text:
# Natural Sisal Fiber, Nesting, Rope Making, Paper Making, 100 Grams, 3.52 Ounces,

• \$5.00
Unit price per
Shipping calculated at checkout.

Natural golden colored sisal fiber in a flat roving like section that easily pulls apart. You can see it in the images. You are purchasing this by the 100 gram amount. Sisal comes from the sisal plant and is sometimes referred to as "sisal hemp". The sisal fibre is traditionally used for rope and twine, and has many other uses, including paper, cloth, footwear, hats, bags, carpets, and nesting material for birds.

Sisal is truly an eco-friendly option as not only is it a sustainable resource but it is also grown with minimal ƒ?? if any ƒ?? pesticides and herbicides, thus further reducing damage to the environment. ... Not only is sisal sustainable and 100% biodegradable.

format: plain


text:
# How to properly scale a TikZ/PGF picture which has a \begin{axis}…\end{axis}

Applying a scale option to a TikZ picture works great. It keeps the text the proper size and scales everything else. I would like to have the same behavior when using a \begin{axis}...\end{axis}.

Consider the following in which the first graphic does not use the {axis} environment. Changing the scale works great. However, try changing the scale option of the second \begin{tikzpicture} to scale=1.5. This has the effect that the x and y axis labels are moved far away form the axis.

Furthermore, I can see no apparent change in the scale option applied to \begin{axis}. I don't see any error messages complaining about it, so it must be a valid option.

\documentclass{article}
\usepackage{pgfplots}

\begin{document}
\begin{tikzpicture}[scale=1.0]
\draw [->][gray, thin](-3,0) -- (3,0)
node[blue, right] {$x$};

\draw [->][gray, thin] (0,-3) -- (0,3)
node[blue, above] {$y$};
\end{tikzpicture}

\begin{tikzpicture}[scale=1.0]
\begin{axis}[scale=1.5 % Has no effect
minor tick num=0,
axis y line=center,
axis x line=middle,
xmin=-3,
xmax=3,
ymin=-3,
ymax=3,
xlabel=$x$,
ylabel=$y$
]
\end{axis}
\end{tikzpicture}
\end{document}

Also, one other minor nit-pick now that I posted this: How do I got about getting the x,y labels within {axis} to be to the right and on top as in the first example?

-
Seeing you posted a new question saying you didn't get a good answer to this one: Maybe you could clarify what it is exactly you're trying to achieve here? If the question is "How do I make a PGFplots larger without changing line widths and font sizes?", then the answer is given below: Use widthand height (or the development version's scale option). If the question is "How do I scale a tikzpicture which, among other things, contains a PGFplots axis, while keeping line widths and font sizes", then the answer is: You can't; the axis environment is jus too complex for this. –  Jake May 20 '11 at 10:03
@Jake: Sorry, I did not mean to imply that the postings here not good. They are extremely helpful, I just have to wait until these changes make it to the released versions. The other question has similar behavior but without the scale adjustment - I referenced this question in case the problem is related. –  Peter Grill May 20 '11 at 16:50

In order to adjust the size of a plot without changing the text size, you should use the width and height options. There is no scale option for this, unfortunately: You'll have to multiply the width and height yourself.

The development version contains the options x post scale and y post scale (which are marked EXPERIMENTAL in the source code, but seem to work fine). This will in effect increase the width and height of the plot, so it's not really just scaled: Depending on whether you increase or decrease the size of the plot, you will get more or less tick lines. This is a sensible thing, I believe, but maybe not what you need?

To install the development version using TeXLive, you can use the command sudo -i tlmgr update pgfplots --repository http://tlcontrib.metatex.org/2010

You can then define a new scale style that sets all post scale options to the same value simultaneously:

\pgfkeys{/pgfplots/scale/.style={
x post scale=#1,
y post scale=#1,
z post scale=#1}
}

In order to place the x label and y label at the ends of the axes, you can use the following options:

xlabel style={at={(current axis.right of origin)}, xshift=1.5ex, anchor=center},
ylabel style={at={(current axis.above origin)}, yshift=1.5ex, anchor=center}

Here's a complete example:

\documentclass{minimal}
\usepackage{pgfplots}

\begin{document}
\pgfkeys{/pgfplots/scale/.style={
x post scale=#1,
y post scale=#1,
z post scale=#1}
}

\pgfkeys{/pgfplots/axis labels at tip/.style={
xlabel style={at={(current axis.right of origin)}, xshift=1.5ex, anchor=center},
ylabel style={at={(current axis.above origin)}, yshift=1.5ex, anchor=center}}
}

\begin{tikzpicture}
\begin{axis}[
axis y line=center, axis x line=middle, xmin=-3, xmax=3, ymin=-3, ymax=3, xlabel=$x$,ylabel=$y$,
axis labels at tip
]
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
scale=0.6,
axis y line=center, axis x line=middle, xmin=-3, xmax=3, ymin=-3, ymax=3, xlabel=$x$,ylabel=$y$,
axis labels at tip
]
\end{axis}
\end{tikzpicture}
\end{document}

-
I am getting an Package pgfkeys Error: I do not know the key '/tikz/x post scale' and I am going to ignore it. Perhaps you misspelled it.. I had done a complete update about 3 weeks ago so should have the latest. The log file lists pgfplots 2010/07/14 Version 1.4.1. Is there a later version? –  Peter Grill May 5 '11 at 17:17
@Peter: Oh yes, sorry, it's only been added to the development version. I'll change the answer to reflect this. –  Jake May 6 '11 at 8:11
thanks for the investigation with that style. I'll include it in pgfplots for the moment as this appears to be a bug. –  Christian Feuersänger May 9 '11 at 19:33
This now seems to be working as advertised with TeXLive2011. –  Peter Grill Nov 3 '11 at 22:48


text:
# Mean occurrences of letters in complete strings given by a Bernoulli scheme

Suppose one has an alphabet of $K$ letters, from which we draw sequentially letters; assume that the $n$-th letter occurs with a fixed probability $p_n$ independently of the others and of the previous drawn letters and that $\sum_{n=1}^K p_n =1$; this model should be known in the literature as Bernoulli scheme.

Suppose moreover that the letters are not chosen uniformly at random, i.e. there exists a letter (without loss of generality the first one) whose corresponding probability $p_1$ satisfies the property $p_1 \geq p_n$ for all $2 \leq n \leq K$ and this inequality is strict for at least one letter.

We keep sampling letters according to this scheme until we create a string $s$ where all $K$ letters appear at least once.

How many occurrences of the first letter do one have on average in such string? If an exact result is impossible, can one bound this mean or say something its magnitude?

Any hint, partial solution or reference for this problem would be really appreciated.

-
this question belongs on math stackexchange - it's more a homework question than a research question – Anthony Quas Mar 6 '12 at 17:54
Actually it is a difficult question that might not have been answered in the literature. – Brendan McKay Mar 7 '12 at 4:35

## 1 Answer

The magic words are "Coupon collector problem". For some generalizations, you can check out

The moment zeta function and applications Authors:
Rivin, Igor Publication:
eprint arXiv:math/0201109 Publication Date:
01/2002

-
Thanks Igor for your suggestions. Do you know whether my specific question was already answered somewhere in the literature? The reformulation of my question using the CCP terms is: what is the mean of the largest number of cupons of the same type, once that the collection is complete? – alezok Mar 6 '12 at 20:27
@Ale: Yes, I understand your question: I believe that something very close to this is covered in "the coupon-collector problem revisited", by A Boneh et al (google finds the PDF). – Igor Rivin Mar 6 '12 at 21:25
See the list on page 7 in the paper... – Igor Rivin Mar 6 '12 at 21:27


text:
# How do you solve the quadratic 2/(x+5)-x/(x-5)=1 using any method?

Jan 5, 2017

#### Answer:

Stipulate that $x \ne \pm 5$. Multiply both sides by $\left(x - 5\right) \left(x + 5\right)$. Collect all the terms on one side. Factor or use the quadratic formula. Discard any restricted roots.

#### Explanation:

Stipulate that $x \ne \pm 5$:

2/(x + 5) - x/(x - 5) = 1; x !=+-5

Multiply both sides by $\left(x - 5\right) \left(x + 5\right)$:

2(x - 5) - x(x + 5) = x^2 - 25; x !=+-5

2x - 10 - x^2 - 5x = x^2 - 25; x !=+-5

Collect all of the terms to one side:

2x^2 + 3x - 15 = 0;x !=+-5

Check the discriminant:

$d = {b}^{2} - 4 \left(a\right) \left(c\right) = {3}^{2} - 4 \left(2\right) \left(- 15\right) = 129$

Drop the restriction, because, with 129 under the radical, there is no way to obtain a root equal to $\pm 5$.

Use the quadratic formula:

$x = \frac{- b \pm \sqrt{d}}{2 a}$

$x = \frac{- 3 \pm \sqrt{129}}{2 \left(2\right)}$

$x = \left\{\begin{matrix}\frac{- 3 + \sqrt{129}}{4} \\ \frac{- 3 - \sqrt{129}}{4}\end{matrix}\right.$


text:
# Ratio Measure

Ratio Measure

Subject: Survey Research

• Entry
• Entries A-Z
• Subject Index

• Ratio measure refers to the highest (most complex) level of measurement that a variable can possess. The properties of a variable that is a ratio measure are the following: (a) Each value can be treated as a unique category (as in a nominal measure); (b) different values have order of magnitude, such as greater than or less than or equal to (as in an ordinal measure); (c) basic mathematical procedures can be conducted with the values, such as addition and division (as with an interval measure); and (d) the variable can take on the value of zero. An example of a ratio measure is someone's annual income.

A ratio measure may be expressed as either a fraction or percentage; in addition, a ratio measure may be ...

• [0-9]
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text:
##### Graphene-based Josephson junction microwave bolometer
Sensitive microwave detectors are critical instruments in radioastronomy, dark matter axion searches, and superconducting quantum information science. The conventional strategy towards higher-sensitivity bolometry is to nanofabricate an ever-smaller device to augment the thermal response. However, this direction is increasingly more difficult to obtain efficient photon coupling and maintain the material properties in a device with a large surface-to-volume ratio. Here we advance this concept to an ultimately thin bolometric sensor based on monolayer graphene. To utilize its minute electronic specific heat and thermal conductivity, we develop a superconductor-graphene-superconductor (SGS) Josephson junction bolometer embedded in a microwave resonator of resonant frequency 7.9 GHz with over 99\% coupling efficiency. From the dependence of the Josephson switching current on the operating temperature, charge density, input power, and frequency, we demonstrate a noise equivalent power (NEP) of 7 $\times 10^{-19}$ W/Hz$^{1/2}$, corresponding to an energy resolution of one single photon at 32 GHz and reaching the fundamental limit imposed by intrinsic thermal fluctuation at 0.19 K.
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text:
# 1,484. An Impossible Condition

Hilchos Ishus 6:6

Let’s say that a man tells a woman, “If you don’t give me 200 zuz, then you aren’t consecrated to me, but if you do give them to me, then you are consecrated to me through this dinar,” after which he gives her the dinar. In such a case, the condition is immaterial because the ramification of not fulfilling the condition was stated before the ramification of fulfilling it. In such a case, marriage is immediately effected and she need not give him anything.

