text:
# Is $\sqrt{x^2} = x$? The many different modulus functions.

What is $\sqrt{x^2}$? Most of us will intuitive say "$x$": after all, $\sqrt{9} = \sqrt{3^2} = \sqrt{3}$, for example. However, what is $\sqrt{(-3)^2}$?

It is not $-3$ and is in fact $\sqrt{(-3)^2} = \sqrt{9} = 3$. Hence $\sqrt{x^2} =x$ is only valid if $x$ is non-negative. If $x$ is negative, it turns out that $\sqrt{x^2} = -x$.

The reason for this stems from definition: the symbol $\sqrt{ \cdot}$ is defined to be the "positive square root" when there are actually two possible square roots to every positive real number (this is the reason the equation $x^2 = k$ has two solutions, $\pm \sqrt{k}$, for positive $k$).

A compact way to summarize:

format: plain


text:
# All Questions

212 views

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### Recommendations for Statistical Mechanics book

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### Black hole analog experiment?

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### What differs string theory from philosophy or religion? [duplicate]

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### Did time exist before the creation of matter in the universe?

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### Study Quantum Physics

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### Learn algebra and interpretation of QM

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667 views

### Speed of neutrinos

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### What's the difference between the five masses: inertial mass, gravitational mass, rest mass, invariant mass and relativistic mass?

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### Invariant spacetime - distance - Circular Motion

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### What is a field, really?

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text:
# How do you find the amplitude, period and graph y=3cos(1/2theta)?

Mar 30, 2017

The amplitude is $3$, the period is $4 \pi$, and the graph is shown below.

#### Explanation:

This is a graph in the form $y = a \cos \left(b \left(\theta - h\right)\right) + k$ where $a = 3$, $b = \frac{1}{2}$, and $h = k = 0$

For a graph in this form, the amplitude is $| a |$ and the period is $\frac{2 \pi}{|} b |$. In this case, the amplitude is $| 3 | = 3$ and the period is $\frac{2 \pi}{|} \frac{1}{2} | = 4 \pi$.

To graph it, it is helpful to find certain easy-to-find points. Because $h = k = 0$, there is no horizontal or vertical shift, so we can start with a regular cosine curve.
graph{cos(x) [-10, 10, -5, 5]}

Looking at the graph, the period is $2 \pi$ and the graph is maximum at the start of the period, minimum halfway through the period, and $0$ one quarter and three quarters through. Keeping these rules in mind but expanding the period to the one we want, $4 \pi$, we get:
graph{cos(.5x) [-10, 10, -5, 5]}

Currently the amplitude of the graph is $1$, but we need it to be 3 so we expand the maximum and minimum from $1$ and $\text{-} 1$ to $3$ and $\text{-} 3$:
graph{3cos(.5x) [-10, 10, -5, 5]}

This process is equivalent to performing the appropriate transformations to the parent cosine graph: a vertical stretch by 3 and a horizontal stretch by 2.


text:
# How to approximate a function involving arc-tangent for a sufficiently large argument?

I have a function, $f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)$. Let's assume the argument $x$ is sufficiently high. If so, how can I approximate $f(x)$ in terms of $x$? I have computed $f(x)$ with MATLAB. It looks like that $f(x)$ can be approximated to ${1 \over {2x}}$. That is, ${\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x) \approx {1 \over {2x}}$ for a large $x$. It this right?

Yes. You have $$\tag1 \arctan t=t-o(t^3),$$ and $$\tag2\sqrt{1+x^2}-x=\frac1{\sqrt{1+x^2}+x}.$$Also, $$\tag3 \frac1{\sqrt{1+x^2}+x}-\frac1{2x}=\frac{x-\sqrt{1+x^2}}{2x(\sqrt{1+x^2}+x)} =-\frac1{2x(\sqrt{1+x^2}+x)^2}=o(1/x^3).$$ Combining everything, $$\arctan(\sqrt{1+x^2}-x)=\frac1{\sqrt{1+x^2}+x}+o(x^{-3})=\frac1{2x}+o(x^{-3})$$
If you want more accuracy, you could use Taylor series $$\sqrt{1+x^2}=x+\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16 x^5}+O\left(\frac{1}{x^7}\right)$$ $$\sqrt{1+x^2}-x=\frac{1}{2 x}-\frac{1}{8 x^3}+\frac{1}{16 x^5}+O\left(\frac{1}{x^7}\right)$$ Now, using $$\tan^1(\epsilon)=\epsilon -\frac{\epsilon ^3}{3}+\frac{\epsilon ^5}{5}+O\left(\epsilon ^7\right)$$ you should arrive to $$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\frac{1}{2 x}-\frac{1}{6 x^3}+\frac{1}{10 x^5}+O\left(\frac{1}{x^7}\right)$$ or more generally $$f(x) = {\tan ^{ - 1}}(\sqrt {1 + {x^2}} - x)=\sum_{i=1}^\infty \frac{(-1)^{i-1} }{(4 i-2)x^{2 i-1}}$$
Using $x=5$ (which is quite small), the truncated series given above leads to $\frac{9253}{93750}\approx 0.0986987$ while the "exact" value would be $\approx 0.0986978$.


text:
What is your favorite “data analysis” cartoon?

This is one of my favorites:

One entry per answer. (This is in the vein of the Stack Overflow question What’s your favorite “programmer” cartoon?.)

• @sharpie: are jokes out? We obviously don't want the entire site to be humor, but everyone benefits from a little educational humor in small doses. – Shane Jul 22 '10 at 5:15
• @Sharpie, feel free to close or reopen according to your feelings! I agree with Shane, a bit is ok, but not too much. For example, this question already included a funny cartoon. The jokes question not really a funny joke.... – Peter Smit Jul 22 '10 at 13:58
• These cartoons are useful too; they can be included in a lecture on a particular topic where you are trying to explain a concept (e.g. correlation/causation above). A little humor can help to keep an audience engaged. – Shane Jul 22 '10 at 14:22
• According to the tour, this question should be closed, since it is a question that has "too many possible answers" and since it is "primarily opinion-based". I'm not complaining, just surprised it has stayed open for this long. – Flimm Dec 9 '14 at 10:29
• Data Science analogy to cartoon in OP. Data Scientist: I went to data science bootcamp and learned how to find correlations in big data. Those insights can be converted into big money. Statistician: But many of those correlations are spurious. Correlation does not imply causation. Data Scientist: Don't give me none of that century old statistics mumbo-jumbo. This is big data. That means the data has everything. So by definition, all relationships in the data are correct. I ring the cash register while you snooze and lose, grandpa. – Mark L. Stone Dec 19 '15 at 22:42

Was XKCD, so time for Dilbert:

• Did anyone else notice that the tour guide changes colors between the second and third frames? – Max Aug 22 '12 at 20:41
• On RANDU: "We guarantee that each number is random individually, but we don't guarantee that more than one of them is random." – Iain Samuel McLean Elder Mar 18 '14 at 15:11
• Link not working, was it this one dilbert.com/strip/2001-10-25 ? – Tim Jan 27 '15 at 19:27

Another from XKCD:

Mentioned here and here.

• You can't read this one without the alt text. it said something like "But because of that we're totally breaking up" – generic_user Mar 10 '14 at 18:16

My favourite Dilbert cartoon:

• Definitively my favorite cartoon about Data Mining – steffen Dec 1 '10 at 12:24
• Link not working. Was it this one: dilbert.com/strip/2008-05-07 ? – Tim Jan 27 '15 at 19:27

One more Dilbert cartoon:

...

• This one reminds me of the recent bailout in the States, where they just made up 700 billion number - they said they just wanted a really large number. :) – Roman Luštrik Aug 12 '10 at 8:53
• Fixed. I had to add some dots after the cartoon since SE didn't allow me to submit the changes :-\ – Ching Chong Feb 5 '15 at 11:30

One of my favorites from xckd:

Random Number

RFC 1149.5 specifies 4 as the standard IEEE-vetted random number.

• Now this is hilarious. – StatsStudent Mar 7 '16 at 18:35
• But that isn't even prime! – kjetil b halvorsen Jan 25 '17 at 15:27

From: A visual comparison of normal and paranormal distributions Matthew Freeman J Epidemiol Community Health 2006;60:6. Lower caption says 'Paranormal Distribution' - no idea why the graphical artifact is occuring.

• I think this version of the joke works better (from oneweirdkerneltrick.com), though apparently this version was seven years earlier. – Dougal Feb 3 '15 at 1:06
• this isn't really funny. it's more of a twist on english terms – user46925 Mar 7 '16 at 18:20
• @zero "A twist on English terms" describes a great many jokes – Fomite Mar 7 '16 at 18:22
• Yeah - I think all those jokes suck. There is no underlying statistical humour. This joke should be put on the English stackexchange instead. – user46925 Mar 7 '16 at 18:23
• @Phil: In the 2-dimensional version linked by Dougal, the "paranormal distribution" is indeed a (bizarrely truncated) distribution; so the joke has some statistical content. It doesn't work in one dimension, where your comment certainly applies. – John Bentin Jun 19 '16 at 17:47

'So, uh, we did the green study again and got no link. It was probably a--' 'RESEARCH CONFLICTED ON GREEN JELLY BEAN/ACNE LINK; MORE STUDY RECOMMENDED!'

xkcd: significant

• This is by far my favorite cartoon of all time. It's super educational. It really gets to the heart of the definition of a p-value. In fact, I bet that less than 10% the students who pass a college freshman "intro to stats" class get this joke, and this makes me sad. – WetlabStudent Jan 15 '14 at 3:36
• Maybe so! Fortunately for freshmen, @Glen_b has offered an excellent breakdown here. – Nick Stauner Feb 27 '14 at 1:03
• Great! But yellow appears twice :P – Rodrigo Mar 1 '16 at 0:51
• this is a pretty good joke as it clearly demonstrates why repeated multiple testing is dangerous. For anyone interested check out Bonferi correction to deal with this. – user46925 Mar 7 '16 at 18:21

I just came across this and loved it:

• That's great. The standard way of dealing with outliers. – Shane Jul 26 '10 at 20:06
• Who's the artist? – blubb Sep 25 '14 at 16:01
• This cartoon was drawn by Ben Shabad – ff524 Jan 7 '16 at 20:20

Another from xkcd #833:

And if you labeled your axes, I could tell you exactly how MUCH better.

