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# SB@ in dtx and sty files

When reading several dtx files in order to write my own, I saw the command \SB@... in front of several other commands. What does it mean? I could not find reliable sources as explanation.

It is not in front of the command it is part of the command, in packages @ is a letter so \SB@foobar is just the command with name \SB@foobar package authors often use such a naming scheme to avoid clashes with other packages.
In this case the songs package author has used this naming scheme
 \def\SB@star...

so \SB@star could have been called \zzzstar the @ has no special role here other than for example all longtable package internals are called \LT@... so the chance of two packages clashing is reduced.


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# Math Help - percentage % of a number

1. ## percentage % of a number

okay i'm currently studying for a ASVAB test for the military. and they have questions like the one below,

" what number is 5% of 2000 "

how would i solve for something like that?

or they have

" 35% of what number is 70 "

" if 45 is 120% of a number, what is 80% of the same number? "

" if 6 is 24% of a number, what is 40% of the same number? "

" 25% of $900= " questions like those, please help 2. Originally Posted by lambofgod " what number is 5% of 2000 " how would i solve for something like that? ...questions like those, please help To learn how to set up and solve this sort of exercise, try this lesson on "percent of" word problems. For instance, the above exercise would be set up as "(some number) is (five percent) of (2000)", or x = (0.05)(2000). 3. ok. The trick to percent problems is to move the decimal in the percent two places to the left. Example: 5% becomes 0.05 then if they question says: find 5% of 2000, you would move the decimal two places to the left, and multiply it by 2000. That is your answer. In this case 5% of 2000 is 100. so: (0.05)(2000)= 100 4. Originally Posted by lambofgod okay i'm currently studying for a ASVAB test for the military. and they have questions like the one below, " what number is 5% of 2000 " how would i solve for something like that? or they have " 35% of what number is 70 " " if 45 is 120% of a number, what is 80% of the same number? " " if 6 is 24% of a number, what is 40% of the same number? " " 25% of$900= "

try these formulas:

B=base(original number)
R=rate
P=percentage

if you are looking for the base:

$B=\frac{P}{R}$

ex.
35% of what number is 70

$B=\frac{P}{R}$

$B=\frac{70}{.35}$

$B=200$

percentage

$P=RB$

" what number is 5% of 2000 "

$P=.05(2000)$

$P=100$

rate

$R=\frac{P}{B}$

what % of 2000 is 100

$R=\frac{P}{B}$

$R=\frac{100}{2000}$

$R=5\%$

format: plain


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# Partial Pressure, Ideal Gases and Partial Pressure, Dalton՚s Law, Questions (For CBSE, ICSE, IAS, NET, NRA 2022)

Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.

## What is Partial Pressure?

• Partial Pressure is defined as if a container filled with more than one gas, each gas exerts pressure. The pressure of anyone gas within the container is called its partial pressure
• The pressure that is exerted by one among the mixture of gasses if it occupies the same volume on its own is known as Partial pressure. Every gas exerts certain pressure in a mixture. The total pressure of a mixture of an ideal gas is the sum of partial pressures of individual gases in the mixture, based on the following equation:

• denotes the partial pressure of the particular gas.
• indicates the partial pressure of the gas x.
• denotes the total volume of the mixture.
• indicates the amount of gaseous substance.
• denotes the total pressure of the mixture.
• is the total amount of substance in a mixture.

Partial Pressure

## Ideal Gases and Partial Pressure

• Ideal gas behavior allows gas mixtures to be specified simply. In particular, the ideal gas law holds for each component of the mixture separately. Each component exerts its own pressure referred to as its partial pressure.
• Partial pressure is the measure of the thermodynamic activity of gas molecules. The gasses diffuse and react based on their partial pressures and not concentrations in a gaseous mixture.

## Dalton՚s Law

Dalton՚s law of partial pressures is a gas law which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture. For example, the total pressure exerted by a mixture of two gases A and B is equal to the sum of the individual partial pressures exerted by gas A and gas B.

## Dalton՚s Law of Partial Pressure

According to Dalton՚s law of partial pressures, the total pressure exerted by the mixture of gases is the sum of the partial pressure of every existing individual gas, and every gas is assumed to be an Ideal gas.

Where are the partial pressures of gas 1, gas 2, and gas 3? Since every gas has an independent behavior, the ideal gas law is used to find the pressure of that gas if its number of moles, the volume of container and temperature is known.

The equality arises because the molecules are so wide apart that there is minimal interaction in an ideal gas. For example, a mixture of an ideal gas that consists of Nitrogen, hydrogen, and ammonia.

Where,

• = Partial pressure of nitrogen
• = Partial pressure of hydrogen
• = Partial pressure of Ammonia

Developed by:


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# Math Help - Optimization of Income

1. ## Optimization of Income

Im having a little trouble coming up with an equation to find maximum gross income.

A television company agrees to establish a community antenna in a new suburb at a charge of $25 each for 1000 subscribers or less. To encourage more subscribers, it is agreed to give a discount of 10 cents to each subscriber for every additional 10 subscribers in excess of 1000. For the company to make the most income what number of subscribers should it have? Any help would be greatly appreciated! 2. Let s be the number of subscribers. Then the price is$25 minus 0.1 * (s-1000)/10. So the revenue for the tv company is given by:

$R=s(25-0.1\frac{s-1000}{10})=35s-0.01s^2$

I assume you can find the maximum of this function by setting the derivative to zero. The answer is s=1750.

But that assumes the price is changing continuously with the number of subscribers, which is only an approximation. To get the exact answer, we should calculate some values around 1750. We can reduce the amount of work by recognizing that the answer must end in 9.

s=1739, R=30780.3
s=1749, R=30782.4
s=1759, R=30782.5
s=1769, R=30780.6


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+0

# I am confused with this question

+1
119
2

By the end of her first day of reading, Emily had read 1/4 of her book. By the end of the second day, she had read another 29 pages. There are 136 pages in the book. How many pages did she have to read after the second day to finish the book?

Apr 6, 2021

#1
+484
+1

So in total she read $\frac{1}{4}*136+29=34+29=63$.

She has $136-63=\boxed{73}$ pages left to read

Apr 6, 2021
#2
+320
+2

Hey there, Guest!