Hilchos Ishus 6:7

Similarly, let’s say that a man tells a woman that she is consecrated to him through a dinar on the condition that she ascends to the heavens or descends to the depths, but that if she doesn’t do so, then she is not consecrated, after which he gives her the dinar. Such a condition is immaterial and marriage is immediately effected. It is obvious that she cannot fulfill this condition and he must therefore be joking.


text:
# Why a subvariety of a variety of general type is of general type

It seems a well-known fact that subvarieties of a variety of general type containing a general point are also of general type. This fact is an essential property used to prove some extension theorems of pluricanonical forms on algebraic varieties of general type, see for example the very nice survey on extension of pluricanonical forms

I want to know why this is true? Is there any thing more we can say about the relation between the canonical line bundle of a variety and that of subvarieties of codimension no less than 2.

• I am no algebraic geometer, but still what on Earth are subvarieties of codimension no less than 2 containing a general point? – Leonid Positselski Dec 6 '09 at 22:00
• I think he means that if X is general type, there is a proper closed subvariety Z of X with the property that EVERY closed subvariety of X not of general type is a subscheme of Z. – JSE Dec 6 '09 at 23:54
• As we know, for a codimension 1 subvariety $D$ in $X$, the canonical line bundle $\Omega_D$ on $D$ isomorphic to the restriction of $(\Omega_X\otimes\mathcal{O}_X(D))|_D$. This is called the adjunction formula. But for higher codimension subvarieties, this formula is no longer true. As I know, Kawamata has papers to deal with the higher codimension case. Can someone explain Kawamata's results? – Fei YE Dec 7 '09 at 0:52
• A correction, $\Omega_D$ and $\Omega_X$ should be read as $\omega_D$ and $\omega_X$. – Fei YE Dec 7 '09 at 1:35

Let me reproduce the relevant bit from the reference above (Internet Archive):

Exercise 3.1. Let $$X$$ be a variety of general type. Prove that a subvariety of $$X$$ passing through a general point is of general type.

Bogomolov proved in his paper "Families of curves on a surface of general type" that surfaces of general type satisfying $$c_1^2 > c_2$$ contain at most a finite number of rational and elliptic curves. It is unknown if the same holds true for an arbitrary surface of general type. Thus in the exercise above general point does not mean outside a closed subvariety. Instead it means outside a countable union of closed subvarieties. Of course, the author is considering all his varieties defined over $$\mathbb C$$.

Standard arguments reduces the exercise to the following

Exercise. Let $$X$$ be a variety of general type. If $$Y \rightarrow T$$ is a family of irreducible general type subvarieties of $$X$$ parametrized by a irreducible complex variety $$T$$ then the natural map $$Y\to X$$ is not dominant.

• I don't understand how to reduce the question to the above exercise. Can you say something about the arguments? Thanks! – Fei YE Dec 7 '09 at 1:51
• If through every point you have a variety of special type then an uncountable number of them will be parametrized by the same irreducible component of the Hilbert scheme.... – Jorge Vitório Pereira Dec 7 '09 at 1:59

You need to be careful about what you mean by "a general point". Usually, this means "a point in a certain Zariski open set". So in particular, this statement would say that on a surface of general type all rational and elliptic curves lie in a Zariski closed subset. This is Lang's conjecture, still open I believe (proofs were suggested about 15 yrs ago but then withdrawn).

EDIT: OK, so from the comments and the other answer it appears that it should be a "very general point of $X$", and the statement reduces to showing that if you have a morphism $f:Y\to T$ with irreducible $T$ and general fiber $Y_t$, and a finite dominant morphism $\pi:Y\to X$, then $Y_t$ is also of general type.

The basic reason for that is very simple: if $X$ is of general type then it has lots of pluricanonical forms. You can pull them back to $Y$ (they are differential forms, after all) and get lots of pluricanonical forms on $Y$. Then you can restrict them to $Y_t$ and get lots of pluricaninical forms on $Y_t$.

For a more precise answer, I suggest you look at old papers by Kawamata, Viehweg and Kollar, search for "additivity of Kodaira dimension". There is a whole sequence of $C_{n,m}$ conjectures about the Kodaira dimension of a fibration $Y$ in terms of Kodaira dimensions of $T$ and $Y_t$. Some of them are proved, some are still open.

(Note: general type means "maximal Kodaira dimension", i.e. equal to the dimension of the variety.)

• Thank you! I know nothing about Lang's conjecture. The question I asked is an exercise in GIANLUCA PACIENZA's survey paper PLURICANONICAL SYSTEMS ON PROJECTIVE VARIETIES OF GENERAL TYPE. Exercise 3.1. Let X be a variety of general type. Prove that a subvariety of X passing through a general point is also of general type. I think when he talks about general point, he does mean that a point in some Zariski open set. – Fei YE Dec 7 '09 at 0:57
• Thanks. The explain makes sense. I surely need to read the paper you suggested. – Fei YE Dec 7 '09 at 2:17

Theorem Let $$f:X\rightarrow T$$ be a morphism of smooth projective varieties over $$\mathbb{C}$$, then for general $$t\in T$$, let $$X_{t}=f^{-1}(t)$$, we have $$\kappa(X_{t}) \geq \kappa(X)-\dim T$$

proof: For $$m\in\mathbb{N}$$ sufficiently large, consider Iitaka fibration $$\phi=\phi_{mK_{X}}:X\dashrightarrow Z$$, where $$\dim Z=\kappa(X)$$. Resolving the indeterminant locus of $$\phi$$, we may assume $$\phi$$ is a morphism. \begin{align*} \kappa(X_{t})& \geq \dim(\phi(X_{t}))\\ &\dim X_{t}-\dim(X/Z)\\ &=\dim Z-\dim T\\ &=\kappa(X)-\dim T \end{align*}

Corollary Let $$f:X\rightarrow T$$ be a morphism of smooth projective varieties over $$\mathbb{C}$$. If $$X$$ is of general type, then any general fiber $$X_{t}$$ is of general type.

proof:\begin{align*} \kappa(X_{t})&\geq \kappa(X)-\dim T \\ &=\dim X-\dim T\\ &=\dim X_{t}\\ \end{align*}

The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.

Corollary Let $$f: Z\rightarrow T$$ be a projective morphism and $$g:Z\rightarrow X$$ is a dominant morphism to a variety of general type, then $$Z_{t}$$ is of general type for general $$t\in T$$.

proof: By taking resolution of $$X,Z,T$$, we may assume that they are all smooth. Cutting by hyperplanes on $$T$$, we may assume that $$\dim Z=\dim X$$, hence $$g$$ is generically finite. We have $$K_{Z}=g^{*}K_{X}+R$$ where $$R$$ is effective. Since $$K_{X}$$ is big, it follows that $$K_{Z}$$ is big and $$Z$$ is of general type. So $$Z_{t}$$ is of general type for general $$t\in T$$.

Theorem Let $$X$$ be a projective variety of general type. $$x\in X$$ is a very general point. If $$V$$ is a subvariety containing $$x$$, then $$V$$ is also of general type.

proof: The Hilbert scheme of $$X$$ $$Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$$ contained countably many components. For each $$P\in \mathbb{Q}[t]$$, we have the universal family $$U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$$ Let $$p:U_{P}\rightarrow X$$ be the first projection. Then either the closure of $$U_{P}$$ is equal to $$X$$ or it's a proper closed set of $$X$$. Since our variety is over $$\mathbb{C}$$, an uncountable set, let $$W$$ be the union of all $$\overline{U_{P}}$$ such that $$\overline{U_{P}}\neq X$$, then $$W\neq X$$. For any $$x\in X-W$$, if $$V$$ is a subvariety containing $$x$$, then there is a universal family $$U_{P}$$ for some $$P$$, such $$V$$ is a fiber of $$U_{P}\rightarrow Hilb_{P}(X)$$ and $$p:U_{P}\rightarrow X$$ is dominant. Then the assertion follows form the Corollary above.

• This is a nice write-up. Note that more is actually true: the locus of $t$ in $T$ such that $X_t$ is of general type is a Zariski open. (Your first theorem only says that this locus contains a Zariski open.) A proof of this fact appears in Frank's answer here: mathoverflow.net/questions/321189/… – Ariyan Javanpeykar Mar 20 '19 at 14:20


text:
Factorizing polynomials

1. Oct 10, 2007

Trail_Builder

1. The problem statement, all variables and given/known data

factorise the following as far as possible

1) $$x^3 + y^3$$
2) $$x^4 + x^3 - 3x^2 - 4x - 4$$

2. Relevant equations

3. The attempt at a solution

1) Not quite sure really what to do, lol, only just been taught how to divide polynominals, and the factor and remainder theorem. Havn't been taught how to factorise this case yet. I'm not sure how I could work out the answer myself. Ive tried a few ways but they dont lead anywhere :S.

Any help?

2) I was thinking of rearranging it into $$x^4 - 3x^2 - 4 + x^3 - 4x$$. This way I could maybe solve the first half but substituting $$x^2 = y$$ in so get rid of the 4th degree, and then solve the bit on the right after? or would that just plain not work. Other than that I have no ideas :S.

Any help?

thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 10, 2007

HallsofIvy

Staff Emeritus
You first learned to factor polynomials in order to solve equations, didn't you? If x-a is a factor of P(x) then p(a)= 0 It works the other way too. That's the "factor and remainder theorem you mention. For example, it is obvious (I hope!) that x= -y satisfies $x^3+ y^3= 0$ because $(-y)^3+ y^3= -y^3+ y^3= 0$. That tells you that x-(-y)= x+y is a factor of $x^3+ y^3$. Dividing $x^3+ y^3$ by x+ y should give you a remainder of 0 and a quotient which is a quadratic that you can factor (if necessary by using the quadratic formula to solve the equation).

As far as $x^4+ x^3- 3x^2- 4x- 4$ is concerned, unfortunately solving half an equation doesn't help- unless both parts happen to have the same roots. One thing you can try is the "rational root theorem": if a polynomial (set equal to 0) has a rational root: m/n, then n, the denominator must be a divisor of the leading coefficient and m, the numerator, must be a divisor of the constant term. Here, the leading coefficient is 1 and the constant term is -4. That means any rational root must be a factor of 4: the possibilitiees are 1, -1, 2, -2, 4, and -4. I would reccomend that you try each of those. If any of them make the polynomial equal to 0, then x minus that number is a factor. If none of them do, there are no rational roots and you aren't going to be able to factor in any reasonable way! (I've checked a couple of those- they work!)

