to release the 0.11.0 release of Nu. This release continues to push us forward with new commands, improved engine internals, bugfixes, and more.

# # Where to get it

Nu 0.11.0 is available as pre-built binaries (opens new window) or from crates.io (opens new window). If you have Rust installed you can install it using cargo install nu.

If you want more goodies, you can install cargo install nu --features=stable.

As part of this release, we also publish a set of plugins you can install and use with Nu. To install, use cargo install nu_plugin_<plugin name>.

# # What's new

## # Compound shorthand flags (Amanita-muscaria)

Over the last few releases, we've seen the ability for flags to have shorthand version. You no longer needed to use --full if the flag supported the shorter -f version.

In this release, we've taken it one step further and allowed you to combine multiple shorthand flags into one group. For example, ls --full --short-names can now be written as ls -fs

## # Improvements and cleanup for ls (Bocom, quebin31, lincis, equal-l2)

A lot of improvements went into ls for this release. The updated ls command is now more forgiving of coming across directories the user can't access and will simply show what it can and continue, it can work with named pipes on systems with them, it shows a more reasonable display for zero-length files, and the internals of ls have been cleaned a bit to make way for future improvements.

## # External command output no longer blocks (thegedge)

Continuing our series of improvements to working with external commands, we have a few updates in this release. The first of these updates is an improvement to how content streams out of externals. With 0.11.0, this content will no longer block, which allows content to flow more easily between external and internal commands. As a result, some corner cases that came up with a series of commands in a pipeline not working because one blocks no longer is an issue.

## # Allow externals to output texts or binary (jonathandturner, thegedge)

The second update to externals now allows externals to freely output either text or strings, and for this content to be read in a block at a time into Nu. This allows externals to stream any content they wish, and for this stream to continue for as long as needed.

## # New shuffle command (hirschenberger)

With 0.11.0, we introduce a new shuffle command that will let you shuffle the rows of a table. This allows you to choose to not always process data the same way, or to search through data more stochastically.

❯ help commands | shuffle
─────┬──────────────┬───────────────────────────────────────────────────────────────────────────────────────────────
#   │ name         │ description
─────┼──────────────┼───────────────────────────────────────────────────────────────────────────────────────────────
0 │ textview     │ Autoview of text data.
1 │ pwd          │ Output the current working directory.
2 │ sort-by      │ Sort by the given columns.
3 │ tree         │ View the contents of the pipeline as a tree.
4 │ du           │ Find disk usage sizes of specified items
5 │ table        │ View the contents of the pipeline as a table.

## # New column rename command (andrasio)

You can also now use the new rename command to rename a column to a new name.

❯ help commands | rename action description
─────┬──────────────┬───────────────────────────────────────────────────────────────────────────────────────────────
#   │ action       │ description
─────┼──────────────┼───────────────────────────────────────────────────────────────────────────────────────────────
0 │ append       │ Append the given row to the table
1 │ autoview     │ View the contents of the pipeline as a table or list.
2 │ average      │ Compute the average of a column of numerical values.
3 │ binaryview   │ Autoview of binary data.

## # Merge the env command into the $nu variable (jonathandturner) The $nu variable has grown to include values that were previously part of env, effectively merging the two into the $nu variable. This allows you to easily find out where the history file is stored, where config is stored, and where a good temp directory are all in one place: ❯ echo$nu.home-dir $nu.temp-dir$nu.config-path \$nu.history-path
───┬──────────────────────────────────────────────────────────────────────
# │ <value>
───┼──────────────────────────────────────────────────────────────────────
0 │ C:\Users\joturner
1 │ C:\Users\joturner\AppData\Local\Temp\
2 │ C:\Users\joturner\AppData\Roaming\nu shell developers\nu\config.toml
3 │ C:\Users\joturner\AppData\Local\nu shell developers\nu\history.txt
───┴──────────────────────────────────────────────────────────────────────

## # The format command can now use column paths (andrasio)

The format command is a handy way of putting column data into a string for displaying, but it previously had the restriction that you couldn't use it to print out information that was deeper in the table. With 0.11.0, we've removed that restriction.

❯ open Cargo.toml | format "The edition is {package.edition}"
The edition is 2018

## # Allow help to be printed with -h in more cases (rabisg0)

You should now be able to use -h to reach the help of a command in more cases than was previously allowed, making it a universal way to get to the help text.

## # Improved documentation (coolshaurya, Amanita-muscaria, notryanb)

Documentation for du, uniq, debug, and plugin docs have all been improved in this release.


text:
### Multiplication Magic

Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). The question asks you to explain the trick.

### N000ughty Thoughts

How many noughts are at the end of these giant numbers?

### DOTS Division

Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.

# Expenses

##### Age 14 to 16Challenge Level

What is the largest number which, when divided into each of $1905$, $2587$, $3951$, $7020$ and $8725$, leaves the same remainder each time?


text:
# Changes between Version 8 and Version 9 of WRF4GGridConfiguration

Ignore:
Timestamp:
Oct 24, 2011 11:31:10 AM (11 years ago)
Comment:

--

### Legend:

Unmodified
 v8 '''WRF4G Grid Resources''' currently are defined by [[http://www.gridway.org/doku.php?id=documentation:release_5.8:cg |GridWay Drivers]], therefore if you want to access to Grid Resources, you must configure '''gwd.conf'''. This file is located under $GW_LOCATION/etc. == WRF4G !GridWay Drivers Available == The following drivers are included in WRF4G: * Information Driver: * BDII (gw_im_mad_mds2_glue-bdii) * Transfer Driver: * Dummy Transfer driver (gw_tm_mad_dummy) * Execution Drivers: * GRAM2 (gw_em_mad_gram2) * CREAM (gw_em_mad_cream) == Required Middleware == * gLite UI 3.2 (CREAM-based) == GridWay Drivers == The following drivers are included in WRF4G: * Information Driver: * BDII (gw_im_mad_mds2_glue-bdii) * Transfer Driver: * Dummy Transfer driver (gw_tm_mad_dummy) * Execution Drivers: * GRAM2 (gw_em_mad_gram2) * CREAM (gw_em_mad_cream) == Configuration options Divers == {{{ WRF4G_BASEPATH="gsiftp://$WRF4G_LOCATION/repository/output"


text:
The Raspberry Pi 3 has been released for sale, what's inside it, and what can you do with it?

Four years since the Raspberry Pi was first announced the Raspberry Pi foundation has released its latest version of its single board computer, the Raspberry Pi 3 and it seems like everyone is pretty excited about this announcement. The latest incarnation of the Raspberry Pi is packed with a bunch of new features, and for the same price compared to last years model.

### New Features

The major notable improvements for the new Raspberry Pi 3 include:

• A faster processor for better performance
• Built-in WiFi
• Built-in Bluetooth

### Performance

Benchmarks have already surfaced, showing that the new raspberry Pi, with the 1.2ghz processor, is about 50% faster than the Pi 2. You can find the specs and benchmarks for the Raspberry Pi 3 on their website.

The Pi 2 ran a Broadcom BCM2836 quad-core processor clocked at 900MHz. The latest Pi runs a Broadcom BCM2835 quad core processor clocked at 1.2 GHz. In additional to a faster clock speed the BCM2835 is based around a Newer ARM core with the Cortex A53 architecture, where the BCM2835 is based on the ARM cortex A7 core.

### Wireless

By far the biggest change to the new Pi is the included wireless connectivity. The days of needing a WiFi and Bluetooth dongle are over! The new Pi uses a BMC43438 wireless radio for both WiFi and Bluetooth connectivity. The radio is connected to a small surface mount ceramic antenna.

### Criticisms

Whenever a new version of a product, regardless of what it is, there will be some criticisms about it. Some of the potential criticisms that are facing the Raspberry PI 3 include

• Limited Ram.  The new Pi still has 1GB of ram, just like last years model.
• Single band wifi.  The new Pi only contains a 2.4GHz radio, so 5Ghz wifi won't be an option without a dongle.

### Moving Onward

With these new features on the Raspberry Pi, it now let's hobbyist easily connect the Pi to wireless networks without an aftermarket accessory. Now IOT projects not only will be easier to implement, but they will also be a fair bit cheaper. In addition to the benefits of the wireless connectivity, the extra power of the ARM processor will lend itself to more demanding projects, such as projects involving computer vision or impair processing.

At the end of the day, the Raspberry Pi 3 is a good deal for around $35. With this new and more powerful Pi, more people will use this as a desktop computer replacement, and hopefully, this will get more students interested in engineering. #### Learn More About: ### Comments #### 6 CommentsLogin • dimtass 2016-03-11 Is there anything else in market with better specs at the same price (or even a little bit higher) and the same availability (and also future proof regarding availability)? If not, then it worths it. • fstanley 2016-03-11 Coming to the market is a new board: Pine 64. It is an Android based O/S SBC for$15.  Like the Pi it has headers and “stuff” that will make it great platform for tinkering, school computer labs and embedded systems.  Unlike the Pi, the choice of O/S will initially limit the amount of Open Software.  Check it out on http://www.Pine64.com

• fstanley 2016-03-11

I forgot to add that you can get a very scaled-down version of the Pi for 5 dollars: RPi-Zero.  A single core processor that is about the size of a compact fingernail file.  Also, the quantity of add-ons from third party vendors make the RPi a good first choice for anyone.  I have four of the Pi’s.  I am using one to teach high school freshmen about digital circuits and Python.  Scratch is also available for the Pi and is included in the Raspian distribution.  Also in the distribution is an office productivity suite, a browser and Mathematica.  All-in-all a pretty good sized package for the $35 (or$5) spent.

• dimtass 2016-03-11

That’s great. Thanks. This is something that needs follow up though, as it seems that it’s currently unavailable for the low price. This is why I said that availability is a big plus, so you can buy it anytime you like and even after 5 or 6 years.
Anyway, as I’ve said before it doesn’t matter if another platform is $5 or$10 more expensive as long as it worths it. Pine seems promising but as I’ve seen not mature enough yet. But it will be.

• coopecb1 2016-03-11

ODROID-C2 may be competitor: 64-bit quad core at 2GHz. http://www.hardkernel.com/main/products/prdt_info.php?g_code=G145457216438 .

