text:
# Definition on multiplication in rings [closed]

Let's say I have a ring $$R=(\mathbb{R}, +, *)$$. Is

$$a+a=2*a$$

($$a \in \mathbb{R}$$)

always true for any ring¹? (So that when this equation is not true then I can definitively infer that $$R$$ is not a ring.)

1 ) With "any" I mean any other ring which is also using $$\mathbb{R}$$ as underlying set, thank you for your comment nick.

• If $R = \Bbb{R}$ and the usual number $1 \in \Bbb{R}$ is the identity, then yes. The distributive law in a ring says that $a+a = a(1+1)$. However, when you say "any ring", this makes it confusing, because the underlying set does not always have to be $\Bbb{R}$. – Nick Nov 29 '20 at 23:06
• In particular, the answer depends on what the definition of "2" is in the ring. If it always means "1+1", then the equation is true. – Greg Martin Nov 29 '20 at 23:07
• In any ring, for $n\in\mathbb{Z}$, we define $0a = 0_R$, $(n+1)a=na+a$, and $(-n)a = -(na)$ for $n\gt 0$. This is not “really” multiplication, but rather abbreviation for some addition operations. When the ring has a unity, then one can show that $na = (n1_R)a$, and so we can turn this “abbreviated addition” into actual multiplication in the ring. But one does not normally think of “$2a$” as being a product in an arbitrary ring; instead, it is addition and stands for $a+a$. – Arturo Magidin Nov 29 '20 at 23:39

If your ring has a unit, i.e. a multiplicative identity, (and the definition that almost everybody uses these days requires that a ring does have an identity), then yes.

As the commenters point out, $$2$$ is defined* to be $$1 +1$$, where $$1$$ is the multiplicative identity, and so it follows from the distributive law and the fact that $$1$$ is the multiplicative identity.

The only thing to be careful about is that it is possible that $$2 = 0$$ (e.g. in $$\mathbb Z_2$$), or perhaps $$2 = -1$$ (e.g. in $$\mathbb Z_3$$), so these "integers" inside your ring might not behave the way you expect integers to behave.

BTW, if you are dealing with an algebraic structure that doesn't have a $$1$$, people will often define an "action" of $$\mathbb Z$$ on your elements, and use multiplication to denote it, where

$$n \cdot a = a + a + .... + a \text{ (n times)}$$

Edit: Okay, you added "With 'any' I mean any other ring which is also using $$\mathbb{R}$$ as underlying set", and this needs to be addressed: You can take the underlying set $$\mathbb R$$, and define a wacky new addition and multiplication on it. The simplest is $$a \oplus b = a + b -1$$ and $$a \otimes b = ab - a -b + 2$$.

Let's use the symbol $$S$$ to denote this new ring $$\langle \mathbb R, \oplus, \otimes\rangle$$. Then the number 1 in $$\mathbb R$$ (which I'm going to write as $$1_{\mathbb R}$$) is not the multiplicative identity for the ring $$S$$. $$1_S$$, which is the standard notation for the multiplicative identity in a ring named $$S$$, is, in fact $$2$$, by which I mean good old 2 in good old $$\mathbb R$$, which we might want to write as $$2_{\mathbb R}$$, and yes $$2_{\mathbb R} = 1_{\mathbb R} + 1_{\mathbb R}$$.

But what your question asks is still true in $$S$$, i.e. $$a \oplus a =2_{S} \otimes a$$; however notice that you have to be sure to use the ring operations of $$S$$, and remind yourself that you're using $$2_{S}$$, which is defined to be $$1_{S} \oplus 1_{S}$$. (And corresponds to the underlying real number $$3_{\mathbb R}$$!)

The ring $$S$$ is of course extremely confusing to work with, and I've never seen it used seriously, only to break the brains of undergraduate math majors, to show them how we can define groups, rings, fields, etc. that behave very differently from what they are used to. I.e. $$\langle \mathbb R, \oplus, \otimes\rangle$$ is a cautionary tale, not a commonly used mathematical tool, but the only requirement you put was that $$\mathbb R$$ was the underlying set, and so you left it open to me to define really weird addition and multiplication. I wouldn't spend too much time agonizing over it, but it can be a fun example to contemplate and sharpen your wits.

*If someone uses the symbol "$$2$$" and says it is not equal to $$1+1$$, you get to look at them funny, ask just what the heck they think they are doing, and demand that they explain why they are using that symbol.

• thank you for your detailed explanation! – anion Nov 30 '20 at 11:29

This is basically true by definition, although there are some things you should be aware of.

Some people require that any ring $$(R,+_R,\cdot_R)$$ contains a multiplicative identity $$1_R,$$ and that ring homomorphisms $$f : (R,+_R,\cdot_R)\to (S,+_S,\cdot_S)$$ satisfy $$f(1_R) = 1_S.$$ If you require this condition, then for any ring $$(R,+_R,\cdot_R)$$ there is a unique ring homomorphism $$i_R : (\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R).$$ In this case, even if the set $$R$$ does not literally contain $$2,$$ you may think of $$i_R(2)\in R$$ as being $$2$$ (you might even write $$i_R(2) = 2_R$$). It is true then that for any $$r\in R,$$ $$2_R\cdot_R r = i_R(2)\cdot_R r = r +_R r,$$ because \begin{align*} i_R(2)\cdot_R r &= i_R(1 + 1)\cdot_R r\\ &= (i_R(1) + i_R(1))\cdot_R r\\ &= (1_R + 1_R)\cdot_R r\\ &= r + r. \end{align*} As JonathanZ supports MonicaC notes, it might be the case that $$i_R(2)$$ behaves differently than you'd expect, or looks different than you'd expect. It could be that $$i_R(2) = -1_R$$ or even $$i_R(2) = 0_R$$! See the last paragraph for a particularly outrageous example of this.

If you do not require that your rings have multiplicative identitites and/or that ring homomorphisms need not send multiplicative identities to multiplicative identities, then this is still true to some extent, although we should be careful about what we mean.

Let $$(R,+_R,\cdot_R)$$ be our possibly non-unital ring. In this case, we can't use the unique homomorphism $$i_R :(\Bbb{Z},+,\cdot)\to(R,+_R,\cdot_R)$$ from before -- there might be more than one ring homomorphism now! Additionally, the set $$R$$ might not contain $$2.$$

So, what do we do? Well, remember that any ring has an underlying abelian group $$(R,+_R).$$ An abelian group is the same as a $$\Bbb{Z}$$-module (see here for the definition of a module over a ring if you're not familiar). This means explicitly that we have an action of $$\Bbb{Z}$$ on $$R$$ which interacts nicely with addition. We define this action by setting $$n\cdot r :=\begin{cases} \underbrace{r + \dots + r}_{n\textrm{ times}},&n > 0,\\ 0,&n=0,\\ \underbrace{-r + \dots + -r}_{-n\textrm{ times}}, &n <0. \end{cases}$$ Notice that I'm not writing $$n\cdot_R r$$ -- that's because there's not necessarily an element $$n\in R$$ which behaves like $$n.$$ However, it's still sensible to think of adding the element $$r$$ to itself $$n$$ times, which is what $$n\cdot r$$ means by definition. The $$\cdot$$ refers to the action of $$\Bbb{Z}$$ on the underlying abelian group of $$(R,+_R,\cdot_R),$$ not multiplication in the ring itself. In this sense, the equality $$2\cdot r = r+r$$ always holds, and this is basically by definition!

One last remark. You asked if this is true of any ring which has $$\Bbb{R}$$ as its underlying set. You should be slightly careful here. Consider the following ring structure on $$\Bbb{R}$$: \begin{align*} +' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\ (r,s)&\mapsto r+'s:=\sqrt[3]{r^3 + s^3},\\ \cdot' : \Bbb{R}\times\Bbb{R}&\to\Bbb{R}\\ (r,s)&\mapsto r\cdot's := rs. \end{align*} This is not the standard ring structure on $$\Bbb{R}$$ -- the multiplication is the same, but the addition is "twisted." In this case, $$2\in \Bbb{R}$$, but it is not true that $$2\cdot' r = r +' r.$$ Suppose $$r = 2.$$ Then: \begin{align*} 2 +' 2 &= \sqrt[3]{2^3 + 2^3}\\ &= \sqrt[3]{16}\\ &= 2\sqrt[3]{2}. \end{align*} On the other hand, $$2\cdot'2 = 4.$$ What happened? I'll let you think about this for yourself before revealing the answer below!

What happened here is that $$2\in\Bbb{R}$$ is no longer playing the same role it was before. Our ring $$(\Bbb{R},+',\cdot')$$ still has a multiplicative identity, but our ring homomorphism $$i_{(\Bbb{R},+',\cdot')} : (\Bbb{Z},+,\cdot)\to(\Bbb{R},+',\cdot')$$ now sends $$i_{(\Bbb{R},+',\cdot')}(2) = i_{(\Bbb{R},+',\cdot')}(1) +' i_{(\Bbb{R},+',\cdot')}(1) = 1 +' 1 = \sqrt[3]{2}.$$So there is an element of $$(\Bbb{R},+',\cdot')$$ which behaves like $$2$$ should -- it's $$\sqrt[3]{2}$$. We thus have$$\sqrt[3]{2}\cdot' r = r +' r$$for any $$r\in\Bbb{R}.$$ This is very confusing, because we already have $$2\in\Bbb{R}$$! In this case, it would be very important to distinguish between $$2\cdot r$$ (which is $$2\in\Bbb{Z}$$ acting on $$r,$$ giving $$r +'r$$) and $$2\cdot' r$$ (which as we computed, is not $$r +' r$$ in general). In the notation of the first paragraph, $$2_{(\Bbb{R},+',\cdot')} = \sqrt[3]{2}$$ and $$2\neq 2_{(\Bbb{R},+',\cdot')}$$.