3. Oct 10, 2007

Trail_Builder

thanks for the help!

for 1) I ended up with (using quadratic formula) $$x^3 + y^3 = (x+y)(x-\frac{y+\sqrt-3y^2}{2})(x-\frac{y-\sqrt-3y^2}{2})$$ seems a lil messy, is that right?

and i got 2) right :D thanks man

4. Oct 11, 2007

HallsofIvy

Staff Emeritus
The original problem "factorize as far as possible" is a little ambiguous. Typically, factorization problems expect integer coefficients. What I had in mind was $(x+y)(x^2+ xy+ y^2)$ but, of course, you can solve the quadratic and get down to completely linear terms as you did.

5. Oct 11, 2007

Feldoh

I think you could also solve number two by $$(x-a)(x-b)(x-c)(x-d)=x^4 - 3x^2 - 4 + x^3 - 4x$$ and Des Cartes Rule of signs. Look at the coefficients and you should be able to find the roots...

Last edited: Oct 11, 2007


text:
# Georgi statements about the symmetry breaking of $\rm SO(10)$

Here is a paragraph with some statements about the Gauge Symmetry Breaking from Georgi's book Lie Algebras in Particle Physics 2nd ed -- From Isospin to Unified Theories (Georgi, 1999) p.285.

Georgi wrote:

His claim is too quick. Can some experts explain which symmetry breaking pattern he is thinking of? In particular, he uses three Higgs fields in three representations:

• 24 should be the adjoint representation of SU(5).

• 5 should be the fundamental representation of SU(5).

• 45 should be EITHER the adjoint representation of SO(10) or some (what kind?) representation of SU(5).

question: How do we use representation 24, 5, and 45 to break from $$SO(10)$$ to $$SU(5)$$ to $$SU(3) \times SU(2) \times U(1)$$?

• It's all in Slansky, equation (3.4) and Table 43 but you must learn the language... The WP article on the SO(10) GUT: the antisym 45 of SO(10) breaks it down to SU(5) GG and contains the 24 of GG for further breaking to the SM. I think the higgs doublet of the SM is coming out of the 5 of SU(5) , etc... I'm not sure why you'd care about the 45 of SU(5), though--its YTableau, 4column & 2column is in Table 28. Really, there are no easy summaries... – Cosmas Zachos Jul 25 '20 at 15:25
• Answering my own question of the 45 of SU(5). Georgi and Jarlskog, PhysLett B86 (1979) 297 use it for the Yukawas of SU(5) which yield better mass relations. It has a Young tableau of a 4-story column and a 2-story column next to it. Lots of them in the 120 of SO(10). – Cosmas Zachos Jul 25 '20 at 20:36
• MANY THANKS FOR PRECIOUS COMMENTS - <3 thanks ~ – annie marie heart Jul 25 '20 at 22:19
• any comments math.stackexchange.com/questions/3777000/… ? thanks! – annie marie heart Aug 1 '20 at 20:18
• Frankly, emailing Kephart should produce a superior answer....! There can be a self-dual and an anti-self-dual one... – Cosmas Zachos Aug 1 '20 at 20:27


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# a, b and c are positive integers. If b equals the square root of a, an

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a, b and c are positive integers. If b equals the square root of a, and if c equals the sum of a and b, which of the following could be the value of c?

I. 21
II. 30
III. 72

A. I only
B. II only
C. III only
D. II and III only
E. I, II and III

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Re: a, b and c are positive integers. If b equals the square root of a, an  [#permalink]

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04 Dec 2017, 06:47
c= $$\sqrt{a}$$ + a
a= 25, c= 5+25 = 30
a= 64, c= 8+ 64 = 72
a= 16, c= 4+16 = 20 (1 eliminated)
II and III only possible
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Re: a, b and c are positive integers. If b equals the square root of a, an  [#permalink]

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04 Dec 2017, 06:56
Option D.
a,b,c
=> b2, b, c
c=b*(b+1)

Product of 2 consecutive +ve integers
30 &72

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Re: a, b and c are positive integers. If b equals the square root of a, an  [#permalink]

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04 Dec 2017, 07:05
1
3
Bunuel wrote:
a, b and c are positive integers. If b equals the square root of a, and if c equals the sum of a and b, which of the following could be the value of c?

I. 21
II. 30
III. 72

A. I only
B. II only
C. III only
D. II and III only
E. I, II and III

$$b=\sqrt{a}..............a=b^2$$..
and $$c=a+b = b^2+b =b(b+1)$$
so C is PRODUCT of two consecutive integers
21=3*7 or 1*21 ......NO
$$30 = 5*6$$ ................YES
$$72 = 8*9$$ ................ YES

D
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Re: a, b and c are positive integers. If b equals the square root of a, an  [#permalink]

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12 Jan 2018, 07:12
1
Bunuel wrote:
a, b and c are positive integers. If b equals the square root of a, and if c equals the sum of a and b, which of the following could be the value of c?

I. 21
II. 30
III. 72

A. I only
B. II only
C. III only
D. II and III only
E. I, II and III

We have the following equations:

b = √a

b^2 = a

and

c = a + b

Thus,

c = a + a^2

So c is the sum of an integer + that integer squared.

Statement I is not true, since 4^2 + 4 = 20 and 5^2 + 5 = 30; thus the value of a that satisfies a^2 + a = 21 is not an integer, but some real number between 4 and 5.

Looking at the answer choices we see that II and III can work since 30 = 5 + 25, and 72 = 8 + 64.

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Re: a, b and c are positive integers. If b equals the square root of a, an   [#permalink] 12 Jan 2018, 07:12
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# Preprints

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14:17
 Coherent $\mathrm{ J } / \psi$ photoproduction in ultra-peripheral PbPb collisions at $\sqrt{s_{ \mathrm{NN}}} =$ 2.76 TeV with the CMS experiment The cross section for coherent $\mathrm{ J } / \psi$ photoproduction accompanied by at least one neutron on one side of the interaction point and no neutron activity on the other side, $X_{\mathrm{n}}0_{\mathrm{n}}$, is measured with the CMS experiment in ultra-peripheral PbPb collisions at $\sqrt{s_{ \mathrm{NN}}} =$ 2.76 TeV. [...] CERN-EP-2016-098 ; CMS-HIN-12-009-003. - 2016. Additional information for the analysis - CMS AuthorList - Fulltext

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 Cryogenic rf test of the first plasma etched SRF cavity / Upadhyay, J ; Palczewski, A ; Popović, S ; Valente-Feliciano, A -M ; Im, D ; Phillips, L ; Vušković, L Plasma etching has a potential to be an alternative processing technology for superconducting radio frequency (SRF) cavities. [...] arXiv:1605.06494. - 2016. - 6 p. Preprint

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 Recent progress and review of Physics Dynamics Coupling in geophysical models / Gross, Markus ; Wan, Hui ; Rasch, Philip J ; Caldwell, Peter M ; Williamson, David L ; Klocke, Daniel ; Jablonowski, Christiane ; Thatcher, Diana R ; Wood, Nigel ; Cullen, Mike et al. Geophysical models of the atmosphere and ocean invariably involve parameterizations. [...] arXiv:1605.06480. - 2016. Preprint

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 Initial stage of cavitation in liquids and its observation by Rayleigh scattering / Pekker, M ; Shneider, M N A theory is developed for the initial stage of cavitation in the framework of Zel'dovich-Fisher theory of nucleation in the field of negative pressure, while taking into account the surface tension dependence on the nanopore radius. [...] arXiv:1605.06469. - 2016. Preprint

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 Simulating the interaction of road users: A glance to complexity of Venezuelan traffic / Correa, Juan C (Facultad de Psicología. Fundación Universitaria Konrad Lorenz. Colombia) ; Caicedo, Mario I (Departamento de Física. Universidad Simón Bolívar. Venezuela) ; Bazzan, Ana L C (Instituto de Informática. Universidade Federal do Rio Grande do sul. Brazil) ; Jaffe, Klaus (Centro de Estudios Estratégicos. Universidad Simón Bolívar. Venezuela) Automotive traffic is a classical example of a complex system, being the simplest case the homogeneous traffic where all vehicles are of the same kind, and using different means of transportation increases complexity due to different driving rules and interactions between each vehicle type. [...] arXiv:1605.06462. - 2016. - 10 p. Preprint

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# Region enclosed by the graph

• Apr 22nd 2010, 07:11 PM
downer
Region enclosed by the graph
A problem I had recently on a test has been bothering me for quite some time now and would like to seek assistance for it. There were 3 questions that went along with it.

The problem goes:
Let R be the region enclosed by the graph of y = 2lnx and y = x/2, and the lines x = 2 and x = 8.

a) Find the area of R
b) Set up an integral expression, in terms of a single variable,for the volume of the solid generated when R is revolved about the x-axis.
c)Set up an integral expression, in terms of a single variable,for the volume of the solid generated when R is revolved about the line x = -1.

Any possible help regarding any of the 3 questions is greatly appreciated.
• Apr 22nd 2010, 07:33 PM
bhuang
Do you have the answer to a)?
• Apr 22nd 2010, 07:52 PM
pickslides
Quote:

Originally Posted by downer

The problem goes:
Let R be the region enclosed by the graph of y = 2lnx and y = x/2, and the lines x = 2 and x = 8.

For part a) solve $\int_2^82\ln x - \frac{x}{2}~dx$

What do you get?
• Apr 22nd 2010, 07:59 PM
bhuang
I did x/2 - 2lnx instead because y=x/2 is graph that gives the larger values. I end up getting -7.49, but I think some of my algebra may have been wrong...
• Apr 22nd 2010, 08:06 PM
pickslides
Quote:

Originally Posted by bhuang
I did x/2 - 2lnx instead because y=x/2 is graph that gives the larger values.

I don't agree with this.

In my understanding $2\ln x > \frac{x}{2}, \forall x \in [2,8]$

Quote:

Originally Posted by bhuang
I end up getting -7.49, but I think some of my algebra may have been wrong...

This answer is not correct. If anything it supports my statement above.
• Apr 22nd 2010, 08:12 PM
drumist
I've attached a graph of the two curves so you can see which one is above and which is below. We are only concerned with the region between x=2 and x=8.


text:
## Ground states for nonlinear fractional Choquard equations with general nonlinearities.(English)Zbl 1344.35168

Summary: We study the existence of ground states for the nonlinear Choquard equation driven by fractional Laplacian: $(-\Delta)^su+u=\left(|x|^{-\mu}\ast F(u)\right)\,f(u)\text{ in }\mathbb R^n,$ where the nonlinearity satisfies the general Berestycki-Lions-type assumptions.

