text:
## 109.51 A torsor which is not an fppf torsor

In Groupoids, Remark 39.11.5 we raise the question whether any $G$-torsor is a $G$-torsor for the fppf topology. In this section we show that this is not always the case.

Let $k$ be a field. All schemes and stacks are over $k$ in what follows. Let $G \to \mathop{\mathrm{Spec}}(k)$ be the group scheme

$G = (\mu _{2, k})^\infty = \mu _{2, k} \times _ k \mu _{2, k} \times _ k \mu _{2, k} \times _ k \ldots = \mathop{\mathrm{lim}}\nolimits _ n (\mu _{2, k})^ n$

where $\mu _{2, k}$ is the group scheme of second roots of unity over $\mathop{\mathrm{Spec}}(k)$, see Groupoids, Example 39.5.2. As an inverse limit of affine schemes we see that $G$ is an affine group scheme. In fact it is the spectrum of the ring $k[t_1, t_2, t_3, \ldots ]/(t_ i^2 - 1)$. The multiplication map $m : G \times _ k G \to G$ is on the algebra level given by $t_ i \mapsto t_ i \otimes t_ i$.

We claim that any $G$-torsor over $k$ is of the form

$P = \mathop{\mathrm{Spec}}(k[x_1, x_2, x_3, \ldots ]/(x_ i^2 - a_ i))$

for certain $a_ i \in k^*$ and with $G$-action $G \times _ k P \to P$ given by $x_ i \to t_ i \otimes x_ i$ on the algebra level. We omit the proof. Actually for the example we only need that $P$ is a $G$-torsor which is clear since over $k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots )$ the scheme $P$ becomes isomorphic to $G$ in a $G$-equivariant manner. Note that $P$ is trivial if and only if $k' = k$ since if $P$ has a $k$-rational point then all of the $a_ i$ are squares.

We claim that $P$ is an fppf torsor if and only if the field extension $k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots )/k$ is finite. If $k'$ is finite over $k$, then $\{ \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)\}$ is an fppf covering which trivializes $P$ and we see that $P$ is indeed an fppf torsor. Conversely, suppose that $P$ is an fppf $G$-torsor. This means that there exists an fppf covering $\{ S_ i \to \mathop{\mathrm{Spec}}(k)\}$ such that each $P_{S_ i}$ is trivial. Pick an $i$ such that $S_ i$ is not empty. Let $s \in S_ i$ be a closed point. By Varieties, Lemma 33.14.1 the field extension $\kappa (s)/k$ is finite, and by construction $P_{\kappa (s)}$ has a $\kappa (s)$-rational point. Thus we see that $k \subset k' \subset \kappa (s)$ and $k'$ is finite over $k$.

To get an explicit example take $k = \mathbf{Q}$ and $a_ i = i$ for example (or $a_ i$ is the $i$th prime if you like).

Lemma 109.51.1. Let $S$ be a scheme. Let $G$ be a group scheme over $S$. The stack $G\textit{-Principal}$ classifying principal homogeneous $G$-spaces (see Examples of Stacks, Subsection 94.14.5) and the stack $G\textit{-Torsors}$ classifying fppf $G$-torsors (see Examples of Stacks, Subsection 94.14.8) are not equivalent in general.

Proof. The discussion above shows that the functor $G\textit{-Torsors} \to G\textit{-Principal}$ isn't essentially surjective in general. $\square$

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# The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue).If the total revenue(in rupees) received from the sale of Xunits of a product is given by R(x)=$3x^2+36x+5$,find the marginal revenue,when x=5,and write which value does the question indicate.

This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Toolbox:
• If the Total Revenue is given by $f(x)$, the Marginal Revenue is the rate of change of total revenue and is nothing but the first order derivative of the function $\large \frac{df(x)}{dx}$
Given the total revenue $=R(x)=3x^2+36x+5$.
If the Total Revenue is given by $f(x)$, the Marginal Revenue is the rate of change of total revenue and is nothing but the first order derivative of the function $\large \frac{df(x)}{dx}$
$\Rightarrow$ Differentiating $R(x)$, we get $\large \frac{dR(x)}{dx}$$= 6x+36$
If $x=5$, Marginal revenue $= 6 (5) + 36$
Therefore, Marginal revenue $= 30+36 = 66$.
The concern over the welfare of the employees by the employer.

edited Mar 22, 2013

format: plain


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## LaTeX forum ⇒ Page Layout ⇒ Page number style issue

Information and discussion about page layout specific issues (e.g. header and footer lines, page formats, page numbers).
jorgesisco
Posts: 3
Joined: Wed Mar 16, 2022 6:41 pm

### Page number style issue

I have all my pages in romans numbers until table of content,

from introduction I added to my document:

\input{chapters/1. Introduction}
\pagenumbering{arabic}
\setcounter{page}{10}

and this help me set the numbers in arabic from the introduction page, but since my introduction takes more than one page I get the 1st page of the introduction in romans number and from the second page of the introduction is where I start getting the arabic pagination numbers.

Does anyone knows how can I mitigate this issue?

Attachments
latex.png (145.15 KiB) Viewed 326 times

LaTeXguide.org • LaTeX-Cookbook.net

Bartman
Posts: 283
Joined: Fri Jan 03, 2020 2:39 pm

### Page number style issue

A suggestion for switching from roman to arabic page numbering without resetting the page counter:

\documentclass{report}
\usepackage[nottoc]{tocbibind}
\usepackage{blindtext}

\begin{document}
\pagenumbering{Roman}
\tableofcontents
\listoffigures
\cleardoublepage
\renewcommand{\thepage}{\arabic{page}}
\chapter{Introduction}
\Blindtext[4]
\end{document}

A minimal working example could help to find out whether this also works in your context.


text:
Question

A ray reflected successively from two plane mirrors inclined at a certain angle undergoes a deviation of 300∘. The number of observable images are (Formula for number of images is 360θ−1, where θ is the angle between the mirrors.)