To be even more explicit about what happened, given any set $$X,$$ any ring $$(R,+_R,\cdot_R),$$ and any bijection $$f : X\to R,$$ we can give $$X$$ the structure of a ring by defining addition on $$X$$ by $$x +_X y := f^{-1}(f(x)+_R f(y))$$ and $$x\cdot_X y := f^{-1}(f(x)\cdot_R f(y)).$$ We're taking the ring structure on $$R$$ and transporting it to $$X$$ via the bijection $$f$$: first, take your elements $$x$$ and $$y$$ in $$X,$$ send them over to $$R$$ where you add or multiply them, and then bring them back to $$X.$$ In my example above, I'm using the bijection $$\Bbb{R}\to\Bbb{R}$$ which sends $$x$$ to $$x^3.$$

• Stahl, we wrote very similar answers! In a fair world our points would be pooled and split evenly. I think I like your "weird $\mathbb R$" better -- mine is just the standard conjugating by $\phi(a) = a-1$, but I'd never seen yours before. Having multiplication stay the same while getting strange with addition is a really nice touch. – JonathanZ supports MonicaC Nov 30 '20 at 0:10
• @JonathanZsupportsMonicaC Thanks! I think your answer is very nice -- it's simple and to the point, while I've rambled on a bit. You posted yours while I was in the middle of writing mine, and I thought I would keep going and post in case the extra details helped someone. – Stahl Nov 30 '20 at 0:20
• Yeah, I figured that it was an overlap. These days if I find myself writing an essay, I've taken to posting a few times as I complete sections (I'm still trying to get to 10k rep!). And I ended up rambling too when addressing the OP's edit. But I'm glad you finished and posted yours; I'm going to have to go and have a think about your twisted addition. I think my answer targets the average Fraleigh reader, and yours is good for the Hungerford crowd. – JonathanZ supports MonicaC Nov 30 '20 at 0:28
• thank you for your detailed explanation! – anion Nov 30 '20 at 11:29


text:
Volume 398 - The European Physical Society Conference on High Energy Physics (EPS-HEP2021) - T04: Neutrino Physics
Correlating Muon $g-2$ Anomaly with Neutrino Magnetic Moments
S. Jana*, K.S. Babu, M. Lindner and V. P.K.
Full text: pdf
Pre-published on: March 09, 2022
Published on: May 12, 2022
Abstract
We show that the models that induce neutrino magnetic moments, while maintaining their small masses naturally, also predict observable shifts in the muon anomalous magnetic moment. This shift is of the right magnitude to be consistent with the Brookhaven measurement as well as the recent Fermilab measurement of the muon $g-2$. Thus, it provides the most minimal and standalone framework to solve the decades-old neutrino mass-magnetic moment puzzle and which simultaneously addresses three shortcomings of the SM (a) neutrino mass generation, (b) large neutrino magnetic moment (in order to address the Xenon1T anomaly) and (c) muon $g-2$ anomaly. This is pointing out the direct correlation between the magnetic moment of SM charged lepton and neutral lepton (neutrino) by showing that the measurement of muon $g-2$ by the Fermilab experiment can be an in-direct and novel test of the neutrino magnetic-moment hypothesis, which can be as sensitive as other ongoing-neutrino/dark matter experiments. Such a correlation between muon $g-2$ and the neutrino magnetic moment is generic in models employing leptonic family symmetry to explain a naturally large neutrino magnetic moment. This talk is based on results obtained with K.S. Babu, Manfred Lindner, and Vishnu P.K. and presented in hep-ph 2007.04291 and 2104.03291.
DOI: https://doi.org/10.22323/1.398.0189
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access

format: plain


text:
1. How can we solve this question?

Suppose that y'=v, v'=F(y,v), where F,Fy and Fv are continuous for all y,v.
(a) As t increases, why do all orbits above the y-axis move to the right and orbits below the y-axis move to the left?
(b) Why do nonconstant orbits that cross the y-axis always do so perpendicularly to the axis?

2. Originally Posted by ninano1205
How can we solve this question?

Suppose that y'=v, v'=F(y,v), where F,Fy and Fv are continuous for all y,v.
(a) As t increases, why do all orbits above the y-axis move to the right and orbits below the y-axis move to the left?
"Above the y-axis" means that v>0. Therefore y'>0, which means that orbits are moving in the direction of increasing y (in other words, "to the right").

(b) Why do nonconstant orbits that cross the y-axis always do so perpendicularly to the axis?
On the y-axis, v=0 and so y'=0. That means that the component of velocity in the y-direction is 0, so the orbit must move in the v-direction (that is, "perpendicularly to the [y-]axis").
..


text:
Let $f(x) = \displaystyle \left(2x^{2}-6x+3\right)\!\left(8x^{2}+6x-8\right)$. Evaluate $f'(x)$ at $x = 2$.

$f'( 2 ) =$


text:
# How to escape a quotation mark character/glyph?

I have the same problem as in this question, but the answers just circle around how to best use quotation marks. This is fine as long as you actually want a quotation mark. However, I explicitly don't want it.

And there is one legitimate reason: I want to display code. Verbatim. Without any change. So I put that in \texttt, or in a special command I have defined that also just does this.

MWE:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[ngerman]{babel}

% you may add magic here to make it ignore/escape any special characters
\newcommand{\inlinecode}{\texttt}

% disable " making strange characters, it breaks our code quotations
% \shorthandoff{"} % does not fully work (see below)

\title{shorthandoff}
\date{August 2020}

\begin{document}

\maketitle

\section{Introduction}

This is a JSON snippet:
\inlinecode{"example\_string\_mentioned": "yes\_no"} \\
\inlinecode{"okay"} \\
\inlinecode{"yikes"}\\
\inlinecode{"ohno"}\\

\end{document}

This resullts in this:

This is a JSON snippet: ëxample string mentioned": "yes no"
ökay"
"yikes"
öhno"

Obviously this is bad. Obviously I don't want to use typographically correct quotation marks.

• Escape as {"} (caused display errors and showed errors in the console)¹ or as \" (did not help).
• Actually escape the {e} in example – did not work either.
• \shorthandoff is a good recommendation from the already linked question and helps. Unfortunately, if you uncomment it above, it still does mangle with the e: We do not get ökay anymore, but we do get ëxample.
• This answer pointed me to \verb. While this works, this means I a) cannot put it in \texttt anymore, it refuses that, and worse, b) it prevents any line breaks at the line end.
Edit: It accidentally uses typewriter (same-width) font already, which I want in this case, but which may not be desired in other use cases.

¹ Actually, it seems to force a line break and showed this:

Argument of \language@active@arg" has an extra }.

I basically just want Markdown-like behaviour of the   .

In other words: How do I really escape " in LaTeX?

TexLive 2019

• move \shorthandoff{"} behind \begin{document}, or load babel with \usepackage[ngerman,shorthands=off]{babel}. Aug 29, 2020 at 14:12
• @UlrikeFischer Thanks, but wtf, why does that make a difference? Respectively, why does the placement in the preamble disable it for o, but not for e??? BTW, feel free to submit an answer to this question, so I can accept it and give you the credit for solving this.
– rugk
Aug 29, 2020 at 14:29
• lots of stuff is initialized at begin document. And for me both the ö and ë are there. Btw: the code from @egreg is naturally better, as he disables the shorthand only in the place where you really don't want it. Aug 29, 2020 at 14:36
• why not \verb|example_string_mentioned": "yes_no"| ? Aug 29, 2020 at 15:31
• no you mentioned verbatim in the question by which I assumed you meant \begin{verbatim} I did not understand your comment about not allowing \texttt as \verb uses the texttt font. Aug 29, 2020 at 17:36

You can disable shorthands locally with \languageshorthands{none}:

\documentclass{article}
\usepackage[ngerman]{babel}
\usepackage[T1]{fontenc}
\newcommand{\inlinecode}[1]{{\languageshorthands{none}\ttfamily #1}}

\begin{document}
This is a JSON snippet:
\inlinecode{"example\_string\_mentioned": "yes\_no"} \\
\inlinecode{"okay"} \\
\inlinecode{"yikes"}\\
\inlinecode{"ohno"}
\end{document}

• @DavidCarlisle ups ;-) Aug 29, 2020 at 15:45

Use an indirect method that changes the meaning of the active ", if it is active; if not, the code does nothing harmful.

\documentclass{article}
\usepackage[ngerman]{babel}

\newcommand{\inlinecode}{}% just to warn if it is already defined
\DeclareRobustCommand{\inlinecode}{%
% don't propagate this outside the argument
\begingroup
% define the active quote to yield a normal "
\begingroup\lccode~=" \lowercase{\endgroup\edef~}{\string"}%
% call an auxiliary command to do the typesetting and close the group
\doinlinecode
}
\newcommand{\doinlinecode}[1]{\texttt{#1}\endgroup}

% disable " making strange characters, it breaks our code quotations
% \shorthandoff{"} % does not fully work (see below)

\title{shorthandoff}
\date{August 2020}

\begin{document}

\maketitle

\section{Introduction}

This is a JSON snippet:
\inlinecode{"example\_string\_mentioned": "yes\_no"} \\
\inlinecode{"okay"} \\
\inlinecode{"yikes"}\\
\inlinecode{"ohno"}

\end{document}

• Oh, thanks. This looks complicated though, so I guess the solution by @UlrikeFischer is easier/less complicated.
– rugk
Aug 29, 2020 at 14:34

Even though the question focuses on the " character, it seems to really be about how to display inline verbatim (JSON) code.

The standard way to include inline verbatim text is with \verb, like

In our \textsc{json} we do this with \verb|"otter": 5|.

With verbatim characters are shown as themselves. That includes spaces, which means that there are no linebreaks. (Also these verbatim spaces are not stretchy as normal spaces are.)

Another way to include code is to use packages that are meant for that, like listings. That package includes commands for showing code listings given in the TeX file or taken from an external file, and also for short inline code, like this:

\documentclass{article}
\usepackage{listings}
\lstset{
basicstyle=\ttfamily\small,
}

\begin{document}
In our \textsc{json} we do this with \lstinline|"otter": 5|.
\end{document}

(Of course the basicstyle setting here is just an example.)

Both here and for \verb the delimiting | character can be some other character. Some popular choices are /, + and !, depending on what's in the text.

One big feature of a package like listings is to do code highlighting. It doesn't know about JSON syntax out of the box though. If you are interested in customizing that (even though it wasn't mentioned in the question), you can see for example this and this question.

More to the point here, there is an option to allow linebreaks. Just add

  breaklines=true,

to the \lstset settings.

There is nothing extra you need to do for the straight " character if you use normal typewriter font with the above. But you will find that the straight ' character is still shown as a curly quote! There is a package to avoid that in verbatim text. Add:

\usepackage{upquote}
`


text:
## Algebra 2 Common Core

$$\tan^{2}\theta$$
Use the Pythagorean Identity: $$1+\tan^{2}\theta=\sec^{2}\theta$$ Rearrange terms: $$\sec^{2}\theta-1=\tan^{2}\theta$$ Alternatively, replace $\sec{\theta}$ with $\frac{1}{\cos\theta}$ $$\sec^{2}\theta-1= \frac{1}{cos^{2}\theta}-1$$ Common denominator: $$\sec^{2}\theta-1= \frac{1-cos^{2}\theta}{cos^{2}\theta}$$ By identity, $sin^{2}\theta+cos^{2}\theta=1$ (subtract $cos^{2}\theta$ on both sides) Therefore, replace 1-$cos^{2}\theta$ with $sin^{2}\theta$ $$\sec^{2}\theta-1=\frac{sin^{2}\theta}{cos^{2}\theta}$$ By identity: $\frac{sin\theta}{cos\theta}=tan\theta$ Hence, $$\sec^{2}\theta-1=tan^{2}\theta$$


text:
# ON THE TATE-SHAFAREVICH GROUPS OVER DEGREE 3 NON-GALOIS EXTENSIONS

• Yu, Hoseog (Department of Mathematics, Sejong University)
• Accepted : 2016.01.25
• Published : 2016.03.25
• 94 13

#### Abstract

Let A be an abelian variety defined over a number field K and let L be a degree 3 non-Galois extension of K. Let III(A/K) and III(A/L) denote, respectively, the Tate-Shafarevich groups of A over K and over L. Assuming that III(A/L) is finite, we compute [III(A/K)][III($A_{\varphi}/K$)]/[III(A/L)], where [X] is the order of a finite abelian group X.

