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Hi, I am creating a process map in Visio, is it possible to get Visio to number my boxs on my flowchart. At the moment I create the shape, put the text in but I have to manually put the number before the text, can visio do this automatically. Thanks Glenn In Visio 2003, try Tools > Add-ons > Visio Extras > Number Shapes. The same feature can be found in previous versions under Tools > Macros. -- Mark Nelson Microsoft Corporation This posting is provided "AS IS" with no warranties, and confers no rights. "Glenn Robertson"
However, what I want it to do now, instead of just putting the number into the attribute value, I want the number to be a suffix to the existing default tag value of the tag named "ID" which is "ADDRESS:" for some blocks and "ELEVATION:" for other blocks but the tag name that the routine always has to filter for is "ID". That way the user can pick anywhere on the block - not being forced to pick the tag we want to add numbers to.
Number Pro does not create your raffle ticket or document it just allows an easy way to  number them. Check out Number Pro, take a look at the demo videos and even try the demo of Number Pro to see how it works. Number Pro is a cloud based application meaning there is no download of the application. We access the application online in the members area. They offer a three year membership with unlimited use.

See Word's Numbering Explained by John McGhie, MVP - comprehensive and not pretty (Downloadable pdf file in letter size) - Reading this is vital to anyone attempting to use automatic numbering or bullets in a law office setting or other places where the documents are likely to be reused or heavily edited. See also How to Create a Template with a downloadable template with style-based numbering.

A sequence space is a vector space whose elements are infinite sequences of real or complex numbers. Equivalently, it is a function space whose elements are functions from the natural numbers to the field K, where K is either the field of real numbers or the field of complex numbers. The set of all such functions is naturally identified with the set of all possible infinite sequences with elements in K, and can be turned into a vector space under the operations of pointwise addition of functions and pointwise scalar multiplication. All sequence spaces are linear subspaces of this space. Sequence spaces are typically equipped with a norm, or at least the structure of a topological vector space.
Moreover, the subscripts and superscripts could have been left off in the third, fourth, and fifth notations, if the indexing set was understood to be the natural numbers. Note that in the second and third bullets, there is a well-defined sequence {\displaystyle (a_{k})_{k=1}^{\infty }} , but it is not the same as the sequence denoted by the expression.
I'm producing gift certificates for a restaurant and they need to be numbered sequentially from 0001 to 0250. Is there any way to do this easily as opposed to numbering each manually? I'm sure I could probably work it out with a print shop, but the job was thrust on me last minute and my options are limited by the short turn around time. Any help would be appreciated. Thanks!...
Warning  Making even minor changes to an outline numbering scheme won't necessarily change the initial position you've selected in the gallery, but rather may create a new gallery position, overwriting an existing one. Because of this problem, attaching numbering to styles is strongly recommended. This is covered in greater detail later in this chapter and in the chapter on Styles.

### InDesign allows you to add a page number marker to a master page within the document. The master page functions as a template for every page it's applied to, so the consecutive page numbers appear on every page. InDesign updates the page number automatically as you insert, delete and move pages. To add a page number marker to a master page, create a text box on the master page by going to the Type menu and choosing "Insert Special Character," "Markers" and then "Current Page Number."

Apple has a formalized version number structure based around the NumVersion struct, which specifies a one- or two-digit major version, a one-digit minor version, a one-digit “bug” (i.e. revision) version, a stage indicator (drawn from the set development/prealpha, alpha, beta and final/release), and a one-byte (i.e. having values in the range 0–255) pre-release version, which is only used at stages prior to final. In writing these version numbers as strings, the convention is to omit any parts after the minor version whose value are zero (with “final” being considered the zero stage), thus writing 1.0.2 (rather than 1.0.2b12), 1.0.2 (rather than 1.0.2f0), and 1.1 (rather than 1.1.0f0).
2. Yes, The code should be entered using CodeBuilder. Where you enter it depends on how and when you want to trigger the generation of the next number. If you want to use a button, that works. And no, you don’t use 000 in the NZ() function. If you want to DISPLAY at least 3 digits with leading zeros, then you do that in the Format function. Note, though, you will need to change that when you hit 1000 POs.
I make the design with as many up as I need on the master page, linking the frames where the numbers will go. Then I make the list using Excel, copy paste to ID and apply a paragraph style with "start in next frame" option. Click the outbox on the pasted text to get a loaded cursor and delete the frame. Then just shift-click over the first textframe on a live page to have as many "tickets" added as needed automatically.
Add numbers by inserting the same identifier at another location. Alternatively, copy the first one, select it and press “F9” to update the display. As a third option, you can use the keyboard shortcut “Ctrl-F9” and type “SEQ” and your identifier between the braces that appear. For instance, insert “{SEQ mynum}.” Press “F9” to update the field and enable the number to display.
In this system, the third number (instead of the second) denotes a minor release, and a fourth number (instead of the third) denotes bug-fix/revision releases.[23] Because the first number is always 10, and because the subsequent numbers are not decimal, but integer values, the 11th major version of OS X is labeled “10.10” (rather than 11.0). This number scheme continues above point-10, with Apple releasing macOS 10.13 in 2017.[24]

For whatever reason, AllExperts did not let me post a direct reply to your response re: “Ok, what is the ControlSource of the Fixture Number control? It should be: =cboZone & “-” & Format(Me.FNumber,”000″)” and adding “Me.Refresh” to my code (within the last 10 minutes). It just had the “rate this reply”. I added the Me.Refresh and corrected my location of the =cboZone code and it works correctly now.
Before you complete the merge, preview the merge results to make sure that the tracking numbers will display as you want them to in your publications. You can preview the merge in two ways: While you are refining the layout to review the layout of the individual coupon or gift certificate, or when you are getting ready to print, to preview the arrangement of coupons or gift certificates on the printed sheet.
Ok I guess it is better for me to explain what I am doing. I am in the process of creating an Access Database that an individual has been using an Excel spreadsheet forever and a day. Well the individual has on occasion doubled up numbers, forgot numbers, etc. So on what I have learned from different Access courses and Google searches I am trying to apply my knowledge.
Brand New Replacement Parts, Equipment And Supplies For: AB Dick, Ryobi, Multilith, Chief, Davidson, Hamada, Heidelberg,  T-51, Varn Kompac , Crestline, Baumfolder,  Astro envelope feeders  And Thompson envelope feeder, at below regular manufactures prices. We Also Stock, Prodeco “NO-MARK” Anti-Tracking devices,  Syntac rollers,  blankets,  wash-up blades / bottles, wash-up units / blades,  Rapid 106 staplers,  Airtech spray units,  Kompac Units,  Kompac Seals, Cutting blades, cutting sticks, dampening covers, hollow drill bits, numbering  units,  business card slitters, padding presses, disposable latex gloves, envelope backstops, press lamps, Align-mate , quick change blanket bars, Tilt and Turn  For T-51,  Litho Perf, Litho Score , electric switchs, counters, motors, vacuum pumps, spray powder, press dollies, pad counters, vacuum pump repair kits, gripper bar repair kits, hardware kits, metric hardware kits, V-belts, sucker feet / tubes,  etc.
We use the auto page numbering trick here, the trick is to do it as spreads. You just make your pages the same size as the ticket with crops and bleeds. Then set up the shell on the master page and place your auto page number where you want your numbers. Each pages is the next number. You can also have two sets of numbers for perforated ticket. You then can use the page start options to add to the numbers and keep the same number of total numbers. For example you would do a new page start after number 99. 0001 to 0099 so it doesn't add two "0's" to the front of 100 and you get 00100 when you want 0100. Works great for small quantities, we have done up to 1000+ tickets and once you do it once or twice it is a snap to set up. To avoid having to do 1000 pages just send the tickets in groups. Usually between 25 to 100 at a time works good depending on artwork file size. Just save each time before sending to printer and name the file with what the tickets are. Like "MyTickets-0001-0100.Indd". Makes it easier to go back if you have a problem. You can even do two rolls of tickets if your using 12 x 18 sheets by flipping the sheet and doing one row along the top and one along the bottom on reverse side. That is if your ticket is single Sided. Two sided tickets are a bit trickier. :-)
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Free Version
Moderate

# Angular Momentum of a Launched Projectile

APPHMC-92JSUU

A projectile of mass $m$ is launched with initial speed $v_0$ five times from a small height above the ground, each time at a different angle above the horizontal: $15^{\circ}$, $30^{\circ}$, $45^{\circ}$, $75^{\circ}$, and $90^{\circ}$ (directly upward).