### MSC:

 35R11 Fractional partial differential equations 35J25 Boundary value problems for second-order elliptic equations 35J60 Nonlinear elliptic equations 35A15 Variational methods applied to PDEs
Full Text:

### References:

 [1] Applebaum, Lévy processes from probability to finance and quantum groups, Notices of the American Mathematical Society 51 pp 1336– (2004) · Zbl 1053.60046 [2] Laskin, Fractional quantum mechanics and Lévy path integrals, Physics Letters A 268 pp 298– (2000) · Zbl 0948.81595 [3] Caffarelli, An extension problem related to the fractional Laplacian, Communications in Partial Differential Equations 32 pp 1245– (2007) · Zbl 1143.26002 [4] Cabré, Nonlinear equations for fractional Laplacians: I. Regularity, maximum principle and Hamiltonian estimates, Annales de l’Institut Henri Poincaré. Analyse Non Linéaire 31 pp 23– (2014) · Zbl 1286.35248 [5] Frank RL Lenzmann E On ground states for the L 2 -critical boson star equation 2721 [6] Lieb, Existence and uniqueness of the minimizing solution of Choquard’s nonlinear equation, Studies in Applied Mathematics 57 pp 93– (1976) · Zbl 0369.35022 [7] Lions, The Choquard equation and related questions, Nonlinear Analysis 4 pp 1063– (1980) · Zbl 0453.47042 [8] Ma, Classification of positive solitary solutions of the nonlinear Choquard equation, Archive for Rational Mechanics and Analysis 195 pp 455– (2010) · Zbl 1185.35260 [9] Cingolani, Multiple solutions to a magnetic nonlinear Choquard equation, Zeitschrift für Angewandte Mathematik und Physik 63 pp 233– (2012) · Zbl 1247.35141 [10] Moroz, Groundstates of nonlinear Choquard equations: existence, qualitative properties and decay asymptotics, Journal of Functional Analysis 265 pp 153– (2013) · Zbl 1285.35048 [11] Moroz, Existence of groundstates for a class of nonlinear Choquard equations, Transactions of the American Mathematical Society 367 pp 6557– (2015) · Zbl 1325.35052 [12] Frank, Uniqueness and nondegeneracy of ground states for in R, Acta Mathematica 210 pp 261– (2013) · Zbl 1307.35315 [13] Felmer, Positive solutions of nonlinear Schrödinger equation with fractional Laplacian, Proceedings of the Royal Society of Edinburgh 142 pp 1237– (2012) · Zbl 1290.35308 [14] Lakshmikantham, Basic theory of fractional differential equations, Nonlinear Analysis 69 pp 2677– (2008) · Zbl 1161.34001 [15] Barrios, On some critical problems for the fractional Laplacian operator, Journal of Differential Equations 252 pp 6133– (2012) · Zbl 1245.35034 [16] Brändle, A concave-convex elliptic problem involving the fractional Laplacian, Proceedings of the Royal Society of Edinburgh A 143 pp 39– (2013) · Zbl 1290.35304 [17] Chang, Ground state of scalar field equations involving a fractional Laplacian with general nonlinearity, Nonlinearity 26 pp 479– (2013) · Zbl 1276.35080 [18] Ros-Oton, The Pohozaev identity for the fractional Laplacian, Archive for Rational Mechanics and Analysis 213 pp 587– (2014) · Zbl 1361.35199 [19] Tan, Nonlinear fractional field equations, Nonlinear Analysis 75 pp 2098– (2012) · Zbl 1244.35057 [20] Yang, Fractional Sobolev-Hardy inequality in RN, Nonlinear Analysis 119 pp 179– (2015) · Zbl 1328.35288 [21] Cho, Profile decompositions and blowup phenomena of mass critical fractional Schrödinger equations, Nonlinear Analysis 86 pp 12– (2013) · Zbl 1278.35227 [22] Herr, The boson star equation with initial data of low regularity, Nonlinear Analysis 97 pp 125– (2014) · Zbl 1292.35280 [23] Coti Zelati, Ground states for pseudo-relativistic Hartree equations of critical type, Revista Matemática Iberoamericana 29 pp 1421– (2013) · Zbl 1283.35121 [24] Coti Zelati, Existence of ground states for nonlinear, pseudorelativistic Schrödinger equations, Rendiconti Lincei - Matematica e Applicazioni 22 pp 51– (2011) · Zbl 1219.35292 [25] Lei, On the regularity of positive solutions of a class of Choquard type equations, Mathematische Zeitschrift 273 pp 883– (2013) · Zbl 1267.45010 [26] D’avenia, On fractional Choquard equations, Mathematical Models & Methods in Applied 25 pp 1447– (2015) · Zbl 1323.35205 [27] Berestycki, Nonlinear scalar field equations. I. Existence of a ground state, Archive for Rational Mechanics and Analysis 82 pp 313– (1983) [28] Lieb, ’Analysis’, Gradute Studies in Mathematics, AMS (2001) [29] Willem, Minimax Theorems (1996) [30] Jeanjean, Existence of solutions with prescribed norm for semilinear elliptic equations, Nonlinear Analysis 28 pp 1633– (1997) · Zbl 0877.35091 [31] Jeanjean, A remark on least energy solutions in RN, Proceedings of the American Mathematical Society 131 pp 2399– (2003) · Zbl 1094.35049 [32] Brezis, Remarks on the Schrödinger operator with regular complex potentials, Journal de Mathématiques Pures et Appliquées 58 pp 137– (1979) [33] Palatucci, Local and global minimizers for a variational energy involving a fractional norm, Annali di Matematica Pura ed Applicata Series IV 192 pp 673– (2013) · Zbl 1278.82022 [34] Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Communications on Pure and Applied Mathematics 60 pp 67– (2006) · Zbl 1141.49035 [35] Brock, An approach to symmetrization via polarization, Transactions of the American Mathematical Society 352 pp 1759– (2000) · Zbl 0965.49001
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text:
### Ph.D. Dissertation Thesis Defense: Yi-Peng Wei

Friday, December 7, 2018
3:00 p.m.
2168 AV Williams
Emily Irwin
301 405 0680
eirwin@umd.edu

ANNOUNCEMENT: Ph.D. Dissertation Thesis Defense

Name: Yi-Peng Wei

Committee member:
Professor Sennur Ulukus, Chair
Professor Alexander Barg
Professor Lawrence C. Washington, Dean's Representative

Date/Time: Fri., Dec. 7th, 2018 at 3:00 pm

LocationAVW 2168

Abstract:
This dissertation studies privacy and security problems from an information-theoretic point of view. We study the privacy problem via the private information retrieval (PIR) problem with a focus on its interactions with available side information. We study the security problem via the wiretap channel with a focus on the design of practical coding schemes to achieve information-theoretically achievable random-coding based secrecy rates.

First, we consider the problem of PIR from $N$ non-colluding and replicated databases when the user is equipped with a cache that holds an uncoded fraction $r$ from each of the $K$ stored messages in the databases. We consider the case where the databases are unaware of the cache content. We investigate $D^*(r)$ the optimal download cost normalized with the message size as a function of $K$, $N$, $r$. For a fixed $K$, $N$, we develop converses and achievability schemes for the $D^*(r)$ curve. The largest additive gap between our achievability and the converse bounds is $\frac{1}{6}$. Our results show that the download cost can be reduced beyond memory-sharing if the databases are unaware of the cached content.

Second, we consider the same setting under a more restricted model where the databases know the user cache content partially. The user receives an uncoded fraction $r$ from each of the $K$ stored messages, with the $\frac{r}{N}$ fraction of it coming from the $n$th database. The side information obtained from the $n$th database is known by the $n$th database and is unknown by the remaining databases. We investigate the optimal normalized download cost $D^*(r)$, and develop converses and achievability schemes for $D^*(r)$. The largest additive gap between our achievability and the converse bounds is $\frac{5}{32}$ for this case. We observe that the achievable download cost here is larger than that in the previous case due to the partial knowledge of the databases regarding the cache content.

Third, we consider the problem of PIR with private side information (PSI) when the cache content is partially known by the databases. Here, a cache-enabled user of cache-size $M$ possesses side information in the form of full messages that are partially known by the databases. The user wishes to download a desired message privately while keeping the identities of the side information messages that the user did not prefetch from a database private against that database. We characterize the exact capacity of PIR with PSI under partially known PSI condition. We show that the capacity of PIR with partially known PSI is the same as the capacity of PIR with fully unknown PSI.

Fourth, we consider PIR with PSI under storage constraints where a cache-enabled user of cache-size $S$ possesses side information in the form $M$ messages that are unknown to the databases, where $M>S$. We address the problem of which uncoded parts of $M$ messages the user should keep in its constrained cache of size $S$ in order to minimize the download cost during PIR subject to PSI. We characterize the exact capacity of this PIR-PSI problem under the storage constraint $S$. We show that a uniform caching scheme which caches equal amounts from all messages achieves the lowest normalized download cost.

Fifth, we consider the PIR problem from decentralized uncoded caching databases. Here, the contents of the databases are not fixed a priori, and we design the probability distribution adopted by each database in the decentralized caching phase in order to minimize the expected normalized download cost in the retrieval phase. We characterize the exact capacity of this problem, and show that uniform and random caching results in the lowest normalized download cost.

Next, we focus on security of communication by designing practical coding schemes to achieve the information-theoretically achievable random-coding based secrecy rates. By applying two recently developed techniques for polar codes, namely, universal polar coding and polar coding for asymmetric channels, we propose a polar coding scheme to achieve the secrecy capacity of the general wiretap channel. We then apply this coding scheme to achieve the best-known secrecy rates for the multiple access wiretap channel, and the broadcast and interference channels with confidential messages.


text:
A Python program with vectorization and plotting

We return to the problem where a ball is thrown up in the air and we have a formula for the vertical position $$y$$ of the ball. Say we are interested in $$y$$ at every milli-second for the first second of the flight. This requires repeating the calculation of $$y=v_0t-0.5gt^2$$ one thousand times.

We will also draw a graph of $$y$$ versus $$t$$ for $$t\in [0,1]$$. Drawing such graphs on a computer essentially means drawing straight lines between points on the curve, so we need many points to make the visual impression of a smooth curve. With one thousand points, as we aim to compute here, the curve looks indeed very smooth.

In Python, the calculations and the visualization of the curve may be done with the program ball_plot.py, reading

from numpy import linspace
import matplotlib.pyplot as plt

v0 = 5
g = 9.81
t = linspace(0, 1, 1001)

y = v0*t - 0.5*g*t**2

plt.plot(t, y)
plt.xlabel('t (s)')
plt.ylabel('y (m)')
plt.show()

This program produces a plot of the vertical position with time, as seen in Figure 1. As you notice, the code lines from the ball.py program in the chapter A Python program with variables have not changed much, but the height is now computed and plotted for a thousand points in time!