#### Keywords

Tate-Shafarevich group;abelian varieties;restriction of scalars

#### Acknowledgement

Supported by : National Research Foundation of Korea(NRF)

#### References

1. M. I. Bashmakov, The cohomology of abelian varieties over a number field, Russian Math. Surveys 27, no. 6 (1972), 25-70. https://doi.org/10.1070/RM1972v027n06ABEH001392
2. K. S. Brown, Cohomology of groups, Grad. Texts in Math. 87. Springer-Verlag 1982.
3. J. W. S. Cassels, Arithmetic on curves of genus 1. VII. The dual exact sequence, J. Reine Angrew. Math. 216 (1964), 150-158.
4. J. W. S. Cassels, Arithmetic on curves of genus 1. VIII. On the conjectures of Birch and Swinnerton-Dyer, J. Reine Angrew. Math. 217 (1965), 180-189.
5. J. S. Milne, Arithmetic Duality Theorems, Perspectives in Math. vol. 1. Academic Press Inc. 1986.
6. B. Poonen and M. Stoll, The Cassels-Tate pairing on polarized abelian varieties, Ann. Math. 150 (1999), 1109-1149. https://doi.org/10.2307/121064
7. C. Riehm, The Corestriction of Algebraic Structures, Inven. Math. 11 (1970), 73-98. https://doi.org/10.1007/BF01389807
8. J. Tate, Relations between $K_2$ and Galois cohomology, Inventiones Math. 36 (1976), 257-274. https://doi.org/10.1007/BF01390012
9. J. Tate, WC-group over p-adic fields, In: Seminaire Bourbaki, 1957-58, expose 156.
10. J. Tate, Duality theorem in Galois cohomology over number fields, Proc. Int. Cong. Math., Stockholm (1962), 288-295.
11. A. Weil, Adeles and algebraic groups, Progrss in Math. 23. Birkhauser 1982.
12. H. Yu, On Tate-Shafarevich groups over Galois extensions, Israel J. Math. 141 (2004), 211-220. https://doi.org/10.1007/BF02772219
13. H. Yu, On Tate-Shafarevich groups over cyclic extensions, Honam Math. J. 32 (2010), 45-51. https://doi.org/10.5831/HMJ.2010.32.1.045


text:
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# If m^3-n^2=-300, then the lowest possible value of m is

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If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5
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guerrero25 wrote:
If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5

$$m^3-n^2=-300$$ --> $$m=\sqrt[3]{n^2-300}$$. To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is $$m_{min}=\sqrt[3]{-300}$$.

m is less than -5 (since (-5)^3=-125) and more than -10 (since (-10)^3=-1000).

_________________
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Kudos [?]: 516 [0], given: 325

Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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01 Apr 2013, 03:02
Bunuel wrote:
guerrero25 wrote:
If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5

$$m^3-n^2=-300$$ --> $$m=\sqrt[3]{n^2-300}$$. To minimize m we should minimize n^2. The lowest value of n^2 is 0, thus the lowest value of m is $$m_{min}=\sqrt[3]{-300}$$.

m is less than -5 (since (-5)^3=-125) and more than -10 (since (-10)^3=-1000).

Such an Easy approach , Bunuel . I succumbed to the time pressure . I wish I could think like you
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guerrero25 wrote:
If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5

$$m^3-n^2=-300$$
So, $$m^3 = n^2 - 300$$
For $$m^3$$ to be minimum, $$(n^2 - 300)$$ must be minimum
For $$(n^2 - 300)$$ to be minimum, $$n^2$$ must be minimum, so $$n^2$$ = 0
So $$m^3$$ = -300
So m = -6. .....
So m lies between -10 and -5
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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25 Apr 2013, 06:57
Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)?
I was considering the cube root for the lowest positive value of m
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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25 Apr 2013, 08:34
Expert's post
sdas wrote:
Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)?
I was considering the cube root for the lowest positive value of m

Even roots from negative number is undefined on the GMAT: $$\sqrt[{even}]{negative}=undefined$$, for example $$\sqrt{-25}=undefined$$.

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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25 Apr 2013, 08:53
sdas wrote:
Bunuel, pls help. how are we considering m as minimum with cube root of -300, is that not an unreal number (negative root)?
I was considering the cube root for the lowest positive value of m

To add to what Bunnuel said. Try to think in reverse. You can always multiply a negative number 3 times to get an odd number, but you cannot multiply a negative number 2 times to get a negative number
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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10 Jul 2014, 14:07
The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal -300. For example if I chose m to be -20 than (m^(3)) would be -8000. and I would find a number that for (n2) that is equal to 7970.

Bunuel whats wrong with this logic?
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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11 Jul 2014, 02:07
$$m^3 - n^2 = -300$$

$$n^2 = 300 + m^3$$

$$5^3 = 125; & 10^3 > 300$$

So least value of m should be between -5 & -10

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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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11 Jul 2014, 05:09
bankerboy30 wrote:
The way I intepreted this problem, is m can be any negative number. M does not have to be an integer. I immediately chose the greatest negative range as the answer because I figured I could offset it with some (n^2) to equal -300. For example if I chose m to be -20 than (m^(3)) would be -8000. and I would find a number that for (n2) that is equal to 7970.

Bunuel whats wrong with this logic?

Hi Bankerboy30,

In your case, you would need to find n such that square of n would equal -7700 (300-8000). Now, we know that square of a real number cannot be negative and we don't deal with imaginary numbers in GMAT.

So, you need to go by a limitation that square of n can be minimum ZERO, not less than that. If you use that, you will get the answer as Bunuel got.

Does it help?

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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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05 Aug 2015, 20:56
Thank you
guerrero25 wrote:
If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5

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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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08 Apr 2016, 19:23
guerrero25 wrote:
If m^3-n^2=-300, then the lowest possible value of m is between

(A) -20 and -15
(B) -15 and -10
(C) -10 and -5
(D) -5 and 0
(E) 0 and 5

m will be minimum when n=0, otherwise by deducting a positive number, the negative will get even bigger.
m^3 = 300
ok...
-5x-5x-5=-125..so clearly can be lower than -5. D and E are out.
-10x-10x-10=-1000 clearly not lower than -10. A and B out.
C remains.
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Re: If m^3-n^2=-300, then the lowest possible value of m is [#permalink]

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15 Apr 2016, 02:39
Here the value of M must lie between -6 and -7
Re: If m^3-n^2=-300, then the lowest possible value of m is   [#permalink] 15 Apr 2016, 02:39
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# IntMath Newsletter: Golden spiral, $0 math e-books, CDF format By Murray Bourne, 05 Sep 2011 5 Sep 2011 In this Newsletter: 1. Golden Spiral 2. Math e-Books for$0
3. 13 yr-old designs efficient solar collectors using Fibonacci and trees
4. Wolfram's new Computable Document Format
5. Math Puzzle: Solution and new one
6. Friday math movies
7. Final thought: Powerful weapon

## 1. Golden Spiral

Suitable for: This is an interesting article because there are several important math concepts involved. If you have some basic understanding of trigonometry and exponentials you will get more out of it.

 Golden Spirals involve a lot of interesting math, from exponential curves to polar coordinates, tangents to a curve and approximating curves. Golden Spiral

## 3. 13 yr-old designs efficient solar collectors using Fibonacci sequence and trees

Suitable for: Everyone.

 This is a useful discovery by a 13 yr-old, using math and good observation skills. Trouble is, it turned out to be not as efficient as claimed! (See the comments.) 13 yr-old designs efficient solar collectors using Fibonacci sequence and trees

## 4. Wolfram's new Computable Document Format and Online Conference

We've had PDF format for some time now.

Wolfram, developers of Mathematica and the brilliant Wolfram|Alpha, have released a new type of document which can include interactive animations of math concepts.

The documents it produces are free, but you need to purchase Mathematica if you want to produce such documents. But never mind, there's still a lot of value in this. You'll need to download a small file in order to view, and play with, the CDF files.

Interactive Whiteboard users - this is for you!

Wolfram Mathematica Online Conference

Wolfram is also hosting a free Mathematica Virtual Conference on Sep 26th (for US time zone) and on 27th Sep (Europe time zone - also good for Asia).

For those new to Mathematica, the "Discover Mathematica" session will be good. There's also a session on "Applying Mathematica in Industry and Research" and for the teachers, "Applying Mathematica in Education".

I'll be attending the 27th Sep session - say hello if you see my name in the participant list!

Information and free registration: Wolfram Mathematica Virtual Conference 2011

## 5. Math Puzzle - solution and new one

Last newsletter, the math puzzle was:

If the end of the world should come on the first day of a new century, what is the probability that it will happen on a Sunday?

While it seems the probability should be 1/7, it isn't!

The first day of a century can never fall on a Sunday; nor on a Wednesday or a Friday. See here for an explanation.

New puzzle:

A business man works on the 10th floor of an office building and always takes the elevator down to ground level at the end of the day.

Yet every morning he only takes the elevator to the 7th floor and walks up the stairs to the 10th floor, even when he is in a hurry.

## 6. Friday math movies

I have a back-log of math movies this Newsletter. Enjoy!

## (a) Miss USA 2011 - Should math be taught in schools?

Suitable for: Everyone - brief (2 min) and funny.

This is hilarious - or very disturbing - depending on your point of view.

Friday math movie: Miss USA 2011 - Should math be taught in schools?

## (b) Wireless data from every light bulb

Suitable for: Everyone - about 13 minutes.

Here's a great idea - get your wireless data more efficiently and safely.

Friday math movie: Wireless data from every light bulb

## (c) How algorithms shape our world

Suitable for: Everyone - about 15 minutes.

Done a Google search today? You've used an algorithm that is helping to change the world.

Friday math movie: How algorithms shape our world

## (d) Trial, error and the God complex

Suitable for: Everyone - about 18 minutes.

How can we cope with the 10 billion-odd bits of data that assault us every day? Tim Harford shows how we learn by trial and error.

Friday math movie: Trial, error and the God complex

## 6. Final thought – powerful weapon

Sadly, a lot of students (and some teachers!) mope around complaining about how everything "sucks" and is "sooo boring". But the world is full of so many great and interesting things - we only have to look - and enjoy!

Ferdinand Foch, French soldier and military theorist of the late 19th century, could have been talking to jaded students when he said:

The most powerful weapon on earth is the human soul on fire.
[Ferdinand Foch]

Until next time, enjoy whatever you learn.

### 12 Comments on “IntMath Newsletter: Golden spiral, \$0 math e-books, CDF format”

1. 3arif says:

He is only tall enough to reach the button of 7th floor.

2. Bill says:

He is a midget.

3. Michael Jordan says:

Our subject is vertically challenged - he can only reach as high as the button for the seventh floor.

4. MykeyLane says:

There is no elevator between the 7th and 10th floors. At the end of the day he also has to walk from the 10th floor to the 7th floor to catch the elevator down to the ground floor. There was no mention that he could catch the elevator down from the 10th floor.

5. Tom Polak says:

He is too short to reach the 10th floor button!

6. Steven Nickerson says:

He's too short to reach the button for the tenth floor. Buttons are arranged vertically from 10 to 1

7. sam says:

Mr shorty!

