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# 2D Histograms in R

How to make a 2D histogram in R. A 2D histogram is a visualization of a bivariate distribution.

### New to Plotly?

Plotly's R library is free and open source!
You can set up Plotly to work in online or offline mode.
We also have a quick-reference cheatsheet (new!) to help you get started!

### Version Check

Version 4 of Plotly's R package is now available!
Check out this post for more information on breaking changes and new features available in this version.

library(plotly)
packageVersion('plotly')

## [1] '4.5.2'

#### Basic 2D Histogram

2D histograms require x/y, but in contrast to heatmaps, z is optional. If z is not provided, binning occurs in the browser (see here for a list of binning options).

# install.packages('mvtnorm')
library(plotly)

s <- matrix(c(1, -.75, -.75, 1), ncol = 2)
obs <- mvtnorm::rmvnorm(500, sigma = s)
p <- plot_ly(x = obs[,1], y = obs[,2])
pp <- subplot(
)

# Set up API credentials: https://plot.ly/r/getting-started

#### Colorscale

If z is not provided, the only way to control coloring is through the colorscale attribute

p <- p %>% add_histogram2d(colorscale = "Blues")

# Set up API credentials: https://plot.ly/r/getting-started

#### Z Matrix

If you want more control for the binning algorithm, you can supply a 2D table or matrix to z. In this case, the R package will impose it's colorscale default (and the colors argument can be used to control the colorscale from R):

cnt <- with(diamonds, table(cut, clarity))
p <- plot_ly(diamonds, x = ~cut, y = ~clarity, z = ~cnt) %>%


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# Planck intermediate results XI. The gas content of dark matter halos: the Sunyaev-Zeldovich-stellar mass relation for locally brightest galaxies

* Corresponding author
Abstract : We present the scaling relation between Sunyaev-Zeldovich (SZ) signal and stellar mass for almost 260,000 locally brightest galaxies (LBGs) selected from the Sloan Digital Sky Survey (SDSS). These are predominantly the central galaxies of their dark matter halos. We calibrate the stellar-to-halo mass conversion using realistic mock catalogues based on the Millennium Simulation. Applying a multi-frequency matched filter to the Planck data for each LBG, and averaging the results in bins of stellar mass, we measure the mean SZ signal down to $M_\ast\sim 2\times 10^{11} \Msolar$, with a clear indication of signal at even lower stellar mass. We derive the scaling relation between SZ signal and halo mass by assigning halo properties from our mock catalogues to the real LBGs and simulating the Planck observation process. This relation shows no evidence for deviation from a power law over a halo mass range extending from rich clusters down to $M_{500}\sim 2\times 10^{13} \Msolar$, and there is a clear indication of signal down to $M_{500}\sim 4\times 10^{12} \Msolar$. Planck's SZ detections in such low-mass halos imply that about a quarter of all baryons have now been seen in the form of hot halo gas, and that this gas must be less concentrated than the dark matter in such halos in order to remain consistent with X-ray observations. At the high-mass end, the measured SZ signal is 20% lower than found from observations of X-ray clusters, a difference consistent with Malmquist bias effects in the X-ray sample.
Keywords :
Document type :
Journal articles
Domain :

Cited literature [37 references]

https://hal-cea.archives-ouvertes.fr/cea-01135420
Contributor : EDP Sciences Connect in order to contact the contributor
Submitted on : Wednesday, March 25, 2015 - 12:27:57 PM
Last modification on : Sunday, June 26, 2022 - 12:03:14 PM
Long-term archiving on: : Monday, April 17, 2017 - 10:52:55 PM

### File

aa20941-12.pdf
Publication funded by an institution

### Citation

Planck Collaboration, P. A. R. Ade, N. Aghanim, M. Arnaud, M. Ashdown, et al.. Planck intermediate results XI. The gas content of dark matter halos: the Sunyaev-Zeldovich-stellar mass relation for locally brightest galaxies. Astronomy and Astrophysics - A&A, EDP Sciences, 2013, 557, pp.A52. ⟨10.1051/0004-6361/201220941⟩. ⟨cea-01135420⟩

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Spring 2019 Final Exam Solution

Here are the final numerical results for each section of the exam. You can use them to check your work if you do the exam for practice.  If you have trouble with the problems, or don't get the answers shown here, stop by during office hours or make and appointment and we can go over them.

Question 1

Relative to BAU, the NPV of policy C is $4.285M and the NPV of policy P is$5.105M. Both are better than BAU and policy P is the best.

Question 2

(a) The EV of the startup is $208k. The EU is 280, which is less than the utility of the job, which is 316; the researcher will not launch the startup. The CE is$78,400, which is much less than the EV because the researcher is risk averse.

(b) The EU is now 331 and the researcher will accept the offer. Not required but the CE of the offer is $109.4k, which beats the$100k risk-free job.

(b) The EV of the study is $10B. The government would do it. A few of the intermediate results that are needed along the way: probability of rG = 60%; conditional probability of G given rG = 75%; probability of rB = 40%; conditional probability of B given rB = 87.5%. Question 5 Q = 64; P =$4160; profit = $480. Question 6 Q = 40; P =$8200; profit = $320k. Question 7 (a) Expected NPV = -$283k. The firm would not undertake the project.

(b) PV of CS = $6.818M. Expected PV =$1.909M.

Question 8

New expected NPV to the firm = $293k; the firm would undertake the project. Expected NPV to the government, including the CS, the profits if the vaccine succeeds, and less the grant =$1.332M.

URL: https://wilcoxen.maxwell.insightworks.com/pages/4918.html
Peter J Wilcoxen, The Maxwell School, Syracuse University
Revised 12/10/2019


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# Finding the attractors of the vector field constructed as the gradient of an interpolated 3D scalar field

I have a list of data existing as {{{x, y, z}, f},...} from which I construct a 3D interpolation function using Interpolation[ ]. I am able to construct a vector field from the gradient of this scalar field, I have verified this looks correct by plotting it.

I wish to find the 'attractors' of this vector field.

By 'attractor' I mean the points at which a particle would end up at $t \rightarrow \infty$ by following the field. I.e. they flow down the gradient field to these points. I think attractor is the right name for this, please let me know if I am using the wrong word. I have also seen them referred to as $\omega$ limits.

I would like to use mathematica to find all of these points and here I begin to struggle. My initial attempt was to use:

interp = Table[
{{dat[[i, 1]], dat[[i, 2]], dat[[i, 3]]}, 1/(1 + dat[[i, 7]])},
{i, 1, Length[dat[[All, 1]]]}
]

intf = Interpolation[interp]

intfd[x_,y_,z_] :=Evaluate[D[intf[x,y,z],{{x,y,z}}]]

Table[
NMinimize[
{Norm[intfd[x,y,z]],-4<=x<=4,-4<=y<=4,-4<=z<=4},
{{x,-5,5},{y,-5,5},{z,-5,5}},
],
{i,1,20}
]

To try to find the points of 0 gradient. I'm aware this will also find the $\alpha$ limits (or sources for the field, where the particle is at $t \rightarrow -\infty$) and I'm not even 100% sure that Norm[V[x,y,z]] == 0 is a definition for one of these limits.

Also this is a crappy stochastic method for finding as many as I can. Though I can intuit where the points should be ahead of time to verify I have found them all I would like a general solution.

Can anyone suggest a better method for finding the attractors of the field?

• Perhaps you could use ContourPlot and plot the zero-contours of the norm of the vector field to start with. Again, this will find points other than the "attractors", but it's a start, maybe. Of course, your functions are functions of three variables? Then this might not work. – march Sep 14 '15 at 16:49
• Is there anything in this Q&A that might be of help? Basins of attraction – MarcoB Sep 14 '15 at 16:49
• Wouldn't these points just be the local minima of the function? Why not just use FindMinimum on intf? – Michael Seifert Sep 15 '15 at 15:59
• Finding the Maxima of the function intf turned out to be the most stable way to do this, thanks Michael for the suggestion. – JamesFurness Sep 25 '15 at 12:25

With a fresh head I have constructed a somewhat satisfactory solution to this problem.

Rather than stochastically using a random search we can use the "DifferentialEvolution" NMinimise method and start from a grid of points covering the space. This should then minimize to downhill to all possible attractors.

Continuing from my previous code for the gradient field:

min = ParallelTable[
Table[
Table[
NMinimize[{
Norm[intfd[x, y, z]],
-4 <= x <= 4, -4 <= y <= 4, -4 <= z <= 4
},
{
{x, i - 0.5, i + 0.5},
{y, j - 0.5, j + 0.5},
{z, k - 0.5, k + 0.5}
},
Method -> "DifferentialEvolution"],
{i, -4, 4, 2}],
{j, -4, 4, 2}],
{k, -4, 4, 2}];

caseList = {res_, coord_};

zeros = Cases[Flatten[min, 2], caseList /; res == 0];

We can verify these solutions using:

Show[
Join[
{VectorPlot3D[intfd[x, y, z], {x, -5, 5}, {y, -5, 5}, {z, -5, 5}]},
Table[
Graphics3D[{Green, Sphere[{x, y, z}, 0.25] /. zeros[[i, 2]]}],
{i, 1, Length[zeros]}]
]
]

Clearly this is not ideal, there is a basin of attraction where the gradient is zero, so points are clustered, not identical. A finer grid of data may help with this.

I also expect some more attractors to exist between the ones found. Again, finer data and a finer mesh of minimizer starting points may help this.

I'm still open to more refined suggestions to this problem if anyone has any!

# EDIT:

I've refined the method a bit to a point where I am happy with the result.