By the third trimester, there will be hundreds of babies inside you.

Also from XKCD

This isn't technically a cartoon, but close enough:

• That's definitely my favorite. I always have to stop on this and laugh when scrolling over this page. It's just so bad!! – Henrik Sep 9 '10 at 9:02

this too:

• To be honest, those are the bad physicists. The good physicists stick around and make a name for themselves. – Iterator Aug 6 '11 at 3:17
• Yet it's amazing how often it works... – wnoise Sep 22 '11 at 4:44
• This cartoon...speaks to me. – Fomite Sep 29 '12 at 23:38

There is this one on Bayesian learning:

• What's the source? – Shane Aug 31 '10 at 15:02
• It was taken from Mike West's website: stat.duke.edu/~mw/fineart.html – ebony1 Aug 31 '10 at 16:23

Nice. The importance of variance when thinking about a population.

Saturday Morning Breakfast Cereal

• This has to be my favorite. – Pseudo_Scientist Aug 22 '14 at 21:09
• This cartoon makes me sad. – Flimm Dec 9 '14 at 10:40

And another one from xkcd.

Title: Self-Description

The mouseover text:

The contents of any one panel are dependent on the contents of every panel including itself. The graph of panel dependencies is complete and bidirectional, and each node has a loop. The mouseover text has two hundred and forty-two characters.

Another one from xkcd:

Alt-text:

Hell, my eighth grade science class managed to conclusively reject it just based on a classroom experiment. It's pretty sad to hear about million-dollar research teams who can't even manage that.

• this one is just great :-) – Curious May 12 '12 at 10:39

Here's another one from Dilbert:

• Looks like this one needs an updated image. – Nick Orlando Feb 24 '15 at 20:45

More about design and power than analysis, but I like this one

I liked this one:

This is probably fun to show in class as well...

A classic...

• "Because medical research findings can be difficult to reconcile, are not always pre-digested, and can seem overwhelming to us casual observers, let us make fun of those who dedicate their lives to obtaining them." – rolando2 Feb 9 '14 at 0:53
• @rolando2 As a medical researcher, I find the sensationalist incompetence of mainstream science reporters hilarious. – Superbest Feb 26 '14 at 12:08
• There's a listserv from HealthNewsReview devoted to evaluating media handling of health research findings. – rolando2 Mar 9 '15 at 14:14
• As a statistician, I find the sensationalist incompetence of mainstream "data analysts" hilarious. – StatsStudent Mar 7 '16 at 21:35

I found this from a NoSQL presentation, but the cartoon can be found directly at

http://browsertoolkit.com/fault-tolerance.png

• Can you please explain this cartoon? – Christian Sep 3 '14 at 0:00
• yeah but.... this one isn't true... it mostly depends on how you parameterize the time variable $t$... i guess if you go back far enough, but come on... – William Sep 22 '11 at 17:24

Found this one in the comments on Andrew Gelman's blog.

• This chart is wrong, correct percentage is about fifty-fifty. – juliohm Mar 10 '14 at 16:09

Source: unknown. Posted on flowingdata.com.

• There could be a secondary y-axis labeled "Ballerina chicken with cloaca". – Waldir Leoncio Jan 31 at 9:38

Allright, I think this one is hilarious- but let's see if it passes the Statistical Analysis Miller test.

Fermirotica

I love how Google handles dimensional analysis. Stats are ballpark and vary wildly by time of day and whether your mom is in town.

• Statistical voyeurism? And there we were wondering what to call the site... – walkytalky Jul 23 '10 at 15:48

From xkcd:

This is data analysis in the form of a cartoon, and I find it particularly poignant.

The universe is probably littered with the one-planet graves of cultures which made the sensible economic decision that there's no good reason to go into space--each discovered, studied, and remembered by the ones who made the irrational decision.

• Very funny! (____) – Tal Galili Aug 5 '11 at 19:51

Another one from xkcd:

• Bananas are always tasty. – Anonymous Type Dec 1 '10 at 4:00
• @AnonymousType and easy! – tdc Feb 23 '12 at 13:07
• Where are persimmons? – EngrStudent May 18 '13 at 3:18
• @EngrStudent: they're simultaneously off both ends of the tasty scale. – naught101 Aug 21 '13 at 1:35
• I'm surprised Durian isn't on here... – Frank H. Oct 20 '15 at 17:41

protected by Community♦Oct 7 '16 at 1:31

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


text:
## 116b- Homework 7

February 25, 2008

Homework 7

Due Tuesday March 4 at the beginning of lecture.

## 116b- Lecture 14

February 21, 2008

We defined translations of a language in another and interpretations between theories. These notions allow us to state a more general version of the incompleteness results.

We then defined Turing machines.

## 116b- Homework 6

February 20, 2008

Homework 6

Due Tuesday February 26 at the beginning of lecture.

## 116b- Lecture 13

February 19, 2008

We proved the Kanamori-McAloon result on regressive functions as a corollary of Ramsey’s theorem, and proved, using the method of indicators, that it cannot be proven in $\mathsf{PA}$.

## 116b- Lecture 12

February 15, 2008

We sketched the method of indicators and the hierarchy of fast-growing functions as techniques to prove independence in $\mathsf{PA}$. Then we proved Ramsey’s theorem. We will use the method of indicators to show the Kanamori-McAloon theorem that the version of Ramsey’s theorem for regressive functions is not provable in $\mathsf{PA}$.

## 116b- Lecture 11

February 14, 2008

We presented the provability conditions and showed how a theory with a provability predicate satisfies the second incompleteness theorem. We showed several remarks about the incompleteness results, including Rosser’s proof of the first incompleteness theorem, and Löb’s  theorem on sentences asserting their own provability.

We closed with some examples of “mathematically meaningful” sentences independent of $\mathsf{PA}$.

## 116b- Homework 5

February 13, 2008

Homework 5

Due Tuesday February 19 at the beginning of lecture.

## 116b- Lecture 10

February 13, 2008

We showed that $\mathsf{ACA}_0$ proves the completeness theorem. We then proved the second incompleteness theorem using the result on completeness, the fact that if a model of $\mathsf{ACA}_0$ thinks that it contains a model of $\mathsf{ACA}_0$ then we can really see this latter model, and the diagonal lemma.

## Francisco Bem

February 8, 2008

Born December 27, 2007. In the picture here, he is about a month old.

## Cardinal preserving elementary embeddings

February 8, 2008

For a while now I have been thinking about structural restrictions that elementary embeddings must satisfy in set theory. Some of my findings were reported in Oaxaca during the XIII SLALM. An updated version of these results has been accepted for publication in the Proceedings of the Logic Coloquium 2007.  You can find it in my papers page.


text:
# Does this “flipping lexicographic” ordering have a standard name?

For sets $\x, \y \in \binom{\N}{m+1}$, write $\x = \{x_0 < \ldots < x_m\}$, $\y = \{y_0 < \ldots < y_m\}$.

Definition. $\x \lfl \y$ if $\x$ and $\y$ differ first in the $i$th place, and

• $i$ is even, and $x_i < y_i$; or
• $i$ is odd, and $y_i < x_i$. (This is the flip!)

As for ordinary lex, there’s also a nice inductive characterisation: Write $\x = \{x_0\} \cup \x^{\geq 1}$, and $\y = \{y_0\} \cup \y^{\geq 1}$, similarly to above. Then $\x \lfl \y$ if and only if either $x_0 < y_0$, or $x_0 = y_0$ and $\y^{\geq 1} \lfl \x^{\geq 1}$. (Again, note the flip.)

Does this ring any bells with anybody?

(Of course, $\lfl$ has obvious generalisations beyond $\binom{\N}{m+1}$; I’m sticking to that case here partly for definiteness, mainly since that’s the specific case I’m interested in.)

Background: I’ve been playing around with implementing the algorithms from Ross Street’s “The Algebra of Oriented Simplices” (and related papers) in Haskell/Agda, and this ordering turns out to make a computationally convenient stand-in for his $\lhd$ order, in places.

-
Reminds me of boustrophedon. –  Stephen S Feb 5 '11 at 9:16
+1 for 'boustrophedon order' –  ndkrempel Feb 5 '11 at 13:03
The first 4-letter word in the boustrophedonic dictionary is "waxy", since the only things that would beat it are $\{a < x < y < z\}$ and $\{a < w < x < z\}$, neither of which has any anagrams. –  Tracy Hall Feb 5 '11 at 21:04

Ah, thankyou very much — indeed, googling "alternating lexicographic" turns up quite a number of results, which (from a small sample) seem to describe the right ordering. This looks like the answer I was after. Although it now seems so pedestrian compared to ‘flipping’ and ‘boustrophedonic’… –  Peter LeFanu Lumsdaine Feb 5 '11 at 22:39


text:
# A theorem on multiplicative cell attachments with an application to Ravenel’s X(n) spectra

@article{Beardsley2017ATO,
title={A theorem on multiplicative cell attachments with an application to Ravenel’s X(n) spectra},
author={Jonathan Beardsley},
journal={Journal of Homotopy and Related Structures},
year={2017},
pages={1-14}
}
We show that the homotopy groups of a connective $$\mathbb {E}_k$$Ek-ring spectrum with an $$\mathbb {E}_k$$Ek-cell attached along a class $$\alpha$$α in degree n are isomorphic to the homotopy groups of the cofiber of the self-map associated to $$\alpha$$α through degree 2n. Using this, we prove that the $$2n-1$$2n-1st homotopy groups of Ravenel’s X(n) spectra are cyclic for all n. This further implies that, after localizing at a prime, $$X(n+1)$$X(n+1) is homotopically unique as the… Expand
2 Citations
Higher chromatic Thom spectra via unstable homotopy theory
We investigate implications of an old conjecture in unstable homotopy theory related to the Cohen-Moore-Neisendorfer theorem and a conjecture about the $\mathbf{E}_{2}$-topological HochschildExpand
Left Bousfield localization without left properness
• Mathematics
• 2020
Given a combinatorial (semi-)model category $M$ and a set of morphisms $C$, we establish the existence of a semi-model category $L_C M$ satisfying the universal property of the left BousfieldExpand