8. Peter Evans says:

9. Murray says:

@Peter: There are 3 possible situations in Google Books:

(1) The book is in the public domain (out of copyright, and usually old), and you can view all of the book in Google Books and can download it in a variety of formats (eg Here is Treasure Island in available formats). PDF is the simplest for desktop or laptop computers.

(2) The book is copyrighted, but the publishers have given permission for Google to display "some" pages (although I usually find they show "most" pages). There is no download link, but links to the publishers (fair enough, they own the rights)

(3) There is strict copyright, and Google only provides information about the book, and you can't view any of it except maybe a few "snippet" sentences.

Hope that helps!

10. wilson says:

wow the answers here are quite logical, I feel bad for not thinking this way....... but no matter, i want to ask; if the problem was really his height don't you think he would have found a way to overcome it by tagging along with someone who works on the same floor as he does? hence this question won't even exist in the first place. so I'm thinking it's something else, I was thinking he collects the day's newspaper every morning on the 7th floor and decides to walk the rest may be cos the elevator is busy or something.....Of course I'm not debunking anyone's answer since I'm the only one with a different answer here. So can anyone explain to me what i asked initially?

11. Michael Jordan says:

"Yet every morning he only takes the elevator to the 7th floor and walks up the stairs to the 10th floor, even when he is in a hurry."

By this statement I would gather that if he had some sort of alternate method available (i.e. by "tagging along with someone who works on the same floor as he does") he would avail himself of this at least on occasions where he was in a hurry. This leads me to the (rather presumptuous, admittedly) conclusion that he has no other available method, except the one that limits him somewhat.

12. wilson says:

thanx jordan

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## An analogue of Hilbert's Theorem 90 for infinite symmetric groups

Let K be a field and G be a group of its automorphisms. If K is algebraic over the subfield KG fixed by G then, according to Hilbert's Theorem 90, any smooth (i.e. with open stabilizers) K-semilinear representation of the group G is isomorphic to a direct sum of copies of K. If K is not algebraic over KG then there exist non-semisimple smooth semilinear representations of G over K, so Hilbert's Theorem 90 does not hold. Let now G be the group of all permutations of an infinite set S acting naturally on the field k(S) freely generated over a subfield k by the set S. The goal of this note is to present three examples of G-invariant subfields K\subseteq k(S) such that the smooth K-semilinear representations of G of {\sl finite length} admit an explicit description, close to Hilbert's Theorem 90. Namely, (i) if K=k(S) then any smooth K-semilinear representation of G of finite length is isomorphic to a direct sum of copies of K, (ii) if K\subset k(S) is the subfield of rational homogeneous functions of degree 0 then any smooth K-semilinear representation of G of finite length splits into a direct sum of one-dimensional K-semilinear representations of G, (iii) if K\subset k(S) is the subfield generated over k by x-y for all x,y\in S then there is a unique isomorphism class of indecomposable smooth K-semilinear representations of G of each given finite length.


text:
# Limit value using definition of derivative

1. Nov 1, 2011

### karkas

1. The problem statement, all variables and given/known data
I have encountered a limit that I wasn't really sure whether I was solving right, it's limit (a). I thought that I could find it considering a function f(x) (b) , find the derivative of that function and substitute x->1 so that I can find the value of the limit. Is this correct thinking???

2. Relevant equations

(a) lim(x->1) [ sin(x*pi/4) - sin(pi/4) ] / x-1

(b) f(x) = sin(x *pi/4) and so f ' (x) = pi/4 cos(x* pi/4)

3. The attempt at a solution
That would mean that (a) = f ' (1) = pi/4 * (√2)/2 , right?

2. Nov 1, 2011

### lurflurf

yes that is right

3. Nov 1, 2011

### karkas

Ok, excellent, thanks!


text:
# Bounded operator

In functional analysis (a branch of mathematics), a bounded linear operator is a linear transformation L between normed vector spaces X and Y for which the ratio of the norm of L(v) to that of v is bounded by the same number, over all non-zero vectors v in X. In other words, there exists some M > 0 such that for all v in X,

[itex]\|L(v)\|_Y \le M \|v\|_X.\,[itex]

The smallest such M is called the operator norm [itex]\|L\|_{op}[itex] of L.

Let us note that a bounded linear operator is not necessarily a bounded function; the latter would require that the norm of L(v) is bounded for all v. Rather, a bounded linear operator is a locally bounded function.

It is quite easy to prove that a linear operator L is bounded if and only if it is a continuous function from X to Y.

## Examples

• Any linear operator between two finite-dimensional normed spaces is bounded, and such an operator may be viewed as multiplication by some fixed matrix.
[itex]K:[a, b]\times [c, d]\to {\mathbf R}[itex]
is a continuous function, then the operator [itex]L,[itex] defined on the space [itex]L^1[a, b][itex] of Lebesgue integrable functions with values in the space [itex]L^1[c, d][itex]
[itex](Lf)(y)=\int_{a}^{b}\!K(x, y)f(x)\,dx,[itex]
is bounded.
[itex]\Delta:H^2({\mathbf R}^n)\to L^2({\mathbf R}^n)[itex]
(its domain is a Sobolev space and it takes values in a space of square integrable functions) is bounded.
• The shift operator on the space of all sequences (x0, x1, x2...) of real numbers with [itex]x_0^2+x_1^2+x_2^2+\cdots < \infty,[itex]
[itex]L(x_0, x_1, x_2, \dots)=(x_1, x_2, x_3,\dots)[itex]
is bounded. Its norm is easily seen to be 1.

One can prove, by using the Baire category theorem, that if a linear operator L has as domain and range Banach spaces, then it will be bounded. Thus, to give an example of a linear operator which is not bounded, we need to pick some normed spaces which are not Banach. Let X be the space of all trigonometric polynomials defined on [−π, π], with the norm

[itex]\|P\|=\int_{-\pi}^{\pi}\!|P(x)|\,dx.[itex]

Define the operator L:XX which acts by taking the derivative, so it maps a polynomial P to its derivative P′. Then, for

[itex]v=e^{in x}[itex]

with n=1, 2, ...., we have [itex]\|v\|=2\pi,[itex], while [itex]\|L (v)\|=2\pi n\to\infty[itex] as n→∞, so this operator is not bounded.

## Further properties

A common procedure for defining a bounded linear operator between two given Banach spaces is as follows. First, define a linear operator on a dense subset of the domain, such that it is locally bounded. Then, extend the operator by continuity to a continuous linear operator on the whole domain (see continuous linear extension).

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# Resistance

A resistor having an electrical resistance of 1.5 k ohms passes an electrical current of 0.1 A. Calculate what voltage is between the terminals of the resistor.

Result

U =  150 V

#### Solution:

$R=1.5 \cdot \ 1000=1500 \ Ω \ \\ I=0.1 \ \text{A} \ \\ U=R \cdot \ I=1500 \cdot \ 0.1=150 \ \text{V}$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Be the first to comment!

## Next similar math problems:

1. Effective and mean voltage
A voltage divider consisting of resistors R1 = 103000 Ω and R2 = 197000 Ω is connected to the ideal sine wave voltage source, R2 is connected to a voltmeter which measures the mean voltage and has an internal resistance R3 = 200300 Ω, the measured value i
2. Resistance
Determine the resistance of the bulb with current 200 mA and is in regular lamp (230V).
3. Theorem prove
We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?
4. Functions f,g
Find g(1) if g(x) = 3x - x2 Find f(5) if f(x) = x + 1/2
5. 6 terms
Find the first six terms of the sequence. a1 = 7, an = an-1 + 6
6. Volleyball
8 girls wants to play volleyball against boys. On the field at one time can be six players per team. How many initial teams of this girls may trainer to choose?
7. Two equations
Solve equations (use adding and subtracting of linear equations): -4x+11y=5 6x-11y=-5
8. Factory and divisions
The factory consists of three auxiliary divisions total 2,406 employees. The second division has 76 employees less than 1st division and 3rd division has 212 employees more than the 2nd. How many employees has each division?
9. Trees
A certain species of tree grows an average of 0.5 cm per week. Write an equation for the sequence that represents the weekly height of this tree in centimeters if the measurements begin when the tree is 200 centimeters tall.
10. Linear system
Solve a set of two equations of two unknowns: 1.5x+1.2y=0.6 0.8x-0.2y=2
11. One half
One half of ? is: ?
12. Nineteenth member
Find the nineteenth member of the arithmetic sequence: a1=33 d=5 find a19
13. Powers
Express the expression ? as the n-th power of the base 10.
14. Percentage increase
Increase number 400 by 3.5%
15. First man
What is the likelihood of a random event where are five men and seven women first will leave the man?
16. Examination
The class is 21 students. How many ways can choose two to examination?
17. Average
If the average(arithmetic mean) of three numbers x,y,z is 50. What is the average of there numbers (3x +10), (3y +10), (3z+10) ?


text:
# mainul's profile - activity

2014-07-21 14:26:39 -0500 asked a question Changing color space in opencv without changing color of the image

Hi, I am working on a project where I read an image load the data into Mat data type. The I do some operations on it. All my operations are done assuming the color space is RGB (BGR as opencv stores in that way). Everuything is working fine. I was doing experiment on converting the output image to YUV format. But when I transform the output image from BGR2YUV using the following command cvtColor(img,out,CV_RGB2YCrCb); I found that the resulting image color is changed completely.

How can I change the color space from RGB to YUV without changing the colors in the image? For example, my output RGB image is green. WHen I convert this to YUV format and show the resulting image I found it blue and NOT green. I want a way to convert so that the output also become green.

2014-07-07 14:42:01 -0500 asked a question solution to CL_EXEC_STATUS_ERROR_FOR_EVENTS_IN_WAIT_LIST error

Hi,

I am trying to run face detection on a 5 minute 720P video. I am using opencv API and opencl kernels to do my job. When I try to run the program, it runs fine for intial few number of frames. But after, some frames it gives me an error

OpenCV Error: Gpu API call (CL_EXEC_STATUS_ERROR_FOR_EVENTS_IN_WAIT_LIST) in openCLMemcpy2D, file /opencv-2.4.6.1/modules/ocl/src/initialization.cpp, line 466

Sometimes I get this error after processing 200 frames, sometimes I get this error after 2000 frames. Can anyone suggest me why I am facing this error and what could be a possible solution? I searched internet but did not find anythng.

Here is my psuodo code:

Main()
{
for (;;)
{
grab a frame;
detectface(frame);
}
}

detectface(frame)
{
tranfer frame to oclMatrix;

while some conditions are true:
while some conditions are true:
kernel1; //parameter: one oclmatrix for source, one for output, two temporary ocl matrix for calculation
kernel2;//parameter: one oclmatrix for source, one for output, one temporary ocl matrix for calculation
kernel3;//parameter: one oclmatrix for source, one for output, two temporary ocl matrix for calculation
kernel4;//parameter: one oclmatrix for source, one for output, one temporary ocl matrix for calculation
end;
kernel5; //this does the detection and gives a output. parameter: one oclmatrix for source, one for output, 4 temporary ocl matrix for cascade classifier.
end;

}

2014-06-03 11:41:38 -0500 asked a question How to get device list from context

Hi,

I am developing a project using opencv and opencl. Right now, the kernels and the buffer runs using the opencv specified functions. For some reasons, I want to use OpenCL's native function instead of opencv function with least number of changes. I have two files. In the main file I created the context and pass it to the function to the second file. In the second file I need to build and execute kernel. If I would use OpenCV's function I can use execute kernel with the context that has been passed to the second file. But as I am planning to use OpenCL's native one, I need to bild the kernel and for that I need device list. My question is, is there any way I can get the device list from opencv created context? or how can I get the device list without passing the info from the main file?