Let us take a look at the differences between the new program and our previous program. From the top, the first difference we notice are the lines

from numpy import *
from matplotlib.pyplot import *

You see the word import here, so you understand that numpy must be a library, or module in Python terminology. This library contains a lot of very useful functionality for mathematical computing, while the matplotlib.pyplot module contains functionality for plotting curves. The above import statement constitutes a quick way of populating your program with all necessary functionality for mathematical computing and plotting. However, we actually make use of only a few functions in the present program: linspace, plot, xlabel, and ylabel. Many computer scientists will therefore argue that we should explicitly import what we need and not everything (the star *):

from numpy import linspace
from matplotlib.pyplot import plot, xlabel, ylabel

Others will claim that we should do a slightly different import and prefix library functions by the library name:

import numpy as np
import matplotlib.pyplot as plt
...
t = np.linspace(0, 1, 1001)
...
plt.plot(t, y)
plt.xlabel('t (s)')
plt.ylabel('y (m)')

We will use all three techniques, and since all of them are in so widespread use, you should be familiar with them too. However, for the most part in this book we shall do

from numpy import *
from matplotlib.pyplot import *

for convenience and for making Python programs that look very similar to their Matlab counterparts.

The function linspace takes $$3$$ parameters, and is generally called as

linspace(start, stop, n)

This is our first example of a Python function that takes multiple arguments. The linspace function generates n equally spaced coordinates, starting with start and ending with stop. The expression linspace(0, 1, 1001) creates 1001 coordinates between 0 and 1 (including both 0 and 1). The mathematically inclined reader will notice that 1001 coordinates correspond to 1000 equal-sized intervals in $$[0,1]$$ and that the coordinates are then given by $$t_i = i/1000$$ ($$i = 0, 1, ..., 1000$$).

Figure 1: Plot generated by the script ball_plot.py showing the vertical position of the ball at a thousand points in time.

The value returned from linspace (being stored in t) is an array, i.e., a collection of numbers. When we start computing with this collection of numbers in the arithmetic expression v0*t - 0.5*g*t**2, the expression is calculated for every number in t (i.e., every $$t_i$$ for $$i = 0, 1,..., 1000$$), yielding a similar collection of 1001 numbers in the result y. That is, y is also an array.

This technique of computing all numbers "in one chunk" is referred to as vectorization. When it can be used, it is very handy, since both the amount of code and computation time is reduced compared to writing a corresponding for or while loop (the chapter Basic constructions) for doing the same thing.

The plotting commands are simple:

1. plot(t, y) means plotting all the y coordinates versus all the t coordinates
2. xlabel('t (s)') places the text t (s) on the $$x$$ axis
3. ylabel('y (m)') places the text y (m) on the $$y$$ axis
At this stage, you are strongly encouraged to do Exercise 4: Volumes of three cubes. It builds on the example above, but is much simpler both with respect to the mathematics and the amount of numbers involved.


text:
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15 results found
 no title author magazine year volume issue page(s) type 1 Asymptotic behavior of generalized quasiisometries at a point Kovtonyuk, D. A. 2011 63 4 p. 555-563 article 2 Best m-term approximation of the classes$$B_{\infty, \theta }^r$$ of functions of many variables by polynomials in the haar system Stasyuk, S. A. 2011 63 4 p. 638-645 article 3 Conditions of smoothness for the distribution density of a solution of a multidimensional linear stochastic differential equation with lévy noise Bodnarchuk, S.V. 2011 63 4 p. 501-515 article 4 Landau–Kolmogorov problem for a class of functions absolutely monotone on a finite interval Skorokhodov, D. S. 2011 63 4 p. 617-637 article 5 Multidimensional random motion with uniformly distributed changes of direction and erlang steps Pogorui, A. A. 2011 63 4 p. 665-671 article 6 On hyperholomorphic functions of the space variable Herus, O. F. 2011 63 4 p. 530-537 article 7 On positive solutions of one class of evolutionary inclusions of the subdifferential type Kapustyan, O.V. 2011 63 4 p. 544-554 article 8 On some criteria of convexity for compact sets Zelins’kyi, Yu. B. 2011 63 4 p. 538-543 article 9 On some imbedding relations between certain sequence spaces Mursaleen, M. 2011 63 4 p. 564-579 article 10 On the fredholm theory of a planar problem with shift for a pair of functions Lysenko, Z. M. 2011 63 4 p. 658-664 article 11 On the mean value of the function $${\bar{S}_k}(n)$$ Varbanets, P.D. 2011 63 4 p. 516-529 article 12 On the reconstruction of the variation of the metric tensor of a surface on the basis of a given variation of christoffel symbols of the second kind under infinitesimal deformations of surfaces in the euclidean space E3 Potapenko, I. V. 2011 63 4 p. 609-616 article 13 On uniqueness theorems for holomorphic curves sharing hypersurfaces without counting multiplicity Phuong, Ha Tran 2011 63 4 p. 646-657 article 14 Properties of a certain product of submodules Nikmehr, M. J. 2011 63 4 p. 580-595 article 15 Resonance elliptic variational inequalities with discontinuous nonlinearities of linear growth Pavlenko, V. N. 2011 63 4 p. 596-608 article
15 results found

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text:
# Are two rational vector spaces having the same cardinal number are isomorphic?

Discuss the following assertion:if two rational vector spaces have the same cardinal number(i.e., if there is some one-to-one correspondence between them), then they are isomorphic(i.e., there is linearity-preserving one-to-one correspondence between them).

-
Is this homework? The imperative wording looks like it. (Also: consider $\mathbb Q$ versus $\mathbb Q^2$). –  Henning Makholm Feb 6 '12 at 18:38
No Bud i am myself working through Paul Halmos's Finite Dimensional vector spaces book and this is a question in there. Since i was not sure about the answer. I thought of seeking help from those who know better. Yes you are right Q versus \mathbb Q^2Q2 also form infinite dimensional vector spaces and should have the same cardinal number, although i am unsure how we would show such a one-to-one correspondence of a vector space perhaps we could use tuples ?? –  Hardy Feb 6 '12 at 18:51
@Hardy: Please provide context in the post, not in comments; some of us consider posts in the imperative to be rather rude, so perhaps you may try, in the future, not to simply post as if you were issuing orders to the readers. –  Arturo Magidin Feb 6 '12 at 19:10
Definitely not true if the two spaces are countable. $\mathbb Q^k$ is countable for any finite $k$. –  Thomas Andrews Feb 6 '12 at 19:27
@ArturoMagidin Thanks very much for the tip my goal was n't to be rude. I mearly copied the question as was in the book. I would try rephrase in future with this in mind. –  Hardy Feb 6 '12 at 20:15

In general assuming the axiom of choice then there is a unique cardinality for all maximal linearly independent sets, i.e. bases.

This means that the idea of "dimension" is well defined, and two vector spaces whose dimension is the same are isomorphic. Infinities are quite varied and there are more than one size of infinity.

So there can be two vector spaces over a field $F$, both have an infinite dimension, but alas they are not isomorphic.

However, do note that any nonzero vector space contains a copy of the field, $F$. If this $F$ is infinite (such examples are $\mathbb Q$, $\mathbb R$, and other fields) then the vector space is infinite, as a set. The dimension, however, may still be finite.

-
Thank you very much for that. I have n't studied axiom of choice or about different sizes of infinity. Will do ASAP. That was very insightful. –  Hardy Feb 6 '12 at 19:11
@Hardy: The axiom of choice is not needed for developing the theory of finite dimensional linear algebra, but is indeed used in infinitely dimensional linear algebra. As for different sizes of infinity, this is an important fact. For example $\mathbb Q$ and $\mathbb R$ do not have the same cardinality. –  Asaf Karagila Feb 6 '12 at 19:28
thanks for the clarification actually i was aware of Q and \mathbb RR not having the same cardinality fact, but not sure if i have come across yet what they are or how they are represented in mathematical literature. –  Hardy Feb 6 '12 at 20:18
No. Any two finite-dimensional vector spaces over $\mathbb{Q}$ have the same cardinality, but two such vector spaces are isomorphic if and only if they have the same dimension (which in general they don't).
@Hardy: $\mathbb{Q}^n$ for every positive integer $n$. –  Brian M. Scott Feb 6 '12 at 19:12


text:
Problem with my proof that NP = coNP?

Is there a problem with this proof that NP = coNP?

It suffices to show that Satisfiability can be solved efficiently with at most a polynomial number of queries to an oracle for Tautology. The algorithm for a problem P is, pick a variable X in P and fix it to T to obtain a problem P'. Submit ~P' to the oracle for Tautology. If ~P' is a tautology, then X is a "no go" for the problem P. If the oracle accepts ~P', then fix X to T in the result, otherwise fix X to F. Then proceed to the next variable. This requires linear queries in the size of the problem. An analogous result reduces Tautology to Satisfiability. Therefore NP = coNP.

My only doubt is that maybe when you're using an oracle you only get to find out if it accepts, not if it rejects. That would block this proof.

• Therefore NP=coNP. Although I am not a complexity expert, this is false to me. Try providing an argument for that also. – bellpeace Aug 16 '14 at 17:22
• Of course there's a problem with it! Even without reading what you wrote, it's very, very, very unlikely that you really managed to solve one of the most important open problems in theoretical CS in a single paragraph using techniques that any undergraduate should know about. – David Richerby Aug 16 '14 at 17:45
• You are claiming to have a solution for a well-known, difficult open problem. This is an extraordinary claim requiring extraordinary evidence. You have not provided such so there is not much to talk about. Even if you had, this would not be a good post for SE; it is not our goal here to make broad advances to science in a single post. See here for a related discussion. – Raphael Aug 16 '14 at 18:32

You are confusing Turing reductions with mapping reduction, in a sense. The problem with your argument is that if you allow an oracle $O$ to an NP-complete (or coNP-complete) problem, you immediately get that $P^O=co-P^O$. That is, if you already have an oracle for coNP, then you cannot distinguish NP from coNP (but that does not mean that they are equal!).
Here is a shorter version of your argument, to make things clear: suppose we have an oracle for Tautology. It is true that $\phi$ is satisfiable iff $\neg \phi$ is not a tautology. Thus, in order to test if $\phi$ is satisfiable, just run the oracle on $\neg \phi$, and invert the answer. But this only shows that $NP\subseteq P^{coNP}$. Your assumption that $P^{coNP}=coNP$ is incorrect (or at least, it's an open question).


text:
# Abstract Nonsense

## The Dimension Theorem

Point of post: In this post we prove what is called the ‘dimension’ theorem which in essence says that the degree of any irrep of a finite group divides the order of the group.

Motivation

So far we’ve obtained some interesting information about the degrees of the irreps of a finite group $G$. We’ve proven that the sum of the degrees squared must equal the order of the group. Also, we’ve proven that the number of degree one irreps of $G$ is equal to the order of the abelinization of $G$. In this post we’ll prove the supremely interesting result that the degree of any irrep must divide the order of the group. One of many uses for this will be that we will be able to prove some interesting results about finite groups.