#### References

SHOWING 1-10 OF 26 REFERENCES
Relative Thom Spectra Via Operadic Kan Extensions
We show that a large number of Thom spectra, i.e. colimits of morphisms $BG\to BGL_1(\mathbb{S})$, can be obtained as iterated Thom spectra, i.e. colimits of morphisms $BG\to BGL_1(Mf)$ for some ThomExpand
Kan extensions and the calculus of modules for $∞$-categories
• Mathematics
• 2017
Various models of $(\infty,1)$-categories, including quasi-categories, complete Segal spaces, Segal categories, and naturally marked simplicial sets can be considered as the objects of anExpand
A simple universal property of Thom ring spectra
• Mathematics
• 2014
We give a simple universal property of the multiplicative structure on the Thom spectrum of an $n$-fold loop map, obtained as a special case of a characterization of the algebra structure on theExpand
An ∞‐categorical approach to R‐line bundles, R‐module Thom spectra, and twisted R‐homology
• Mathematics
• 2014
We develop a generalization of the theory of Thom spectra using the language of infinity categories. This treatment exposes the conceptual underpinnings of the Thom spectrum functor: we use a newExpand
A cellular construction of BP and other irreducible spectra
In this note we construct and derive the basic properties of the Brown-Peterson spectrum BP by attaching cells to the sphere spectrum S o (localized at a prime p) so as to kill the odd-dimensionalExpand
The Method of Infinite Descent in Stable Homotopy Theory II
• Mathematics
• 2018
This paper is a continuation of the version I of the same title, which intends to clarify and expand the results in the last chapter of `the green book' by the second author. In particular, we giveExpand
$\infty$-Operads as Analytic Monads
• Mathematics
• 2017
We develop an $\infty$-categorical version of the classical theory of polynomial and analytic functors, initial algebras, and free monads. Using this machinery, we provide a new model forExpand
CATEGORICAL HOMOTOPY THEORY
This paper is an exposition of the ideas and methods of Cisinksi, in the context of A-presheaves on a small Grothendieck site, where A is an arbitrary test category in the sense of Grothendieck. TheExpand
Commutative –algebras of prime characteristics and applications to unoriented bordism
The notion of highly structured ring spectra of prime characteristic is made precise and is studied via the versal examples S//p for prime numbers p. These can be realized as Thom spectra, andExpand
Complex Cobordism and Stable Homotopy Groups of Spheres
An introduction to the homotopy groups of spheres Setting up the Adams spectral sequence The classical Adams spectral sequence $BP$-theory and the Adams-Novikov spectral sequence The chromaticExpand


text:
Periodic solutions for a class of Lorenz-lagrangian systems
Annales de l'I.H.P. Analyse non linéaire, Volume 5 (1988) no. 3, p. 211-220
@article{AIHPC_1988__5_3_211_0,
author = {Toland, J. F.},
title = {Periodic solutions for a class of Lorenz-lagrangian systems},
journal = {Annales de l'I.H.P. Analyse non lin\'eaire},
publisher = {Gauthier-Villars},
volume = {5},
number = {3},
year = {1988},
pages = {211-220},
zbl = {0657.34042},
mrnumber = {954471},
language = {en},
url = {http://www.numdam.org/item/AIHPC_1988__5_3_211_0}
}

Toland, J. F. Periodic solutions for a class of Lorenz-lagrangian systems. Annales de l'I.H.P. Analyse non linéaire, Volume 5 (1988) no. 3, pp. 211-220. http://www.numdam.org/item/AIHPC_1988__5_3_211_0/

[1] H. Hofer and J. Toland, On the Existence of Homoclinic Heteroclinic and Periodic Solutions for a Class of Indefinite Hamiltonian Systems, Math. Annolen, Vol. 268, 1984, pp. 387-403. | MR 751737 | Zbl 0569.70017

[2] V.V. Kozlov, Calculus of Variations in the Large and Classical Mechanics, Russ. Math. Surveys, Vol. 40, 1985, pp. 37-71. | MR 786086 | Zbl 0579.70020

[3] J.F. Toland, Hamiltonian Systems Withmonotone Trajectories, Proc. Centre Math. Anal., Vol. 8, 1964, pp. 51-63. | Zbl 0622.58013

[4] J.F. Toland, An Index for Hamiltonian Systems with a Natural Order Structure. In Nonlinear Functional Analysis and its Applications, D. Riedel, 1986, pp. 147-161. | MR 852574 | Zbl 0606.58023


text:
# Tree doesn't show scroll bar when using vflex="1"

gkleinoscheg
1

I have got a simple tree component on my page similar to the following:

<tree id="treeOriginal" droppable="false" vflex="1" multiple="false" checkmark="false" >
<treecols sizable="true">
<treecol hflex="1" />
<treecol width="32px" />
<treecol width="32px" />
<treecol width="32px" />
<treecol width="44px" />
</treecols>
</tree>

The tree is later filled with content. Generally, the component is working well. However, when I'm expanding one of the nodes within the tree. A vertical scroll bar is displayed for flashlike - i.e. I can scroll with the scroll wheel of my mouse but after the flashlike display of the scroll bar I don't see it any longer.

Most interestingly, the problem exists in Chrome but not in Firefox (haven't tested Edge so far). Any ideas how to work around this problem?

delete retag edit
Be the first one to answer this question!
[hide preview]


text:
# zbMATH — the first resource for mathematics

Subgroup lattices and symmetric functions. (English) Zbl 0813.05067
Mem. Am. Math. Soc. 539, 160 p. (1994).
Let $$G$$ be an Abelian group of type $$\lambda$$, i.e. $$G\simeq \mathbb Z/(p^{\lambda_ 1}\times\cdots\times \mathbb Z/(p^{\lambda_ I})$$. The famous Hall polynomial $$g^{\lambda}_{\mu,\nu}(p)$$ counts the number of exact sequences $$0\to H\to G\to G/H\to 0$$, where $$H$$ has type $$\mu$$ and cotype $$\nu$$. It is well known that the Hall polynomial is defined over $$Z$$. The paper under review shows that a necessary and sufficient condition for the Hall polynomial $$g^{\lambda}_{\mu,\nu}(p)$$ always to have nonnegative coefficients is that no two parts of $$\lambda$$ differ by more than one, i.e. $$\lambda= i^ a(i+ 1)^ b$$, $$i\in \mathbb N$$. The author first uses a generalization of Knuth’s study of subgroup lattices to obtain some combinatorial formula for the Hall polynomial $$g^{\lambda}_ {\mu,\nu}(p)$$ in this case. She then employs the theory of Hall-Littlewood polynomials (cf. I. G. Macdonald [Symmetric functions and Hall polynomials. Oxford: Clarendon Press (1979; Zbl 0487.20007)]) to attack the necessary condition, where she also studies and examines the Lascoux-Schützenberger proof of the nonnegativity for the $$p$$-Kostka polynomial. Finally the author gives a conjecture on the relation between two $$(q,t)$$-polynomials $$K_{\lambda,\mu}$$ and $$K_{\lambda,\nu}$$, which are defined by the Macdonald polynomial as in the case of the Hall-Littlewood polynomial.

##### MSC:
 05E05 Symmetric functions and generalizations 20D30 Series and lattices of subgroups 20K01 Finite abelian groups 06A11 Algebraic aspects of posets 11B65 Binomial coefficients; factorials; $$q$$-identities
Full Text:


text:
# Prevent a page break across two different environment

I need to tie two different environments, i.e., prevent a page break from happening. I tried \nopagebreak, \nobreak, \begin{samepage}-\end{samepage}, but nothing worked.

\begin{enumerate}
\item ....
\item týden
Pondělí:
\begin{enumerate}
\begin{multicols}{2}
\item $\frac{\frac{1}{a-1}+\frac{2}{a-1}+\frac{3}{a-1}}{\frac{1}{a^{2}-1}}$
\vysl{$6(a+1)$ PODM: $a\neq\pm1$}
\item $(y^{2}-2)^{3}$
\vysl{$y^{6}-6y^{4}+12y^{2}-8$}
\end{multicols}
\item $A-B=?;A=\{x\in\R;-3\leq x\leq 10\};B=\{x\in\R;-4<x<8\}$
\vysl{$A-B=\{x\in\R;8 \leq x\leq 10\}$}
\end{enumerate}

Page breaks somewhere between word "Pondělí" and environment multicols, but I need it together.

Thanks for all answers and excuse my English, I hope everything is understandable.

• Welcome to TeX.SE. Have you looked into the needspace package? – Mico Sep 28 '15 at 19:42

You should try the needspace package. Since it looks like 5 to 6 lines are occupied by the material that needs to be kept together, you might try:

\documentclass{article}
\usepackage{multicol,needspace}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
%% how is the "\vysl" macro defined?

\begin{document}

\begin{enumerate}
\item \dots
\item týden
\needspace{6\baselineskip}
Pondělí:
\begin{enumerate}
\begin{multicols}{2}
\item $\frac{\frac{1}{a-1}+\frac{2}{a-1}+\frac{3}{a-1}}{\frac{1}{a^{2}-1}}$
\vysl{$6(a+1)$ PODM: $a\neq\pm1$}
\item $(y^{2}-2)^{3}$
\vysl{$y^{6}-6y^{4}+12y^{2}-8$}
\end{multicols}
\item $A-B=?;A=\{x\in\R;-3\leq x\leq 10\};B=\{x\in\R;-4<x<8\}$
\vysl{$A-B=\{x\in\R;8 \leq x\leq 10\}$}
\end{enumerate}

\end{document}

(I'm afraid I can't post a screenshot as I don't know how the \vsyl macro is defined.)


text:
7

An important problem of teaching methods of solving math problems is to find ways to improve the learning process, activation of cognitive activity of students. The solution to this Mymathlab solver involves strengthening the ideological aspect of training, improvement of methods of realization of applied and practical teaching of the math problem solver. Among the possibilities of solving the problem is seen in the practical work of teachers teaching math through the tasks.

Here we have some loyal allies to the study of numbers and you can solve math problems, but not only, and deal as well with more serenity a day at school, university or work.

If you were a student of the medium or higher today probably wouldn’t have much fear because there are apps that show you the algebra, geometry, and physics in a truly complete.