Instead of NMinimize on the norm of the gradient I now have a recursive gradient search function to walk along the gradients until the norm is 0. This is faster and has the advantage that the $\alpha$ points can also be found by supplying a multiplier of -1 to the step. The function is:

recurWalk[vec_, sPos_, mult_, tol_, maxIt_] :=
If[Or[
Norm[vec[sPos[[1]], sPos[[2]], sPos[[3]]]] <= tol,
maxIt - 1 <= 0,
sPos[[1]] < -5,
sPos[[1]] > 5,
sPos[[2]] < -5,
sPos[[2]] > 5,
sPos[[3]] < -5,
sPos[[3]] > 5
],

(* Return {Norm, {x,y,z}} *)
{Norm[vec[sPos[[1]], sPos[[2]], sPos[[3]]]], {x -> sPos[[1]],
y -> sPos[[2]], z -> sPos[[3]]}},

(* Recursively call the next point *)
recurWalk[vec,
{
sPos[[1]] + vec[sPos[[1]], sPos[[2]], sPos[[3]]][[1]]*mult,
sPos[[2]] + vec[sPos[[1]], sPos[[2]], sPos[[3]]][[2]]*mult,
sPos[[3]] + vec[sPos[[1]], sPos[[2]], sPos[[3]]][[3]]*mult
},
mult, tol, maxIt - 1]
]

The successes are filtered from the failures in the same way as before.

This appears to be much faster and I can now produce:

Still comments and improvements are welcomed.


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# The element calcium has the atomic number of 20 and the mass number of 40, how many neutrons does it have?

Nov 2, 2015

20

#### Explanation:

There are few simple rules to follow.

(1) The atomic number is equal to the number of protons.

$Z$ = number of protons

(2) In neutrally charged elements, the number of electrons is the same as the number of protons.

$Z$ = number of protons = number of electrons (no charge)

Otherwise, positive charge means that the element lost an electron and negative charge means it gained an electron.

(3) The atomic mass is equal to the sum of the number of protons and number of neutrons.

$A$ = number of protons + number neutrons

or

$A$ = $Z$ + number neutrons

So if you say that $Z$ = 20 and $A$ = 40, then

$A$ = $Z$ + number neutrons

40 = 20 + number of neutrons

40 - 20 = number of neutrons

Therefore,

number of neutrons = 20

Nov 2, 2015

An atom of the calcium-40 isotope has $20$ neutrons in its nucleus.

#### Explanation:

Isotopes are named for their mass numbers, for example, carbon-14. The mass number of an isotope is the sum of the number of protons and neutrons in the atomic nucleus. The atomic number is the number of protons.

The calcium-40 isotope has a mass number of $40$, and an atomic number of $20$, therefore it has $20$ protons.

$\text{Number of neutrons"="(mass number)"-"(atomic number)}$

$\text{Number of neutrons} = 40 - 20 = 20$


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23rd August 2020, 15:51   #9
ryerman
Major Dude

Join Date: Mar 2010
Location: Canada
Posts: 726
Quote:
 Originally Posted by aleksey_WinAmp No, I need to implement my own logic to choose the next song, using C# or C++ or JScript, etc.
If you can do that before playing the first song, then you could create a playlist file (M3U or M3U8) containing a list of the paths to the songs, in the order you want them to play.
For the command "/ADD <file>", file can be a M3U or M3U8 file. No special formatting is required and the file specs should not be surrounded by quotes.

code:"C:\Program Files (x86)\Winamp\winamp.exe" /ADD "C:\path\to\list of files\Files.m3u8"

Windows 10 Home, 64 bit, Winamp 5.666, Bento Skin


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# System of equations with tag which has subscript

Suppose I have an equation

u_t = F(x)
u(0) = u_0

I want to put a brace around on the left hand side of both lines and give it a tag with a subscript (eg. P_w) AND I want to be able to refer to this later (by using \eqref) by giving it a label.

Problems:

1. I don't know how to tag something with a subscript in it
2. I don't know how to put it on the LHS

Help appreciated.

• What have you tried? Look here for hints, tex.stackexchange.com/questions/98397/enumerate-formulas/…, at the cases construction for the brace. – Ethan Bolker Apr 25 '13 at 12:06
• To assign an arbitrary "tag" instead of a number to an equation, load the amsmath package and use the command \tag{...}. E.g., you could write <body of equation> \tag{$P_w$} \label{eq:pw}; note that the argument of \tag is in text mode by default, hence the use of $...$ to create the subscripted material. Elsewhere in your document, you can then create a cross-reference to this equation with the command \eqref{eq:pw}. – Mico Apr 25 '13 at 13:00
• Thank you so much, I actually tried putting the dollar signs in the tag but inexplicably got an error. It works now. – mark.a Apr 25 '13 at 13:07

Equation tag on the LHS is done by the option leqno to the article class.

Arbitrary Tags can be set with the commands \tag{} and \tag*{} (the first has parentheses, the second not).

If you want to refer to it later, you have use $…$ in the \tag{} command instead of $$…$$.

Lastly, if you want to treat a set of equations as single object, for example if you want to tag it, you should use the aligned environment from amsmath package.

The code:

\documentclass[leqno,fleqn]{article}
\usepackage[showframe]{geometry}
\usepackage{amsmath}

\begin{document}
\tag{\text{P}_\text{W}}\label{one} \left\{ \begin{aligned} u_t&=F(x)\\ u(0)&=u_0 \end{aligned} \right.
\paragraph{Some Text} Hello World! \eqref{one} is a d*** good pair of equations.
\end{document}

The subscript in the tag can be created by several ways. For example, if you use xelatex, you can also use the xltxtra package with its \textsubscript command, in order to circumvent the use of math inside the \tag.


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# Typesetting permutations with latex

I split the question into two parts and included some answers.

# 1. Alternatives to the permute-package

The permute-package is very old (version 0.2, 1999) and can lead to problems in connection with other packages.

Is there an up-to-date package (or possibility) that generates similarly appealing results with the same ease?

Consider the following small working example, which demonstrates just a few possibilies of permute:

\documentclass{article}
\usepackage{permute}
\begin{document}

That's what I want to have:

% $\pmt{(123)(45)}$ without star: no spacing between numbers

% $\pmt*{(123)(45)}$ with star: no difference (see note)

\renewcommand*\pmtseparator{\;} % thickspace, but see note below

% $\pmt{(123)(45)}$ without star: no spacing between numbers

$\pmt*{(123)(45)}$ with star: spaces between numbers

I do \emph{not} want to write something horrific like
$\begin{array}{llllll}(1&2&3)(4&5)\end{array}$.

Of course I could write $(123)(45)$ or $(1,2,3)(4,5)$,
but I would like to have an apperance different from that of tuples,
and numbers greater than 10 rule out the first possibility anyway.
\end{document}

Notes:

• Concerning \pmtseparator. The separator is included if the *-variant of \pmt is used. It is predefined as a space  . Hence, it only appears in text mode. Redefining it to \ , as the manual suggests, makes the space visible both in text and math mode. If you use \pmt only in math mode, it is better, as Philippe Goutet pointed out, to use \, for a thinspace or \; for a thickspace. (More about lengths.)
• Verbose permutations can be created with \pmtv.

Solutions so far:

• There seems to be no alternative, but the package permute works.
• Alternatively, if you do not insist on a special appearance, you use one of the two possibilities from the last paragraph.

(However, the permute-package has some limitations concerning the order of multiplication and the use of numbers greater than 9, but this could be discussed later and in another question.)

# 2. Typesetting permutations with LaTeX

Or what do typographers or other authors say about (1 2 3)(4 5) vs. (1,2,3)(4,5) vs. (123)(45)?

• I always enjoyed it being "written verbose as an explicit sequence of preimage/image pairs" like on p 2 of the permute documentation. – Werner Aug 1 '12 at 18:15
• @Werner preimage/image pairs explain the permutation clearly but cycle notation is what you need if you want to do serious permutation arithmetic. – Ethan Bolker Aug 1 '12 at 18:24
• can you give an example of a problem, otherwise just using that package seems to be your preferred option. If you could make an example with it failing after loading another package it would give us something to work towards – David Carlisle Aug 1 '12 at 19:48
• As long as you use a consistent notation, there's nothing against commas; however I'd prefer spaces, probably \, instead of \ – egreg Aug 1 '12 at 19:56

To answer your second question, when you must choose a notation for an object, the main point is non-ambiguity: this rules out writing (123) as this could be, not knowing the context, either (1,2,3) or (12,3) or (1,23). This means that there should be a separation between the numbers.

The simplest choice is the comma (followed by a thin space, automatically added by TeX) but some people prefer avoiding any punctuation.

In this case, the rule is to use a space at least one rank above the smallest possible (in this case, that would be a thick space (which is \; in TeX), as the smallest used space in formulas is a thin one); this is to take into account the fact that there is a loss of readability without the comma to mark the separation. Using a thick space is what the Acta Mathematica (by Lansburgh, 1964) and Oxford (by Chaundy, 1957) math style guides recommend for enumerations without commas.

Another perfectly valid line of thought is that a permutation written without any commas is akin to a matrix, which would use an em space (\quad in TeX). As an em space is clearly too much for inline formulas, this would mean using a space one rank below (i.e. an en space, \enspace in TeX).

Here is a quick comparison of these solutions along with the word space one (which is \  in TeX, but you normally don't use it in formulas not containing text):

Of course, the simplest solution to implement is with commas as you can just write the formula as it will appear.

• Thanks for this answer. I have included a note on spacing commands in the question. – One Aug 2 '12 at 14:04


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5

# What are the differences between sexual and natural selection? Define and explain each (including the basic principles of evolutionary changes and natural selection...

## Question

###### What are the differences between sexual and natural selection? Define and explain each (including the basic principles of evolutionary changes and natural selection) with examples from the readings and class.Name two species with non-genetic sex determination and describe the effective mechanism_Describe the key evolutionary transitions that emerged as sexual organisms transitioned from unimorphic t0 dimorphic and how this led to sexual selection. Explain with examples from readings and articles

What are the differences between sexual and natural selection? Define and explain each (including the basic principles of evolutionary changes and natural selection) with examples from the readings and class. Name two species with non-genetic sex determination and describe the effective mechanism_ Describe the key evolutionary transitions that emerged as sexual organisms transitioned from unimorphic t0 dimorphic and how this led to sexual selection. Explain with examples from readings and articles mentioned in class:

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text:
# A.C for power transmission

Why is an alternating current used for long distance transmission of electrical power ?


text:
## Wednesday, August 10, 2016 ... /////

### Naturalness, a null hypothesis, hasn't been superseded

Quanta Magazine's Natalie Wolchover has interviewed some real physicists to learn

What No New Particles Means for Physics
so you can't be surprised that the spirit of the article is close to my take on the same question published three days ago. Maria Spiropulu says that experimenters like her know no religion so her null results are a discovery, too. I agree with that. I am just obliged to add that if she were surprised she is not getting some big prizes for the discovery of the Standard Model at $\sqrt{s}=13\TeV$, it's because her discovery is too similar to the discovery of the Standard Model at $\sqrt{s}=1.96\TeV$, $\sqrt{s}=7\TeV$, and $\sqrt{s}=8\TeV$, among others. ;-) And the previous similar discoveries were already done by others.

