text:
# How to draw a Turing Machine for Even-Even

A TM to accept EVEN-EVEN (a collection of all strings with an even number of a's and an even number of b's) can be based on the algorithm:

1. Move up the string, changing a’s to A’s
2. Move down the string, changing b’s to B’s

Modified algorithm: To avoid the problem of crashing on the way down the $Tape$, change the letter in the first cell to an $X$ (if it's an $a$) and to a $Y$ (if it's a $b$). This way we can recognize the left end of the tape.

So far I have this but am not sure how to implement the modified algorithm (note: the e's from start to halt mean empty):

Any help is appreciated!

-
It seems that your language is regular. –  A.Schulz Nov 20 '12 at 6:12
I don't see where the problem is, your machine appears to work fine (the one in the diagram), although your description of the algorithm doesn't match the diagram (and as A.Schulz suggests, the language is regular). –  Luke Mathieson Nov 20 '12 at 7:59
Note that the usual assumption is that unused cells on the TM's tape contain some gap symbol like $\#$ or $\square$. Therefore, you can always recognise the end of your input without rewriting it. –  Raphael Nov 20 '12 at 9:27

From what I understand, your first algorithm intends to process the $a$ and $b$ inputs separately, which doesn't seem necessary. Your difficulty seems to come from formalizing a TM, which you can do in a couple ways.

Direct Conversion (Courtesy of Raphael in the comments)

You can view an NFA as TM that can not write and move only to the right. In other words, you don't have to do anything at all for the conversion, just add $R$ to transitions that consume a symbol and $N$ to $ε$-transitions, and then you basically have your equivalent TM.

Multitape Approach - Suppose we know that a multitape TM is equivalent to a TM, so we can have two tapes:

• An input tape (with a string $x\in(a|b)^*$), which we'll read through exactly once.
• A computation tape which will have one of four numbers $1$, $2$, $3$, or $4$, reflecting the state that the analogous finite automata (for which you have an image) might have. If we didn't want to expand the alphabet, we could instead have $\{\epsilon\epsilon,a\epsilon,\epsilon b,ab\}$ to indicate the parity of each letter.

As we proceed through the input, we update the computation tape accordingly: if we read an $a$, and we're in state $1$, we swtich to state $3$, etc. To reduce this back to one tape, we need another symbol $\#$ to separate tapes, perhaps $\hat{a},\hat{b}$ to designate the location of the heads and pre/append our computation tape (for which we only need two characters of space).

Formal Approach - we're going to define the set of states $Q$, the tape alphabet $\Gamma$, the blank symbol $\epsilon$, the input symbols $\Sigma$, the initial state $q_0$, the accepting states $F$ and the transition function $\delta$.

$$\langle Q=\{q_{\epsilon\epsilon},q_{a\epsilon},q_{\epsilon b},q_{ab}\},\Gamma=\{a,b\},b=\epsilon,\Sigma=\{a,b\},q_0=q_{\epsilon\epsilon},F=\{q_\epsilon\},\delta\rangle$$ $$\delta= \begin{vmatrix} & q_{\epsilon\epsilon} & q_{a\epsilon} & q_{\epsilon b} & q_{ab} \\ a & q_{a\epsilon} & q_{\epsilon\epsilon} & q_{ab} & q_{\epsilon b} \\ b & q_{\epsilon b} & q_{ab} & q_{\epsilon\epsilon} & q_{a\epsilon} \\ \end{vmatrix}$$

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Why don't we just encode the NFA's state in the TM's state? –  Raphael Nov 26 '12 at 0:49
If that's permissible, than that would certainly be more elegant; constructing TMs for regular languages is not typical. –  Merbs Nov 26 '12 at 3:04
I don't see why it would not be permissable; you can view an NFA as TM that can not write and move only to the right. In other words, you don't have to do anything at all for the conversion, just add $R$ to transitions that consume a symbol and $N$ to $\varepsilon$-transitions, and then you basically have your equivalent TM. –  Raphael Nov 26 '12 at 11:42
One technical note regarding that: depending on how you define a deciding TM, you may have to write something after all (and maybe even delete the input); usually that would be $0$ or $1$, depending on whether you accept or not. This is an easy update to the "naive" translation. For common definitions, entering an accepting (or non-accepting) state will be enough "output" for a decider, so the "naive" translation works just fine. –  Raphael Nov 26 '12 at 18:04

I ended up figuring it out, thanks all for the help.

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text:
Use of # in align block

I have a chemfig command which involves the use of #. It compiles properly on itself.

\chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----#(,0pt)\chembelow[2.5em]{}{+}(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)}

But when I put it into the align block, pdfLaTeX gives error.

\begin{align*}
\chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----#(,0pt)\chembelow[2.5em]{}{+}(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)}
\end{align*}

The error message is here: http://pastebin.com/3FLzX4RP

MWE:

\documentclass[10pt]{article}
\usepackage[usenames]{color} %used for font color
\usepackage{amssymb} %maths
\usepackage{amsmath} %maths
\usepackage[utf8]{inputenc} %useful to type directly diacritic characters
\usepackage{chemfig}

\pagestyle{empty}
\begin{document}

\begin{align*}
\chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----#(,0pt)\chembelow[2.5em]{}{+}(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)}
\end{align*}

\end{document}

In chemfig, # is used to adjust the bond shortening. The following example is from the chemfig documentation, pg. 6

I've tried using \#(0pt), but instead of adjusting the bond, it outputs the string directly.

• Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}.What does # do in chemfig? Perhaps it's required literally, i.e. \# within align? – user31729 Feb 22 '15 at 11:28
• @ChristianHupfer , thank you for the reminder. I have updated the the question. – Eana Hufwe Feb 22 '15 at 11:54
• Related, for the plus sign. By the way, aren't the angles 150,390? – Manuel Feb 22 '15 at 12:04

The \chemfig command relies on category code changes, in particular it changes # to an “other character” (category code 12). Such commands cannot appear in the argument to another command.

While equation and equation* are safe on this respect, align* isn't: the multiline alignment environments of amsmath absorb their content as the argument to a command. You'd have the same problem trying \fbox{\chemfig{...}}.

Using align* as a substitute for equation* is not recommended, so in the example case equation* would do. But if you really need a multiline environment, here's a trick:

\documentclass{article}
\usepackage{amsmath,chemfig}

\makeatletter
\newcommand{\chemfigon}{\begingroup\@makeother\#}
\let\chemfigoff\endgroup
\makeatother

\begin{document}

\chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----#(,0pt)\chembelow[2.5em]{}{+}(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)}

\chemfigon
\begin{align*}
\chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----#(,0pt)\chembelow[2.5em]{}{+}(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)}
\end{align*}
\chemfigoff

\end{document}

This makes # an “other character” before the input is grabbed as an argument.

• \chemfig{**[150,390,dash pattern=on 2pt off 2pt,late options={name=arccenter,label=center:+}]6(----#(,0pt)(-#(0pt)[3]H)(-#(0pt)[1]NO_3)-#(0pt)-)} this is more elegant in my opinion. – Manuel Feb 22 '15 at 12:20
• And this one, without #, should be prefered: \chemfig{**[135,405,dash pattern=on 2pt off 2pt]6(----(-[3]H)(-[1]NO_3)(-[6,,,,draw=none]+)--)} – unbonpetit Feb 22 '15 at 15:34

In theory, \chemfig{<code>} can be written in the argument of a macro since \CF@chemfig@iv does a \scantokens of the <code>. Unfortunately, there is a bug because via \scantokens, # becomes ## in the argument of a macro and this behaviour is not taken into account by \chemfig.

For example, if you write \fbox{\chemfig{A-#(0pt)B}}, the code of the molecule becomes A-##(0pt)B, where # is catcode 12. The bug is in \CF@grab@bondoffset: this macro expects a ( juste after the first #. And this is not the case when \chemfig is in the argument of a macro.

Here is a fix:

\documentclass{article}
\usepackage{chemfig}
\makeatletter
\def\CF@grab@bondoffset#1(#2)#3\@nil{%
\ifx\@empty#2\@empty
\else
\CF@if@instr{#2},%
{\CF@analyse@bondoffset#2\@nil}%
{\def\CF@start@offset{#2}}%
\fi
\def\CF@remain@afterbond{#3}%
}
\makeatother
\begin{document}
\fbox{\chemfig{A-#(0pt)B}}

\chemfig{A-#(0pt)B}
\end{document}

format: plain


text:
# cheb2ap

Chebyshev Type II analog lowpass filter prototype

## Syntax

``[z,p,k] = cheb2ap(n,Rs)``

## Description

example

````[z,p,k] = cheb2ap(n,Rs)` returns the zeros, poles, and gain of an order `n` Chebyshev Type II analog lowpass filter prototype with `Rs` dB of ripple down from the passband peak value in the stopband.```

## Examples

collapse all

Design a 6th-order Chebyshev Type II analog lowpass filter with 70 dB of ripple in the stopband. Display its magnitude and phase responses.

```[z,p,k] = cheb2ap(6,70); % Lowpass filter prototype [num,den] = zp2tf(z,p,k); % Convert to transfer function form freqs(num,den) % Frequency response of analog filter```

## Input Arguments

collapse all

Filter order, specified as an integer.

Data Types: `single` | `double`

Stopband ripple, specified as a scalar in decibels.

Data Types: `single` | `double`

## Output Arguments

collapse all

Zeros of the filter, returned as an `n`-length column vector. If `n` is odd, `z` has length `n`–1.

Poles of the filter, returned as an `n`-length column vector.

Gain of the filter, returned as a scalar.

## Algorithms

Chebyshev Type II filters are monotonic in the passband and equiripple in the stopband. The pole locations are the inverse of the pole locations of the `cheb1ap` function, whose poles are evenly spaced about an ellipse in the left half plane. The Chebyshev Type II stopband edge angular frequency ω0 is set to 1 for a normalized result. This value is the frequency at which the stopband begins. The filter has a magnitude response of 10–Rs/20.

Chebyshev Type II filters are sometimes called inverse Chebyshev filters because of their relationship to Chebyshev Type I filters. The `cheb2ap` function is a modification of the Chebyshev Type I prototype algorithm:

1. `cheb2ap` replaces the frequency variable ω with 1/ω, turning the lowpass filter into a highpass filter while preserving the performance at ω = 1.

2. `cheb2ap` subtracts the filter transfer function from unity.

