text:
# Reflexive Reduction of Relation Compatible with Group Operation is Compatible

## Theorem

Let $\struct {S, \circ}$ be a group.

Let $\RR$ be a relation on $S$ which is compatible with $\circ$.

Let $\RR^\ne$ be the reflexive reduction of $\RR$.

Then $\RR^\ne$ is compatible with $\circ$.

## Proof

By definition of reflexive reduction, for all $a, b \in S$:

$a \mathrel {\RR^\ne} b$ if and only if $a \mathrel \RR b$ but $a \ne b$.

By definition of the diagonal relation $\Delta_S$:

$a \ne b$ if and only if $\tuple {a, b} \notin \Delta_S$

Thus, considered as subsets of $S \times S$, we have:

$\RR^\ne = \RR \setminus \Delta_S$

By Diagonal Relation is Universally Compatible, $\Delta_S$ is compatible with $\circ$.

Thus by Set Difference of Relations Compatible with Group Operation is Compatible, $\RR^\ne$ is compatible with $\circ$.

$\blacksquare$

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Kinetic energy of an electron

1. May 13, 2008

Cowtipper

1. The problem statement, all variables and given/known data
If an electron has a measured wavelength of 0.850 x 10^-10 m, what is its kinetic energy?

2. Relevant equations
I'm not sure.

3. The attempt at a solution
And once again, I'm not too sure. Where do I start?

2. May 13, 2008

Redbelly98

Staff Emeritus
Find an equation in your textbook that deals with the wavelength of electrons or particles in general. That would be a start.

3. May 13, 2008

Cowtipper

Well, I've got this:

I'm not sure if it's one to use.

E = hc/$$\lambda$$

Similarly, I have this one:

K=1/2MV$$^{2}$$

However, I'm not sure where or how to use it in this instance, or if I have to use it at all.

Man, this stuff is getting tough. I was doing well there for a while too...

4. May 13, 2008

hage567

No, this equation will not work. Look back in your notes and try to figure out why.

Look up "de Broglie wavelength."

5. May 13, 2008

Cowtipper

Aha! That is all I needed to know.

Thanks a lot!


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Published on Fri Mar 08 2019

### Random Matrix-Improved Estimation of the Wasserstein Distance between two Centered Gaussian Distributions

,

This article proposes a method to consistently estimate functionals. A consistent estimate of the Wasserstein distance between centered Gaussian distributions is derived. The new estimate is shown to largely outperform the classical sample-based plug-in' estimator.

0
0
0
###### Abstract

This article proposes a method to consistently estimate functionals of the eigenvalues of the product of two covariance matrices based on the empirical estimates ($\hat C_a=\frac1{n_a}\sum_{i=1}^{n_a} x_i^{(a)}x_i^{(a){{\sf T}}}$), when the size and number of the (zero mean) samples are similar. As a corollary, a consistent estimate of the Wasserstein distance (related to the case ) between centered Gaussian distributions is derived. The new estimate is shown to largely outperform the classical sample covariance-based plug-in' estimator. Based on this finding, a practical application to covariance estimation is then devised which demonstrates potentially significant performance gains with respect to state-of-the-art alternatives.


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## 170- Quiz 4

Quiz 4 is here. Please remember that the first midterm is this Wednesday.

Solutions follow.

Problem 1 asks for the derivative of the function $6/\root 3\of {x^5}$.

To solve this, remember the following laws of exponents:

• $\root 3\of t=t^{1/3}$,
• $(t^a)^b=t^{ab}$, and
• $1/t^a=t^{-a}$.

Using these laws we quickly obtain that $6/\root 3\of {x^5}=6/x^{5/3}=6x^{-5/3}$. Now, to take the derivative of this function, we apply two of the rules covered on lecture:

• $(cf)'=cf'$ when $c$ is a constant, and
• $(x^k)'=kx^{k-1}$ when $k$ is a constant.

We have $(6x^{-5/3})'=6\cdot(-5/3)\cdot x^{-8/3}=-10x^{-8/3}$.

Problem 2 asks to use the product rule to find the derivative of $(x^2+5x-3)(x^5 -6x^3+3x^2-7x+1)$.

The product rule says that

$(fg)'=f'g+fg'$,

and in this case it gives us that the derivative is

$(2x+5)(x^5-6x^3+3x^2-7x+1)$ $+(x^2+5x-3)(5x^4-18x^2+6x-7)$.


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# Differentiating cosine squared of x

I am trying to differentiate with respect to $x$, $y = \cos^2{x}$

Using the chain rule and my working out is this:

\begin{align*}\frac{dy}{dx} &= 2 \cos(x)(-\sin(x)) \\ &= -2 \sin x \end{align*} I am not sure how to get to the correct answer of $-\sin{(2x)}$.

Should I be using the chain rule or maybe the product rule?

• I have typeset your question with LaTeX. Please double-check that I correctly transcribed it. – Neal Oct 28 '12 at 16:50
• Your differentiation is correct. $2\cos x(-\sin x) \ne -2 \sin x$ but $2\cos x(-\sin x)=-2\sin{x}\cos{x}=-\sin{2x}$ – M. Strochyk Oct 28 '12 at 16:51
• You can also use that $\cos^2(x)=\frac 1 2 (1+\cos(2x))$ and differentiate that. – Mark Bennet Oct 28 '12 at 16:54

$2\cos (x)(-\sin(x))=-2\cos(x)\sin(x)=-\sin(2x)$

• Thanks for clearing that up, what trig identity have you used.. thanks – user866190 Oct 28 '12 at 18:39
• $\sin(2x)=2\sin(x)\cos(x)$ – Mark Bennet Oct 28 '12 at 19:50

## Chain Rule

To differentiate $y = \cos^2x$ with respect to $x$, one must apply the chain rule as shown:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Firstly, $\,\,let \,\,\, u = \cos x \,\,$
One can then differentiate this with respect to $x$ such that $$\frac{du}{dx} = -sinx$$ Then, $\,\,let \,\,\, y = u^2$
Differentiate $y$ with respect to $u$ such that $\frac{dy}{du} = 2u$

Next, one can substitute $u$ back in to make $$\frac{dy}{du} = 2\cos x$$

Thus, $$\frac{dy}{dx} = -2\sin x \cdot \cos x$$

## Double Angle Formula Simplification

Using the formula: $$\sin(2u) = 2\sin u\cos u$$
We can simplify to:

# $$\frac{dy}{dx} = -\sin(2x)$$

You did a mistake there - you misuse the trigonometric identity. Look at http://www.sosmath.com/trig/Trig5/trig5/trig5.html

It's $$2 \cdot \cos(x)\cdot \sin(x) = \sin(2x)$$ Not $$\cos(x)\cdot \sin(x) = \sin(s)$$


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• Subject:

...

• Topic:

...

An underwater bubble with radius of  5 cm at the bottom of tank , where the temperature is 5oC and pressure is 3 atm rises to the surface, where temperature is 25 oC and pressure is 1 atm. What will be the radius of bubble when it reaches to the surface?

1. 74 cm
2. 49 cm
3. 94 cm
4. 13 cm


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SMS scnews item created by Daniel Daners at Mon 17 Mar 2014 1135
Type: Seminar
Distribution: World
Expiry: 17 Mar 2014
Calendar1: 17 Mar 2014 1400-1500
CalLoc1: AGR Carslaw 829
Auth: daners@como.maths.usyd.edu.au

# The p-Dirichlet-to-Neumann operator

### Hauer

Daniel Hauer
University of Sydney
17 March 2014 14:00-15:00, Carslaw Room 829 (AGR)

## Abstract

In this talk we are interested in the Dirichlet-to-Neumann operator associated with the $p$-Laplace operator on a bounded Lipschitz domain in ${ℝ}^{d}$, where $1 and $d\ge 2$. If $p\ne 2$, then the Dirichlet-to-Neumann operator becomes nonlinear and not much was known so far. We outline how one obtains well-posedness and Hölder-regularity of weak solutions of some elliptic problems associated with the Dirichlet-to-Neumann operator. Further, we show that the semigroup generated by the negative Dirichlet-to-Neumann operator can be extrapolated on all ${L}^{q}$-spaces and enjoys an interesting ${L}^{q}-{C}^{0,\alpha }$-smoothing effect. Moreover, we outline how the part of the Dirichlet-to-Neumann operator in the space of continuous functions on the boundary is $m$-accretive and give a sufficient condition to ensure that the negative operator generates a strongly continuous semigroup on this space. We conclude this talk by stating some results to the large time stability of the semigroup and give decay rates.

Check also the PDE Seminar page. Enquiries to Daniel Hauer or Daniel Daners.

If you are registered you may mark the scnews item as read.
School members may try to .


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## Integral bases of the Gaussian integers

Let $E = \mathbb{Q}(i)$ and let $a,b,c,d \in \mathbb{Z}$. Prove that $B = \{a+bi, c+di\}$ is an integral basis for $E$ if and only if $ad-bc = \pm 1$.

Recall that the discriminant of $\mathbb{Q}(i)$ is $\mathsf{det} \left( \left[ \begin{array}{cc} 1 & i \\ 1 & -i \end{array} \right] \right)^2 = -4$, and that the conjugates of $a+bi$ are $a\pm bi$.

Note that the discriminant of $B$ is $\mathsf{det} \left( \left[ \begin{array}{cc} a+bi & c+di \\ a-bi & c-di \end{array} \right] \right)^2 = -4(ad-bc)^2$. By Theorem 6.10 in TAN, $B$ is an integral basis if and only if $-4(ad-bc)^2 = -4$, if and only if $ad-bc = \pm 1$.


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# Metrizable Space and continuous extension function

Let A be a closed subspace of a metrizable space $X$, then for every continuous function $f:A\rightarrow I$ and every metric $\rho$ on the space $X$ , I want to show that the formula

$$F(x) = \begin{cases} \inf \left\{f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1: a\in A \right\}, & \textrm{ if x\in X\backslash A  ,} \\\\ f(x), & \textrm{ if x\in A}, \\ \end{cases}$$

define a continuous extension $F$ of $f$ over $X$.

Remark:- What I know I have to use TIETZE-URYSOHN THEOREM,which states:-

Every continuous function from a closed subspace $M$ of a normal space $X$ to $I$ or $\mathbb{R}$ is continuously extendable over $X$. .

From the comment I figure out this remark is not useful.

• You don't have to use Tietze. You have to show that $F$ is continuous, then you have an explicit extension of $f$, while Tietze/Urysohn only gives existence [and the proof a way to construct one, but very cumbersome]. This shows that metric spaces are particularly nice. – Daniel Fischer Nov 30 '14 at 20:32
• Now, I understand we don't need to use the Tietze/Urysohn theorem. thanks – Birkary Nov 30 '14 at 20:51
• Historical aside: this extension formula first appeared in 1919 paper by Hausdorff, following the 1915 paper by Tietze from which the extension result originates (although Tietze proved it for metric spaces only). References here – user147263 Dec 1 '14 at 0:27
• More importantly, I think you have to assume $f$ is bounded from below. For example, suppose $A$ is the closed unit ball of a Hilbert space: then there is a continuous real-valued $f$ with $\inf_A f= -\infty$, hence $F(x)\equiv -\infty$ according to this definition. – user147263 Dec 1 '14 at 0:32

In order for this extension to work, $f$ must be assumed bounded from below. Indeed, if $A$ is the $x$-axis of $\mathbb R^2$ and $f(x,0)=-x^2$ , the extension gives $F(x)=-\infty$ for every $x\in \mathbb R^2\setminus A$. Such examples exist even if $A$ is a bounded closed set.