She and others at the LHC are doing a wonderful job and tell us the truth but the opposite answer – new physics – would still be more interesting for the theorists – or any "client" of the experimenters. I believe that this point is obvious and it makes no sense to try to hide it.

Nima Arkani-Hamed says lots of things I appreciate, too, although his assertions are exaggerated, as I will discuss. It's crazy to talk about a disappointment, he tells us. Experimenters have worked hard and well. Those who whine that some new pet model hasn't been confirmed are spoiled brats who scream because they didn't get their favorite lollipop and they should be spanked.

Yup. But when you look under the surface, you will see that there are actually many different opinions about naturalness and the state of physics expressed by different physicists. If you're strict enough, many of these opinions almost strictly contradict each other.

Nathaniel Craig whom I know as a brilliant student at Harvard says that the absence of new physics will have to be taken into account and addressed but he implicitly makes it clear that he will keep on thinking about theories such as his "neutral naturalness". Some kind of naturalness will almost certainly be believed and elaborated upon by people like him in the future, anyway. I think that Nathaniel and other bright folks like that should grow balls and say some of these things more clearly – even if they contradict some more senior colleagues.

Aside from saying that the diphoton could have been groundbreaking (yup), Raman Sundrum said:
Naturalness is so well-motivated that its actual absence is a major discovery.
Well, it is well-motivated but it hasn't been shown not to exist. This claim of mine contradicting Sundrum's assertion above was used in the title of this blog post.

What does it mean that you show that naturalness doesn't exist? Well, naturalness is a hypothesis and you want to exclude it. Except that naturalness – while very "conceptual" – is a classic example of a null hypothesis. If you want to exclude it, you should exclude it at least by a five-sigma deviation! You need to find a phenomenon whose probability is predicted to be smaller than 1 in 1,000,000 according to the null hypothesis.

We are routinely used to require just a 2-sigma (95%) exclusion for particular non-null hypotheses that add some new particles of effects. But naturalness is clearly not one of those. Naturalness is the null hypothesis in these discussions. So you need to exclude it by the five-sigma evidence. Has it taken place?

Naturalness isn't a sharply defined Yes/No adjective. As parameters become (gradually) much smaller than one, the theory becomes (gradually) less natural. When some fundamental parameter in the Lagrangian is fine-tuned to 1 part in 300, we say that $\Delta=300$ and the probability that the parameter is this close to the special value (typically zero) or closer is $p=1/300$.

(The precise formula to define $\Delta$ in MSSM or a general model is a seemingly technical but also controversial matter. There are many ways to do so. Also, there are lots of look-elsewhere effects that could be added as factors in $\Delta$ or removed from it. For these reasons, I believe that you should only care about the order of magnitude of $\Delta$, not some precise changes of values.)

The simplest supersymmetric models have been shown to be unnatural at $\Delta \gt 300$ or something like that. That means that some parameters look special or fine-tuned. The probability of this degree of fine-tuning is $p=1/300$ or so. Does it rule out naturalness? No because we require a five-sigma falsification of the null hypothesis e.g. $p=1/1,000,000$ or so. We're very far from it. Superpartners at masses comparable to $10\TeV$ will still allow naturalness to survive.

Twenty years ago, I wasn't fond of using this X-sigma terminology but my conclusions were basically the same. If some parameters are comparable to $0.01$, they may still be said to be of order one. We know such parameters. The fine-structure constant is $\alpha\approx 1/137.036$. We usually don't say that it's terribly unnatural. The value may be rather naturally calculated from the $SU(2)$ and $U(1)_Y$ electroweak coupling constants and those become more natural, and so on. But numbers of order $0.01$ only differ from "numbers of order one" by some 2.5 sigma.

My taste simply tells me that $1/137.036$ is a number of order one. When you need to distinguish it from one, you really need a precise calculation. For me, there's pretty much "no qualitative realm in between" $1$ and $1/137.036$. Numbers like $0.01$ have to be allowed in Nature because we surely know that there are dimensionless ratios (like $m_{\rm Planck}/m_{\rm proton}$) that are vastly different from one and they have to come from somewhere. Even if SUSY or something else stabilizes the weak scale, it must still be explained why the scale – and the QCD scale (it's easier) – is so much lower than the Planck scale. The idea that everything is "really" of the same order is surely silly at the end.

OK, assuming that $\Delta\gt 300$ has been established, Sundrum's claim that it disproves naturalness is logically equivalent to the claim that any 3-sigma deviation seen anywhere falsifies the null hypothesis, and therefore proves some new physics. Well, we know it isn't the case. We had a 4-sigma and 3-sigma diphoton excess. Despite the fact that the Pythagorean combination is exactly 5 (with the rounded numbers I chose), we know that it was a fluke.

Now, the question whether naturalness is true is probably (even) more fundamental than the question whether the diphoton bump came from a real particle. But the degree of certainty hiding in 3-sigma or $p=1/300$-probable propositions is exactly the same. If you (Dr Sundrum) think that the observation that some $\Delta \gt 300$ disproves naturalness, then you're acting exactly as sloppily as if you consider any 3-sigma bump to be a discovery!

A physicist respecting that particle physics is a hard science simply shouldn't do so. Naturalness is alive and well. 3-sigma deviations such as the observation that $\Delta \gt 300$ in some models simply do sometimes occur. We can't assume that they are impossible. And we consider naturalness to be the "basic story to discuss the values of parameters" because this story looks much more natural or "null" than known competitors. If and when formidable competitors were born, one could start to distinguish them and naturalness could lose the status of "the null hypothesis". But no such a convincing competitor exists now.

As David Gross likes to say, naturalness isn't a real law of physics. It's a strategy. Some people have used this strategy too fanatically. They wanted to think that even $\Delta \gt 10$ was too unnatural and picked other theories. But this is logically equivalent to the decision to follow research directions according to 1.5-sigma deviations. Whenever there's a 1.5-sigma bump somewhere, such a physicist would immediately focus on it. That's simply not how a solid physicist behaves in the case of specific channels at the LHC. So it's not how he should behave when it comes to fundamental conceptual questions such as naturalness, either.

Naturalness is almost certainly a valid principle but when you overuse it – in a way that is equivalent to the assumption that more than 3-sigma or 2-sigma or even 1.5-sigma deviations can't exist – you're pretty much guaranteed to be sometimes proven wrong by Mother Nature because statistics happens. If you look carefully, you should be able to find better guides for your research than 1.5-sigma bumps in the data. And the question whether $\Delta\lt 10$ or $\Delta \gt 10$ in some model is on par with any other 1.5-sigma bump. You just shouldn't care about those much.

While I share much of the spirit of Nima's comments, they're questionable at the same moment, too. For example, Nima said:
It’s striking that we’ve thought about these things for 30 years and we have not made one correct prediction that they have seen.
Who hasn't made the correct predictions? Surely many people have talked about the observation of "just the Standard Model" (Higgs and nothing else that is new) at the LHC. I've surely talked about it for decades. And we had discussions about it with virtually all high-energy physicists who have ever discussed anything about broader physics. I think that the Standard Model was by far the single most likely particular theory expected from the first serious LHC run at energies close to $\sqrt{s}=14\TeV$.

The actual adjective describing this scenario wasn't "considered unlikely" but rather "considered uninteresting". It was simply not too interesting for theorists to spend hours about the possibility that the LHC sees just the Standard Model. And it isn't too interesting for theorists now, either. A hard-working theorist would hardly write a paper in 2010 about the "Standard Model at $\sqrt{s}=13\TeV$". There's simply nothing new and interesting to say and no truly new calculation to be made. But the previous sentence – and the absence of such papers – doesn't mean that physicists have generally considered this possibility unlikely.

I am a big SUSY champion but even in April 2007, before the LHC was running, I wrote that the probability was 50% that the LHC would observe SUSY. Most of the remaining 50% is "just the Standard Model" because I considered – and I still consider – the discovery of all forms of new physics unrelated to SUSY before SUSY to be significantly less likely than SUSY.

So I think that the statement that "we haven't made any correct prediction" to be an ill-defined piece of sloppy social science. The truth value depends on who is allowed to be counted as "we" and how one quantifies these voters' support for various possible answers. When it came to a sensible ensemble of particle physicists who carefully talk about probabilities rather than hype or composition of their papers and who are not (de facto or de iure) obliged to paint things in more optimistic terms, I am confident that the average probability they would quote for "just the Standard Model at the LHC" was comparable to 50%.

I want to discuss one more quote from Nima:
There are many theorists, myself included, who feel that we’re in a totally unique time, where the questions on the table are the really huge, structural ones, not the details of the next particle. We’re very lucky to get to live in a period like this — even if there may not be major, verified progress in our lifetimes.
You know, Nima is great at many things including these two:
1. A stellar physicist
2. A very good motivational speaker
I believe that the quote above proves the second item, not the first. Are we in a totally unique time? When did the unique time start? Nima has already been talking about it for quite some time. ;-) If the young would-be physicists believe it, the quote may surely increase their excitement. But aren't they being fooled? Is there some actual evidence or a rational reason to think that "physics is going through a totally unique time when structural paradigm shifts are around the corner"?

I think that the opposite statement is much closer to be a rational conclusion of the available evidence. Take the statements by Michelson or Kelvin over 100 years. Physics is almost over. All that remains is to measure the values of the parameters with a better precision.