The transfer function is given by

`$H\left(s\right)=\frac{z\left(s\right)}{p\left(s\right)}=k\frac{\left(s-z\left(1\right)\right)\left(s-z\left(2\right)\right)...\left(s-z\left(n\right)\right)}{\left(s-p\left(1\right)\right)\left(s-p\left(2\right)\right)...\left(s-p\left(n\right)\right)}.$`

## References

[1] Parks, Thomas W., and C. Sidney Burrus. Digital Filter Design. New York: John Wiley & Sons, 1987.

## Version History

Introduced before R2006a


text:
# solving the wave equation using separation of variables

How can I solve the following Wave equation using separation of variables? I am interested in a general way of solving all problems of this type, not some sort of tricks that for some reason happen to work on this problem only (not sure if there are any).

$\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}$

$0<x<L$, $t>0$

With the initial and boundary conditions:

$u(x,0)=f(x)=5cos(\frac{10x\pi}{L})$

$u_t(x,0)=0$

$u_x(0,t)=u_x(L,t)$

What I have done:

Assume: $u(x,t)=X(x)T(t)$

=> $\frac{X''}{X} = \frac{T''}{T}=-\lambda$

Now I can solve $X''+\lambda X=0$ I get the conditions $X'(0)=X'(L)=0$

case1: $\lambda>0$

Characteristic equation $r^2 + \lambda =0 <=> r_1=\sqrt\lambda i, r_2=-\sqrt\lambda i$

General solution: $X(x)=c_1 cos(\sqrt\lambda x) + c_2 sin(\sqrt\lambda x)$

$X'(x)=-c_1 \sqrt\lambda sin(\sqrt\lambda x) + c_2 \sqrt\lambda cos(\sqrt\lambda x)$

Now from the conditions $X'(0)=X'(L)=0$ I get

$0=0 + c_2 \sqrt\lambda *1 => c_2=0$ and

$0=-c_1 \sqrt\lambda sin(L\sqrt\lambda)$

Therefore, the eigenvalues are the values of $lambda$ for which $L\sqrt\lambda=n*\pi$ (has to do with when $sin(L\sqrt\lambda)=0$

The eigenvalue is $\lambda_n = \frac{n^2 \pi ^2}{L^2}$ for $n=1,2,...$ and corresponding eigenfunction is $cos(\frac{xn\pi}{L})$

case2: $\lambda=0$

The eigenfunction is 1.

case3: $\lambda <0$

no eigenvalues.

Note: If I let $\lambda_n=\frac{n^2 \pi^2}{L^2}$ for $n=0,1,2,...$ Then case1 and case2 are both accounted for.

$T''+\lambda_n T=0 => T_n(t)=k_1 cos(\frac{tn\pi}{L})+k_2 sin(\frac{tn\pi}{L})$

$T_n'(t)=-k_1 \frac{n\pi}{L}sin(\frac{tn\pi}{L}) + k_2 \frac{n\pi}{L} cos(\frac{tn\pi}{L})$

using the initial condition $T'(t)=0$ I get $k_2=0$ so the fundamental solutions are $u_n(x,t)=X_n(x)T_n(t)$ (the two eigenfunctions multiplied together).

The solution is:

$u(x,t)=\sum_{n=0}^\infty a_n cos(\frac{tn\pi}{L})cos(\frac{xn\pi}{L})$

Now what I am wondering, first of all is what I have done so far fine, I suppose to have more of a complete answer I should have probably explained a bit about lambda <0. Secondly what is the logic in how to find $a_n$ ? How is finding $a_n$ generally done on these type of problems? I understand that I have to use the last initialvalue condition (that I haven't used yet), but how?

The solution you got looks okay.

In general, to get the coefficient $a_n$, plug in the initial condition $u(x,0) = f(x)$,then multiply by $\cos(x n' \pi/L)$.

$$\sum_{n=0}^\infty a_n \cos( \frac{n \pi x}{L}) \cos(\frac{n' \pi x}{L}) = f(x)\cos(\frac{n' \pi x}{L})$$

Now integrate from $0$ to $L$

$$\sum_{n=0}^\infty a_n \int_0^L \cos( \frac{n \pi x}{L}) \cos(\frac{n' \pi x}{L})dx = \int_0^L f(x)\cos(\frac{n' \pi x}{L})dx$$

If $n \ne n'$ the integral is $0$. In the case that $n = n'$, we get

$$a_n \int_0^L \cos(\frac{n \pi x}{L})^2 dx = \int_0^L f(x) \cos(\frac{n \pi x}{L}) dx$$

So if you divide by the integral on the left (because it is non-zero) we get an equation for $a_n$. (Note that the expression is of a different form if $n = 0$)

However, this is not necessary here, because of the form of $f(x)$

$$\sum_{n=0}^\infty a_n \cos(\frac{n \pi x}{L}) = 5 \cos(\frac{10 \pi x}{L})$$

This implies that $a_{10} = 5$ and $a_n = 0$ for $n \ne 10$

• @stat333: I updated to show the general approach – infinitylord Jul 15 '17 at 4:37
• @stat333 Yes, you would just integrate from $-L$ to $L$ instead, because that is the region that $f$ is defined. Though it's note worthy that if $f$ is an even function, this reduces back to the $0$ to $L$, and if $f$ is odd, the integral is $0$, so there is no solution with cosines. I don't know what you mean by the period of the problem. $\cos(n \pi x/L)$ is $2 L/n$, so each term has a different period (that is the nature of Fourier series). And yes, that is how you treat the case that it depends on $\sin$ – infinitylord Jul 15 '17 at 5:56
• if I have $u(x,t)=\sum sinnx(\alpha_n cos(2nt) + \beta_n sin(2nt)$ and the initial value conditions: $u(x,0)=sin(3x)-4sin(10x)$ and $\frac{\partial u}{\partial t}(x,0)=2sin(4x)+sin(6x)$ and $0<x<\pi$ how would it be then? for instance solving for $\alpha_n$ using the first condition, how would that work when it comes to writing up the formula for $\alpha_n$ ? I am able to see by inspection that $\alpha_3 =1$ and $\alpha_{10} = -4$ by comparing terms. – stat333 Jul 16 '17 at 3:56
• @stat333: You would write it as a piecewise function. Say $\alpha_3 = 1$, $\alpha_{10} = -4$, and $\alpha_n = 0$ if $n \ne 3, 10$. Writing up the general function would be the same. Plug in the initial conditions, multiply by the appropriate trig function, integrate to isolate the coefficient – infinitylord Jul 17 '17 at 1:56


text:
# Solve for: limit as x approaches 3 of (x^2-9)/(x+3)

## Expression: $\lim _{x\to 3}(\frac{x^{2}-9}{x+3})$

Plug in the value $x=3$

$=\frac{3^{2}-9}{3+3}$

Simplify

$=0$

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text:
Volume of a cylinder

A battery in the shape of a perfect cylinder has a volume of 15 cm³. Its height is 6 cm. Find the radius of the battery.

h= 6 cm

volume= 15 cm³

We know the formula for the volume of a cylinder is:

$h\pi { r }^{ 2 }$

So, $15=6\pi { r }^{ 2 }$, then we divide both sides by 6. $\frac { 15 }{ 6\pi } =\frac { 6\pi }{ 6\pi } { r }^{ 2 }$
This simplifies to: $.7958={ r }^{ 2 }$. Taking the square root of both sides gives us: $.89={ r }$. In other words, the radius of the battery is a little less than one centimeter.


text:
# tcs math – some mathematics of theoretical computer science

## April 13, 2008

### Planar multi-flows, L_1 embeddings, and differentiation

Filed under: Math — Tags: , , , , — James Lee @ 9:18 am

In this post, I’ll discuss the relationship between multi-flows and sparse cuts in graphs, bi-lipschitz embeddings into $L_1$, and the weak differentiation of $L_1$-valued mappings. It revolves around one of my favorite open questions in this area, the planar multi-flow conjecture.

### The max-flow/min-cut theorem

Let $G=(V,E)$ be a finite, undirected graph, with a mapping $\mathsf{cap} : E \to \mathbb R_+$ assigning a capacity to every edge. If $\mathcal P_{s,t}$ is the set of all paths from $s$ to $t$, then an $s\!-\!t$ flow is a mapping $F : \mathcal P_{s,t} \to \mathbb R_+$ which doesn’t overload the edges beyond their capacities: For every edge $e \in E$,

$\displaystyle \sum_{\gamma \in \mathcal P_{s,t} : e \in \gamma} F(\gamma) \leq \mathsf{cap}(e).$

The value of the flow F is the total amount of flow sent: $\mathsf{val}(F) = \sum_{\gamma \in \mathcal P_{s,t}} F(\gamma)$. A cut in $G$ is a partition $V = S \cup \bar S$ which we will usually write as $(S, \bar S)$. Naturally, one defines the capacity across $(S, \bar S)$ by $\mathsf{cap}(S,\bar S) = \sum_{xy \in E} \mathsf{cap}(x,y) |\mathbf{1}_S(x)-\mathbf{1}_S(y)|$, where $\mathbf{1}_S$ is the characteristic function of $S$. Since one can only send as much flow across a cut as there is capacity to support it, it is easy to see that for any valid flow $F$ and any cut $(S, \bar S)$ with $s \in S$ and $t \in \bar S$, we have $\mathsf{val}(F) \leq \mathsf{cap}(S, \bar S)$. The classical max-flow/min-cut theorem says that in fact these upper bounds are achieved:

$\max_{F} \mathsf{val}(F)=\min_{(S, \bar S)} \mathsf{cap}(S, \bar S)$

where the maximum is over all $s\!-\! t$ flows, and the minimum is over all cuts $(S, \bar S)$ with $s \in S$ and $t \in \bar S$.

### Multi-commodity flows and sparse cuts

We will presently be interested in multi-commodity flows (multi-flows for short), where we are given a demand function $\mathsf{dem} : V \times V \to \mathbb R_+$ which requests that we send $\mathsf{dem}(x,y)$ units of flow from $x$ to $y$ for every $x,y \in V$. In this case, we’ll write the value of a valid flow (i.e. one which doesn’t try to send more total flow than an edge can carry) as

$\displaystyle \mathsf{val}(F) = \min_{x,y \in V} \frac{\sum_{\gamma \in \mathcal P_{x,y}} F(\gamma)}{\mathsf{dem}(x,y)}$

where the minimum is over pairs with $\mathsf{dem}(x,y) \neq 0$. So if the value of the flow is $\frac12$, it means that every pair gets at least half the flow it requested (well, demanded).