Here is the snippet of main file:

     vector<ocl::Info> oclinfo;
int devnums = ocl::getDevice(oclinfo);
if( devnums < 1 )
{
std::cout << "no device found\n";
return -1;
}
ocl::setBinpath("./");
cv::ocl::Context* clCxt = ocl::Context::getContext();

ScanKernel( Mat& img,string kernelsrc,string kernelName,cv::ocl::Context* clCxt)

here is the code snippet of the second file:

const char * source = kernelsrc.c_str(); size_t sourceSize[] = { strlen(source) };

cpProgram = clCreateProgramWithSource((cl_context)clCxt->oclContext(), 1, &source, sourceSize, &ciErr1);
if (ciErr1 != CL_SUCCESS) {
printf("Error in clCreateProgramWithSource, Line %u in file %s %d !!!\n\n", __LINE__, __FILE__,ciErr1);

}
else
{       printf("*** Got createprogramwithsource\n");

**ciErr1 =  clBuildProgram(cpProgram, NULL, NULL, NULL, NULL, NULL);**
if (ciErr1 != CL_SUCCESS) {
printf("Error in building, Line %u in file %s error NO: %d!!!\n\n", __LINE__, __FILE__,ciErr1);

}

I need help on getting the device list for clBuildProgram function.

2014-05-11 12:25:11 -0500 asked a question Weird behavior of ocl matrix in opencv

Hi,

I am faciong a qweird problem. I have two matrix, input and output. For each element i in a row r in the input matrix, it sums all element before i in that row and put the sum in i th column and r th row in output matrix. When I do this for small matrix, it gives me no problem. But for a big matrix (604x454), the output matrix contains GRARBAGE value in all output matrix rows except the first row.

"__kernel void "
"  v2_integral_cols_sum(__global uchar *src,\n"
"int rows,int cols,__global int *lm_sum, int pixels,int steps,int o_steps)\n"
"{\n"

"int gid=get_global_id(0);\n"
"if(gid>=pixels)"
"return;\n"
"else \n"
"{"
"int x = gid % steps;\n"
"int y = gid / steps;\n"
"int sum=0;\n"

"for (int i=0;i<=x;i++)"
"{\n"
"sum=sum+src[y * steps + i];\n"
"}\n"

"lm_sum[y*o_steps + x]=sum;\n"

"}"

"}\n"

interestingly if I use the printf statement after the output assignment statement as in this block,I get correct output matrix. I do not find any explanation of this weird behavior. Anyone can help me?

"lm_sum[y*o_steps + x]=sum;\n"
"if(gid==640)\n"
"{"

"printf(\"lm_sum [%d %d %d] %d  \",x,y,gid,lm_sum[y * o_steps+ x]);\n"

"}\n"

2014-04-24 11:02:06 -0500 asked a question data transfer between ocl matrix of type CV_32FC1 and CV_8UC1

I have created to ocl matrix of same size (rows and columns) bu different types. In my opencl code I am trying to transfer the data from CV_8UC1 matrix to CV_32FC1 matrix. I can see the values are correct when I used printf from kernel. But when I downloaded the output ocl matrix and printed the values in file I got wrong values. Here is my kernel:

"__kernel void " " v2_integral_cols_sum(__global uchar *src,\n" "int rows,int cols,__global int *lm_sum, int pixels,int steps,int o_steps)\n" "{\n"

                "int gid=get_global_id(0);\n"
"if(gid>=pixels)"
"return;\n"
"else \n"
"{"
"int x = gid % steps;\n"
"int y = gid / steps;\n"
"int sum=0;\n"

"for (int i=0;i<=x;i++)"
"{\n"

"sum=sum+src[y * steps + i];\n"
"}\n"

"lm_sum[y * o_steps + x]=sum;\n"

"}"

"}\n"

Here is the host code:

ocl::oclMat lm_sum_mem; lm_sum_mem.create(src.rows,src.cols, CV_32FC1); args.push_back( make_pair( sizeof(cl_mem) , (void *)&src.data )); args.push_back( make_pair( sizeof(cl_int) , (void *)&src.rows )); args.push_back( make_pair( sizeof(cl_int) , (void *)&src.cols )); args.push_back( make_pair( sizeof(cl_mem) , (void *)&lm_sum_mem.data)); args.push_back( make_pair( sizeof(cl_int) , (void *)&npixels)); args.push_back( make_pair( sizeof(cl_int) , (void *)&src.step )); args.push_back( make_pair( sizeof(cl_int) , (void *)&lm_sum_mem.step ));

            size_t gt[3] = {1728, 1, 1}, lt[3] = {32, 1, 1};

if(calTime)
{
//double dtKERNL1=get_wall_timelbp();
gettime_tocalc2[0]=get_wall_timelbp();
}

//ocl::openCLExecuteKernel(clCxt, &lbpobjectdetect, "v2_integral_cols", gt, lt, args, -1, -1,build_options);
ocl::openCLExecuteKernel(clCxt, &lbpobjectdetect, "v2_integral_cols_sum", gt, lt, args, -1, -1,build_options);

Here is the file printing: Mat t_sumhost; lm_sum_mem.download(t_sumhost); std::stringstream strFname3; printf("loop count %d",loopCount); strFname3 << "Integral_Cols"<<loopcount<<".txt"; const="" char*="" filenames="strFname3.str().c_str();" file="" *fp="fopen(filenames," "w");<="" p="">

            fprintf(fp,"Columnn %2d Rows: %d\n: ", t_sumhost.cols,t_sumhost.rows);
for (int i = 0; i < t_sumhost.rows; i++)
{
fprintf(fp,"\nrow %2d: ", i);
for (int j = 0; j < t_sumhost.cols; j++)
{
fprintf(fp," %5d ", t_sumhost.at<int>(i,j));
}
}
fclose(fp);
`
2014-03-21 11:34:42 -0500 commented question How the lbpcascade_frontalface.xml work?

@StevenPuttemans I tried to search for answers. And, I could not get two clear answers on how lbpcascade_frontalface.xml works let alone 100 answers. May be you understand the algorithm and usage of the xml file very well. For you, answering this question might be a time waste. But I do not understand the algorithm and how the cascade (nodes, leaves, ntrees, features etc) are used in the algorithm. For me, finding answers for this question is top most priority. If you can please help me.

2014-03-21 10:40:07 -0500 commented answer LBP based Face Detection

Can you please provide some link on LBP papers or documents that gives a better understanding of features and face detection?

2014-03-21 10:03:38 -0500 asked a question LBP algorithm of OpenCV

HI Everyone,

I am working on LBP based face detection on openCV. I wanted to understand the algorithm clearly but I could not find any clear documents on these algorithm or their implementation. I have some very specific questions. Here they are:

1. In lbpCascade_frontal_face.xml: what is maxCatCount, maxWeakCount, minHitRate, maxFalseAlarm, weightTrimRate, maxDepth, internalNodes, leafValues, tree etc? I know they are used in the algorithm. But I do not understand the meaning of each one of them. For example, why we take a particular feature and not the other for a particular stage? how we are deciding which feature/ node to choose?

2.What is feature values in the LBP_frontal_face_classifier.xml? I know they are a vector of 4 integers. But how should I use this features? I though stage 0 access to the first feature but access is not in this pattern. What is the access pattern to this features?

1. All the papers in literature give a high level overview only. Their descriptions mainly consist of LBP calculation from neighborhood pixels. But how this LBP values is used against those elements in the classifier?

2. How does intergral image help in calculating LBP value of a pixel? Please do not give reference/answer using HAAR features. I know how HAAR is used. I need to understand LBP.

I read some papers, articles. But none clearly describes how LBP based face detection works or the algorithm in details. If someone wants to develop a face detection program on his own,what are the steps he should follow- no document describes that.

If anyone can help me with the answers it would be a great help for me. -you can answer the questions or you can direct me to someone who knows the answers or you can give me some link of articles/documents/papers that would help me to understand this algorithm.

-Hassan

2013-10-30 09:53:13 -0500 asked a question OpenCV compilation error with OpenCL 2.4.6.1

Hi,

I compiled OpenCV 2.4.6.2 from git earlier with opencl enabled for arm architecture and everything worked fine. Then I changed in the initialization file and added some printf statements and tried recompiling the opencv. But could not compile and got a number of errors. Then I downloaded another opencv from sourgeforge (version 2.4.6.1) and tried to compile it. It compiled successfully. But when I tried to recompile it with OpenCL-after 87% of compilation I got this error:

## /opencv/modules/ocl/src/imgproc.cpp: In function 'void cv::ocl::copyMakeBorder(const cv::ocl::oclMat&, cv::ocl::oclMat&, int, int, int, int, int, const Scalar&)': /opencv/modules/ocl/src/imgproc.cpp:748:9: internal compiler error: output_operand: invalid expression as operand Please submit a full bug report, with preprocessed source if appropriate. See <file: usr="" share="" doc="" gcc-4.6="" readme.bugs=""> for instructions. Preprocessed source stored into /tmp/cconMG0K.out file, please attach this to your bugreport. make[2]: * [modules/ocl/CMakeFiles/opencv_ocl.dir/src/imgproc.cpp.o] Error 1 make[1]: [modules/ocl/CMakeFiles/opencv_ocl.dir/all] Error 2 make: ** [all] Error 2

I have not changed any file. It was a fresh copy of opencv. But I do not know what's wrong. However I tried to download a copy from git but the git version is so changed that I have to do a lot of things/changes in my code to cope with the changes made in git version.

Can anyone kindly show me a path why I am getting that error and how to solve that?

2013-10-30 09:45:29 -0500 commented answer Passsing size of a parameter to the kernel

I did it in different way. I changed the kernel string before execution. And inside kernel I used a macro for size.

2013-10-30 09:44:37 -0500 commented answer Opencv OpenCL Compilation error: x was not declared in this scope

I do not know how to mark it as solved. I dont see any icon for that.

2013-09-26 13:10:36 -0500 asked a question Printing from inside module/src/initialization.cpp file

Hi, I am running codes with a opencl kernel on an arm architecture. When I run the program, it stops while calling kernel and gets timeout. I would like to know which function related to kernel is getting the error. So I want to have some printf statements inside the initialization file. I added some lines and recompile opencv but while running the program nothing is printed from initialization file. can anyone help me how I can make initialization file to print something?

2013-09-26 13:06:34 -0500 commented answer Opencv OpenCL Compilation error: x was not declared in this scope

Thank you. I passed the location of opencl while compiling the codes.