In which of these launches does the projectile have the most angular momentum with respect to the ground directly underneath it?

A

$15^{\circ}$

B

$30^{\circ}$

C

$45^{\circ}$

D

$75^{\circ}$

E

$90^{\circ}$


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Leaching

(Redirected from Leaching (mobilization))

Definition

1. Leaching (mobilization) is the process when a solid material is dissolved into an aqueous solution. In other words, metals bound in minerals are transformed into metal ions that are released into an aqueous solution, i.e. immobilized metals become mobilized.
2. Extraction of metals from ores to aqueous solutions by means of chemicals.

Leaching occurs naturally but may also be applied as a preparatory step for metal recovery.

Rock which is unexposed and exposed to leaching. By courtesy of Barrie Johnson

Applied leaching

The process of extracting a component from a mixture by treating the mixture with a solvent which will dissolve the component but has no effect on the remaining portions of the mixture. Acid is commonly used to dissolve soluble metals contained in mineral ores. The resulting solution is subjected to further treatment to recover the desired metals. The non-soluble residue obtained is usually discarded.

Leaching can be done on primary raw materials like metal containing ores and concentrates or on secondary resources like metallic scrap or metal containing by-products. In many cases the raw materials are pre-treated by grinding, flotation, roasting etc. to enhance leaching yields and rates. Leaching is done with a number of different leaching reagents and methods. The metal rich aqueous solution obtained is subsequently processed for recovery of the metal.

Leaching can broadly be classified into two types, namely non-oxidative and oxidative leaching.

Non-oxidative leaching involves a chemical dissolution process using water, acid or an alkali as reagent. Some examples are the dissolution of oxidized copper ore (CuSiO3.2H2O) or zinc calcine (ZnO) using sulfuric acid (H2SO4) as reagent and the leaching of bauxite with sodium hydroxide (NaOH):

$CuSiO_3 . 2H_2O + H_2SO_4 \rightarrow CuSO_4 + SiO_2 + 3 H_2O$
$CuSiO_3 . 2H_2O + H_2SO_4 \rightarrow CuSO_4 + SiO_2 + 3 H_2O$
$ZnO + H_2SO_4 \rightarrow ZnSO_4 + H_2O$
$Al(OH)_3 + NaOH \rightarrow Na^+ + Al(OH)_4^-$

Oxidative leaching involves the use of oxidizing agents such as O2, Cl2, Fe3+, Cu2+ etc. For example Fe3+ is used as oxidant in the leaching of uranium from uraninite (UO2) and copper from chalcocite (Cu2S) respectively:

$UO_2 + 2Fe^{3+} \rightarrow UO_2^{2+} + 2Fe^2+$
$Cu_2S + 4Fe^{3+} \rightarrow 2Cu^{2+} + 4Fe^{2+}+S$

Under suitable conditions the ferric iron (Fe3+) can be regenerated by iron oxidising microorganisms as is practised in bioleaching operations.

Natural leaching

Leaching also occurs naturally, for example when rocks weather in contact with water resulting in the release of metals into the surroundings. One special case is the natural leaching of sulfide minerals. Leaching of sulfide minerals may be enhanced by microbes up to 100 000 times (Singer & Stumm, 1970)[1]. Without the microbes, natural leaching of sulfide minerals is a slow process.

Leaching Theory and Kinetics

Leaching is a heterogeneous reaction that takes place at the interface between a solid and liquid phase and sometimes also a gaseous phase. At the boundary between the two phases a diffusion layer is formed. In the case of a solid in an aqueous phase this layer consists of a stationary aqueous layer. The diffusion layer can be thinned by vigorous stirring but never be completely removed. Typical thickness of the diffusion layer in a well stirred system is in the range of 1-10 μm.

1. Diffusion of reagent through the diffusion layer
2. Adsorption of reagent on surface
3. Reaction on the surface
4. Desorption of product from surface
5. Diffusion of product through the diffusion layer

The slowest step in the leaching reaction is the rate-controlling step. Depending on which process is rate-controlling, three different type reactions may be obtained, i.e. reaction controlled leaching, diffusion controlled leaching and intermediate controlled leaching:

Factors affecting leaching kinetics

The equations arrived at for reaction controlled and diffusion controlled leaching have implications on operating costs. Such costs are for example size reduction by grinding, leaching temperature and agitation rate. Depending on the leaching mechanism, i.e. if the leaching process is reaction- or diffusion-controlled the leaching kinetics are influenced differently by variations of these parameters.

References

1. SINGER, P. C. & STUMM, W. (1970) Acid mine drainage: The rate determening step. Science, 167, 1121-1123.


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# Graph each polynomial function. Factor first if the expression is not in factored form. f(x)=(4x+3)(x+2)^{2}

Graph each polynomial function. Factor first if the expression is not in factored form. $f\left(x\right)=\left(4x+3\right){\left(x+2\right)}^{2}$
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Step 1 $f\left(x\right)=\left(4x+3\right){\left(x+2\right)}^{2}$ The function has two zeros $-\frac{3}{4},-2$ So, the graph of f(x) crosses the x-axis at $\left(\frac{4}{3},0\right)$ and $\left(-2,0\right)$ To find the y-intercept, substitute 0 for x in f(x) $f\left(x\right)=\left(4x+3\right){\left(x+2\right)}^{2}$
$f\left(0\right)=\left(4\left(0\right)+3\right){\left(0+2\right)}^{2}$ Substitute 0 for x Simplify $=\left(3\right){\left(2\right)}^{2}$
$=12$ So, the function f(x) crosses the y-axis at (0,12) Step 2 $4x\cdot {\left(x\right)}^{2}=4{x}^{3}$ The leading coefficient is 4 Since the leading coefficient is positive and the function f(x) of degree 3 (odd degree) So, the end behavior is $x\to \mathrm{\infty },f\left(x\right)\to \mathrm{\infty }$
$x\to -\mathrm{\infty },f\left(x\right)\to -\mathrm{\infty }$ See the graph below


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Adding contents list to the beginning of each chapter with minitoc

I am trying to use the minitoc package to add contents to the beginning of each chapter. It dosen't seem to work, I have seen in the documentation that there might be problems when using starred chapters or

\addcontentsline{toc}{chapter}...

But whatever i try to include it dosen't seem to work... any suggestions ? maybe I am misplacing the \dominitoc or missing something here

\documentclass[a4paper,12pt,oneside]{report}

\usepackage{hyperref}

\usepackage{minitoc}

% --------------------------------------------- %

\setcounter{secnumdepth}{3}
\setcounter{tocdepth}{3}
\pagenumbering{roman}

\begin{document}
\newcounter{rom}

\begin{titlepage}
Some text
\end{titlepage}

\newpage\thispagestyle{plain}
\newpage\thispagestyle{plain}
\setcounter{page}{3}

\chapter*{Declaration}
\chapter*{Acknowledgements}

\chapter*{Abstract}

\cleardoublepage
\phantomsection
\dominitoc
{
\tableofcontents
}

\newpage\thispagestyle{plain}~
\clearpage
\pagenumbering{arabic}

\chapter{Introduction}
\minitoc
\section{Motivation}
\newpage\cleardoublepage

\chapter{background}
\section{Motivation}
\newpage\cleardoublepage

\end{document}

For some reason adding \adjustmtc before the \minitoc solves the prolem.