Well, I think that the evidence is rather strong that this statement would actually be extremely appropriate for the present situation of physics and the near future! This surely looks like a period in which no truly deep paradigm shifts are taking place and none are expected in coming months or years. I think that Nima's revolutionary proposition reflects his being a motivational speaker rather than a top physicist impartially evaluating the evidence.

We should really divide the question of paradigm shifts to those that demand an experimental validation; and those that may proceed without an experimental validation. These two "realms of physics" have become increasingly disconnected – and this evolution has always been unavoidable. And it's simply primarily the first one, the purely theoretical branch, where truly new things are happening. When you only care about theories that explain the actual doable experiments, the situation is well-described by the Michelson-Kelvin quote about the physics of decimals (assuming that Michelson or Kelvin become a big defenders of the Standard Model).

Sometimes motivational speeches are great and needed. But at the end, I think that physicists should also be grownups who actually know what they're doing even if they're shaping their opinions about big conceptual questions.

For years, Nima liked to talk about the unique situation and the crossroad where it's being decided whether physics will pursue the path of naturalness or the completely different path of the anthropic reasoning. Well, maybe and Nima sounds persuasive but I have always had problems with these seemingly oversimplified assertions.

First, the natural-vs-anthropic split is just a description of two extreme philosophies that may be well-defined for the theories we already know but may become inadequate for the discussion of the future theories in physics. In particular, it seems very plausible to me that the types of physics theories in the future will not "clearly fall" into one of the camps (natural or anthropic). They may be hybrids, they may be completely different, they may show that the two roads discussed by Nima don't actually contradict each other. At any rate, I am convinced that when new frameworks to discuss the vacuum selection and other things become persuasive, they will be rather quantitative again: they will have nothing to do with the vagueness and arbitrariness of the anthropic principle as we know it today. Also, they will be careful in the sense that they will avoid some of the naive strategies by the "extreme fans of naturalness" who think that 2-sigma deviations or $\Delta \gt 20$ are too big and can't occur. Future theories of physics will be theories studied by grownups – physicists who will avoid the naivite and vagueness of both the extreme naturalness cultists as well as the anthropic metaphysical babblers.

Even if some new physics were – or had been – discovered at the LHC, including supersymmetry (not sure about the extra dimensions, those would be really deep), I would still tend to think that the paradigm shift in physics would probably be somewhat less deep than the discovery of quantum mechanics 90 years ago.

So while I think that it's silly to talk about some collapse of particle or fundamental physics and similar things, I also think that the talk about the exceptionally exciting situation in physics etc. has become silly. It's surely not my obsession to be a golden boy in the middle. But I am in the middle when it comes to the question whether the future years in physics are going to be interesting. We don't know and the answer is probably gonna be a lukewarm one, too, although both colder and hotter answers can't be excluded. And I am actually confident that the silent majority of physicists agrees that the contemporary physics and fundamental physics in the foreseeable future is and will be medium-interesting. ;-)


text:
02-4   Berichtsreihe des Mathematischen Seminars der Universität Kiel

## On proper analytic maps with one critical point

Let U \subset C be a domain containing
{ z \in \C : \re z > \sigma, ¦ \arg z ¦ < \eta }
for some \sigma,\eta>0 and let \$f:U ->U be a proper holomorphic map satisfying f(z) = z+1+a/z+o\left(1/¦z¦\right) as z\to\infty, ¦\arg z¦<\eta, with a \in \C. We show that if U contains only one critical point of f, and this critical point is simple, then \re a \geq \frac14. This slightly generalizes a previous result concerning critical points in Leau domains. We also show that the condition \re a \geq \frac14 is sharp.

Mathematics Subject Classification (1991): Primary 37F10; Secondary 30D05, 37F45

Bibliographical note: erscheint in einem Tagungsbericht: Reihe "Contemp. Math." der AMS


text:
# asinf Function

Calculates the trigonometric arc sine function of a single precision floating-point value.

Include

<math.h>

Prototype

float asinf(float x);

Argument

 x value between -1 and 1 for which to return the arc sine

Return Value

Returns the arc sine in radians in the range [-π/2, +π/2] radians (inclusive) or NaN if a domain error occurs.

Remarks

If x is less than -1 or greater than 1, a domain error will occur and errno will be set to EDOM.

Example

See the notes at the beginning of this chapter or section for information on using printf() or scanf() (and other functions reading and writing the stdin or stdout streams) in the example code.

#include <math.h>
#include <stdio.h>
#include <errno.h>

int main(void)
{
float x, y;

errno = 0;
x = 2.0F;
y = asinf(x);
if (errno)
perror("Error");
printf("The arcsine of %f is %f\n", x, y);

errno = 0;
x = 0.0F;
y = asinf(x);
if (errno)
perror("Error");
printf("The arcsine of %f is %f\n", x, y);
}

Example Output

Error: Domain error
The arcsine of 2.000000 is nan
The arcsine of 0.000000 is 0.000000


text:
# 8.2 A single population mean using the student t distribution

 Page 1 / 21

In practice, we rarely know the population standard deviation . In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.

William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution . The name comes from the fact that Gosset wrote under the pen name "Student."

Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ .

If you draw a simple random sample of size n from a population that has an approximately a normal distribution with mean μ and unknown population standard deviation σ and calculate the t -score t = $\frac{\overline{x}–\mu }{\left(\frac{s}{\sqrt{n}}\right)}$ , then the t -scores follow a Student's t-distribution with n – 1 degrees of freedom . The t -score has the same interpretation as the z -score . It measures how far $\overline{x}$ is from its mean μ . For each sample size n , there is a different Student's t-distribution.

The degrees of freedom , n – 1 , come from the calculation of the sample standard deviation s . In [link] , we used n deviations $\left(x–\overline{x}\text{values}\right)$ to calculate s . Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df).

## Properties of the student's t-distribution

• The graph for the Student's t-distribution is similar to the standard normal curve.
• The mean for the Student's t-distribution is zero and the distribution is symmetric about zero.
• The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.
• The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution.
• The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.

what do they mean in a question when you are asked to find P40 and P88
Mani
hi
Mehri
you're asked to find page 40 and page 88 on that particular book.
Joseph
hi
ravi
any suggestions for statistics app better than this
ravi
sorry miss wrote the question
omar
No problem) By the way. I NEED a program For statistical data analysis. Any suggestion?
Mani
Eviews will help u
Hello
Okonkwo
arey there any data analyst and working on sas statistical model building
ravi
IMAGESNEWSVIDEOS A Dictionary of Computing. measures of location Quantities that represent the average or typical value of a random variable (compare measures of variation). They are either properties of a probability distribution or computed statistics of a sample. Three important measures are the mean, median, and mode.
define the measures of location
IMAGESNEWSVIDEOS A Dictionary of Computing. measures of location Quantities that represent the average or typical value of a random variable (compare measures of variation). They are either properties of a probability distribution or computed statistics of a sample. Three important measures are th
Ahmed
what is confidence interval estimate and its formula in getting it
discuss the roles of vital and health statistic in the planning of health service of the community
given that the probability of
BITRUS
can man city win Liverpool ?
There are two coins on a table. When both are flipped, one coin land on heads eith probability 0.5 while the other lands on head with probability 0.6. A coin is randomly selected from the table and flipped. (a) what is probability it lands on heads? (b) given that it lands on tail, what is the Condi
0.5*0.5+0.5*0.6
Ravasz
It should be a Machine learning terms。
Mok
it is a term used in linear regression
Saurav
what are the differences between standard deviation and variancs?
Enhance
what is statistics
statistics is the collection and interpretation of data
Enhance
the science of summarization and description of numerical facts
Enhance
Is the estimation of probability
Zaini
mr. zaini..can u tell me more clearly how to calculated pair t test
Haai
do you have MG Akarwal Statistics' book Zaini?
Enhance
Haai how r u?
Enhance
maybe .... mathematics is the science of simplification and statistics is the interpretation of such values and its implications.
Miguel
can we discuss about pair test
Haai
what is outlier?
outlier is an observation point that is distant from other observations.
Gidigah
what is its effect on mode?
Usama
Outlier  have little effect on the mode of a given set of data.
Gidigah
How can you identify a possible outlier(s) in a data set.
Daniel
The best visualisation method to identify the outlier is box and wisker method or boxplot diagram. The points which are located outside the max edge of wisker(both side) are considered as outlier.
Akash
@Daniel Adunkwah - Usually you can identify an outlier visually. They lie outside the observed pattern of the other data points, thus they're called outliers.
Ron
what is completeness?
I am new to this. I am trying to learn.
Dom
I am also new Dom, welcome!
Nthabi
thanks
Dom
please my friend i want same general points about statistics. say same thing
alex
outliers do not have effect on mode
Meselu
also new
yousaf
I don't get the example
ways of collecting data at least 10 and explain
Example of discrete variable
Gbenga
I am new here, can I get someone to guide up?
alayo
dies outcome is 1, 2, 3, 4, 5, 6 nothing come outside of it. it is an example of discrete variable
jainesh
continue variable is any value value between 0 to 1 it could be 4digit values eg 0.1, 0.21, 0.13, 0.623, 0.32
jainesh
hi
Kachalla
what's up here ... am new here
Kachalla
sorry question a bit unclear...do you mean how do you analyze quantitative data? If yes, it depends on the specific question(s) you set in the beginning as well as on the data you collected. So the method of data analysis will be dependent on the data collecter and questions asked.
Bheka
how to solve for degree of freedom
saliou
Quantitative data is the data in numeric form. For eg: Income of persons asked is 10,000. This data is quantitative data on the other hand data collected for either make or female is qualitative data.
Rohan
*male
Rohan
Degree of freedom is the unconditionality. For example if you have total number of observations n, and you have to calculate variance, obviously you will need mean for that. Here mean is a condition, without which you cannot calculate variance. Therefore degree of freedom for variance will be n-1.
Rohan
data that is best presented in categories like haircolor, food taste (good, bad, fair, terrible) constitutes qualitative data
Bheka
vegetation types (grasslands, forests etc) qualitative data
Bheka
I don't understand how you solved it can you teach me
solve what?
Ambo
mean
Vanarith


text:
# American Institute of Mathematical Sciences

• Previous Article
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• DCDS Home
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Boundedness and large time behavior in a two-dimensional Keller-Segel-Navier-Stokes system with signal-dependent diffusion and sensitivity
July  2018, 38(7): 3617-3636. doi: 10.3934/dcds.2018156

## Global dynamics in a two-species chemotaxis-competition system with two signals

 1 College of Mathematics and Statistics, Chongqing University, Chongqing 401331, China 2 Depart. of Appl. Math., Chongqing Univ. of Posts and Telecommun., Chongqing 400065, China 3 College of Economic Math., Southwestern Univ. of Finance and Economics, Chengdu 611130, China

* Corresponding author: Xinyu Tu

Received  November 2017 Revised  February 2018 Published  April 2018

Fund Project: The first author is partially supported by the China Scholarship Council (201706050065). The second author is partially supported by National Natural Science Foundation of China (Grant Nos: 11771062, 11371384, 11571062), the Basic and Advanced Research Project of CQCSTC (Grant No: cstc2015jcyjBX0007). Fundamental Research Funds for the Central Universities (Grant Nos. 10611CDJXZ238826). The third author is partially supported by National Natural Science Foundation of China (Grant Nos: 11601053, 11526042). The fourth author is partially supported by Chongqing Scientific & Technological Talents Program (Grant No. KJXX2017006).