Two equations can solve this problem for us here.

$$\frac{136}{4}=34$$

$$136 - 34 - 29 =\left [ 73\right ]$$


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On the Necessity of Enumerating All Programs

In a tutorial on average-case complexity that I gave at FOCS 2008, I mentioned four of my favorite open questions in the theory. I have some progress to report on one of the questions (#3 in this post), and on two related problems.

The question is whether a certain exponential loss is necessary in Levin’s proof that there are problems that are “NP-complete on average.” The answer is yes, it is necessary, unless the polynomial hierarchy collapses. The two related problems are whether related exponential losses in the analysis of Levin’s universal search algorithm and of Levin’s universal one-way function are also necessary. The answers are, respectively, yes unless ${P=NP}$, and yes relative to an oracle.

This is one of those now fashionable papers whose arguments are “obvious in retrospect.” (I would like to claim, however, that the proof of the result for Levin’s average-case completeness, while obvious in hindsight, is not so obvious without hindsight.) In all three cases, the exponential loss is in terms of the code length of certain programs. The idea in each case is to construct programs that contain in their code information about a specific instance of a problem of interest.

CS 276 Lecture 11: One-Way Functions

Summary

Today we begin a tour of the theory of one-way functions and pseudorandomness.

The highlight of the theory is a proof that if one-way functions exist (with good asymptotic security) then pseudorandom permutations exist (with good asymptotic security). We have seen that pseudorandom permutations suffice to do encryption and authentication with extravagantly high levels of security (respectively, CCA security and existential unforgeability under chosen message attack), and it is easy to see that if one-way functions do not exist, then every encryption and authentication scheme suffers from a total break.

Thus the conclusion is a strong “dichotomy” result, saying that either cryptography is fundamentally impossible, or extravagantly high security is possible.

Unfortunately the proof of this result involves a rather inefficient reduction, so the concrete parameters for which the dichotomy holds are rather unrealistic. (One would probably end up with a system requiring gigabyte-long keys and days of processing time for each encryption, with the guarantee that if it is not CCA secure then every 128-bit key scheme suffers a total break.) Nonetheless it is one of the great unifying achievements of the asymptotic theory, and it remains possible that a more effective proof will be found.

In this lecture and the next few ones we shall prove the weaker statement that if one-way permutations exist then pseudorandom permutations exist. This will be done in a series of four steps each involving reasonable concrete bounds. A number of combinatorial and number-theoretic problems which are believed to be intractable give us highly plausible candidate one-way permutations. Overall, we can show that if any of those well-defined and well-understood problems are hard, then we can get secure encryption and authentication with schemes that are slow but not entirely impractical. If, for example, solving discrete log with a modulus of the order of ${2^{1,000}}$ is hard, then there is a CCA-secure encryption scheme requiring a ${4,000}$-bit key and fast enough to carry email, instant messages and probably voice communication. (Though probably too slow to encrypt disk access or video playback.)

CS276 Lecture 11 (draft)

Summary

Today we begin a tour of the theory of one-way functions and pseudorandomness.

The highlight of the theory is a proof that if one-way functions exist (with good asymptotic security) then pseudorandom permutations exist (with good asymptotic security). We have seen that pseudorandom permutations suffice to do encryption and authentication with extravagantly high levels of security (respectively, CCA security and existential unforgeability under chosen message attack), and it is easy to see that if one-way functions do not exist, then every encryption and authentication scheme suffers from a total break.

Thus the conclusion is a strong “dichotomy” result, saying that either cryptography is fundamentally impossible, or extravagantly high security is possible.

Unfortunately the proof of this result involves a rather inefficient reduction, so the concrete parameters for which the dichotomy holds are rather unrealistic. (One would probably end up with a system requiring gigabyte-long keys and days of processing time for each encryption, with the guarantee that if it is not CCA secure then every 128-bit key scheme suffers a total break.) Nonetheless it is one of the great unifying achievements of the asymptotic theory, and it remains possible that a more effective proof will be found.

In this lecture and the next few ones we shall prove the weaker statement that if one-way permutations exist then pseudorandom permutations exist. This will be done in a series of four steps each involving reasonable concrete bounds. A number of combinatorial and number-theoretic problems which are believed to be intractable give us highly plausible candidate one-way permutations. Overall, we can show that if any of those well-defined and well-understood problems are hard, then we can get secure encryption and authentication with schemes that are slow but not entirely impractical. If, for example, solving discrete log with a modulus of the order of ${2^{1,000}}$ is hard, then there is a CCA-secure encryption scheme requiring a ${4,000}$-bit key and fast enough to carry email, instant messages and probably voice communication. (Though probably too slow to encrypt disk access or video playback.)


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• # question_answer When chlorine water is exposed to sunlight, ${{O}_{2}}$ is liberated. Hence [AFMC 1989] A) Hydrogen has little affinity to ${{O}_{2}}$ B) Hydrogen has more affinity to ${{O}_{2}}$ C) Hydrogen has more affinity to $C{{l}_{2}}$ D) It is a reducing agent

Solution :

$\underset{\text{Exposed to air}}{\mathop{{{H}_{2}}O+C{{l}_{2}}}}\,\,\to \,HCl+HClO$ $HClO\,\to \,HCl+[O]$ or $2HClO\,\to \,2HCl+{{O}_{2}}$

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# Rabinowitsch in Translation

One of our projects in number theory led us to thinking about the class number problem, which has a story too long and interesting to recount here. See the survey of Goldfeld from an old AMS Bulletin to get an overview. Briefly, the question is about the degree to which unique factorization holds in quadratic extensions of the rational numbers.  In any case, one of the first important results in the program is an old theorem of G. Rabinowitsch (Rabinowitz), which gives a testable criterion for whether the number field , with a negative integer, possesses unique factorization. In modernish language (cf. Theorem 6 in the linked-to document), keeping this framework we have:

Theorem (Rabinowitsch, 1913): The field is a unique factorization domain (UFD) if and only if is prime for all , where .

The paper was published in German in Crelle’s journal over a century ago, and was somewhat hard to find on its own. We could not locate any other source where the content has been rewritten since then, so we translated and typeset the article (with the aid of various online translation tools). To preserve the spirit and style of the writing, some outdated and perhaps idiosyncratic (to author or translator, as the case may be) jargon has been allowed to survive. These designations are hopefully all made clear within the article so it is readable. Some notation has been modified for clarity.