• gnagy 2016-03-19

The tiny RAM kills it for me.


text:
# Unix for Oracle DBAs Pocket Reference

Author: Don Burleson
Publisher: O'Reilly Media

Year: 2001
Pages: 110
Language: English
Format: PDF (very good quality), 1 Mb
ISBN:

The Unix for Oracle DBAs Pocket Reference puts within easy reach the commands that Oracle database administrators need most when operating in a Unix environment. If you are an Oracle DBA moving to Unix from another environment such as Windows NT or IBM Mainframe, you know that these commands are far different from those covered in most beginning Unix books. To jump start your learning process, Don Burleson has gathered together in this succinct book the Unix commands he most often uses when managing Oracle databases. You'll be able to reach into your pocket for the answer when you need to know how to:

• Display all Unix components related to Oracle, identify the top CPU consumers on your server, and even kill processes when necessary
• Stack Unix commands into powerful scripts that can perform vital DBA functions
• Monitor Unix filesystems, and automatically manage your trace files, dump files, and archived redo log files
• Use essential server monitoring commands such as top, sar, and vmstat


text:
# Hide_Plus Crack Free (April-2022)

While you can minimize almost any window to the Taskbar, and sometimes send them to the system tray, it is not normally possible to hide them from your desktop completely and keep them running only in the background.
Hide_Plus is a lightweight application that can come to your aid. It enables you to hide any program window, which will continue to run normally in the background, and then reveal them again whenever they are needed.
Unobtrusive program that can be deployed effortlessly
As far as installing and configuring the software goes, things could hardly be simpler. The setup process is remarkably straightforward, and there are no settings to tinker with once you have installed the application.
You can find Hide_Plus in the system tray, and you just need to right-click the icon whenever you wish to hide or reveal a window. To make things easier in the future, you can even have it launch with the OS automatically.
Every active window will have an entry in the system tray menu, and the hidden ones will be separated from the rest to make it easier to find them. Once a window has been hidden, you will only be able to access it again by revealing it from the tray menu.
Working with Hide_Plus is very simple, as you can hide or reveal windows with just a couple of mouse clicks. You can use the Shift+Ctrl key combination to perform this action, but it would have been great if this hotkey was customizable.
Handy utility for running applications in the background
All in all, Hide_Plus does its job well, enabling you to make any application window invisible while letting it run in the background. It lacks a classic GUI, as it is controlled entirely from a system tray menu, and it is very easy to use.

## Hide_Plus Crack + Free

Hide_Plus is an unobtrusive utility that enables you to hide any window that comes to your aid. Hide_Plus is easy to use and superfast. You can drag windows to the Hide_Plus tray icon to automatically hide them. You can also select which windows you want to hide and uninstall Hide_Plus at any time.
Hide_Plus System Requirements:
– Windows 2000/XP/Vista
– 2 GHz processor
– 2.6 GB of free hard disk space
– 1024×768 screen resolution
Hide_Plus is absolutely free. you can get it from our website

Hide_Plus is an unobtrusive utility that enables you to hide any window that comes to your aid. Hide_Plus is easy to use and superfast. You can drag windows to the Hide_Plus tray icon to automatically hide them. You can also select which windows you want to hide and uninstall Hide_Plus at any time.
Hide_Plus Description:
Hide_Plus is an unobtrusive utility that enables you to hide any window that comes to your aid. Hide_Plus is easy to use and superfast. You can drag windows to the Hide_Plus tray icon to automatically hide them. You can also select which windows you want to hide and uninstall Hide_Plus at any time.
Hide_Plus System Requirements:
– Windows 2000/XP/Vista
– 2 GHz processor
– 2.6 GB of free hard disk space
– 1024×768 screen resolution
Hide_Plus is absolutely free. you can get it from our website are here

International Bank Drafts

What are International Bank Drafts and how are they sent?

International Bank Drafts are used by the international trading community for many purposes. They are a paperless method of payment and are used as an advance against goods or services. The recipient gets the money and forwards the draft for collection from the corresponding bank.

The major purpose for using International Bank Drafts is to promote trade and as such the international banking community are quick to accept them.

International Bank Drafts are accepted by all major international banks in any currency, including the Euro, the Pound, the Dollar, the Swiss Franc, or any other type of currency. Therefore, anyone can use them to pay for goods, services or debts,

## Hide_Plus License Key Free X64 (2022)

Hide_Plus is a lightweight and unobtrusive utility that enables you to run any program window in the background, so you won’t have to click on the taskbar to access it.
Hide windows from the desktop and taskbar
Show windows as they are launched
Show recently opened windows
Hide all windows without using the taskbar
Active windows only
Multiple instances of the application can be run at the same time
Unobtrusive, easy to use
Minimize all to tray
Uninstall

Proving $\log\frac{x-y}{x+y}y>0$

I was studying logarithm identity:
$$\log(ab) = \log(a) + \log(b)$$
And I came up with this question. How can we show that:
$$\log\frac{x-y}{x+y}y>0 A:$$\log\frac{x-y}{x+y} = \log(x-y) – \log(x+y) + \log2\log2 y>0$, then$x-y>0$and$x+y>0\$.

[Informed consent: personal medical history].
The issue of informing patients about their disease, which is not a form of treatment, was first discussed in 1954 in a French article, Le Consentement Respecting a Clinique Discipline. The study aims to examine how this topic is addressed in medical texts and to assess whether this subject is adequately considered. Fourteen medical texts, from the Grégoire’s Encyclopedia of Medicine to National Academy Press’ handbooks, were analysed. The majority of works (11 out of 14)
09e8f5149f

## Hide_Plus [Latest]

This is one of the best, and many would say the most important, Windows utility. It does so much, so well, that there are not many reasons to not install it. It’s just the right size, and there are no settings to tinker with once you’ve installed the application.
It is easy to use. It runs unobtrusively in the background, and it can be launched with the OS through an easily accessible system tray menu.
It works with all major Windows programs, so you don’t have to do anything special to use it. If you want to use it, simply right-click the Hide_Plus icon and then click Hide or Unhide.
Many would say it provides an easy-to-use GUI.
It’s a must-have utility for your Windows toolkit.
Hide_Plus Review:

The Office Suite is the most popular and wide-spread computing application on the planet, and it is available for many different platforms. The main selling point of the suite is that it runs the Microsoft office suite applications and it can run some of the most demanding applications as well.
With Office, there’s more than just the familiar Microsoft Office applications. It is a complete suite with other applications such as Word, Excel, PowerPoint, Access, Outlook, One Note, etc.
The new apps that are added to this version of Office almost make the old versions obsolete. It just felt like a collection of add-ons that just made the suite feel more like it’s primary purpose is to be a unified package for all of the Microsoft office applications.
Before you comment on the new features, let me also point out the major drawbacks with the apps that were already here. The Microsoft word editor has always left a lot to be desired. It is not extremely versatile, and it has problems with tables. There isn’t a professional PDF creator anywhere.
Another mistake was the lack of an Office Styles app for formatting purposes. If you know an app that will create the same formatting and graphics for you, then you are out of luck. The only app that comes close to being able to pull off even simple formatting is Publisher, but it doesn’t integrate with the rest of the suite.
The suite would benefit greatly from a

## What’s New in the Hide_Plus?

Hide_Plus is a useful utility that can be used to hide every window on your desktop. As windows are normally kept on the Taskbar and sometimes even in the system tray, you need to keep some in mind when hiding them. Thus, Hide_Plus enables you to keep all the windows you need visible without hiding them permanently.

How to get Hide_Plus? You can get Hide_Plus from its official website. I`m very grateful if you give me some comments or appreciation.

On the inner side of a barcode/QR code printed on an in-app product, there is a small custom message you can set. You might be able to get unique information about a person or just some text you want to print on your product. This is the number of pixels between the lines that the custom code is printed, and this represents the font size. For more information about the code, please visit this link:

A:

I use this for QRCode (more)
(I use the C# assembly with the C# binary serializer, and a small C#-manipulated memory stream)

Source :

A:

I have the exact same problem and no, i don’t think i found a solution yet. AFAIK this issue is not really known to developers. Here’s what i found out about it.
Q: Is there a way to display a text beside the already printed barcode?
A: The short answer is no.
Here’s the key.jpg

A:

In the “How to make a barcode printer PDF from a PDF file”, there is a way to display text beside the barcode using the beamer class.

Membrane signaling properties of a cation-selective permeability mutant of Torpedo electroplax.
The mutant TsJIM5, a Torpedo marmorata expressed P-type ionic channel with relatively low KCa, was used to examine the intrace

## System Requirements For Hide_Plus:

Minimum:
OS: Windows 7 / 8 / 10 (64-bit) Processor: 1.6 GHz or faster RAM: 1 GB Video Memory: 256 MB DirectX: Version 9.0
Recommended:
OS: Windows 7 / 8 / 10 (64-bit) Processor: 2 GHz or faster RAM: 2 GB Video Memory: 1 GB DirectX: Version 9.0
DirectX: Version 9.0Q:
How to install a large number of modules?
In my server, it requires a lot of modules such as


text:
# Tag Info

3

Let's count the number of permutations with a cycle of length $\ge n/2$. It can only have one such cycle, for obvious reasons. First, pick a natural $n/2\le k\le n$. Next, pick a $k$-cycle from $S_n$, and finally the rest of the cycle decomposition can be determined by choosing any permutation on the set of elements in $\{1,\cdots,n\}$ not already present in ...

3

$$\log a_n = \sum_{k=0}^{n-1} \left(\log((3k+1)!) - \log((n+k)!\right)$$ Next use Stirling's approximation. Note e.g. that $$\sum_{k=1}^n k \log(k) \approx \int_1^n k \log(k)\; dk \approx \dfrac{n^2 \log(n)}{2} - \dfrac{n^2}{4}$$ EDIT: Ah, the Robbins numbers: OEIS sequence A005130. It seems (R.W. Gosper's approximation) $$a_n \sim ... 3 We can write$$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$which leads to your$$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$but we can also write$$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$and that gives us$$\begin{align} \sum_{n\leqslant x} \tau_{r+1}(n) &= ...