A
15
B
12
C
11
D
13

Solution

The correct option is D 11Given: Angle of deviation, δ=300∘ We know that, the angle of deviation when the ray is successively reflected by two plane mirrors is given by, δ=360∘−2θ, where θ is the angle between the mirrors.  ⇒300∘=360∘−2θ ⇒θ=360∘−300∘2=30∘ 30∘ is the angle at which both mirrors are inclined. So, n=360∘θ=360∘30∘=12 As n is even, hence, the number of images,  =n−1=12−1=11Physics

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5. HEUR CONSTRAINT PENALTY

# HEUR CONSTRAINT PENALTY

Controls the constraint penalty heuristic. There’s nothing fancy about this one, it will simply reformulate the problem into:

$$\min~f(x)+\alpha g(x)^2 + \beta h(x)^2$$

The coefficient vectors for $$\alpha$$ and $$\beta$$, as well as the details for specific symbolic special cases are calculated automatically.

Works for all problem types, including MINLP.

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## Factoring Trinomials of the Form x² + bx + c when c is Negative

Read this article. Pay attention to the overview of the steps for these problems. Read the examples and note where factoring is not possible, like in the last example. After you have reviewed the materials, complete a few practice problems until you feel comfortable. Be sure to check your answers when you are done.

Here, you will practice factoring trinomials of the form $x^{2}+bx+c$, where $b$ and $c$ are integers, and $c< 0$.
That is, the constant term is negative.

Recall that the integers are: ..., $-3$, $-2$, $-1$, $0$, $1$, $2$, $3$ ...

As discussed in Basic Concepts Involved in Factoring Trinomials, you must first find two numbers that add to $b$ and that multiply to $c$, since then:

$x^{2}+bx+c=x^{2}+\overset{=b}{\overbrace{(f+g)}}x+\overset{=c}{\overbrace{fg}}=(x+f)(x+g)$

Since $c$ is negative in this exercise, one number will be positive, and the other will be negative.

How can two numbers multiply to give a negative result? One must be positive, and the other negative. That is, the numbers will have different signs.

When you add numbers that have different signs, then in your head you actually do a subtraction problem.

For example, to mentally add $(-5)+3$, in your head you would compute $5-3$, and then assign a negative sign to your answer.

Think of it this way: Start at zero on a number line. Walk $5$ units to the left, and $3$ units to the right. You end up at $-2$. You walked farther to the left than you did to the right, so your final answer is negative.

The sign of $b$ (the coefficient of the $x$ term) determines which number will be positive, and which will be negative:
If $b> 0$, then the bigger number (the one farthest from zero) will be positive.
If $b< 0$, then the bigger number (the one farthest from zero) will be negative.
In other words, the biggest number takes the sign (plus or minus) of $b$.

These results are summarized below:

Factoring Trinomials of the Form $x^{2}+bx+c$, $c$

• Check that the coefficient of the square term is $1$.

• Check that the constant term $c$ is negative.

• It is easier to do mental computations involving only positive numbers.
So, you will initially ignore all minus signs and just work with the numbers $\left | b \right |$ and $\left | c \right |$.

• Find two numbers whose DIFFERENCE is $\left | b \right |$ and whose PRODUCT is $\left | c \right |$.
That is, find two numbers that subtract to give $\left | b \right |$ and that multiply to give $\left | c \right |$.

• Now (and only now), you will use the actual plus-or-minus sign of $b$.
If $b> 0$, then the bigger of your two numbers is positive; the other is negative.
If $b< 0$, then the bigger of your two numbers is negative; the other is positive.
That is, the biggest number takes the sign (plus or minus) of $b$.

• Use these two numbers to factor the trinomial, as illustrated in the examples below.

### Examples

Question: Factor: $x^{2}-5x-6$

Solution: Thought process: Is the coefficient of the $x^{2}$ term equal to $1$? Check!

Is the constant term negative? Check!

Find two numbers whose difference is $5$ and whose product is $6$.

That is, find two numbers that subtract

to give $5$ and that multiply to give $6$.

The numbers $1$ and $6$work, since $6-1=5$ and $6\cdot 1=6$.

Since the coefficient of $x$ is positive, the bigger number $(6)$ will be positive, and the other will be negative.

The desired numbers are therefore $6$ and $-1$.

Then

$x^{2}+5x-6=x^{2}+\overset{=5}{\overbrace{(6+(-1))}}x+\overset{=-6}{\overbrace{6\cdot (-1)}}=(x+6)(x-1)$

Check: $(x+6)(x-1)=x^{2}-x+6x-6=x^{2}+5x-6$

Question: Factor: $x^{2}-5x-6$

Solution: Thought process: Is the coefficient of the $x^{2}$ term equal to $1$?Check!
Is the constant term negative? Check!

Find two numbers whose difference is $5$ and whose product is $5$.

That is, find two numbers that subtract to give $5$ and that multiply to give $6$.

The numbers $1$ and $6$ work, since $6-1+5$ and $6\cdot 1=6$.

Since the coefficient of $x$ is negative, the bigger number $(6)$ will be negative, and the other will be positive.
The desired numbers are therefore $-6$ and $1$.

Then,

$x^{2}-5x-6+x^{2}+\overset{=-5}{\overbrace{((-6)+1)}}x+\overset{=-6}{\overbrace{(-6)\cdot 1}}=(x-6)(x+1)$

Check: $(x-6)(x+1)=x^{2}+x-6x-6=x^{2}-5x-6$

Question: Factor: $x^{2}+x-1$

Solution: There are no integers whose difference and product are both $1$.

Thus, $x^{2}+x-1$ is not factorable over the integers.

Source: Tree of Math, https://www.onemathematicalcat.org/algebra_book/online_problems/fac_tri_one_cneg.htm


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How To Prove It Velleman Solutions Manual Pdf

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text:
Once, I inadvertantly used the letter$$p$$ for two different things in a (difficult, at the time, for me) group theory proof. The professor wrote nothing on it except, "There are twenty-five other letters of the alphabet."