## MyMathLab: Student Access Kit

MyMathLab is a series of text-specific, easily customizable online courses for Pearson Addison-Wesley and Pearson Prentice Hall textbooks in Mathematics. MyMathLab gives you the tools you need to deliver all or a portion of your course online, whether your students are in a lab setting or working from home.

MyMathLab provides a rich and flexible set of course materials, featuring free-response exercises correlated directly to the textbook that instructors can assign for online homework, quizzes, and tests. These exercises regenerate algorithmically for unlimited practice and mastery, and in homework and practice modes, each exercise is accompanied by an interactive guided solution and a sample problem.

MyMathLab provides students additional with multimedia resources, such as video lectures, animations, and an eBook, to independently improve their understanding and performance. MyMathLab’s online grade book automatically tracks all student results and gives the instructor control over how to calculate final grades. Because MyMathLab courses are delivered via the CourseCompass online learning platform, instructors also have the ability to customize their course using a variety of course-management tools

## Math Problem Solver with steps

### Math problem solver

1. What is a MATH PROBLEM SOLVER?First, talk to your child about whether she understood what is a math problem, how it is built and what it’s for.– Questions that help to understand the text of the problem.– Questions that help you find (or calculate) the information you are missing.– The final question is also the solution to the problem.

The first thing to do is and then help your child distinguish the different questions and their contents.

2. UNDERSTAND the MATH PROBLEMThe second stage is to ensure that your child has understood what he read. Make sure the child understands all the problem. Try to identify words that do not belong to the usual vocabulary of your child and ask them to explain the order of solving.In addition to words, the child may not understand the whole problem, simply because it fails to “see” and to represent the situation. Try to do it with him through words or with the help of a drawing or outline.

3. UNDERSTAND the sense of REQUESTThe situation is understood. The child must now understand what is asked of him. In other words, it must identify:The final question. The questions that will help you to answer the problem result.

4. FIND the order of CLUESExplain to the child that all the information you need to find the solution is contained in the statement but that is often hidden or encoded. We must, therefore, become a detective to identify all resources used to be able to answer questions. If the child has understood the ultimate question, ask him to find in the text of the proposition: All response elements.

### Math online problem solver

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### Math problem solver with steps

The tasks used to prepare students for the study of the mathematical theory should be selected so that their staging has led not only to the need to acquire new knowledge and skills but also to the use of acquired under the influence of this necessary knowledge to math solver, address along with this and a wide range of other tasks.

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Often useful to preface the study of new math problem solver fact decision known disciples and tasks, using the analogy to justify the preservation method of its solutions for new challenges, that is not possible due to lack of mathematical knowledge.

## How to Solve It

Although we had promised not to give you any mathematical “good advice” in abstract terms, the book by Georg Polya, How to Solve It (Princeton UniversityPress, 1973), contains a procedural approach to problem solving, which can be useful sometimes. The following summary is lifted verbatim from the book preface:

### UNDERSTANDING THE PROBLEM

First. You have to understand the problem. What is the unknown? What are the data? What is the condition? Is it possible to satisfy the condition?Is the condition suﬃcient to determine the unknown? Or is it insuﬃcient? Or redundant? Or contradictory? Draw a ﬁgure. Introduce suitable notation. Separate the various parts of the condition. Can you write them down?

### DEVISING A PLAN

Second. Find the connection between the data and the unknown. You may be obliged to consider auxiliary problems if an immediate connection can not be found. You should obtain eventually a plan of the solution. Have you seen it before? Or have you seen the same problem in a slightly diﬀerent form? Do you know a related problem? Do you know a theorem that could be useful? Look at the unknown! And try to think of a familiar problem having the same or a similar unknown. Here is a problem related to yours and solved before. Could you use it? Could you use its result? Could you use its method? Should you introduce some auxiliary elements in order to make its use possible? Could you restate the problem? Could you restate it still diﬀerently? Go back to deﬁnitions. If you cannot solve the proposed problem try to solve ﬁrst some related problems. Could you imagine a more accessible related problem? A more general problem? A more special problem? An analogous problem? Could you solve a part of the problem? Keep only a part of the condition, drop the other part; how far is the unknown then determined, how can it vary? Could you derive something useful from the data? Could you think of other data appropriate to determine the unknown? Could you change the unknown or data, or both if necessary, so that the new unknown and the new data are nearer to each other? Did you use all the data? Did you use the whole condition? Have you taken into account all the essential notions involved in the problem?

### CARRYING OUT THE PLAN

Third. Carry out your plan. Carrying out your plan of the solution, check each step. Can you see clearly that the step is correct? Can you prove that it is correct?

### LOOKING BACK

Fourth. Examine the solution obtained. Can you check the result? Can you check the argument? Can you derive the solution diﬀerently? Can you see it at a glance? Can you use the result, or the method, for some other problem?

• Sam says:

6/2(1+2)= 9 🙁

• Ami Gayu says:

6 ÷ 2(2 + 1)
=> 6 ÷ 2 * 3
since only multiplication and division remain, we work left to right,
=> 3 * 3 = 9
(Microsoft excel and most other calculators will give this answer)

• engvarta says:

Very useful details provided by you. I hope it would be useful for many of the seekers. Thank you for sharing details here with us.

• andrey says:

• Mike says:


text:
# Howto Calculate An Error's Partial Derivative in ANN

This is a follow-on question from this post I made, "Multilayer Perceptron (Neural Network) for Time Series Prediction", a few months back.

I'm constructing a feed-forward artificial neural network, using resilient propagation training. At the moment, I'm trying to implement an individual neuron input's weight update algorithm. For the life of me, I can't seem to find a clear and straightforward answer on how to calculate the partial derivative of the error for a given weight. The only thing I can find on the web, is the fact that a neuron's weight update is a function of $\frac{dE}{dW}$ (cf. the original Paper [p. 2 & 3], or this one [p. 4]).

However none of these papers actually outlines how to calculate this.

I understand the concept of a partial derivative in a mathematical sense. And I assume that the current neuron input's weight change calculation is the operation at hand, while all other neuron input values are held constant.

So for each of these neurons below, I calculate each inputs' individual error by taking a total error ( -0.3963277746392987 ), that's been multiplied by that neuron input's weight (each :calculated-error is the sum of the individual inputs' error).

For both neurons, what would be the weight change for each input?

Here is my data structure:

:input-layer
({:calculated-error -1.0991814559154283,
:calculated-value 0.9908633780805893,
:inputs
({:error -0.07709937922001887,
:calculated 0.4377023624017325,
:key :avolume,
:value 2.25,
:weight 0.19453438328965889,
:bias 0}
{:error -0.19625185888745333,
:calculated 1.4855269156904067,
:key :bvolume,
:value 3.0,
:weight 0.4951756385634689,
:bias 0}
{:error -0.3072203938672436,
:calculated 1.0261589301119642,
:value 1.32379,
:weight 0.7751674586693994,
:bias 0}
{:error -0.36920086975057054,
:calculated 1.2332848282147972,
:key :bid,
:value 1.3239,
:weight 0.9315543683169403,
:bias 0}
{:error -0.14940895419014188,
:calculated 0.5036129016361643,
:key :time,
:value 1.335902400676,
:weight 0.37698330460468044,
:bias 0}),
:id "583c10bfdbd326ba525bda5d13a0a894b947ffc"},
...)

:output-layer
({:calculated-error -1.1139741279964241,
:calculated-value 0.9275622253607013,
:inputs
({:error -0.2016795955938916,
:calculated 0.48962608882549025,
:input-id "583c10bfdbd326ba525bda5d13a0a894b947ffb",
:weight 0.5088707087900713,
:bias 0}
{:error -0.15359996014735702,
:calculated 0.3095962076691644,
:input-id "583c10bfdbd326ba525bda5d13a0a894b947ffa",
:weight 0.38755790024342773,
:bias 0}
{:error -0.11659507401745359

:calculated 0.23938733624830652,
:input-id "583c10bfdbd326ba525bda5d13a0a894b947ff9",
:weight 0.2941885012312543,
:bias 0}
{:error -0.2784739949663631,
:calculated 0.6681581686752845,
:input-id "583c10bfdbd326ba525bda5d13a0a894b947ff8",
:weight 0.7026355778870271,
:bias 0}
{:error -0.36362550327135884,
:calculated 0.8430641676611533,
:input-id "583c10bfdbd326ba525bda5d13a0a894b947ff7",
:weight 0.9174868039523537,
:bias 0}),
:id "583c10bfdbd326ba525bda5d13a0a894b947ff6"})

-

It's a bit unclear to me what you're trying to do, and maybe a better place to ask your question is Stats.SE but I would encourage you to go and have a look at this online class on machine learning which provides an implementation of the backpropagation algorithm.

You can either register or hit preview and go to the NN:Learning chapter, but I would recommend you to register and do the programming exercise on NN. Maybe this is not exactly what you are trying to do, but I think you might find a basic example of how to compute the partial derivative of the error term.

-
Thanks for this. Andrew Ng is actually the guy whose videos I first started following, when I was getting into AI. But this course seems to emphasize using NNs for classification problems (discreet values). I want to compute a continuous value in a time series. But I will dig further into this series. Thanks again. –  Nutritioustim Oct 5 '12 at 3:25
There is another class specifically for NN if you look on the site. –  SRKX Oct 5 '12 at 8:50

The calculation of the gradient depends on the activation functions of your neurons (and, more specifically, their derivatives). Moreover, the gradient of any individual weight depends on weights and activations appearing later in the network -- that's the key idea behind the backpropagation algorithm, which takes advantage of that fact to compute gradients. You can find a very detailed description of backprop in this paper. Note that the examples use sigmoid activation functions, but the method is applicable to any differentiable activation.

Note also that the principal difference between backprop and rprop is that backprop requires the value of the weight gradient and rprop only requires its sign. Though rprop may seem simpler in that regard, I advise you to make sure you completely understand backprop first.