The Dimension Theorem

Let as always $G$ be a finite group and suppose that we’ve chosen particular representatives $\displaystyle D^{(\alpha)}$ and thus irreducible characters $\chi^{(\alpha)}$ have the representation $\displaystyle \chi^{(\alpha)}(x)=\sum_{j=1}^{n}D^{(\alpha)}_{j,j}(x)$. Our first theorem will show the connection between irreducible characters and algebraic integers. Namely:

Theorem: Let $G$ be a finite group then for every $\alpha\in\widehat{G}$ and every $g\in G$ one has that $\chi^{(\alpha)}(g)\in\mathbb{A}$.

Proof: We merely note that since $|G|<\infty$ one has that $g^{|G|}=e$ and so $D^{(\alpha)}(g)^|G|=D^{(\alpha)}\left(g^|G|\right)=D^{(\alpha)}(g)=I$ and so by basic matrix analysis every eigenvalue of $D^{(\alpha)}$ is a $|G|$-root of unity and thus trivially an algebraic integer. Thus, $\chi^{(\alpha)}(g)$ being the sum of these algebraic integers (it of course, being the trace of $D^{(\alpha)}(g)$) is an algebraic integers (since they form a ring). $\blacksquare$

We now use this to show that $d_\alpha\mid |G|$ for every $\alpha\in\widehat{G}$.

Theorem: Let $G$ be a finite group and $\alpha\in\widehat{G}$. Then, $d_\alpha\mid |G|$.

Proof: Choose an ordering for $G$ so that we can list $G$ as $(g_1,\cdots,g_{n})$ where $|G|=n$. We then define a matrix $A$ by $A_{i,j}=\chi^{(\alpha)}\left(g_i g_j^{-1}\right)$. We then define $v\in\mathbb{C}^n$ by $v=(\chi^{(\alpha)}(g_1),\cdots,\chi^{(\alpha)}(g_n))^{\top}$. We note then by the convolution relations that

\displaystyle \begin{aligned}Av &=\left(\sum_{k=1}^{n}\chi^{(\alpha)}\left(g_1 g_k^{-1}\right)\chi^{(\alpha)}(g_k),\cdots,\sum_{k=1}^{n}\chi^{(\alpha)}\left(g_n g_k^{-1}\right)\chi^{(\alpha)}\left(g_k\right)\right)^{\top}\\ &= \left(\frac{|G|}{d_\alpha}\chi^{(\alpha)}(g_1),\cdots,\frac{|G|}{d_\alpha}\chi^{(\alpha)}(g_n)\right)^{\top}\\ &= \frac{|G|}{d_\alpha}v\end{aligned}

and thus $\displaystyle \frac{|G|}{d_\alpha}$ is an eigenvalue for $A$. That said, by the previous theorem we know that every entry of $A$ is an algebraic integer and thus by previous theorem we may conclude that $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{A}$. But, since evidently $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{Q}$ we may conclude that $\displaystyle \frac{|G|}{d_\alpha}\in\mathbb{Z}$ and thus $d_\alpha\mid |G|$ as desired. $\blacksquare$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

March 3, 2011 -

1. […] our last post we proved the incredible fact that the degree of any irrep divides the order of the group. We now […]

Pingback by Representation Theory: The Dimension Theorem (Strong Version) « Abstract Nonsense | March 7, 2011 | Reply

2. […] This follows from the fact that the values of every character of a group are algebraic […]

Pingback by Representation Theory: Character Tables « Abstract Nonsense | March 21, 2011 | Reply

3. […] the sum runs over all with . We recall next that for every we must have that so that but since this implies that . Thus, with these […]

Pingback by Representation Theory: Character Table of S_3 By Finding the Irreducible Representations « Abstract Nonsense | March 22, 2011 | Reply

4. […] the dimension theorem tells us that each and so . But, since there is always the degree of which is we may conclude […]

Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply

5. […] irrep of is complex. Thus, for every in one has that and . But, since is odd and we know that we may conclude that for some . Thus, since we know that the cardinality of is (where ) we […]

Pingback by Representation Theory: The Number of Conjugacy Classes of a Finite Group of Odd Order is Equivalent to The Order of The Group Modulo Sixten « Abstract Nonsense | April 23, 2011 | Reply


text:
# How do you solve -16d+4=d?

Dec 22, 2016

$d = \frac{4}{17}$

#### Explanation:

The first step to solving this equation is to do the necessary mathematics to isolate the $d$ terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

$\textcolor{red}{16 d} - 16 d + 4 = \textcolor{red}{16 d} + d$

$\left(\textcolor{red}{16 d} - 16 d\right) + 4 = \textcolor{red}{16 d} + d$

$0 + 4 = 16 d + d$

$4 = 16 d + d$

Next, we can combine like terms:
Hint: remember ($\textcolor{g r e e n}{k = 1 k}$)

$4 = \left(16 + 1\right) d$

$4 = 17 d$

Now, we can solve for $d$ while keeping the equation balanced:

$\frac{4}{\textcolor{red}{17}} = \frac{17 d}{\textcolor{red}{17}}$

$\frac{4}{17} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{17}}} d}{\textcolor{red}{\cancel{\textcolor{b l a c k}{17}}}}$

$\frac{4}{17} = d$ or $d = \frac{4}{17}$


text:
MathSciNet bibliographic data MR344479 30A78 Ahern, P. R.; Clark, D. N. On inner functions with \$H\sp{p}\$$H\sp{p}$-derivative. Michigan Math. J. 21 (1974), 115–127. Article

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text:
# TCS model Placement Questions - 5

1. Ray writes a two digit number.  He sees that the number exceeds 4 times the sum of its digits by 3.  If the number is increased by 18, the result is the same as the number formed by reversing the digits.  Find the number.
a) 35
b) 42
c) 49
d) 57
Solution: Let the two digit number be xy.
4(x + y) +3 = 10x + y .......(1)
10x + y + 18 = 10 y + x ....(2)
Solving 1st equation we get 2x - y = 1 .....(3)
Solving 2nd equation we get y - x = 2 .....(4)
Solving 3 and 4, we get x = 3 and y = 5

2. a, b, c are non negitive integers such that 28a+30b+31c = 365. a + b + c = ?

a) Greater than 14
b) less than or equal to 11
c) 13
d) 12
In a calender,
Number of months having 28 days = 1
Number of months having 30 days = 4
Number of months having 31 days = 7
28 x 1 + 30 x 4 + 31 x 7 = 365
Here, a = 1, b = 4, c = 7.
a+b+c = 12

3. George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours.  George, paul and hari start the same work at 9 am, while george stops at 11 am, the remaining two complete the work. What time will the work complete?

a) 11.30 am
b) 12 noon
c) 12.30 pm
d) 1 pm
Let the total work = 120 units.
As George completes this entire work in 8 hours, his capacity is 15 units /hour
Similarly, the capacity of paul is 12 units / hour
the capacity of Hari is 10 units / hour
All 3 started at 9 am and worked upto 11 am. So total work done upto 11 am = 2 x (15 + 12 + 10) = 74
Remaining work = 120 - 74 = 46
Now this work is to be done by paul and hari.  46 / (12 + 10) = 2 hours (approx)
So work gets completed at 1 pm

4. If x^y denotes x raised to the power y, Find last two digits of (1141^3843) + (1961^4181)

a) 02
b) 82
c) 42
d) 22
Remember 1 raised to any power will give 1 as unit digit.
To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
So the Last two digits of the given expression = 21 + 61 = 82

5. J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days.  H alone can dig the well in How many days?
a) 32
b) 48
c) 96
d) 24
Assume the total work = 48 units.
Capacity fo J = 48 / 16 = 3 units / day
Capacity of P = 48 / 24 = 2 units / day
Capacity of J, P, H = 48 / 8 = 6 units / day
From the above capacity of H = 6 - 2 - 3 = 1
So H takes 48 / 1 days = 48 days to dig the well

6. If a lemon and apple together costs Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a  lemon.  What is the cost of lemon?
L + A = 12 ...(1)
T + L = 4 .....(2)
L + 8 = A
Taking 1 and 3, we get A = 10 and L = 2

7. 3 mangoes and 4 apples costs Rs.85. 5 apples and 6 peaches costs 122.  6 mangoes and 2 peaches costs Rs.144.  What is the combined price of 1 apple, 1 peach, and 1 mango.
a) 37
b) 39
c) 35
d) 36
Sol: Note: It is 114 not 144.
3m + 4a = 85 ..(1)
5a + 6p = 122 ..(2)
6m + 2p = 114 ..(3)
(1) x 2 => 6m + 8a = 170
(3) =>  6m + 2p = 114
Solving we get 8a - 2p = 56 ..(4)
(2) =>  5a + 6p = 122
3 x (4) = 24a - 6p = 168
Solving we get a = 10, p = 12, m = 15
So a + p + m = 37

8. An organisation has 3 committees, only 2 persons are members of all 3 committee but every pair of committee has 3 members in common. what is the least possible number of members on any one committee?
a) 4
b) 5
c) 6
d) 1
Sol:
Total 4 members minimum required to serve only on one committee.

9. There are 5 sweets - Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday. A person eats one sweet a day, based on the following constraints.
(i) Ladu not eaten on monday
(ii) If Jamun is eaten on Monday, Ladu should be eaten on friday.
(iii) Peda is eaten the day following the day of eating Jilebi
(iv) If Ladu eaten on tuesday, kaju should be eaten on monday
based on above, peda can be eaten on any day except
a) tuesday
b) monday
c) wednesday
d) friday
From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on monday.

10. If YWVSQ is 25 - 23 - 21 - 19 - 17, Then MKIGF
a) 13 - 11 - 8 - 7 - 6
b) 1 - 2-3-5-7
c) 9 - 8 - 7 - 6 - 5
d) 7 - 8 - 5 - 3
MKIGF = 13 - 11 - 9 - 7 - 6
Note: this is a dummy question. Dont answer these questions

11. Addition of 641 + 852 + 973 = 2456 is incorrect.  What is the largest digit that can be changed to make the addition correct?
a) 5
b) 6
c) 4
d) 7
Sol:
641
852
963
------
2466
Largest among tens place is 7, so 7 should be replaced by 6 to get 2456

12. Value of a scooter depriciates in such a way that its value at the end of each year is 3/4th of its value at the beginning of the same year. If the initial value of scooter is 40,000, what is the value of the scooter at the end of 3 years.
a) 23125
b) 19000
c) 13435
d) 16875
value of the scooter at the end of the year = $40000 \times {\left( {\dfrac{3}{4}} \right)^3}$ = 16875

13. At the end of 1994, R was half as old as his grandmother.  The sum of the years in which they were born is 3844.  How old R was at the end of 1999
a) 48
b) 55
c) 49
d) 53
In 1994, Assume the ages of GM  and R = 2k, k
then their birth years are 1994 - 2k, 1994 - k.
But given that sum of these years is 3844.
So 1994 - 2k + 1994 - k = 3844
K = 48
In 1999, the age of R is 48 + 5 = 53

14. When numbers are written in base b, we have 12 x 25 = 333, the value of b is?
a) 8
b) 6
c) None
d) 7
Let the base = b
So, (b+2)(2b+5) = $(b + 2)(2b + 5) = 3{b^2} + 3b + 3$
$2{b^2} + 9b + 10 = 3{b^2} + 3b + 3$
${b^2} - 6b - 7 = 0$
Solving we get b = 7 or -1
So b = 7

15. How many polynomials of degree >=1 satisfy $f({x^2}) = {[f(x)]^2} = f(f(x)$
a) more than 2
b) 2
c) 0
d) 1
Sol:Let f(x) = ${x^2}$
$f({x^2}) = {[{x^2}]^2} = {x^4}$
${\left( {f(x)} \right)^2} = {[{x^2}]^2} = {x^4}$
$f(f(x)) = f({x^2}) = {[{x^2}]^2} = {x^4}$
Only 1

16. Figure shows an equilateral triangle of side of length 5 which is divided into several unit triangles.  A valid path is a path from the triangle in the top row to the middle triangle in the bottom row such that the adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle.  An example is given below. How many such valid paths are there?
a) 120
b) 16
c) 23
d) 24

Sol:
Number of valid paths = (n-1) ! = (5-1)! = 24

17. In the question, A^B means, A raised to power B. If x*y^2*z < 0, then which one of the following statements must be true?
(i) xz < 0 (ii) z < 0  (iii) xyz < 0
a) (i) and (iii)
b) (iii) only
c) None
d) (i) only
As y^2 is always positive,  x*y^2*z < 0 is possible only when xz < 0. Option d is correct.