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# Lesson 5 Discussion Question -- Emergency Planning Lesson 5 Discussion Question -- Emergency Planning

###### Question:

Lesson 5 Discussion Question -- Emergency Planning Lesson 5 Discussion Question -- Emergency Planning

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# Math Help - Binary Operation

1. ## Binary Operation

A binary operation $*$ is defined on the set $S=\{a,b,c\}$ by $x*y=x$ for all $x,y\inS$. Determine which of the properties (Associative, Identity, Inverse, Commutative) are satisfied by $(S,*).$

Solution:
Let $x,y$, and $z$ be any three elements of $S$ (distinct of not). Then $x*(y*z) = x*y=x$, while $(x*y)*z=x*z=x$. Thus (S,*) is associative. Now (S,*) has no identity since for every element $e \in S$, it follows that $e*a=e*b=e$ and so it is impossible for $e*a=a$ and $e*b=b$. Since (S,*) has no identity, the question of inverses does not apply here. Certainly, $(S,*)$ is not commutative since $a*b=a$ while $b*a=b$.

Question:
I suppose $e$ is an identity element. Since (S,*) has no identity why it contradict itself by saying "(S,*) has no identity since for every element " $e\in S$"?

$e*a=a$ and $e*b=b$ make sense, but what is $e*a=e*b=e$?

2. Originally Posted by novice
A binary operation $*$ is defined on the set $S=\{a,b,c\}$ by $x*y=x$ for all $x,y\inS$. Determine which of the properties (Associative, Identity, Inverse, Commutative) are satisfied by $(S,*).$

Solution:
Let $x,y$, and $z$ be any three elements of $S$ (distinct of not). Then $x*(y*z) = x*y=x$, while $(x*y)*z=x*z=x$. Thus (S,*) is associative. Now (S,*) has no identity since for every element $e \in S$, it follows that $e*a=e*b=e$ and so it is impossible for $e*a=a$ and $e*b=b$. Since (S,*) has no identity, the question of inverses does not apply here. Certainly, $(S,*)$ is not commutative since $a*b=a$ while $b*a=b$.

Question:
I suppose $e$ is an identity element. Since (S,*) has no identity why it contradict itself by saying "(S,*) has no identity since for every element " $e\in S$"?
$e$ is not an identity element, it is just a letter. You can think of it as "candidate for identity element." The text says "for every element $e \in S$", not "for an identity element $e \in S$". The text does not contradict itself.

Originally Posted by novice
$e*a=a$ and $e*b=b$ make sense, but what is $e*a=e*b=e$?
Look at the definition of identity element. The logic here is for left identity element in particular. The point is that if $e$ is an identity element then $e*a = a$ and $e*b = b$, therefore $e*a\ne e*b$. This produces a contradiction since we know $e*a=e*b=e$. Therefore, no matter what choice of $e$ we make, $e$ is not an identity element.

3. Originally Posted by undefined
$e$ is not an identity element, it is just a letter. You can think of it as "candidate for identity element
If $e$ is only a possible candidate, the what operational property is there that can produce $e*a=e*b=e$, or is this just non-sense created for making the point.

4. Originally Posted by novice
If $e$ is only a possible candidate, the what operational property is there that can produce $e*a=e*b=e$, or is this just non-sense created for making the point.
$e$ is an element of $S$. Direct application of the definition of $*$ gives $e*a=e*b=e$, regardless of what $e$ is. It is not nonsense.

5. Thank you sir. All is clear now.


text:
# Ideal transformer model (2 ODEs) warning “NDSolve will try solving the system as differential-algebraic equations”

I have a simple electrical circuit of an ideal transformer:

I model it in Mathematica 11.3 with the idea to solve for secondary current and potential difference. Here is my code:

ClearAll["Global*"]

time = 0.001;

r1 = 0.5;
r2 = 1000;
l1 = 0.001;
l2 = 0.1;
m = Sqrt[l1*l2];

vSource[t_] = 20 Sin[10000*Pi* t];

eq1 = vSource[t] == i1[t]*r1 + l1*i1'[t] - m*i2'[t];
eq2 = m*i1'[t] == i2[t]*r2 + l2*i2'[t];

ic1 = i1[0] == 0;
ic2 = i2[0] == 0;

sol = NDSolveValue[{eq1, eq2, ic1, ic2}, i2, {t, 0, time}];
Plot[sol[t], {t, 0, time}, PlotLabel -> "Current through r2"]
Plot[sol[t]*r2, {t, 0, time}, PlotLabel -> "Potential difference across r2"]

And here is the plotted solution:

which corresponds to my LTSpice solution:

Now my problem is that I get the following warning:

Despite getting the solution, warning signifies that I am doing something wrong. Could anyone please help solve my problem properly and explain why I get the warning?

• If you try to bring {eq1,eq2} in an explicit form you'll recognize some problems. That's the error message MMA shows! – Ulrich Neumann Nov 14 '18 at 11:09
• The system of equations is degenerate because l1 l2 - m^2=0. Therefore Mathematica gives a warning. Do not pay attention, it is still a solution to the equations in this case. – Alex Trounev Nov 14 '18 at 11:36
• If you want the equations not to degenerate, you can write m = 0.99 Sqrt[l1 l2]. Then there is no warning anymore. The solver doesn't commute to the DAE Solver, which is a good thing because the DAE solver is not very robust (and useless in this case), and physically it is realistic. – andre314 Nov 14 '18 at 11:59
• @andre, that's a neat trick that makes sense physically. In fact, this solved my problem in SystemModeler. Unfortunately, I get weird results if I increase load resistance. Could you try to switch r2 to 1,000,000 and see what you get? – space bobcat Nov 14 '18 at 12:27
• @andre, please disregard my comment. I was receiving awkward result because I had AccuracyGoal set to a low value. In SystemModeler, unfortunately, my solution gets weird if I increase r2. – space bobcat Nov 14 '18 at 12:32

A warning only signifies that you may have done something wrong, but you've just done nothing wrong in this case. This can be verified by comparing the numeric solution with the analytic solution:

analyticsol = i2 /. DSolve[{eq1, eq2, ic1, ic2}, {i1, i2}, t][[1]]

Plot[{analyticsol@t, sol[t]}, {t, 0, time}, PlotStyle -> {Automatic, {Red, Dashed}},
PlotLegends -> {"Analytic Solution", "Numeric Solution"}]

So, why does the warning pop up? As already explained in the comments above, it's because NDSolve is trying to transform the ODE system to an explicit form i.e. right hand side (RHS) of the resulting system consists no derivative term, and left hand side (LHS) of every equation is only a single 1st order derivative term whose coefficient is $$1$$, but your system happens to be one that can't be transformed to that form:

Solve[{eq1, eq2}, {i1'[t], i2'[t]}]
(* {} *)

The output is {}, which indicates there's no solution.