The idea of this formula is that even though the infimum allows $a$ to be any point of $A$, the "far-away" points have large ratio $\rho(x,a)/\rho (x,A)$ which effectively drops them from the competition for the infimum. In contrast, when $a$ is an almost-nearest point of $A$ to $x$, we have $\rho(x,a)\approx \rho(x,A)$ and so the "penalty" $\frac{\rho(x,a)}{\rho(x,A)}-1$ drops down to nearly zero.

### Continuity at $x_0\in X\setminus A$

Focus on the ball of radius $\frac12 \rho(x_0,A)$ centered at $x_0$. Within this ball, $\rho(x,A)\ge \frac12 \rho(x_0,A)$. The ratio of two Lipschitz continuous functions is Lipschitz continuous, provided that the denominator is bounded away from $0$. More precisely, the function $$g_a(x) = f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1$$ is Lipschitz on this ball with constant $4/\rho(x_0,A)$, independent of $a$. Taking the infimum preserves this property.

### Continuity at $x_0\in A$

Given $\epsilon>0$, take $R>0$ such that $|f(a)-f(x_0)|<\epsilon$ when $\rho(a,x_0)<R$. If $x\notin A$ and $\rho(x,x_0)<r\ll R$ ($r$ much smaller than $R$), then every point $a\in A$ with $\rho(a,x_0)\ge R$ has $$f(a)+\frac{\rho(x,a)}{\rho(x,A)}-1 \ge \inf_A f+\frac{R}{r}-1$$ So, choose $r$ small enough so that $$\inf_A f+\frac{R}{r}-1> f(x_0)$$ Use this to conclude that $F(x)\ge f(x_0)-\epsilon$ when $\rho(x,x_0)<r$.

The inequality $F(x)\le f(x_0)+\epsilon$ follows by considering $a\in A$ such that $\rho(x,a)/\rho(x,A)$ is very close to $1$; note that such $a$ will be within $B(x_0,R)$.


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# Weird formatting in a smaller created empty set symbol

A while back I created my own sort of emptyset symbol because the "not equal" slash didn't line up and it annoyed me. So I defined my own emptyset via \DeclareMathOperator{\vempty}{{\centernot{\Circle}}}. However, when I try to use it in a subscript (as demonstrated below), the slash becomes unaligned with the circle. Does anyone know how to remedy this? Here's my MWE:

\documentclass[12pt]{letter}
\usepackage{amsmath,amssymb,amsthm,centernot,cases}
\usepackage{wasysym}

\DeclareMathOperator{\vempty}{{\centernot{\Circle}}}
\DeclareMathOperator{\vempty}{{\text{\O}}}

\begin{document}
Example of difference between emptyset, varnothing, and vempty:
\begin{equation*}
\end{equation*}

Example of pathological behavior:
\begin{equation*}
\int_{\varnothing} f dm \quad \int_{\vempty} f dm
\end{equation*}
\end{document}

Symbols from wasysym are defined as

\mbox{\wasyfamily\symbol{<number>}}

so they don't change size in subscripts. Add \text around it.

Oh, and it's not an operator, but an ordinary symbol.

\documentclass[12pt]{letter}
\usepackage{amsmath,amssymb,amsthm,centernot,cases}
\usepackage{wasysym}

\newcommand{\vempty}{{\centernot{\text{\Circle}}}}

\begin{document}

Example of difference between emptyset, varnothing, and vempty:
\begin{equation*}

Personal note: I consider \varnothing a symbol good for plumbers. Your variation is even uglier. ;-)
• Is there a way to create a symbol which is more aesthetic and has the same slash angle as \neq? I tried, for example, \centernot{0}, but the slash was misaligned inside of the zero. – Isomorphism Jan 16 '19 at 0:04


text:
Archived

This topic is now archived and is closed to further replies.

Quake like console design question

This topic is 5273 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

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This is more of a design question than a programming one, but it''s not a game design question as such either, so I didn''t put it in that forum. I''m designing a Quake like console type of thing. However, instead of all commands being in one long list, I''m considering placing related commands into groups. Like this :
Console // Handles commands for clearing the console buffer and moving from group to group, and what not
Graphics // For graphics related commands
JokeOfTheDayGenerator // etc.

Groups can contain subgroups too. This is a bit like a directory tree structure, where there are commands instead of files, and groups instead of directories. Like directories, there can only ever be one current one. e.g. Console.SomeSubgroup> My questions is this: What commands should be accessible from the current group? At the moment, my group structure looks like this:
Console
|
Graphics
|     |
|     Textures
|
Sound
|
Sounds
|
etc.

In other words, Console is like the root. When you''re in Console.Graphics, the availability options could be: Commands in... a) Console.Graphics only b) Console, and Console.Graphics c) Console.Graphics, and Console.Graphics.Textures d) Console.Graphics, and also Console because Console is always available, no matter where you are Guidance, anyone? Thanks, Ro_Akira

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Just my .02, but I think something like that would work great for engine settings (Engine.Graphics.Resolution=800x600), but for commands I think you''re better keeping the hierarchy flat. It would be annoying to have to pay attention to where you were in order to send a chat message, for example.

bpopp (bpopp.net)

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In my opinion, this looks unintuitive, unclear, annoying and harder to manage.

If you want categories, I find it better to just use postfix names, like Quake 3.

con_clear
con_cmdlist
con_varlist
eng_dedicated
eng_quit
eng_restart
map_flush
map_timelimit
net_rate
net_name
net_hostname
net_connect
net_reconnect
net_disconnect
snd_effectvol
snd_musicvol
snd_restart
vid_mode
vid_colordepth
vid_fullscreen
vid_restart

The fact is, you won''t have enough commands for some group system to be worth it. You also have to think that the players want console options to be easy to use, people would hate constantly changing "groups" to access other things, it would make your console unattractive.

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quote:

Just my .02, but I think something like that would work great for engine settings (Engine.Graphics.Resolution=800x600), but for commands I think you''re better keeping the hierarchy flat. It would be annoying to have to pay attention to where you were in order to send a chat message, for example.

Agreed. Similarly, if you want to do something like clear the console, it''d be inconvenient.

What prompted me to think along these lines was:
a) Quake''s long list of commands isn''t very handy to scroll. Through. QIII addressed some of these problems with the ability to list commands that matched the partial command name you gave (QII doesn''t appear to have it). QIII also doesn''t list the cmds in alphabetic order, but the prefixing of cvars with r_ and s_, and the like helps. Anyway, better organisation is one reason.
b) You can get a list of related commands easily.
c) Fewer name conficts. Using fully qualified names, Console.Graphics.restart is different from Console.Sound.restart. And not because it has an r_, instead of an s_ prefixed.

The way I have it designed, there aren''t any cvars as such. Each module has it''s own ''command interpreter'' which can be loaded from a .dll. Each module ''has a'' command interpreter object.

For example, my Graphics class loads up a ''graphics'' command interpreter. The command interpreter has access to the Graphics class. The command interpreter might implement any commands it wants. "setresolution" might take 2 parameters, and call the Graphics::SetResolution (...) method of the Graphics object with which the command interpreter is associated with.

Is ''command interpreter'' too long? My fingers seem to think so...! :\

quote:

The fact is, you won''t have enough commands for some group system to be worth it. You also have to think that the players want console options to be easy to use, people would hate constantly changing "groups" to access other things, it would make your console unattractive.

Agreed. I hate changing directories at a DOS prompt. Why would I think users of a console think any different?!

Ro_Akira

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quote:
Quake''s long list of commands isn''t very handy to scroll. Through. QIII addressed some of these problems with the ability to list commands that matched the partial command name you gave.

I''m planning on trying to implement an auto-type feature in mine. As you type, it would start trying to guess which command you want. So if you typed /s it would show a dimmed ''ay'' after the cursor. If you hit tab, it would auto complete the command for you, add a space, and move the cursor to the end of the line. You could even implement a pop-down list of matched commands (similar to the way IE autofills URL''s in the address field). This would give you a functionality similar to what you were going for I think.

quote:
For example, my Graphics class loads up a ''graphics'' command interpreter. The command interpreter has access to the Graphics class. The command interpreter might implement any commands it wants. "setresolution" might take 2 parameters, and call the Graphics::SetResolution (...) method of the Graphics object with which the command interpreter is associated with.

Interesting. I''m working on something very similar for parsing network messages. I wrote a question about it yesterday and got some interesting comments (a couple anyway).

bpopp (bpopp.net)

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quote:

I saw that question alright, but then I spotted the word ''pattern'' amongst the posts. I didn''t run and hide immediately, just soon after, when I followed the link to what they were talking about. Curiously, this type of thing is what I''ve ended up with in my code also:
class CoinInsertedEvent : public TEvent

That''s from the link ''typed message pattern'' provided by antareus in one post. I didn''t/doesn''t look ''right'', but it works. I''ll have another look at that thread, since you discribed what you are working on as ''very similar''.

quote:

I''m planning on trying to implement an auto-type feature in mine. As you type, it would start trying to guess which command you want. So if you typed /s it would show a dimmed ''ay'' after the cursor. If you hit tab, it would auto complete the command for you, add a space, and move the cursor to the end of the line. You could even implement a pop-down list of matched commands (similar to the way IE autofills URL''s in the address field). This would give you a functionality similar to what you were going for I think.

A combination of what you''re suggesting would be cool. A listbox would be really necessary I think. If you type just ''s'' for example, it could match a good number of commmands. Rather than the I.E. approach of the listbox appearing separate from the line you''re typing in, one could combine it with the ''greyed out rest of command'' approach. I''ll try some illustration:
salad   --  <-greysalad       |salamander  | <-listboxsalute      |

Just imagine that without the gap with the ''--'', and ''ad'' in grey. I''m unsure whether salad should be repeated in the listbox as shown?

Regardless, this is the type of thing that might make my obtuse hierarchical system quite useable. I already have a ''history'' system, like that provided by a Win2k console, or DOSKey.

Oh, and one can scroll to the very top of the console buffer by using ''Ctrl-Pgup'', and to the very bottom by using ''Ctrl-Pgdn''. The latter is something that drove me mad using the Quake console. Scolling all the way up to see something, and then having to scroll all the way down again. I''m sure I''ve just missed the key to do it?!

Ro_Akira

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Hmmm. Looks like I might have to use d) instead of a) (see above). The console command interpreter implements such feats as clearing the console buffer, and moving from group to group. However, if I go from Console> to Console.Graphics>, suddenly I''m not able to go from group to group, because I no longer have access to the command!