The problem was arising after using \addcontentsline and so I solved the problem by adding just as many \adjustmtc. So for the code provided in my question, the solution would look like this:

...
\chapter{Introduction}


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# Simulations are finaly working!

Finally got the long-awaited simulation data. Left part show that when radial mode is totally cold, simulations agree with 1D theory. Right part shows that 1/f noise leads to some departure from the model. I don’t know if we see this.


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# Find the distribution of $X_1^2 + X_2^2$? [duplicate]

Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ which means (mean = 0, variance = $\sigma^2$) random variables. What is the distribution of $X_1^2 + X_2^2$?

My approach is that $X_1\sim N(0, \sigma^2)$ and $X_2\sim N(0, \sigma^2)$.

Transforming $X_1$ and $X_2$ into standard normal, $X_1/\sigma\sim N(0, 1)$ and $X_2/\sigma\sim N(0, 1)$.

Then $X_1^2/\sigma$ and $X_2^2/\sigma$ have chi-squared distribution with 1 degree of freedom.

Then I found the moment-generating function for $X_1^2$ and $X_2^2$;$$m_{X_1^2} = (1-2t)^{-1/2}$$ and $$m_{X_2^2} = (1-2t)^{-1/2}$$

So the moment generating function for $X_1^2 + X_2^2$ is $$m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}$$

So $X_1^2 + X_2^2$ has a chi-squared distribution with 2 degrees of freedom. My question can I treat $X_1^2/\sigma$ + $X_2^2/\sigma$ as $X_1^2$ + $X_2^2$ like I did above?

## marked as duplicate by Did, heropup, user61527, Claude Leibovici, user91500May 23 '14 at 5:52

• You mean $\frac{X_1^2}{\sigma^2}$ and $\frac{X_2^2}{\sigma^2}$ have chi-squared distributions with 1 degree of freedom. – afedder May 14 '14 at 21:11
• yes, it is what I mean – user111548 May 14 '14 at 21:12
• This question is a duplicate of this one by a different user or possibly the same user posting under a pseudonym because the work shown is exactly the same. – Dilip Sarwate May 14 '14 at 21:13
• So try deriving the moment-generating function with these scaled random variables (by the factor $\frac{1}{\sigma^2}$)...shouldn't be too hard if you already know how to derive the moment-generating function normally. – afedder May 14 '14 at 21:14
• @afedder: Do you think it should be $X_1/\sigma$~$N(0, 1)$ and $X_2/\sigma$~$N(0, 1)$ or $X_1/\sigma^2$~$N(0, 1)$ and $X_2/\sigma^2$~$N(0, 1)$.? – user111548 May 14 '14 at 21:16

Hint: Recall that if $X_1,X_2,...,X_n$ are independent and identically distributed as $\chi_{1}^{2}\,$, then $$\sum_{i=1}^{n} X_{i} \sim \chi_{n}^{2}$$ and that if $Z \sim N(0,1)\,$, then $$Z^2 \sim \chi_{1}^{2} \,\,.$$ Additional hint/spoiler: By the above, for independent $X,Y \sim \chi_{1}^{2}$ , it follows that $X+Y \sim \chi_2^2 \,.$ It can be shown that $X+Y \equiv W$ for $W \sim \text{Exp}(\frac{1}{2})$. You should verify this and then you are basically finished. To do this, prove that $X \sim \chi_n^2$ has density given by $$f(x \mid n) = \frac{1}{2^{n/2}\Gamma(n/2)}x^{n/2-1}e^{-x/2} \,\,\,\,\text{for x>0}\,.$$ Then, see that the density of $X \sim \chi_2^2$ is $$f(x \mid 2) = \frac{1}{2}e^{-\frac{1}{2}x} \,\,\,\,\text{for x>0}\,,$$ and this is the density of a random variable from an exponential distribution with parameter $\frac{1}{2}$.  Another relevant derivation: Suppose $X$ is a random variable and $Z = aX$ for some $a \in \mathbb{R} \backslash \{0\}$. Then the cumulative distribution function of $Z$ is given by $$F_Z(z) =\mathbb{P}(Z \leq z)=\mathbb{P}(aX \leq z)=\mathbb{P}\left(X\leq \frac{z}{a}\right)=F_X\left(\frac{z}{a}\right)\,\,,$$ where $F_X$ is the cumulative distribution function of $X$. Now, we derive the density function of $Z$, which we will denote $f_Z$, in the case that $a>0$ (this is the only case that applies here since $\sigma^2>0$): $$f_Z(z)=\frac{\text{d}}{\text{d}z}F_Z(z)=\frac{\text{d}}{\text{d}z}F_X\left(\frac{z}{a}\right)=\frac{1}{a}f_X\left(\frac{z}{a}\right) \,,$$ where $f_X$ is the density function of $X$. Final hint: The above hints are in the order of usage.  The following is a solution based on the above hints for future readers' benefit. Notice that we can standardize $X_1$ and $X_2$, so that $$\frac{X_i - 0}{\sigma} = \frac{X_i}{\sigma} \sim N(0,1) \,\,\,\,\,\text{for} \,\,i=1,2\,\,.$$ It follows that $$\left(\frac{X_i}{\sigma}\right)^2 \sim \chi_1^2 \,\,\,\,\,\text{for}\,\,i=1,2\,\,,$$ so that $$\left(\frac{X_1}{\sigma}\right)^2 + \left(\frac{X_2}{\sigma}\right)^2=\frac{1}{\sigma^2}(X_1^2 + X_2^2) \sim \chi_2^2\,\,.$$ Also, it is not difficult to verify that a $\chi_2^2$ random variable is equivalent in distribution to an $\text{Exp}(\frac{1}{2})$ random variable, so $$\frac{1}{\sigma^2}(X_1^2 + X_2^2) \sim \text{Exp}\left(\frac{1}{2}\right)\,.$$ Now, let $X = \frac{1}{\sigma^2}(X_1^2 + X_2^2)$ and $a=\sigma^2>0$, so that $Z=aX=X_1^2 + X_2^2$, and apply the last hint. We know that the density function of $X$ is given by $$f_X(x)=\frac{1}{2}e^{-\frac{1}{2}x},$$ and it follows that the density function of $Z$ is $$f_Z(z)= \frac{1}{\sigma^2}f_X\left(\frac{z}{\sigma^2}\right)=\frac{1}{2\sigma^2}e^{-\frac{1}{2\sigma^2}z} \,\,.$$ We recognize this as the density for a random variable from an exponential distribution with parameter $\frac{1}{2\sigma^2}$. In other words, $$X_1^2 + X_2^2 \sim \text{Exp}\left(\frac{1}{2\sigma^2}\right)\,\,.$$ @DilipSarwate @FelixMarin @user3001408, you might be interested in this derivation.

• @DilipSarwate: That's why I said to the OP to check what will happen to the distribution if you scale by $1/\sigma^{2}$ using the MGF derivation – afedder May 14 '14 at 21:44
• From the above, we know that $$\frac{1}{\sigma^{2}}(X_1^2 + X_2^2) \sim \chi_{2}^{2}.$$ This is certainly useful information, although it is not the entire solution. – afedder May 14 '14 at 21:51
• So $cU \sim \chi_{2}^{2}$ for a constant $c>0$ and a random variable $U$. Can we figure out the distribution of $U$? The answer is, of course, at least in this case. – afedder May 14 '14 at 21:54
• He pretty much spoon fed you the answer...did the entire calculation, all you have to do is conclude the distribution. He tells you the cumulative distribution for a general exponential random variable with parameter $\lambda$...all you have to do is compare the constants in his answer to $\lambda$ in the general function and you will figure out the parameter of this exponential distribution (he has already confirmed for you that it is exponential, I wouldn't have even done this much after going through the calculation). Do some work on your own homework, man. @user111548 – afedder May 17 '14 at 9:05
• The second equality is using the fact that the random variables are independent and the density function of a normal distribution with mean $0$ and variance $\sigma^{2}$...the third equality is a shift from rectangular to polar coordinates - this type of substitution is learned in a multivariable calculus course (usually required for an intro probability theory course). Recall that $\text{d}x\text{d}y$ is substituted with $r\text{d}r\text{d}\theta$ and that $r^2 = x^2+y^2$. – afedder May 17 '14 at 9:25

You have $$X_{i}^{2}=\sigma^{2}(Z^{2})=\sigma^{2}\Gamma(\frac{1}{2},2)=\Gamma(\frac{1}{2},2\sigma^{2})$$

Therefore we have $$X_{1}^{2}+X_{2}^{2}=\Gamma(1,2\sigma^{2})$$

where we used property of $\Gamma$-distribution.