If you don't want to see the warning, add Method -> {"EquationSimplification" -> "Residual"} / SolveDelayed -> True to NDSolve.

BTW, though the DAE solver works properly in this case, you may need to force NDSolve to use ODE solver in certain situations because ODE solver of NDSolve` is more robust compared to the DAE solver. (Andre has already shown you one possible approach in his comment. )

You may check this answer for more details.


text:
# Increase labelwidth MSC package

I'm using the MSC package to draw a sequence chart. I only need one instance, with ingoing and outgoing (lost and found) messages. I've set up a working example suing the LaTeX compiler.

http://texdoc.net/texmf-dist/doc/latex/msc/manual.pdfhttp://texdoc.net/texmf-dist/doc/latex/msc/manual.pdf

\documentclass{article}
\usepackage{msc}
\begin{document}
\begin{msc}{test}
\declinst{a}{}{$A$}
\found{label}{}{a}
\end{msc}
\end{document}

which gives this result

The label i need is quite long and complicated, so I need to make the distance between the circle and the arrow longer, so there's place. I've looked at the documentation, and tried inserting various values at the pos, labelpos and placement optionals, but without any luck.

How can I make the label longer?

After some time, a few hours actually, I finally figured out how to do it!

\documentclass{article}
\usepackage{msc}
\begin{document}
\begin{msc}{test}
\drawframe =
\declinst{a}{}{$A$}
\selfmesswidth = 5cm

%First outgoing
\lost[1]{label}{}{a}
\nextlevel
\nextlevel

%First receiving
\found[1]{label2}{}{a}
\nextlevel
\nextlevel

%Second outgoing
\lost[1]{label3}{}{a}

\end{msc}

\end{document}

With the setting \setlength{<some parameter>}{<length>}, you can change the default settings as you wish. Here, I changed the \selfmesswidth parameter:

\documentclass{article}
\usepackage{msc}
\begin{document}
\begin{msc}{test}
\setlength{\selfmesswidth}{3cm}
\setlength{\envinstdist}{3.5cm} % According to the above one
\declinst{a}{}{$A$}
\found{Very long label}{}{a}
\end{msc}
\end{document}

Note that the distance between environments and nearest instance line should be adjusted accordingly as I did in \setlength{\envinstdist}{3.5cm}.

• What is the difference between \setlength{\selfmesswidth}{3cm} and \selfmesswidth = 5cm ? :) – Andersnk May 11 '15 at 16:07


text:
jAMM module for mediation models

Draft version, mistakes may be around

The module estimates simple, multiple, and conditional mediation models with maximum likelihood regression. It encompasses models with continuous and categorical independent variables.

# Variable role definition

To estimate a model, the user should define: a dependent variable, one or more mediators, one or more independent variables. Independent variables can be continuous (going in Covariates) or categorical (going in Factors). In case one or more variables is a moderator, it should first be included in the independent variables fields, and then defined as moderator (in Moderators tab).

# Model building

When one defines the roles of the variable, jAMM tries to guess what model the user is estimating, with very simple rules. The independent variables are all considered parallel indepenent variables. The mediators are all inserted in the model as parallel mediators. All effects from the independent variables through the mediators are estimated. In other words, by default the module builds a multiple IV, multiple mediators model, like the example in the Figure with 2 IV and 2 mediatiors. In the simple case of three variables model (simple mediation), one obtains the standard mediation trianglular model.

When there are more independent variables and mediatiators, the rules described above apply. For two IV and two ME, the default model is as in the following Figure.

The structure of the model can be changed in the Mediators models and Full model tabs. See jAMM: Model building for examples on building models.

## Mediators models

The tab shows the mediators models, the models with mediators as dependent variable. On the right (Models for mediators) one field for each mediator is shown. There, the user can define the predictors of each mediator. See jAMM: Model building for examples on building models.

## Full model

The tab shows the full model, that is the model predicting the dependent variable. On the right (Model terms), oen can put all the model terms required. By default, jAMM puts in the model terms all the mediators and the mediators predictors. See jAMM: Model building for examples on building models.

# Results

## Models Info

When variables roles are yet not defined, the table shows some suggestions to get the user started. When the variables and the models are defined, the table shows the linear equations needed to estimate the model, and the indirect (mediated) effects involved in the model. If something goes wrong with the model specification, warnings are issued. The main reason a warning is issued is the specification of an incomplete or absurd model. The path diagram also signals problems in the model.

## Diagram

Upon defining variables and models terms, the path diagram is generated. This type of diagram is often called model diagram or conceptual model as opposed to the working model, which shows all terms and all estimated statistical paths. Without moderators, the model diagram corresponds to the working model but does not show the covariances between independent variables.

The diagram shows the model implied by the Mediators models and the Full model terms. The estimated paths (coefficients) are shown in black. Required variables or required paths are shown in red. A required path, for instance, can be a path from the IV to the mediator. The diagram also shows impossible paths in purple. Impossible paths are paths that do not make sense in terms of mediation, not necessarily in terms of statistical estimability. For instance, a model with two mediators m1 and m2, in which the user defines a path from m1 to m2 and a path from m2 to m1 is deemed as impossible, because this king of models are not mediation models. An example is in the Figure.

The diagram also shows suggested paths (in green). Those are paths whose absence does not invalidate the model, but may require extra care in interpreting the results. Suggested paths may be hidden in the Mediation options tab.

## Effect estimates

For models without interactions, jAMM produces a table showing the Indirect effects (mediated), the Direct effects, and the Total effects. The example below shows results for a model with one IV (x1), two parallel mediators (m1, m2) and a DV (y).

Indirect effects are the mediated effects. The Effect column indicates which mediated effect is shown in the corresponding row. Estimate reportes the estimation of the mediated effect. The columns are self-explanatory. See jAMM: Model estimates for details.

The direct effects are the effects computed keeping the mediators constant, thus the un-mediated effects. The total effects are the effects computed without the mediators, or, equivalently, the sum of the indirect and the direct effects.