I''ve just had a thought though (while typing this), since any commands can be referenced from any group by using a full path, I can still get around by using Console.cd ...

...and another thing. This console system seems to be really a half-baked scripting system. Especially when you can put a group of commands into a file, and execute them using ''exec'' in QIII for example. If I implement ''if'' statements, and return values, I''m half way there. Oh, and I''d need to treat commands more like functions. i.e. >Console.Graphics.SetResolution(800,600) instead of >Console.Graphics.SetResolution 800 600

Ro_Akira

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This is similiar to what bf1942 does, assuming you don''t actually mean you can "change directories" forcing you to back out of graphics should you wish to get to sound - that''d be asinine.

[My site|SGI STL|Bjarne FAQ|C++ FAQ Lite|MSDN|Jargon]
Ripped off from various people

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If you did autocomplete, it could be nice to arrange the commands in a heierarchial tree. For example you type "Con." and a menu pops up with all the commands controling the console, things like "Speed", "Color", "TextColor", etc. It would be just like in quake where you type "con_", just that it looks a little more familiar to programmers and has a nice popup menu =-)
Also, maybe you could have a menu up top with the choices spelled out fully, like "Console" (which inserts "Con." and pops up the list of choices), "Graphics" ("Gfx."), etc. Its really only good for those that don''t know the commands well, but can remeber what it effects. Want to increase texture brightness? pick gfx, then tex, then brightness

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wild_pointer,
In that case, I think it''s more asinine, than what bf1942 does. I don''t like things being asinine, what does bf1942 do exactly? To be fair, it''s still up in the air as to what''s accessible from where. Which is why I''m still talking here!

Extrarius,
That''s the type of thing I was thinking about.

quote:

...Also, maybe you could have a menu up top with the choices spelled out fully, like "Console" (which inserts "Con." and pops up the list of choices)...

You mean, if you want to type it out, you only need Con.Log.Enable, i.e. shorthand, but the menu will display the long version that can easily be understood. That''s good.

An complete alternative/useful addition would be to use a mouse driven listbox for changing variables. A bit like Unreal (the original, anyway. I haven''t played any since then) come to think of it. I think that might make a useful addition rather than an alternative. The command line based approach makes to easy for variables to be read from a text file and what not. More work of course.

Ro_Akira

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# Combinatorics problem using inclusion/exclusion on matching pairs

There are 5 different pairs of shoes. How many ways are there for 5 people to each choose 2 shoes with no one getting a matching pair?

So I think this is a inclusion/exclusion problem.

I think that the universe $|U|=5!$

Then I'm not sure how to continue.

I guess I have to add up the number of ways for 1 person to get a matching pair. Then 2 people and so forth to all 5 people getting matching pairs.

$5! - [5C1 * 4! - 5C2 * 3! + 5C3 * 2! - 5C4 * 1! + 5C5 * 0!] = 44$

So I'm pretty sure this is wrong and the $|U|=\frac{10!}{2^5}$. However I have trouble setting up the $P_i,P_i\cap P_j,...$.

I tried taking the number of ways that a person gets a matching pair as

$\binom{5}{1} * \frac{8!}{2^4}$ but it doesn't work.

• @N.Shales the first time i tried this I had $\frac{10!}{2^5} - [(5C1)*8!/2^4 - (5C2)*6!/2^3 + (5C3)*4!/2^2 - (5C4)*2!/2^1 + (5C5)*0!=101644]$ – HiPolyEraser Apr 11 '17 at 3:21
• You nearly had it, but I think you need to make clear in your mind what $|P_i|$ is. Where do the terms $5C1$, $5C2$ etc come from? They should be attributed to the number of ways of choosing $1,2,\ldots$ overlaps from $5$ sets, I think you have interpreted them as something else. p.s. I feel we should delete these comments soon, before they get reported as discussion. – N. Shales Apr 11 '17 at 3:29
• @N.Shales I think I have it written down as was as $5C2$ is the number of ways to take the intersection of $2P$ sets and so forth with the rest. I was interpreting it as the number of ways to pick a matching pair. – HiPolyEraser Apr 11 '17 at 3:37
• Okay, I'll just answer. I understand how this confusion can arise. – N. Shales Apr 11 '17 at 3:45

Firstly, the universal set $U$ of all shoe distributions has cardinality

$$|U|=\frac{10!}{2!^5}$$

because there are $10$ distinct shoes to split into $5$ pairs.

Now call set $P_i$ the set of shoe distributions such that person $i$ has a matching pair, then

$$|P_i|=5\cdot \frac{8!}{2!^4}$$

because there are $5$ types of shoe for person $i$ to pick a pair from and for each the remaining $8$ shoes can be distributed to the $4$ other people in $\frac{8!}{2!^4}$ ways.

The intersection $P_{i_1}\cap P_{i_2}$ is the set of shoe distributions such that person $i_1$ and person $i_2$ both get a matching pair each hence

$$|P_{i_1}\cap P_{i_2}|=5\cdot 4\cdot\frac{6!}{2!^3}$$

because person $i_1$ can pick a shoe type for a matching pair in $5$ ways then for each of those person $i_2$ can pick a matching pair type in $4$ ways then for each the remaining $6$ shoes can be distributed in $\frac{6!}{2!^3}$ ways.

The remaining intersections are calculated in the same way.

The crucial formula here is inclusion-exclusion which says that your desired count is

$$|U|-S_1+S_2-S_3+S_4-S_5$$

where

$$S_k=\binom{5}{k}|P_{i_1}\cap P_{i_2}\cap\cdots P_{i_k}|$$

hence

$$\text{desired count}=\frac{10!}{2!^5}-\binom{5}{1}\cdot 5\cdot\frac{8!}{2!^4}+\binom{5}{2}\cdot 5\cdot 4\cdot\frac{6!}{2!^3}-\binom{5}{3}\cdot 5\cdot 4\cdot 3\cdot\frac{4!}{2!^2}+\binom{5}{4}\cdot 5\cdot 4\cdot 3\cdot 2\cdot\frac{2!}{2!^1}-\binom{5}{5}\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot\frac{0!}{2!^0}$$

• (+1) This gives $65280$ – robjohn Apr 11 '17 at 4:28
• Thank you very much for explaining. – HiPolyEraser Apr 11 '17 at 4:40
• My pleasure! Thank you for the question. – N. Shales Apr 11 '17 at 9:09


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The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus says, roughly, that the following processes undo each other:

The first process is differentiation, and the second process is (definite) integration. To say that the two undo each other means that if you start with a function, do one, then do the other, you get the function you started with.

In equation form, you can say

This equation is the key to evaluating definite integrals. It says that if I can find an antiderivative for , then I can compute the definite integral by plugging the limits a and b into .

Another way of putting it is: Finding slopes of tangents and finding rectangle sums should be related in this way.

Theorem. ( The Fundamental Theorem of Calculus (first version)) Suppose f is integrable on , and that for some differentiable function F defined on . Then

The Fundamental Theorem of Calculus says that I can compute the definite integral of a function f by finding an antiderivative F of f.

Example. Compute .

Example. Compute .

But note that

And

Definite integrals may be positive, negative, or 0.

Example. Compute .

Example. Compute .

If you do an integral using a substitution, you can either use the substitution to change the limits of integration, or put the original variable back at the end.

Example. Compute .

Alternatively,

Example. Compute .

If the velocity of a particle at time t is , the change in position from to is

Example. A particle's velocity is

Find the change in position from to .

There is another version of the Fundamental Theorem which says in a direct way that "integration and differentation are opposites".

Theorem. ( Fundamental Theorem, Second Version) Suppose f is continuous on an interval . Then

This says that if you start with a function (" "), integrate (" "), then differentiate (" "), you get what you started with (" "). This is another way of saying that differentiation and integration are opposite processes.

Proof. I'll prove the second version of the Fundamental Theorem using the first version.

By the definition of the derivative,

Using properties of definite integrals, I can swap the limits on the second integral, then combine the two integrals into one:

Suppose that is an antiderivative of , so . Applying the first version of the Fundamental Theorem, I get

However, the last expression is just the limit definition of the derivative of . Since , I get

Putting all the equalities together, I have

Example. (a) Compute .

(b) Compute .

(a)

Note that the 3 is irrelevant; the answer would be the same if 3 was replaced by (say) 42.

(b) I can't apply the theorem as is, because the thing I'm differentiating with respect to ("x") doesn't match the upper limit of the integral (" "). Hence, I must apply the Chain Rule first:

Notice that in applying the Chain Rule, I got the thing I was differentiating with respect to (" ") to match the upper limit of the integral (" ").

Contact information


text:
What constitutes a low number of views in the description of the Tumbleweed badge?

-
An unofficial thread on Meta.SO suggests that the number is in the range 11-15, but SE leadership never disclosed the exact number. – user53153 Jan 17 at 4:19
Could you make this an answer so that I may accept it? – Michael Albanese Jan 19 at 14:38

An unofficial thread on Meta.SO suggests that the number is in the range 11-15, but SE leadership never disclosed the exact number.

...and I can add that it's less than 43 on meta.

-
since I just earned it for this question... – draks ... Jan 20 at 11:18
... but maybe I've just hit the 7 days requirement today... – draks ... Jan 20 at 14:18
That's quite an honor; only the 4th Tumbleweed in the history of meta.Math.SE. – user53153 Jan 21 at 3:18
@5PM a trophy for the soul of a badge hunter...$\phantom{test}$ – draks ... Jan 21 at 8:15
@5PM will I get an entry in the history post? – draks ... Jan 21 at 10:16
@draks... So how will you try to hunt for Unsung Hero? – Hagen von Eitzen Jan 22 at 16:46
@HagenvonEitzen I'm already more unsung than you...;-) – draks ... Jan 22 at 23:10

Please don't take the fun out of this by disclosing the numbers...

-
 This is a comment, not an answer. – user53153 Jan 21 at 2:53 Can't comment (yet)... – vonbrand Jan 21 at 2:56 You just did... – user53153 Jan 21 at 2:57 I can comment on my answers, but not on random stuff. – vonbrand Jan 21 at 3:00 The privilege comment anywhere requires 50 reputation; you have more than that already. – user53153 Jan 21 at 3:04 Just got it ;-) – vonbrand Jan 21 at 3:12 +1 for fun...$\phantom{lord}$ – draks ... Jan 21 at 9:20


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# Toronto Math Forum

## APM346-2012 => APM346 Math => Misc Math => Topic started by: Thomas Nutz on September 27, 2012, 04:23:33 PM

Title: Lecture 6 example
Post by: Thomas Nutz on September 27, 2012, 04:23:33 PM
Hello,
I was going through the 6. lecture notes, where in the end an example is brought up that leads to an integral
$$\frac{1}{4}\int_0^tcos(t')(cos(x-ct+ct')-cos(x+ct-ct'))dt'$$
I was trying to do that integral, but the only way that I could do it was to write out the cosines as complex exponentials, which lead me to eight terms in the end... Is there a cleverer way to do this integral?
Thanks!
Title: Re: Lecture 6 example
Post by: Victor Ivrii on September 27, 2012, 04:49:15 PM
Hello,
I was going through the 6. lecture notes, where in the end an example is brought up that leads to an integral
$$I=\frac{1}{4}\int_0^t\cos(t')\bigl(\cos(x-ct+ct')-\cos(x+ct-ct')\bigr)dt'$$
I was trying to do that integral, but the only way that I could do it was to write out the cosines as complex exponentials, which lead me to eight terms in the end... Is there a cleverer way to do this integral?
Thanks!