2

Since $f$ is real valued, we have, for $0 < t \leqslant \tau$, $$0 < \tau^2 \leqslant \tau^2 + f(\tau)^2,$$ and therefore $$0 < \int_t^\infty \frac{d\tau}{\tau^2 + f(\tau)^2} \leqslant \int_t^\infty \frac{d\tau}{\tau^2} = \frac{1}{t}.$$

2

$$\frac{k}{\log k}\to\infty\implies\frac{n^{0.1}}{(\log n)^2}=\frac1{400}\cdot\left(\frac{n^{0.0.5}}{\log(n^{0.05})}\right)^2\to\infty$$

2

In the context of branching processes, the two statements are indeed equivalent. This is due to the following facts: The value of each random variable $X(t)$ is almost surely a nonnegative integer. The events $[X(t)=0]$ are nondecreasing with respect to $t$. Thus, $P(X(t)=0)\to P(A)$ when $t\to\infty$, where $A=\bigcup\limits_t[X(t)=0]$. On the other ...

1

more basic looking, by L'hospital $$\lim_{x \rightarrow + \infty} \frac{x}{\log x} = \infty$$ Next $$\lim_{x \rightarrow + \infty} \frac{x}{(\log x)^2 } =\lim_{x \rightarrow + \infty} \frac{x}{2 \log x } = \infty$$ The general induction step is $$\lim_{x \rightarrow + \infty} \frac{x}{(\log x)^k } =\lim_{x \rightarrow + \infty} \frac{x}{k (\log ... 1 For the N(0,1) distribution we have that:$$\mathbb{P}[X>k]=\frac{1}{\sqrt{2\pi}}\int_{k}^{+\infty}e^{-x^2/2}dx =\frac{e^{-k^2/2}}{\sqrt{2\pi}}\int_{0}^{+\infty}e^{-kx}\cdot e^{-x^2}\,dx<\frac{e^{-k^2/2}}{k\sqrt{2\pi}}.\tag{1}$$You can easily adapt this argument to the N(\mu,\sigma) distribution. For an efficient continued fraction approximation ... 1 Define$$ \lambda:=\frac{\sum_{i\not\in P(i)} X_i(n)}{\sum_{i\in P(i)} X_i(n)} $$and note that \sum_{i\not\in P(i)} X_i(n)=\mathcal{O}(X(n)/t) and \sum_{i\in P(i)} X_i(n)=\Omega(X(n)/t). Your desired result requires that \lambda=o(1). But this only necessarily follows if \mathcal{O}(f)/\Omega(f)=o(1) for all functions f, which isn't true. ... 1 Basically you should just mimic the proof of Mertens' formula. We have that $\sum_{p \leq x} \log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) = \frac{4}{3} \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x}\left(\log\left(1 + \frac{4}{3p} + \frac{C}{p^{3/2}}\right) - \frac{4}{3p}\right),$ and the first term is asymptotic to \frac{4}{3} \log \log x, while ... 1 Actually the relation you wanted to write down, via Merten's 3rd theorem, should be$$\prod_{p\le x}\left(1-\frac{1}{p}\right)^{4/3}\left(1+\frac{4}{3p}+\frac{C}{p^{3/2}}\right) \ll 1. Hint: we know $(1+x)^\alpha=1+\alpha x+O(x^2)$ (for $x\approx0$) and when $\prod(1-p^{-s})$ converges.

1

$\log(1-1/x)=-1/x-O(1/x^2)$ and $\pi(x)=x/\log x+O(x/\log^2x)$ so their product is $-(1/x+O(1/x^2))(x/\log x+O(x/\log^2x))=-1/\log x-O(1/\log^2x).$

1

Necessary conditions: $\liminf_{n\to\infty}f(n)\geqslant1$ and $\limsup\limits_{n\to\infty}f(n)\leqslant2$. Sufficient conditions: $\liminf\limits_{n\to\infty}f(n)\gt1$ and $\limsup\limits_{n\to\infty}f(n)\lt2$. Examples: every constant $f(n)=\alpha$, provided $1\lt\alpha\lt2$.

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


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$n-\pi^{\ast}$ SINGLET STATES OF THALAZINE.

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/15401

Files Size Format View
1967-Y-03.jpg 142.4Kb JPEG image

 Title: $n-\pi^{\ast}$ SINGLET STATES OF THALAZINE. Creators: Marzzacco, Charles Issue Date: 1967 Publisher: Ohio State University Abstract: Phthalazine differs from other ortho-diazines, pyridazine and diazaphenanthrene, in having a lowest $\pi^{\ast}$orbital of $b_{2}$ rather than $a_{2}$ symmetry. This results in a symmetry forbidden $A_{2}\leftarrow A_{1}$ lowest excited promotion. Polarized absorption studies of single crystals of phthalazine and of phthalazine in naphthalene have enabled the locations and assignments of the lowest two excited $n\pi^{\ast}$ state of phthalazine to be made. In the crystal a weak 0.0 band is found at $25 365 cm^{-1} (0)$ followed by a stronger vibronic origin $(0^{\prime})$ at $25 875 cm^{-1}$. Most of the spectrum of the $A_{2}\leftarrow A_{1}$ transition is built on this false origin $(0^{\prime})$. A more intense band is found at $0+2168 cm^{-1}$ [or $0^{1}+1658 cm^{-1}$] which is not due to a totally symmetric mode built on $0^{\prime}$. Both $0^{\prime}$ and $0^{\prime}+1658$ act as origins for comparable spectra. Mixed crystal studies indicate in plane polarization for both systems. The strong $0+2168 cm^{-1}$ band is interpreted as part of a higher $n\pi^{\ast}$ state $B_{2}\leftarrow A_{1}$. The polarization and weakness of the transition are discussed. Description: Author Institution: Laboratory for Research on Molecular Structure, University of Pennsylvania URI: http://hdl.handle.net/1811/15401 Other Identifiers: 1967-Y-3


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67 views

### Should incorrect answers be flagged to delete? [duplicate]

If someone posts an answer that is mathematically incorrect, should we flag it to delete, or just downvote and comment? See this answer as an example.
604 views

### Is it in the Moderator's Job Description to patrol for correctness?

A user flagged a comment of another user, with the specific request to delete the other user's comment because it is "factually incorrect". Now, I'm inclined to just ignore flags like this, as I've ...
960 views

### Capture the Flag: FAQ on Flagging

What is flagging for moderator attention? How does the flagging process work? What are the different flags, and what are they useful for?
795 views

### Is sometimes keeping wrong answer reasonable?

My memory might be fooling me, but I believe I have seen comments or discussions an meta regarding this matter. I am unable to find it now, so I will ask about your opinion. If I answer some question ...
289 views

### Is it OK to flag an accepted answer because contains a significant, conceptual error?

In this question: What does $dx$ mean? someone asked what the "$dx$" in an indefinite integral meant. In one of the answers, which was accepted, the answer began by saying "Formally, $dx$ does not ...
625 views

There are one or two users that I have come across who continually and persistently post fake answers. Why do I say fake? Because they always do one of the following: Give a mathematically completely ...
920 views

### Deletion of on-topic answers only because they contain mathematical errors.

Are mathematical errors in an answer that is otherwise reasonable and on-topic ever a sufficient reason for users (other than the answer-author) to delete that answer? For a recent answer, the ...
534 views

I was told " flags should not be used to indicate technical inaccuracies, or an altogether wrong answer" and then I found Don't flag wrong answers? A claim here made by a user named Neal, asked ...
865 views

### What to do with blatantly false answers?

Recently, I came across this answer to an old question, which attempts to prove Fermat's Last Theorem (edit: the answer has now been deleted). It takes almost no effort to see why the argument there ...
248 views

Please clarify what can be done by people with low reputation points? I'm questioning about answers which don't answer the question, not questioning if the math is correct in these posts. Some of ...
125 views

### Flagging comments as useless and for wrong guidance

Sometimes people are posting comments which are completely wrong as happened, for instance, here, comment #2, or offering a wrong guidance as happened here, also comment #2. If I flag these comments ...


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# What do the tensors here equal to?

In the process $$e^+e^- \rightarrow \gamma \gamma$$ for which the amplitude can be written as: $M= \epsilon^*_{1\nu}\epsilon^*_{2\mu}(A^{\mu\nu}+\tilde{A}^{\mu\nu})$, where $\epsilon_i$ is the polarization vector of a photon. What do the tensors $A^{\mu\nu}$ & $\tilde{A}^{\mu\nu}$ equal to?


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# How to prepare KF/Al2O3 i.e. potassium fluoride on alumina (/on aluminum oxide)

It's surprisingly hard to find an answer to this. One book I found referenced a method wherein one would simply mix (with a mortar I assume?) equal (another source said 40:60) portions of Al2O3 and KF and that's it.

Alumina (30 g, neutral alumina, type 90 for column chromatography, activity I, 70-230 mesh, Merck Jpn.) was mixed with $\ce{KF}$ (20 g) in 200 mL of water and the water was removed at 50-60 °C in a rotary evaporator. The impregnated alumina was then dried in a vacuum drying oven at 75 °C for several hours.
In another publication, J. Chem. Soc., Perkin Trans 2, 1986, 1133-1139, by the same main author, which focuses on the properties of the $\ce{KF}/\ce{Al2O3}$ system, a "dry preparation" by grinding $\ce{KF}$ and $\ce{Al2O3}$ under a nitrogen atmosphere is mentioned.


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# Kalkülüs

### İlgili konuları keşfedin

Parent Functions

Visual Log LawsBy Brittany Bordewyk, as a great response to Prof. Wilson’s visual exponent laws. Did I somehow never post these?

*FREE* Conic Section Posters

Japanese Multiplication 101 #Multiplication #JapaneseMultiplication

Inverses of Matrices

Tweet Tweet Math is not everyone’s cup of tea, but learning new math shortcuts can be a lot of fun. There are many of them to learn. This infographic happens to cover 6 math shortcuts: [via: Center of Math]

Calculus - Derivatives and Limits

Trig functions on the unit circle

If you need a note card for Integral Calculus - Imgur

LOVE this! Calculus becomes soo much easier when you realize you are just finding the slope!!

Pinterest


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Seminars

# Joint Instrumentation Seminar: VELOPix and the LHCb VELO Upgrade

## by Paula Collins (CERN, Geneva, Switzerland)

Friday, March 25, 2011 from to (Europe/Berlin)
at Bldg. 28c, seminar room
 Description With the successful start up of the LHC the experiments are already planning the upgrades which are necessary to maintain and enhance the physics output over the next decade and beyond. At the LHCb experiment there is an opportunity to massively increase the data taking capabilities, independently of the planned LHC luminosity increases, by modifying the electronics architecture to be able to readout the entire detector at 40 MHz and perform all trigger algorithms in software. One of the implications of this scheme is the need to completely replace the silicon vertex detector of LHCb, or VELO. For this a new readout ASIC is being developed, dubbed VELOPix, which is based on the Timepix/Medipix family of chips. This seminar introduces the LHC upgrade landscape and the motivation for the LHCb upgrade, and describes the R&D steps which are currently in place to develop the new chip and the associated upgraded pixel vertex detector. The installation of the LHCb upgrade is planned for the long LHC shut down in 2017/2018, and the LHCb Upgrade LOI has been recently presented to the LHCC and can be viewed at https://indico.cern.ch/conferenceDisplay.py?confId=130706.