Again, cuts give a natural obstruction to flows. If we define $\Phi(S, \bar S) = \displaystyle \frac{\mathsf{cap}(S, \bar S)}{\mathsf{dem}(S, \bar S)}$, where $\mathsf{dem}(S, \bar S) = \sum_{x,y \in V} \mathsf{dem}(x,y) |\mathbf{1}_S(x)-\mathbf{1}_S(y)|$ is the total demand requested across $(S, \bar S)$, then $\mathsf{val}(F) \leq \Phi(S, \bar S)$ for any valid flow $F$. Unfortunately, it is no longer true in general that $\max_F \mathsf{val}(F) = \min_{(S, \bar S)} \Phi(S)$. (It is true as long as the demand is supported on a set of size at most 4, while it stops being true in general when the demand is supported on a set of size 5 or larger.)

In fact, the gap between the two can be arbitrarily large, as witnessed by expander graphs (this rather fruitful connection will be discussed in a future post). For now, we’ll concentrate on a setting where the gap is conjectured to be at most $O(1)$.

### The planar multi-flow conjecture

It has been conjectured that there exists a universal constant $C \geq 1$ such that if $G=(V,E)$ is a planar graph (i.e. can be drawn in the plane without edge crossings), there is a $C$-approximate multi-commodity max-flow/min-cut theorem: For any choice of $\mathsf{dem}$ and $\mathsf{cap}$, one has a

$\displaystyle \frac{1}{C} \min_{(S, \bar S)} \Phi(S, \bar S) \leq \max_F \mathsf{val}(F) \leq \min_{(S, \bar S)} \Phi(S, \bar S)$ (1)

The conjecture first appeared in print here, but was tossed around since the publications of Linial, London, and Rabinovich and Aumann and Rabani, which recast these multi-flow/cut gaps as questions about bi-lipschitz mappings into $L_1$ (discussed next). Perhaps the most compelling reason to to believe the conjecture is the beautiful result of Klein, Plotkin, and Rao which shows that $C = O(1)$ for any planar instance where $\mathsf{dem}(x,y) = 1\,\,\forall x,y \in V$ (this is called a uniform multi-flow instance).

It is relatively easy to see that we cannot take $C=1$, as the following example of Okamura and Seymour shows.

In the example, the non-zero demands are $\mathsf{dem}(s_i,t_i) = 1$ for $i \in \{1,2,3,4\}$. It is easy to check that every cut has $\Phi(S,\bar S) \geq 1$, but the value of the maximum flow is only $3/4$, implying that $C \geq 4/3$. If we use the same pattern on the complete bipartite graph $K_{2,n}$ (instead of $K_{2,3}$), taking $n \to \infty$ shows that we must have $C \geq 3/2$ in (1). (Later a differentiation argument applied to a different family of graphs will show that we need $C \geq 2$ in (1).)

For any graph $G$, define $\mathsf{gap}(G)$ as the smallest value $C$ such that (1) holds for every choice of capacities and demands on $G.$ We will now see how $\mathsf{gap}(G)$ is determined by the geometric properties of path metrics defined on $G.$

### Bi-Lipschitz embeddings into $L_1$

Consider a metric space $(X,d)$, and the space of absolutely integrable functions $L_1 = L_1([0,1])$, equipped with the $\|\cdot\|_1$ norm. A mapping $f : X \to L_1$ is said to be $C$-bi-lipschitz if there exist constants $A,B > 0$ such that for all $x,y \in X$,

$\frac{1}{B} d(x,y) \leq \|f(x)-f(y)\|_1 \leq A\,d(x,y)$

and $C = A \cdot B$ (so if $C=1$, then $f$ is an isometry up to scaling). The infimal value of $C$ for which $f$ is $C$-bi-lipschitz is called the distortion of $f$. Define $c_1(X)$ as least distortion with which $(X,d)$ maps into $L_1$.

We will now relate $\mathsf{gap}(G)$ to the $L_1$-distortion of the geometries that $G$ supports. Any set of non-negative weights $w : E \to \mathbb R_+$ on the edges of $G$ gives rise to a distance $\mathrm{dist}_w$ on $V$ defined by taking shortest-paths. Define $c_1(G)$ to be the maximum of $c_1(V, \mathrm{dist}_w)$ as $w$ ranges over all such weightings. (Strictly speaking, $\mathrm{dist}_w$ is only a pseudometric.)

Here is the beautiful connection referred to previously (see this for a complete proof).

Theorem (Linial-London-Rabinovich, Aumann-Rabani): For any graph $G$, $\mathsf{gap}(G) = c_1(G)$.

Metric obstructions. To give an idea of why the theorem is true, we mention two facts. It is rather obvious how a cut obstructs a flow. A more general type of obstruction is given by a metric $\mathrm{dist}$ on $V$. For any valid flow $F$,

$\displaystyle \mathsf{val}(F) \leq \frac{\sum_{xy \in E} \mathsf{cap}(x,y)\,\mathrm{dist}(x,y)}{\sum_{x,y \in V} \mathsf{dem}(x,y)\,\mathrm{dist}(x,y)}$ (2)

To see why this holds, think of giving every edge $(x,y) \in E$ a “length” of $\mathrm{dist}(x,y)$, and having a cross-sectional area of $\mathsf{cap}(x,y)$. Then the numerator in (2) is precisely the “volume” of the network, while the denominator is the total “volume” required to satisfy all the demands: By the triangle inequality, sending one unit of flow from $x$ to $y$ requires volume at least $\mathrm{dist}(x,y)$. Observe that the bound (2) is a generalization of the cut upper bound when we define the pseudometric $\mathrm{dist}(x,y) = |\mathbf{1}_S(x)-\mathbf{1}_S(y)|$ for some $S \subseteq V$ (this is called a cut pseudometric).

It turns out (by linear programming duality) that in fact metrics are the correct dual objects to flows, and maximizing the left hand side of (2) over valid flows, and minimizing the right-hand side over metrics yields equality (it is also easy to see that the minimal metric is a shortest-path metric). One might ask why one continues to study cuts if they are the “wrong” dual objects. I hope to address this extensively in future posts, but the basic idea is to flip the correspondence around: minimizing $\Phi(S,\bar S)$ is NP-hard, while maximizing $\mathsf{val}(F)$ can be done by linear programming. Thus we are trying to see how close we can get to the very complex object $\Phi$, by something which is efficiently computable.

The cut decomposition of $L_1$. Now that we see why metrics come into the picture, let’s see how $L_1$ presents itself. Define a cut measure on a finite set $X$ as a mapping $\mu : 2^X \to \mathbb R_+$ which satisfies $\mu(S) = \mu(\bar S)$ for every $S \subseteq X$.

Fact: Given $F : X \to L_1$, there exists a cut measure $\mu$ on $X$ such that $\|F(x)-F(y)\|_1 = \sum_S \mu(S) |\mathbf{1}_S(x)-\mathbf{1}_S(y)|$ for every $x,y \in X$. Conversely, for every cut measure $\mu$, there exists a mapping $F : X \to L_1$ for which the same equality holds.

Thus every $L_1$-distance on a finite set is precisely a weighted sum of cut pseudometrics (and, of course, cuts are where this story began). Proving the fact is straightforward; first check that it holds for $F : X \to \mathbb R$, and then integrate. As an example, consider these two isometric embeddings, presented via their cut measures (the graphs are unit-weighted):

In the 4-cycle, each cut has unit weight. The corresponding embedding into $(\mathbb R^2, \|\cdot\|_1)$ maps the four vertices to $\{(0,0),(0,1),(1,0),(1,1)\}$. In the second example, the dashed cuts have weight 1/2, and the dotted cut has weight 1. (Exercise: Verify that every tree embeds into $L_1$ isometrically.)

Two comments are in order; first, when $X$ is finite, one can equivalently take the target space to be $\mathbb R^{|X| \choose 2}$ equipped with the $\|\cdot\|_1$ norm (exercise: prove this using Caratheodory’s theorem). Thus one might ask why we would introduce a function space $L_1$ in the first place. One reason is that the dimension is irrelevant; a deeper reason is that when $X$ is infinite (in which case an appropriately stated version of the fact holds), $L_1$ is the proper setting (and not, e.g. $\ell_1$). This arises, for instance, in arguments involving ultralimits in the passage between the finite and infinite settings.

### The planar embedding conjecture

Using the connection with $L_1$-embeddings, we can now state the planar multi-flow conjecture in its dual formulation.

Planar embedding conjecture: There exists a constant $C \geq 1$ such that every shortest-path metric on a planar graph $G=(V,E)$ admits a $C$-bi-lipschitz embedding into $L_1$.

By a relatively simple approximation argument in one direction, and a compactness argument in the other, one has the following equivalent conjecture.

Riemannian version: There exists a constant $C \geq 1$ such that every Riemannian metric on the 2-sphere admits a $C$-bi-lipschitz embedding into $L_1$.

Recently, Indyk and Sidiropoulos proved that if it’s true for the 2-sphere, then it’s true for every compact surface of genus $g \geq 1$, where the constant $C$ depends on $g$ (as it must).

We can now state some known results in the dual setting of embeddings. Let $G=(V,E)$ be a planar graph, and let $\mathrm{dist}$ be a shortest-path metric on $V$. The previously mentioned theorem of Klein, Plotkin, and Rao can be stated as follows: There exists a non-expansive mapping $f : V \to \mathbb R$ such that

$O(1) \sum_{x,y \in V} |f(x)-f(y)| \geq \sum_{x,y \in V} \mathrm{dist}(x,y)$.

In Gromov’s language, the observable diameter of planar graph metrics is almost as large as possible (i.e. no forced concentration of Lispchitz mappings).

A classical theorem of Okamura and Seymour is stated in the embedding setting as: Let $F \subseteq V$ be a face in some drawing of $G$ in the plane. Then there exists a non-expansive mapping $f : V \to L_1$ such that $f|_F$ is an isometry. In the flow setting, this says that if the demand function is supported on the pairs belonging to a single face, then (1) holds with $C=1$.

Finally, we mention the bound of Rao which shows that $c_1(V,d) = O(\sqrt{\log |V|})$ for planar metrics. A slightly more general version says that whenever the demand is supported on at most $k$ pairs, the flow/cut gap can be at most $O(\sqrt{\log k})$.