Perhaps one day this material can be condensed and fully modernised. The article consists mostly of pleasant and clean elementary number theory, situated near the headwaters of one of the important achievements of 20th century number theory. It seems worthy of further propagation, so we post here our translation (no warranty included).

# Spectral Sequences III

McCleary introduces the concept of a differential graded algebra in section 1.3 (Definition 1.6, p. 11). These are algebras (over a field ), which tend to be -graded, and importantly carry with them a map called a differential which is -linear, shifts the degree of elements (in the grading) up by one, and satisfies a “Leibniz rule:”

for in our algebra . This is a twisted version of what is usually called Leibniz’ rule in calculus (which is basically just product rule), which coincides with how the differential works in the algebra of differential forms.

This idea is easily extended to the notion of a differential bigraded algebra , where now the elements are graded (for the time being, later we’ll have ), but remains a total-degree 1 mapping. That is,

and still satisfies the Leibniz rule

(1)

where .

A standard construction is to form a bigraded algebra by tensoring two graded algebras together. This would work with just component-wise multiplication, but to get a working differential that satisfies our version of the Leibniz rule 1 as well, we introduce an extra sign: we mean, supposing and are differential graded algebras, then we can assign , and furthermore

(2)

Then if we define a differential on by

(3)

then satisfies the Leibniz rule 1. It is clarifying to check this, so we’ll record it here. Switching notation a bit, we will write instead of . To satisfy 1 we need

we then apply 3 to the individual terms on the right side above to get

Now applying the multiplication rule 2 and distributing, we find

(4)

To check the rule holds, we perform this computation by instead multiplying first and then applying the differential. That calculation looks like

Finally, remarking that and shows that terms of the last line above match with those of 4, so everything checks out and becomes a differential bigraded algebra.

## A Chain Rule

Given the length and detail of section 1.3, surprisingly we find no glaring errors in this section, but the use of the differential becomes somewhat muddled in calculation in section 1.4. Again, perhaps as an undesirable side effect of the fact that we remain at the “informal stage,” it’s always difficult to keep track of what assumptions we’re working with in each example. Case in point, example 1.H, p. 20. The paragraph preceding definition 1.11 seems to indicate that all graded algebras are assumed to be graded commutative — at least for the rest of the section, one guesses, though the language is vague. Let’s try this here with a bit more force.

Assumption: All graded algebras are graded commutative for the rest of the post. This is to say, for all in any , we have . Now let’s have a look at the example. We suppose a spectral sequence of algebras with , converging to the graded algebra which is in degree 0 and in all others.  The example asserts that if is a graded commutative polynomial algebra in one generator/variable, then is a graded commutative exterior algebra in one generator, and vice versa.

The first confusion appears in a restatement of the Leibniz rule near the bottom of page 20, except this time there are tensors involved. This appears to be a mixed use/abuse of notation, which was slightly different in the first edition of the book, but not more consistent. The idea is as follows. and embed into under the maps and .  Then one can also write an element (mind the inexplicable inconsistent choice of letters) as

(5)

since the degree of 1 is zero in each graded algebra. Note that this also allows us to regard as graded commutative with the tensor product as multiplication between pure and pure elements, writing

One can apply Leibniz rule to the product in 5 so that if comes with a differential , we get

The thing is we really need not write the tensor product ; it is just as correct to write on it’s own, as we often do with polynomial algebras and so on. Then the above can be written instead as

as McCleary does near the bottom of page 20. What makes this confusing is that up to this point we had only seen differentials acting on tensors by defining the bigraded differential from tensoring two differential graded algebras together, seen above. In this context, the differential of the bigraded algebra must act on an element of the algebra coming from , it cannot act on just one side of the tensor. What’s different here is that the tensor product is actually the multiplication operation on each page of the spectral sequence. Thus, the restatement of the familiar rule with new notation.

Nevertheless, the next equality is also a bit confounding at first, partly because McCleary, goes back to writing the extra in the tensor, suggesting that we need to pay attention to its effect. He says that if , then

(6)

which looks sort of reasonable as it resembles something like a chain rule, . It is presented as if it should follow immediately from the Leibniz rule stated before. But this seems weird when the degree of is odd. To be totally transparent about this, let’s illustrate the case where , suppressing the subscript on the differential again, but maintaining the tensorial notation.

where the last line follows since has total degree , so the sign inside the sum there has exponent which is even. We see that if has odd degree then, these terms cancel and we get 0. So you say “wait a minute, that’s not right, I wan’t my chain rule looking thing” until you eventually realize that if has odd degree, since it’s sitting in a graded commutative algebra, is actually zero! And the same goes for all higher powers of . Then, makes complete sense. Meanwhile, if has even degree, the terms will pile up with positive sign and we get the chain rule looking thing that was claimed. So the statement 6 is in fact true, though it really breaks down into two distinct cases.

Going forward in the example, McCleary only really seems to use the chain rule (liberally mixing in the described sort of abuse of notation) on terms of even degree, so it’s tempting to think that it only applies there, but it is sort of “vacuously true” in odd degree as well. Oh well. Onwards.

# Spectral Sequences II

## Two Stripes

The next thing to address in McCleary is an apparent mistake on p. 9 of section 1.2. Here we again assume a first quadrant spectral sequence converging to a graded vector space . This is mentioned at the beginning of the section, but it’s easy to forget that when a bold-faced and titled example (1.D) seems to be presenting a reset of assumptions, rather than building upon prior discussion. Furthermore, in this example, McCleary seems to be working again with the assumption from example 1.A that for . On the other hand, this can be seen as a consequence of the fact that our spectral sequence is limited to the first quadrant, provided the filtration is finite in the sense of Weibel’s Homological Algebra, p. 123 ( for some ). But then it would be unclear why McCleary took this as an additional assumption rather than as a consequence of prior assumptions in the first case. : /

The new part of this example is the assumption that unless or , so all terms of the spectral sequence are to be found just in two horizontal stripes. In particular is only possibly non-zero in these stripes, and since these correspond to filtration quotients, the filtration takes a special form.