In this example $c=2$ helps a bit but you could sea; without it.

The 1-st line -> 2nd (just integration) -> 3rd (formula $\cos (\alpha)-\cos (\beta)= 2\sin \bigl((\beta+\alpha)/2\bigr)\cdot \sin \bigl((\beta-\alpha)/2\bigr)$ and we ge the 4-th line.
Then formula $2\sin( \alpha) \cdot \cos(\beta)=\sin (\alpha+\beta)+\sin (\alpha-\beta)$ and integration.


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# Recursion Theory I

 Title: Recursion Theory I Author: Michael Nedzelsky Submission date: 2008-04-05 Abstract: This document presents the formalization of introductory material from recursion theory --- definitions and basic properties of primitive recursive functions, Cantor pairing function and computably enumerable sets (including a proof of existence of a one-complete computably enumerable set and a proof of the Rice's theorem). BibTeX: @article{Recursion-Theory-I-AFP, author = {Michael Nedzelsky}, title = {Recursion Theory I}, journal = {Archive of Formal Proofs}, month = apr, year = 2008, note = {\url{https://isa-afp.org/entries/Recursion-Theory-I.html}, Formal proof development}, ISSN = {2150-914x}, } License: BSD License Used by: Minsky_Machines Status: [ok] This is a development version of this entry. It might change over time and is not stable. Please refer to release versions for citations.


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# Enhanced Auditor Reporting

Effective for periods ending on or after December 2018, the audit report has undergone some significant changes.The revised wording is expected to increase the auditor's use of professional skepticism and the quality of audit work performed.

On demand, 1 hour(s)

$49 ($0)

()

## Course Description

It was not too long ago that we adopted the Canadian Auditing Standards (CAS) and gave up the three-paragraph auditor’s report. Now, less than 10 years later, we are changing the format of the auditor’s report again.

The audit report has undergone some significant changes.  Learn about the new organization of the report, the enhanced descriptions of management's and the auditor's responsibilities, the increased emphasis of the going concern assumption, increased focus on financial statement disclosures and the introduction of key audit matters (KAM).

Sample report wording and additional resources will be provided.

## Course Walkthrough

• Course: Enhanced Auditor Reporting
• CPD Quiz
• CPA Canada: Implications of Financial Reporting
• Slides: Enhanced Auditor Reporting


text:
## Elementary Statistics (12th Edition)

a) Fail to reject the null hypothesis. b) $\mu_1-\mu_2$ is between 4.6832 and 7.4038.
a) Null hypothesis:$\mu_1=\mu_2$, alternative hypothesis:$\mu_1$ is more than $\mu_2$. Hence the value of the test statistic: $t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{\sqrt{s_1^2/n_1+s_2^2/n_2}}=\frac{(69.825-63.7815)-(0)}{\sqrt{0.79974^2/10+2.59665^2/40}}=12.533.$ The degree of freedom: $min(n_1-1,n_2-1)=min(10-1,40-1)=9.$ The corresponding P-value by using the table: p is more than 0.1. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is more than $\alpha=0.01$, because it is more than 0.1, hence we fail to reject the null hypothesis. b) The corresponding critical value using the table: $t_{\alpha/2}=t_{0.01}=2.821.$ The margin of error: $E=t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=2.821\sqrt{\frac{0.7994^2}{10}+\frac{2.59665^2}{40}}=1.3603.$ Hence the confidence interval $\mu_1-\mu_2$ is between $\overline{x_1}-\overline{x_2}-E$=(69.825-63.7815)-1.3603=4.6832 and $\overline{x_1}-\overline{x_2}+E$=(69.825-63.7815)+1.3603=7.4038.


text:
Opuscula Math. 27, no. 1 (2007), 151-165

Opuscula Mathematica

Unfolding spheres size distribution from linear sections with B-splines and EMDS algorithm

Zbigniew Szkutnik

Abstract. The stereological problem of unfolding spheres size distribution from linear sections is formulated as a problem of inverse estimation of a Poisson process intensity function. A singular value expansion of the corresponding integral operator is given. The theory of recently proposed $$B$$-spline sieved quasi-maximum likelihood estimators is modified to make it applicable to the current problem. Strong $$L^2$$-consistency is proved and convergence rates are given. The estimators are implemented with the recently proposed EMDS algorithm. Promising performance of this new methodology in finite samples is illustrated with a numerical example. Data grouping effects are also discussed.

Keywords: inverse problem, singular value expansion, stereology, discretization, quasi-maximum likelihood estimator.

Mathematics Subject Classification: 62G05, 45Q05.

Full text (pdf)


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# The guru of a yogi lives in a Himalyan cave, 1000 km away from the house of the yogi . The yogi claims that whenever he thinks about his guru, the guru immediately knows about it. Calcualte the minimum possible tiem interval between the yogi thinking about the guru and the guru knowing about it.

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

Updated On: 20-3-2021

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3.7 K+

3.3 K+

Text Solution

Solution :
Here S = 1000 km = 10^(6) m
 The process requires minimum possible time if the velocity is maximum. We know that maximum velocity can be that of light (i.e. 3xx 10^(8) m /s)
  Time = Distance / Speed
  = 10^(6) / 3 xx 10^(8) = 1 /300 s


text:
Then, going by the developer instructions in the Sage wiki, I would presume (I haven't tried myself) that you can get access to the sage sources by going to the <drive letter>:\\home\sage\sage\devel\sage\sage\ directory. Essentially, <drive letter>:\\home\sage\sage should be the root folder of the Sage installation.
Then, going by the developer instructions in the Sage wiki, I would presume (I haven't tried myself) that you can get access to the sage sources by going to the <drive letter>:\\home\sage\sage\devel\sage\sage\ directory. Essentially, <drive letter>:\\home\sage\sage should be the root folder of the Sage installation.


text:
• ### Transverse polarization of $\Lambda$ hyperons from quasireal photoproduction on nuclei(1406.3236)

Oct. 3, 2014 hep-ex, nucl-ex
The transverse polarization of $\Lambda$ hyperons was measured in inclusive quasireal photoproduction for various target nuclei ranging from hydrogen to xenon. The data were obtained by the HERMES experiment at HERA using the 27.6 GeV lepton beam and nuclear gas targets internal to the lepton storage ring. The polarization observed is positive for light target nuclei and is compatible with zero for krypton and xenon.
• ### Beam-helicity asymmetry in associated electroproduction of real photons $ep \to e\gamma \pi N$ in the $\Delta$-resonance region(1310.5081)

Feb. 11, 2014 hep-ex, nucl-ex
The beam-helicity asymmetry in associated electroproduction of real photons, $ep\to e\gamma \pi N$, in the $\Delta$(1232)-resonance region is measured using the longitudinally polarized HERA positron beam and an unpolarized hydrogen target. Azimuthal Fourier amplitudes of this asymmetry are extracted separately for two channels, $ep\to e\gamma \pi^0 p$ and $ep\to e\gamma \pi^+ n$, from a data set collected with a recoil detector. All asymmetry amplitudes are found to be consistent with zero.
• ### The HERMES Recoil Detector(1302.6092)

May 6, 2013 hep-ex, physics.ins-det
For the final running period of HERA, a recoil detector was installed at the HERMES experiment to improve measurements of hard exclusive processes in charged-lepton nucleon scattering. Here, deeply virtual Compton scattering is of particular interest as this process provides constraints on generalised parton distributions that give access to the total angular momenta of quarks within the nucleon. The HERMES recoil detector was designed to improve the selection of exclusive events by a direct measurement of the four-momentum of the recoiling particle. It consisted of three components: two layers of double-sided silicon strip sensors inside the HERA beam vacuum, a two-barrel scintillating fibre tracker, and a photon detector. All sub-detectors were located inside a solenoidal magnetic field with an integrated field strength of 1 T. The recoil detector was installed in late 2005. After the commissioning of all components was finished in September 2006, it operated stably until the end of data taking at HERA end of June 2007. The present paper gives a brief overview of the physics processes of interest and the general detector design. The recoil detector components, their calibration, the momentum reconstruction of charged particles, and the event selection are described in detail. The paper closes with a summary of the performance of the detection system.
• ### Multiplicities of charged pions and kaons from semi-inclusive deep-inelastic scattering by the proton and the deuteron(1212.5407)

April 24, 2013 hep-ex, nucl-ex
Multiplicities in semi-inclusive deep-inelastic scattering are presented for each charge state of \pi^\pm and K^\pm mesons. The data were collected by the HERMES experiment at the HERA storage ring using 27.6 GeV electron and positron beams incident on a hydrogen or deuterium gas target. The results are presented as a function of the kinematic quantities x_B, Q^2, z, and P_h\perp. They represent a unique data set for identified hadrons that will significantly enhance our understanding of the fragmentation of quarks into final-state hadrons in deep-inelastic scattering.
• ### Beam-helicity asymmetry arising from deeply virtual Compton scattering measured with kinematically complete event reconstruction(1206.5683)

Nov. 28, 2012 hep-ex
The beam-helicity asymmetry in exclusive electroproduction of real photons by the longitudinally polarized HERA positron beam scattering off an unpolarized hydrogen target is measured at HERMES. The asymmetry arises from deeply virtual Compton scattering and its interference with the Bethe--Heitler process. Azimuthal amplitudes of the beam-helicity asymmetry are extracted from a data sample consisting of $ep\rightarrow ep\gamma$ events with detection of all particles in the final state including the recoiling proton. The installation of a recoil detector, while reducing the acceptance of the experiment, allows the elimination of background from $ep\rightarrow eN\pi\gamma$ events, which was estimated to contribute an average of about 12% to the signal in previous HERMES publications. The removal of this background from the present data sample is shown to increase the magnitude of the leading asymmetry amplitude by 0.054 +/- 0.016 to -0.328 +/- 0.027 (stat.) +/- 0.045 (syst.).
• ### Measurement of double-spin asymmetries associated with deeply virtual Compton scattering on a transversely polarized hydrogen target(1106.2990)