1. Planar multi-flows, $L_1$ embeddings, and differentiation


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# ΛQCDΛQCD from the beta function of QCD coupling constant

spaghetti3451
The beta function for the strong coupling ##g_3## is given by

##\displaystyle{\mu \frac{\partial g_{3}}{\partial\mu}(\mu) = - \frac{23}{3} \frac{g_{3}^{3}(\mu)}{16\pi^{2}},}##

with

##\alpha_{3}(\mu = M_{Z}) = 0.118.##

We can use separation of variables to solve the beta function equation:

##\displaystyle{\int \frac{dg_{3}}{g_{3}^{3}} = - \frac{23}{48\pi^{2}} \int \frac{d\mu}{\mu}}##

##\displaystyle{\frac{1}{g_{3}^{2}} = \frac{23}{24\pi^{2}}\ln\left(\frac{\mu}{\Lambda_{\text{QCD}}}\right)}##

##\displaystyle{\frac{1}{\alpha_{3}} = \frac{23}{6\pi}\ln\left(\frac{\mu}{\Lambda_{\text{QCD}}}\right).}##

Using the physical condition, we then find that

##\displaystyle{\Lambda_{\text{QCD}} = \left(91\ \text{GeV}\right) e^{-6\pi/2.714}.}##

--------------------------------------------------------------------------------------------------------------------------------------

What is the physical significance of ##\Lambda_{\text{QCD}}## as obtained by solving for the the strong coupling ##g_3##?

Gold Member
I am not sure I understand your question, but LambdaQCD introduces a natural (lower) cut-off for your theory, as shown by the equation of 1/α3 (or g32)... ? you already used that when you integrated your energy scale.

spaghetti3451
Does this mean that the theory of QCD is undefined below the QCD scale?

I thought that QCD does not have a Landau pole.

Gold Member
Does this mean that the theory of QCD is undefined below the QCD scale?

I think it's known though that QCD is non-perturbative at low energies? And your LambdaQCD you found (~90MeV) is the energy scale where α3 = 1 (which already whispers that it's not a good idea to expand into powers of α3).
That's why there is Lattice QCD.
Of course there can be more, but I am not very confident to write, for example I think LambdaQCD appears only in some renormalization schemes...

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# Metaphilosophy

Metaphilosophy is a website product that describes and visualizes the state of academic philosophy. This document provides product and technical specifications for its development. I'm still researching this product. If you can build it faster and better than me, feel free to do so, but please let me know about it.

## Product specification

The following sections describe the purpose and scope of the Metaphilosophy product.

### Objective

The objective of Metaphilosophy is to provide anyone interested in academic philosophy with descriptions and visualizations of who academic philosophers are, which topics they write and talk about, and how those philosophers and topics are related to each other. The descriptions and visualizations help provide clarity on the state of academic philosophy, which helps determine the future of philosophy as an academic discipline.

### User stories

The following stories describes potential users for Metaphilosophy and their use cases.

#### Interested persons

People interested in philosophy can use Metaphilosophy to understand trends in academic philosophy and how topics relate to each other. They can use trends and relations to help them discover new topics or decide which problems are worth thinking more about.

• As a person interested in philosophy, I can see which topics academic philosopher are writing about.
• As a person interested in philosophy, I can see how topics are related to each other.

Academics include current and prospective teachers and students, such as professors and students of high school, undergraduate institutions, or graduate programs.

• As an academic, I can see which academic departments have many people thinking about the topics that I'm interested in.
• As an academic, I can see the number of faculty and graduate students at each academic department.
• As an academic, I can see the distribution of seniority rankings among the faculty at each academic department, such as "Assistant Professor", "Associate Professor", and "Professor".

#### Out of scope

The following stories are valuable and worth pursuing, but are likely out of the scope of this Metaphilosophy product because of the difficulty of obtaining such data.

• As an teacher or professor, I can see which academic departments have faculties that are diverse in gender, race, religion, and research topics.
• As an teacher or professor, I can see that I'm being paid equally to others of my level despite differences in gender, race, religion, and research topics.
• As an prospective student, I can see the employment outcomes of people who graduated from each academic department in philosophy.

### Methodology

The methodology for Metaphilosophy is as follows:

1. Scrape relevant websites for text.
2. Process and analyze the text.
3. Create visualizations of the text.
4. Display the visualizations on an accessible website.

The analysis of the text can be standard data analysis and natural language processing. The visualizations can be scatter plots, word clouds, mind maps, bar graphs, histograms, pie charts, distribution curves, geographical maps, and other common and understandable visualizations. The visualizations can be in any file format suitable for websites, such as .svg, .png, or .jpg.

#### Sources

Sources for the text data include:

### Success metrics

The objective of Metaphilosophy is to improve awareness of the state of academic philosophy in hopes to interested persons discover philosophy better and drive greater diversity in the academic philosophy. The simplest success metric for improving awareness is increasing unique page views of the website.

## Technical specification

The following sections describe how to build the Metaphilosophy product. As a guiding principle, build the product in a way that you can make its source public for others to repeat the research and recreate visualizations that Metaphilosophy provides.

### Technologies

Python and TypeScript and promising technologies for building Metaphilosophy. Python has many libraries to obtain, analyze, and visualize text data, and some libraries can output the visualizations in file formats suitable for the web, such as .svg, .png, or .jpg. TypeScript and React simplify the process to build a simple website for the visualizations.

Consider the following Python libraries:

• pandas
• scattertext
• wordcloud
• matplotlib
• seaborn
• plotly
• bokeh
• pygal
• geoplotlib

### Architecture

There's very little about the Metaphilosophy product that's distinct to or dependent on the academic discipline of philosophy. You can ask the same questions and write the same user stories about other academic subjects. In case you want to create similar visualizations about other academic subjects, you should build the product in a way that's agnostic to philosophy as possible.


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GR 8677927796770177 | # Login | Register

GR9277 #21
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{21}
A soap film with index of refraction greater than air is formed on a circular wire frame that is held in a vertical plane. The film is viewed by reflected light from a white-light source. Bands of color are observed at the lower parts of the soap film, but the area near the top appears black. A correct explanation for this phenomenon would involve which of the following?
I. The top of the soap film absorbs all of the light incident on it; none is transmitted.
II. The thickness of the top part of the soap film has become much less than a wavelength of visible light.
III. There is a phase change of 180 degrees for all wavelengths of light reflected from the front surface of the soap film.
IV. There is no phase change for any wavelength of light reflected from the back surface of the soap film.

1. I only
2. II and III only
3. III and IV only
4. I, II, and III
5. II, III, and IV

Optics$\Rightarrow$}Thin Film

Elimination time.

I. Can't be this, since one knows from basic thin-film theory that choice IV is right. (None of the letter choices allow for both choices I and IV.)

II. Thin film theory has $2t=\lambda/2$ for constructive interference and $2t=\lambda$ for destructive interference. Thus, the thickness of the film is smaller than that of the light. (Search on the homepage of this site for more on thin film theory---it is explained in the context of other problems.)

III. This phase change allows for the half-integer constructive interference.

IV. Phase change only occurs when light travels from a medium with lower index of refraction to a medium with higher index of refraction. Since at the back surface, the light would be going from higher to lower index of refraction, there is no phase change.

Thus, choice (E).

Alternate Solutions
 djh1012014-08-29 15:59:03 All you need to know to solve this problem is that reflection off of a surface with larger n causes a phase shift (you actually don't even need to remember whether larger or smaller n causes the phase shift, just that one of them does. From this: A. Why would the top of the film absorb everything while the bottom doesn't? B. If this is true, then waves reflected from the front and back would have a negligible phase shift (excluding the flip) relative to each other. When you add in the phase flip of the front reflected waves, well, you now have destructive interference. C. Follows from B. D. Follows from B.Reply to this comment
Yaroslav Shustrov
2017-09-10 09:59:21
https://physics.stackexchange.com/questions/333368/why-are-thin-soap-bubbles-colorless/333419#333419\r\n\r\nThis link is extremely useful in order to understand the question in more details.
socolenco_
2015-11-11 14:20:40
For me it\'s a bit confusing the phrase in IV: \"reflected from the back surface of the soap film\". Do they mean \"refracted\"??
djh101
2014-08-29 15:59:03
All you need to know to solve this problem is that reflection off of a surface with larger n causes a phase shift (you actually don't even need to remember whether larger or smaller n causes the phase shift, just that one of them does. From this:

A. Why would the top of the film absorb everything while the bottom doesn't?
B. If this is true, then waves reflected from the front and back would have a negligible phase shift (excluding the flip) relative to each other. When you add in the phase flip of the front reflected waves, well, you now have destructive interference.
C. Follows from B.
D. Follows from B.
segfault
2009-09-03 12:14:04
To clear up some confusion about II: "much less than a wavelength" does not mean it's still some half-integer multiple of $\lambda$. If we have an infinitely thin film, then the *reflected* light will destructively interfere by default, because the light reflected off of the front surface of the film sees a 180 phase shift, while the light reflected off the back surface is not phase shifted; thus there is a total of a 180 phase shift--> destructive interference-->"top appears black"
Jeremy
2007-10-30 17:06:39
When I first solved this problem, statement II bothered me because of the "much less than" phrase. The problem was, like all the other posts here, I used $t=\lambda/2$ (first minima) and the reasoning that goes along with it. But this leads to problems pointed out in the other posts in addition to implicitly saying $1/2 \ll 1$. Typically much less means at least an order of magnitude, but I knew that some thickness argument had to be used to explain the top part being black, so I went with it.

The fault with the first minima approach is that it only succeeds in complete destructive interference for ONE wavelength, while the problem uses a white light source, so this cannot be the proper explanation. It turns out that "much less than" really means what it says. In fact, the film thickness is so small that the path difference for light reflecting off the back of the film is negligible, and the only thing that matters is the $180^{\circ}$ phase change for the light reflected at the front surface. Thus, regardless of the wavelength, there is always destructive interference.
 madfish2007-10-31 16:06:50 Perhaps (II) implies that there is not enough high index material to even create a thin film?
sharpstones
2006-12-01 09:58:45
I don't understand the explanation for II. According to your explanation the thickness will be less then the wavelength of light for both constructive and destructive interference. In fact for the constructive interference case the thickness is even smaller. So how is this an explanation for why the top is dark and the middle is bright?
 Zephyr3.142007-04-13 20:13:07 I think it is because it is white light (parts of all colors of visible light) and $t = \frac{\lambda}{2}$ so $t << \lambda_{visible}$. This describes the thickness of the soap, which would be integral in "a correct explanation for this phenomenon."
 Richard2007-10-17 18:39:30 I think maybe Yosun meant to be more rigorous and include the general formula for constructive interference: $t = (n+\frac{1}{2})\lambda$ So usually the thickness will be greater than a wavelength. ??

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$


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Home > Latex Error > \requirepackage Or \loadclass In Options Section

# \requirepackage Or \loadclass In Options Section

## Contents

Processing can get really weird if the file changes while it is still being processed. mode A # character is where it should not be. This can't happen ( ! Now latex2html has a nasty habit of leaving blank lines when it removes the labels from equations.