-
I conceptually understand what you mean by "the gradient of any individual weight depends on weights and activations appearing later in the network". But I don't understand how that bears out when calculating weight changes. I've been using this paper as a guide to backprop. Anyways, I'll definitely go through the paper you provided. Thanks. –  Nutritioustim Oct 5 '12 at 3:33
Ahh, I'm getting what you mean by that statement. Starting to sink in slowly :) –  Nutritioustim Oct 5 '12 at 3:59
Hi @Nutritioustim, sorry if I wasn't clear -- an example will help. Say you're building a 2-layer neural network. Now suppose all the weights in your second layer are 0. Then the gradient for all the weights in your first layer must be 0, because no matter what you set them out, their output is going to be multiplied by 0 in the second layer, eliminating their impact. That's why backprop is such an important algorithm -- it takes the error back through each layer. Note that from a mathematic perspective, it's basically just a chain rule application. –  jlowin Oct 6 '12 at 12:10
In your reference, note how the "error signal" at each neuron is the weighted sum of the error signals in the following layer. That error signal gets multiplied by the neuron's derivative and its input to find the weight update -- therefore the total gradient (i.e. the weight update) for that neuron is dependent on results in later layers. –  jlowin Oct 6 '12 at 12:14
Yeah, I'm starting to get the jist of from the paper you provided - it's a doozey. I'm on p.16, "Steps of the algorithm". I'm sure I'll have 1 or 2 more Qs. It really comes down to the implementation details. As I'm trying to use this to build a neural network (see here) –  Nutritioustim Oct 8 '12 at 2:22


text:
# GATE2013-48

A reverse osmosis unit treats feed water (F) containing fluoride and its output consists of a permeate stream (P) and a reject stream (R). Let C${_F}$, C${_p}$, and C${_R}$ denote the fluoride concentrations in the feed, permeate, and reject streams, respectively. Under steady state conditions, the volumetric flow rate of the reject is 60% of the volumetric flow rate of the inlet stream, and C${_F}$ = 2 mg/L and C${_P}$ = 0.1mg/L.
The value of C${_R}$ in mg/L, up to one digit after the decimal point, is $\_\_\_\_\_\_\_$


text:
# I Would Like to Add a Line to a Table Plot with Tikz

I am plotting data with pgfplots. I would like to add a line to this plot. My code is as follows

\documentclass{article}
\usepackage{pgfplots}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\begin{axis}[]
0   90
1   72
2   57
24  -43
25  -26
26  0
27  0
};
\end{axis}
\end{tikzpicture}
\end{document}

I tried to add a line with its plot coordinates, to no avail

\documentclass{article}
\usepackage{pgfplots}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\begin{axis}[]
0   90
1   72
2   57
24  -43
25  -26
26  0
27  0
};
\end{axis}

\draw [dashed] (24, -50) -- (24,60);
\end{tikzpicture}
\end{document}

How can I add this line ?

just add this line with \addplot:

\documentclass{article}
\usepackage{pgfplots}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\begin{axis}
0   90
1   72
2   57
24  -43
25  -26
26  0
27  0
};
\addplot[dashed] coordinates {(24, -50) (24,60)};
\end{axis}
\end{tikzpicture}
\end{document}

The reason is pgfplots having different (implicitly and automatically) scaling. You may draw with plain TikZ to get explicit scaling.

\documentclass[tikz,border=5mm]{standalone}
\begin{document}
\begin{tikzpicture}[x=2mm,y=.4mm]
\def\datapoints{
(0,90) (1,72) (2,57) (24,-43) (25,-26) (26,0) (27,0)}

\draw[densely dotted] (24,-50)--(24,60);

\draw[blue,mark=ball] plot coordinates{\datapoints};

\foreach \i in {0,5,...,25}
\draw[gray] (\i,-55) node[below,black]{$\i$}--+(90:1.5mm);
\foreach \j in {-50,0,50,100}
\draw[gray] (-3,\j) node[left,black]{$\j$}--+(0:1.5mm);
\draw (-3,-55) rectangle (30,105);
\end{tikzpicture}
\end{document}

There are two options. Either you use a reasonably new compatibility mode (1.11 or higher) or preprend the coordinates with axis cs:. I illustrate the first option. Compared to using \addplot his has the advantage that you can use all the usual constructions like, say, bend left, you do not have problems with legends and so on.

\documentclass{article}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}

\begin{document}
\begin{tikzpicture}
\begin{axis}[]


text:
# Question about the soundness of using a pairing map in the Kate Polynomial Commitment Scheme

I am looking at the paper on Kate Polynomial Commitments.

On Page 7, VerifyEval, the verifier checks the following to verify commitment.

$$e(\mathcal C, g) \stackrel {?}{=} e(w_i, \frac {g(\alpha)}{g(i)})e(g, g)^{\phi(i)}$$

In the next line, the paper explains why this equality will be true if the commitment is in fact honest.

I understand the completeness part of the proof, but I am not convinced about the soundness part.

Let the pairing map be

$$e: G\space \times \space G \mapsto G_T$$

Let $$h_1, h_2, h_3, h_4$$ be elements of $$G$$

Let $$r_1$$ & $$r_2$$ be elements of $$G_T$$.

$$e(h_1, h_2) = r_1$$

$$e(h_3, h_4) = r_2$$

If $$h_1 = h_3$$ & $$h_2 = h_4$$, then

$$r_1 = r_2$$

What we are checking in Commitment Scheme's VerifyEval is if $$r_1 \stackrel {?}{=} r_2$$

I agree this proves the completeness of the commitment.

However, I am not sure if it's proves the soundness.

Can't 2 different sets of elements when used as input to the pairing map end up with the same output element?

In the case where

$$h_1 \ne h_3$$ & $$h_2 \ne h_4$$, can't the output of the 2 mappings still be the same?

i.e. can't this be true still?

$$e(h_1, h_2) = e(h_3,h_4)$$

What property of elliptic curve pairings proves that the probability of $$e(h_1, h_2)$$ being equal to $$e(h_3,h_4)$$ is negligible in case $$h_1 \ne h_3$$ & $$h_2 \ne h_4$$?

## 1 Answer

The chance of $$e(h_1,h_2)$$ equalling $$e(h_3,h_4)$$ for $$h_1$$, $$h_2$$, $$h_3$$ and $$h_4$$ selected independently and uniformly at random from $$\mathbb G$$ is $$1/p$$ where $$p$$ is the prime order of the groups in the pairing. This is because the output is uniform.

More germanely, if an adversary could reliably construct $$\hat w$$ and $$\hat\phi$$ such that $$e\left(\hat w,\frac{g^\alpha}{g^{i}}\right)e(g,g)^{\hat\phi}=e\left( w_i,\frac{g^\alpha}{g^{ i}}\right)e(g,g)^{\phi(i)}$$ then they would be able to convert this conctruction into a means to solve the Strong Diffie-Hellman (SDH) problem for the group $$\mathbb G$$. Solving this problem is believed to be hard in the group used in pairing-based cryptography. A demonstration of this is given in the appendix section C.1 in the Evaluation Binding subsection on pages 19 and 20.

• The notations used in your reply & in the document itself is a little confusing. Am I right in assuming that there are now two polynomials $\phi(x)$ & $\hat \phi(x)$ & both are evaluated at $\alpha$ & $i$ or is there also a different sampling value $\hat i$? Because your denominator of the left hand side, you also use $g$ raised to $\hat i$. In the Kate document, I don't see this. Jan 18 at 6:55
• @user93353 Apologies, the adversary would not have control of the $i$ value and so it should be the same on both sides. Their goal would be to create fake values for $\phi(i)$ and $w_i$ which I have labelled $\hat\phi$ and $\hat w$. I've edited out the $\hat i$. Jan 18 at 7:39
• Thank you for the quick response. Jan 18 at 7:41


text:
Concentration inequality for sum of iid random variables that involve KL distance

Conider $X \in \mathbb{R}^d$ and $Y \in \{0,1\}$, and a joint distribution $p_{XY}(x,y)$, and a set of $N$ i.i.d. samples $\{(X_i,Y_i)\}_{i=1}^{N}$. Define $p_{X0} = p_{XY}(x,0)$ and $p_{X1} = p_{XY}(x,1)$. Define deterministic function $f: \mathbb{R}^d \mapsto \{0,1\}$. Define binary variable $Z_i = \mathbb{1}_{\{Y_i \neq f(X_i)\}}$, where $\mathbb{1}_{\{\cdot\}}$ is an indicator function. Consider metric
$$\Pr \left(\frac{1}{N}\sum_{i=1}^{N} Z_i \geq \mathbb{E} \left[Z\right] + \varepsilon \right) \:.$$ Concentration inequalities bounds it based on only $N$ and $\varepsilon$. Is there any upper bound that involves the KL distance or total variation of $p_{X0}$ and $p_{X1}$?

• Suppose $p_{X0}=p_{X1}$. Presumably, this is the "best" case, in which both the total variation and the KL are zero. What sort of an improved bound would you expect in this case? – Aryeh Kontorovich Nov 28 '17 at 17:06
• In that very special case (not sure if we can call it "best"), perhaps nothing. I am actually interested in $p_{X0} \neq p_{X1}$ or even $p_{X0} \gg p_{X1}$, in which the bound may be improved. – Jeff Nov 28 '17 at 17:22
• OK, let's consider another extreme: $p_{X0}=0$ everywhere. What kind of an improvement would you expect then? – Aryeh Kontorovich Nov 28 '17 at 17:26
• Let's consider this toy example. $p_{X0} \sim \mathcal{N}(0,\Sigma)$ and $p_{X1} \sim \mathcal{N}(\mu,\Sigma)$ be multivariate normal distributed for some constant vector $\mu$ and covariance matrix $\Sigma$. Now, I would expect to concentrate faster when we increase $\| \mu \|_2^2$, like when we increase "distance" of two distributions. – Jeff Nov 28 '17 at 17:49
• Indeed, there are results that show that classification becomes easier when the two classes are well-separated. But the inequality in your question is just a concentration result for a Bernoulli random variable, which is indifferent to the separation properties. – Aryeh Kontorovich Nov 28 '17 at 17:52