• Remember, $\Gamma(\alpha,\lambda)$ is special for $\alpha=1$. – afedder May 17 '14 at 12:43
• Also, this isn't a formal proof whatsoever. – afedder May 17 '14 at 12:46
• No, but I believe it should be readable to OP and he/she should be able to do it himself. I have no intention to fill in detail for his/her homework. – Bombyx mori May 17 '14 at 12:59
• Also, most intro probability classes don't teach the properties of the gamma distribution to the extent that you are using here. - the most important property for this is the constant multiple of a chi-squared random variable of $n$ degrees of freedom (or $\Gamma(\frac{n}{2},\frac{1}{2})$ random variable), which you seem to use as if it is a trivial leap. – afedder May 17 '14 at 13:14
• 2nd and 3rd equalities (and even 1st in terms of notation) would be very confusing to me at the level of the OP – afedder May 17 '14 at 13:19

Not really - recall that the mgf is $$m_X(t) = \mathbb{E}\left[e^{tX}\right]$$ and if you rescale $X$ by a constant $\sigma$, what happens to the result?

• Based on what you said, you think $X_1/\sigma$~$N(0, 1)$ and $X_2/\sigma$~$N(0, 1)$ is wrong? – user111548 May 14 '14 at 21:17
• @user111548 no. I meant that the mgf of $X$ and mgf of $X/\sigma$ are not the same thing. – gt6989b May 14 '14 at 21:18
• ok...Before I got into the mgf. Do you agree that $X_1 /σ$ ~$N(0,1)$ and $X_2 /σ$ ~$N(0,1)$ and $X_1^2/\sigma^2$ has a chi-squared distribution with 1 degree of freedom? – user111548 May 14 '14 at 21:21
• @user111548 Yes. – gt6989b May 14 '14 at 21:22

\begin{align} &\color{#00f}{\large\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} {\expo{-X_{1}^{2}/\pars{2\sigma^{2}}} \over \root{2\pi}\sigma} \,{\expo{-X_{2}^{2}/\pars{2\sigma^{2}}} \over \root{2\pi}\sigma} \delta\pars{X - X_{1}^{2} - X_{2}^{2}}\,\dd X_{1}\,\dd X_{2}} \\[3mm]&={1 \over 2\pi\sigma^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{-\pars{X_{1}^{2} + X_{2}^{2}}/\pars{2\sigma^{2}}} \delta\pars{X - X_{1}^{2} - X_{2}^{2}}\,\dd X_{1}\,\dd X_{2} \\[3mm]&=\Theta\pars{X}\,{\expo{-X/\pars{2\sigma^{2}}} \over 2\pi\sigma^{2}}\times \\[3mm]&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \!\!\!\!\!\Theta\pars{\root{X} - \verts{X_{1}}}\bracks{% {\delta\pars{X_{2} + \root{X - X_{1}^{2}}} \over 2\verts{X_{2}}} +{\delta\pars{X_{2} - \root{X - X_{1}^{2}}} \over 2\verts{X_{2}}}} \,\dd X_{1}\,\dd X_{2} \\[3mm]&=\Theta\pars{X}\,{\expo{-X/\pars{2\sigma^{2}}} \over 2\pi\sigma^{2}} \int_{-\root{X}}^{\root{X}}{\dd X_{1} \over \root{X - X_{1}^{2}}} =\Theta\pars{X}\, {\expo{-X/\pars{2\sigma^{2}}} \over 2\pi\sigma^{2}}\bracks{2\arcsin\pars{1}} \\[3mm]&=\color{#00f}{\large\Theta\pars{X}\, {\expo{-X/\pars{2\sigma^{2}}} \over 2\sigma^{2}}} \end{align}

$\ds{\Theta\pars{x}}$ is the Heaviside Step Function. $\ds{\delta\pars{x}}$ is the Dirac Delta Function.

• What is $\Theta(X)$? – Dilip Sarwate May 15 '14 at 3:32
• @DilipSarwate I added two links at the answer end. I guess we are not allowed to put more than one link in a comment. Thanks a lot. – Felix Marin May 15 '14 at 3:56
• Thanks for clarifying the notation. You might want to take a look at this answer of mine to a question which is a duplicate of this question. – Dilip Sarwate May 15 '14 at 4:14
• This is like killing a kitten with a bazooka, in my opinion. – afedder May 15 '14 at 6:26
• @afedder It's likely true but it's a nice technique in more complicated situations. It avoids to think about limits since those functions take care of that. For me it's quite useful since I'm pretty bad with drawing. It can save your life when the real messy shows up. I saw other solutions somewhere with inequalities and a derivative to the end. The $\Theta$ can handle the inequalities and the derivative yields the $\delta$ such that we avoid one step and "the drawing". We have a saying in spanish for your idea ( translated ): "To kill a cockroach with a machine gun". Thanks a lot. – Felix Marin May 15 '14 at 6:51

You can approach it like this:

1) Calculate the distribution of $X_1^2$ and $X_2^2$, individually. Call them $f(X_1)$, and $g(X_2)$.

2) Now the second step. Now you calculate the distribution of the sum of $f(X_1)$ and $g(X_2)$ via convolution integral.

This can be one way of calculating what you are asking!

Note:

To calculate the distribution of $X_1^2$ (e.g.), you can use the CDF method. For example $F_{X_1^2}(x)=P(X_1^2 \le x)$. Now you express it in terms of $X_1$, and then differentiate to get the PDF.

• What a perfectly dreadful way of finding the answer where no advantage is being taken of the fact that $X_1$ and $X_2$ are zero-mean independent normal random variables with the same variance. This approach would "work" in general for arbitrary independent random variables but why bother trying to touch your ear after wrapping your arm around your head two times? – Dilip Sarwate May 14 '14 at 23:04
• I would be interested to learn if you have something better to add. – user3001408 May 14 '14 at 23:09
• Sure. Look at this answer to the question which I have already pointed (see my comment on the main question) as a duplicate of this one. – Dilip Sarwate May 14 '14 at 23:22
• @dilip Yes, it is good! I agree convolution integral is terrible when you have to add, say 5, or 50 independent random variables. But convolution integral is somewhat intuitive to start with and understand what is going on. In addition, it has a nice application if you can connect with Fourier transform!!! – user3001408 May 15 '14 at 16:02


text:
### back_slash's blog

By back_slash, history, 9 months ago, ,

I am writing this blog mainly for 2 reasons. One is to discuss the problems. I want to know what is the logic behind C Large. And also if we can get a count of how many people got full points in the round.