In this paper, we consider a chemotaxis-competition system of parabolic-elliptic-parabolic-elliptic type
 $\begin{eqnarray*}\label{1}\left\{\begin{array}{llll}u_t = Δ u-χ_{1}\nabla·(u\nabla v)+μ_{1}u(1-u-a_{1}w), &x∈ Ω, ~~~t>0, \\0 = Δ v-v+w, &x∈Ω, ~~~t>0, \\w_t = Δ w-χ_{2}\nabla·(w\nabla z)+μ_{2}w(1-a_{2}u-w), &x∈ Ω, ~~~ t>0, \\0 = Δ z-z+u, &x∈Ω, ~~~t>0, \\\end{array}\right.\end{eqnarray*}$
with homogeneous Neumann boundary conditions in an arbitrary smooth bounded domain
 $Ω\subset R^n$
,
 $n≥2$
, where
 $χ_{i}$
,
 $μ_{i}$
and
 $a_{i}$
 $(i = 1, 2)$
are positive constants. It is shown that for any positive parameters
 $χ_{i}$
,
 $μ_{i}$
,
 $a_{i}$
 $(i = 1, 2)$
and any suitably regular initial data
 $(u_{0}, w_{0})$
, this system possesses a global bounded classical solution provided that
 $\frac{χ_{i}}{μ_{i}}$
are small. Moreover, when
 $a_{1}, a_{2}∈ (0, 1)$
and the parameters
 $μ_{1}$
and
 $μ_{2}$
are sufficiently large, it is proved that the global solution
 $(u, v, w, z)$
of this system exponentially approaches to the steady state
 $\left(\frac{1-a_{1}}{1-a_{1}a_{2}}, \frac{1-a_{2}}{1-a_{1}a_{2}}, \frac{1-a_{2}}{1-a_{1}a_{2}}, \frac{1-a_{1}}{1-a_{1}a_{2}}\right)$
in the norm of
 $L^{∞}(Ω)$
as
 $t\to ∞$
; If
 $a_{1}≥1>a_{2}>0$
and
 $μ_{2}$
is sufficiently large, the solution of the system converges to the constant stationary solution
 $\left(0, 1, 1, 0\right)$
as time tends to infinity, and the convergence rates can be calculated accurately.
Citation: Xinyu Tu, Chunlai Mu, Pan Zheng, Ke Lin. Global dynamics in a two-species chemotaxis-competition system with two signals. Discrete & Continuous Dynamical Systems - A, 2018, 38 (7) : 3617-3636. doi: 10.3934/dcds.2018156
##### References:

show all references

##### References:
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# Cauchy-Product of non-absolutely convergent series

While grading some basic coursework on analysis, I read an argument, that a Cauchy product of two series that converge but not absolutely can never converge i.e. if $\sum a_n$, $\sum b_n$ converge but not absolutely, the series $\sum c_n$ with $$c_n= \sum_{k=0}^n a_{n-k}b_k$$ diverges.

Although we didn't have any theorem in the course stating something like this, it made me wonder if it was true.

-
I'm probably missing something, but if we take $a_k:=\frac{(-1)^k}k=:b_k$ for $k\geq 1$ and $a_0=b_0=0$, we have $c_n=2\frac{(-1)^n}n\sum_{k=1}^{n-1}\frac 1k$, and this series is converging (but not absolutely). – Davide Giraudo Jun 10 '12 at 10:12
@Davide Giraudo I get $c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}.$ – Ragib Zaman Jun 10 '12 at 10:20
@RagibZaman: They're the same. – anon Jun 10 '12 at 10:26
@anon ahh I see. Thanks. – Ragib Zaman Jun 10 '12 at 10:28
ha ... you're right Davide. Taking the same series I also got to $c_n=2\frac{(-1)^n}n\sum_{k=1}^{n-1}\frac 1k$ but then didn't see that $\frac1n\sum_{k=1}^{n-1}\frac 1k$ is converging to zero. Thanks for helping out. – Peter Patzt Jun 10 '12 at 11:21

As Davide Giraudo has said in the comments, we can find a counter example by using $a_k = b_k = (-1)^k/k$ for $k\geq 1$ and $a_0=b_0=0.$ In that case we compute $$c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$
$$= \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{n-k+k}{k(n-k)} = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left( \frac{1}{k} + \frac{1}{n-k}\right) = 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.$$
We show $\sum c_n$ converges by applying the Leibniz criterion. $c_n \to 0$ is clear, so we need only verify that $d_n = \frac{1}{n} \sum_{k=1}^{n-1}$ is monotonically decreasing for sufficiently large $n.$
We compute $$d_{n+1}- d_n = \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k} - \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{k}= \frac{1}{n(n+1)} - \frac{1}{n(n+1)}\sum_{k=1}^{n-1} \frac{1}{k}.$$
Since $\displaystyle \sum_{k=1}^{n-1} \frac{1}{k} \geq 1$ for all $n\geq 2$ so $d_n$ is indeed monotone.


text:
# problem adding additional column under multicolumn option in table [duplicate]

This question already has an answer here:

First part of the question: I have a table

\begin{document}
\begin{tabular}{|*9{c|}}\hline
A & \multicolumn{2}{c|}{B} & \multicolumn{2}{c|}{C} & \multicolumn{2}{c|}{D} & \multicolumn{2}{c|}{E} \\\cline{2-9}
& \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} \\\cline{2-9}
& U & V &  U & V & U & V & U & V \\\hline
y  & 7.23 & 6.39  & 7.76 &  6.93 & 2.81 &  2.54 & 0.59 &  0.55  \\ \hline
z  & 2.5503 &  2.2658  & 2.5345 &  2.3741 & 1.85 &  1.64 & 0.46 &  0.37   \\ \hline
$\sum{E^{i}_{U/V}}$  & & & & &  & & $\sim 1$ & $\sim 1$   \\ [1ex]  \hline
\end{tabular}

\end{document}

which nicely re-produce what I want.

But as I want to add 3rd column under B, it does not work and keep giving me error "illegal character in array.arg".

\begin{document}

\begin{tabular}{|*10{c|}}\hline
A & \multicolumn{3}{c|}{B} & \multicolumn{2}{c|}{C} & \multicolumn{2}{c|}{D} & \multicolumn{2}{c|}{E} \\\cline{2-10}
& \multicolumn{3}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} \\\cline{2-10}
& U & V & W &  U & V & U & V & U & V \\\hline
y  & 7.23 & 6.39 & X & 7.76 &  6.93 & 2.81 &  2.54 & 0.59 &  0.55  \\ \hline
z  & 2.5503 &  2.2658 & X  & 2.5345 &  2.3741 & 1.85 &  1.64 & 0.46 &  0.37   \\ \hline
$\sum{E^{i}_{U/V}}$  & & & & & &  & & $\sim 1$ & $\sim 1$   \\ [1ex]  \hline
\end{tabular}

\end{document}

can anyone help me solve the issue please.

Second part of the question:

How can I remove the column lines from the last row for the columns with empty entries?

Any help will be highly appreciated, thanks!

## marked as duplicate by David Carlisle tables StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 9 '15 at 15:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• *10 is 1 copy of 0. You meant *{10} for 10 copies. The documented syntax always has braces, although you can sometimes omit them in some cases eg single digits. – David Carlisle Jun 9 '15 at 15:08
• Please always make the examples complete starting from \documentclass{article} so people can easily reproduce the error. – David Carlisle Jun 9 '15 at 15:09
• wow, this is surprizing, *9 works but *10 doesn't !!! thanks! any insights for the second part of the question? – SAS Jun 9 '15 at 15:11
• Why is that surprising? It is same as \fbox ab just boxes a you need \fbox{ab} to box more than one token. – David Carlisle Jun 9 '15 at 15:12
• \multicolumn{1}{c}{} will make an entry with no vertical rule. – David Carlisle Jun 9 '15 at 15:13

## 1 Answer

The main issue here was that *10 is interpreted as "one time". Commands in LaTeX always look for the next character. If you need to handle more then one character, you have to put it in braces. *{10}. You can see the same for \textbf ab.