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Section2.9Using C Programs to Explore Data Formats

Before writing any programs, I urge you to read Appendix A on writing Makefiles, even if you are familiar with them. Many of the problems I have helped students solve are due to errors in their Makefile. And many of the Makefile errors go undetected due to the default behavior of the make program.

We will use the C programming language to illustrate these concepts because it takes care of the memory allocation problem, yet still allows us to get reasonably close to the hardware.

You probably learned to program in the higher-level, object-oriented paradigm using something like C++, Java, or Python. C does not support the object-oriented paradigm. It is a procedural programming language. The program is divided into functions. Since there are no classes in C, there is no such thing as a member function. The programmer focuses on the algorithms used in each function, and data items are explicitly passed to functions or accessed globally.

Compiling and running the program in Listing 2.9.1 on my Raspberry Pi gave:

pi@rpi3:~/chp02 $gcc -Wall -o intAndFloat intAndFloat.c pi@rpi3:~/chp02$ ./intAndFloat
The integer is 19088743 and the float is 19088.742188
pi@rpi3:~/chp02 \$

Yes, the float really is that far off. This will be explained in Chapter 16.

The program source code starts with some documentation that gives the name of the file, a very brief description of what the program does, the author's name, and the date it was written. Everything between the /* and */ is considered a comment. It is there for the human reader and has no effect on the program itself.

The first operation that actually affects the program is to include another file, the stdio.h header file. This particular header file, stdio.h, defines the interface to many of the functions in the C Standard Library, which allows the compiler to know what to do when any of these functions are encountered in our source code.

Next you see the definition of a C function. All C programs are made up of functions which have the general format:

return-data-type function-name( parameter-list ) {
function-body
}

When a C program is executed, the operating system first sets up a C Runtime Environment, which sets up the resources on your computer to run the program. The C Runtime Environment then calls the main function, so the program you write must begin with a function whose function-name is main. The main function can call other functions, but program control normally ends up back in the main function, which then returns to the C Runtime Environment. When the program has completed execution, it should return a single integer to the C Runtime Environment, so the return-data-type is int. In our example main function here, no parameters are passed to it, so the parameter-list is void.

Our main function has two variables, an integer value (int) and a floating point value (float), which are listed at the beginning of the function-body. Most modern programming languages allow you to introduce new variables anywhere in the code, but C requires that they be listed at the beginning. (There are some exceptions to this rule, but they are beyond the scope of this book.) Think of it as being like listing the ingredients for a cooking recipe before the instructions on how to use them. This program provides values for each of the variables.

This program only prints the values of the two variables on the screen. It does this by calling a function in the C Standard Library, printf. The printf function allows for a great variety of formatting, but our use will be quite simple.

The first argument passed to the printf function is essentially a template, surrounded by quotes, for what will be printed on the screen. It is simply the text that you want printed, with the % character, immediately followed by some conversion code characters, at each place that you want the value of a variable substituted.

This text string is followed by a comma-separated list of the names of the variables to be substituted in the same order that their respective % codes appear in the template. You should enter the program above, compile it, and then run it. I am convinced that you will see how printf works in this simple program.

Common conversion codes (used after the % character) are:

 u unsigned decimal integer d or i signed decimal integer f float x hexadecimal

The conversion codes may include other characters to specify properties like the field width of the display, whether the value is left or right justified within the field, etc. We will not cover the details here. You can read man page 3 for printf to learn more.

Our next program illustrates how to read numbers from the keyboard.

The primary new concept introduced in Listing 2.9.2 is the scanf function from the C Standard Library. After the program calls printf to write a message on the screen, it calls scanf to read what the user types on the keyboard. Since the user is asked to type an unsigned decimal integer, the conversion code is %u. But with scanf we need to prefix the variable name, unsignedInteger, with the & character. In C the & is the “address of” operator, so the memory address where the variable is stored in memory is passed as an argument to scanf. The address of the variable is needed because scanf will read the keystrokes from the keyboard, convert the number to its equivalent binary representation, and store this bit pattern in memory at this address. You will get a chance to do this conversion in Chapter 14.

The second call to printf writes a message on the screen asking the user to enter a bit pattern in hexadecimal, so scanf needs the %x conversion code. After both numbers are read from the keyboard printf is called to write each of them on the screen twice. Enter the program, compile and run it. Now you have your own decimal-to-hexadecimal and hexadecimal-to-decimal conversion program.

The %u conversion code that is used when reading the integers with scanf and writing them with printf indicates that these are unsigned integers. The %#010x conversion code shows some of the more complex formatting that can be done with printf. The # together with the x tells printf to write the number in hexadecimal and prefix it with 0x. The first 0 tells printf to zero pad the displayed number so that it is the full width of the integer, that is, $$32$$ bits. Thirty-two bits require eight hexadecimal digits to display, and the prefix is two more characters, so the 10 tells printf to write $$10$$ characters on the screen. Feel free to play around with this conversion code in your program to see the effects of changing it. You won't break anything.

The program in Listing 2.9.2 demonstrates a very important concept—hexadecimal is used as a human convenience for stating bit patterns. A number is not inherently binary, decimal, or hexadecimal. A particular value can be expressed in a precisely equivalent way in each of these three number bases. For that matter, it can be expressed equivalently in any number base. But since a computer is made up of binary switches, all numbers are stored in binary in the computer.


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# encoding.parallel¶

• Current PyTorch DataParallel Table is not supporting mutl-gpu loss calculation, which makes the gpu memory usage very in-balance. We address this issue here by doing DataParallel for Model & Criterion.

Note

This code is provided together with the paper

• Hang Zhang, Kristin Dana, Jianping Shi, Zhongyue Zhang, Xiaogang Wang, Ambrish Tyagi, Amit Agrawal. “Context Encoding for Semantic Segmentation” The IEEE Conference on Computer Vision and Pattern Recognition (CVPR) 2018

Encoding Data Parallel

## DataParallelModel¶

class encoding.parallel.DataParallelModel(module, device_ids=None, output_device=None, dim=0)[source]

Implements data parallelism at the module level.

This container parallelizes the application of the given module by splitting the input across the specified devices by chunking in the batch dimension. In the forward pass, the module is replicated on each device, and each replica handles a portion of the input. During the backwards pass, gradients from each replica are summed into the original module. Note that the outputs are not gathered, please use compatible encoding.parallel.DataParallelCriterion.

The batch size should be larger than the number of GPUs used. It should also be an integer multiple of the number of GPUs so that each chunk is the same size (so that each GPU processes the same number of samples).

Parameters: module – module to be parallelized device_ids – CUDA devices (default: all devices)
Reference:
Hang Zhang, Kristin Dana, Jianping Shi, Zhongyue Zhang, Xiaogang Wang, Ambrish Tyagi, Amit Agrawal. “Context Encoding for Semantic Segmentation. The IEEE Conference on Computer Vision and Pattern Recognition (CVPR) 2018

Example:

>>> net = encoding.nn.DataParallelModel(model, device_ids=[0, 1, 2])
>>> y = net(x)

## DataParallelCriterion¶

class encoding.parallel.DataParallelCriterion(module, device_ids=None, output_device=None, dim=0)[source]

Calculate loss in multiple-GPUs, which balance the memory usage for Semantic Segmentation.

The targets are splitted across the specified devices by chunking in the batch dimension. Please use together with encoding.parallel.DataParallelModel.

Reference:
Hang Zhang, Kristin Dana, Jianping Shi, Zhongyue Zhang, Xiaogang Wang, Ambrish Tyagi, Amit Agrawal. “Context Encoding for Semantic Segmentation. The IEEE Conference on Computer Vision and Pattern Recognition (CVPR) 2018

Example:

>>> net = encoding.nn.DataParallelModel(model, device_ids=[0, 1, 2])
>>> criterion = encoding.nn.DataParallelCriterion(criterion, device_ids=[0, 1, 2])
>>> y = net(x)
>>> loss = criterion(y, target)

forward(inputs, *targets, **kwargs)[source]

Defines the computation performed at every call.

Should be overridden by all subclasses.

Note

Although the recipe for forward pass needs to be defined within this function, one should call the Module instance afterwards instead of this since the former takes care of running the registered hooks while the latter silently ignores them.

## allreduce¶

encoding.parallel.allreduce(*inputs)[source]

Cross GPU all reduce autograd operation for calculate mean and variance in SyncBN.


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# Additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.

A function $T:V \rightarrow W$ is additive if $T(x+y) = T(x) + T(y)$ for every $x, y \in V$. Prove that there exists an additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.

My attempt: Let $T$ be the function $T: \mathbb{R}$ (over the field $\mathbb{Q}$) $\rightarrow \mathbb{R}$ (over the field $\mathbb{R}$). The set $\{1, \sqrt{2}\} \subseteq \mathbb{R}$ is linearly independent for the vector space $\mathbb{R}$ over $\mathbb{Q}$. Then, there must exist a linearly independent set $W \subseteq \mathbb{R}$ (over $\mathbb{Q}$) such that $\{1, \sqrt{2}\} \subseteq W \subseteq \text{span}(W) = \mathbb{R}$ (over $\mathbb{Q}$).

I have been told that the function defined as $T(1) = 1$ and $T(w) = 0$ for all $w \in \text{span}(W) \setminus \{1\}$ is additive but not linear, but I cannot see why this is? I can see why it is not linear, clearly $T(\sqrt{2} \cdot 1) = 0$ but $\sqrt{2} T(1) = \sqrt{2}$. But, why is $T$ additive? For example, $T(1+1) = T(2) =0$ but $T(1) + T(1) = 1+1 = 2$? Is there a mistake somewhere?