•
• +25
•

 » 9 months ago, # |   +24 HintMatrix Exponentiation hinttry to find probabilities of starting at each vertex after some x rounds. And now with small trick(not hard) to take sum during exponentiation it is done.I leave tiny details for you to try out.
•  » » 9 months ago, # ^ |   0 Hint for B please ?
•  » » » 9 months ago, # ^ |   0 It's Greedy. First sort the given array. Then check if you can dance. If you can't dance, check if you can recruit. From the available teams, for dancing, dance with the team which has least energy. And for recruiting, recruit the team which has highest energy. Main logic sort(a.begin(), a.end()); int res = 0; int gres = 0; while(a.size()){ if(e <= a[0]){ if(res){ --res; e += a.back(); a.erase(--a.end()); } else break; } else{ ++res; e -= a[0]; gres = max(gres, res); a.erase(a.begin()); } }
•  » » » 9 months ago, # ^ |   +6 The following observations were important: You can process the teams in any order (ie. decide to dance, recruit or truce) because if he current one isn't the one you want to process you can just make an excuse Recruiting is most beneficial when done against highest energy teams Dancing is most beneficial when done against lowest energy teams
•  » » 9 months ago, # ^ | ← Rev. 2 →   0 Can you elaborate a bit more please(for C large) ?
 » 9 months ago, # |   +15 My solutionsProblem 1. AnsRecursion. It's true if the list is empty or when pivoting as described, one of the sublists is empty and the recursive call is true. def solve(N, A): if N == 0: return True j = (len(A) - 1) // 2 B, C, P = [], [], A[j] for i, x in enumerate(A): if i == j: continue (B if x <= P else C).append(x) if B and C: return False return solve(N-1, B or C) Problem 2. AnsBecause we can excuse or ignore teams, we can choose any subset of the teams in any order. Also, if we are to beat a team, we should beat the weakest one; and if we are to recruit a team, we should recruit the strongest one. Thus our strategy is going to take the form of: "Defeat teams until we can't, then recruit a team" and we will repeat this. def solve(E, N, A): A.sort() ans = honor = 0 while A: while A and E > A[0]: E -= A.pop(0) honor += 1 if honor == 0: break ans = max(ans, honor) if A: honor -= 1 E += A.pop() return ans Problem 3. AnsFirst, use Floyd's algorithm to find dist[i][j] (0 indexed). Now, notice that if you are not home, you are in every other location with equal probability. If you are at home (location 0) then the expected travel time to your destination is sum(dist[0]) / (N-1). If you are away (ie., not at home), then the expected travel time to your destination is sum(dist[1:]) / (N-1)^2. Also, we can figure out the transition function: if you are home with probability x, then on the next monster you are home with probability (1-x)/(N-1).To solve the problem for large, you need one extra fact. For the function f(x) = (1-x)/(N-1), getting within 1e-9 of the limit of the orbit lim_{i ->inf} f^i(1) = 1/N is done very quickly. Indeed, let's look at abs(x — f(x)) = abs(Nx — 1)/(N-1), suggesting f^i(1) -> 1/N. To prove it, abs(f(x) — f(f(x))) = abs(Nf(x) — 1)/(N-1) = abs(N(1-x) — (N-1))/(N-1)^2 = abs(Nx — 1)/(N-1)^2. So the differences are getting smaller by a factor of 1/(N-1) each time. def solve(N, M, P, A): dist = [[float('inf')] * N for _ in xrange(N)] for u, v, w in A: u -= 1; v -= 1 dist[u][v] = dist[v][u] = w dist[u][u] = dist[v][v] = 0 for k in xrange(N): for i in xrange(N): for j in xrange(N): d = dist[i][k] + dist[k][j] if dist[i][j] > d: dist[i][j] = d val_home = sum(dist[0]) / float(N - 1) val_away = sum(sum(row) for row in dist[1:]) / float((N-1)**2) ans = 0 home = 1.0 MAGIC = 200 for _ in xrange(min(MAGIC, P)): delta = val_home * home + val_away * (1.0 - home) ans += delta home = (1.0 - home) / float(N-1) return ans + delta * max(0, P - MAGIC) Problem 4. AnsStraightforward DP. _issq = lambda N: int(N**.5) ** 2 == N issq = map(_issq, range(10001)) memo = {0: 0} def solve(N): if N not in memo: ans = 4 for x in xrange(1, N+1): if issq[x]: ans = min(ans, 1 + solve(N-x)) memo[N] = ans return memo[N] map(solve, range(10000))
 » 9 months ago, # |   -27 For problem B, my solution is incorrect. Can some body help here?My code #include using namespace std; int energy, N; vector vec; int ans; int funcc(vector vv, int ind, int en){ if(ind == vv.size()) return 0; int x = vv[ind]; int ret = INT_MIN; //dance if(en - x > 0) ret = 1 + funcc(vv, ind + 1, en - x); //truce ret = max(ret, funcc(vv, ind + 1, en)); //recruit ret = max(ret, funcc(vv, ind+1, en + x) - 1); //delay if(ind < vec.size()){ vv.push_back(x); ret = max(ret, funcc(vv, ind + 1, en)); } return ret; } int main(){ freopen("B-small-attempt1.in","r",stdin); freopen("output_file_name.out","w",stdout); int tc; cin>>tc; for(int t = 1; t <= tc; t++){ cin>>energy>>N; vec.clear(); int x; for(int i=0; i>x; vec.push_back(x); } ans = 0; ans = funcc(vec, 0, energy); cout<<"Case #"<
 » 9 months ago, # |   +8 What is with the ranking? It shows my rank when I am logged in but not otherwise??


text:
# Does digit $6$ always lead to $\ 25921=161^2\$?

Consider prime numbers with the property that the product of the factorials of the digits plus $$1$$ is a perfect square, for example the prime $$30241$$ leads to the square $$3!\cdot 0!\cdot 2!\cdot 4!\cdot 1!+1=17^2$$

If the prime number starts with digit $$6$$, the only square seems to be $$\ 25921=161^2\$$ which appears if the prime contains of one digit $$6$$ , two digits $$3$$ and the rest digits $$0$$ or $$1$$.

Is another square possible, when the prime starts with digit $$6$$ ? To this, we would need factorials below $$10!$$ , at least one of them $$6!$$ , such that their product plus $$1$$ is a perfect square , other than $$6!\cdot 3!\cdot 3!+1$$

• Is there any sort of evidence except for the fact that you did not find any example ? – Captain Lama Apr 17 at 19:49
• In your last paragraph you correctly note that the equation only depends on the digits of the number, not their order, so it makes sense to search on collections of digits. The ten digits can be ordered in $10!,$ about $3$ million ways, so just checking them in one order can save a lot of time. As you get lots of digits it is very likely that some permutation of them will be prime, so that seems the last thing to check. Demanding the number be prime seems strange as it is unrelated to whether the product of factorials is one less than a square, but it is your question. – Ross Millikan Apr 17 at 20:25
• @RossMillikan I did that and according to my program, we need at least $30$ factorials to get another perfect square. I do not think that the primality is a great restriction because we can insert ones and zeros arbitarily. – Peter Apr 17 at 20:29

A comment on strategy too long for the comment section, but not a complete answer:

You want to find $$6!\prod a_i! +1=x^2$$ where $$a_i$$ are single digits. Note that the single digits are $$0,1,2,3,2^2,5,2\cdot 3,7,2^3,3^2$$, so the factorials will all either be $$1$$ or will have only the prime factors $$2,3,5,7$$.

Rearranging, $$6!\prod a_i!=x^2-1=(x-1)(x+1)$$. LHS is even, so $$(x-1)\ \text{and}\ (x+1)$$ are both even. Successive even numbers have only a single factor of $$2$$ in common.

$$2^2\cdot3^2\cdot5\cdot\prod a_i!=\frac{(x-1)}{2}\frac{(x+1)}{2}$$. RHS are two consecutive numbers; hence their gcd is $$1$$. So the first problem is to find consecutive numbers which only have prime factors of $$2,3,5,7$$, and none in common.

$$161$$ leads to a solution because $$\frac{161-1}{2}=80=2^4\cdot 5;\ \frac{161+1}{2}=81=3^4$$.

$$17$$ leads to a solution (although not for the $$6$$ question) because $$\frac{17-1}{2}=8=2^3;\ \frac{17+1}{2}=9=3^2$$.

Finding candidate consecutive numbers will not necessarily lead to a solution, as the numbers of various prime factors will have to be distributable among the $$a_i!$$, which might prove difficult if (for example) the number of factors of $$2$$ is less than the number of factors of $$3$$ in the consecutive numbers. If consecutive numbers can be found that satisfy that requirement, then identifying suitable $$a_i$$ will be possible, and as OP points out, it should then be possible to generate a prime starting number by strategic insertion of digits $$0,1$$.