% arara: pdflatex

\documentclass{article}
\usepackage{showframe} % remove this. It just shows that the second table is too large.

\begin{document}
\noindent
\begin{tabular}{|*9{c|}}\hline
A & \multicolumn{2}{c|}{B} & \multicolumn{2}{c|}{C} & \multicolumn{2}{c|}{D} & \multicolumn{2}{c|}{E} \\\cline{2-9}
& \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} \\\cline{2-9}
& U & V &  U & V & U & V & U & V \\\hline
y  & 7.23 & 6.39  & 7.76 &  6.93 & 2.81 &  2.54 & 0.59 &  0.55  \\ \hline
z  & 2.5503 &  2.2658  & 2.5345 &  2.3741 & 1.85 &  1.64 & 0.46 &  0.37   \\ \hline
$\sum{E^{i}_{U/V}}$  & & & & &  & & $\sim 1$ & $\sim 1$   \\ [1ex]  \hline
\end{tabular}

\noindent % the table is still a bit to wide for this page...
\begin{tabular}{|*{10}{c|}}\hline % important to set 10 into braces. (and every other multi digits number)
A & \multicolumn{3}{c|}{B} & \multicolumn{2}{c|}{C} & \multicolumn{2}{c|}{D} & \multicolumn{2}{c|}{E} \\\cline{2-10} % 2-10, not 2-9, I guess...
& \multicolumn{3}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} & \multicolumn{2}{c|}{x} \\\cline{2-10}
& U & V &  W &  U & V & U & V & U & V \\\hline
y  & 7.23 & 6.39 & X & 7.76 &  6.93 & 2.81 &  2.54 & 0.59 &  0.55  \\ \hline
z  & 2.5503 &  2.2658 & X  & 2.5345 &  2.3741 & 1.85 &  1.64 & 0.46 &  0.37   \\ \hline
$\sum{E^{i}_{U/V}}$  & \multicolumn{2}{c}{}  & & \multicolumn{1}{c}{} & &\multicolumn{1}{c}{} & & $\sim 1$ & $\sim 1$   \\ [1ex]  \hline
\end{tabular}
\end{document}

It would not be helping with the horizontal width issue, but you could get rid of all those lines with use of the package booktabs. Here is a quick solution which might need some more tweaking.

\documentclass{article}
\usepackage{siunitx}
\usepackage{booktabs}

\begin{document}
\scriptsize\noindent
\begin{tabular}{c*{2}S[table-format=1.4]c*{2}S[table-format=1.4]*{4}S[table-format=1.2]}\toprule
A & \multicolumn{3}{c}{B} & \multicolumn{2}{c}{C} & \multicolumn{2}{c}{D} & \multicolumn{2}{c}{E} \\
& \multicolumn{3}{c}{x} & \multicolumn{2}{c}{x} & \multicolumn{2}{c}{x} & \multicolumn{2}{c}{x} \\\addlinespace
& U & V &  W &  U & V & U & V & U & V \\\cmidrule(lr){2-4}\cmidrule(lr){5-6}\cmidrule(lr){7-8}\cmidrule(lr){9-10}
y  & 7.23 & 6.39 & X & 7.76 &  6.93 & 2.81 &  2.54 & 0.59 &  0.55  \\
z  & 2.5503 &  2.2658 & X  & 2.5345 &  2.3741 & 1.85 &  1.64 & 0.46 &  0.37   \\\midrule
$\sum{E^{i}_{U/V}}$  & &  & &  & & & & {$\sim 1$} & {$\sim 1$}  \\\bottomrule
\end{tabular}
\end{document}


text:
## ABSTRACT

Until now, we have not found it necessary to assume any distributional form for the errors ". However, if we want to make any confidence intervals or perform any hypothesis tests, we will need to do this. The common assumption is that the errors are normally distributed. In practice, this is often, although not always, a reasonable assumption. We have already assumed that the errors are independent and identically distributed (i.i.d.) with mean 0 and variance #2, so we have " ~ N(0, #2I). Now since y=X!+", we have y ~ N(X!, #2I). which is a compact description of the regression model. From this we find, using the fact that linear combinations of normally distributed values are also normal, that:

3.1 Hypothesis Tests to Compare Models

Given several predictors for a response, we might wonder whether all are needed. Consider a larger model, #, and a smaller model, ', which consists of a subset of the predictors that are in ( If there is not much difference in the fit, we would prefer the smaller model on the principle that simpler explanations are preferred. On the other hand, if the fit of the larger model is appreciably better, we will prefer it. We will take ' to represent the null hypothesis and ( to represent the alternative. A geometrical view of the problem may be seen in Figure 3.1.


text:
## a neat (theoretical) Monte Carlo result

Posted in Books, Statistics, University life with tags , , , , on December 19, 2014 by xi'an

Mark Huber just arXived a short paper where he develops a Monte Carlo approach that bounds the probability of large errors

$\mathbb{P}(|\hat\mu_t-\mu|>\epsilon\mu) < 1/\delta$

by computing a lower bound on the sample size r and I wondered at the presence of μ in the bound as it indicates the approach is not translation invariant. One reason is that the standard deviation of the simulated random variables is bounded by cμ. Another reason is that Mark uses as its estimator the median

$\text{med}(S_1R_1,\ldots,S_tR_t)$

where the S’s are partial averages of sufficient length and the R’s are independent uniforms over (1-ε,1+ε): using those uniforms may improve the coverage of given intervals but it also means that the absolute scale of the error is multiplied by the scale of S, namely μ. I first thought that some a posteriori recentering could improve the bound but since this does not impact the variance of the simulated random variables, I doubt it is possible.

## R midterms

Posted in Kids, Linux, R, Statistics, University life with tags , , , , , , , , , , , on November 9, 2012 by xi'an

Here are my R midterm exams, version A and version B in English (as students are sitting next to one another in the computer rooms), on simulation methods for my undergrad exploratory statistics course. Nothing particularly exciting or innovative! Dedicated ‘Og‘s readers may spot a few Le Monde puzzles in the lot…

Two rather entertaining if mundane occurences related to this R exam: one hour prior to the exam, a student came to my office to beg for being allowed to take the solution manual with her (as those midterm exercises are actually picked from an exercise booklet, some students cooperated towards producing a complete solution manual and this within a week!), kind of missing the main point of having an exam. (I have not seen yet this manual but I’d be quite interested in checking the code they produced on that occasion…) During the exam, another student asked me what was the R command to turn any density into a random generator: he had written a density function called mydens and could not fathom why rmydens(n) was not working. The same student later called me as his computer was “stuck”: he was not aware that a “+” prompt on the command line meant R was waiting for him to complete the command… A less comical event that ended well is that a student failed to save her R code (periodically and) at the end of the exam and we had to dig very deep into the machine to salvage her R commands from \tmp as rkward safeguards, as only the .RData file was available at first. I am glad we found this before turning the machine off, otherwise it would have been lost.


text:
# Set notation what does the bar symbol mean?

In set notation could somebody explain the meaning of $$\mid$$ in the equation below please? How does it read?

I read it as $$s$$ and $$j$$ are an element of $$E$$ but what does the $$j \mid$$ mean?

Similarly in this equation what does the comma mean?

• Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$. – Steven Wagter Dec 26 '18 at 12:48
• – Mauro ALLEGRANZA Dec 26 '18 at 12:53
• @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.) – John Bentin Dec 26 '18 at 13:03

Suppose the $$m_{j,i}$$'s are elements of the set $$M$$, and that each pair $$(j,i) \subset I \subset \Bbb{N}^2$$ is an index of an element in $$M$$, where $$I$$ is an indexing set. Then

$$Inputs(i) = \sum_{j|(j,i) \in E} m_{j,i}$$ means "Given some fixed input value $$i$$, sum those elements in $$M$$ whose index $$(j,i)$$ satisfies $$(j,i) \in E$$.

Here $$E$$ is some subset of $$\Bbb{N}^2$$. So we loop over all the $$j$$'s ($$i$$ is fixed) and add the terms corresponding to those $$j$$'s which satisfy $$(j,i) \in E$$.

In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.

EDIT - in response to Shaun:

$$I$$ is some set used for indexing the elements in $$M$$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $$2\times2$$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $$a_{1,2}$$. But what is really going on, is that you have the index set $$I = \{ (1,1),(1,2),(2,1),(2,2)\}$$, where there is a one-to-one correspondence between elements in $$I$$ and the set of matrix elements.

For example, returning to our situation, taking the same indexing set $$I$$ as above and letting $$E = \{(1,1),(2,1),(2,2)\}$$, then $$Inputs(2) = m_{2,2}$$.

Another example of indexing sets: $$1/2+1/4+1/8 +... = \sum_{n \in \Bbb{N} } \frac{1}{2^n}$$

Here $$\Bbb{N}$$ is the indexing set.

• Where did the $I$ come from? Don't you mean $E$? – Shaun Dec 28 '18 at 5:32
• @Shaun added some clarification, does it make more sense now? – Steven Wagter Dec 28 '18 at 8:42
• Yes. Thank you, @StevenWagter :) – Shaun Dec 28 '18 at 8:44


text:
# How do you solve the inequality x^2+3x-28<0?

Apr 4, 2018

$- 7 < x < + 4$

#### Explanation:

${x}^{2} + 3 x - 28 = {x}^{2} + 7 x - 4 x - 48 = x \left(x + 7\right) - 4 \left(x + 7\right) = \left(x - 4\right) \left(x + 7\right)$

So, the inequality is

$\left(x - 4\right) \left(x + 7\right) < 0$

Now. the left hand side vanishes for $x = - 7$ and $x = + 4$.

If $x < - 7$, then both the factors $\left(x + 7\right)$ and $\left(x - 4\right)$ are negative, and the product is positive. On the other hand, $\mathmr{if} x > + 4$, both factors are positive, leading once again to a positive product.

Thus the inequality is only satisfied when

$- 7 < x < + 4$

where the facor $\left(x + 7\right)$ is positive and $\left(x - 4\right)$ is negative.

You can see this in the graph below

graph{x^2+3x-28 [-10, 10, -40, 40]}


text:
# How do you find the slope and intercept of y=3x-7?

Arrange the equation in to the form of $y = m x + c$ and equate values: here, gradient = 3 and ($y$) intercept = -7
As it happens, this equation is already in the correct form of $y = m x + c$. This makes finding the gradient and $y$ intercept easy, as in this form, gradient = $m$ and y intercept = $c$
Hence, $m$ = 3 and $y$ = -7


text:
# Tag Info

0

Sodium trioxocarbonate (iv) pentahydrate, a.k.a washing soda, is an efflorescent crystal containing 10 molecules of water of crystallization in its molecular formula.The molar mass of washing soda comprises a: carbonate portion of 106 g + pentahydrate portion of 180 g = 286 g/mol. i.e. 1 mole or 286 g of washing soda contains 106 g of ...