If requested in the Mediation options tab by ticking Components one can see in the table also the components of the indirect effects. The components are the coefficients that are multiplied to obtain the indirect effects. In simple mediation, for instance, if the models are $$m=a \cdot x$$ and $$y= b \cdot m + c \cdot x$$, the mediated effect is $$IE=a \cdot b$$ and $$a$$ and $$b$$ are its components.

In the example shown in the table above, asking the Components yields:

# Moderators

Moderators are variables expected to moderate the indirect effects. In the tab, one can define one or more moderators for each mediator. That is equivalent to declaring one or more moderators for each IE. When a moderator is defined, jAMM updates the Mediators models and the Full model to include the interactions required to build the model. The user can then tweak the model, but also completely change it, by changing the terms in the Mediators models and the Full model tabs.

Declaring the moderator(s), however, triggers jAMM to nicely display the moderator in the model diagram and to compute simple indirect effects, that is, indirect effects computed at different levels of the moderators. What levels are used is ruled by Factor coding and Covariates scaling options.

Two important caveats should be noted:

To add a moderator, there should be at leat another independent variable in the medatiators models. For example, if the user defines x1 and x2 in the mediators models, jAMM allows one of them (say x1) to be a moderator. However, if the user tries to add also x2 as moderator, a warning is issued because with two moderators, there are no IV left to be moderated. An example is in the Figure shows the warning that appears if the user tries to add x2 as a moderator of the model.

See jAMM: Building conditional models for examples on building models with moderators.

### 2) Interactions without moderators

jAMM allows to define interactions in the model without specifying a moderator (see example in the following Figure).

This feature allows users to define an interaction effect that will be treated as an additional term of the mediation model. Thus, no simple indirect effects will be computed. Nontheless, the model may be perfectly fine and the user may have good reasons to insert an interaction without explicitely define a moderator. The feature also allows estimating very complex models with several interactions, not all of them interesting from the point of view of the mediated effects.

See jAMM: Building conditional models for examples on building models with moderators.

# Factors coding

It allows to code the categorical variables according to different coding schemas. The coding schema applies to the parameter estimates, simple effects and plots. The Omnibus Tests (ANOVA) table uses always the “deviation” scheme, that is “-1 0 1” coding, which results in a correct Type III analysis.

Note that all contrasts but dummy guarantee to be centered to zero (intercept being the grand mean), so when involved in interactions the other variables coefficients can be interpret as (main) average effects. If contrast dummy is set, the intercept and the effects of other variables in interactions are estimated for the first group of the categorical IV.

Contrasts definitions are provided in the estimates table. More detailed definitions of the comparisons operated by the contrasts can be obtained by selecting “Show contrast definition table”.

Differently to standard R naming system, contrasts variables are always named with the name of the factor and progressive numbers from 1 to K-1, where K is the number of levels of the factor.

In reading the contrast labels, one should interpret the (1,2,3) code as meaning “the mean of the levels 1,2, and 3 pooled toghether”. If factor levels 1,2 and 3 are all levels of the factor in the samples, (1,2,3) is equivalent to “the mean of the sample”. For example, for a three levels factor, a contrast labeled 1-(1,2,3) means that the contrast is comparing the mean of level 1 against the mean of the sample. For the same factor, a contrast labeled 1-(2,3) indicates a comparison between level 1 mean and the subsequent levels means pooled together.

More details and examples .

# Covariates Scaling

Continuous variables can be centered, standardized or used as they are (none). The default is centered because it makes our lives much easier when there are interactions in the model. If one needs to unscale the variable, simply select none.

Covariates conditioning rules how the model is conditioned to different values of the continuous independent variables when one or more moderators are defined.

• Mean+SD: means that the IV is conditioned to the $$mean$$, to $$mean+k \cdot sd$$, and to $$mean-k\cdot sd$$, where $$k$$ is ruled by the white field below the option. Default is 1 SD.

• Percentile 50 +offset: means that the IV is conditioned to the $$median$$, the $$median+k P$$, and the $$median-k\cdot P$$, where $$P$$ is the offset of percentile one needs. Again, the $$P$$ is ruled by the white field below the option. Default is 25%. The default conditions the model to:

• $$50^{th}-25^{th}=25^{th}$$ percentile

• $$50^{th}$$ percentile

• $$50^{th}+25^{th}=75^{th}$$ percentile

The offset should be within 5 and 50.

Covariates labeling decides which label should be associated with the estimates and table titles:

• Labels produces strings of the form $$Mean \pm SD$$

• Values uses the actual values of the variables

• Labels+Values produces labels of the form $$Mean \pm SD=XXXX$$, where XXXX is the actual value.

[to be continued]

# Examples

Some worked out examples of the analyses carried out with jamovi jAMM are posted here (more to come)

If you are intested in checking the convergence of jAMM results with other software results, please visit:


text:
# fastpos

The R package fastpos provides a fast algorithm to calculate the required sample size for a Pearson correlation to stabilize within a sequential framework (Schönbrodt & Perugini, 2013, 2018). Basically, one wants to find the sample size at which one can be sure that $\dpi{110}&space;\bg_white&space;1-\alpha$ percent of many studies will fall into a specified corridor of stability around an assumed population correlation and stay inside that corridor if more participants are added to the study. For instance, find out how many participants per study are required so that, out of 100k studies, 90% would fall into the region between .4 to .6 (a Pearson correlation) and not leave this region again when more participants are added (under the assumption that the population correlation is .5). This sample size is also referred to as the critical point of stability for the specific parameters.

This approach is related to the AO-method of sample size planning (e.g. Algina & Olejnik, 2003) and as such can be seen as an alternative to power analysis. Unlike AO, the concept of stability incorporates the idea of sequentially adding participants to a study. Although the approach is young, it has already attracted a lot of interest in the psychological research community, which is evident in over 800 citations of the original publication (Schönbrodt & Perugini, 2013). Still, to date, there exists no easy way to use the stability approach for individual sample size planning because there is no analytical solution to the problem and a simulation approach is computationally expensive with $\dpi{110}&space;\bg_white&space;\mathcal{O}(n^2)$. The presented package overcomes this limitation by speeding up the calculation of correlations and achieving $\dpi{110}&space;\bg_white&space;\mathcal{O}(n)$. For typical parameters, the theoretical speedup should be at least around a factor of 250. An empirical benchmark for a typical scenario even shows a speedup of about 460, paving the way for a wider usage of the stability approach.