2013-09-26 12:52:42 -0500 commented answer Best way to add new projects/files using Opencv and OCL

Thanks for your answer. I did it in another way. I will do that in your way too. On a different note I have one question. I am running codes on an arm architecture. It stops while calling kernel and got timeout. I would like to know which function is getting the error. So I want to have some printf statements inside the initialization file. I added some lines and recompile opencv but when running the program nothing printed from initialization file. Do you know how I can make initialization file print something?

2013-09-24 12:09:15 -0500 asked a question Best way to add new projects/files using Opencv and OCL

Hi,

I have written some custom codes for facedetection that uses Opencv's libraries and opencl function. I copied my files in the sample folders and tried to compile but I got several errors. Most of them occurs because it cannot link the OpenCV files and OPencl files. So, I manually put the full path of the files in the headers of my files as well as Opencv's files. For example, I got an error that cl_mem is not declared in the scope. To overcome this I had to put the full path of CL.h to all the dependent files. This process worked and the program compiled successfully.

But I think this is not a good approach to manually put the path of dependent files in the header. The samples provided by opencv can easily link all the necessary files. So, there must be way that will automatically link any new files with opencv and opencl's dependent files.

So, I need your suggestion regarding the best way of adding new projrcts/files so that they can automatically link all the OpenCL and OpenCV files.

2013-09-22 18:28:45 -0500 asked a question Opencv OpenCL Compilation error: x was not declared in this scope

Hi I created a program using OpenCV in windows with visual studio. My program has two cpp files. Both of them include many library files. The main function is in detect.cpp file and the other file is lbp.cpp. In windows the program runs just fine and can get every library files and header files. But when I tried to compile the program in Linux it did not compile and gave me error. The file cannot find opencl's function and data type. It gives me this error:

‘cl_command_queue’ was not declared in this scope ‘cl_mem’ was not declared in this scope ‘cl_int’ was not declared in this scope

There are many of them. Anyone has an idea how to solve them? I checked other sample files and used the same headers as they used. But my program just does not compile. Please help.

2013-09-19 15:55:55 -0500 commented question Getting error while compiling opencv sample program

oh I forgot to mention that. I am using Ubuntu 12.04. I followed these two pages to install: http://docs.opencv.org/doc/tutorials/introduction/linux_install/linux_install.html http://www.samontab.com/web/2012/06/installing-opencv-2-4-1-ubuntu-12-04-lts/

Then I tried to compile and run the sample codes. I tried to run the CPP ones also but got the same error.

2013-09-19 14:09:04 -0500 asked a question Getting error while compiling opencv sample program

Hi,

I have installed opencv following the steps mentioned in opencv wiki. Then I tried to compile and build the sample programs provided in the sample folder I got error. I tried to run my code but that did not work so I tried to run the sample code first and I got the following error:

rt: No such file or directory pthread: No such file or directory m: No such file or directory dl: No such file or directory

I removed those rt,pthread,m,dl from opencv.pc file. Then I tried to compile the sample codes again. Now I got many more errors. For example:

compiling bgfg_codebook.cpp bgfg_codebook.cpp: In function ‘int main(int, const char*)’: bgfg_codebook.cpp:108:39: error: ‘cvCaptureFromCAM’ was not declared in this scope bgfg_codebook.cpp:113:57: error: ‘cvCreateFileCapture’ was not declared in this scope bgfg_codebook.cpp:128:46: error: ‘cvQueryFrame’ was not declared in this scope bgfg_codebook.cpp:145:37: error: ‘cvNamedWindow’ was not declared in this scope bgfg_codebook.cpp:171:42: error: ‘cvShowImage’ was not declared in this scope bgfg_codebook.cpp:177:25: error: ‘cvWaitKey’ was not declared in this scope bgfg_codebook.cpp:235:32: error: ‘cvReleaseCapture’ was not declared in this scope bgfg_codebook.cpp:236:28: error: ‘cvDestroyWindow’ was not declared in this scope compiling blobtrack_sample.cpp blobtrack_sample.cpp: In function ‘int RunBlobTrackingAuto(CvCapture, CvBlobTrackerAuto, char, char*)’: blobtrack_sample.cpp:140:64: error: ‘cvWaitKey’ was not declared in this scope blobtrack_sample.cpp:152:33: error: ‘cvQueryFrame’ was not declared in this scope blobtrack_sample.cpp:175:50: error: ‘CV_FOURCC’ was not declared in this scope blobtrack_sample.cpp:177:27: error: ‘cvCreateVideoWriter’ was not declared in this scope.

I installed and reinstalled opencv three times. But still no luck. I tried to run the sample codes from this location: Opencv/samples/c/

I do not know what to do now. Can anyone help me with some idea?

2013-09-18 13:40:33 -0500 commented question Compiling prgram with OpenCV in Linux

The original haar.cpp file has that header file and as my lbp.cpp also quite similar to haar I kept it. However, removing the header line or placing lbp.cpp file in teh same location as haar.cpp gives me some different error. It says that some functions are not declared in the scope. My question is if you have a main file and a function file both of which use OpenCV and OpenCL, how will you compile that code? Any help will be appreciated.

2013-09-18 12:24:51 -0500 asked a question Compiling prgram with OpenCV in Linux

Hi I created a program using OpenCV in windows with visual studio. My program has two cpp files. Both of them include many library files. The main function is in detect.cpp file and the other file is lbp.cpp. In windows the program runs just fine and can get every library files and header files. But when I tried to compile the program in Linux it did not compile and gave me error. For simplicity I put the both the cpp files in the same directory and placed this directory in opencv samples folder. But when I tried to compile I got this error:

fatal error: precomp.hpp: No such file or directory

But this file is there in /modules/core/src folder. I am not sure why it cannot find the file. How can I pass the file location while compiling? can anyone suggest me some tutorial how I can compile my code? I searched some articles but they are not helping.

Thanks

2013-09-16 18:15:01 -0500 asked a question Passsing size of a parameter to the kernel

Hi,

I have a kernel that has a local variable __local int* sbuff as parameter. I want to set its size dynamically. In normal opencl we can do that using setkernelArg function. But in OpenCV they provides a wrapper to execute kernel and passing parameter. For example, we use openCLExecuteKernel to execute kernel and args.push_back to pass parameter.

My question is how I can pass the size of the local variable to the kernel using OpenCV's functions?

2013-09-05 13:33:20 -0500 commented answer Problems with the opencv document

ok, I will. I thought this topic has answer that contains area that resembles mine, so I asked here thinking that I might get an answer quickly. I will ask a new question.

2013-09-05 13:05:47 -0500 commented answer Problems with the opencv document

What is CV_EXPORTS? I see this macros in so many places. But could not really understand what it is. There is no information about it also. I also want to know what NCV_STATUS and NCV_EXPORTS is? what is the difference between CV and NCV? can anyone enlighten me?

2013-09-05 13:05:17 -0500 answered a question Problems with the opencv document

What is CV_EXPORTS? I see this macros in so many places. But could not really understand what it is. There is no information about it also. I also want to know what NCV_STATUS and NCV_EXPORTS is? what is the difference between CV and NCV? can anyone enlighten me?

2013-09-05 10:20:56 -0500 asked a question What is nppiStIntegralGetSize_8u32u , NCV, NCV_Exports

Hi, I have starting working on OpenCV recently. While working on it, I stumbled upon some functions about which I know nothing about. I was checking the LBP facedetect codes for CUDA. There I found a function nppiStIntegralGetSize_8u32u. I found no implementation of this file or any substantial information on it. Can anyone tell me what this is? what is NCV or NCV Exports. Please let me know something.


text:
# Difference between sodium ion and a transition metal ion dissolving in water?

In class, I learned that when $\ce{NaCl}$ is dissolved in water, the partially negative oxygen part of the water molecules surround the $\ce{Na+}$ ion and energy is given off (hydration enthalpy)

Likewise, when a transition metal ion like $\ce{Fe^2+}$ is dissolved in water, water molecules surround the ion. However, in this case, we say that a complex ion is formed.

Can someone please tell if there is really a difference in the two cases and if yes, could someone specify them.

• I think that in transition metal complexes the donor atom (the center transition metal ion in the complex) has a large amount of ligands and that sodium has much less ligands (however (Na(NH3)6)+ is a complex) – user2117 Dec 11 '13 at 12:55

There is a slight difference between solvated ions and aqua complexes: The nature of the bond.

When solvating a sodium ion, the electron pairs of the oxygen in water never really interact with the sodium in a way that creates a bond of some sort: It's all just electrostatic interactions.

However, when a transition metal ion gets surrounded with these water molecules, the lone pair electrons of the oxygen form a bond with the ion. This is because they interact with the $d$ orbitals of the transition metal (which usually also contains some electrons), thus changing the energetic state of the ion itself. First and second group elements don't have that energy-stabilizing effect because they don't have electrons in the orbitals that get split up by the formation of the complex.

It is not unimaginable that there then exist further electrostatic interactions between the transition metal aqua complex and the surrounding water molecules, thus leading to the solvation of said complex.


text:
Show that in the sequence $b,2b,3b…,mb$ there are exactly $\gcd(b,m)$ numbers divisible by $m$.

"Where $b$ and $m$ are integers and $m>1$, show that in the sequence $b,2b,3b...,mb$ there are exactly $\gcd(b,m)$ numbers divisible by $m$.

I'm having a real hard time proving that... can anyone help me?

• $m$ needn't be bigger than one: when $m = 1$, in the sequence $b, 2b, 3b, \dots mb$, there is exactly 1 integer, and it is divisible by 1 :) – Ben Millwood Oct 22 '14 at 22:22

The given sequence is the first $m$ multiples of $b$. Scanning from left to right, when will we find the the first multiple of $b$ that is also a multiple of $m$? By definition, this will occur at $\ell = \text{lcm}(b, m)$. Letting $\ell = kb$, notice that $\ell$ is the only multiple of $m$ in the first $k$ multiples of $b$.

Now observe that the next common multiple of $b$ and $m$ will be $2\ell = 2kb$, which shows up after listing out $k$ more multiples of $b$ after $\ell$. Now since $\ell$ divides $mb$, we can continue in this manner so that the list of all common multiples of $b$ and $m$ is: $$\ell, 2\ell, \ldots, g\ell$$ where $g\ell = mb$. [Remark: At this step, we are using the fact that the least common multiple of $m$ and $b$ must divide any common multiple of $m$ and $b$.] But then since: $$\gcd(b, m) \cdot \text{lcm}(b, m) = mb$$ we have that $g = \gcd(b, m)$, as desired.

• oh, interesting, thank you! i'm curious, though, the book i'm using places this question before the question that asks you to prove that gcd(a,b) * lcm(a,b) = |a*b|. i wonder what's the other way to do it... – violeta Oct 22 '14 at 23:02
• also, b can be negative... hmm.. – violeta Oct 22 '14 at 23:31

So the sequence is the following:

$$b,2b,3b,..,mb$$

Let $g= \gcd(b,m)$ , so $ng=m$ and $gd=b$

Lets consider the following members of the sequence:

$$I= \{x \in \mathbb{N} : x=bjn =\frac{bjm}{g}\ \text{ such that } j \in \mathbb{N} \land 0<j\leq g \}$$

So all the members $I$ are divisible by $m$. The number of members is equal to $g$. Now I want to prove that this are the only members of the sequence that are divisible by $m$.