If stumped, try the general tricks. ! Just shows a * and sits there. Is there something wrong in the column specifications? Unless you are defining a very long list, you probably brought this upon yourself somehow. (Who else is readily available to be blamed?) If you are really defining a long list, http://tex.stackexchange.com/questions/203387/declare-package-options-and-loading-packages-leads-to-options-section-error

## \requirepackage Or \loadclass In Options Section

Hot Network Questions Were students "forced to recite 'Allah is the only God'" in Tennessee public schools? Display math should end with . Then after having saved your corrections so far, reprocess. Suggested solution: install the txfonts package.

You did not close something properly. Missing font identifier ! LaTeX Error: Too deeply nested. ! It Attempts To Load `pdftexcmds' In The Options Section Missing control sequence inserted A control sequence means a LaTeX command, like \backslash.

To fix this, put the \label{...} in equations at the end, followed by a %. Aiaa Journal Latex Template There is something not matching something else, like maybe a \fi, \or, or \else not matching an earlier \if. A misplaced \protect (before something like a length or a number) can also produce this error. Converting a counter > 26 to a letter a-z can also cause this error.

Now packages by themselves do not have errors. Latex Requirepackage As Jubobs commented, we'd need to see a complete minimal class to help further. –Nicola Talbot Jun 15 '14 at 15:29 4 @Jubobs: The difference between \RequirePackage and \usepackage is In general, make sure that you put your pictures in the same folder as index.tex. LaTeX Error: Illegal character in array arg.

## Aiaa Journal Latex Template

Please use ... ! http://tex.stackexchange.com/questions/19919/whats-the-difference-between-requirepackage-and-usepackage more hot questions question feed lang-tex about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation \requirepackage Or \loadclass In Options Section Read up on arrays, tables (tabular), or whatever on http://en.wikibooks.org/wiki/LaTeX/. Latex Error File Elsarticle Cls Not Found Example: $\alpha\beta=0\mbox{ if$\beta$is zero}$ needs the \$ signs around the second \beta as shown.

What is the meaning of the so-called "pregnant chad"? Or try hitting c while holding down Ctrl. Ambiguous; you need another { and } Compare the items on double subscripts and superscripts. Like it says?

How to create a company culture that cares about information security? Aiaa Latex Template Number too big ! Missing } inserted !

## If stumped, try the general tricks. !

Maybe another error message can clarify the problem. Reduce its size. While defining columns, there is a p not followed by {width} argument. How To Install Elsarticle.cls In Latex The following code in the main exam document works: \usepackage{listings} \usepackage{probsoln} When I look up information sources I gather the information that RequirePackage and usepackage are identical with the difference that

Duplicate pattern I assume this refers to hyphenation patterns. Immidiately get an error3Error with \DeclareRobustCommand, \usepackage{hyperref,xcolor} and \usepackage{graphicx}3Error when typing Matrix and Binomial coefficients3Why is adding \usepackage{physics} causing an error with \let\left\relax2I keep getting an error message that report.cls was If you leave a single blank line in an equation, it will generate the error message ! Extra \.... !

These errors will then show up in its math images processing. share|improve this answer answered Jun 15 '14 at 15:55 Nicola Talbot 21.2k4076 Thank you for this answer. Note that in math, ' is a superscript too. (It is really the same as ^\prime.) If stumped, try the general tricks. ! is undefined (character ...) In math mode, a special font is needed.

Argument of ... Type H for immediate help. ... If stumped, try the general tricks. ! Check the value of \pdfengine or \webengine. !

A * or [ immediately behind \\ will cause problems. Font ... Equation which has to be solved with logarithms 2002 research: speed of light slowing down? Paragraph ended before ...


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# LINE has started to donate to Prettier – why companies should support OSS

My name is Hanatani (@potato4d). I am a member of the Front-End Dev.9 team at the Front-end Development Center (commonly called UIT).

In this post, let me introduce the sponsorship activities of the front-end domain led by UIT.

## Circumstances of OSS sponsoring related to the front-end domain of LINE

As of January 2022, LINE has been supporting the following OSS related to the front end.

#### Vue.js

• Ongoing support since April 2021

#### Bootstrap

• Ongoing support since June 2021
• $100 payment to the account of the development team via Open Collective every month The organizational awareness of sponsoring OSS has increased and we have been supporting OSS since 2021. This month, we have decided to support Prettier. ## Why we sponsor OSS Some of you may be skeptical about for-profit companies sponsoring OSS, as most OSS is provided free of charge and non-profit, and can be cloned and modified if necessary. As a for-profit company, we may not find it reasonable to support non-profit OSS at a cost. However, behind the scenes, did you know that this situation can be a risk? Much of the OSS in the Node.js ecosystem comes from individuals and volunteers, and is undoubtedly essential to us front-end engineers. Despite this, it is not always the case that these OSS are fully sustainable. Recently, there are many OSS such as Next.js and Hasura that are developed and operated by companies with an eye to monetization, but the overwhelming majority of OSS are not directly linked to revenue. In the midst of all of this, there has been a lot of uncertainty in the Node.js ecosystem lately. Since the hijacking of UAParser.js last year, there have been a number of problems targeting packages, and even if they are not, there are an increasing number of cases where maintainers have stopped maintenance. Under such circumstances, if a bug or vulnerability occurs that is difficult to fix except by the developers themselves, it can be a big risk. As a commercial enterprise that takes its responsibilities seriously, we cannot afford to overlook such risks. The lack of profitability of OSS has been supported by the service of individuals, but we must not turn a blind eye and provide adequate support. Now that the distortions of the past have become apparent, we are entering an era in which there are sufficient reasons for companies to support OSS, even if they are judged only from the viewpoint of risk hedging, without considering sympathy and support for OSS culture. ## Intent of the support for Prettier and in-house usage condition The following blog post was the catalyst for our support. The article was written by vjeux, who is one of the main maintenance personnel of Prettier. The article and the references say that currently Prettier is supported by only two maintenance personnel and$1,500 per month is rewarded for their work.

In addition, as for the UIT-related activities of LINE, we had Mr. Suzuki explain the current situation of Prettier at UIT INSIDE ep.37 “Learn Mechanism of Prettier through Contribution” and UIT INSIDE ep.100 “OSS Free Conversation,” we understood that the maintenance was unsustainable.

Prettier is an essential tool for modern front-end projects and a must for maintaining high front-end productivity both inside and outside the company. The discussion of the sustainability of OSS is a recent trend. We decided to support Prettier considering that continued maintenance of Prettier would be very important for front-end developers.

## Usage condition of Prettier in LINE

Not only from the perspective of the whole industry, Prettier is popular in-house as well. In LINE, the shares of React and Vue are about the same size and there are various projects with a wide range of scales from a traditional JavaScript project to a modern TypeScript project. Of course, the library and infrastructure used in each project changes, but what does not change are Linter and formatter.

In our company, ESLint and Prettier are introduced regardless of the language and framework and it is strongly recommended to use Prettier in the in-house front-end development guideline.

In general, Prettier is a must for any of our projects. Therefore, we decided to support Prettier considering that Prettier becoming sustainable will be a great benefit for us.

### Details of the support

We decided to donate $3,000. This is equivalent to the amount of one payment to the currently active maintenance personnel. The reason behind this is that although we had been providing mainly continuous support of$100, we determined that one-shot monetary support with a sizable sum of money to be important under the circumstances of this OSS.

We recognized that, unlike Vue.js and Bootstrap, Prettier is still affected significantly by the motivations of the few maintenance personnel and is hardly sustainable. Under such circumstances, the existence of a large-scale enterprise like us who provide support at the appropriate timing will have a great positive effect on the motivations of maintenance personnel and establishment of cases for other enterprises to take similar actions in the future.

This is the reason why we have decided to provide support of $3,000 in advance while considering continuous support. ## Key points for recommending companies to support OSS So far, we introduced the intent of sponsoring OSS and the case of LINE’s support for Prettier. Lastly, let me introduce the key points for companies with motivations to support OSS. ### Clarify the motivation to support OSS Even if the intention to support the OSS community as developers is the original purpose, a sufficient reason is required for organizational support. Therefore, it is recommended to consider the main purpose of support by the organization in advance. Usually, the reasonable reason for supporting OSS must be any of the following points, so an organization should consider which of the following points are the reasons to support the OSS community: • Minimize the risk of collapse of OSS; • If the library or framework the organization is currently using has no alternative, • Other reasons such as lack of security updates can also be a big risk; • Use the support to promote the company in recruitment; • If the in-house policy on establishing an impression related to a specific technology has been established, • When there are already cases of holding tech events, sharing knowledge through engineer blogs, and so on aiding OSS projects will further establish a positive reputation of investment towards that technology. • Cultivate a development culture in the organization; • When aiming to become a company or developer actively supporting OSS, support can naturally intensify interests in OSS and attract highly interested developers, which will contribute to the development of the organization ### Investigate the support plan of OSS and its return in advance When considering support for OSS, the procedure will be smooth if you investigate a list of plans and returns of such plans in advance. For example, there are an abundance of plans to support Vue.js via Evan’s Patreon page as follows. Various perks such as access to banners or PV are made available depending on the membership level, so if you put much value on branding, recruitment, and so on, the return will be an important factor in the decision of whether to sponsor OSS or not. To decide donation with sufficient reasons, it is recommended to research in advance. ### Obtain qualitative metrics Lastly, it is recommended to be sure to obtain qualitative metrics. Though this recommendation is opposite to the quantitative indicators such as PV mentioned above for return, in reality, branding involves abstract elements, so you cannot measure it with quantitative indicators alone. Therefore, it will be desirable to determine whether the effect is sufficient mainly based on qualitative information such as the voices of developers inside and outside the organization. With quantitative evaluation alone, you might overlook the hidden effect, and you might stop sponsoring. It is recommended to sufficiently explain the characteristics of OSS sponsorship and determine the effect mainly based on qualitative information. ### Understand that the amount is not so large Lastly, it is important to understand that a donation of about$100 is not so significant for the organization.

Of course, individuals may hesitate to donate \$100 to multiple pieces of OSS every month, but there is a big gap between the budget of individuals and that of corporates.

It may be important to make suggestions without thinking too much assuming that they will succeed as long as there are sufficient reasons.


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2,934 views

Consider a network with three routers $\text{P, Q, R}$ shown in the figure below. All the links have cost of unity.