• Be careful, you want the $\mathbb Q-$homogeneity ! Therefore $T(\sqrt 2 \cdot 1)\neq \sqrt 2 T(1)$ since it's not $\mathbb R-$homogeneous. – Surb Jul 31 '17 at 12:46

You can indeed construct an additive, non-linear function $T:\Bbb R\to \Bbb R$ (as vetor spaces over $\Bbb R$, since additive maps of $\Bbb Q$-vector spaces necessarily become linear automatically) by starting with what you have: $T(p) = p$ for $p\in \Bbb Q$, and $T(r) = 0$ for all $r$ with $r/\sqrt2 \in \Bbb Q$. However, constructing it is not easy, and requires the so-called axiom of choice in order to be finalised.

So, we start out with what we already have: some values that $T$ sends to themself, and some values that $T$ sends to $0$. By additivity, this forces $T$ to be defined for a slightly greater class of numbers, namely $T(p + q\sqrt2) = p$, for all rational $p, q$.

But we haven't defined $T$ for all real numbers yet. Note that whatever else we define $T$ to be, it already cannot be linear. So we just build on it as best we can and ensure that it stays additive. We do this by picking a number for which $T$ isn't already defined, like $\pi$ or $\sqrt 3$ or $\ln 2$ or $e$ or anything else. I'll pick $\pi$ for now. We can make $T(\pi)$ to be whatever we want it to be. I'll pick $3$ because I like $3$. By additivity, $T$ is now forcibly defined to be $T(p+q\sqrt2+r\pi) = p+3r$ for any rational $r$.

However, we are not done definining $T$ yet; there are still many more real numbers to cover. And this is where the axiom of choice comes in: there is no finite way, or even countably infinite way to finalise this definition of $T$. We just have to keep going and going, picking one new real number each step, until we've covered the entire real number line, which requires an uncountably infinite number of steps. The axiom of choice is exactly what allow us to say that even though we cannot finalise our $T$, some finalised $T$ does indeed exist somewhere out there.

Such a function cannot be defined exmplicitly. It only can be shown (assuming the axiom of choice), that such a function exists. First note that each such function $T:\mathbb R\to\mathbb R$ is a solution to the so called Cauchy functional equation. It can be shown that $T$ is automatically linear, if $T$ is Lebesgue-measurable or continuous or bounded. In particular, non-linear examples have to be really pathological.

For a more detailed discussion and a proof of the existence of a nonlinear example see

https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation

You need some key elements here.

First of all, note that if we consider $\mathbb R$ as a vector space over $\mathbb Q$, then it is infinite dimensional. This means that a "Hamel" basis $E$ for $\mathbb R$ over $\mathbb Q$ is at least countably infinite, and exists by the axiom of choice. Note : Without the axiom of choice, this is a rather interesting question, one that I don't know much about.

Choose two elements $e_1,e_2$ from this Hamel basis. Define the "linear transformation" $f$ from $\mathbb R \to \mathbb R$ as vector spaces over $\mathbb Q$, by defining $f(e_1) = 1$ and $f(e) = 0$ for all $e \neq e_1 \in E$. Note that because we are extending $f$ from it's basis construction, it obviously is additive, but then it isn't of the said form, since $f(e_2) = 0 \neq \frac{e_2}{e_1}f(e_1)$.

Hence $f$ is additive, but not linear.

• Yes, that is right. It is to be corrected. Indeed, I meant the base fields to be the same, but only indexed the first one to indicate that the base field of the second will be the same. – астон вілла олоф мэллбэрг Jul 31 '17 at 13:13
• Some clarification: (correct me if I've misunderstood your intent) $f$ is linear when $\Bbb R$ is seen as a vector space over $\Bbb Q$ (any additive function over $\Bbb Q$-vector spaces is automatically linear), but not when seen as a vector space over $\Bbb R$. However, additivity is independent of the base field of the vector space, and thus $f$ is still additive. – Arthur Jul 31 '17 at 13:16
• Aslo, do you mean to define $f(e_1) = 1$? Because the way it stands now, your $f$ is actually the constant zero function. – Arthur Jul 31 '17 at 13:19
• Oh, many errors. Let me edit it and get back to you. Thank you. – астон вілла олоф мэллбэрг Jul 31 '17 at 13:19
• Okay, so after corrections : You must clarify what this statement means, because it's confusing me : "any additive function over $\mathbb Q$-vector spaces is automatically linear". What is the difference between linear and additive, in the context of $\mathbb Q$- vector spaces? Because this notion of $f(x) = kx$ seems to make sense only in $\mathbb R$, right? Again, I must be wrong, so please correct me, it can only be helpful. – астон вілла олоф мэллбэрг Jul 31 '17 at 13:25


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# What functions $f:\mathbb{R}\to\mathbb{R}$ satisfy $(a+b)f(a+b)=af(a)+bf(b)$?

The relative error transformation $T(f)=\dfrac{f^\prime}{f}$ for differentiable functions $f:\mathbb{R}\to\mathbb{R}$ satisfies the properties

1. $T(fg)=T(f)+T(g)$
2. $T\left(\dfrac{f}{g}\right)=T(f)-T(g)$
3. $T\left(f^n\right)=nT(f)$
4. $(f+g)T(f+g)=f\,T(f)+g\,T(g)$

The first three properties are shared with logarithmic functions on $\mathbb{R}^+$, but not the fourth.

Suppose $f:\mathbb{R}\to\mathbb{R}$ and for $a,\,b\in\mathbb{R}$

$$(a+b)f(a+b)=af(a)+bf(b)\tag{1}$$

Clearly, every constant function defined on $\mathbb{R}$ satisfies this property. If $f(x)=c$ then we have

\begin{eqnarray} (a+b)f(a+b)&=&(a+b)c=ac+bc\\ af(b)+bf(b)&=&ac+bc \end{eqnarray}

Are there other functions $f:\mathbb{R}\to\mathbb{R}$ that satisfy property $(1)$ ?

• There are other solutions, but these are not continuous anywhere. See e.g. en.wikipedia.org/wiki/Cauchy's_functional_equation – Winther Jan 9 '17 at 19:53
• I used to ask my freshmen algebra students to show that if $f,g$ satisfy $(1)$ and $h=c_1f+c_2g$, $c_1,c_2\in\mathbb{R}$ then $h$ satisfies $(1)$. I asked my calculus students to show that if $f$ were differentiable, and if $f$ satisfied $(1)$ then $f$ is a constant function. – John Wayland Bales Jan 9 '17 at 20:10

Let $g(x) = xf(x) \implies g(a+b) = g(a) + g(b)$. The general solution for this functional equation is that $g(x) = cx$. Thus $f(x) = c$ is the only function which meets this condition.
• Good argument. I was unfamiliar with this argument and could only show it true if $f$ were differentiable since for differntiable $f$ it must be the case that $f^{\prime}(x)=0$. – John Wayland Bales Jan 9 '17 at 19:52
• There are plenty of other solutions to $g(a+b) = g(a) + g(b)$ besides $g(x) = cx$ on $\mathbb R$! en.wikipedia.org/wiki/Cauchy's_functional_equation – Joshua Mundinger Jan 9 '17 at 19:53


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## Animated Logical Graphs • 22

The step of controlled reflection we just took can be repeated at will, as suggested by the following series of forms:

Written inline, we have the series ${}^{\backprime\backprime} \texttt{(} a \texttt{)} {}^{\prime\prime},$ ${}^{\backprime\backprime} \texttt{(} a \texttt{,} b \texttt{)} {}^{\prime\prime},$ ${}^{\backprime\backprime} \texttt{(} a \texttt{,} b \texttt{,} c \texttt{)} {}^{\prime\prime},$ ${}^{\backprime\backprime} \texttt{(} a \texttt{,} b \texttt{,} c \texttt{,} d \texttt{)} {}^{\prime\prime},$ and so on, whose general form is ${}^{\backprime\backprime} \texttt{(} x_1 \texttt{,} x_2 \texttt{,} \ldots \texttt{,} x_k \texttt{)} {}^{\prime\prime}.$  With this move we have passed beyond the graph-theoretical form of rooted trees to what graph theorists know as rooted cacti.

I will discuss this cactus language and its logical interpretations next.

### 1 Response to Animated Logical Graphs • 22

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# Diversions in Mathematics #2: Hilbert’s Hotel

In this instalment, I introduce the concept of infinity in a simple and (hopefully) entertaining way, which puts into practice the counting concepts introduced in the previous Diversion. In fact, Hilbert‘s infinite hotel was one of the ‘stories’ that got me seriously interested in mathematics in the first place, and so it is a pleasure to share it here. This is a very well-known piece of story-driven mathematics. I hope that experienced mathematicians who happen to come across this blog do not tire of hearing (reading) it again, and that they see the value in telling the story to the general public.

Just before we start: I assume knowledge of the definitions and notations introduced in the previous instalment, namely, the very basics of set theory.

# Diversions in Mathematics #1: How to Count like a Pure Mathematician

If you want to know what this is all about, read my introduction to the series.

If you have already read the introduction, then welcome to the first Diversion in Mathematics! I emphasise, as I did in the introduction, that I will write with the general public in mind, so don’t be worried if you don’t consider yourself a “fan” of mathematics, or if you’ve totally suppressed all memories of maths classes from high school. And if you are a keen mathematician, whether recreationally or studying seriously at college/university, hopefully you will also find these blogs to be of some interest.

# Diversions in Mathematics #0

Introduction to the series

I have always been interested in maths, and not only in the subject itself but also the ways in which maths is explained and taught. In general, a crucial part of studying and researching is to be able to communicate one’s findings to other people, who may or may not be knowledgeable in your field. For this reason, I’m all for popular science books and magazines, which (provided that it is done well) serve to explain scientific research in an accessible way, and to promote scientific awareness and appreciation amongst the general public. However, in my opinion, popular science books too often simplify, and even completely skip the mathematics behind the science. There is certainly a cultural aversion to mathematics — at least from my perspective as an Australian, and from my awareness of similar attitudes in the US and UK — which may be part of the reason for the lack of ‘real’ mathematics in popular science writing. Here is an anecdote: apparently Stephen Hawking’s publisher advised the great scientist that every equation he included in his A Brief History of Time would result in reduced sales. (There is one equation though: Einstein’s $E = mc^2$). Of course, this is one of the bestselling science books ever, and sits atop many a coffeetable, but I wonder how many people have seriously read it…


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# From eigenvectors to eigentensors. Is eigentensor (hypermatrix) theory developed and useful?