First, we might look at the filtration on where .  Note that the spectral sequence terms that give information about are those along the diagonal line where .  Since , the only place where anything interesting might happen is when this line crosses the -axis, i. e. when . This forces , so the only possible nonzero filtration quotient is

working with the assumption that . So on the one hand, we get no interesting filtration of for , but on the other hand we can see exactly what it is from the spectral sequence limit.

Now we treat the case of , where . I find this awkward notation again, preferring to reserve for a pure arbitrary spectral sequence index, but since we are trying to address the mistake in this notation, we should keep it for now. The filtration of this vector space/cohomology is interesting when and , where the quotients are given by

Every where else, successive quotients are 0, meaning the filtration looks like…

In the filtration on page 9, McCleary puts one of the (possibly) non-trivial quotients at instead of at where it should be.  That’s all I’m saying.

This situation is modeled on a spectral sequence for sphere bundles i.e. bundles where the fibers are spheres of a given dimension. The stripes coincide with the fact that a sphere has nontrivial cohomology only at and . This sort of computation is famous enough that it has a name: the Thom-Gysin sequence (or just Gysin sequence).

As a final remark on section 1.2, McCleary says that the sequence in example 1.C is the Gysin sequence. Example 1.C doesn’t exist, we mean example 1.D : )

# Spectral Sequences I

A goose chase through homological algebra etc. has led us to start reading McCleary’s A User’s Guide to Spectral Sequences. The book seems like a nice introduction for those that know their way around graduate topology and geometry, but haven’t yet encountered cause to pull out this extra machinery to compute (co)homology. First published by Spivak’s Publish or Perish Press in the 80’s, a second edition was released by Cambridge University Press in 2000. Though it sounds like the second edition is rather improved, there seem to be a number of mistakes remaining which may frustrate those trying to learn a notoriously complicated subject for the first time. Pending an official list of errata, we may as well collect some of them here.

## Section 1.1 – Notation

The first comment worth making is regarding some confusing notation, largely an overuse of the letter . The first use comes on p. 4 (Section 1.1), given a graded vector space and a filtration of , by defining

Here, the symbol seems to designate an endofunctor on (graded) vector spaces — it eats one and gives back another; transporting morphisms through the filtration and quotient shouldn’t be a problem either. It isn’t really clear what the subscript is supposed to indicate at this point, but the reader sits tight expecting the truth to be revealed.

However, on the very same page, McCleary twists things by making the assignment

(1)

where , the graded piece of the filtration. Now, with the extra index , is a vector space on it’s own. The notation doesn’t indicate reference to , though in this case it really depends on . For instance, McCleary indicates that we should write something like

The definition immediately afterwards (Definition 1.1) indicates is to be used to designate a vector space in a spectral sequence which is irrespective of any for all . The typical way to relate a spectral sequence , to a graded vector space is the situation of convergence (Definition 1.2, p. 5) where instead

The right hand side above has nothing to do with the spectral sequence (since we take in our definition), it is just an instance of the definition from equation 1… but with distinct use of notation… oh. So on the one hand, should be a standalone vector space, like the other ‘s, but also it needs to come from an so one should really write as in Definition 1.2. Wha? Shoot. Couldn’t we have used like an instead or something?

Perhaps there is good reasoning for all of this to be discovered once we get further in. Also, it seems so far that initial terms are usually . Why not ? And why don’t we allow -pages? In these cases the differentials would be vertical and horizontal (resp.) instead of diagonal, which feels less interesting somehow, though this doesn’t seem like it would be totally frivolous… TBD.

## Splicing Short Exact Sequences

Finishing out the first section, we address what seems to be a typo in example 1.A (p. 6). McCleary’s expository style consists of many statements which are not obvious, though usually not difficult to work out. This is perhaps for the best, as the community seems to indicate that the only real way to learn spectral sequences (make that: all math?) is by working them out. Nevertheless, it is a bit discouraging to find yourself at odds with the author at the first example…

We have assumed a first quadrant spectral sequence with initial term converging to with a filtration satisfying for all . Then we have a filtration on in particular, given by

since, by the assumption, etc., and by definition. By convergence, then,

so is a submodule of . But also because lies on the -axis (depicted as what is usually the -axis) and our spectral sequence has only first quadrant terms, must be the zero map for all . Furthermore, is too close to the -axis to get hit by any differential , thus survives as the kernel of every , mod the image of a from a zero vector space in the second quadrant. This is all to say

We then have part of the short exact sequence McCleary gives for as

How can we describe the third term using the spectral sequence? Well, from our definitions, . The book seems to be indicating that but this is not necessarily the case! It also doesn’t make sense with how the short exact sequences are spliced later on.

Let’s address the first claim first. Because lies on the -axis, and the differentials point “southeast” towards the empty fourth quadrant, is the zero map for any , but it can’t be hit by anything so we have now

The denominator is the image of a map from a zero vector space, so it is zero, and thus is a subspace of , but this latter space can be larger! This is all to say, the short exact sequence for is misprinted, and should go

(2)

One can confirm this by examining the SES given just below, where we see injecting into :

(3)

This is a standard decomposition of the map in the middle: for any morphism (in an abelian category at least, we suppose) there is a SES

It remains to see that . Because of where sits on the -axis, it is again the kernel of for all . Further, it can only possibly be hit by , so in fact survives through all further terms to give the desired equality

To splice all this together, we recall that we can connect

as

where . We maintain exactness since and .

Performing this surgery on sequences 2 and 3 yields the main exact sequence claimed by the example, namely

(4)

Stay tuned for more clarifications from Chapter 1.

# Hilbert Schemes IV: A Definition

At this point in the project, we begin to understand why so many books and papers are written in dry, terse, often difficult mathematical prose. Discursiveness which is beneficial is risky, time consuming and hard to pull off! As we continue to hurtle preliminary posts, maybe a bit of motivation is in order.

At the highest level, far-out, hardcore algebraic geometry is finding application in strange places these days. We will just cite the three instances which are close to the heart of mathlab: Hodge Theory in Combinatorics, Geometric Complexity Theory, and the Macdonald Positivity Conjecture where Hilbert schemes played a starring role. Understanding this last one is what we are heading for in these posts. Moving on…

So let’s just do this the way where you put a loaded definition out front and then try to tease it apart, following Nakajima‘s exposition, repeated in seminar.