Consider the sub-probability measures $\mu_0$ and $\mu_1$ defined by the conditions $\mu_0(A):=P(Y=0,X\in A)$ and $\mu_1(A):=P(Y=1,X\in A)$ for Borel sets $A\subseteq\mathbb R$, so that $\mu:=\mu_0+\mu_1$ is the probability distribution of $X$. Let \begin{equation*} p:=p_f:=P(Y\ne f(X))=EZ. \end{equation*} Then, by Chebyshev's inequality, \begin{equation*} P\Big(\frac1n\sum_{i=1}^n Z_i \ge EZ + \varepsilon\Big)\le\frac{p(1-p)}{n\varepsilon^2} =\frac1{n\varepsilon^2}\,[\tfrac14-(p-\tfrac12)^2]. \end{equation*} So, it is enough to give a lower bound on $|p-\tfrac12|$ in terms of the total variation norm $\|\mu_0-\mu_1\|$.
First here, it is easy to see that, without any restrictions on the function $f$ (and assuming such a mild condition as the probability measure $\mu$ being non-atomic), the only lower bound on $|p-\tfrac12|$ is the trivial one, $0$. Indeed, let $$C_0:=C_{0,f}:=\{x\in\mathbb{R}^d\colon f(x)=0\},\quad C_1:=C_{1,f}:=\{x\in\mathbb{R}^d\colon f(x)=1\}. \tag{0}$$ Then $p=\mu_0(C_1)+\mu_1(C_0)$ and $\mu_0(C_0)+\mu_0(C_1)+\mu_1(C_0)+\mu_1(C_1)=1$, whence $$2(p-\tfrac12)=\mu_0(C_1)-\mu_1(C_1)+\mu_1(C_0)-\mu_0(C_0). \tag{1}$$ Letting $f(x)=1$ for all $x$, we have $C_0=\emptyset$, $C_1=\mathbb{R}^d$, and hence $2(p-\tfrac12)=\mu_0(\mathbb{R}^d)-\mu_1(\mathbb{R}^d)=:\delta$. Vice versa, letting $f(x)=0$ for all $x$, we have $C_1=\emptyset$, $C_0=\mathbb{R}^d$, and hence $2(p-\tfrac12)=\mu_1(\mathbb{R}^d)-\mu_0(\mathbb{R}^d)=-\delta$. So, if $\mu$ is indeed non-atomic, then for some (bad enough classification rule) $f\colon \mathbb{R}^d \mapsto \{0,1\}$ we will have $2(p-\tfrac12)=\frac12\,\delta+\frac12\,(-\delta)=0$, so that $|p-\tfrac12|=0$.
Now suppose that the classification rule $f\colon \mathbb{R}^d \mapsto \{0,1\}$ is chosen optimally, as follows: \begin{equation*} f(x)=f_*(x):=I\{g_1(x)>g_0(x)\} \end{equation*} for all $x\in \mathbb{R}^d$, where $I$ is the indicator function and for each $i=1,2$ the function $g_i$ is the density of $\mu_i$ with respect to (say) the measure $\mu=\mu_0+\mu_1$. Then, by (1) and (0), \begin{equation*} 2(p_{f_*}-\tfrac12)=-|\mu_0(C_1)-\mu_1(C_1)|-|\mu_1(C_0)-\mu_0(C_0)|=-\|\mu_0-\mu_1\|, \end{equation*} so that \begin{equation*} 2|p_{f_*}-\tfrac12|=\|\mu_0-\mu_1\|. \end{equation*} (On the other hand, again by (1), \begin{equation*} 2|p_f-\tfrac12|\le|\mu_0(C_1)-\mu_1(C_1)|+|\mu_1(C_0)-\mu_0(C_0)|\le\|\mu_0-\mu_1\| \end{equation*} for all $f$, which shows that the choice $f=f_*$ is indeed optimal in the sense that it maximizes $|p_f-\tfrac12|$; moreover and more importantly, this choice of $f$ is also optimal in the sense that it minimizes the misclassification probability $p_f$.) Thus, for $f=f_*$ we have indeed an upper bound in terms of the total variation norm $\|\mu_0-\mu_1\|$:
\begin{equation*} P\Big(\frac1n\sum_{i=1}^n Z_i \ge EZ + \varepsilon\Big)\le\frac{1-\|\mu_0-\mu_1\|^2}{4n\varepsilon^2}. \end{equation*} This bound can be vastly improved, again for $f=f_*$. E.g., bound (2.2) in Hoeffding 1963 yields
\begin{equation*} P\Big(\frac1n\sum_{i=1}^n Z_i \ge EZ + \varepsilon\Big)\le \exp\{-n\varepsilon^2 h(\|\mu_0-\mu_1\|)\}, \end{equation*} where $h(u):=\frac1u\,\ln\frac{1+u}{1-u}$, which is increasing from $2$ to $\infty$ in $u\in(0,1)$.
In order to exploit the structure in the marginal distribution over $X$ for better generalization, you need a "smart" learning algorithm that knows something about this structure. In your example of two Gaussians, the learner would estimate the means of $p_{X0}$ and $p_{X1}$, and the more well-separated these are, the better of a generalization guarantee one can give. The random variable $Z_i$ is just a Bernoulli random variable, however, and its empirical mean exploits nothing about the structure of the marginal distribution of $X$, and hence the latter cannot be used to obtain sharper concentration bounds.


text:
Latino Elongated, XeTeX and small capitals

I'm using the Latino Elongated LET font with XeTeX and I can't get small capitals to display:

\documentclass{scrbook}

\usepackage{fontspec}
\newfontfamily\lelong{Latino Elongated LET}
\newfontfamily\lelongsc[SmallCapsFeatures={Letters=SmallCaps}]{Latino Elongated LET}

\begin{document}

\lelong \scshape Small capitals

\lelongsc \scshape Small capitals

\newfontinstance\scshape[Letters=SmallCaps]{Latino Elongated LET}

\lelong \scshape Small capitals

\end{document}

gives me three lines with normal characters:

How can I get small capitals? It is possible that this font is lacking sc functionalities?

Note:

The third attempt (\newfontinstance) gives a warning in the logs:

Package fontspec Warning: OpenType feature 'Letters=SmallCaps' (+smcp) not available for font "Latino Elongated LET Plain:1.0/ICU" at 10.95pt, with script 'Latin', and language 'Default'.

-
I get the message that small caps are not present in the font. You can't have what is not in the font. –  egreg Jun 8 '11 at 15:15
have you checked if the font really has the small caps feature with a otfinfo -f /path/to/font? –  henrique Jun 8 '11 at 15:15
@egreg: OK, so maybe I need another Latino Elongated font? –  ℝaphink Jun 8 '11 at 15:18
@henrique: the file is a ttf, so I can't run that. Is there an equivalent for true type/T1 fonts? –  ℝaphink Jun 8 '11 at 15:19
@Raphink a few truetype fonts do have opentype features that can be displayed with otfinfo... If it doesn't, the font simply lacks the functionality and you can't have smallcaps with it. (Some commercial fonts come in bundles with a separated file for smallcaps... if that is the case, you'll need to create a \newfontface to use it.) –  henrique Jun 8 '11 at 15:30

3 Answers

The font is a TrueType font that doesn't have "small caps" functionality. That is, the font just doesn't have any small caps.

-
Are you sure it doesn't contain small caps in a private Unicode area? –  Andrey Vihrov Jun 8 '11 at 16:14
@Andrey: that was actually my question, I'd love to know how to know that. –  ℝaphink Jun 8 '11 at 20:01
@Raphink Open the font with fontforge or another utility (I have OTM_Light) and look at the glyphs. I can confirm that there is nothing in the PUA. –  egreg Jun 8 '11 at 20:12
OK, thanks. I don't have fontforge. I thought fontconfig had a tool for that, maybe fc-cat can tell me this? –  ℝaphink Jun 8 '11 at 20:17
@Raphink fc-cat only lists fonts known to the font library of fontconfig. Look at fontforge.sourceforge.net –  egreg Jun 8 '11 at 20:22

You can cheat:

\newcommand\fakesc[1]{\scalebox{0.7}{\MakeUppercase{#1}}}

Some things to note:

• This requires the graphicx package
• Uppercase letters in the fakesc scope get lost. (Although I'm sure some TeX-fu can fix that...
• The actual amount of scaling should probably depend on the exact font used...
-
I've actually cheated already in another way. Eventually, I'll see about purchasing the OTF version of the font which I believe has proper sc. –  ℝaphink Jun 8 '11 at 20:02

I found a cleaner solution:

• open the TTF in fontforge;
• create small caps;
• save the font as OTF;
• use the new font;
• remember to not distribute non-free modified fonts on github ;-)
-
So long as the font licence allows you to do this. –  egreg Jul 20 '11 at 9:19
The font doesn't allow me to redistribute, but can it prevent me from making private changes? Isn't that the same as stretching it in inkscape for example? –  ℝaphink Jul 20 '11 at 9:30


text:
# physics

• October 18th 2007, 02:03 PM
jenjen
physics
Hi,

I am trying to help my brother do this problem but I can't figure it out! can anyone please help me...

A 3.45 kg object is moving in a plane, with its x and y coordinates given by x = 7t² - 3 and y = 4t³ + 3, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.35 s.
• October 18th 2007, 02:31 PM
topsquark
Quote:

Originally Posted by jenjen
Hi,

I am trying to help my brother do this problem but I can't figure it out! can anyone please help me...

A 3.45 kg object is moving in a plane, with its x and y coordinates given by x = 7t² - 3 and y = 4t³ + 3, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.35 s.

$\sum \bold{F} = m \bold{a} = m \frac{d^2\bold{x}}{dt^2}$

So the x component of the force would be:
$F_x = m(14)$

and the y component would be...?

-Dan


text:
Inventor General Discussion

## Inventor General Discussion

Valued Contributor
Posts: 71
Registered: 06-21-2013
Message 1 of 5 (550 Views)

# 2D equation curve for a parabolic spiral

550 Views, 4 Replies
06-22-2013 05:19 PM

Can anyone please advise me what the 2D or 3D equation curve for a parabolic spiral, otherwise known as a Fermat's spiral would be?

I have found this info but do not know how to apply it:

Fermat equation

Therefore in terms of the affine plane its equation is

The Fermat curve is non-singular and has genus

I have also found this: $r^{2}=a^{2}\theta$

*Expert Elite*
Posts: 28,046
Registered: 04-20-2006
Message 2 of 5 (519 Views)

# Re: 2D equation curve for a parabolic spiral

06-23-2013 07:00 AM in reply to: CJ.10E-10mm

Are you familiar with using Equation Curves in general?