Sept. 15, 2011 hep-ex
Double-spin asymmetries in exclusive electroproduction of real photons from a transversely polarized hydrogen target are measured with respect to the product of target polarization with beam helicity and beam charge, and with respect to the product of target polarization with beam helicity alone. The asymmetries arise from the deeply virtual Compton scattering process and its interference with the Bethe--Heitler process. They are related to the real part of the same combination of Compton form factors as that determining the previously published transverse target single-spin asymmetries through the imaginary part. The results for the double-spin asymmetries are found to be compatible with zero within the uncertainties of the measurement, and are not incompatible with the predictions of the only available GPD-based calculation.
• Hadron multiplicities in semi-inclusive deep-inelastic scattering were measured on neon, krypton and xenon targets relative to deuterium at an electron(positron)-beam energy of 27.6 GeV at HERMES. These ratios were determined as a function of the virtual-photon energy nu, its virtuality Q2, the fractional hadron energy z and the transverse hadron momentum with respect to the virtual-photon direction pt . Dependences were analysed separately for positively and negatively charged pions and kaons as well as protons and antiprotons in a two-dimensional representation. Compared to the one-dimensional dependences, some new features were observed. In particular, when z > 0.4 positive kaons do not show the strong monotonic rise of the multiplicity ratio with nu as exhibited by pions and negative kaons. Protons were found to behave very differently from the other hadrons.
• ### Measurement of azimuthal asymmetries associated with deeply virtual Compton scattering on a longitudinally polarized deuterium target(1008.3996)

May 2, 2011 hep-ex
Azimuthal asymmetries in exclusive electroproduction of a real photon from a longitudinally polarized deuterium target are measured with respect to target polarization alone and with respect to target polarization combined with beam helicity and/or beam charge. The asymmetries appear in the distribution of the real photons in the azimuthal angle $\phi$ around the virtual photon direction, relative to the lepton scattering plane. The asymmetries arise from the deeply virtual Compton scattering process and its interference with the Bethe-Heitler process. The results for the beam-charge and beam-helicity asymmetries from a tensor polarized deuterium target with vanishing vector polarization are shown to be compatible with those from an unpolarized deuterium target, which is expected for incoherent scattering dominant at larger momentum transfer. Furthermore, the results for the single target-spin asymmetry and for the double-spin asymmetry are found to be compatible with the corresponding asymmetries previously measured on a hydrogen target. For coherent scattering on the deuteron at small momentum transfer to the target, these findings imply that the tensor contribution to the cross section is small. Furthermore, the tensor asymmetry is found to be compatible with zero.
• ### Effects of transversity in deep-inelastic scattering by polarized protons(1006.4221)

June 22, 2010 hep-ex
Single-spin asymmetries for pions and charged kaons are measured in semi-inclusive deep-inelastic scattering of positrons and electrons off a transversely nuclear-polarized hydrogen target. The dependence of the cross section on the azimuthal angles of the target polarization (phi_S)and the produced hadron (phi) is found to have a substantial sin(phi+phi_S) modulation for the production of pi+, pi- and K+. This Fourier component can be interpreted in terms of non-zero transversity distribution functions and non-zero favored and disfavored Collins fragmentation functions with opposite sign. For pi0 and K- production the amplitude of this Fourier component is consistent with zero.
• ### Exclusive Leptoproduction of Real Photons on a Longitudinally Polarised Hydrogen Target(1004.0177)

April 1, 2010 hep-ex
Polarisation asymmetries are measured for the hard exclusive leptoproduction of real photons from a longitudinally polarised hydrogen target. These asymmetries arise from the deeply virtual Compton scattering and Bethe-Heitler processes. From the data are extracted two asymmetries in the azimuthal distribution of produced real photons about the direction of the exchanged virtual photon: A_UL with respect to the target polarisation and A_LL with respect to the product of the beam and target polarisations. Results for both asymmetries are compared to the predictions from a generalised parton distribution model. The sin(phi) and cos(0*phi) amplitudes observed respectively for the A_UL and A_LL asymmetries are compatible with the sizeable predictions from the model. Unexpectedly, a sin(2*phi) modulation in the A_UL asymmetry with a magnitude similar to that of the sin(phi) modulation is observed.


text:
#### • Class 11 Physics Demo

Explore Related Concepts

# economic profit formula

#### Best Results From Wikipedia Yahoo Answers Youtube

From Wikipedia

Profit (accounting)

In accounting, profit can be considered to be the difference between the purchase price and the costs of bringing to market whatever it is that is accounted as an enterprise (whether by harvest, extraction, manufacture, or purchase) in terms of the component costs of delivered goods and/or services and any operating or other expenses.

## Definition

There are several important profit measures in common use which will be explained in the following. Note that the words earnings, profit and income are used as substitutes in some of these terms (also depending on US vs. UK usage), thus inflating the number of profit measures.

Gross profitequals sales revenue lesscost of goods sold (COGS), thus removing only the part of expenses that can be traced directly to the production of the goods. Gross profit still includes general (overhead) expenses like R&D, S&M, G&A, also interest expense, taxes and extraordinary items.

Operating profit equals gross profit less all operating expenses. This is the surplus generated by operations. It is also known as earnings before interest and taxes (EBIT), operating profit before interest and taxes (OPBIT) or simply profit before interest and taxes (PBIT).

(Net)profit before tax (PBT) equals operating profit less interest expense (but before taxes). It is also known asearnings before taxes (EBT), net operating income before taxes or simply pre-tax Income.

Net profit equals profit after tax (unless some distinction about the treatment of extraordinary expenses is made). In the US the term net incomeis commonly used. Income before extraordinary expenses represents the same but before adjusting for extraordinary items.

Net income less dividends becomes retained earnings.

There are several additional important profit measures, notably EBITDA and NOPAT.

To accountants, economic profit, or EP, is a single-period metric to determine the value created by a company in one period - usually a year. It is the net profit after tax less the equity charge, a risk-weighted cost of capital. This is almost identical to the economist's definition of economic profit.

There are commentators who see benefit in making adjustments to economic profit such as eliminating the effect of amortized goodwill or capitalizing expenditure on brand advertising to show its value over multiple accounting periods. The underlying concept was first introduced by Schmalenbach, but the commercial application of the concept of adjusted economic profit was by Stern Stewart & Co. which has trade-marked their adjusted economic profit as EVA or Economic Value Added.

Some economists define further types of profit:

Optimum Profitâ€”This is the "right amount" of profit a business can achieve. In business, this figure takes account of marketing strategy, market position, and other methods of increasing returns above the competitive rate.

Accounting profits should include economic profits, which are also called economic rents. For instance, a monopoly can have very high economic profits, and those profits might include a rent on some natural resource that firm owns, where that resource cannot be easily duplicated by other firms.

Profit maximization

In economics, profit maximization is the (short run) process by which a firm determines the price and output level that returns the greatest profit. There are several approaches to this problem. The total revenue–total cost method relies on the fact that profit equals revenue minus cost, and the marginal revenue–marginal cost method is based on the fact that total profit in a perfectly competitive market reaches its maximum point where marginal revenue equals marginal cost.

## Basic definitions

Any costs incurred by a firm may be classed into two groups: fixed costs and variable costs. Fixed costs are incurred by the business at any level of output, including zero output. These may include equipment maintenance, rent, wages, and general upkeep. Variable costs change with the level of output, increasing as more product is generated. Materials consumed during production often have the largest impact on this category. Fixed cost and variable cost, combined, equal total cost.

Revenue is the amount of money that a company receives from its normal business activities, usually from the sale of goods and services (as opposed to monies from security sales such as equity shares or debt issuances).

Marginal cost and revenue, depending on whether the calculus approach is taken or not, are defined as either the change in cost or revenue as each additional unit is produced, or the derivative of cost or revenue with respect to quantity output. It may also be defined as the addition to total cost or revenue as output increase by a single unit. For instance, taking the first definition, if it costs a firm 400 USD to produce 5 units and 480 USD to produce 6, the marginal cost of the sixth unit is approximately 80 dollars, although this is more accurately stated as the marginal cost of the 5.5th unit due to linear interpolation. Calculus is capable of providing more accurate answers if regression equations can be provided.

## Total revenue - total cost method

To obtain the profit maximising output quantity, we start by recognizing that profit is equal to total revenue (TR) minus total cost (TC). Given a table of costs and revenues at each quantity, we can either compute equations or plot the data directly on a graph. Finding the profit-maximizing output is as simple as finding the output at which profit reaches its maximum. That is represented by output Q in the diagram.

There are two graphical ways of determining that Q is optimal. First, the profit curve is at its maximum at this point (A). Secondly, at the point (B) the tangent on the total cost curve (TC) is parallel to the total revenue curve (TR), meaning that the surplus of revenue net of costs (B,C) is at its greatest. Because total revenue minus total costs is equal to profit, the line segment C,B is equal in length to the line segment A,Q.

Computing the price at which to sell the product requires knowledge of the firm's demand curve. The price at which quantity demanded equals profit-maximizing output is the optimum price to sell the product.

## Marginal revenue-marginal cost method

An alternative argument says that for each unit sold, marginal profit (MÏ€) equals marginal revenue (MR) minus marginal cost (MC). Then, if marginal revenue is greater than marginal cost, marginal profit is positive, and if marginal revenue is less than marginal cost, marginal profit is negative. When marginal revenue equals marginal cost, marginal profit is zero. Since total profit increases when marginal profit is positive and total profit decreases when marginal profit is negative, it must reach a maximum where marginal profit is zero - or where marginal cost equals marginal revenue. If there are two points where this occurs, maximum profit is achieved where the producer has collected positive profit up until the intersection of MR and MC (where zero profit is collected), but would not continue to after, as opposed to vice versa, which represents a profit minimum. In calculus terms, the correct intersection of MC and MR will occur when:

\frac{dMR}{dQ} < \frac{dMC}{dQ}

The intersection of MR and MC is shown in the next diagram as point A. If the industry is perfectly competitive (as is assumed in the diagram), the firm faces a demand curve (D) that is identical to its Marginal revenue curve (MR), and this is a horizontal line at a price determined by industry supply and demand. Average total costs are represented by curve ATC. Total economic profit are represented by area P,A,B,C. The optimum quantity (Q) is the same as the optimum quantity (Q) in the first diagram.

If the firm is operating in a non-competitive market, minor changes would have to be made to the diagrams. For example, the Marginal Revenue would have a negative gradient, due to the overall market demand curve. In a non-competitive environment, more complicated profit maximization solutions involve the use of game theory.

## Maximizing revenue method

In some cases a firm's demand and cost conditions are such that marginal profits are greater than zero for all levels of production. In this case the MÏ€ = 0 rule has to be modified and the firm should maximize revenue. In other words the profit maximizing quantity and price can be determined by setting marginal revenue equal to zero. Marginal revenue equals zero when the marginal revenue curve has reached its maximum value. An example would be a scheduled airline flight. The marginal costs of flying the route are negligible. The airline would maximize profits by filling all the seats. The airline would determine the \Pi_max conditions by maximizing revenues.