Now all the members that are not in $I$ are of the form:

$$S=\{t \in \mathbb{N} : t=b(jn+c) =\frac{bjm}{g}+bc \ \text{ such that } j \in \mathbb{N} \land 0<j< g \land 0<c<n \}$$

$$m|t \text{ such that } t\in S$$ $$m|b(jn+c)$$

by definition:

$$mk=jbn+bc \tag{1}$$ $$mk=jb\frac{m}{g} + bc$$ $$m(k-\frac{jb}{g})=bc$$ $$k-jd=\frac{bc}{m} \tag{2}$$ $$k=\frac{dc}{n}+jd$$ $$kn=njd+dc$$ $$kn=d(jn+c)$$

Now $d|k$ by the Euclid Lemma, since $(n,d)=1$ . This will be useful in a moment:

Now as $c<n$ we can obtain: $$\frac{c}{m}<\frac{1}{g} \Rightarrow \frac{bc}{m}<d$$

From $(2)$we can obtain:

$$k-jd<d$$ $$k<d+dj$$ $$k<d(1+j) \tag{3}$$

As $d|k$ then $d\hat{k}=k$ , replacing in $(3)$

$$d\hat{k}<d(1+j) \Rightarrow \hat{k}<(j+1)$$

So $\hat{k}\leq j$ then $mk\leq djm$, now in $(1)$

$$djm \geq jbn+bc$$ $$djm \geq djm+bc$$ $$0 \geq bc$$

As $b$ and $c$ are positive, $bc >0$, so contradiction.

So no element of $S$ is divisible by $m$, now only the elements of $I$ are divisible by $m$. As the number of members is $g= \gcd(b,m)$ , the statement is proven.


text:
How do you write the standard form equation of the parabola  y = x^2 - 6x + 14?

Feb 9, 2017

$y = {\left(x - 3\right)}^{2} + 5$

Explanation:

The equation is already in standard form, seeing as it is ordered from higher to lower exponent. However, to find the vertex form of the equation, we need to convert the $x$-containing terms into a factorable expression, namely the square of a difference. We need to remember that:

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

and vice versa. Now, let's take a look at the terms containing an $x$ in the equation:

${x}^{2} - 6 x$

Not as much as a perfect square, but if $a = x$, then the second term will be $2 x b$, so $b$ is $3$. To complete the square, we require a "$+ {b}^{2}$" in the end, so we need to add $9$.

However, we can't just do that, since that changes the equation. To "keep the balance", we need to also subtract $9$. This becomes:

$y = {x}^{2} - 6 x + \textcolor{red}{9} + 14 - \textcolor{red}{9}$.

Using the expression above, factor the first three terms into a difference of squares, then do the subtraction $14 - 9 = 5$:

$y = {\left(x - 3\right)}^{2} + 5$

In general, the vertex form helps us find the vertex of a parabola. If the vertex form is

$y = a {\left(x - b\right)}^{2} + c$, where $a$ is the "coefficient" of ${x}^{2}$ in the original equation, then the vertex is the point $\left(b , c\right)$.


text:
# Transfer Function - Opamp

My work:

$$\text{I}_{\text{in}}(t)+\text{I}_{\text{p}}(t)=0\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\text{I}_{\text{in}}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}$$

$$\text{I}_{\text{p}}(t)=\text{I}_{\text{C}}(t)+\text{I}_{\text{R}_2}(t)\Longleftrightarrow$$ $$\text{I}_{\text{p}}(t)=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}$$

$$-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}=\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\Longleftrightarrow$$ $$\mathcal{L}_t\left[-\frac{\text{V}_{\text{in}}(t)}{\text{R}_1}\right]_{(s)}=\mathcal{L}_t\left[\text{C}\text{V}'_{\text{out}}(t)+\frac{\text{V}_{\text{out}}(t)}{\text{R}_2}\right]_{(s)}\Longleftrightarrow$$ $$-\frac{\text{V}_{\text{in}}(s)}{\text{R}_1}=\text{C}s\text{V}_{\text{out}}(s)+\frac{\text{V}_{\text{out}}(s)}{\text{R}_2}\Longleftrightarrow$$ $$\color{red}{\frac{\text{V}_{\text{out}}(s)}{\text{V}_{\text{in}}(s)}=-\frac{\text{R}_2}{\text{R}_1\left(1+\text{C}\text{R}_2s\right)}}$$

• Yes - looks good. However, there is a simpler (and quicker) way for arriving at the transfer function: Calculate from the beginning in the s-domain (without the necessity to perform the Laplace transformation).
– LvW
Feb 20, 2016 at 16:34
• @LvW Thanks for your response, can you show me that method? Feb 20, 2016 at 17:10
• @JanEerland Here's a brief introduction s-domain circuit analysis. Basically, you can transform R, L, and C into complex impedances in terms of s and then all the math becomes much easier. Feb 20, 2016 at 23:00

The solution you have arrived at is correct. The circuit is a practical integrator. The resistor in parallel with capacitor limits low frequency gain and minimizes variations in output. Here is a simpler and quicker solution:

Since the opamp is in inverting configuration, the transfer function is:

$$Av = \frac{-Z2(s)}{Z1(s)}$$ Note that all impedances are in s-domain. Z2(s) happens to be the parallel combination of R2 and 1/sC $$Z2(s) =\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}$$ $$Z1(s) =R1$$ $$\frac{vo(s)}{vin(s)} = -\frac{\frac{R2\cdot\frac{1}{sC}}{R2+\frac{1}{sC}}}{R1}$$

Which on simplification reduces to: $$\frac{vo(s)}{vin(s)} = -\frac{R2}{R1\cdot(1+R2Cs)} = -\frac{\frac{R2}{R1}}{1+R2Cs}$$

So the gain at low frequencies is -R2/R1 which without R2 would have quickly rolled off.


text:
# networkx.algorithms.coloring.strategy_random_sequential¶

strategy_random_sequential(G, colors, seed=None)[source]

Returns a random permutation of the nodes of G as a list.

G is a NetworkX graph. colors is ignored.

seedinteger, random_state, or None (default)

Indicator of random number generation state. See Randomness.


text:
Automated counting applications for production lines are designed and developed to track, identify, separate and count products, and all sorts of objects in a bounded image area, and provide fast and highly accurate results. Many of today’s systems are built following traditional approaches to image processing which lowers their efficiency and accuracy. The automated counting system developed by A-Grade IT dedicated developers team for the Bakery factory is designed to locate, identify and count objects in the input image we get from video processing. One of the major advantages of this system is that it’s capable of separating touching products prior to counting, thus the results are highly accurate 99.5%.

Today’s post from one of our image processing experts describes the problem of traditional approaches and introduces a new one that eliminated  the issues that would arise otherwise, and touches upon other areas it could be applied to identify and count products.

Generally speaking image processing is considered as any form of signal processing with an image input, such as photo or video frames. The counting problem in image/video processing is the estimation of the number of objects in a still image or video frame. It arises in many real-world applications including cell counting in microscopic images, monitoring crowds in surveillance systems, or in our case it’s a problem of counting the baked products on the production lines.

The initial version of the production counting system was designed and developed in MATLAB, and had certain limits and restrictions.

1. Rather low counting accuracy that was dropping even more, if the bakery products were touching and in uneven rows on the production line.
2. Poorly optimized, the system required too many resources.
3. Limited to processing only 4 video channels at once.

So, how can we solve it to get an accurate result?

Fig.1. Montage image built from first 25 video frames

## I. Traditional image processing approach

Here we cover object counting following a traditional approach to image processing, and what challenges we may face, if we were to do it.

The first thing we need to do is to separate foreground and background. The easiest way would be to use color segmentation, however then we need to resolve the problem of uneven illumination. To avoid it, it is better to apply adaptive thresholding with Otsu segmentation algorithm for simple cases, using RGB channel. As a result, we get a binary image (where every pixel equals 1 to refer an object, and 0 otherwise). This binary image will have a lot of artificial objects with a little square (about 1-10 pixels) that is why we have to remove them using median filtering or methods of the mathematic morphology.  In our case, for counting objects on the production line, the morphological opening operation will be a better choice.

At this stage, it is crucial to separate all objects into individual 4-connected areas. We define a pixel set as a 4-connected area where every pixel has at least one northern, southern, eastern or western neighbor.

Now that we have cleared the binary image that corresponds to the given frame, we move to the issue of counting the objects themselves. If we were to have one video frame, we could use a wave algorithm to count the products. However, we need to count them during a certain time; hence every object will be present at several sequent frames. Therefore, we have to count every object that crosses a certain imaginary line which is perpendicular to the moving direction.

Fig.2. Image processing:

a) source color image; b) red channel of the source image;

c) Otsu threshold applied; d) cleared binary image

The main challenge for the traditional approach is to identify every object and track it to determine whether it has crossed given a line or not. In other words, instead of counting the objects we need to track the object trajectories. And to do that you can apply one of the standard tracking techniques, Kalman filter etc., though it will require large computational resources and results are not that accurate. For example, the change in illumination causes errors in tracking or object recapturing. Those are clear disadvantages of the standard approach.

Sometimes they suggest using rectangular areas of interest instead of lines. The rectangle height is bigger than the object height, allowing us to count objects that are inside of the area completely. Still, this method works for counting similar objects that lay in the single row only. Therefore, we have to look for another way to count the objects on the production line.

Fig.3. A line of counting (a line of interest)

## II. Our approach to image processing for object counting

Let’s consider the video as a sequence of frames which are arranged in a stack of paper sheets. As it was mentioned before, we have to count every object that crosses the given line.  Therefore, only pixels that are on this line for each frame should be taken into account. That is, we need to consider the frames cross-sectioning along the line. This method is known as a slit-scan camera algorithm and is widely used in the sport for registering a winner crossing the finish line, as well as for artistic purposes.

### How does it work?

1) During a certain period of time, we just copy pixels along the given line from every frame and append them to the resulting image. Thus, we should place the camera in a way that would record objects moving strictly horizontally or vertically. In this case, it is enough to copy a row (or column, respectively) from every frame.

Fig.4. Slit-scan image formation

From the object counting perspective the resulting image is equivalent to the source video and can be considered as its hash, and it is something quite peculiar.

Fig.5. Resulting image

2) The binary image is built from the resulting image in a way that was described above. Now it is easy to estimate background as a mean or median for each resulting image column.  Every column is the image of the same point, but in different points in time. To clear binary image and separate every object we can use the usual way.

Fig.6. Image processing: a) resulting color image; b) red channel of the resulting image; c) Otsu threshold applied; d) cleared binary image

3) Now we have to find 4-connected areas amount using a wave algorithm.

Fig.7. Labeled 4-connected areas

And it wraps it up! We have a solution for objects counting on production lines.

In practice, though, we face some additional problems such as stitching of the resulting images. It is necessary to take into account the objects that are between resulting images. This task is solved efficiently using mathematical morphology, too.