Just as eigenvectors are invariant vectors up to a multiplicative constant $Au=\lambda u$, I wondered if eigentensor theory is equally developed for any arbitray tensor or hypermatrix array, i.e., $AB=\lambda B$, where A and B are suitable hyperobjects, and, maybe, even $\lambda$ could be an eigentensor as well.

Related: Why should we encounter eigentensors naturally? Is the eigentensor/hypermatrix theory hard? Any nice reference for current status of the subject including something else beyond hyperdeterminants (Kapranov et alii)?

• You might be interested in the theory of eigenoperators. Jan 28, 2018 at 19:13
• That thing...Does really exist? LoL Academia is obsolete, really! ;) Any reference? Jan 28, 2018 at 19:29
• It could also be called an eigenvector, just in a "vector space" of tensors. For example, on the space of $n \times n$ real matrices ${\text M}_n(\mathbf R)$ the transpose operation $B \mapsto B^\top$ is a linear operator of order 2 and the symmetric matrices are all eigenvectors of this operator. The function $e^{\lambda x}$ is an eigenvector of the differentiation operator $d/dx$, with eigenvalue $\lambda$, although in the setting of analysis these are called eigenfunctions instead of eigenvectors, but it's the same thing.
– KCd
Jan 28, 2018 at 20:26
• @riemannium I've only seen it scattered across papers, but not in books. I'd suggest searching around and finding papers that seem relevant. Jan 28, 2018 at 21:05


text:
# Logarithms

## Recommended Posts

Solve:

log5(x-1) + log5(x-2) - log5(x+6) = 0

I feel like this is really simple but I cannot get the answer.

This is what I attempted

Since they have the same base, I multiplied the top

log5[(x-1)(x-2)] - log5(x+6) = 0

This gives me

log5(x2-3x+2) - log5(x+6) = 0

At this point I did not now what to do, I tried 2 methods,

1.
Since they are the same base again, I divided the top this time

log5(x2-3x+2) / (x+6) = 0

but I don't know how to continue

2.

I moved - log5(x+6) to the other side

log5(x2-3x+2) = log5(x+6)

Since they have the same base I ignored them so

x2-3x+2 = x+6

Which gave me

x2 - 4x - 4

Which cannot be factored.

Edited by Acnhduy
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What methods, other than factorisation do you know for solving quadratic equations?

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There are basically three ways to solve a quadratic eguation.

1. by factorisation

2. completing the square method

3. using the almighty formula.

I see no lapses with you manipulations. you are nearly there.

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log5(x-1) + log5(x-2) - log5(x+6) = 0

[(x-1)(x-2)]/(x+6) = 0

(x-1)(x-2) = 0

therefore: x = 1; 2

Since the denominator can never be zero.

Edited by ComplexCalcBro
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[(x-1)(x-2)]/(x+6) = 0
That is not correctly derived. Exp(0) = 1, not 0. The OP has it right, in the part where they "ignored the base".

For many people, btw, the quadratic formula is more easily remembered by sound - memorized as a one would a short poem or limerick - rather than by sight. If you are one of those people, say it as you are writing it - by hand, with a pencil - a few times, and it will stick.

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$\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)=0$

There are probably more efficient ways, but the first thing I thought of doing was...

$5^{\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)}=5^0$

Take 5 to the power of both sides. After some simplification...

$\frac{(x-1)(x-2)}{x+6}=1$

From there, you can bring out the denominator to the other side and rearrange the equation into standard quadratic form.

Edited by Amaton

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text:
# Transversal wave speed derivation for small amplitudes

The above is a derivation for the wave speed equation in my physics textbook. However, I've read online that this equation is only true for waves with small amplitudes. I do not see where this assumption is made in the derivation, so why is the equation only true for small amplitudes?

The above picture shows the vertical restoring force should be 2*T*sin(phi)

The explanation is not a very full one. As you correctly note, you're taking a limit, so the assumption $\sin\theta \to\theta$ as $\delta z\to0$ becomes exact. So Eq 16-23 contains no approximation.

The assumption creeps in subtly when one assumes that the force calculated in Eq 16-23 is at right angles to the $z$ axis. That is, that $\mathrm{d}y/\mathrm{d} z$ is small, so that the normal to the tangent to the curve stays approximately vertical in the diagram. The best way to understand all this is to work out a more accurate equation; then the vertical component of the force restoring the small length $\mathrm{d}\,s$ of string is

$$T\,\partial\theta \,\cos\theta = \mathrm{d}s\, T\,\frac{\partial\theta}{\partial s}\,\cos\theta = \mathrm{d}s\, T\,\frac{\partial^2y}{\partial z^2}\,\frac{1}{\left(1+\left(\frac{\partial y}{\partial z}\right)^2\right)^2}$$

(recalling that $\frac{\partial\theta}{\partial s}$ is the curvature of the string and then using the formula for the curvature) and THEN you approximate that $\frac{\partial y}{\partial z}\ll 1$ and, equivalently, that $\mathrm{d}z = \mathrm{d}s$. The small amplitude approximation is then indirect: we're directly assuming small gradients, which imply and are implied by small amplitudes, given that we know the wavelength is limited.

• How did you get the starting equation Tdthetacos(theta)? For the last step, should the denominator be raised to the power of 3/2? Did you replace cos(theta) with cos(0) since theta is small? Can't we make the force at right angles to the z-axis by making ds small? We know that the tangent at the middle is flat, so the tension force will be approximately flat if we stay near the middle, right? – roobee May 23 '16 at 18:42
• The formula is simply the same as 16-23 in your text with a $\cos\theta$ added to account for the fact that the force is not perfectly vertical but skewed by the string's gradient. And you're right about the missing $3/2$ power, I've now put it back in all its glory (it multiplies the $\cos\theta$ factor to become the power of two now in the denominator on the RHS). Also, you can't make the force at right angles at all points: recall that the string is assumed to be perfectly bendable, which means that it can resist no shear and can only impart tension along its tangent. – WetSavannaAnimal May 24 '16 at 13:01
• doesn't the sin(theta) in the textbook already account for the skew? – roobee Jun 3 '16 at 3:52
• @roobee No, it's simply measuring the arclength. I should probably have used a different symbol from $\theta$; in the book its simply the angle subtended by the section of string at the center of curvature, and we should use a second symbol, say $\phi$, to measure the skew. – WetSavannaAnimal Jun 3 '16 at 4:15
• I've added a picture in my question. I've replaced theta with phi. It shows that the vertical component of the force is 2*T*sin(phi), so doesn't that show that sin(phi) accounted for the skew? – roobee Sep 16 '16 at 17:49

Right off the bat in eq (16-23) it's assumed that the restoring force is linear in the displacement. That's only true for small displacements.

• Are you referring to where sin(theta) changes to theta? If so, I thought that was because theta was small due to the segment being small, and not because the amplitude was small. – roobee May 23 '16 at 2:38
• No, I'm not. I was referring to the approximation that $\tau$ is taken to be a constant, whereas we know that the tension will increase as the string is stretched. However, as you can now see, there's more to it than that. – garyp May 23 '16 at 11:49


text:
# zbMATH — the first resource for mathematics

Ruin probabilities. (English) Zbl 0960.60003
Advanced Series on Statistical Science & Applied Probability. 2. Singapore: World Scientific. x, 385 p. (1997).
This book is devoted to risk theory and ruin probabilities in particular. Risk theory in general and ruin probabilities in particular are traditionally considered as part of insurance mathematics, and have been an active area of research from the days of Lundberg all the way up today. However, the practical relevance of the area has been questioned repeatedly. And, as the author writes: “One reason for writing this book is a feeling that the area has in the recent years achieved a considerable mathematical maturity, which has in particular removed one of the standard criticisms of the area, that it can only say something about very simple models and questions. Apart from these remarks, I have deliberately stayed away from discussing the practical relevance of the theory; if the formulations occasionally give a different impression, it is not by intention. Thus, the book is basically mathematical in its flavour.”
The book contains Preface, eleven chapters, Appendix, Bibliography with 384 items and Index. In brief: Chapters I and II are Introduction and Some general tools and results, respectively; Chapters III–VII introduce some of the main models and give a first derivation of some of their properties; Chapters IX and X then go in more depth with some of the special approaches for analyzing specific models and add a number of results on the models in Chapters III–VII; Chapter XI is Miscellaneous topics; Appendix contains some necessary results in renewal theory, Wiener-Hopf factorization, matrix-exponentials, some linear algebra and complements on phase-type distributions. The end of each chapter contains notes and references.
The Introduction contains descriptions of the risk process, claim size distributions, the arrival process, and a summary of main results methods, models and topics to be studied in the rest of the book. In this chapter, a risk reserve process and ruin probabilities with infinite and finite horizon for it are introduced. They are the main topics of study of the present book. Chapter II collects and surveys some topics which repeatedly show up in the study of ruin probabilities. The martingale approach in Section 1 is essentially only used here. The likelihood ratio approach in Section 2 is basic for most of the models under study. The duality results in Section 3 are not crucial for the rest of the book, however, the topic is fundamental and the probability involved is rather simple and intuitive. Sections 4, 5 on random walks and Markov additive processes can be skipped until reading Chapter VI on the Markovian environment model. The ladder height formula in Theorem 6.1 is basic for the study of the compound Poisson model in Chapter III. Chapter III is devoted to the study of the compound Poisson model with Pollaczeck-Khinchin formula, change of measure via exponential families, Lundberg conjugation and various approximations for the ruin probabilities and estimation of the adjustment coefficient.
Chapter IV is concerned with the finite time ruin probabilities. Only the compound Poisson case is treated. Generalizations to other models are either discussed in the notes and references or in relevant chapters. Chapter V studies the risk process with renewal arrivals. The renewal model is often referred to as Sparre Andersen process, after E. Sparre Andersen whose 1959 paper was the first to treat renewal assumptions in risk theory in more depth. Chapter VI contains the risk theory in a Markovian environment. The author assumes that arrivals are not homogeneous in time but determined by a Markov process with finite state space. This latest process describes the environmental conditions for the risk process. In Chapter VII the author assumes that accumulated claims or total claims process is a compound Poisson process, however, the premium charged is assumed to depend upon the current reserve process $$R_{t}$$ so that the premium rate is $$p(r)$$ when $$R_{t}=r.$$ Ruin probabilities for such models are studied. Chapter VIII deals with matrix-analytic methods in the risk theory. Ruin probabilities in the presence of heavy tails are considered in Chapter IX. Chapter X gives a summary of some basic methodology in simulation and Monte Carlo methods. Miscellaneous topics, beginning from Bernoulli random walk, Brownian motion and two-barrier problem and ending with reinsurance, are considered in Chapter XI.
The book is an excellent handbook in the risk theory and brilliant encyclopedia in ruin probabilities results. This book will be useful both for graduate students who deal with stochastic models in insurance mathematics, experts in risk processes and also for those specialists who apply the methods of risk theory in practice.