0

The molar mass of copper (ii) sulfate penta-hydrate is $\pu{249.69 g/mol}$ and that for anhydrous copper(ii) sulfate is $\pu{159.61 g/mol}$. Thus, in $1$ mole of the hydrated substance there are $\pu{159.61 g}$ of copper (ii) sulfate and $5 \times \pu{18.00 g}$ (i.e. $\pu{90.00 g}$) of molecules of water of crystallization. $\pu{5M}$ copper (ii) sulfate ...

1

$\ce{P2O5}$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70\% \ \ce{P2O5}$ could just be expressed as a percentage of $\ce{H3PO4}$ but there are times when a higher $\ce{P2O5}$ concentration is desired. For example ...

5

Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two ...

5

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. Always write the values with the correct units and carry the units through the calculation. Do ...

-7

$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$. Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is ...

4

As Ed said you are following the correct protocol by diluting the sample. The key reason is for doing so is to avoid a non-linear region in the calibration curve. This is why one would dilute the sample rather than make more concentrated standards. Several interesting effects happen with alkali metals at high concentrations in the flame or plasma. One of ...

1

A solution by definition is a homogeneous mixture, so its "concentration" is same throughout the bulk. Let us have a look as to how one would prepare a solution of a given concentration, say in 1 L. We would weigh out a known weight of a solute and dissolve it ~ 500 mL of solvent, with mechanical stirring or shaking. We would add more solvent until the ...

Top 50 recent answers are included


text:
# How to count bytes in languages with autoformatters [duplicate]

In QBasic, VBA, AppleScript, etc., there is an autoformatter that adds spaces and expands some syntactic sugar. For example, if I type this code:

?x*2

QBasic will expand it to

PRINT x * 2

Should I count this code as 4 bytes or 11?

Edit: Nathan Merrill brings up How to count bytes in macro heavy languages as a closely related question. I agree that it's related, but I don't think it's a duplicate. The reason why I asked this question is because the normal usage of these programming languages adds spaces automatically. It's not something that happens when the programmer chooses to expand macros--it's something that happens without any action on the programmer's part. It would be easy to conclude that the fully spaced version is the correct syntax of the language, and therefore that the condensed version is illegal for code golf submissions.

Here's a link to the answer that prompted me to ask this question. Also pertinent is this comment on my QBasic quine, which suggests that a quine using condensed syntax isn't valid because "the autoformatter can't be turned off" in QBasic 1.1.

• Related: scoring in Vim. – DLosc Oct 7 '16 at 20:17
• I don't really see how that's related... – James Oct 7 '16 at 20:20
• @DJMcMayhem This question essentially asks, "Are keystrokes directly comparable to bytes," like the Vim question does. – DLosc Oct 7 '16 at 20:26
• Its really related to this – Nathan Merrill Oct 7 '16 at 20:27
• It's been a long time since I've used Basic variants, but can't you turn off the auto formatting in any sane IDE? – Geobits Oct 7 '16 at 20:31
• @Geobits Hm, that's an interesting question. I know you can in QB64, and I just found where you can in VBA. I don't recall such an option in DOS QBasic, but I may not have ever looked in the right menu. – DLosc Oct 7 '16 at 20:47
• @Geobits According to at least one source, QBasic 1.1 for DOS was not a sane IDE. – DLosc Oct 10 '16 at 4:18
• @DLosc Hmm. IIRC you can run directly from the command line, something like qbasic /run mything.bas. Now I wonder if that works with "unformatted" code. If so, I'd say it's still pretty obvious that it should count that way. – Geobits Oct 10 '16 at 4:21
• @Geobits Idk how to do it in Visual Studio, and it autoformats .NET code. Edit: A short look shows you can. stackoverflow.com/a/5493003/2415524 – mbomb007 Oct 10 '16 at 18:53
• "It's not something that happens when the programmer chooses to expand macros--it's something that happens without any action on the programmer's part." Just to clarify: macros in Lisp are automatically expanded too. In fact it makes no sense to evaluate a form without performing macroexpansion first. The user can call macroexpand to control when it happens, but that's all. – coredump Oct 13 '16 at 11:08

## Count code size before autoformatting

Rationale: You can type or paste unformatted code into the interpreter, hit run, and it works. The fact that the interpreter adds whitespace or expands shortcuts is irrelevant. If you typed 4 bytes, the size of your program is 4 bytes.

Rationale 2: Geobits points out that many IDEs have an option to turn autoformatting off. In this case, it's even clearer that the unformatted code is legitimate code that the interpreter can run directly.

• Even if the code is dependent on the IDE, you can still count it that way, but the technical "language" would then be the IDE+interpreter. – Nathan Merrill Oct 10 '16 at 15:32
• Oh good I don't have to go back and change all of my AppleScript answers to correctly format their bytecount. relieved – Addison Crump Oct 11 '16 at 7:59

# Count file size

Why not use the original scoring mechanism and just use the byte count of the source file?

This can be quite effective due to the fact that BASIC languages often use tokenisation, shortening the length of the file.

• QBasic-based response, may apply to other languages too: This could work as long as you write the file in a text editor other than the QBasic IDE. (I've done this once.) Otherwise, the autoformatter kicks in before you can save the file, meaning the source file will have all the extra spaces in it. And even in QB64 with the autoformatter turned off, the file will always have a trailing newline, which I'm not in the habit of counting in my submissions. – DLosc Oct 10 '16 at 4:29
• ... This not to mention the Windows \r\n newline convention, which I'm also not in the habit of counting in my submissions. You can get 1 byte per newline by counting keystrokes OR by writing the code in Notepad++ with Unix-style newlines. I've successfully run such code files in QB64. However, I haven't been able to test this approach against DOS QBasic. – DLosc Oct 10 '16 at 4:34
• What do you mean here? Do you mean that we need to count after autoformatting? – Erik the Outgolfer Oct 12 '16 at 15:44
• This would effectively kill VBA for code-golf - a blank .xlsm file called a.xlsm with one worksheet and one VBE code module has a file size of 10,970 bytes – Taylor Scott Dec 28 '16 at 20:02


text:
# [tex-eplain] eplain and Context

Karl Berry karl at freefriends.org
Fri Apr 30 15:23:45 CEST 2004

    \def^^L{\par}%

I just wanted to make the definition non-\outer, it seems, so I could be
used in \@setuplisting.

\catcode\^^L=13 should be added before it, to avoid the error.

`


text:
# A question about lattices and their being partially-ordered sets, rather than total-ordered sets

Am I correct in saying that, given a and b are elements of S, the join(a, b) and meet(a, b) won't be either a or b, if there is no order between a and b?

-
In fact, $a\land b = a$ if and only if $a\leq b$; $a\lor b = a$ if and only if $a\geq b$. So if $a$ and $b$ are not comparable in the lattice, then you are correct, $a\land b\notin\{a,b\}$ and $a\lor b\notin\{a,b\}$. –  Arturo Magidin Nov 20 '11 at 6:33

Yes, you are correct. If $a$ and $b$ are comparable, say $a\leq b$, then $a\land b = a$ and $a\lor b= b$; conversely, since $a\land b\leq a$ and $a\land b\leq b$ for all $a$ and $b$, if $a\land b\in\{a,b\}$, then $a$ and $b$ are comparable (similar if $a\lor b\in{a,b}$).


text:
# Building the period-finding circuit for Shor's Algorithm & the classical complexity of finding the period

I've been trying to learn about Shor's algorithm by writing out implementations of the circuit for modular exponentiation, $${ a }^{ x }\; ({ mod }\; N)$$, to find the period $$r$$ for small numbers such as:

$$(N=15,\quad a=11)\quad \longrightarrow \quad r=2$$,

$$(N=35,\quad a=13)\quad \longrightarrow \quad r=4$$,

$$(N=21,\quad a=5)\quad \longrightarrow \quad r=6$$.

I know these are incredibly small numbers, but I realised that in order to actually start building the cicuit at all (using the binary exponentiation method), I needed to calculate the values of $${ a }^{ { 2 }^{ k } }\; ({ mod }\; N)$$ for all $$0\;\le\;k \; <\; 2\; \lceil { \log _{ 2 }{ N } }\rceil$$, which pretty much immediately made the values of $$r$$ obvious before I'd even started building the circuit.

For example, take the least trivial case from the above, where $$r=6$$. I needed to calculate:

$${ 5 }^{ 0 }\; ({ mod }\; 21)=1$$,

$${ 5 }^{ 1 }\; ({ mod }\; 21)=5$$,

$${ 5 }^{ 2 }\; ({ mod }\; 21)=4$$,

$${ 5 }^{ 4 }\; ({ mod }\; 21)=16$$,

$${ 5 }^{ 8 }\; ({ mod }\; 21)=4$$,

$${ 5 }^{ 16 }\; ({ mod }\; 21)=16$$.

Simply by inspection, I can already see what the value of $$r$$ must be, because the result is the same when $$x$$ increases by six or twelve, but not after increasing by two or three.

I therefore have two questions:

1. Is calculating these modular exponentials of $$a$$ to powers of two required when building these circuits, or is there some way of not even having to do that?