Fig.8. Image stitching example:

a, b) two sequential resulting images, parts to be stitched are marked by color; c) result of the stitching

## Conclusion

1. Low computational cost
2. High accuracy: the counting accuracy reaches 99.5% , as it is capable of separating touching products in uneven rows
3. Flexible and universal solution: our automated counting system can be applied to counting other manufactured goods on the production lines, including Pharmaceutical products, Food and beverages, Can counting, Part and component counting, etc.

Overall, the system developed with this approach proved to be simple, flexible and highly accurate and could be applied to solve issues in other areas, such as road traffic identification and monitoring. And here is a little sneak peek into our future post that touches upon video processing, and how can we apply it to road traffic analysis.

Fig.9. Example of video processing for road traffic:

a) a video frame with the red line as a line of interest;

b) resulting image; c) background image estimation; d) detected foreground objects, namely cars, other objects were disregarded.

If you found this interesting, or have some questions to our expert Alex M., you can ask them in our social media listed below.

$${}$$


text:
# A circle has a center that falls on the line y = 8/7x +2  and passes through  ( 7 ,8 ) and (3 ,9 ). What is the equation of the circle?

Jul 19, 2016

$20 {x}^{2} + 20 {y}^{2} - 189 x - 296 y + 1431 = 0$. graph{20x^2+20y^2-189x-296y+1431=0 [-10, 10, -5, 5]}

#### Explanation:

Suppose that eqn. of circle is, $: S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$.

We know that the Centre $C$ of $S$ is $C \left(- g , - f\right)$, and, it is given that $C$ lies on the line $: y = \frac{8}{7} x + 2$, giving,

$- f = - \frac{8}{7} g + 2 , \mathmr{and} , f = \frac{8}{7} g - 2. \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

We are also given that the pts.$\left(7 , 8\right) \mathmr{and} \left(3 , 9\right)$ lie on $S$. Hence, these co-ords. must satisfy the eqn. of $S$. Accordingly,

$49 + 64 + 14 g + 16 f + c = 0. \ldots \ldots \ldots \ldots \left(2\right)$, and,

$9 + 81 + 6 g + 18 f + c = 0. \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$.

We Solve eqns. $\left(1\right) , \left(2\right) , \mathmr{and} , \left(3\right)$.

$\left(2\right) - \left(3\right) \Rightarrow 23 + 8 g - 2 f = 0 \Rightarrow 2 f = 8 g + 23. \ldots \left(4\right)$.

By $\left(1\right)$, then, $2 \left(\frac{8}{7} g - 2\right) = 8 g + 23 , i . e . , \frac{16}{7} g - 8 g = 23 + 4$

$\therefore - \frac{40}{7} g = 27 \Rightarrow g = - 27 \cdot \frac{7}{40} = - \frac{189}{40}$.

Using $\left(1\right) , f = \frac{8}{7} \cdot \left(- 27 \cdot \frac{7}{40}\right) - 2 = - \frac{27}{5} - 2 = - \frac{37}{5}$.

Finally, $\left(3\right)$ gives, $90 - 6 \cdot \frac{189}{40} - 18 \cdot \frac{37}{5} + c = 0$.

$\therefore c = 3 \cdot \frac{189}{20} + 18 \cdot \frac{37}{5} - 90 = \frac{567}{20} + \frac{666}{5} - 90$.

$\therefore c = \frac{567 + 2664 - 1800}{20} = \frac{1431}{20}$

Hence, $S : {x}^{2} + {y}^{2} - \frac{189}{20} x - \frac{74}{5} y + \frac{1431}{20} = 0$,

$S : 20 {x}^{2} + 20 {y}^{2} - 189 x - 296 y + 1431 = 0$.


text:
## Stackelberg Pricing is Hard to Approximate within 2−ε

Author Parinya Chalermsook, Bundit Lekhanukit, Danupon Nanongkai

Conference:

Abstract:

Stackelberg Pricing Games is a two-level combinatorial pricing problem studied in the Economics, Operation Research, and Computer Science communities. In this paper, we consider the decade-old shortest path version of this problem which is the first and most studied problem in this family. The game is played on a graph (representing a network) consisting of fixed cost edges and pricable or variable cost edges. The fixed cost edges already have some fixed price (representing the competitor’s prices). Our task is to choose prices for the variable cost edges. After that, a client will buy the cheapest path from a node $s$ to a node $t$, using any combination of fixed cost and variable cost edges. The goal is to maximize the revenue on variable cost edges.

In this paper, we show that the problem is hard to approximate within $2-\epsilon$, improving the previous APX-hardness result by Joret [to appear in Networks]. Our technique combines the existing ideas with a new insight into the price structure and its relation to the hardness of the instances.

Update History

[v1] Oct 2, 2009 (Manuscript posted on arXiv)


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# Hardness of Approximation between P and NP

Wednesday, November 30, 2016
12:00 PM - 1:00 PM
Location: Baxter B125
Aviad Rubenstein, Computer Science PhD Candidate, UC Berkeley

The first question we computer scientists ask when facing a new algorithmic challenge is: is it NP-hard, or is it in P? Surprisingly, for many important problems, the answer is "neither!" Rubenstein will discuss recent progress towards understanding the complexity of those problems.

Series: Linde Institute/Social and Information Sciences Laboratory Seminar Series (SISL)


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# Comprehensive Guide on Sample Mean

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To estimate the population mean, we can collect some observations from the population to form a sample. We would expect the average of these observations, which is called the sample mean, to be a good estimate of the true population mean.

Theorem.

# Sample mean

The mean of a sample of observations $(x_1,x_2,...,x_n)$ is computed by:

$$\bar{x}=\frac{1}{n} \sum^n_{i=1}x_i$$

$\bar{x}$ is known as the sample mean.

Example.

## Computing the sample mean

Compute the sample mean of the following sample:

$$(2,4,8,2)$$

Solution. The size of the sample is $n=4$. Using the formula for the sample mean:

\begin{align*} \bar{x}&=\frac{1}{4}\sum^4_{i=1}x_i\\ &=\frac{1}{4}(2+4+8+2)\\ &=4 \end{align*}

Therefore, the mean of our sample is $4$.

# Samples with different distribution can have the sample mean

The sample mean is a measure of the central tendency of a sequence of values. However, the sample mean itself is not sufficient to describe the distribution of the values since vastly different distributions can have the same sample mean. For instance, consider the following frequency distribution of two samples:

Here, their distributions look different but they have the same sample mean. Therefore, when numerically describing a sample, we should additionally quote other statistical measures such as the sample variance.

# Properties of the sample mean

Definition.

## Expected value of the sample mean

The expected value of the sample mean $\bar{X}$ is equal to the population mean $\mu$, that is:

$$\mathbb{E}(\bar{X})=\mu$$

Equivalently, we say that the expected value is an unbiased estimator of the population mean.

Proof. From the definition of sample mean, we have that:

$$\bar{X}=\frac{1}{n} \sum^n_{i=1}x_i$$

Taking the expected value of both sides and using the properties of expected values to simplify:

\begin{align*} \mathbb{E}(\bar{X})&= \mathbb{E}\Big(\frac{1}{n} \sum^n_{i=1}X_i\Big)\\ &=\frac{1}{n} \sum^n_{i=1}\mathbb{E}(X_i) \end{align*}

Since the expected value of $X_i$ is the population mean $\mu$, we have that:

\begin{align*} \mathbb{E}(\bar{X})&= \frac{1}{n} \sum^n_{i=1}\mu\\ &= \frac{1}{n} \cdot{n\mu}\\ &=\mu \end{align*}

This completes the proof.

Theorem.

## Variance of the sample mean

The variance of the sample mean is:

$$\mathbb{V}(\bar{X})=\frac{\sigma^2}{n}$$

Where:

• $\sigma^2$ is the population variance.

• $n$ is the sample size.

Proof. The variance of the sample mean is:

\begin{align*} \mathbb{V}(\bar{X}) &=\mathbb{V}\Big(\frac{1}{n}\sum^{n}_{i=1}X_i\Big)\\ &=\frac{1}{n^2}\cdot\mathbb{V}\Big(\sum^{n}_{i=1}X_i\Big)\\ \end{align*}

Because $X_1,X_2,\cdots,X_n$ are independent, we know from theoremlink that we can swap the position of the summation and variance to get:

\begin{align*} \mathbb{V}(\bar{X})&=\frac{1}{n^2}\sum^{n}_{i=1}\mathbb{V}(X_i)\\ &=\frac{1}{n^2}\sum^{n}_{i=1}\sigma^2\\ &=\frac{1}{n^2}(n\sigma^2)\\ &=\frac{\sigma^2}{n}\ \end{align*}

This completes the proof.

Theorem.

## Standard error of the sample mean

The standard error of the sample mean, which is defined as the standard deviation of the sample mean, is given by:

$$\text{SE}(\bar{X})= \sqrt{\mathbb{V}(\bar{X})} =\frac{\sigma}{\sqrt{n}}$$

Where:

• $\sigma$ is the population standard deviation.

• $n$ is the sample size.

Proof. We already know that the variance of the sample mean is:

$$\mathbb{V}(\bar{X})=\frac{\sigma^2}{n}$$

To obtain the standard error, we take the square root of the variance:

$$\text{SE}(\bar{X})=\frac{\sigma}{\sqrt{n}}$$

This completes the proof.

# Computing the sample mean in Python

Computing the sample mean is easy using Python's numpy library. Import the library, and use the mean(~) method:

 import numpy as npnp.mean([2,4,8,2]) 4.0
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# Why decision boundary is of (D-1) dimensions?

I am reading Bishop - Pattern Recognition and Machine Learning book. In Chapter 4 first page, there is a statement that

In this chapter, we consider linear models for classification, by which we mean that the decision surfaces are linear functions of the input vector x and hence are defined by (D − 1)-dimensional hyper planes within the D-dimensional input space.

If the input vector is of D dimensions then the decision boundary should also be of D dimensions. Why is it D-1? For example, in 2-D plane, the line separating the points is also a straight line which is of 2 dimensions.

Am I thinking something terribly wrong? Can anyone explain me how is it D-1 dimensions for decision boundary?

• Not so; the line is considered 1 dimensional. – Nick Cox Jan 8 '18 at 13:49
• There is a unique $D$-dimensional linear (or affine) subspace of any (finite) $D$-dimensional space: the space itself. That would be useless as a decision boundary, wouldn't it? – whuber Jan 8 '18 at 14:02

In the images below the decision boundaries for 2D and 3D input space are shown in blue. The orange lines illustrate that these decision boundaries have D$-1$ dimensions themselves.
• I agree that a straight line is a 1D object and a plane is a 2D object. But if the input space has two co-ordinates or features then we represent the straight line or decision boundary as ax+by+c=0 i.e., by using two dimensions. Is the author in this context discussing only about their physical appearance or their representation in vector form? – bharadwaj Jan 9 '18 at 15:13
• @bharadwaj $ax+by+c=0$ is because it exists in 2D space, but it is itself a 1D slice of that 2D space. In my images, the number of black lines (i.e. the axes) show the dimensionality of the space but the orange lines show the dimensionality of the boundary and it is always one less than that of the space. – Dan Jan 9 '18 at 15:26