18. The marked price of a coat was 40% less than the suggested retail price.  Eesha purchased the coat for half the marked price at the fiftieth anniversary sale.  What percentage less than the suggested retail price did Eesha pay?
a) 60
b) 20
c) 70
d) 30
Let the retail price is Rs.100. then market price is (100-40) % of 100 = 60.  Eesha purchased the coat for half of this price. ie., 30 only. which is 70 less than the retail price. So Option C is correct.


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01cr56n350m
DC FieldValueLanguage
dc.contributor.authorBhattacharya, Pathikrit-
dc.contributor.otherGeosciences Department-
dc.date.accessioned2017-04-12T20:40:06Z-
dc.date.available2017-04-12T20:40:06Z-
dc.date.issued2017-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/dsp01cr56n350m-
dc.description.abstractThe laboratory derived rate-state friction (RSF) relationships are the most widely used constitutive equations for fault friction in numerical models of fault mechanics. But even after more than three decades of these being first proposed, we are far from certain about the identity of the ‘proper’ set of these equations which describe all laboratory friction data. In fact, the two most popular choices of the ‘state’ evolution component of RSF represent two end-member physical pictures of how frictional strength evolves – with time even without slip (Aging law) or only with slip (Slip law). Yet both these view points have traditionally been inferred to be independently supported by different classes of friction experiments which (sometimes) access similar portions of the RSF parameter space. We present a set of comprehensive studies which establish, both theoretically and with inversion of laboratory data, that in fact all the widely used experimental protocols provide evidence that friction dominantly evolves with slip even when the interface is sliding at the lowest slip rates accessed by these experiments. We examined these state evolution laws under a diverse range of sliding conditions – up to 3.5 orders of velocity steps on both initially bare rock and gouge, up to 3X10^4 s long holds on initially bare rock performed using machine stiffnesses differing by 1.5 orders of magnitude and 5% normal stress steps on initially bare rock carried out at an order of magnitude different sliding rates. For all of these experimental regimes, the widely used Aging law generally performed worse than the Slip law, even in those parts of the parameter space where conventional RSF wisdom would have predicted it to find strong support. Additionally, across all these experiments, more recent prescriptions of state evolution were generally found to fit the data only as well as the Slip law even with the freedom of extra parameters. We argue that these findings contradict the traditional view that the state variable is a proxy for the ‘quantity’ of true contact area alone, it is likely that some measure of the ‘quality’ of contacts contributes significantly to state evolution as well.-
dc.language.isoen-
dc.publisherPrinceton, NJ : Princeton University-
dc.relation.isformatofThe Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: <a href=http://catalog.princeton.edu> catalog.princeton.edu </a>-
dc.subjectEarthquake Nucleation-
dc.subjectFault mechanics-
dc.subjectFriction-
dc.subjectRate and State friction-
dc.subject.classificationGeophysics-
dc.subject.classificationSecondary education-
dc.titleExamination of the rate-state friction equations under large perturbations from steady sliding: A theoretical and experimental study.-


text:
# Math Help - differentiating through the integral

1. ## differentiating through the integral

The problem is as follows:
Solve for f(x), given

$f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}$

What I did:
1. Differentiate both sides wrt x, and partially differentiate through the integral.

$f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}$
$f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}$
$f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}$
$f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}$

2. Differentiate both sides wrt x again.
$f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}$
$f''(x) = - xf'(x) - f(x) + f(x)$
$f''(x) = - xf'(x)$

This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)

2. Originally Posted by compliant
The problem is as follows:
Solve for f(x), given

$f(x) = x + \int^{x}_{0}{(x-2t)f(t) dt}$

What I did:
1. Differentiate both sides wrt x, and partially differentiate through the integral.

$f'(x) = 1 + \frac{d}{dx} \int^{x}_{0}{(x-2t)f(t) dt}$
$f'(x) = 1 + \int^{x}_{0}\frac{\partial}{\partial x}{(x-2t)f(t) dt}$
$f'(x) = 1 + (x-2x)f(x) + \int^{x}_{0}{f(t) dt}$
$f'(x) = 1 - xf(x) + \int^{x}_{0}{f(t) dt}$

2. Differentiate both sides wrt x again.
$f''(x) = - xf'(x) - f(x) + \frac{d}{dx}\int^{x}_{0}{f(t) dt}$
$f''(x) = - xf'(x) - f(x) + f(x)$
$f''(x) = - xf'(x)$

This is an ODE, yes, but how would I solve for this? (i.e. is there a less messy way to calculate this other than using series solutions, which I hate with a passion?)
Let y = f(x). Then you have

$\frac{d^2 y}{dx^2} = - x \frac{dy}{dx}$.

Substitute $\frac{dy}{dx} = p$ .... (1):

$\frac{dp}{dx} = - x p$ .... (2)

DE (2) is seperable. Solve it for p as a function of x. Then substitute your expression for p into DE (1) and solve for y as a function of x.


text:
# Lbp

## 1-bit news

OK, it’s been a while again, but here are the reasons for being silent. 1-bit LBD reconstruction is out! Remember all those posts about LBDs, FREAKs, BRIEFs, and reconstructing images ? The latest version of tis work is out with a major feature: 1-bit quantized LBDs reconstruction. Yes, 1-bit ! To achieve this, we leverage some results from 1-bit Compressed Sensing thanks to my co-author Laurent Jacques. So, the pre-print is on arxiv and on infoscience, and you can get the code from github.

## Linearized model of LBPs

LBP model An LBP can be decomposed into two tiers: first, a real description vector, obtained by convolution-then-difference then the quantization (binarization) operation. Mathematically, the real i-th component of the descriptor is computed with the formula: $${\mathcal L}(p)i = \langle{\mathcal G}{x_i, \sigmai} , p \rangle - \langle G{x_i’, \sigmai’} , p\rangle,$$ where ${\mathcal G}{x_i, \sigmai}, {\mathcal G}{x’_i, \sigma’_i}$ are two Gaussians. The variety of the LBP family comes from the choice of these Gaussians : they can have a fixed size but random positions (a la BRIEF), fixed sizes and positions (a la BRISK)1… The choice in FREAK was :

## FREAK reconstruction as an inverse problem

In a previous post, I have briefly introduced our FREAK descriptor, which belongs to the more general family of the LBPs. In this post, I will state mathematically the problem of the reconstruction of an image patch given its descriptor, i.e. answering the question:

Can you guess which image part created this particular description vector ?

## FREAK makes it into OpenCV trunk

Integrating our lab’s FREAK in your vision workflow is getting easier and easier!

## Introducing FREAKs

OK, starting to mix posts in English and French since the English version of the blog is still buggy.


text:
# Set of Finite Subsets of an Infinite Set (Enderton, Chapter 6.32)

Exercise 6.32 (Enderton, Elements of Set Theory) (answer to same question) states:

Let $\mathsf{F}(A)$ be the set of all finite subsets of $A$. Show that if $A$ is infinite then $A \approx \mathsf{F}(A)$.

For the purposes of this exercise, I could use the approach cited. However, I want to know what is wrong with the attempt (2) I made. I will describe it informally.

1. If $|A| = \aleph_0$ then the injection $f(x) = \{x\} : A \rightarrow \mathsf{F}(A)$ gives $|A| \leq |\mathsf{F}(A)|$. Conversely, the bijections $f : \omega \rightarrow A$ and $g : \bigcup_{i \in \omega} \omega^i \rightarrow \omega$ (i.e., a tupling function) can be composed to obtain an injection $f : \mathsf{F}(A) \rightarrow A$. So $|\mathsf{F}(A)| \leq |A|$. By the Schroder-Bernstein Theorem, $|A| = |\mathsf{F}(A)|$ and $A \approx \mathsf{F}(A)$.
2. If not, I want to do the following: (i) define a set $H$ of bijections from subsets of $\mathsf{F}(A)$ to subsets of $A$; (ii) show that $H$ is closed under union of chains ($\bigcup B : \mathsf{F}(A) \rightarrow A$ is bijective where $B$ is a chain in $H$); (iii) employ Zorn's Lemma to obtain a maximal function $f \in H$; (iv) conclude that $\mathsf{F}(A) \approx A$. This approach would be very easy, because I could appeal to prior theorems regarding unions of functions and so on.

After step (iii), I need to show that $dom(f) = \mathsf{F}(A)$ and $ran(f) = \mathsf{F}(A)$. So I suppose not and obtain a contradiction due to $f$'s maximality in $H$. However, this is problematic. Why? Because in each attempt that I have made, I can replace the "finite powerset" operator by the powerset operator throughout and obtain a false statement. So I am missing something about $\mathsf{F}(A)$. I'm not sure what this is.

-

The maximality of $f$ isn't enough to show that the domain of the function is $F(A)$. It it enough to tell you just the fact that either the domain is $F(A)$ or the range is $A$.

This is because if you can find $x\in A$ and $y\in F(A)$ such that $y\notin dom(f)$ and $x\notin ran(f)$ then you can extend $f$ to $f\cup\{(y,x)\}$. But it may be possible that the domain of this maximal function is a proper subset of $F(A)$ while its range is $A$ and therefore it is impossible to extend this function inside $H$ since you require the functions to be $1-1$.

As Asaf points out the problem (assuming choice) is very easy. You just observe that $\kappa\times\kappa=\kappa$ for all cardinals and by using that you get that $\kappa^n=\kappa$ and $\aleph_0\times\kappa=\kappa$ which is enough to show what you want.