• That's a fast method indeed, because all 7-smooth numbers up to $x$ can be checked in $\log^4(x)$ time. I have checked using this so far up to $x=10^{100}$ (within which there are exactly 51,428,828 7-smooth numbers) and no other solutions can be found – Shieru Asakoto May 5 at 2:12

# The answer is: $$25921=161^2$$ is the only possible square.

This solution is based on Keith's partial solution on the requirements of the solution if one exists, and Strømer's Theorem, which is a theorem on the finitude of smooth pairs.

Keith pointed out that the necessary condition for a solution $$x^2$$ fulfilling the conditions of the question is that both $$\frac{x-1}{2}$$ and $$\frac{x+1}{2}$$ are 7-smooth. Another condition will be that, if both $$\frac{x-1}{2}$$ and $$\frac{x+1}{2}$$ are not divisible by 7, then $$180\mid\frac{(x-1)(x+1)}{4}$$, or otherwise $$907200\mid\frac{(x-1)(x+1)}{4}$$.

Strømer's Theorem states that there are only finitely many pairs of consecutive smooth numbers up to any specific degree of smooth. An algorithm invented by Strømer, later simplified by Lehmer, can be used to find all pairs of consecutive smooth numbers. As a result, we can inspect all 7-smooth pairs in reasonable time for such a solution.

The algorithm states that we need only $$2^{\pi(k)}-1$$ Pell equations to find all $$k$$-smooth pairs. Let $$P$$ be the set of prime numbers which are at most $$k$$, then the equations we need to solve are the following:

$$x^2-2(\prod_{q\in Q} q)y^2=1\text{ for every }Q\in (\mathcal{P}(P)\setminus\{2\})$$

We only need to inspect the first $$\max(3, \frac{k+1}{2})$$ solutions per equation, and each solution $$(x,y)$$ for the Pell equation gives a pair $$(\frac{x-1}{2},\frac{x+1}{2})$$ which may or may not be both $$k$$-smooth. For a Pell equation with the form $$x^2-ay^2=1$$, by analyzing the continued fraction of $$\sqrt{a}$$ the solutions are easy to find.

In this case, we need to find all 7-smooth pairs, so we need to solve $$2^{\pi(k)}-1=15$$ equations, and for each equation we need $$\max(3, \frac{7+1}{2})=4$$ pairs of solutions:

• $$x^2-2(1)y^2=1$$.
• $$\sqrt{2}=[1;2,2,\cdots]$$, the first 4 solutions are $$(3,2)$$, $$(17,12)$$, $$(99,70)$$ and $$(577,408)$$. These generate 3 7-smooth pairs: $$(1,2)$$, $$(8,9)$$, $$(49,50)$$.
• $$x^2-2(3)y^2=1$$.
• $$\sqrt{6}=[2;2,4,2,4,\cdots]$$, the first 4 solutions are $$(5,2)$$, $$(49,20)$$, $$(485,198)$$ and $$(4801,1960)$$. These generate 3 7-smooth pairs: $$(2,3)$$, $$(24,25)$$, $$(2400,2401)$$.
• $$x^2-2(5)y^2=1$$.
• $$\sqrt{10}=[3;6,6,\cdots]$$, the first 4 solutions are $$(19,6)$$, $$(721,228)$$, $$(27379,8658)$$ and $$(1039681,328776)$$. These generate 1 7-smooth pair: $$(9,10)$$.
• $$x^2-2(2\cdot 3)y^2=1$$.
• $$\sqrt{12}=[3;2,6,2,6,\cdots]$$, the first 4 solutions are $$(7,2)$$, $$(97,28)$$, $$(1351,390)$$ and $$(18817,5432)$$. These generate 2 7-smooth pairs: $$(3,4)$$, $$(48,49)$$.
• $$x^2-2(7)y^2=1$$.
• $$\sqrt{14}=[3;1,2,1,6,1,2,1,6,\cdots]$$, the first 4 solutions are $$(15,4)$$, $$(449,120)$$, $$(13455,3596)$$ and $$(403201,107760)$$. These generate 2 7-smooth pairs: $$(7,8)$$, $$(224,225)$$.
• $$x^2-2(2\cdot 5)y^2=1$$.
• $$\sqrt{20}=[4;2,8,2,8,\cdots]$$, the first 4 solutions are $$(9,2)$$, $$(161,36)$$, $$(2889,646)$$ and $$(51841,11592)$$. These generate 2 7-smooth pairs: $$(4,5)$$, $$(80,81)$$.
• $$x^2-2(2\cdot 7)y^2=1$$.
• $$\sqrt{28}=[5;3,2,3,10,3,2,3,10,\cdots]$$, the first 4 solutions are $$(127,24)$$, $$(32257,6096)$$, $$(8193151,1548360)$$ and $$(2081028097,393277344)$$. These generate 1 7-smooth pair: $$(63,64)$$.
• $$x^2-2(3\cdot 5)y^2=1$$.
• $$\sqrt{30}=[5;2,10,2,10,\cdots]$$, the first 4 solutions are $$(11,2)$$, $$(241,44)$$, $$(5291,966)$$ and $$(116161,21208)$$. These generate 1 7-smooth pair: $$(5,6)$$.
• $$x^2-2(3\cdot 7)y^2=1$$.
• $$\sqrt{42}=[6;2,12,2,12,\cdots]$$, the first 4 solutions are $$(13,2)$$, $$(337,52)$$, $$(8749,1350)$$ and $$(227137,35048)$$. These generate 2 7-smooth pairs: $$(6,7)$$, $$(4374,4375)$$.
• $$x^2-2(2\cdot 3\cdot 5)y^2=1$$.
• $$\sqrt{60}=[7;1,2,1,14,1,2,1,14,\cdots]$$, the first 4 solutions are $$(31,4)$$, $$(1921,248)$$, $$(119071,15372)$$ and $$(7380481,952816)$$. These generate 1 7-smooth pair: $$(15,16)$$.
• $$x^2-2(5\cdot 7)y^2=1$$.
• $$\sqrt{70}=[8;2,1,2,1,2,16,2,1,2,1,2,16,\cdots]$$, the first 4 solutions are $$(251,30)$$, $$(126001,15060)$$, $$(63252251,7560090)$$ and $$(31752504001,3795150120)$$. These generate 1 7-smooth pair: $$(125,126)$$.
• $$x^2-2(2\cdot 3\cdot 7)y^2=1$$.
• $$\sqrt{84}=[9;6,18,6,18\cdots]$$, the first 4 solutions are $$(55,6)$$, $$(6049,660)$$, $$(665335,72594)$$ and $$(73180801,7984680)$$. These generate 1 7-smooth pair: $$(27,28)$$.
• $$x^2-2(2\cdot 5\cdot 7)y^2=1$$.
• $$\sqrt{140}=[11;1,4,1,22,1,4,1,22,\cdots]$$, the first 4 solutions are $$(71,6)$$, $$(10081,852)$$, $$(1431431,120978)$$ and $$(203253121,17178024)$$. These generate 1 7-smooth pair: $$(35,36)$$.
• $$x^2-2(3\cdot 5\cdot 7)y^2=1$$.
• $$\sqrt{210}=[14;2,28,2,28,\cdots]$$, the first 4 solutions are $$(29,2)$$, $$(1681,116)$$, $$(97469,6726)$$ and $$(5651521,389992)$$. These generate 1 7-smooth pair: $$(14,15)$$.
• $$x^2-2(2\cdot 3\cdot 5\cdot 7)y^2=1$$.
• $$\sqrt{420}=[20;2,40,2,40,\cdots]$$, the first 4 solutions are $$(41,2)$$, $$(3361,164)$$, $$(275561,13446)$$ and $$(22592641,1102408)$$. These generate 1 7-smooth pair: $$(20,21)$$.