##### MSC:
 60-02 Research exposition (monographs, survey articles) pertaining to probability theory 60J25 Continuous-time Markov processes on general state spaces 91B30 Risk theory, insurance (MSC2010) 60K05 Renewal theory 60K15 Markov renewal processes, semi-Markov processes 60G44 Martingales with continuous parameter 60G50 Sums of independent random variables; random walks 60J60 Diffusion processes 60F10 Large deviations 60K37 Processes in random environments


text:
### Martynas's blog

By Martynas, history, 5 months ago, ,

This weekend the best coders from Lithuanian high schools will participate in the final stage of the Lithuanian Olympiad in Informatics. As it was the case in the previous years, you are invited to take part in an online mirror. This IOI-style competition will take place on two days:

• +46

 » 5 months ago, # |   +3 The dates should be 30 and 31 (not 29 and 30).
•  » » 5 months ago, # ^ |   0 Uh, sorry for the OBOE. At least the timeanddate.com link was correct. Thanks!
 » 5 months ago, # |   +11 It doesn't seem like I can register, there's just a login form.
•  » » 5 months ago, # ^ |   +8 Yes, I was trying to register too ...
•  » » 5 months ago, # ^ |   +8 Apparently there's a delay, the registration should start soon at https://cms.lmio.lt/vyr-eng/
•  » » 5 months ago, # ^ |   +8 Now it's possible to register!
•  » » 5 months ago, # ^ |   0 You can register
 » 5 months ago, # | ← Rev. 5 →   +3
 » 5 months ago, # |   +30 Solutions: AWe must find for each star the time frame in which it's inside the circle. Then we sort those to find the time in which we could see the most stars.To find the intersections, if they exist, I changed every star to a function and found intersection points with the circle by a formula on paper. Because of imprecision, after I find the intersection points and the times, I just iterate on the close integers (+- 3) to find the exact time frame. BWe will never want to change a height below 0 anyway (exactly 0 is at least as good), so we can ignore this restriction.Let's subtract from position $1 \le i \le n$ the value $i * M$. Now the sequence of heights we iterate over is non-increasing. When we adjust the height of some pole, it's best to change it to the height of the one before it. We can view it as ignoring all the poles which we picked to change the height of. Since we're looking to minimize the number of ignored poles, we're looking to maximize the number of unchanged poles, which form a non-increasing subsequence. So, the task is to find the longest non-increasing subsequence of the array (and make sure it begins with a 0 before the start of the array). CLet's define $C_i = A_i - B_i$, as the number of extra fertilizers position $i$ has. Interestingly, we do not care whether our operations for some solution make some position have a negative amount of fertilizers; we will always be able to reorder the operations to satisfy this restriction. So we can ignore it.Also, we can regard every operation as taking 1 fertilizer from position $i$ and moving it to position $i-1$ or $i+1$, with the cost of 1.Let's transform $C$ to its prefix sum. Our operations become to change a value in the array from $x$ to $x+1, x-1$ with the cost of 1 = changing $x$ to $y$ with the cost of $|x - y|$. We can only do this operation for all indicies except the last (which is the sum of all $C$), and the goal is for the array to be nondecreasing, and starting from (at least) 0. So first, for every value that is less than 0 or more than the sum of the array, adjust it to fit. Notice that this already solves subtask 1. We can now ignore the last value and the restriction with beginning with 0.The following part is the complex one (making the array sorted). Our process will be to iterate on values from small to large, and in every iteration we have a set of "changeable" values — values which will be at least the current ones in the answer sequence, and we keep incrementing them until it's not beneficial. There will also be a group of values which we already fixed — they will also form a prefix since we fix values from small to large. All values which we have not iterated on are considered unknowns (but we know they are larger than what we have now). Also notice that changeable values will always be the same (they become different when we fix some of them).Suppose the case in which we have one changeable value, and two unknowns to its left. If we decide to fix this value now, then both unknowns to the left must become this value. Observe that since our value is the smallest between those 3, it will be at least as good to increment our changeable value until it's equal to the smaller between the other two (in other words, we decide to keep it changeable). On the contrary, if we had 2 changeables and 1 unknown to their left, then it's better to decrease that unknown towards them instead of increasing them. Also, this makes us fix those 3 positions.Let's formulate this; imagine the array, in which every fixed value (in the prefix) implies a 0 in its position. Every changeable value is a +1, and every unknown is a -1. If there exists a prefix, whose sum is positive, then we should fix all the values in this prefix. We can implement this array with a lazy segment tree — when we change a value between $-1, 0, 1$, we update a suffix by adding some value to it. Then a query is the maximum value and its position.Fixing values can be done by bruteforce, since overall we fix $n$ values.
•  » » 9 days ago, # ^ |   +8 C can be done with a priority queue (submission). Some other problems which can be solved in a similar fashion can be found here:https://github.com/bqi343/USACO/blob/master/Implementations/06%20-%20DP%20(3)/README.md#6Is this trick well-known?
 » 5 months ago, # | ← Rev. 6 →   +8 I only participated in Day 2, got 2nd in Day 2 only results by taking 277.1 points. Here's my solution: Problem A (Cards)Let $X_i$ the number of placements that can be done by $i$ ways. First, calculate the number of unordered pairs of ("??ab", "ab??"), ("?a?b", "a?b?") which two are different. It can be calculated by brute-forcing a and b, and multiply number of ways for two strings. Let this number of ways $C_1$. Second, calculate the number of unordered pairs of ("abcd", "cdab"), ("abcd", "badc"), ("abb?", "?aab"), ("?bba", "baa?"), ("ab?a", "b?ab"), ("a?ba", "ba?b") which two are different. We can brute-force them. Let this total number of ways $C_2$. Third, calculate the number of unordered pairs of ("abba", "baab") which two are different. We can also brute-force them. Let this number of ways $C_4$. Then, we can say that $(X_1, X_2, X_3, X_4) = (C_1, C_2 \div 2, C_3 \div 2 \times 4, C_4 \div 2)$. Finally we can print $X_1 - X_2 + X_3 - X_4$ by inclusion-exclusion principle. Source Code (C++) Problem B (Coins, 77.1 points)Let $seq_i$ the number of zeros in $i$-th row. We want to optimize this sequence with hill-climbing algorithm, but since this problem is required to process "biased" (not random) image data, I thought that we should do good initial $seq$ for it. So, how to produce a good initial $seq$? I've thought the following ad-hoc approach. First, don't think about sum of silver/copper coins. Let's just think about the "cost", which is defined by silver -> copper: cost x, copper -> silver: cost 1-x, where $x$ is a real value in $[0, 1]$. Then, we can easily compute the minimum cost solution by DP in $O(n^2)$. The number of silver coins in minimum cost solution will be an increasing function of $x$. Therefore, we can binary-search $x$ that the number of silver/copper coins will be nearest to true value. But it is not perfect because there is some error in number of silver/copper coins. We adjust the number by random greedy approach. Then, we got 50.2 points, with getting better than perfect score in testcase 5. And then, we can do hill-climbing because the solution is near enough to the optimal. The neighborhood in hill-climbing is: do [1, 14] process of seq[i]++, seq[j]-- for random valid (i, j). If the score improves or stays, we will update the solution. If the score deteriorates by one, we will update the solution by $1/2$ probability. Otherwise, we will not update the solution. Then, finally, we could get 77.1 points by getting better than perfect score in testcase 1, 3, 4, 5 and 7! Source Code (C++) Problem C (Tax Evasion)If the coin is in depth $d$, this coin can climb up to depth $\lfloor (d-1)/2 \rfloor$, so the range which a coin can move (and finally stop) will form a subtree where the tree is rooted from vertex $0$. So, if we do "Euler Tour", we can reduce this problem into interval problem — the range where the coin finally stops will be like $l_i \leq x \leq r_i$. Let's think about binary-searching the answer. If the answer is $a$, all coins must stop at the different spot which the depth is at least $a-1$. We can judge if the answer is $a$ or more, by greedy algorithm, with priority queue, and it takes $O(n \log n)$ time complexity. Therefore, we can solve the problem by $O(n \log^2 n)$ time complexity. Source Code (C++)I was completely beaten by scanhex because I got zero points on testcase 2 of "Coins". Now I especially want to hear scanhex's solution of "Coins" :)
•  » » 5 months ago, # ^ | ← Rev. 2 →   +8 Problem C can also be solved with the same idea, but without binary searching. Incrementally add nodes as possible sinks from the bottom up. We iterate over the coins (after pulling them upwards to their subtree root) in decreasing order of depth (this implies the ranges the nodes cover as we keep going cannot be contained by previous ranges, which is used to prove the greedy), and add a node as another necessary sink, while the current coin doesn't have a free sink in its subtree. Once there is a free sink we can remove it (we maintain sinks with a set). $\mathcal{O}(n \log n)$.Also, you can solve the first two tests of B optimally with dp in $\mathcal{O}(n^4)$.
•  » » 5 months ago, # ^ |   +8 Well. How does an answer look like? Let's consider black coins. On each horizontal line, they form a prefix of it. Also, these prefixes are not increasing. The output of our program is the number of white coins initially lying on these prefixes.This observation can give us a precise solution. Let's calculate dp[i][j][k], which will be minimal number of white coins lying on our prefixes, assuming that we already chose first $i+1$ of them and the last ($i$-th) has a length of $\geq j$, while the sum of lengths of prefixes equals to $k$. Transitions would be pretty trivial, I won't describe them here for now.Of course, this DP has approximately N*N*(N*N/2) states and it is terrible. So, let's store only $MAGIC_1$ best states for each pair of $i$ and $j$. I say that state dp[i][j][k] is better than dp[i][j][l] if $k-dp[i][j][k]>l-dp[i][j][l]$. Also, it will likely result in storing only states with small $k$, so let's store only a set of states with disctinct $k/MAGIC_2$. That's pretty much it, constants differ from one test to another, but for me $MAGIC_1$ is generally equals to $70$ and $MAGIC_2$ vary from $100$ to $280$. With regards to the second test case, it is clear that this solution is particularly good on it, because you can store almost all possible states.
•  » » 5 months ago, # ^ |   +8 By the way, our solutions combined would have approximately score of 98. What was your approach?
•  » » » 5 months ago, # ^ |   +8 Now I wrote my solution of "Coins" in comment above. I actually struggled to finalize the way to select, "DP + heuristic-like improvement" or "Making good initial solution + improvement by hill-climbing". And I finally selected the latter. Your solution is based on the former and you won against me :) Obviously, seeing the scores for each results, your solution is good at some tests, and my solution is good at other tests. I think it's because your solution is based on heuristic-like DP approach and my solution is based on optimization (marathon match)-like approach. For example, for testcase 2, your solution is strong because its base solution is designed to be "absolutely optimal". For me, its base solution is not designed to be absolutely optimal. Thus, I got score $[577, 585]$, not the optimal $576$, and it varies by submission. But my solution is based on optimization-like approach, so, for some testcase, my solution is even better than $K_1$, which want to get more than 10 points :)
 » 9 days ago, # |   +8 You can submit all problems here: https://oj.uz/problems/source/431


text:
## Cracks in the Witten’s index theorem?