The routers exchange distance vector routing information and have converged on the routing tables, after which the link $\text{Q-R}$ fails. Assume that $\text{P}$ and $\text{Q}$ send out routing updates at random times, each at the same average rate. The probability of a routing loop formation (rounded off to one decimal place) between $\text{P}$ and $\text{Q},$ leading to count-to-infinity problem, is _______________.

Few points about Count to Infifnity problem:

1. This problem appears in DVR only not in LSR (Link state Routing)

This is a Possibility not a definite case . (why??)                                                                                P-----Q-------X

let this is a connection that is already stabilised

P: [(0,X),(1,Q),2]

now some how Q-----X is disconnected .

DVR of Q : [1,0,infinity], [generally DVR is applicable for 15 or lesser nodes therefore 16 refers infinity this is by convention as per “Foruzan”]

Here Comes two possible ways:

case a:

P sends its distance vector to Q  as : [0,1,2]

After receiving this DV to Q

Q will start to count to infinity

and hence the DV of Q will be like: [1,0,3] as Q thinks P may have some path to X

because in DV we send the values only.

Case b:

Q sends it DV to P and P recognizes that there is no link between Q----R

so, DV of P is updates and understood that there is no way to go X

So, among this two cases

MORE IMPORTANT POINTS:

1. There are two ways to recover from this

a.      Split Horizon

b.   Poison reverse

NOTE:
Two nodes instability can be Solved by using the two approaches that are enlisted above but three node instability may or may not solvable by using this approaches too.

Refer this book

“Computer Networks: A Top Down Approach” ---

B.Forouzan

F.Mosharraf

Answer : $0.5$

Once Q-R fails then Q will immidiately update its distance to R to $\infty$. But P will still be having some finitie value (which is 2).

Now it depends on P and Q, who is sending distance vector first.

if Q sends then system becomes stable immediately but if P sends first then it will be count to infinity. Please understand that count to infinity is not some wrong thing, it is just it takes some time to stable.

Since it is given in question that both have same average rate hence probability is also $\frac{1}{2}$ that P sends first than Q. Hence the answer.

But we can calculate answer more mathematically considering that time is continious variable.

We are intrested in probability represented by shaded are.

Which will be  = $\frac{\text{Area of Tringle}}{\text{Area of Square}} = \frac{\frac{1}{2}t^2}{t^2} = \frac{1}{2}$

Also refer forouzan snapshot of count to infinity

why we have not taken a case when both sends routing packet simultaneously....

case1 when P sends first

case 2 when Q sends first

case 3 when both sends simultaneously

edited

@Prakhar_shrimal

Time is continuous, $\color{red}{\text{in case of continuous probability, there is a zero prob for any EXACT point}}$.

If both are choosing to send exactly at 10:00am then you need to make sure that it is 10:00000000000 for both and not 10:00000000000am for one and 10:00000000001 am for other.

Choosing both 10:00000000000000000am is zero probability. because it is 1/inf. inf is because of infinite possible values of time.

edited by

Respected @Sachin Mittal 1 sir, I was having one doubt. The above is a screen shot from Forouzan. But Forouzan just discusses the thing in words, without rigorous mathematics.  Especially the line:

If A can send its table to B immediately, everything is fine. However, the system becomes unstable if B sends its routing table to A before receiving A’s routing table.

Sir, here the important point is, to avoid the loop B should send its DV only after it has received the DV from A. So to work out this gate problem more formally, I tried this way. Let $T_p$ be the propagation time and $T_t$ be transmission time of the DV.

Now suppose sir, $Q$ sends it packets out at time $x$ then it shall reach $P$ only at time $y=x+T_p+T_t$. So $P$ should transmit only after $x+T_p+T_t$ so that there is no loop.

Let us consider the time span of $[0,t]$. Then the required probability is lies in the area shaded in yellow.

$$\text{Probability}=\frac{\text{Area of yellow triangle}}{\text{Area of the rectangle}}$$

$$=\frac{\frac{1}{2}[t-(T_p+T_t)]^2}{t^2} =\frac{1}{2}\left[1-\frac{T_p+T_t}{t}\right]^2$$

If we assume $T_p+T_t=0$ then only we get $probability=0.5$

Otherwise in the most general case, $probability<0.5$

Please can you check my analysis once sir...

[PS: Or is it so that since the $t$ is arbitrarily chosen by us, so to allow for this arbitrary thing to be reflected in our answer, we should account for infinitely large $t$, i.e. $\lim_{t\to \infty} probability = 0.5$ ?]

Since it is given in question that both have same average rate hence probability is also 1/2 that P sends first than Q. Hence the answer.

can you explain precisely the meaning of “same average rate”.

0.5 because it depends on Node Q or node P sharing information first.

If node Q shares it first there is no looping.

If node P does it there will definitely be a count to infinity problem.


text:
# Perturbation of a function in the Orlicz class by a constant ($\int_\Omega\Phi(|u|)<\infty$ implies $\int_\Omega \Phi(|u|+\alpha)<\infty$?).

Assume that $\Phi:[0,\infty)\to [0,\infty)$ is an N-function, i.e., $\Phi(0)=0$, $\Phi$ is convex, strictly increasing, $\Phi(t)/t\to 0$ if $t\to 0$ and $\Phi(t)/t\to \infty$ if $t\to \infty$, or equivalently, $\Phi$ can be written as $$\Phi(t)=\int_0^t \phi(s)ds,$$

where $\phi(t)=0$ if and only if $t=0$, $\phi$ is right continuous, $\phi$ is nondecreasing and $\phi(t)\to \infty$ if $t\to \infty$.

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain ans suppose that $u:\Omega\to \mathbb{R}$ is a measurable function satisfying $$\tag{1}\int_\Omega \Phi(|u|)<\infty.$$

My question: is it true that for small $\alpha>0$, we have that $$\int_\Omega \Phi(|u|+\alpha)<\infty?$$

The class of functions for which $(1)$ is true, is called Orlicz class and it is denoted by $\mathcal{L}_\Phi(\Omega)$. It is a convex space and it is know that (for some N-functions) there are functions $u\in\mathcal{L}_\Phi(\Omega)$ with the property that $\lambda u\notin \mathcal{L}_\Phi(\Omega)$ for some $\lambda>0$, so it is not a vector space.

If $\Phi$ satisfies the $\Delta_2$ condition, i.e., if there are constants $t_0\ge 0$ and $k>0$ such that $\Phi(2t)\le k\Phi(t)$ for $t\ge t_0$ then, the Orlicz class $\mathcal{L}_\Phi(\Omega)$ is a vector space, so we can assume that $\Phi$ does not satisfies the $\Delta_2$ condition.

If $\Phi$ can be compared with a polynomial near the infinity then the above is true, so we can even assume that $\Phi$ grows really fast at the infinity.

My motivation to understand this problem is to calculate the derivative of the functional $\int_\Omega \Phi(|u|)$ in the direction $v$, where $v$ is a bounded function.

Any idea or references is appreciated.

Let us consider $$\Phi\left(t\right)=e^{e^{t}}-e-et.$$ We note that (for $t$ large), we have $$\frac{1}{2}e^{e^{t}}\geq e+et$$ and thus $\Phi\left(t\right)\geq\frac{1}{2}e^{e^{t}}$ for $t$ large.
Now, let $\Omega=\left(0,c\right)$ for some $0<c\ll1$ and define $$f\left(x\right):=\ln\ln\left[\sum_{n\in\mathbb{N}}\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}\cdot c_{n}}\right]$$ with $c_{n}:=\left\Vert x\mapsto1/x^{1-\frac{1}{n}}\right\Vert _{L^{1}\left(\left(0,1\right)\right)}$. This idea is borrowed from one possible way of constructing a function $f$ with $f \in L^p$ for all $p>1$, but $f \notin L^1$.
We have \begin{eqnarray*} \int_{0}^{c}\Phi\left(\left|f\left(t\right)\right|\right)\, dt & \leq & \int_{0}^{c}e^{e^{f\left(t\right)}}\, dt\\ & = & \int_{0}^{c}\sum_{n\in\mathbb{N}}\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}\cdot c_{n}}\, dx\\ & = & \sum_{n\in\mathbb{N}}\frac{1}{2^{n}\cdot c_{n}}\cdot\int_{0}^{c}\frac{1}{x^{1-\frac{1}{n}}}\, dx\\ & \leq & \sum_{n\in\mathbb{N}}\frac{1}{2^{n}}<\infty. \end{eqnarray*} But for arbitrary $\alpha>0$, we have \begin{eqnarray*} \int_{0}^{c}\Phi\left(\left|f\left(t+\alpha\right)\right|\right)\, dt & \geq & \int_{0}^{c_{\alpha}}\Phi\left(\alpha+\ln\ln\left(\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}c_{n}}\right)\right)\, dx\\ & = & \int_{0}^{c_{\alpha}}e^{e^{\alpha+\ln\ln\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)}}\, dx\\ & = & \int_{0}^{c_{\alpha}}e^{e^{\alpha}\cdot\ln\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)}\, dx\\ & = & \int_{0}^{c_{\alpha}}\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)^{e^{\alpha}}\, dx\\ & = & c_{n,\alpha}\cdot\int_{0}^{c_{\alpha}}\frac{1}{x^{\left[1-\frac{1}{n}\right]\cdot e^{\alpha}}}\, dx \end{eqnarray*} for each $n\in\mathbb{N}$, for some suitable $c_{\alpha}\in\left(0,c\right)$. But since $\alpha>0$, we have $e^{\alpha}>1$, so that there is some $n=n\left(\alpha\right)\in\mathbb{N}$ with $\left(1-\frac{1}{n}\right)\cdot e^{\alpha}>1$. Thus, the above integral diverges.


text:
# How do Microwaves Work?

## Recommended Posts

I understand a few things about microwaves that i am not sure if my source of information is really correct. Food is heated in the microwaves when water molecules in the food vibrate. Could you guys give me more information on how Microwaves work and why the reject materials that conduct electricity? Help will be appreciated

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ugochukwu said in post # :

I understand a few things about microwaves that i am not sure if my source of information is really correct. Food is heated in the microwaves when water molecules in the food vibrate. Could you guys give me more information on how Microwaves work

That, i can help u with.

ugochukwu said in post # :

why the reject materials that conduct electricity? Help will be appreciated

That I can't

Well, when microwaves are beamed down onto the food, like you said, the water molecules obsorb it and start to vibrate. This causes an energy change from kinetic (movement) to heat. This is what warms ur food up.

If yo put an empty plastic bowl in the mocrowave, there is no water in it, which means that it does not heat up. It may heat p slightly however, because mostly everything u have contains a tiny bit of water in it.