From the calculations above, it turns out that there are only 23 pairs of consecutive 7-smooth numbers. 18 of them are obviously too small to fulfill the second condition, so in order to save space I will take them out, remaining 5 candidate pairs:

• $$15, 16$$: $$180\nmid15\cdot16$$
• $$24, 25$$: $$180\nmid24\cdot25$$
• $$\color{red}{80, 81}$$: This is the case of $$6!\cdot 3!\cdot 3!+1$$.
• $$2400, 2401$$: $$907200\nmid2400\cdot2401$$*
• $$4374, 4375$$: $$907200\nmid4374\cdot4375$$*

*: One of the numbers is divisible by 7.

Most of the solutions are ruled out because of the second condition stated in the second paragraph. As there is only one solution fulfilling $$2^2\cdot 3^2\cdot 5\prod a_i!=\frac{(x-1)(x+1)}{4}$$ as in Keith's partial answer, your $$6$$ always leads to $$25921=161^2$$.

Footnote: I ran the exhaustion search up to $$x=10^{425}$$ using Keith's method before I found how to reach the conclusion of having no other solutions.


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## CryptoDB

### Thomaz Oliveira

#### Publications

Year
Venue
Title
2019
JOFC
In this work, we retake an old idea that Koblitz presented in his landmark paper (Koblitz, in: Proceedings of CRYPTO 1991. LNCS, vol 576, Springer, Berlin, pp 279–287, 1991 ), where he suggested the possibility of defining anomalous elliptic curves over the base field ${\mathbb {F}}_4$ F 4 . We present a careful implementation of the base and quadratic field arithmetic required for computing the scalar multiplication operation in such curves. We also introduce two ordinary Koblitz-like elliptic curves defined over ${\mathbb {F}}_4$ F 4 that are equipped with efficient endomorphisms. To the best of our knowledge, these endomorphisms have not been reported before. In order to achieve a fast reduction procedure, we adopted a redundant trinomial strategy that embeds elements of the field ${\mathbb {F}}_{4^{m}},$ F 4 m , with m a prime number, into a ring of higher order defined by an almost irreducible trinomial. We also suggest a number of techniques that allow us to take full advantage of the native vector instructions of high-end microprocessors. Our software library achieves the fastest timings reported for the computation of the timing-protected scalar multiplication on Koblitz curves, and competitive timings with respect to the speed records established recently in the computation of the scalar multiplication over binary and prime fields.
2016
CHES
2014
EPRINT
2014
EPRINT
2013
CHES
2010
EPRINT
Luffa is a new hash algorithm that has been accepted for round two of the NIST hash function competition SHA-3. Computational efficiency is the second most important evaluation criteria used to compare candidate algorithms. In this paper, we describe a fast software implementation of the Luffa hash algorithm for the Intel Core 2 Duo platform. We explore the use of the perfect shuffle operation to improve the performance of 64-bit implementation and 128-bit implementation with the Intel Supplemental SSSE3 instructions. In addition, we introduce a new way of implementing Luffa based on a Parallel Table Lookup instruction. The timings of our 64-bit implementation (C code) resulted in a 16 to 32% speed improvement over the previous fastest implementation.


text:
# cirq.flatten_with_params¶

cirq.flatten_with_params(val: Any, params: Optional[Union[cirq.ParamResolver, Dict[Union[str, sympy.core.symbol.Symbol], Union[float, str, sympy.core.basic.Basic]]]]) → Tuple[Any, Dict[Union[str, sympy.core.symbol.Symbol], Union[float, str, sympy.core.basic.Basic]]][source]
Creates a copy of val with any symbols or expressions replaced with
new symbols. val can be a Circuit, Gate, Operation, or other
type. Also transforms a dictionary of symbol values for val to an
equivalent dictionary mapping the new symbols to their evaluated values.
flatten_with_params goes through every parameter in val and does the
following:
• If the parameter is a number, don’t change it.

• If the parameter is a symbol, don’t change it and use the same symbol with the same value in the new dictionary of symbol values.

• If the parameter is an expression, replace it with a symbol and use the new symbol with the evaluated value of the expression in the new dictionary of symbol values. The new symbol will be sympy.Symbol('<x + 1>') if the expression was sympy.Symbol('x') + 1. In the unlikely case that an expression with a different meaning also has the string 'x + 1', a number is appended to the name to avoid collision: sympy.Symbol('<x + 1>_1').

Parameters
• val – The value to copy and substitute parameter expressions with

• symbols. (flattened) –

• params – A dictionary or ParamResolver where the keys are sympy.Symbols used by val and the values are numbers.

Returns

The tuple (new value, new params) where new value is val with flattened expressions and new params is a dictionary mapping the new symbols like sympy.Symbol(‘<x + 1>’) to numbers like params[‘x’] + 1.


text:
PL EN

Preferencje
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## Commentationes Mathematicae

2008 | 48 | 2 |
Tytuł artykułu

### A spectral synthesis property for $$C_b(X, \Beta)$$

Autorzy
Treść / Zawartość
Warianty tytułu
Języki publikacji
EN
Abstrakty
EN
Let $$(C_b (X) , \Beta)$$ be the algebra of all continuous bounded real or complex valued functions defined on a completely regular Hausdorff space $$X$$ with the usual algebraic operations and with the strict topology $$\Beta$$. It is proved that $$(C_b (X) , \Beta)$$ has a spectral synthesis, i.e. every of its closed ideals is an intersection of closed maximal ideals of codimension 1. We give one necessary and two sufficient conditions over $$X$$ in order that $$(C_b (X) , \Beta)$$ has no proper non-zero closed principal ideals. Moreover if $$X$$ satisfy any of these two conditions and is also a k-space, then any non zero element of $$C_b(X)$$ is invertible or a topological divisor of zero.
Słowa kluczowe
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Tom
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2008
online
2017-12-19
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text:
# Math Help - quadratic question

the profit, P(x), of a video company, in thousands of dollars, is given by P(x)=-5x^2 + 550x - 5000. where x is the amount spent on advertising, in thousands.

a. determine the maximum profit that the company can make. i got $16,125.00 b. determine the amount spent on advertising that will result in a profit. i keep gettin a negative number which i know cannot b rite.bt i am using the quadratic formula. c. determine the amounts spent on advertising that will result in a profit of at least$4,000,000.00

2. Originally Posted by justine_0612
the profit, P(x), of a video company, in thousands of dollars, is given by P(x)=-5x^2 + 550x - 5000. where x is the amount spent on advertising, in thousands.

a. determine the maximum profit that the company can make. i got $16,125.00 this is incorrect. find the x-coordinate for the vertex and plug it into the formula. the result for P(x) is the desired profit. recall: For a quadratic $ax^2 + bx + c = 0$, the x-coordinate of the vertex is given by: $x = \frac {-b}{2a}$ b. determine the amount spent on advertising that will result in a profit. i keep gettin a negative number which i know cannot b rite.bt i am using the quadratic formula. again, you are way off here. BOTH roots are positive, so you did something wrong. recall: for a quadratic $ax^2 + bx + c = 0$, the roots of the quadratic are given by: $x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$ try again c. determine the amounts spent on advertising that will result in a profit of at least$4,000,000.00
are you sure you didn't mean to write 4000? because that is impossible

3. nope its 4,000,000.00
so the sheet seys...
bt im not quite sure wat u were talkin abt b4. can u elaborate or explain thouroughly. i'm still tryin to learn this.

4. I think there's a lot of confusion here over the 'in thousands' comment.
This is how I interpret it:

$-5x^2+550x-5000$

Where x represents money spent on advertising 'in thousands'.

Therefore, any output will be 'in thousands'.

So if you spend $50,000 on advertising, then x = 50 $-5(50)^2+550(50)-5000=10000$ So you gain$10,000,000 profit.

But remember this is just my interpretation.

For Question A complete the square on $P(x)=-5x^2+550x-500$. This will give you the turning point. And from its graph you will find the turning point is the place of highest profit. Remember that it is 'in thousands'.