18/06/2019

In these days, a rather interesting paper (see here for the preprint) appeared on Physical Review Letters. These authors study a Wess-Zumino model for ${\cal N}=1$, the prototype of any further SUSY model, and show that there exists an anomaly at one loop in perturbation theory that breaks supersymmetry. This is rather shocking as the model is supersymmetric at the classical level and, in agreement with Witten’s index theorem, no breaking of supersymmetry should ever be observed. Indeed, the authors, in the conclusions, correctly ask how the Witten’s theorem copes with this rather strange behavior. Of course, Witten’s theorem is correct and the question comes out naturally and is very much interesting for further studies.

This result is important as I have incurred in a similar situation for the Wess-Zumino model in a couple of papers. The first one (see here and here)  went published and shows how the classical Wess-Zumino model, in a strong coupling regime, breaks supersymmetry. Therefore, I asked a similar question as for the aforementioned case: How quantum corrections recover the Witten’s theorem? The second one is remained a preprint (see here). I tried to send it to Physics Letters B but the referee, without any check of mathematics, just claimed that there was the Witten’s theorem to forbid my conclusions. The Editor asked me to withdraw the paper in view of this identical reason. This was a very strong one. So, I never submited this paper again and just checked the classical case where I was more lucky.

So, my question is still alive: Has supersymmetry in itself the seeds of its breaking?

This is really important in view of the fact that the Minimal Supersymmetric Standard Model (MSSM), now in disgrace after LHC results, can have a dark side in its soft supersymmetry breaking sector. This, in turn, could entail a wrong understanding of where the superpartners could be after the breaking. Anyway, it is really something exciting already at the theoretical level. We are just stressing Witten’s index theorem in search for answers.

## That strange behavior of supersymmetry…

07/12/2013

I am a careful reader of scientific literature and an avid searcher for already published material in peer reviewed journals. Of course, arxiv is essential to accomplish this task and to satisfy my needs for reading. In these days, I am working on Dyson-Schwinger equations. I have written on this a paper (see here) a few years ago but this work is in strong need to be revised. Maybe, some of these days I will take the challenge. Googling around and looking for the Dyson-Schwinger equations applied to the well-known supersymmetric model due to Wess and Zumino, I have uncovered a very exciting track of research that uses Dyson-Schwinger equations to produce exact results in quantum field theory. The paper I have got was authored by Marc Bellon, Gustavo Lozano and Fidel Schaposnik and can be found here. These authors get the Dyson-Schwinger equations for the Wess-Zumino model at one loop and manage to compute the self-energies of the involved fields: A scalar, a fermion and an auxiliary bosonic field. Their equations are yielded for three different self-energies, different for each field. Self-energies are essential in quantum field theory as they introduce corrections to masses in a propagator and so enters into the physical part of an object that is not an observable.

Now, if you are in a symmetric theory like the Wess-Zumino model, such a symmetry, if it is not broken, will yield equal masses to all the components of the multiplet entering into the theory. This means that if you start with the assumption that in this case all the self-energies are equal, you are doing a consistent approximation. This is what Bellon, Lozano and Schaposnik just did. They assumed from the start that all the self-energies are equal for the Dyson Schwinger equations they get and go on with their computations. This choice leaves an open question: What if do I choose different self-energies from the start? Will the Dyson-Schwiner equations drive the solution toward the symmetric one?

This question is really interesting as the model considered is not exactly the one that Witten analysed in his famous paper  on 1982 on breaking of a supersymmetry (you can download his paper here). Supersymmetric model generates non-linear terms and could be amenable to spontaneous symmetry breaking, provided the Witten index has the proper values. The question I asked is strongly related to the idea of a supersymmetry breaking at the bootstrap: Supersymmetry is responsible for its breaking.

So, I managed to numerically solve Dyson-Schwinger equations for the Wess-Zumino model as yielded by Bellon, Lozano and Schaposnik and presented the results in a paper (see here). If you solve them assuming from the start all the self-energies are equal you get the following figure for coupling running from 0.25 to 100 (weak to strong):

It does not matter the way you modify your parameters in the Dyson-Schwinger equations. Choosing them all equal from the start makes them equal forever. This is a consistent choice and this solution exists. But now, try to choose all different self-energies. You will get the following figure for the same couplings:

This is really nice. You see that exist also solutions with all different self-energies and supersymmetry may be broken in this model. This kind of solutions has been missed by the authors. What one can see here is that supersymmetry is preserved for small couplings, even if we started with all different self-energies, but is broken as the coupling becomes stronger. This result is really striking and unexpected. It is in agreement with the results presented here.

I hope to extend this analysis to more mundane theories to analyse behaviours that are currently discussed in literature but never checked for. For these aims there are very powerful tools developed for Mathematica by Markus Huber, Jens Braun and Mario Mitter to get and numerically solve Dyson-Schwinger equations: DoFun anc CrasyDSE (thanks to Markus Huber for help). I suggest to play with them for numerical explorations.

Marc Bellon, Gustavo S. Lozano, & Fidel A. Schaposnik (2007). Higher loop renormalization of a supersymmetric field theory Phys.Lett.B650:293-297,2007 arXiv: hep-th/0703185v1

Edward Witten (1982). Constraints on Supersymmetry Breaking Nuclear Physics B, 202, 253-316 DOI: 10.1016/0550-3213(82)90071-2

Marco Frasca (2013). Numerical study of the Dyson-Schwinger equations for the Wess-Zumino
model arXiv arXiv: 1311.7376v1

Marco Frasca (2012). Chiral Wess-Zumino model and breaking of supersymmetry arXiv arXiv: 1211.1039v1

Markus Q. Huber, & Jens Braun (2011). Algorithmic derivation of functional renormalization group equations and
Dyson-Schwinger equations Computer Physics Communications, 183 (6), 1290-1320 arXiv: 1102.5307v2

Markus Q. Huber, & Mario Mitter (2011). CrasyDSE: A framework for solving Dyson-Schwinger equations arXiv arXiv: 1112.5622v2

17/08/2013

Edward Witten is one of the greatest living physicists and also ranks high with mathematicians. He set the agenda for theoretical physics in several areas of research. He is mostly known for championing string theory and being one of few people that revolutionized the field. One of his major contributions to supersymmetry has been a deep understanding of its breaking. In a pair of famous papers (here and here) he put the foundations to our current understanding on the way supersymmetry can break and introduced the well-known Witten index. If a supersymmetric theory breaks supersymmetry then its Witten index is 0. This index is generally very difficult to compute and only perturbative or lattice computations can come to rescue. An important conclusion from Witten’s paper is that the well-known Wess-Zumino model in four dimensions does not break supersymmetry. Witten could rigorously justify this conclusion at small coupling but, at that time, an approach for strong coupling was missing and here Maldacena conjecture cannot help. Anyhow, he concluded that this should be true also for a strongly coupled Wess-Zumino model. Checks to this model in such a regime are rare. After I submitted a paper on arxiv last year (see here) I become aware of an attempt using Dyson-Schwinger equations that confirmed Witten conclusions for small coupling (see here). I have had an interesting mail exchange with one of the authors and this seems a promising approach, given authors’ truncation of Dyson-Schwinger hierarchy. Other approaches consider the Wess-Zumino model in two dimensions on the lattice. So, this appears a rather unexplored area , given the difficulties to cope with a strongly coupled theory, and Witten’s words appear like nails on a coffin to this theory.

I have worked out a lot of techniques to cope with strongly coupled theories and everywhere there is a perturbation going to infinity in a differential equation of any kind and so, I applied these ideas also to this famous model of supersymmetry. The idea is to prove that “supersymmetry has inside itself the seeds of its breaking“. The real issue at stake here is a correct understanding of the way supersymmetry breaks and recover in this way models that now appear to be defeated by data from LHC simply because the idea of symmetry breaking must be applied differently.

Of course, I do not aim to present a claim against the beautiful results given by Witten decades ago but just open up an interesting scientific question. So, considering that the Wess-Zumino model is just a theory of two scalar fields coupled to a Majorana spinor, its equations can be treated classically and so solved both for a strong and a weak coupling limit. I did this in a paper of mine (see here) and this paper has been accepted in these days in the Journal of Nonlinear Mathematical Physics as a letter. The classical solutions contradict the expectations giving a surviving of the supersymmetry at small coupling (as expected from Witten index for the quantum theory) while this does not happen for a strong coupling (formal limit of the coupling going to infinity). This is  a paradox, the Witten paradox, because classical solutions seem to break supersymmetry while the quantum theory does not.  So, we are left with a deep question: How is supersymmetry recovered by quantum corrections?

Marco Frasca (2012). Chiral Wess-Zumino model and breaking of supersymmetry arXiv arXiv: 1211.1039v1

A. Bashir, & J. Lorenzo Diaz-Cruz (1999). A study of Schwinger-Dyson Equations for Yukawa and Wess-Zumino Models J.Phys.G25:1797-1805,1999 arXiv: hep-ph/9906360v1

Marco Frasca (2012). Classical solutions of a massless Wess-Zumino model arXiv arXiv: 1212.1822v2