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Since he is using Zorn's llama to trek up the mountain of functions, he is already using choice. So it's no big deal. Without choice I am fairly certain it is a simple exercise to generate $A$ such that $|A|\lneqq |Fin(A)|$. –  Asaf Karagila May 11 '11 at 16:41
@Asaf: Indeed Zorn's lemon is equivalent to choice (and sour), I just wanted to emphasize the use of choice. And without choice you cannot even guarantee that $|[A]^2|=|A|$ (I think). –  Apostolos May 11 '11 at 17:13
+1 for both comments, and for you clever use of llama, lemon (aka lemma!) –  amWhy May 11 '11 at 19:47

First of all, for the countable case, one does not need this much of a hassle, simply because there is a very nice well ordering of $Fin(\omega)$, namely:

$$\{x_1,\ldots,x_n\}\mapsto\sum_{i=1}^n 2^{x_i}$$

As for the higher cardinalities:

By induction we have $\lambda^n=\lambda$ for all $n<\omega$, and therefore $\lambda\le\lambda\cup\lambda^2\cup\ldots \le\lambda\times\lambda = \lambda$ and clearly every finite subset of $\lambda$ can be found within some $\lambda^n$.

Note, however this requires the axiom of choice, and will not work without it as not always $\lambda+\lambda=\lambda$ and let alone $\lambda\times\lambda$.

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text:
# Spring force

1. Jul 30, 2006

### teng125

A spring with length l = 8 cm hangs in
vertical direction somewhere close the earth's surface. When a mass of m = 10 kg is attached
to the spring its rest-position is at y0 = 11 cm. (a) Determine the spring constant k.

i try to use F=-kx = mg where x=11cm - 8cm
therefore,i would like to know whether my steps are correct or i used the wrong formula

somebody pls help

2. Jul 30, 2006

### teng125

part b) The
mass is pulled out of its rest-position by a distance Δy = 5 cm. What is the acceleration acting
on the mass in the moment of its release?

i got the answer of a=16.35m/s^2

am i right??

3. Jul 30, 2006

### sdekivit

You're in the right direction :) Remember that at the spring its rest position $$F_{gr.} = F_{spring} = C \cdot u$$ where C = spring constant.

So you need to solve: $$m \cdot g = C \cdot u$$ with u = distance of strecthing of the spring (3 cm)

4. Jul 30, 2006

### sdekivit

foir b i would say that you use the second law of Newton: $$F_{res} = m \cdot a$$

5. Jul 30, 2006

### Staff: Mentor

You are correct. (Except for that minus sign. Set the magnitude of the spring force equal to the magnitude of the weight.)

6. Jul 30, 2006

### Staff: Mentor

7. Jul 30, 2006

### sdekivit

agrees with Doc Al :)


text:
0
research-article

Performance Analysis and Working Fluid Optimization of a Cogenerative Organic Rankine Cycle Plant

[+] Author and Article Information
D. Micheli, R. Taccani

Department of Engineering and Architecture,
University of Trieste,
via Valerio 10,
34100 Trieste, Italy

P. Pinamonti

Department of Electrical and Mechanical
Engineering and Management,
University of Udine,
via delle Scienze 208,
33100 Udine, Italy
e-mail: piero.pinamonti@uniud.it

Contributed by the Advanced Energy Systems Division of ASME for publication in the Journal of Energy Resources Technology. Manuscript received November 10, 2011; final manuscript received May 29, 2012; published online January 10, 2013. Assoc. Editor: Sarma V. Pisupati.

J. Energy Resour. Technol 135(2), 021601 (Jan 10, 2013) (11 pages) Paper No: JERT-11-1144; doi: 10.1115/1.4023098 History: Received November 10, 2011; Revised May 29, 2012

Abstract

The paper presents the results of a research regarding the application of cogeneration plants, based on organic rankine cycle (ORC), fed with wood residuals. In the first part of the paper an energy audit of the companies in a furniture industry district, located in the North-East of Italy, is presented. On the basis of these data a typical electricity/thermal demand profile dependent on the number of employees has been determined. In order to evaluate the potential savings achievable with an ORC cogeneration plant, a numerical simulation model has been developed to analyse the energy balances of the components as well as the whole ORC power plant performance. The effects on the system energy and exergy efficiencies of different binary and ternary mixtures of polisiloxane as working fluid and of different operating modes (cogeneration or pure electricity production) have been analysed, also in off-design conditions.

<>

Figures

Fig. 1

Different fuels utilized in the industrial district firms

Fig. 2

Installed thermal power as a function of employees number

Fig. 3

Installed electric power as a function of employees number

Fig. 4

Power to thermal ratio as a function of installed electric power

Fig. 5

Simplified scheme of a cogenerative ORC plant

Fig. 8

Heat transfer diagram for some of the considered working fluids

Fig. 7

ORC model flow sheet in Aspen Plus®

Fig. 6

T-S diagram of the ORC cycle with MDM

Fig. 10

Electric efficiency as a function of inlet thermal load, at various condensing pressures (kPa). Non-cogenerative operation.

Fig. 9

Electric efficiency as a function of inlet thermal load, at various condensing pressures (kPa). Cogenerative operation.

Fig. 11

Electricity to heat ratio as a function of inlet thermal load, at various condensing pressures (kPa). Cogenerative operation.

Fig. 12

Gross electric power as a function of inlet thermal load, at various condensing pressures (kPa). Non-cogenerative operation.

Fig. 13

Plant components allocation of exergy introduced with diathermic oil. Cogenerative operation, pCON = 14 kPa, Qin = 6000 kW.

Fig. 14

Plant components allocation of exergy introduced with diathermic oil. Non-cogenerative operation, pCON = 4 kPa, Qin = 6000 kW.

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text:
## Tohoku Mathematical Journal 2001 June SECOND SERIES VOL. 53, NO. 2

 Tohoku Math. J. 53 (2001), 183-202

Title DENSITY ESTIMATE IN SMALL TIME FOR JUMP PROCESSES WITH SINGULAR LÉVY MEASURES

Author Yasushi Ishikawa

(Received July 12, 1999, revised July 31, 2000)
Abstract. We consider the asymptotic behaviour of the transition density for processes of jump type as the time parameter $t$ tends to 0. We use Picard's duality method, which allows us to obtain the lower and upper bounds of the density even for the case where the support of Lévy measure is singular. The main result is that, under certain restrictions, the density behaves in polynomial order or may decrease in exponential order as $t \to 0$ according to geometrical conditions of the objective points.

2000 Mathematics Subject Classification. Primary 60J75; Secondary 60J25.


text:
# How many random permutations to cover all possible permutations?

I have code that generates a random permutation. In my case, a permutation consists of N binary features, and each of the N features is set or unset randomly. How many times must I generate a random permutation in order to be reasonably assured that I have covered all possible permutations? I'm not sure how to define "reasonably assured" (50%, 95%?).

For example, let's say that N = 10, so there are 1024 possible permutations, how many times should I randomly generate a permutation to be 50% confident that I have generated all 1024 permutations?

It seems to me that this is related to the 36.8% duplicate hit rate when drawing a sample with resampling, but I'm not a statistician.

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If you have the computing power to generate all permutations (or some well-defined set of re-assignments of the data), why not systematically generate each exactly once? That would normally be far more efficient and give you exact, rather than approximate, answers. (This idea is identical to the difference between sampling a small finite population with and without replacement: why sample with replacement when you can obtain a complete census and eliminate all sampling error?) –  whuber Oct 11 '12 at 21:28
My goal is not to generate all permutations, but rather to find a way to quantify how "good" a randomly generated sample is. –  Tyro Oct 11 '12 at 22:59
Huh? If you can exhaustively generate all permutations, then you can answer any question of that sort with perfect accuracy. Generating, and re-generating, far more permutations than you need is just a complete waste of computing time and will give you less accurate answers. –  whuber Oct 12 '12 at 14:29

Permutation usually refers to something else, so it's probably better to call your problem "random binary words" or something similar.

The question of how long it takes to get at least one representative of each type is called the Coupon Collector Problem. If you assume that all binary words of length $N$ are equally likely, then there are $2^N$ types of coupons. You can write the time to collect all coupons as a sum of the times to collect the $i$th new coupon, a sum of independent geometric random variables. So, the expected number of coupons it takes to collect them all is $2^N \sum_{i=1}^{2^N} 1/i \sim 2^N \log 2^N$, or more precisely $2^N \log 2^N + 2^N \gamma + 1/2 + o(1)$. For $N=10$ this is about $7689$. The variance is $2^{2N} \sum_{i=1}^{2^N} 1/i^2 \approx 2^{2N} \frac {\pi^2}{6}$, so the standard deviation is about $2^N \frac{\pi}{\sqrt{6}}$. For $N=10$ this is about $1313$. Note that a normal approximation is NOT appropriate here.

One crude bound is Chebyshev's inequality, which says that the chance that a random variable is more than $k$ standard deviations away from the mean is at most $1/k^2$, and the similar Cantelli's inequality is that the chance that a random variable is at least $k$ standard deviations above the mean is at most $1/(k^2+1)$. This gives you an upper bound of about $7689 + 1313 \approx 9002$ for the median, and $7689 + 1313\sqrt{19} \sim 13412$ for the $95$th percentile.

If these bounds are not good enough, there are more precise but more complicated asymptotics known. Another approach, suitable perhaps up to $N = 25$, is to compute the exact distribution numerically using the representation as a sum of independent geometric distributions.

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I don't follow where the sqrt(19) came from for the bound for the 95th percentile. –  Tyro Oct 11 '12 at 23:28
If $k=\sqrt{19}$ then $1/(k^2+1) = 1/20 =5\%$. So, Cantelli's lemma says that the probability that the result is at most $\sqrt{19}$ standard deviations above the mean is at least $95\%$. –  Douglas Zare Oct 11 '12 at 23:58
Thanks for the clarification. –  Tyro Oct 12 '12 at 0:08


text:
# TTP06-29 Equations for Relativistic Particles, Binaries and Ternaries

TTP06-29 Equations for Relativistic Particles, Binaries and Ternaries

TTP06-29 Equations for Relativistic Particles, Binaries and Ternaries

Relativistic differential equations for bound states of $n$ Dirac particles ($n=1,2,3$) are put into forms that depend only on the square of the total cms energy. Instead of coupling $4^n$ Dirac components, they decompose into two equivalent equations for $4^n/2$ components $\psi$ and $\chi$ of total chiralities $+1$ and $-1$, respectively. Their time-dependent versions are of the Klein-Gordon type; they reproduce the relativistic kinematics for the emission of photons or pions. Although the equations are presently extended to mesons and baryons within perturbative QCD only, the necessity of free Klein-Gordon equations for closed systems implies $E^2$-spectra not only for mesons, but also for baryons. For binaries, there exists an intricate transformation which turns the equation for $\psi$ into an effective one-body Dirac equation. The corresponding transformation for $\chi$ is derived here.

 Hartmut Pilkuhn PDF PostScript