2. If we do need to calculate these values, can we not find the period $$r$$ of this modular exponentiation function classically in an efficient way, starting with these values and then using a method analogous to 'interval halving', or an I being misled by using small numbers? [SEE EDIT]

EDIT: As Mark S points out, the results of modular exponentiation have very little structure, and so although perhaps faster than $$O(N)$$, such an 'interval halving' method would not have logarithmic complexity, thus would be slower than Shor's algorithm for large enough $$N$$.
The fast "modular exponentiation by repeated-squaring" part of Shor's algorithm is only one of the many ideas in Shor's paper, and I think a more critical idea/breakthrough is the QFT itself. However, without something equivalent to modular exponentiation by repeated-squaring, you have no way of having a small circuit to evaluate $$f(x)$$, so that you can perform the QFT on $$\vert x\rangle\vert f(x)\rangle$$.
I also think you are being significantly misled by looking at such small numbers. You'll need to prepare a state $$\vert x\rangle\vert f(x)\rangle$$ somehow; here $$f(x)$$ is evaluated with repeated-squaring, which is honestly pretty darned fast already, when $$x$$ gets large. It's not clear if a "binary search" is not already what the repeated-squaring is effectively doing.
• Maybe what I meant by 'binary search' in my second question wasn't clear. What I mean to ask is whether it is possible to quickly find the period r (without any quantum computation) by looking at the values of the modular exponentiation to powers of two, then based on the answers, selectively calculate other values of a^x (mod N) via a form of 'interval halving' until we can deduce what the period is. Surely we don't have to calculate it for every single value of x to do that? I suppose the complexity of doing this must be larger than building and using the cicuit, but I can't see how. May 15, 2020 at 23:03
• But $a^x\bmod N$ doesn't have much structure that you could leverage for interval halving? You'll need at least two different $x$ such that $a^x\bmod N$ is the same, in order to have a chance of knowing the period... May 15, 2020 at 23:51
• Yeh, you must be right, but I just need to convince myself of it. You get quite a lot of tests done each time you calculate a new value, so my original thought was that it would be faster than $O(N)$, but I suppose it isn't logarithmic, which it would have to be to beat Shor's. May 16, 2020 at 2:02


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# RD Sharma Solutions for Class 12 : Maths

## Solutions for all the questions from Maths RD SHARMA, Class 12th

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### RD SHARMA Class 12 | ALGEBRA OF VECTORS

Prove that a necessary and sufficient condition for three vectors  vec a , vec b and  vec c to be coplanar is that there exist scalars l , m , n not all zero simultaneously such that l vec a+m vec b+n vec c= vec0dot

### RD SHARMA Class 12 | ALGEBRA OF VECTORS

Prove that the following vectors are non-coplanar: 3 hat i+ hat j- hat k ,2 hat i- hat j+7 hat k and 7 hat i- hat j+23 hat k  hat i+2 hat j+3 hat k ,2 hat i+ hat j+3 and  hat i+ hat j+ hat k

### RD SHARMA Class 12 | ALGEBRA OF VECTORS

Using vectors show that the points A(-2,3,5),B(7,0,-1)C(-3,-2,-5) and D(3,4,7) are such that A B and C D intersect at the point P(1,2,3)dot

### RD SHARMA Class 12 | ALGEBRA OF VECTORS

Prove that 1,1,1 cannot be direction cosines of a straight line.

### RD SHARMA Class 12 | ALGEBRA OF VECTORS

A vector  vec r is inclined at equal acute angles of x-a xi s , y-a xi s and z-a xi sdot if | vec r|=6 units, find  vec rdot

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# An extension of symplectomorphism group

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sp{Sp}$$Let $$\omega=\sum dx_i\wedge dy_i$$ be the standard symplectic structure of $$\mathbb{R}^{2n}=\mathbb{R}^{n}\times \mathbb{R}^n$$.

We consider the following two
extensions of $$\Sp(2n,\mathbb{R})$$, the group of linear isomorphisms of $$\mathbb{R}^{2n}$$ preserving $$\omega$$:

1) Let $$G$$ be the group of all $$A\in \GL(2n,\mathbb{R})$$ which map all isotropic subspaces to isotropic subspaces. (A closed subgroup of the general linear group containing the symplectic group $$\mathrm{Sp}(2n,\mathbb{R})$$.)

2) Let $$H$$ be the group of all elements in $$\GL(2n,\mathbb{R})$$ which map all symplectic subspaces to symplectic subspaces.

A non-linear construction as above can be introduced on a symplectic manifold $$M$$: $$G(M)$$ is the group of all diffeomorphisms of $$M$$ whose derivatives map isotropic subspaces to isotropic subspace. And similarly $$H(M)$$ is the group of all diffeomorphisms whose linear parts map symplectic subspaces to symplectic subspaces.

Is there a terminology for Lie groups $$G$$, $$H$$, $$\bar{H}$$ and their Lie algebras and also the corresponding structures on symplectic manifolds? Is there any relation between $$G$$ and $$\bar H$$? What can be said about their first fundamental groups, as it is customary to compute the first fundamental groups of classical Lie groups?

Inspired by the concept of symplectic vector fields as Lie algebra of symplectomorphism group, what can be said about the Lie algebra of $$G(M)$$ and some dynamical interpretations on a symplectic manifold?

• Does "symplectic subspace" mean "subspace on which the symplectic form is non-degenerate"? (I originally asked what $\bar H$ was, but deleted it when I realised you meant the closure.) – LSpice May 23 at 21:28
• Yes The symplectic subspace are meant as you mentioned.$H$ is not closed subset of general linear group so we consider its closure $\bar H$ – Ali Taghavi May 23 at 21:32
• Is there an easy example to show that $H$ isn't closed? – LSpice May 23 at 21:33
• +1 for this comment Note that the limit of a sequence of symplectic subspaces is not necessarilly symplectic subspace (In grassmanian topoloy) but the limit of isotropic subspace is again an isotropic subspace. For example The 2 dimensional space $(x,y,\epsilon y, \epsilon x)$ is a symplectic subspace only when $\epsilon \neq 0$ – Ali Taghavi May 23 at 21:40
• @LSpice I think that "diffeomorphisms whose linear part/ derivative" was correct and better than with plural. Googling "functions whose image" yields thrice more occurrences than "functions whose images". – YCor May 23 at 23:39

## 1 Answer

Short answer: $$G = H$$ is the group of conformal symplectic linear maps. What follows is a proof of this (which I've simplified slightly from what I originally wrote):

1) $$G$$ is the group of conformal symplectic linear maps

First, some notation. We write $$\{e_i\} \cup \{f_j\}$$ for the standard basis for $$\mathbb{R}^{2n}$$, i.e. with $$\omega(e_i,e_j) = 0$$, $$\omega(f_i,f_j) = 0$$, and $$\omega(e_i,f_j) = \delta_{i,j}$$.

Suppose $$\phi \in G$$. We begin by singling out the isotropic subspaces $$A = \mathbb{R}^n \times \{0\}$$ and $$B = \{0\} \times \mathbb{R}^n$$. The following lemma is one way you might think about the group $$\mathrm{Sp}(2n,\mathbb{R})$$ in the first place: as being defined by its action on an isotropic subspace and a choice of isotropic complement.

Lemma: Given $$\phi$$, there is a unique element $$\psi \in \mathrm{Sp}(2n,\mathbb{R})$$ such that the following three properties hold:

• $$A = (\psi \circ \phi)(A)$$, with $$(\psi \circ \phi){\big|}_{A} = \mathrm{id} \colon A \rightarrow A$$
• $$B = (\psi \circ \phi)(B)$$

Given this lemma, we have that $$\psi \circ \phi = \begin{pmatrix}\mathrm{id} & 0 \\ 0 & T\end{pmatrix}$$, and it suffices to find conditions on $$T \in \mathrm{GL}(n,\mathbb{R})$$ so that isotropic subspaces are preserved by $$\psi \circ \phi$$. It suffices to consider just two cases:

• For $$i \neq j$$, the span of $$e_i$$ and $$f_j$$ is isotropic, so its image under $$(\psi \circ \phi)$$, the span of $$e_i$$ and $$Tf_j$$, must also be isotropic. This proves that $$T$$ is diagonal.
• Also for $$i \neq j$$, we can use the subspace spanned by $$e_i + f_j$$ and $$e_j + f_i$$, which will give us the condition that the diagonal entries of $$T$$ are all equal. Hence $$T = c \cdot \mathrm{id}$$ for some constant $$c \neq 0$$.

Finally, we see that with $$T = c \cdot \mathrm{id}$$, $$(\psi \circ \phi)^*\omega = c \cdot \omega$$, which clearly preserves isotropic subspaces. So this property completely characterizes $$G$$.

In terms of nomenclature, $$G$$ might be called the group of conformally symplectic linear operators, and the nonlinear theory would fall under the umbrella of (locally) conformal symplectic geometry, which is slightly more general than what you ask (i.e. $$\omega$$ itself does not need to be symplectic, only (locally) conformally symplectic, which is the more natural setting). I know only a little about the theory, and am far from an expert on the history, so I'll say nothing more that there exists literature on the subject.

As for computations of classical invariants, we see that $$G \cong \mathbb{R}^* \times \mathrm{Sp}(2n,\mathbb{R})$$, so there's nothing interesting to say about computing its standard invariants that can't already be said for $$\mathrm{Sp}(2n,\mathbb{R})$$.

2) $$G=H$$

Suppose $$\omega(v,w) \neq 0$$ and $$\omega(v,x) = 0$$. Then $$\omega(v,w+tx) \neq 0$$ for all $$t \in \mathbb{R}$$, so $$v$$ and $$w+tx$$ always span a symplectic subspace. Hence, if $$\phi \in H$$, since $$\phi$$ preserves symplectic subspaces, we require $$0 \neq \omega(\phi(v),\phi(w+tx)) = \omega(\phi(v),\phi(w)) + t\omega(\phi(v),\phi(x))$$ for all $$t \in \mathbb{R}$$. Hence, we must have $$\omega(\phi(v),\phi(x)) = 0$$ whenever $$\omega(v,x) = 0$$. In particular, $$\omega$$ preserves isotropic subspaces, so $$H \subseteq G$$. Meanwhile, every conformal symplectic linear map certainly preserves symplectic subspaces, so $$G \subseteq H$$. Hence $$G=H$$.

Remark: Even though the limit of symplectic subspaces may not be symplectic, as in the comments, this is not enough to show $$H$$ is closed, since if a sequence of elements of $$H$$ realizes this collapsing of a symplectic subspace onto a non-symplectic one, it can (and must by what I've just written) limit to a singular matrix.

• Thank you very much for your attention to my question abd your very interesting answer – Ali Taghavi May 26 at 9:54