A while back I posted some examples that my students submitted.

I think Autodesk employee has also posted some examples.

At one time there were several that were included in the Samples install, there should be a thread here with those attached.

Do a search and come back with more questions if you can't find the examples to help you work it out.

-----------------------------------------------------------------------------------------
Autodesk Inventor 2014 Certified Professional
Certified SolidWorks Professional
Inventor Professional 2015-SP1 64-bit
http://www.autodesk.com/edcommunity
http://home.pct.edu/~jmather/content/DSG322/inventor_surface_tutorials.htm
Valued Contributor
Posts: 71
Registered: 06-21-2013
Message 3 of 5 (505 Views)

# Re: 2D equation curve for a parabolic spiral

06-23-2013 09:17 AM in reply to: CJ.10E-10mm

Thank you, I will search later today.

Equation Curves do appear to a novice like some form of 'black art.' So I am wondering if there is a 'user friendly; App that allows one to configure the equation for any given shape in an Excel Spreadsheet?

Thank you once again.

Distinguished Mentor
Posts: 909
Registered: 04-11-2005
Message 4 of 5 (496 Views)

# Re: 2D equation curve for a parabolic spiral

06-23-2013 11:09 AM in reply to: CJ.10E-10mm
If you can create x,y,z coordinates in a spreadsheet, IV will import them and draw a spline through them.
Valued Contributor
Posts: 71
Registered: 06-21-2013
Message 5 of 5 (478 Views)

# Re: 2D equation curve for a parabolic spiral

06-23-2013 03:11 PM in reply to: rdyson

Thank you Rdyson.

Is there any chance you could assist me with my question on:Variable pitch helical coil with smooth blend please?

Thank you once again.

Post to the Community


text:
# Math Help - M1 need help in a complicated problem

1. ## M1 need help in a complicated problem

Points A,B and C on the wire have coordinates (0,0) , (3,3) , (6,6). A m-kg bead is projected along the wire from A with speed u ms-1 so that it has enough energy to reach B but not C. Prove that 8.16<u<10.95 and that the speed at B >/= 2.58 ms-1.

If u = 10, the bead comes to rest at a point D between B and C. Find
(a) The greatest speed of the bead between B and D.
(b) The coordinates of D.
What happens after the bead reaches D?

The answer for (a) is 6.83ms-1 and for (b) is (5.7 , 5)

2. ## Re: M1 need help in a complicated problem

Originally Posted by abdulrehmanshah
Points A,B and C on the wire have coordinates (0,0) , (3,3) , (6,6).
So the points are on the line y= x.

A m-kg bead is projected along the wire from A with speed u ms-1 so that it has enough energy to reach B but not C. Prove that 8.16<u<10.95 and that the speed at B >/= 2.58 ms-1. The vertical distance from (0, 0) to (3, 3) is $3\sqrt{2}$ so the increase in potential energy is $3\sqrt{2}mg= 3\sqrt{2}(3)(9.8)= 153$ Joules. The vertical distance from (0, 0) to (6, 6) is $6\sqrt{2}$ so the increase in potential energy is $6\sqrt{2}mg= 6\sqrt{2}(3)(9.8)= 306$ Joules. Since the bead has enough kinetic energy, initially, to go between B and C, what bounds must its speed be between? (I presume you know that kinetic energy is $(1/2)mv^2$.)
If u = 10, the bead comes to rest at a point D between B and C. Find
(a) The greatest speed of the bead between B and D.[/quote]
Obviously, the bead is losing speed as it goes up so its greatest speed 'between B and D' is at B. You are given the bead's speed at A so you can calculate its kinetic energy there. It will stop when its potential energy is equal to that kinetic energy (all kinetic energy has converted to potential energy). At B, what is the potential energy? So what is the remaining kinetic energy? And, from that, is the bead's speed at B?

(b) The coordinates of D.
At what height will all of the initial kinetic energy be converted to potential energy?

What happens after the bead reaches D?
Well, that should be obvious! It has come to a stop at D and now gravity is pulling it back down!

The answer for (a) is 6.83ms-1 and for (b) is (5.7 , 5)

3. ## Re: M1 need help in a complicated problem

how do u get vertical distance as 6\sqrt{6} . it should be 6

4. ## Re: M1 need help in a complicated problem

oh sorry i missed something i forgot to write the starting part of the question i.e a smooth wire is bent into the sahpe of graph of y=(1/6)x^3-(3/2)x^2+4x

5. ## Re: M1 need help in a complicated problem

(1/6)*x^3-(3/2)*x^2+4*x


text:
buildName {conjurer} R Documentation

## Build Dynamic Strings

### Description

Builds strings that could be further used as identifiers. This is an internal function and is currently not exported in the package.

### Usage

buildName(numOfItems, prefix)

### Arguments

 numOfItems A number. This defines the number of elements to be output. prefix A string. This defines the prefix for the strings. For example, the function buildCust uses this function and passes the prefix "cust" while the function buildProd passes the prefix "sku"

### Details

This function is used by other internal functions namely, buildCust and buildProd to produce the alphanumeric identifiers for customers and products respectively.

### Value

A character with the alphanumeric strings is returned. These strings use the prefix that is mentioned in the argument "prefix"

[Package conjurer version 1.5.0 Index]


text:
# Pendulum clocks

A pendulum clock uses the oscillations of a pendulum, a swinging weight, to keep track of time. When the temperature rises and the length of the pendulum increases, what will happen to the clock?

×


text:
# Alpha, Significance Level of Test

Alpha, Significance Level of Test

Subject: Survey Research

• Entry
• Entries A-Z
• Subject Index

• Alpha is a threshold value used to judge whether a test statistic is statistically significant. It is chosen by the researcher. Alpha represents an acceptable probability of a Type I error in a statistical test. Because alpha corresponds to a probability, it can range from 0 to 1. In practice, 0.01, 0.05, and 0.1 are the most commonly used values for alpha, representing a 1%, 5%, and 10% chance of a Type I error occurring (i.e. rejecting the null hypothesis when it is in fact correct). If the p-vahie of a test is equal to or less than the chosen level of alpha, it is deemed statistically significant; otherwise it is not.

The typical level of alpha is 0.05, but this is simply a custom and ...

• [0-9]
• A
• B
• C
• D
• E
• F
• G
• H
• I
• J
• K
• L
• M
• N
• O
• P
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• R
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• X
• Y
• Z


text:
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# Guests at a recent party ate a total of fifteen hamburgers.

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Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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18 Nov 2009, 21:47
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Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.
[Reveal] Spoiler: OA

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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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19 Nov 2009, 05:46
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kairoshan wrote:
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

Good question.

Question Stem :
Total number of people attending the party , x = students + vegetarians + neither - both
Also, vegetarians = x/2 (this implies that non vegetarians also = x/2)
And, NVnonstudents = 15

St. (1) : Vstudents/ Vnonstudents = 2/3 ; NVstudents/NVnonstudents = 4/3
The second ratio gives us NVstudents = 20
Therefore total Non Vegetarians = 20 + 15 = 35
This accounts for half the number of people at the party.
This total number of people = 70
Hence, Sufficient.

St. (2) : 30% were Vnonstudents.
By itself, this statement gives us nothing.
Hence, Insufficient.

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09 Nov 2010, 12:15
First of all, we need to be able to find a solid number,the total number of guest, as an answer.

The only solid number we are given to work with is 15, the number of hamburgers eaten by the guests.

From the question we could see that the guests can be broken down into 4 categories.

VEGETARIAN STUDENT (V & S)
NON-VEGETARIAN STUDENT (NV & S)
VEGETARIAN NON-STUDENT (V & NS)
NON-VEGETARIAN NON-STUDENT (NV & NS)

Looking @ # of hamburgers eaten,

The question states that (NV & NS) ate exactly 1 hamburger and that
no hamburger was eaten by any guest who was a student, hamburger eaten by (V & S) = (NV & S) = 0; a vegetarian, hamburger eaten by (V & S) = (V & NS) =0; or both, hamburger eaten by (V & S) = 0.

So from this we can conclude the # of (NV & NS) = 15.

The last piece of information given is that 1/2 Total = V, which also means 1/2 Total = NV, where NV + V = Total and NS + S = Total. Drawing a table can help understand this relationship.

Statement 1:

The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

I think this should be reworded to ratio instead of rate.

Anyways this just means that for every 2 (V & S) there are 3 (V & NS), which is half the ratio of S to NS for NV. Therefore for every 4 (NV & S) there are 3 (NV & NS), which means $$\frac{4}{3}=\frac{(NV & S)}{(NV & NS)}$$

Since we know that (NV & NS) = 15. We can solve for (NV & S) and Find Total because 1/2* Total = (NV & NS) + (NV & S)

Sufficient.

Statement 2:

30% of the guests were vegetarian non-students.

This give us no way to link 15 to the total number of guest. So insufficient.

This is more clear if you draw a table to help visualize things.

I'm sure someone will come up with a better explanation later, but I hope this can help till then.

Last edited by chaoswithin on 09 Nov 2010, 12:24, edited 2 times in total.

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Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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09 Nov 2010, 12:19
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Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

We have 4 groups of guests:
1. Vegetarian students;
2. Vegetarian non-students;
3. Non-vegetarian students;
4. Non-vegetarian non-students.

Now, as guests ate a total of 15 hamburgers and each guest who was neither a student nor a vegetarian (group #4) ate exactly one hamburger and also as no hamburger was eaten by any guest who was a student, a vegetarian, or both (groups #1, #2 and #3) then this simply tells us that there were 15 non-vegetarian non-students at the party (group #4 = 15).

Make a matrix:

Note that we denoted total # of guests by $$x$$ so both vegetarians and non-vegetarians equal to $$\frac{x}{2}$$.

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians --> $$\frac{vegetarian \ students}{vegetarian \ non-students}=\frac{2}{3}$$ --> if the rate X (some fraction) is half of the rate Y (another fraction), then Y = 2*X --> $$\frac{non-vegetarian \ students}{non-vegetarian \ non-students}=2*\frac{2}{3}=\frac{4}{3}$$ --> so, non-vegetarian non-students compose 3/7 of all non vegetarians: $$non-vegetarian \ non-students = 15 = \frac{3}{7}*\frac{x}{2}$$ --> $$x=70$$. Sufficient.