## Installation

You can install the released version of fastpos from CRAN with:

install.packages("fastpos")

You can install the development version from GitHub with devtools:

devtools::install_github("johannes-titz/fastpos")

## Using fastpos

Since you have found this page, I assume you either want to (1) calculate the critical point of stability for your own study or (2) explore the method in general. If this is the case, read on and you should find what you are looking for. Let us first load the package and set a seed for reproducibility:

library(fastpos)
RNGkind("L'Ecuyer-CMRG")
set.seed(20220812)
parallel::mc.reset.stream()

In most cases you will just need the function find_critical_pos which will give you the critical point of stability for your specific parameters.

Let us reproduce Schönbrodt and Perugini’s quite famous and oft-cited table of the critical points of stability for a precision of 0.1. We set the number of studies to 100k and use multicore support (only works under GNU/Linux) so that it runs fairly quickly.

n_cores <- parallel::detectCores()
find_critical_pos(rho = seq(.1, .7, .1), sample_size_max = 1e3,
n_studies = 1e5, n_cores = n_cores)
#> Warning in find_critical_pos(rho = seq(0.1, 0.7, 0.1), sample_size_max = 1000, :
#> 448 simulation[s] did not reach the corridor of stability.
#> Increase sample_size_max and rerun the simulation.
#>   rho_pop pos.80% pos.90% pos.95% sample_size_min sample_size_max lower_limit
#> 1     0.1   257.0     362     478              20            1000         0.0
#> 2     0.2   244.2     355     467              20            1000         0.1
#> 3     0.3   213.0     306     403              20            1000         0.2
#> 4     0.4   182.0     265     352              20            1000         0.3
#> 5     0.5   142.0     217     288              20            1000         0.4
#> 6     0.6   108.0     159     209              20            1000         0.5
#> 7     0.7    64.0     100     134              20            1000         0.6
#>   upper_limit n_studies n_not_breached precision_absolute precision_relative
#> 1         0.2     1e+05            192                0.1                 NA
#> 2         0.3     1e+05            192                0.1                 NA
#> 3         0.4     1e+05              0                0.1                 NA
#> 4         0.5     1e+05             64                0.1                 NA
#> 5         0.6     1e+05              0                0.1                 NA
#> 6         0.7     1e+05              0                0.1                 NA
#> 7         0.8     1e+05              0                0.1                 NA

The results are very close to Schönbrodt and Perugini’s table (see https://github.com/nicebread/corEvol). Note that a warning is shown, because in some simulations the corridor of stability was not reached. As long as this number is low, this should not affect the estimates much. But if you want to get more accurate estimates, then increase the maximum sample size.

If you want to dig deeper, you can have a look at the functions that find_critical_pos builds upon. simulate_pos is the workhorse of the package. It calls a C++ function to calculate correlations sequentially and it does this pretty quickly (but you know that already, right?). A rawish approach would be to create a population with create_pop and pass it to simulate_pos:

pop <- create_pop(0.5, 1e6)
pos <- simulate_pos(x_pop = pop[,1],
y_pop = pop[,2],
n_studies = 1e5,
sample_size_min = 20,
sample_size_max = 1e3,
replace = T,
lower_limit = 0.4,
upper_limit = 0.6,
progress = FALSE)
hist(pos, xlim = c(0, 1e3), xlab = c("Point of stability"),
main = "Histogram of points of stability for rho = .5+-.1")

quantile(pos, c(.8, .9, .95), na.rm = T)
#> 80% 90% 95%
#> 144 208 276

Note that no warning message appears if the corridor is not reached, but instead an NA value is returned. Pay careful attention if you work with this function, and adjust the maximum sample size as needed.

create_pop creates the population matrix by using a method described on SO (https://stats.stackexchange.com/questions/15011/generate-a-random-variable-with-a-defined-correlation-to-an-existing-variables/15040#15040). This is a much simpler way than Schönbrodt and Perugini’s approach, but the results do not seem to differ. If you are interested in how population parameters (e.g. skewness) affect the point of stability, you should instead refer to the population generating functions in Schönbrodt and Perugini’s work.

## Parallelization

Since version 0.4.0 fastpos supports multiple cores. My first attempts to implement this were quite unsuccessful because of several reasons: (1) Higher-level parallelism in R makes it difficult to show progress in C++, which is where the important and time-demanding calculations happen (2) some parallelizing solutions do not work on all operating systems (e.g. mcpbapply) (3) overhead can be quite large, especially for a small number of simulation runs.

I thought the best solution is is to directly parallelize in C++. I tried to do it with RcppThread, but in the end this was even slower than singlethreading. Maybe the overhead of setting up multiple threads outweighs sharing the work.

My intermediate solution was quite simple and pragmatic: to use futures. I divided the number of studies by the available cores $\dpi{110}&space;\bg_white&space;n$, then started $\dpi{110}&space;\bg_white&space;n-1$ simulations via futures in a multisession plan. Meanwhile the main R process also started a simulation, which showed a progress bar in C++. All simulations ended at approximately the same time, the progress bar finished and the futures resolved. The points of stability were combined and the rest of the program worked as for the singlethreaded version.

Unfortunately this solution did not bring any substantial speed benefits. Since version 0.5.0 I switched to the parallel package and pbmclapply. This means there is no multicore support for Windows but the implementation is simple and shows clear speed benefits.