Definition Let be a projective scheme over an algebraically closed field . The contravariant Hilbert functor is the functor that associates to a scheme the set of closed subschemes such that the natural morphism from to is flat.

From the top, we are choosing an algebraically closed field so the “projective” here is with respect to , as described in the last post. We would like to take this opportunity to express our regret for not really describing or the fiber product in these posts even though we earlier gestured towards thoroughness. Hopefully the references serve.

Moving on, an arbitrary scheme is said to be projective (over ) if it is of the form for a finitely generated graded -algebra. A subset of a scheme is said to be a closed subscheme iff:

• The inclusion map is an affine morphism, meaning that for each open affine subset , the inverse image for some ring . “The inverse image of affines is affine.”
• Additionally, for each of these pairs, the map is surjective.

\underline{Note}: In case is not a subset of but we still have a map which satisfies the above properties, that map is said to be a closed embedding (or, less tastefully, a closed immersion) of schemes.

Back in the definition, the map from to is the composition of the inclusion and the projection map . Remember, this isn’t set-wise projection, our product is in the category of schemes, where we can also think of it as the fibered product over the terminal object, . It is guaranteed to exist as in section 9.1 of The Rising Sea.

Finally we come to flatness, which is sort of exciting because this is the first time I’ve come across it in the wild since studying for my algebra qual, but still sort of mysterious. Apparently Mumford agrees, even as he reassures us that it is “technically the answer to many prayers,” in The Red Book (III.10). Remember that an module is said to be flat if tensoring with preserves short exact sequences. More loosely, flat modules are pretty close to (i.e. are direct limits of) free modules.

That the map is flat means that for all , the induced map on the stalks makes into a flat -module. For motivation on why flatness is the “right” condition in these sorts of circumstances, see [EH] II.3.4.

An important theorem of Grothendieck says that with as above is representable by a scheme which is “locally of finite type” called a *gasp* HILBERT SCHEME! A version of the proof with additional discussion and background is here. It is also explained in section 9 there that we can weaken the requirements so that need only be quasi-projective. In fact, one of the most interesting cases for us will be where is ( in particular). Further on, we will refine this by introducing Hilbert polynomials. Yay!

# Hilbert Schemes III

Having established that to each scheme we can associate its functor of points , we continue our introduction by mentioning some eminent examples.

First the projective space . This is sort of the basic example of a moduli space, which is also what were after with Hilbert Schemes. While (over your ring of choice) has a few different constructions, it’s points correspond to other geometric objects: lines in the affine space . This is the essence of moduli, and their value for “counting” objects of a given type.

We also have a description of the functor of points for projective space, , recalling that for a given scheme . See Vakil, Theorem 16.4.1, or also Eisenbud and Harris [EH], Theorem III-37

Theorem Morphisms are in bijection with the data (up to isomorphism) of where is a line bundle (invertible sheaf) on and the are global sections which do not vanish simultaneously.

By “up to isomorphism” here we mean that the value of the section at a given point on is depends on the trivialization used, but it will differ according to multiplication by a scalar which multiplies all simultaneously. Thus, the ensemble of all sections together effectively gives a point in (or a fiber of the tautological bundle, if you prefer), provided they are not all zero. Ranging over all of we get a morphism of schemes.

You might wonder which projective space we are working with here, i.e. over which field/ring. As will usually be the case, it doesn’t much matter because maps behave nicely under base change, but if we want to be as general as possible we should probably use . This is because is the initial object in the category , which makes it the final object in the category of affine schemes. Initialness comes from the fact that there is a unique morphism from to any other commutative ring with identity which just takes 1 to itself.

Then, is also final in because we have this ring morphism (which is an morphism in reverse) into all of the open affine covering neighborhoods of a given scheme, and these glue up nicely. So, whether we construct from a graded ring as or as the gluing of a bunch of , we again have a canonical scheme morphism . This together with lets us construct the fiber product ; see [EH] III.2.5 for a more thorough treatment of all of this. I’m still essentially just recording the things I had to remind myself of in jumping into this so it’s maybe more garbled than pedagogical.

[EH] here also gives the description of as subsheafs of which are locally of rank , which we think of as the kernels of the surjection from the direct sum sheaf , surjection being guaranteed by the non-vanishing of at least one of the sections above (at every point). This leads naturally into our other key example, the Grassmannian.

According to the Yoneda philosophy (which is often attributed to Grothendieck as well) we should actually think of the Grassmannian (and the special case of projective space) not just as the “space of -planes in -space,” but as a moduli space for certain bundles (locally free sheaves in AG-speak) and their sections. Really, it is a functor which can be applied to other schemes, and returns the collection of morphism which define the scheme that represents the functor. : S

The change with the Grassmann functor is that now we look at surjections from where is locally free of rank . At neighborhoods in a locally trivializing cover, we identify with and we can treat as a matrix. The Grassmannian will be covered by open sets which are the nonvanishings of the determinant of the maximal minors. This leaves the other entries of the matrix free, so they can vary and together determine a map in the way the sections gave a map to before. Thus, we can cover by a bunch of affine spaces which glue together. In the parlance, these are open subfunctors represented by the schemes .

It’s worth noting that the full Grassman functor is represented by a bona fide scheme of the form where denotes a subset of of size , which indexes the variables, an is a particular homogeneous ideal given by the Plücker equations. So we have a more traditional geometric object to think of as well. We may come back to this in more detail later if we need to. For now it suffices to say that applying the exterior power functor to gives a surjection from a locally free sheaf of rank onto a line bundle. So by the prior discussion, we have a map from the Grassmannian into which is the Plücker embedding.

Phew, okay I think in the next post we might actually define a Hilbert Scheme.

# Hilbert Schemes II

Next, we introduce the Hilbert scheme functorially following notes by Maclagan. See also Bertram’s course notes, who points out that we (as in Grothendieck) are trying to develop a category-theoretic framework for counting objects with a given property. First we should say a little bit about morphisms of schemes.

To do this, one should indicate that schemes (like manifolds) actually carry the additional structure of a locally ringed space. This means that when you take a kind of (filtered co-)limit on the structure sheaf over all the open neighborhoods containing some point , then we get a local ring . In this case of an affine scheme it works out that is actually the localization , as is a prime ideal of , and localization satisfies exactly the universal property prescribed by the colimit.