## Changes in total costs and profit maximization

A firm maximizes profit by operating where marginal revenue equal marginal costs. A change in fixed costs has no effect on the profit maximizing output or price. The firm merely treats short term fixed costs as sunk costs and continues to operate as before. This can be confirmed graphically. Using the diagram illustrating the total cost total revenue method the firm maximizes profits at the point where the slope of the total cost line and total revenue line are equal. A change in total cost would cause the total cost curve to shift up by the amount of

From Yahoo Answers

Question:Tanya imports sweaters from Peru and sells them from her home. She collects $400,000 in revenue a year, and spends$200,000 on the sweaters and shipping costs, as well as $25,000 on accounting services and utilities. She used to make$100,000 per year working for an advertising agency. Now she works out of the basement of her house, for which she doesn't have any other marketable use. She has no other expenses. Tanya's economic profit is: A. -$25,000 B.$100,000 C. $75,000 D.$175,000 I know that accounting profit is \$175,000 and the equation for economic profit is economic profit = accounting profit - opportunity cost. I'm just unsure which counts as opportunity cost bc my answer is not one of the choices >< THANKS! SORRY!! The TITLE of the question is wrong...i meant ECONOMIC PROFIT!

Answers:Tanya would get 100'000 in ad agency as opportunity profit. But she gets 175'000 instead. So just calculate difference 175'000-100'000=75'000 Correct answer is "C" Economic profit is 75'000.

Question:1. Promote the equal distribution of real assets and wealth 2. Achieve full employment and price level stability 3. Contribute to a more equal distribution of income 4. reallocate resources from less desired to more desired uses. I think 4 but am looking for a second opinion...please help.

Answers:4 is correct. free market increases inequality, so 1&3 are out. and as recession demonstrates, it does not always achieve full employment and price level stability

Question:For my A2 economics coursework i have to look at a companies financial performance and this involves using two major ratios the Acid Test Ratio and the Asset turnover ratio, can someone please help explain what these two ratios show, in english please, and also an easy wy of calculating them both as i am very confused right now.

Answers:A stringent test that indicates whether a firm has enough short-term assets to cover its immediate liabilities without selling inventory. The acid-test ratio is far more strenuous than the working capital ratio, primarily because the working capital ratio allows for the inclusion of inventory assets. Calculated by: Cash+Accounts Received+Short -Term Investments over the Current Liabilities What Does Asset Turnover Mean? The amount of sales generated for every dollar's worth of assets. It is calculated by dividing sales in dollars by assets in dollars. Formula: Asset Turnover=Revenue __________ Assets Asset turnover measures a firm's efficiency at using its assets in generating sales or revenue - the higher the number the better. It also indicates pricing strategy: companies with low profit margins tend to have high asset turnover, while those with high profit margins have low asset turnover.

Question:3. last year, David Nash, a computer programmer received two bonuses from IBM. He was surprised at the end of the year that his yearly salary was much less than he had expected it would be. Because of the bonuses, IBM had deducted more money from his checks to pay additional _____________ a. bonus taxes b. income taxes c. sales taxes d. excise taxes 4. The Phoenicians received silver in exchange for the products they bartered with Spain. Soon Phoenicia was very wealthy. When a nation receives goods from another country, the goods are: a. exported b. a windfall c. imported d. transported 5. Which of the following is NOT a consumer responsibility? a. report defective goods b. report wrongs c. maintain adequate credit d. Do Not Steal e. give correct information 6. The government lowered the income tax to stimulate the economy. This is an example of __________ a. monetary policy b. military policy c. corporate policy d. fiscal policy e. welfare policy 7. the graph comparing the income a consumer makes over a lifetime with the expenses over a lifetime is called the time probability _____ 8. Which of the following are types of product trade-offs? a. energy consumption b. price c. required maintenance d. availability e. all of the above 9. Items which satisfy needs for peer group acceptance are called _______ needs products 10. What happens when a company is monopolized? a. that industry s product prices are reduced b. free enterprise is destroyed c. business generally prospers d. all of these things happen 1. All of the following are examples of growing consumer awareness of responsibilities, EXCEPT _________ a. turning thermostats down to conserve energy b. recycling papers, cans, and glass c. decreasing gasoline consumption to reduce oil imports d. increase consumption of throw away bottles e. refusing to buy aerosol spray cans that contain fluorocarbons that damage ozone layer 2. Which of the following is a relatively scarce resource to the individual in question? a. sand to children at the beach b. timber to a lumberjack in organ c. water to a camper in death valley d. salt of homemaker shopping for the week s groceries e. water to people fishing in atlantic ocean 3. If the money supply declines while the amount of goods remain stable, will each until of money be more or less valuable? a. less b. more 4. Division of Labor is important to the production process because __________ a. it determines human wants b. it reduces dependency on specialization c. it allows labor to be most productive d. it allows smaller nations to compete effectively with larger nations e. it allows workers to be unskilled 5. Economics refer to the process of making something as production. a. false b. true 6. Gasoline is an example of ________ a. basic raw material b. essential service c. major factor of production d. scarce goods e. all of the above 7. The identification of responsibilities and obligations of partners to one another and to the children is called ___________ commitment a. contract b. economic c. social d. moral e. legal 8. people consume particular goods or use particular services, because the goods or services have _________ a. capital b. abundance c. utility d. comparative advantage e. stated value 9. the use of goods and services to satisfy human wants and needs is known as ______ a. general welfare b. value c. rebate d. consumption e. standard of living 10. In a trading economy, the society soon has goods left over after it has paid all its expenses. This surplus is called ____________ a. sound economy b. oversupply of goods c. profit d. good management

Answers:ask a specific question. wat dont u get?

From Youtube

Profit Maximization :This video deals with how firms maximize profits (MC=MR), a very important concept in economics. Please rate and comment!

Video Marketing Success: A 5 Step Formula to Endless Profits :MyInsiderReport.com


text:
After gaining a basic understanding of TikZ/PGF it’s time to move on to what I set out to do in the first place: drawing a spherical triangle.

Before jumping straight to a spherical triangle we need to do a bit more ground work on how to draw 3D diagrams?
Before delving into the 3D world I will go over some useful things like the basic mathematical operations that can be used in TikZ/PGF:

The PGF Mathematical Engine:p, li { white-space: pre-wra

1. \begin{tikzpicture}[scale = 4, rotate=-20]
2.      See previous post for missing code.
3.      % Graph Content
4.      \coordinate (point1) at (1/2, 3*2-5.5);
5.      \draw (0,0) circle (\R);
6.      \draw (0,0) — (point1);
7.      \filldraw [fill = blue!50, draw = black] (point1) circle (\R/30);
8.      \foreach \ang in {0,10,…,80} {
9.           \pgfmathparse{\R*2/2*sin(90)};
10.           \filldraw [fill = green!\ang, draw = black] (\ang:\pgfmathresult) circle (\R/30);
11.           \pgfmathsetmacro{\XVal}{\R*cos(-\ang)};
12.           \pgfmathsetmacro{\YVal}{\R*sin(-\ang)};
13.           \filldraw [fill = yellow!\ang, draw = black] (\XVal, \YVal) circle (\R/30);
14.      }
15. \end{tikzpicture}
• The first and most obvious thing you will probably note is that one image is rotated. Whilst I have applied this transform to the entire figure/diagram a transform can be applied to any primitive.
• Although touched on briefly, this example shows how to save a coordinate. And just as important that the coordinates can be entered as mathematical expressions that will be evaluated with PGF’s mathmatical engine.
• What happens if we want a more complicated expression? It turns out we can do this too. But I found that they couldn’t be placed inline, and instead has to be parsed using \pgfmathparse{}. The result is then stored in the \pgfmathresult variable/macro.
• But what happens if we need two complicated expressions? After some reading and experimenting I found you can use this function: \pgfmathsetmacro{\YVal}{expression} to save an answer to a variable which can the be used later!

Ok, so back to drawing 3D stuff…

Transformations:

It seems the key to drawing in 3D is to use transformations. We will be rotating the x-y plane (drawing canvas) using the  following transformation matrices:

We can also combine these matrices using Rzyx = Rz*Ry*Rx, which yields:
In TikZ/PGF the default plane is the right handed x-y plane with the x-axis going to the right and the y-axis going up (being right handed the z-axis comes out of the page). This default plane can be transformed using the following function/drawing option:
“[cm={a,b,c,d,(dx,dy)}]”
This represents the following transformation:

In the first example we are rotating a 2D shape in the default plane; this means we want to rotate about the z-axis. This is the Rz matrix:

We need to project into the x-y plane, so we can remove the third row and column as this will only affect a z-value:
Lets have a look at this simple example which uses rotations about the z-axis. First a triangle Δ(0,0)(1,0)(1,1) is drawn. The second figure then shows the triangle rotated 45deg. And the final example shows the original triangle rotated -90deg and translated by (1,1). Note: you can think about this by either translating and then rotating, or rotating and then translating.

1. \begin{tikzpicture}
2.      \DrawAxis{0}{0}{2}{2};
3.      \draw [fill = yellow] (0,0) — (1,1) — (1,0);
4.      \begin{scope}[xshift=3cm]
5.          \def\Ang{45};
6.          \DrawAxis{0}{0}{2}{2};
7.          \draw [cm={cos(\Ang),sin(\Ang),-sin(\Ang),cos(\Ang),(0,0)}
8.           (0,0) — (1,1) — (1,0);
9.      \end{scope}
10.      \begin{scope}[xshift=6cm]
11.          \DrawAxis{0}{0}{2}{2};
12.          \draw [cm={cos(\Ang),sin(\Ang),-sin(\Ang),cos(\Ang),(1,1)}
13.           (0,0) — (1,1) — (1,0);
14.      \end{scope}
15. \end{tikzpicture}

Drawing in 3D:

I will start be drawing a parabola, which we will then rotate about the y-axis creating a cone-like shape.

We will now add some ‘circles’ and planes rotated about the x-axis and offset in the y-direction to add some depth to the figure:

p, li { white-space: pre-wra

1. \begin{tikzpicture}
2.      \def\fcn{\x*\x};
3.      \def\phi{10};
4.      \def\xmin{-1.5}; \def\xmax{1.5};
5.      \draw[color=black, domain=\xmin:\xmax]
6.           plot (\x, \fcn) node [above right] {$y = x^2$};
7.      \foreach \x in {0, .25, …, \xmax} {
8.           \tikzset{xyplane/.estyle={cm={1, 0, 0, cos(90 + \phi),(0, \fcn)}}}
9.           \draw[xyplane, color=black] (0, 0) circle (\x);
10.      }
11. \end{tikzpicture}
• \phi is the view angle. We rotate by 90deg to get the plane perpendicular to the page and then adjust a little.