For Question B use the quadratic formula as Jhevon indicated. From the shape of the graph you can see that the 'region of profit' will be in between the two roots. It is very similar to my solution to Question C, only this time set $-5x^2+500x-5000 \geq 0$

For Question C, set

$-5x^2 + 550x - 5000 \geq 4000$

$-5x^2+550x-9000 \geq 0$

$x^2-110+1800 \geq 0$

I would just cut straight to the quadratic formula here

$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-110) \pm \sqrt{(-110)^2 - 4 \times 1800\times 1}}{2 \times 1}$

$x=\frac{110\pm 70}{2}$

$x=55\pm 35$

$x=20$ or $x=90$ "thousands" on advertising.

So the places when x gives 4,000,000 profit is at x = 20 and x = 90. If you look at the shape of the graph of $x^2-110+1800$, you will see that it is an inverted parabola. This can be deduced from the fact that the $x^2$ coefficient is negative. Then it is greater than zero between x = 20 and x = 90.

Therefore, if you want a profit of at least 4,000,000

Then $20 \geq x \geq 90$

5. i understand everything u have sed there...bt how did u interpret x=50?? i neva had a value of 50,000

6. Originally Posted by justine_0612
i understand everything u have sed there...bt how did u interpret x=50?? i neva had a value of 50,000
the units for the formula is thousands. so x = 50 means 50,000 was spent on advetising. now the last question makes more sense. the 4000000 should be 4000 for the "scale" we are using.

the formula i gave you for the vertex calculates the turning point a lot quicker and easier than completing the square does. but do it whichever way you understand


text:
# Of elitists and laypeople

## Spoilers Game of Thrones

Spoilers Game of Thrones.

I have been watching Game of Thrones with great interest the past few weeks. It has very strongly highlighted a struggle that has been gripping my mind for a while now: That between elitists and laypeople. And I find myself in a strange in-between.

For those not in the know, the latest season of Game of Thrones is a bit controversial to say the least. If you skip past the internet vitriol, you’ll find a lot of people disliking the season for legitimate reasons: The battle tactics don’t make any sense, characters miraculously survive after the camera cuts away, time and distance stopped being an issue in a setting that used to take it slow, and there’s a weird, forced conflict that would go away entirely if these two characters that are already in love would simply marry. And the list goes on, I’m sure.

But on the other hand, there appears to be a large body of laypeople who watch and enjoy the series. Millions of people tune in every week to watch fictional people fight over a fictional throne, and they appear to be enjoying themselves. And me? Sure, I’m enjoying myself as well. I had muscle aches from the tension of watching The Long Night, and nothing gripped me more than the half-botched assassination attempt at the end of the episode.

So what gives? On the one hand enthusiasts are rightly criticising the writers for some very strange decisions, but on the other hand millions of people are enjoying the series all the same.

## Of power users and newbies

I frankly don’t know the answer to the posed question, but I do know an analogy that prompted me to write this blog post. I am a humble contributor to the GNOME Project, chiefly as translator for Esperanto, but also miniscule bits and bobs here and there. GNOME faces a similar problem with detractors: They have their complaints about systemd, customisability, missing power user features, themes breaking, and so forth. And I’m sure they have some valid points, but GNOME remains the default desktop environment on many distributions, and many people use and love GNOME as I do.

These detractors often run some heavily customised Arch Linux system with some unintuitive-but-efficient window manager, and don’t have any editor other than Vim installed. Or in other words: They run a system that the vast majority of people could not and do not want to use.

And I understand these people, because in one aspect of my digital life, I have been one of them. For at least two years, I ran Spacemacs as my primary editor. For a while I even did my e-mail through that program, and I loved it. Kind of. Sure, everything was customisable, and the keyboard shortcuts were magically fast, but the mental overhead of using that program was slowly grinding me down. Some menial task that I do infrequently would turn out to involve a non-intuitive sequence of keys that you just simply need to know, and I would spend far too long on figuring that out. Or I would accidentally open Vim inside of the Emacs terminal emulator, and :q would be sent to Emacs instead of the emulator. Sure, if you know enough Emacs wizardry, you could easily escape this situation, but that’s the point, isn’t it? The wizardry involved takes effort that I don’t always want to put in, even if I know that it pays off. Kind of.

These days I use VSCodium, a Free Software version of Visual Studio Code. I like it well enough for a multitude of reasons, but mainly because the mental overhead of using this editor is a lot lower. Even so, is Emacs a better editor? Probably. If I could be bothered to maintain my Emacs wizarding skills, I am fairly certain that it would be the perfect editor. But that’s a big if. So that’s why I settle for VSCodium. And the same line of reasoning can be extended to why I use and love GNOME.

## Back to Westeros

Having made that analogy, can it be mapped onto the kerfuffle surrounding Game of Thrones? Is it a matter of a small group wanting an intricate, advanced plot and a larger group wanting a simple, rudimentary story, because they can’t or don’t want to deal with a complicated story?

It seems that way, but the damnedest thing is that I don’t know. I like the latest season of Game of Thrones for what it is: An archetypical fight of good versus evil. The living gathered together to fight an undead army, and the living won. Such a story is a lot easier to get into as a layperson, and there is nothing wrong with enjoying simple, archetypical stories.

But that’s not what Game of Thrones is. Game of Thrones is the derivative of an incredibly intricate series of books with so many details and plots, and the TV series stayed faithful to that for a long time. The latest season is a huge diversion from its roots. It is, as far as I can tell, replacing vi with nano. There is nothing wrong with either, but there is a good reason why the two are separate.

## Who are these laypeople, anyway?

This question is difficult to answer, because the layperson isn’t me. It can’t possibly be, because here I am writing about the subject. The layperson must be someone who isn’t particularly interested or informed. I imagine that they just turn on the telly and enjoy it for what it is. No deep thoughts, no deep investment.

But why don’t these laypeople care? Why should we care about laypeople? Must we really dumb everything down for the lowest common denominator? Why can’t they just get on my level? This is really frustrating!

Enter cars.

I have a driving license, but I don’t really care about cars. I know how to work the pedals and the steering wheel, and that’s pretty much it. I don’t know why I don’t care about cars. I just want to get from my home to my destination and back. If I can put in as little effort as possible to do that, I’m happy. I just don’t have the time or desire to learn all the intricacies of cars.

And knowing that, I suppose that I’m the layperson I was so frustrated over a moment ago. When I walk into the garage with a minor problem, I like to imagine that I’m the sort of person who shows up at tech support because I can’t log in: I accidentally pressed Caps Lock.

So the layperson is me. Sometimes.

## Then who are the elitists?

Having said all of that, something throws a wrench in the works: Game of Thrones was also immensely popular when it had all the intricacies and inter-weaving plots of the first few seasons. That appears to indicate to me that laypeople aren’t allergic to the kind of story that the enthusiasts want. But they aren’t allergic to the story that is being told in season 8, either, unlike the elitists.

So why do the elitists care? Why can’t they just appreciate the same things that laypeople do? Why must it always be so complex? Why should the complaints of a few outweigh the enjoyment of many?

And this is where I get stuck. Because frankly, I don’t know. Shouldn’t everything be as accessible as possible? The more the merrier? Why should vi exist when nano suffices?

But you can take my vi key bindings from my cold, dead hands. And I love what Game of Thrones used to be, and am sad that it morphed into an archetypical story that used to be its antithesis. I want complex things, even though I switched from Spacemacs to VSCodium and use GNOME instead of i3. Not for the sake of difficulty, but because complexity gives me something that simplicity cannot.

So the elitist is me. Sometimes.

## Fin

I’m still in a limbo about this clash between elitists and laypeople. Maybe the clash is superficial and the two can exist side-by-side or separately. Maybe the writers of Game of Thrones just aren’t very good and accidentally made the story for laypeople instead of their target audience of elitists. Maybe it’s a sliding scale instead of a binary.

I don’t really know. I just wanted to get these thoughts out of my head and into a text box.