For small simulations there is no speed benefit because fastpos is already too fast:

n_cores <- parallel::detectCores()
onecore <- function() {
find_critical_pos(0.5, n_studies = 1e4, progress = FALSE)
}
multicore <- function() {
find_critical_pos(0.5, n_studies = 1e4, n_cores = n_cores, progress = FALSE)
}
microbenchmark::microbenchmark(onecore(), multicore(), times = 10)
#> Unit: milliseconds
#>         expr      min       lq     mean   median       uq      max neval
#>    onecore() 790.5424 806.0623 880.9228 898.5692 914.1386 995.4131    10
#>  multicore() 492.5743 507.3975 532.6609 519.9019 551.4526 617.7952    10

Increasing the number of studies to 1,000,000 makes the speed benefit more visible:

onecore <- function() {
find_critical_pos(0.5, n_studies = 1e6, progress = FALSE)
}
multicore <- function() {
find_critical_pos(0.5, n_studies = 1e6, n_cores = n_cores, progress = FALSE)
}
microbenchmark::microbenchmark(onecore(), multicore(), times = 10)
#> Unit: seconds
#>         expr       min        lq     mean   median       uq      max neval
#>    onecore() 57.639320 58.015838 58.85051 58.97313 59.24430 60.16749    10
#>  multicore()  9.254999  9.753195 11.80759 10.91989 11.91047 18.94043    10

The maximum sample size also affects the speed benefit:

onecore <- function() {
find_critical_pos(0.5, n_studies = 1e4, sample_size_max = 1e4,
progress = FALSE)
}
multicore <- function() {
find_critical_pos(0.5, n_studies = 1e4, sample_size_max = 1e4,
n_cores = n_cores, progress = FALSE)
}
microbenchmark::microbenchmark(onecore(), multicore(), times = 10)
#> Unit: seconds
#>         expr      min       lq     mean   median       uq      max neval
#>    onecore() 5.578969 5.670123 5.740891 5.733023 5.820454 5.971168    10
#>  multicore() 1.387514 1.411387 1.660138 1.444922 2.142137 2.184948    10

Using more than around 10 cores does not seem to bring additional speed benefits:

cores <- function(n_cores) {
bquote(find_critical_pos(0.5, n_studies = 1e5, sample_size_max = 5e3,
n_cores = .(n_cores),
progress = FALSE))
}

function_list <- lapply(1:n_cores, cores)
names(function_list) <- 1:n_cores
mb <- microbenchmark::microbenchmark(list = function_list, times = 10,
unit = "s")
mb
#> Unit: seconds
#>  expr       min        lq      mean    median        uq       max neval
#>     1 27.456555 27.570735 27.818499 27.745047 27.998901 28.552284    10
#>     2 14.386780 14.762170 14.799019 14.846158 14.891730 15.063332    10
#>     3 10.079521 10.272135 10.398187 10.449643 10.493651 10.690283    10
#>     4  7.786282  8.043653  8.123207  8.137474  8.174227  8.430836    10
#>     5  6.394230  6.457601  6.562601  6.559623  6.609138  6.833768    10
#>     6  5.493741  5.529108  5.677393  5.598711  5.857732  6.020727    10
#>     7  4.857490  4.957679  5.010991  5.000947  5.095433  5.150672    10
#>     8  4.501819  4.587342  4.653510  4.633522  4.721233  4.805759    10
#>     9  4.310806  4.513649  4.535217  4.531089  4.575051  4.703378    10
#>    10  4.302939  4.430716  4.556707  4.584788  4.648966  4.776502    10
#>    11  4.355230  4.508450  4.644879  4.641629  4.752381  5.003795    10
#>    12  4.141649  4.423738  4.629711  4.631150  4.817210  5.073758    10
#>    13  4.190796  4.631710  4.703205  4.683888  4.884239  4.962240    10
#>    14  4.389556  4.602093  4.745577  4.769075  4.890196  5.188714    10
#>    15  4.469471  4.506780  4.713059  4.686050  4.877884  5.059185    10
#>    16  4.172919  4.647434  4.666889  4.674721  4.853175  4.959688    10
#>    17  4.396954  4.569385  4.693250  4.604388  4.814465  5.140367    10
#>    18  4.448788  4.735557  4.902612  4.940298  5.105664  5.359574    10
#>    19  4.524529  4.860833  4.993857  4.930301  5.044446  5.557180    10
#>    20  4.903637  4.956342  5.144836  5.080159  5.318339  5.487320    10
#>    21  4.649297  4.967157  5.164920  5.219511  5.281292  5.579557    10
#>    22  4.590055  4.900174  5.218636  5.212350  5.616280  5.841419    10
#>    23  4.735541  5.040456  5.291030  5.239354  5.659753  5.688474    10
#>    24  4.634694  4.751584  5.039350  5.088603  5.243414  5.608231    10
#>    25  4.832792  5.021327  5.216380  5.202682  5.397915  5.566497    10
#>    26  4.510734  4.954355  5.111006  5.107665  5.315212  5.908580    10
#>    27  5.085392  5.151006  5.395888  5.369852  5.663524  5.772946    10
#>    28  4.883048  5.207972  5.294213  5.271934  5.550639  5.631980    10
#>    29  4.838274  5.086868  5.431193  5.409469  5.821201  5.960367    10
#>    30  4.970278  5.046102  5.434072  5.480559  5.805406  6.055113    10
#>    31  4.925469  5.105361  5.487184  5.509620  5.662650  6.186354    10
#>    32  5.014134  5.193149  5.376090  5.405113  5.561297  5.659123    10
plot(mb)

The test was done on a server with 32 cores. In general, the multicore support in fastpos is not perfect because the optimal split of work heavily depends on the specific parameters. In the future I hope to find a better way to implement multiple cores.

## How fast is fastpos?

In the introduction I boldly claimed that fastpos is much faster than the original implementation of Schönbrodt and Perugini (corEvol). The theoretical argument goes as follows:

corEvol calculates every correlation from scratch. If we take the sum formula for the correlation coefficient

we can see that several sums are calculated, each consisting of adding up $\dpi{110}&space;\bg_white&space;n$ (the sample size) terms. This has to be done for every sample size from the minimum to the maximum one. Thus, the total number of added terms for one sum is:

On the other hand, fastpos calculates the correlation for the maximum sample size first. This requires to add $\dpi{110}&space;\bg_white&space;n$ numbers for one sum. Then it subtracts one value from this sum to find the correlation for the sample size $\dpi{110}&space;\bg_white&space;n-1$, which happens repeatedly until the minimum sample size is reached. Overall the total number of terms for one sum amounts to:

The ratio between the two approaches is:

For the typically used $\dpi{110}&space;\bg_white&space;n_\mathrm{max}$ of 1000 and $\dpi{110}&space;\bg_white&space;n_\mathrm{min}$ of 20, we can expect a speedup of about 250. This is only an approximation for several reasons. First, one can stop the process when the corridor is reached, which is done in fastpos but not in corEvol. Second, the main function of fastpos was written in C++ (via Rcpp), which is much faster than R. At the same time, the algorithms involve many more steps than just calculating correlations. For instance, setting up the population with a specific $\dpi{110}&space;\bg_white&space;\rho$ takes some time since it usually consists of a million value pairs. The interface functions to setup the simulations also play an important role. Obviously, there is a lower time limit that cannot be beat. Thus, it is necessary to study the speed benefit empirically.