(2) 30% of the guests were vegetarian non-students --> just says that # of $$vegetarian non-students$$ equal to $$0.3x$$ --> insufficeint, to calculate $$x$$.

[Reveal] Spoiler:
Attachment:

Stem.PNG [ 2.33 KiB | Viewed 21164 times ]

Attachment:

2.PNG [ 2.38 KiB | Viewed 21149 times ]
Attachment:

1.PNG [ 2.59 KiB | Viewed 21343 times ]

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10 Nov 2010, 19:16
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mrinal2100 wrote:
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

can someone explain in detail

This is what I drew when I read the Question stem. Half of the guests were vegetarians so Total/2 stands for the complete vegetarian circle. All outside Vegetarian circle are Total/2.
Attachment:

Ques.jpg [ 17.29 KiB | Viewed 20516 times ]

Statement 1: Veg students : Veg non students = 2:3
Let me say they are 2x and 3x in number.

Non veg students : Non veg non-students = 4:3 (Since veg's ratio is half of non veg's ratio)
Let me say they are 4y and 3y in number.
So now my diagram looks like this:
Attachment:

Ques1.jpg [ 18.33 KiB | Viewed 20508 times ]

3y = 15 hence y = 5
Since 7y is half of the total, 35 is half of the total. So total number of students is 70. Sufficient.

Statement 2: We get that 30% of the guests were veg non students and we already know that 50% of the guests are veg so 20% of the guests are veg students. Essentially, we have got the 3:2 ratio of above. But we do not have the 4:3 ratio of above hence we cannot equate 15 to anything. Therefore, statement 2 is not sufficient alone.
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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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27 Feb 2011, 19:24
Good one!
Stumbled in the test. I like the MGMAT solution.

Non veg - Non Students is 15.

Since Nonveg Student to non student ratio is 4:3, therefore non veg students will be 20.
4/3=x/15
Therefore x= 20. That makes 35 NV and 35 Veg so total 70

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Re: Guests at a recent party ate a totaal of [#permalink]

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20 Oct 2012, 23:59
nitzz wrote:
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

St 1: Sufficient: Veg attended in 2:3 so the Non veg attended in ratio 4:3. We now from Question stem Non veg - Non student ate 15 burgers. therefore Non veg attended the party in 20:15 (4:3). Total no of Non Veg = 35 nos. And total guests = 2*35=70 (from Question Stem).
St 2: Not sufficient: cant calculate other group of non-veg and only %age is provided.

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Last edited by SOURH7WK on 21 Oct 2012, 09:36, edited 1 time in total.

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Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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01 Dec 2014, 05:54
Bunuel, could you explain this statement: " --> so, non-vegetarian non-students compose 3/7 of all non vegetarians." Don't see how you get there. Thanks!

Last edited by Krage17 on 01 Dec 2014, 07:34, edited 1 time in total.

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Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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01 Dec 2014, 06:14
Krage17 wrote:
Bunuel, could you explain this statement: " --> so, non-vegetarian non-students compose 3/7 of all non vegetarians." Don't see how you there. Thanks!

Since non-vegetarian students plus non-vegetarian non-students equals all non-vegetarians, then (non-vegetarian non-students/(total) = 3/(4 + 3) = 3/7.

Hope it's clear.
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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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01 Dec 2014, 09:35
It is. Took me a while to grasp it.

Bunuel wrote:
Krage17 wrote:
Bunuel, could you explain this statement: " --> so, non-vegetarian non-students compose 3/7 of all non vegetarians." Don't see how you there. Thanks!

Since non-vegetarian students plus non-vegetarian non-students equals all non-vegetarians, then (non-vegetarian non-students/(total) = 3/(4 + 3) = 3/7.

Hope it's clear.

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Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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12 Dec 2015, 04:51
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Hi Math Experts, I have one question regarding my approach for this question:

This is a rule From MGMAT

If a Data Sufficiency question asks for the concrete value of one element of a ratio, you will need
BOTH the concrete value of another element of the ratio AND the relative value of two elements
of the ratio

So, after reading Statement 1, I've marked Answer A, because Statement 1 gives us the relationship among all the groups + we have a concrete value from the question stem, hence, one can find the required value --> I've just pretty much followed the rule from MGMAT....
Do you think it's a valid approach for such kind of questions ?
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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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30 Apr 2017, 04:01
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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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28 May 2017, 19:23
kairoshan wrote:
Guests at a recent party ate a total of fifteen hamburgers. Each guest who was neither a student nor a vegetarian ate exactly one hamburger. No hamburger was eaten by any guest who was a student, a vegetarian, or both. If half of the guests were vegetarians, how many guests attended the party?

(1) The vegetarians attended the party at a rate of 2 students to every 3 non-students, half the rate for non-vegetarians.

(2) 30% of the guests were vegetarian non-students.

The first step is to set up an overlapping sets table and insert the one known value: 15 non-vegetarian non-students. It's worded in a confusing way, but clearly the stimulus states that each relevant student at 1 hamburger each and only 15 hamburgers were eaten.

1) The ratio of vegetarian students to vegetarian non-students is 2:3, or half the ratio of non-vegetarians. So we know that non-vegetarians are in the ratio 4:3. 3x = 15, since we already know the number of non-vegetarian non-students, so x = 5. If x = 5, then the total number of vegetarians is 7(5) = 35. Since vegetarians are half of the guests, the total number of guests is 70. Sufficient.

2) This provides the relationship of the vegetarian non-student number to the total number of guests, but we don't know any other values. Not sufficient.

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Re: Guests at a recent party ate a total of fifteen hamburgers. [#permalink]

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25 Oct 2017, 07:31
main catch is 1 statement rate

Veg
2 student ---> 3 non student
1 s-------1.5 ns

half of it is for Non veg
1s-------1.5/2 ns

1s---------3/4 ns
4s => 3ns
4:3 ratio

=> 3x/7 = 15
x= 35
2x is total members => 70 is answer A.
since 2x are members x are veg and x are non veg
so took 4:3 s:ns for nonveg and since total number is x
so 3x/7 => x= 35 and 2x =70

and clearly 2 is insufficient
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Re: Guests at a recent party ate a total of fifteen hamburgers.   [#permalink] 25 Oct 2017, 07:31
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text:
# Moments of a random variable involving numerous Binomial Distributions

#### BobbyRichards

##### New Member
Mean, Variance, Skew, and Kurtosis of X

I am editing my post because it might have been unclear.

Suppose X is a random variable involving 4 binomially distributed random variables:

X = y1(p, 1/2 (a + b)) - y2((1 - p), 1/2 (a + b)) + y3((1 - p),1/2 (a - b)) - y4(p, 1/2 (a - b))

I wish to find the mean, variance, skew, and kurtosis of X, and I believe I have found the correct mean using the formula for the mean of a binomially distributed random variable, np.

E(X) = (p 1/2 (a + b)) - ((1 - p) 1/2 (a + b)) + ((1 - p) 1/2 (a - b)) - (1/2 p (a - b))
E(X) = b (-1 + 2 p)

Now I am using the formula for variance, p(1-p), to find Var(X), but I am not sure I am correct:

Var(X) = (p (1 - p) 1/2 (a + b)) + ((1 - p) (1 - (1 - p)) 1/2 (a + b)) + ((1 - p) (1 - (1 - p)) 1/2 (a - b)) + (p (1 - p) 1/2 (a - b))
Var(X) = -2 a (-1 + p) p

I am not sure my calculation is correct, because I have a reference who believes the correct answer for Var(X) is -4 a (-1 + p) p.

Is there anyone here on Talk Stats who is able to help me? I would really appreciate it.

Thank you.

Last edited:

#### JesperHP

##### TS Contributor
Re: Mean, Variance, Skew, and Kurtosis of X

Two things you can do when this situation occurs
1) use symbolic solver https://www.symbolab.com/solver/
copy paste the large equation you have and get the result you have $$-2p^2a+2pa=2ap(1-p)$$
2) simulate X and estimate variance and compare it to the result of you're formula

#### BobbyRichards

##### New Member
Thank you Jesper,

I have physical access to Mathematica, but mentally not much (yet!). Do you know if Mathematica can solve my problem?

Have a great day!

#### JesperHP

##### TS Contributor
I do not know Mathematica, but according to the solver i linked to your answer is correct, so unless you have another problem as well I would consider it solved.

#### BobbyRichards

##### New Member
Hi Jesper,

While the calculation simplifies to what you and I both got, the formula for variance that I have used is wrong. I simulated via MATLAB and got twice the answer, which is exactly what my reference has.

So the problem remains. What is the correct formula for variance?


text:
Lemma 15.29.4. Let $f_1, \ldots , f_{r - 1} \in R$ be a sequence and $f, g \in R$. Let $M$ be an $R$-module.

1. If $f_1, \ldots , f_{r - 1}, f$ and $f_1, \ldots , f_{r - 1}, g$ are $M$-$H_1$-regular then $f_1, \ldots , f_{r - 1}, fg$ is $M$-$H_1$-regular too.

2. If $f_1, \ldots , f_{r - 1}, f$ and $f_1, \ldots , f_{r - 1}, f$ are $M$-Koszul-regular then $f_1, \ldots , f_{r - 1}, fg$ is $M$-Koszul-regular too.

Proof. By Lemma 15.28.11 we have exact sequences

$H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, f) \otimes M) \to H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, fg) \otimes M) \to H_ i(K_\bullet (f_1, \ldots , f_{r - 1}, g) \otimes M)$

for all $i$. $\square$

Comment #5363 by MAO Zhouhang on

A typo in the hypothesis of the second statement: the second sequence should be $f_1,\dots,f_{r-1},g$, not $f_1,\dots,f_{r-1},f$.

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• 2 comment(s) on Section 15.29: Koszul regular sequences

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text:
# A Collaborative Search Strategy to Solve Combinatorial Optimization and Scheduling Problems

Azizi, Nader and Zolfaghari, Saeed and Liang, Ming (2010) A Collaborative Search Strategy to Solve Combinatorial Optimization and Scheduling Problems. In: Cakaj, S., ed. Modelling, Simulation and Optimization - Tolerance and Optimal Control. InTech. ISBN 978-953-307-056-8. (doi:10.5772/9030)

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