Hope that helps

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when you put a metal conductor in a microwave, you create a large potential difference between different points on the surface, causing electricity arcs between them.

the frequency of microwaves is basically the same as the resonance frequency of water.

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Originally posted by aommaster

This causes an energy change from kinetic (movement) to heat. This is what warms ur food up.

Hope that helps

heat is kinetic energy,

you can actually measure the length of the microwaves if ya want this is how ya do it.

1) stop the glass dish inside from rotating when the things turned on

2) get a hole bunch of marsh mellows and put them closely packed next to each other and then leave it on for bout minute i think

3) measure the distances between the highly metled points and thats the lenght

=)

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Doesn't the kinetic energy CHANGE to heat energy? Kinetic energy is movement, and heat is.... heat!

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How do you explain heat energy in terms of molecules without it involving movement?

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well...:shrug:...

ok. You got me!

But then, why are we taught that it was a different form of energy? And what about Infrared radiation, THAT is heat as well!

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Crash said in post # :

you can actually measure the length of the microwaves if ya want this is how ya do it.

=)

Easiest way is to read the manufacturer's booklet or sometimes it's on the side of microwave.

This method also prevents you having to clean up a rather large mess from exploded/molten marshmellows

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What difference is there if the microwaves are shorter/longer?

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You may be familar with a very famous equation, E = hf, where E is the energy of a photon, h is Planck's constant (6.63x10^-34) and f is the frequency of the photon. We also know that c = f*lambda (c = speed of light, lambda = wavelength). So E = h*c/lambda.

So for microwaves with a shorter wavelength, the energy they have is going to increase, and for microwaves with a longer wavelength, their energy will decrease.

(Please note: I haven't done Physics for about 6 months now, so my definitions and explanations of the equations might not be entirely right )

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oh ok. Does that mean that they heat up the food faster, because they have more energy in them?

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In microwave cooking, the radio waves penetrate the food and excite water and fat molecules pretty much evenly throughout the food. There is no "heat having to migrate toward the interior by conduction". There is heat everywhere all at once because the molecules are all excited together. There are limits of course. Radio waves penetrate unevenly in thick pieces of food (they don't make it all the way to the middle), and there are also "hot spots" caused by wave interference, but you get the idea. The whole heating process is different because you are "exciting atoms" rather than "conducting heat".

In a microwave oven, the air in the oven is at room temperture, so there is no way to form a crust. That is why foods like "Hot Pockets" come with a little cardboard/foil sleeve. You put the food in the sleeve and then microwave it. The sleeve reacts to microwave energy by becoming very hot. This exterior heat lets the crust become crispy as it would in a conventional oven.

oh ok! Thanx

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yw look at website for futher info.

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I use that website quite a lot, amazing that i hadn't had a look at it this time

Something out of topis, i had answered ur maths challenge, please check whether it is correct or not!

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you may even forget that its microwaves, consider it Photons

as you know Infrared light (photons at that frequency) cause a surface warming effect, like when you sit in front of a fire

then think of Radio waves, they can transmit energy through alot of materials and if enough power is used, it will create a heating effect.

NOW, think about a good mix between the 2, because frequencies higher than normal radio and lower than infrared puts you right in the middle of the ....... you guessed it MICROWAVE band

and so you get the best of both worlds, a good heating effect due to the packet energy (quanta) and the penetrating power of radio waves

a typical MW runs at about 2.4 Ghz, I cant comment on the molecular vibration of boiling water as to its accuracy however.

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2.4 ghz? That's my phone frequency! I doubt my phone can microwave things.

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In large enough amounts, thats why there was that scare about phones causeing brain cancer cause they microwave your brain

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Cap'n Refsmmat said in post #17 :

2.4 ghz? That's my phone frequency! I doubt my phone can microwave things.

I believe that's also why there is static on your phone if you're talking on it at the same time that your microwave is running.

As for your question about heat being different from kinetic energy, the temperature of something is the average kinetic energy of the molecules... so that must be why heat and kinetic energy are essentially the same thing.

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text:
## Separation of Mixture

Category : 5th Class

### Separation of mixture

Interconnection of the different states of the matter

SOLID                        LIQUID                      GAS

Releasing heat                                   releasing (heat)

Melting (heat)                               vaporization (heat)

The above mentioned examples are physical change that can be brought back either by giving or loosing heat- Physical change is reversible. Burning of paper is a chemical change as ash can't be brought back to paper by any means. Chemical change is irreversible.

Rusting of Iron                                     Burning Paper

Growth from an infant to an adult is an irreversible change.

Which of the following plays an important role In the process of changing one state of matter into another?

(a) Heat

(b) Volume

(c) Mass

(d) Density

(e) None of these

Explanation

One state of matter is converted into another form either by taking or releasing heat energy. Therefore, option (a) is correct and rest of the options is incorrect.

Which one of the following is an example of liquid state of matter?

(a) Air

(b) Soil

(c) Water

(d) All of these

(e) None of these


text:
# Differentiating arctan

#### ACLerok

http://www.eden.rutgers.edu/~cjjacob/images/arctan.gif [Broken]

I was given this problem yesterday. It's asking me to differentiate a function containing arctan. Should I find a common denominator and combine them into one term, should I just differentiate each term separately? Please I need help with this thing! I've been told it comes out to 0 but I'm not sure if he is correct or not. Thanks for replying if you choose to!

Last edited by a moderator:
Related Introductory Physics Homework Help News on Phys.org

#### Doc Al

Mentor
ACLerok said:
Should I find a common denominator and combine them into one term, should I just differentiate each term separately?
Should you try to combine the two arctan terms into one? Yikes!

No, just differentiate each term separately. If you know how to differentiate arctan(x) and how to use the chain rule, this should be a snap.

#### ShawnD

According to maple, f'(x) is this

$$\left( 1+{x}^{2} \right) ^{-1}-{\frac {1}{{x}^{2} \left( 1+{x}^{-2} \right) }}$$

I can't help with how to get that though; I've never dealt with arctan before.

It simplifies to 0.

Last edited:

Staff Emeritus
Gold Member

#### Janitor

Draw a right triangle with a base of length 1 and a side of length x. Note that one of the acute angles is equal to arctan(x) and the other acute angle is equal to arctan(1/x). Since the three interior angles of any triangle add up to pi radians, and since the remaining angle in this triangle is pi/2, then the sum arctan(x) + arctan(1/x) has to equal the constant value pi - pi/2 = pi/2. Its derivative is thus zero.

$$y = \arctan{x}$$
$$\tan{y} = x$$
$$\frac{dy}{dx}\sec^2{y} = 1$$
$$\frac{dy}{dx} = \cos^2{y}$$

Draw a triangle based on the second line, tan(y) = x. This means that the opposite side is x and the adjacent side is 1, giving the hypotenuse a value of $\sqrt{x^2 + 1}$. This means that cos(y) will be 1 over the hypotenuse, i.e.

$$\cos{y} = \frac{1}{\sqrt{x^2 + 1}}$$

which means that our derivative is

$$\frac{dy}{dx} = \frac{1}{x^2 + 1}$$
$$\frac{d}{dx}\Big(\arctan{x}\Big) = \frac{1}{x^2 + 1}$$

The rest follows from the chain rule.

#### ACLerok

$$y = \arctan{x}$$
$$\tan{y} = x$$
$$\frac{dy}{dx}\sec^2{y} = 1$$
$$\frac{dy}{dx} = \cos^2{y}$$

Draw a triangle based on the second line, tan(y) = x. This means that the opposite side is x and the adjacent side is 1, giving the hypotenuse a value of $\sqrt{x^2 + 1}$. This means that cos(y) will be 1 over the hypotenuse, i.e.

$$\cos{y} = \frac{1}{\sqrt{x^2 + 1}}$$

which means that our derivative is

$$\frac{dy}{dx} = \frac{1}{x^2 + 1}$$
$$\frac{d}{dx}\Big(\arctan{x}\Big) = \frac{1}{x^2 + 1}$$

The rest follows from the chain rule.

ok i think I can see what you have done. can you explain how you got from tan(y)=x to the line below it? did you just differentiate both sides with respect to y?

And then at the end are you saying to just substitute the equation for the derivative of arctan into the orginal equation?

Thanks for all your help guys!

Last edited:

Yes. The third line is reached by differentiating the second with respect to x.

The derivative of arctan(1/x) is found in a similar manner, or you can differentiate it using a substitution and the chain rule. Both methods require just about the same amount of work.

#### Janitor

I had a buddy in high school who liked to use non-calculus techniques to get the answer to homework problems that the calculus teacher assigned. The kid knew darn well that the teacher would not give credit for getting the right answer by "devious" means, but that didn't stop him from doing that. I think he had some sort of persecution complex, and he loved to wail to anyone who would listen about how unfair the teacher was.

#### Janitor

Thanks, Cookiemonster! But I had a guilty conscience since I figured the teacher wouldn't give him any credit for it, since it doesn't teach him how to do derivatives.

Haha, why not? He still has to differentiate pi/2. =]

True!

. .

#### ACLerok

Yes. The third line is reached by differentiating the second with respect to x.

The derivative of arctan(1/x) is found in a similar manner, or you can differentiate it using a substitution and the chain rule. Both methods require just about the same amount of work.

ok so the derivative of arctan x is 1/[(x^2)+1] but what is the derivative of arctan(1/x) do i just plug 1/x into 1/[(x^2)+1]? it seems like the logical thing to do but then again, i may be totally wrong.. Thanks.

As I said in my previous post, you can figure it out using the same method as for arctan(x), i.e.

tan(y) = 1/x

differentiate both sides implicitly to get

dy/dx = -cos^2(y)/x^2

then we build a triangle based on tan(y) = 1/x, which means the opposite side is 1 and the adjacent side is x, so the hypotenuse is sqrt(x^2 + 1), so the cos^2(y) term is x/sqrt(x^2 + 1), meaning

dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

Alternatively, you could have made a substitution u = 1/x into the derivative, so we have

d/dx(arctan(u)) = df(u)/du * du/dx = 1/(1 + 1/x^2)*-1/x^2

which can be shown to be equivalent with a little bit of algebra.