The theoretical difference is so big that it should suffice to give a rough benchmark for which the following parameters were chosen: rho = .1, sample_size_max = 1000, sample_size_min = 20, n_studies = 10000.

Note that corEvol was written as a script for a simulation study and thus cannot be simply called via a function. Furthermore, a simulation run takes a lot of time and thus it is not practical to run it too many times. If you want to experiment with the benchmark, I have forked the original corEvol repository and made a benchmark branch (note that this will only work on GNU/Linux, since here I am using git through the bash):

git -C corEvol pull || git clone --single-branch --branch benchmark \
https://github.com/johannes-titz/corEvol

For corEvol, two files are “sourced” for the benchmark. The first file generates the simulations and the second is for calculating the critical point of stability. I turned off all messages produced by these source files.

library(microbenchmark)
setwd("corEvol")
corevol <- function() {
source("01-simdata.R")
source("02-analyse.R")
}
fastpos <- function() {
find_critical_pos(rho = .1, sample_size_max = 1e3, n_studies = 10e3,
progress = FALSE)
}
bm <- microbenchmark(corevol = corevol(), fastpos = fastpos(), times = 10,
unit = "s")
bm
#> Unit: seconds
#>     expr         min          lq        mean      median          uq
#>  corevol 363.2943526 365.8412867 366.8502931 366.7658715 367.2626313
#>  fastpos   0.7604255   0.7855696   0.8164049   0.7999461   0.8071823
#>          max neval
#>  371.6549927    10
#>    0.9890917    10

For the chosen parameters, fastpos is about 460 times faster than corEvol, for which there are two main reasons: (1) fastpos is built around a C++ function via Rcpp and (2) this function does not calculate every calculation from scratch, but only calculates the difference between the correlation at sample size $\dpi{110}&space;\bg_white&space;n$ and $\dpi{110}&space;\bg_white&space;n-1$ via the sum formula of the Pearson correlation (see above). There are some other factors that might play a role, but they cannot account for the large difference found. For instance, setting up a population takes quite long in corEvol (about 20s), but compared to the ~6 min required overall, this is only a small fraction. There are other parts of the corEvol code that are fated to be slow, but again, a speedup by a factor of 460 cannot be achieved by improving these parts. The presented benchmark is definitely not comprehensive, but only demonstrates that fastpos can be used with no significant waiting time for a typical scenario, while for corEvol this is not the case.

## FAQ

### What does fastpos do if the corridor of stability is not reached for a simulation study?

In this case fastpos will return an NA value for the point of stability. When calculating the quantiles, fastpos will use the maximum sample size, which is a more reasonable estimate than ignoring the specific simulation study altogether.

### Why does fastpos produce different estimates to corEvol?

If the same parameters are used, the differences are rather small. In general, differences cannot be avoided entirely due to the random nature of the whole process. Even if the same algorithm is used, the estimates will vary slightly from run to run. The other more important aspect is how studies are treated where the point of stability is not reached: corEvol ignores them, while fastpos assumes that the corridor was reached at the maximum sample size. Thus, if the parameters are the same, fastpos will tend to produce larger estimates, which is more accurate (and more conservative). But note that if the corridor of stability is not reached, then you should increase the maximum sample size. Previously, this was not feasible due to the computational demands, but with fastpos it usually can be done.

## Issues and Support

If you find any bugs, please use the issue tracker at:

https://github.com/johannes-titz/fastpos/issues

If you need answers on how to use the package, drop me an e-mail at johannes at titz.science or johannes.titz at gmail.com

## Contributing

Comments and feedback of any kind are very welcome! I will thoroughly consider every suggestion on how to improve the code, the documentation, and the presented examples. Even minor things, such as suggestions for better wording or improving grammar in any part of the package, are more than welcome.

If you want to make a pull request, please check that you can still build the package without any errors, warnings, or notes. Overall, simply stick to the R packages book: https://r-pkgs.org/

## References

Algina, J., & Olejnik, S. (2003). Sample size tables for correlation analysis with applications in partial correlation and multiple regression analysis. Multivariate Behavioral Research, 38, 309–323. https://doi.org/10.1207/S15327906MBR3803_02
Schönbrodt, F. D., & Perugini, M. (2013). At what sample size do correlations stabilize? Journal of Research in Personality, 47, 609–612. https://doi.org/10.1016/j.jrp.2013.05.009
Schönbrodt, F. D., & Perugini, M. (2018). Corrigendum to “At What Sample Size Do Correlations Stabilize?” [J. Res. Pers. 47 (2013) 609–612]. Journal of Research in Personality, 74, 194. https://doi.org/10.1016/j.jrp.2018.02.010


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## Found 105 Documents (Results 1–100)

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HiggsTools

Title Authors arxiv ref Node Supervisor ESRs
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Search for a standard model Higgs boson produced in association with a top-quark pair and decaying to bottom quarks using a matrix element method CMS Collaboration 1502.02485
Combined Measurement of the Higgs Boson Mass in pp Collisions at $\sqrt{s}$=7 and 8 TeV with the ATLAS and CMS Experiments ATLAS, CMS Collaborations ATLAS and CMS Collaborations 1503.07589
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Measurement of differential cross sections for Higgs boson production in the diphoton decay channel in pp collisions at $\sqrt{s}$=8 TeV CMS Collaboration 1508.07819
The HiggsTools Handbook: Concepts and observables for deciphering the Nature of the Higgs Sector M. Boggia, J. M. Cruz-Martinez, H. Frellesvig, N. Glover, R. Gomez-Ambrosio, G. Gonella, Y. Haddad, A. Ilnicka, S. Jones, Z. Kassabov, F. Krauss, T. Megy, D. Melini, D. Napoletano, G. Passarino, S. Patel, M. Rodriguez-Vazquez, T. Wolf 1711.09875