First, remember that schemes are ringed spaces, and that a morphism of ringed spaces is a continuous map of topological spaces together with a morphism of sheaves from to the push-forward . This induces a map on stalks

where and supposing sends the maximal ideal of one to the other, then we have a morphism of locally ringed spaces, and it is THIS, my friends, which will be a scheme morphism when the locally ringed spaces happen to be schemes… oh hell, this can’t be useful for anyone as exposition; just seriously go skim the first six chapters of The Rising Sea to clear things up. It’s only 200 pages.

Now we have objects and morphisms so we can talk about the category and functors therefrom. We are interested in the confusingly named functor of points for a given scheme , which is really the one that comes up in many versions of the Yoneda Lemma. The idea is that if category is locally small, meaning the collection of morphisms (arrows) between any two objects forms a set, then we can understand an object by the morphisms going into (or out of) it. Bearing in mind that a contravariant functor between categories is the same as a covariant functor , we have the following.

Definition Suppose is an object of , the category of schemes. Then there is a contravariant functor such that on objects we have , and for a given morphism we have

In general, functors that come from such ‘s are said to be representable. Elements of the set are called the Y-valued points of X, which is weird, because on sight they are just maps. Further, by laziness, in case for some ring (or field) A, then we will call theses maps the “-valued points of X.”

Endnote: for an interesting discussion on whether scheme theory can be taught using just , the maximal ideals of a ring, instead of the primes, that then digresses into thinking of schemes as their functors of points instead of the whole locally ringed space shebang, as well as other AG/Berkeley cultural inside jokes by a bunch of dudes, see the post and extensive comments here. Other interesting background can be found on nLab.

# Hilbert Schemes I

In this series of posts, we will report and editorialize the goings-on of the algebraic geometry (AG) and combinatorics learning seminar. The goal is to learn something about Hilbert Schemes and the relevant combinatorics.

First, since we are working with the category of schemes, we should indicate what those are and where to find out more about them. This feels like joining the chorus or preaching to the choir, but it’s hard to overstate the value of Vakil’s The Rising Sea/FOAG, for the background category and scheme theory: an innovation in mathematical text writing.

Okay so a scheme is like a variety, which is to say it’s like a manifold, which is to say it’s locally like other spaces we know and love and go under the general banner affine. The central metaphor here is that it comes from gluing simpler things together along the places they match up, and that the information about the spaces themselves really comes from information about the functions on the space… Ahem, made precise by the following gesture at a definition which really takes scores of pages to build up properly, and relies on a multitude of other definitions:

Definition 1 A Scheme is a ringed space such that every point in has a neighborhood such that is an affine scheme.

is a sheaf (of rings). That is, it’s a (contravariant) functor from the category of open subsets of , sometimes written , to , the category of commutative rings (with identity) and their homomorphisms. As elsewhere in life, if you find the category theoretic phrasing more obnoxious than illuminating, please ignore and seek the alternative. Or spend some time discovering the joy of cats, and join us. : D

So what’s an affine scheme? Turns out the affine scheme category is nothing but ! You know, the category with the same objects but with the arrows reversed? OK, I shut up. Just take your favorite ring . The collection of prime ideals of is (hopefully) a set, denoted , which can be given a topology by taking sets of the form

as basis for the open sets, where is some element of . We get the sheaf (also called a structure sheaf) for an affine scheme by taking

where the right hand side denotes localization at the multiplicative subset generated by .

Why use the prime ideals? I guess because localization works nicely for prime ideals. But also if we have an algebraically closed field , then Hilbert’s Nullstellensatz gives an identification of the points in with maximal ideals of the polynomial ring .

I always groan when I see people spend a page introducing like a semester’s worth of algebraic geometry, but, well, now I’ve done it too. Maybe this is another one of those rites of passage. Next up: some representable functors.


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# Stochastic process for interest rates allowing negative values

The Cox-Ingersoll-Ross process for the short term interest rate r(t) does not allow r(t) to become negative, but short-term rates are negative in much of the developed world. To account for this, do you use a CIR process for a shadow rate r'(t) that equals r(t) + c, where c = 0.01 if you think short-term rates cannot get more negative than 1%? Has there been research on this?

Yes, people have looked into that alreaday, for instance here and here (for lognormal models). Brigo and Mercurio took the CIR short rate $$(x_t)$$ and added a deterministic shift $$\vartheta(t)$$ to it in order to obtain the short rate process $$(r_t)$$ via $$r_t=x_t+\vartheta(t)$$. The function $$\vartheta$$ serves to guarantee a perfect fit with observed discount factors and hence, can lead to negative short rate. In their book, you can read about their CIR extension (named CIR++) in Section 3.9 which addresses the issue of positivity explicitly in 3.9.3.


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# At 298 K, the reaction below was started with 0.050 M ofABm3. At equilibrium, the molar concentration ofBm2 was 0.0075

###### Question:

At 298 K, the reaction below was started with 0.050 M of ABm3. At equilibrium, the molar concentration of Bm2 was 0.0075 M. 2ABm3(aq) â‡Œ 2ABm2(aq) + Bm2(aq) a) What is the change in molar concentration of Bm2 (i.e., the 'change' for Bm2 in an 'ICE' table)? b) Calculate Kc for this reaction at this temperature.

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# American Institute of Mathematical Sciences

May  2013, 12(3): 1431-1443. doi: 10.3934/cpaa.2013.12.1431

## Convexity of the free boundary for an exterior free boundary problem involving the perimeter

 1 Xi'an Jiaotong-Liverpool University, Mathematical Sciences, 111 Ren'ai Road, Suzhou 215123, Jiangsu Prov., China 2 Institutionen för Matematik, Kungliga Tekniska Högskolan, 100 44 Stockholm, Sweden