Now we’re starting to have something that looks 3-dimensional. Lets now rotate the parabola around the y-axis:

p, li { white-space: pre-wra

1. \begin{tikzpicture} [scale =2]
2.      \def\fcn{\x*\x};
3.      \def\phi{10};
4.      \def\xmin{-1.5}; \def\xmax{1.5};
5.      \foreach \x in {0, 0.25, …, \xmax} {
6.           \tikzset{xyplane/.estyle={cm={1, 0, 0, cos(90 + \phi),(0, \fcn)}}}
7.           \draw[xyplane, color=black] (0, 0) circle (\x);
8.      }
9.      \foreach \theta in {0, 30, …, 360} {
10.           \tikzset{xyplane/.estyle={cm={cos(\theta), 0, 0, 1,(0, 0)}}}
11.           \draw[xyplane, color=black, domain=\xmin:\xmax]
12.                plot (\x, \fcn);
13.      }
14. \end{tikzpicture}\\

Close, but no dice. We need to account for the viewing angle. So to get the correct plane, we need to rotate about the y-axis and the x-axis:

p, li { white-space: pre-wra

1. \begin{tikzpicture} [scale =2]
2.      \def\fcn{\x*\x};
3.      \def\phi{10};
4.      \def\xmin{-1.5}; \def\xmax{1.5};
5.      \foreach \x in {0, 0.25, …, \xmax} {
6.           \tikzset{xyplane/.estyle={cm={1, 0, 0, cos(90 + \phi),(0, \fcn)}}}
7.           \draw[xyplane, color=black] (0, 0) circle (\x);
8.      }
9.      \foreach \theta in {0, 30, …, 360} {
10.           \tikzset{xyplane/.estyle={cm={cos(\theta), sin(\theta)*sin(\phi), 0,                cos(\phi),(0, 0)}}}
11.           \draw[xyplane, color=black, domain=\xmin:\xmax]
12.           plot (\x, \fcn);
13.      }
14. \end{tikzpicture}\\

Close but the top circle does not quite line up with the rotated parabolas. It turns out this is also to do with the viewing angle (If we look at the Rx rotation matrix and project into the x-y plane we can see that the y value of a coordinate is multiplied by cos(\phi)).
Because we are looking from above the heights appear shorter by a factor of cos(\phi), so when we draw the circles, instead of offsetting the plane by x^2, we need to offset it by cos(\phi)*x^2:

The next section will look at drawing a sphere with lines of latitude and longitude.

References:


text:
# Laser in front of a wall - mathemathical physics

1. Mar 8, 2014

### skrat

1. The problem statement, all variables and given/known data
In front of a long, straight, white wall a laser is hanged. Laser is parallel to the wall when we decide to rotate it with constant angular velocity for $\pi$ so that the laser spot travels over the entire wall. The duration of the turn lasts just as long as it takes for the laser light to travel to the wall and back to the laser. The wall radiates the light equally in all directions.

An observer standing next to the laser watches the light on the wall using a high speed camera. When will he first notice a bright spot and from which direction? How fast does the spot on the wall move at $\pi / 4$ (angle next to the perpendicular line)?

2. Relevant equations

3. The attempt at a solution

Let's start by saying that the laser is distance $a$ from the wall. The rotation therefore lasts for $t=\frac{2a}{c_0}$, where $c_0$ is of course the speed of light.

$\pi =\omega t=\omega \frac{2a}{c_0}$ and from here $\omega =\frac{\pi c_0}{2a}$.

So finally $\varphi (t)=\frac{\pi c_0}{2a}t$ for $t\in [0, \frac{2a}{c_0}]$.

I assume, that every spot on the wall that laser hits while rotating, becomes a light source that radiates light in all directions. Now I THINK I am somehow supposed to use Lagrangian multiplier to find the extreme value of time that light needs to come back to the laser. But I am having some really big troubles to find out what $u$ and $v$ are in $min(u+\lambda v)$.

2. Mar 8, 2014

### voko

With respect to what is the angle measured? And, given the angle, what is the duration of the round trip? And what is the time when that round trip ends?

3. Mar 8, 2014

### skrat

Am... I guess i wasn't clear enough.
At $t=0$ the laser is parallel to the wall, therefore the angle of rotation is $\varphi =0$ and there is no bright spot on the wall. The whole rotation of the laser for $\pi$ will take $t= \frac{2a}{c_0}$ and is calculated above.

For example at $t/2$ the laser will be perpendicular to the wall.

I hope that answers your question or maybe I completely missunderstood you. If that is the case, I will prepare a sketch of the problem as soon as i get to the computer.

4. Mar 8, 2014

5. Mar 8, 2014

### skrat

The duration of the round trip is equal to the duration that light needs if it travels from the laser, to the closest point on the wall and back. That is the same as if $\varphi = \pi /2$. Let's mark that distance with $a$.

The time is than $t=\frac{2a}{c_0}$. But that is exactly what i have written in the previous two posts so i guess you want to hear something else.

6. Mar 8, 2014

### voko

$$t = \frac {2a} {c_0}$$ only when the laser is illuminating the wall at the right angle with the wall. At any other time, the round trip distance is not $2a$ and so the round trip duration is different. What is it at an arbitrary angle?

7. Mar 8, 2014

### skrat

Aaaaa, you mean round trip od the light and not the laser. :D hehe, sorry, english is not my mother language but i try.

Of course $t_L=\frac{2a}{c_0sin(\varphi (t))}$.

8. Mar 8, 2014

### voko

So when does the light emitted by the laser at time $t$ comes back to its origin? And what is minimal value of that?

9. Mar 8, 2014

### skrat

$t_L=\frac{2a}{c_0sin(\varphi (t))}$ later.

Which is minimum for $sin\varphi =1$.

10. Mar 8, 2014

### voko

$t_L$ is the time later. What is the total time of that?

11. Mar 8, 2014

### skrat

$t_0=t_L+t$ where $t$ is time measured from the beginning of rotation and $t\in [0,\frac{2a}{c_0}]$.

12. Mar 8, 2014

### voko

And the minimum is?

13. Mar 8, 2014

### skrat

Hmmm... I really hope I am not writing stupidity right now:

$t_0=t_L+t$

$t_0=\frac{2a}{c_0sin(\varphi (t))}+t$

$\frac{\partial t_0}{\partial t}=\frac{2acos(\varphi (t))dt}{c_0sin^2(\varphi (t))}+1=0$

$\frac{2acos(\frac{\pi c_0}{2a}t)\frac{\pi c_0}{2a}}{c_0sin^2(\frac{\pi c_0}{2a}t)}+1=0$

$cos(\frac{\pi c_0}{2a}t)=\frac{1}{\pi }sin^2(\frac{\pi c_0}{2a}t)$

I can probably assume that $t$ is very close to zero which hopefully gives me a green light to use Taylor series...

$1-\frac{1}{2}(\frac{\pi c_0}{2a})^2t^2=\frac{1}{\pi }(\frac{\pi c_0}{2a})^2t^2$

$t=\sqrt{\frac{1}{(\frac{\pi c_0}{2a})^2(\frac{1}{2}+\frac{1}{\pi })}}$

:D
Right now I just hope I didn't do anything embarrassing. :D

14. Mar 8, 2014

### voko

Sign error here, and the derivative of $\phi$ is missing. Recall the chain rule.

15. Mar 8, 2014

### skrat

Sorry, I have the right sign in the notes. There should also stand $\frac{d\varphi }{dt}$ and not $dt$.

$\frac{\partial t_0}{\partial t}=-\frac{2acos(\varphi (t))\frac{d\varphi }{dt} }{c_0sin^2(\varphi (t))}+1=0$

Where $\varphi (t)=\frac{\pi c_0}{2a}t$ and $\frac{d\varphi }{dt}= \frac{\pi c_0}{2a}$

However, the the rest should be fine.

16. Mar 8, 2014

### voko

You do not have to assume that $t$ is small. And, even if it is small, the angle need not be small.

You can solve the equation exactly.

17. Mar 8, 2014

### skrat

Sorry to ask, but why would the angle have to be small? How did you guess that?

18. Mar 8, 2014

### voko

I am not guessing. I am saying that you do not know a priory that it will be small. And there is a way to solve that exactly, so there is no need for that guesswork.

P.S. Sometimes you cannot solve an equation exactly. Then you would have to think how it can be approximated in a physically valid way. But that is not guessing, and not required for this problem anyway.

19. Mar 8, 2014

### skrat

Understood.

Well, the reason I thought about approximation is because I have no idea at all on how to exactly solve this... Could you give me a hint or two?

$\pi cos(\frac{\pi c_0}{2a}t)=sin^2(\frac{\pi c_0}{2a}t)$

20. Mar 8, 2014

### skrat

Or... I can:

$\pi cos(\gamma t)=sin^2(\gamma t)$ for $\gamma =\frac{\pi c_0}{2a}$.

$\pi cos(\gamma t)= 1-cos^2(\gamma t)$ which is quadratic equation for $cos(\gamma t)$.

For $A=cos(\gamma t)$ the equation is $A^2+\pi A-1=0$.

$A_{1,2}=\frac{-\pi \pm \sqrt{\pi^2+4}}{2}$

$\gamma t=arccos[\frac{-\pi \pm \sqrt{\pi^2+4}}{2}]$

$t=\frac{1}{\gamma}arccos[\frac{-\pi \pm \sqrt{\pi^2+4}}{2}]$


text:
# A Brief Introduction to Bayesian Inference

Neil D. Lawrence, University of Sheffield
Magnus Rattray, University of Manchester

Chapter 5 in Learning and Inference in Computational Systems Biology

#### Abstract

Computational systems biology aims to develop algorithms that uncover the structure and parameterization of the underlying mechanistic model—in other words, to answer specific questions about the underlying mechanisms of a biological system—in a process that can be thought of as learning or inference. This volume offers state-of-the-art perspectives from computational biology, statistics, modeling, and machine learning on new methodologies for learning and inference in biological networks. The chapters offer practical approaches to biological inference problems ranging from genome-wide inference of genetic regulation to pathway-specific studies. Both deterministic models (based on ordinary differential equations) and stochastic models (which anticipate the increasing availability of data from small populations of cells) are considered. Several chapters emphasize Bayesian inference, so the editors have included an introduction to the philosophy of the Bayesian approach and an overview of current work on Bayesian inference. Taken together, the methods discussed by the experts in Learning and Inference in Computational Systems Biology provide a foundation upon which the next decade of research in systems biology can be built.