#### ACLerok

As I said in my previous post, you can figure it out using the same method as for arctan(x), i.e.

tan(y) = 1/x

differentiate both sides implicitly to get

dy/dx = -cos^2(y)/x^2

then we build a triangle based on tan(y) = 1/x, which means the opposite side is 1 and the adjacent side is x, so the hypotenuse is sqrt(x^2 + 1), so the cos^2(y) term is x/sqrt(x^2 + 1), meaning

dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

Alternatively, you could have made a substitution u = 1/x into the derivative, so we have

d/dx(arctan(u)) = df(u)/du * du/dx = 1/(1 + 1/x^2)*-1/x^2

which can be shown to be equivalent with a little bit of algebra.

so the derivative of the problem simplifies to -1/(x^2 + 1)? and that equals 0?

Yes. Both Janitor's solution and brute force show the solution to be 0. You wanted it to be 0 in the first place anyway, didn't you?

#### ACLerok

Yes. Both Janitor's solution and brute force show the solution to be 0. You wanted it to be 0 in the first place anyway, didn't you?

alright, i understand everything now except this one line:
dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

how did you get this? i got up to dy/dx = -x^2+1/x^2
whatd you do to get that line? i tried using the quotient rule to differentiate it but i got 2x/x^4 but that is incorrect.

i never wanted the answer to be zero, i was just told that it is. i guess. thanks alot!

Last edited:

#### Doc Al

Mentor
ACLerok said:
so the derivative of the problem simplifies to -1/(x^2 + 1)? and that equals 0?
That's the derivative of the second term in your original expression. Add that to the derivative of the first term and the sum will equal zero.

#### Doc Al

Mentor
ACLerok said:
how did you get this? i got up to dy/dx = -x^2+1/x^2
whatd you do to get that line? i tried using the quotient rule to differentiate it but i got 2x/x^4 but that is incorrect.
I assume you can differentiate arctan(x)? That will be the first term of your answer.

Now use the chain rule to differentiate arctan(u), where u = 1/x. (What's the derivative of 1/x?) That will be the second term making up your answer.

#### ACLerok

for differentiating arctan 1/x, i got up to drawing the triangle to find cos(y) and i got dy/dx = -cos^2(y)/x^2. know what do i do with these two piece of information? do i just substitute sqrt(x^2+1) into dy/dx = -cos^2(y)/x^2 or what? please this is due tomorrow and i dont know how to finish it!

cos(y) can be expressed in terms of x. What does the triangle you drew tell you it should be?

#### ACLerok

sqrt(x^2+1). i dont know how you differentiated it. can you just explain the steps you took?

Differentiated what? I had two entire posts on the method to differentiate arctan(x) and arctan(1/x). Which parts are giving you trouble?

#### ACLerok

i dont understand what you did here and how you got this:
dy/dx = 1/x^2 * -x^2/(1+x^2) = -1/(x^2 + 1)

this is from differentiating arctan(1/x)
i know that you just substituted sqrt(x^2+1) into dy/dx = -cos^2(y)/x^2 but I have no clue how you went about finding the derivative or which method(s) you used. (quotient rule/chainrule/both/etc.) i know i'm giving you a hard time but calc isnt my piece of cake. sorry.

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text:
A link in a quantum network is built from a hybrid atomic system, connected by fiber optics.

Networks provide an essential tool for many communication and computational tasks by connecting different regions of space to allow remote systems to talk to one another. In a quantum network, a link represents shared entanglement between remote systems. Establishing such a link involves the creation, distribution, storage, and retrieval of entanglement.

Writing in Physical Review Letters, Matthias Lettner and colleagues from the Max Planck Institute for Quantum Optics in Garching, Germany, report they are able to create such a link with a notably high fidelity of $95%$. (Fidelity characterizes the reproducibility of preparing a quantum state.) They entangle two systems that sit in different laboratories located $13$ meters apart. In one laboratory, they send a laser pulse onto a single Rubidium atom inside an optical cavity to create a photon that is entangled with the atom. The photon is allowed to depart the cavity to travel through $30$ meters of optical fiber to another laboratory. There, a Bose-Einstein condensate (BEC) of Rubidium atoms captures the photon and stores it in the multi-atom wave function, creating an entangled state between the single atom and the atomic ensemble. Two final steps map the atom-BEC entanglement onto photon-photon entanglement: retrieving the stored photon from the BEC and, meanwhile, producing a second photon from the single atom.

This hybrid system, which demonstrates the creation, distribution, storage, and retrieval of entanglement across remote locations, is an essential link within, and a step towards, a real-world quantum network. – Sonja Grondalski

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text:
# Quantum Forest

### notes in a shoebox

#### Month: October 2012

(This post continues discussing issues I described back in January in Academic publication boycott)

Some weeks ago I received a couple of emails the same day: one asking me to submit a paper to an open access journal, while the other one was inviting me to be the editor of an ‘special issue’ of my choice for another journal. I haven’t heard before about any of the two publications, which follow pretty much the same model: submit a paper for $600 and—if they like it—it will be published. However, the special issue email had this ‘buy your way in’ feeling: find ten contributors (i.e.$6,000) and you get to be an editor. Now, there is nothing wrong per-se with open access journals, some of my favorite ones (e.g. PLoS ONE) follow that model. However, I was surprised by the increasing number of new journals that look at filling the gap for ‘I need to publish soon, somewhere’. Surprised until one remembers the incentives at play in academic environments.

If I, or most academics for that matter, want to apply for academic promotion I have to show that I’m a good guy that has a ‘teaching philosophy’ and that my work is good enough to get published in journals; hopefully in lots of them. The first part is a pain, but most people can write something along the lines ‘I’m passionate about teaching and enjoy creating a challenging environment for students…’ without puking. The second part is trickier because one has to really have the papers in actual journals.

Personally, I would be happier with only having the odd ‘formal’ publication. The first time (OK, few times) I saw my name in a properly typeset paper was very exciting, but it gets old after a while. These days, however, I would prefer to just upload my work to a website, saying here you have some ideas and code, play with it. If you like it great, if not well, next time I hope it’ll be better. Nevertheless, this doesn’t count as proper publication, because it isn’t peer reviewed, independently of the number of comments the post may get. PLoS ONE counts, but it’s still a journal and I (and many other researchers) work in many things that are too small for a paper, but cool enough to share. The problem: there is little or no credit for sharing so Quantum Forest is mostly a ‘labor of love’, which counts bugger all for anything else.

These days as a researcher I often learn more from other people’s blogs and quick idea exchanges (for example through Twitter) than via formal publication. I enjoy sharing analysis, ideas and code in this blog. So what’s the point of so many papers in so many journals? I guess that many times we are just ‘ticking the box’ for promotions purposes. In addition, the idea of facing referees’ or editors’ comments like ‘it would be a good idea that you cite the following papers…’ puts me off. And what about authorship arrangements? We have moved from papers with 2-3 authors to enough authors to have a football team (with reserves and everything). Some research groups also run arrangements where ‘I scratch your back (include you as a coauthor) and you scratch mine (include me in your papers)’. We break ideas into little pieces that count for many papers, etc.

Another related issue is the cost of publication (and the barriers it imposes on readership). You see, we referee papers for journals for free (as in for zero money) and tell ourselves that we are doing a professional service to uphold the high standards of whatever research branch we belong to. Then we spend a fortune from our library budget to subscribe to the same journals for which we reviewed the papers (for free, remember?). It is not a great deal, as many reasonable people have pointed out; I added a few comments in academic publication boycott.

So, what do we need? We need promotion committees to reduce the weight on publication. We need to move away from impact factor. We can and need to communicate in other ways: scientific papers will not go away, but their importance should be reduced.

Some times the way forward is unclear. Incense doesn’t hurt (Photo: Luis).

Making an effort to prepare interesting lectures doesn’t hurt either.
These days it is fairly common editors ‘suggesting’ to include additional references in our manuscripts, which just happen to be to papers in the same journal, hoping to inflate the impact factor of the journal. Referees tend to suggest their own papers (some times useful, many times not). Lame, isn’t it?

PS. 2012-10-19 15:27 NZST. You also have to remember that not because something was published it is actually correct: outrageously funny example (via Arthur Charpentier). Yep, through Twitter.

I’ve ignored my quantitative geneticist side of things for a while (at least in this blog) so this time I’ll cover some code I was exchanging with a couple of colleagues who work for other organizations.

It is common to use diallel mating designs in plant and tree breeding, where a small number of parents acts as both males and females. For example, with 5 parents we can have 25 crosses, including reciprocals and selfing (crossing an individual with itself). Decades ago this mating design was tricky to fit and, considering an experimental layout with randomized complete blocks, one would have something like y = mu + blocks + dads + mums + cross + error. In this model dads and mums were estimating a fraction of the additive genetic variance. With the advent of animal model BLUP, was possible to fit something like y = mu + blocks + individual (using a pedigree) + cross + error. Another less computationally demanding alternative (at least with unrelated parents) is to fit a parental model, overlaying the design matrices for parents with something like this y = mu + blocks + (dad + mum) + cross + error.

The following code simulate data for a a diallel trial with three replicates and runs a parental model with ASReml. Later I expand the analysis building the matrices by hand.

How is the matrix overlay working? We can replicate the calculations used by ASReml by building the matrices from scratch and reusing the variance components, so we avoid the nastiness of writing code for residual maximum likelihood. Once I build the basic matrices I use the code from my How does a linear mixed model look like? post.

The overlay matrix Zparents is double the actual value we should use:

Repeating the analyses ‘by hand’ using Zparents2 to build the Z matrix results in the same overall mean and block effects, but the predicted breeding values for parents when using Zparents are 0.7 of the predicted breeding values for parents when using Zparents2. I may need to refit the model and obtain new variance components for parents when working with the correct overlaid matrix.

Gratuitous picture: walking in Quail Island (Photo: Luis).

I have written a few posts discussing descriptive analyses of evaluation of National Standards for New Zealand primary schools.The data for roughly half of the schools was made available by the media, but the full version of the dataset is provided in a single-school basis. In the page for a given school there may be link to a PDF file with the information on standards sent by the school to the Ministry of Education.

I’d like to keep a copy of the PDF reports for all the schools for which I do not have performance information, so I decided to write an R script to download just over 1,000 PDF files. Once I can identify all the schools with missing information I just loop over the list, using the fact that all URL for the school pages start with the same prefix. I download the page, look for the name of the PDF file and then download the PDF file, which is named school_schoolnumber.pdf. And that’s it.

Of course life would be a lot simpler if the Ministry of Education made the information available in a usable form for analysis.

## Can you help?

It would be great if you can help me to get the information from the reports. The following link randomly chooses a school, click on the “National Standards” tab and open the PDF file.

Then type the achievement numbers for reading, writing and mathematics in this Google Spreadsheet. No need to worry about different values per sex or ethnicity; the total values will do.

Gratuitous picture: a simple summer lunch (Photo: Luis).