Received  November 2011 Revised  March 2012 Published  September 2012

We prove that if the given compact set $K$ is convex then a minimizer of the functional \begin{eqnarray*} I(v)=\int_{B_R} |\nabla v|^p dx+ Per(\{v>0\}), 1 < p < \infty, \end{eqnarray*} over the set $\{v\in W^{1,p}_0 (B_R)| v\equiv 1 \ \text{on} \ K\subset B_R\}$ has a convex support, and as a result all its level sets are convex as well. We derive the free boundary condition for the minimizers and prove that the free boundary is analytic and the minimizer is unique.
Citation: Hayk Mikayelyan, Henrik Shahgholian. Convexity of the free boundary for an exterior free boundary problem involving the perimeter. Communications on Pure & Applied Analysis, 2013, 12 (3) : 1431-1443. doi: 10.3934/cpaa.2013.12.1431
##### References:
 [1] A. Acker, On the existence of convex classical solutions for multilayer free boundary problems with general nonlinear joining conditions,, Trans. Amer. Math. Soc., 350 (1998), 2981. doi: 10.1090/S0002-9947-98-01943-6. Google Scholar [2] F. J. Jr. Almgren, Existence and regularity almost everywhere of solutions to elliptic variational problems with constraints,, Mem. Amer. Math. Soc., 4 (1976). doi: 10.1090/S0002-9904-1975-13681-0. Google Scholar [3] O. Alvarez, J.-M. Lasry and P.-L. Lions, Convex viscosity solutions and state constraints,, J. Math. Pures Appl., 76 (1997), 265. doi: 10.1016/S0021-7824(97)89952-7. Google Scholar [4] L. Ambrosio, N. Fusco and D. Pallara, "Functions of Bounded Variation and Free Discontinuity Problems,", Oxford Mathematical Monographs. Oxford University Press, (2000). Google Scholar [5] R. Argiolas, A two-phase variational problem with curvature,, Matematiche (Catania), 58 (2003), 131. Google Scholar [6] I. Athanasopoulos, L. A. Caffarelli, C. Kenig and S. Salsa, An area-Dirichlet integral minimization problem,, Comm. Pure Appl. Math., 54 (2001), 479. Google Scholar [7] L. C. Evans and R. F. Gariepy, "Measure Theory and Fine Properties of Functions,", Studies in Advanced Mathematics. CRC Press, (1992). Google Scholar [8] D. Gilbarg and N. Trudinger, "Elliptic Partial Differential Equations of Second Order,", Springer Verlag, (2001). Google Scholar [9] A. Henrot and H. Shahgholian, Existence of classical solutions to a free boundary problem for the $p$-Laplace operator. I. The exterior convex case,, J. Reine Angew. Math., 521 (2000), 85. doi: 10.1515/crll.2000.031. Google Scholar [10] A. Henrot and H. Shahgholian, The one phase free boundary problem for the $p$-Laplacian with non-constant Bernoulli boundary condition,, Trans. Amer. Math. Soc., 354 (2002), 2399. doi: 10.1090/S0002-9947-02-02892-1. Google Scholar [11] D. Kinderlehrer, L. Nirenberg and J. Spruck, Regularity in elliptic free boundary problems I,, J. Analyse Math., 34 (1978), 86. doi: 10.1007/BF02790009. Google Scholar [12] B. Kirchheim and J. Kristensen, Differentiability of convex envelopes,, C. R. Acad. Sci. Paris S閞. I Math., 333 (2001), 725. doi: 10.1016/S0764-4442(01)02117-6. Google Scholar [13] P. Laurence and E. Stredulinsky, Existence of regular solutions with convex levels for semilinear elliptic equations with nonmonotone $L^1$ nonlinearities. Part I,, Indiana Univ. Math. J., 39 (1990), 1081. doi: 10.1512/iumj.1990.39.39051. Google Scholar [14] J. L. Lewis, Capacitary functions in convex rings,, Arch. Rational Mech. Anal., 66 (1977), 201. doi: 10.1007/BF00250671. Google Scholar [15] G. Lieberman, Boundary regularity for solutions of degenerate elliptic equations,, Nonlinear Anal., 12 (1988), 1203. doi: 10.1016/0362-546X(88)90053-3. Google Scholar [16] F. Mazzone, A single phase variational problem involving the area of level surfaces,, Comm. Part. Diff. Eq., 28 (2003), 991. doi: 10.1081/PDE-120021183. Google Scholar [17] I. Tamanini, Regularity results for almost minimal oriented hypersurface in $R^n$,, Quaderni del Dipartimento di Matematica, (1994). Google Scholar

show all references

##### References:
 [1] A. Acker, On the existence of convex classical solutions for multilayer free boundary problems with general nonlinear joining conditions,, Trans. Amer. Math. Soc., 350 (1998), 2981. doi: 10.1090/S0002-9947-98-01943-6. Google Scholar [2] F. J. Jr. Almgren, Existence and regularity almost everywhere of solutions to elliptic variational problems with constraints,, Mem. Amer. Math. Soc., 4 (1976). doi: 10.1090/S0002-9904-1975-13681-0. Google Scholar [3] O. Alvarez, J.-M. Lasry and P.-L. Lions, Convex viscosity solutions and state constraints,, J. Math. Pures Appl., 76 (1997), 265. doi: 10.1016/S0021-7824(97)89952-7. Google Scholar [4] L. Ambrosio, N. Fusco and D. Pallara, "Functions of Bounded Variation and Free Discontinuity Problems,", Oxford Mathematical Monographs. Oxford University Press, (2000). Google Scholar [5] R. Argiolas, A two-phase variational problem with curvature,, Matematiche (Catania), 58 (2003), 131. Google Scholar [6] I. Athanasopoulos, L. A. Caffarelli, C. Kenig and S. Salsa, An area-Dirichlet integral minimization problem,, Comm. Pure Appl. Math., 54 (2001), 479. Google Scholar [7] L. C. Evans and R. F. Gariepy, "Measure Theory and Fine Properties of Functions,", Studies in Advanced Mathematics. CRC Press, (1992). Google Scholar [8] D. Gilbarg and N. Trudinger, "Elliptic Partial Differential Equations of Second Order,", Springer Verlag, (2001). Google Scholar [9] A. Henrot and H. Shahgholian, Existence of classical solutions to a free boundary problem for the $p$-Laplace operator. I. The exterior convex case,, J. Reine Angew. Math., 521 (2000), 85. doi: 10.1515/crll.2000.031. Google Scholar [10] A. Henrot and H. Shahgholian, The one phase free boundary problem for the $p$-Laplacian with non-constant Bernoulli boundary condition,, Trans. Amer. Math. Soc., 354 (2002), 2399. doi: 10.1090/S0002-9947-02-02892-1. Google Scholar [11] D. Kinderlehrer, L. Nirenberg and J. Spruck, Regularity in elliptic free boundary problems I,, J. Analyse Math., 34 (1978), 86. doi: 10.1007/BF02790009. Google Scholar [12] B. Kirchheim and J. Kristensen, Differentiability of convex envelopes,, C. R. Acad. Sci. Paris S閞. I Math., 333 (2001), 725. doi: 10.1016/S0764-4442(01)02117-6. Google Scholar [13] P. Laurence and E. Stredulinsky, Existence of regular solutions with convex levels for semilinear elliptic equations with nonmonotone $L^1$ nonlinearities. Part I,, Indiana Univ. Math. J., 39 (1990), 1081. doi: 10.1512/iumj.1990.39.39051. Google Scholar [14] J. L. Lewis, Capacitary functions in convex rings,, Arch. Rational Mech. Anal., 66 (1977), 201. doi: 10.1007/BF00250671. Google Scholar [15] G. Lieberman, Boundary regularity for solutions of degenerate elliptic equations,, Nonlinear Anal., 12 (1988), 1203. doi: 10.1016/0362-546X(88)90053-3. Google Scholar [16] F. Mazzone, A single phase variational problem involving the area of level surfaces,, Comm. Part. Diff. Eq., 28 (2003), 991. doi: 10.1081/PDE-120021183. Google Scholar [17] I. Tamanini, Regularity results for almost minimal oriented hypersurface in $R^n$,, Quaderni del Dipartimento di Matematica, (1994). Google Scholar
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# mtable borders etc