  @InCollection{lawrence-licsbbayes10, title = {A Brief Introduction to Bayesian Inference}, author = {Neil D. Lawrence and Magnus Rattray}, booktitle = {Learning and Inference in Computational Systems Biology}, year = {2010}, editor = {Neil D. Lawrence and Mark Girolami and Magnus Rattray and Guido Sanguinetti}, address = {Cambridge, MA}, month = {00}, publisher = {MIT Press}, edit = {https://github.com/lawrennd//publications/edit/gh-pages/_posts/2010-01-01-lawrence-licsbbayes10.md}, url = {http://inverseprobability.com/publications/lawrence-licsbbayes10.html}, abstract = {Computational systems biology aims to develop algorithms that uncover the structure and parameterization of the underlying mechanistic model—in other words, to answer specific questions about the underlying mechanisms of a biological system—in a process that can be thought of as learning or inference. This volume offers state-of-the-art perspectives from computational biology, statistics, modeling, and machine learning on new methodologies for learning and inference in biological networks. The chapters offer practical approaches to biological inference problems ranging from genome-wide inference of genetic regulation to pathway-specific studies. Both deterministic models (based on ordinary differential equations) and stochastic models (which anticipate the increasing availability of data from small populations of cells) are considered. Several chapters emphasize Bayesian inference, so the editors have included an introduction to the philosophy of the Bayesian approach and an overview of current work on Bayesian inference. Taken together, the methods discussed by the experts in Learning and Inference in Computational Systems Biology provide a foundation upon which the next decade of research in systems biology can be built.}, crossref = {Lawrence:licsb10}, key = {Lawrence:licsbbayes10}, chapter = {5}, OPTgroup = {} }
 %T A Brief Introduction to Bayesian Inference %A Neil D. Lawrence and Magnus Rattray %B %C Learning and Inference in Computational Systems Biology %D %E Neil D. Lawrence and Mark Girolami and Magnus Rattray and Guido Sanguinetti %F lawrence-licsbbayes10 %I MIT Press %P -- %R %U http://inverseprobability.com/publications/lawrence-licsbbayes10.html %X Computational systems biology aims to develop algorithms that uncover the structure and parameterization of the underlying mechanistic model—in other words, to answer specific questions about the underlying mechanisms of a biological system—in a process that can be thought of as learning or inference. This volume offers state-of-the-art perspectives from computational biology, statistics, modeling, and machine learning on new methodologies for learning and inference in biological networks. The chapters offer practical approaches to biological inference problems ranging from genome-wide inference of genetic regulation to pathway-specific studies. Both deterministic models (based on ordinary differential equations) and stochastic models (which anticipate the increasing availability of data from small populations of cells) are considered. Several chapters emphasize Bayesian inference, so the editors have included an introduction to the philosophy of the Bayesian approach and an overview of current work on Bayesian inference. Taken together, the methods discussed by the experts in Learning and Inference in Computational Systems Biology provide a foundation upon which the next decade of research in systems biology can be built.
 TY - CPAPER TI - A Brief Introduction to Bayesian Inference AU - Neil D. Lawrence AU - Magnus Rattray BT - Learning and Inference in Computational Systems Biology PY - 2010/01/01 DA - 2010/01/01 ED - Neil D. Lawrence ED - Mark Girolami ED - Magnus Rattray ED - Guido Sanguinetti ID - lawrence-licsbbayes10 PB - MIT Press SP - EP - UR - http://inverseprobability.com/publications/lawrence-licsbbayes10.html AB - Computational systems biology aims to develop algorithms that uncover the structure and parameterization of the underlying mechanistic model—in other words, to answer specific questions about the underlying mechanisms of a biological system—in a process that can be thought of as learning or inference. This volume offers state-of-the-art perspectives from computational biology, statistics, modeling, and machine learning on new methodologies for learning and inference in biological networks. The chapters offer practical approaches to biological inference problems ranging from genome-wide inference of genetic regulation to pathway-specific studies. Both deterministic models (based on ordinary differential equations) and stochastic models (which anticipate the increasing availability of data from small populations of cells) are considered. Several chapters emphasize Bayesian inference, so the editors have included an introduction to the philosophy of the Bayesian approach and an overview of current work on Bayesian inference. Taken together, the methods discussed by the experts in Learning and Inference in Computational Systems Biology provide a foundation upon which the next decade of research in systems biology can be built. ER -
 Lawrence, N.D. & Rattray, M.. (2010). A Brief Introduction to Bayesian Inference. Learning and Inference in Computational Systems Biology :-


text:
# Template compiling - biber, pdflatex, makeindex required to run in order

I'm using a template called Legrand Orange Book from http://www.latextemplates.com

It compiles fine with `TeXShop` (part of the `MacTeX` distribution), but I'm not sure I'm doing it exactly right. I have to compile it several times to get the index and bibliography (I'm assuming that's expected behaviour). What method should I be using (if any, besides using the compile button) to ensure it compiles correctly (i.e., how do I know for certain how many times to run compile)?

The instructions in the template say:

``````% Compiling this template:
% This template uses biber for its bibliography and makeindex for its index.
% This means that to update the bibliography and index in this template you
% will need to run the following sequence of commands in the template
% directory:
%
% 1) pdflatex main
% 2) makeindex main.idx -s StyleInd.ist
% 3) biber main
% 4) pdflatex main
``````

And I see the following references in the template, so I assume `TeXShop` is calling the necessary programs internally(?):

``````% Bibliography
\usepackage[style=alphabetic,sorting=nyt,sortcites=true,autopunct=true,babel=hyphen,hyperref=true,abbreviate=false,backref=true,backend=biber]{biblatex}

% Index
\usepackage{calc} % For simpler calculation - used for spacing the index letter headings correctly
\usepackage{makeidx} % Required to make an index
\makeindex % Tells LaTeX to create the files required for indexing
``````
-

The cool automation tool `arara` helps in situations like this. If you can afford to use it, here is what should be done.

If you don't have arara installed already (It comes bundled with texlive), download it from github and install. It needs java. Now let us assume that your file is `main.tex`. Put the following just before the `\documentclass{...}`:

``````% arara: pdflatex
% arara: makeindex: { style: StyleInd }
% arara: biber
% arara: pdflatex
\documentclass{...}
..
..
..
..
``````

Now open the command and issue `arara main` (from the same folder where your `main.tex` is and `arara` should be in system path). Entire job will be taken care by `arara` and you should have your pdf file ready.

For details on how to integrate `arara` with texshop see:

and the excellent arara manual.

-
Or may be I didn't understand the question properly! –  Harish Kumar Jul 24 '13 at 8:15

To get the order of compilations correct you should think about the informations the external tool needs to know.

As an example: biber needs only to know all citations you used in your document. This information is stored on the first pdflatex run in a `.bcf` file. So the correct order for biber is:

1. pdflatex % to create the bcf
2. biber % to create the bbl
3. more pdflatex runs (normally at least 2) to use the information in the bbl

`makeindex` on the other side not only needs to know the words you want to put in the index but also their pages! So the compilation order for `makeindex` is much more complicated:

1. enough runs of pdflatex + all tools which can affect page numbers of indexed words (this includes biber as the expansion of citations changes the pages) so that pdflatex can create a correct input file for makeindex
2. makeindex
3. pdflatex to include the index
4. perhaps everything again in case the index itself affected pages numbers of indexed words

So imho the compilation order described in your template is wrong. Naturally if you repeat it two or thrice on your final document it will come out correct at the end (at the small price of having run biber to often).

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text:
The Jacobi Iteration Method

# The Jacobi Iteration Method

So far we have looked at Gaussian Elimination to solve a system of $n$ linear equations in $n$ unknowns (provided that a solution exists) to $Ax = b$. Unfortunately, for very large systems, it is often unrealistic to apply Gaussian Elimination as the computations are long and cumbersome, and error can accumulate rapidly. As a result, we will begin looking at iterative methods to approximate the solution to $Ax = b$, the first iterative method being the Jacobi Iteration Method for solving a system of $n$ linear equations in $n$ unknowns.

Suppose that $x^{(0)} =\begin{bmatrix}x_1^{(0)}\\ x_2^{(0)}\\ \vdots\\ x_n^{(0)} \end{bmatrix}$ is an initial approximation to the solution $x$ of the following system of $n$ equations in $n$ unknowns:

(1)
\begin{align} \quad \begin{matrix} E(1): & a_{11}x_1 & + & a_{12}x_2 & + & \cdots & + & a_{1n}x_n & = & b_1\\ E(2): & a_{21}x_1 & + & a_{22}x_2 & + & \cdots & + & a_{2n}x_n & = & b_2\\ \vdots & \vdots & & \vdots & & \ddots & & \vdots & & \vdots\\ E(n): & a_{n1}x_1 & + & a_{n2}x_2 & + & \cdots & + & a_{nn}x_n & = & b_n & \end{matrix} \end{align}

For the Jacobi Iteration method, from the $i^{\mathrm{th}}$ equation in the system $Ax = b$ we isolate for the each variable $x_i$. Provided that $a_{ii} \neq 0$ for $i = 1, 2, ..., n$ we get that:

(2)

In sigma notation, for each $i = 1, 2, ..., n$, we have that:

(3)
\begin{align} \quad x_i = \frac{b_i - \sum_{j=1}^{n}_{j \neq i} a_{ij}x_j}{a_{ii}} \end{align}

To obtain our first approximation $x^{(1)}$ of the solution $x$ using the Jacobi Iteration Method, we take the isolated equations above and input the values of our initial approximation $x^{(0)}$ to get:

(4)

Or in sigma notation, for $i = 1, 2, ..., n$:

(5)
\begin{align} \quad x_i^{(1)} = \frac{b_i - \sum_{j=1}^{n}_{j \neq i} a_{ij}x_j^{(0)}}{a_{ii}} \end{align}

We can then use our approximation $x^{(1)}$ in a similar manner to obtain another approximation, $x^{(2)}$, and so forth in the hopes that these successive approximations converge to the actual solution $x$ of $Ax = b$. For each $k ≥ 1$, the $k^{\mathrm{th}}$ iteration of the Jacobi Iteration Method yields:

(6)

In sigma notation, we have that each entry of the approximation solution $x^{(k)}$ for $k ≥ 1$ and $i = 1, 2, ..., n$ can be obtained as:

(7)
\begin{align} \quad x_i^{(k)} = \frac{b_i - \sum_{j=1}^{n}_{j\neq i} a_{ij}x_j^{(k-1)}}{a_{ii}} \end{align}

For example, suppose that we want to solve the following system of $3$ linear equations in the three unknowns $x_1$, $x_2$, and $x_3$:

(8)
\begin{align} \quad \begin{matrix} E(1): & 3x_1 & + & x_2 & + & 4x_3 & = & 1 \\ E(2): & x_1 & + & 5x_2 & + & 9x_3 & = & 1\\ E(3): & 2x_1 & + & 6x_2 & + & 5x_3 & = & 1 \end{matrix} \end{align}

Furthermore, suppose that our initial approximation of the solution is $x^{(0)} = \begin{bmatrix} 0.22\\ 0.03\\ 0.06 \end{bmatrix}$. We first isolate the variable $x_i$ from equation $E(i)$ for $i = 1, 2, 3$ to get:

(9)
\begin{align} \quad x_1 = \frac{1 - x_2 - 4x_3}{3} \\ \quad x_2 = \frac{1 - x_1 - 9x_3}{5} \\ \quad x_3 = \frac{1 - 2x_1 - 6x_2}{5} \end{align}

For our first approximation $x^{(1)}$ using the Jacobi Iteration method, we have that:

(10)
\begin{align} \quad x_1^{(1)} = \frac{1 - (0.03) - 4(0.06)}{3} = 0.243 \\ \quad x_2^{(1)} = \frac{1 - (0.22) - 9(0.06)}{5} = 0.048\\ \quad x_3^{(1)} = \frac{1 - 2(0.22) - 6(0.03)}{5} = 0.184 \end{align}

We can then continue the Jacobi Iteration method to get successive approximations. The actual solution to this system is $x = \begin{bmatrix}0.2\bar{3}\\ 0.0\bar{3}\\ 0.0\bar{6}\end{bmatrix}$. Whether or not the Jacobi Iteration Method converges to $x$ will be discussed subsequently.