From: Peter Krautzberger <peter.krautzberger@mathjax.org>
Date: Mon, 10 Nov 2014 23:42:56 +0100
Message-ID: <CABqxo82qYmbE1GowhANtdXY9hmV1huBWWYfmVQDCBS-BdxMKrA@mail.gmail.com>
To: "www-math@w3.org" <www-math@w3.org>
Hi,

The mtable element is limited in terms of columnlines, rowlines, and frame.
In particular, this has come up on the MathJax user group
<https://groups.google.com/forum/#!searchin/mathjax-users/I$20guess$20it$27s$20mostly$20a$20matter$20of$20convenience$2C$20as$20maths$20folks$20are$20usually$20quite$20familiar$20with$20TeX$20but$20not$20so$20with$20HTML.$20%7Csort:relevance/mathjax-users/6vMvWrSCSzA/BH-dfMd5tK4J>
a
few years ago and has resurfaced in a bug report for MediaWiki/Wikipedia
<https://bugzilla.wikimedia.org/show_bug.cgi?id=73169> a few days ago.

Perhaps there's something constructive to be done about it?

Best wishes,
Peter.

Received on Monday, 10 November 2014 22:43:23 UTC

This archive was generated by hypermail 2.3.1 : Tuesday, 6 January 2015 21:27:47 UTC


text:
# All Questions

59,249 questions
13 views

### Understanding Transpose, dot and at

I found this formula Transpose[{Tv1, Tv2}].Inverse@Transpose@{v1, v2}.{x, y} in some post but lost the link. Now I need to understand why this formula solve my ...
11 views

### How reproduce old SplineFit function?

Prior to version 8, Mathematica included a SplineFit function that could be used, for example, in the form: ...
8 views

### How can empty fields in a csv file be set to Missing when importing as a Dataset?

When importing a csv file with empty fields as a Dataset, the empty fields are set to zero-length strings (""). There is an option, ...
15 views

### How to visualize indifference curve in a 2D plot?

I have two functions f(x,y) and g(x,y). I wish to perform two tasks on a 2-dimensional figure: (1) plot the curve where each point satisfies f(x,y)=g(x,y) (2) mark one side of the curve with red ...
14 views

### running a loop to check for multiple congruences

I have the code for a loop to run. Right now it can check if a number is a square and at the same time check if it is congruent 0 mod 47. My questions is, how can I alter the code to see if it is ...
22 views

### Can NDSolve be used to model gravitational motion due to an ellipsoid?

I'm trying to use Mathematica to get a numerical estimate and plot of the motion of a point mass in the gravitational field of a fixed, non-rotating ellipsoid. Ideally, I'd like to plot the motion ...
15 views

### Impose conditions on PDF distribution

Consider some data data = {{x1,y1},{x2,y2},...} I organize a density distribution using ...
31 views

### Why do I get different symbols (a) with the same FontFace (Cambria Math)?

When I create an equation in MS Office, it uses Cambria Math, and I get the following when I use variable 'a'. When I try to use Mathematica to create graphics for my files, I can try to use ...
73 views

### ContourPlot — How do I color by contour curvature?

I'm plotting the stream lines of fluid flow past a cylinder, and I would like the colors to increase with contour curvature (i.e. increase as the velocity of the flow increases. Here's a MWE that ...
26 views

### Element-wise multiplication of matrices with different dimension

I am interested in efficient element-wise multiplication of matrices with different dimension. Here is my solution: Matrix 1 with dim = {3, 4, 4} ...
37 views

### Simplifying long expressions leads to memory allocation failure

I am currently trying to manipulate a number of expressions into a particular form that will allow me to determine their poles and zeros: ...
38 views

### Random integers [on hold]

I'm wondering if there is any way to generate 10 random integers (between 1 and 10) in a way where once an integer is selected, it does not get selected again. I've tried using RandomInteger and then ...
27 views

21 views

### Multiple collection of terms

As the title suggests, I am interested in how to perform multiple collection of terms. This means that for a given expression a*x*b + c*x^2*d+a*x*k, I want to ...
16 views

### Ordinal subset in the set of classes

I need to do an unusual classification, which I probably will not achieve only with Classify[]. Features are quite simple vectors composed of numbers and nominal ...
42 views

### Simplify differential expressions

One can differentiate a two dimensional vector like this, ClearAll["Global*"] u[x_, y_] := {f[x, y], g[x, y]} Div[u[x, y], {x, y}] with output ...
86 views

### Adjusting bounding box of PlotLegends in TimelinePlot

I'd like to align the elements of the PlotLegend in a single horizontal row beneath the TimelinePlot, as there is plenty of room ...
65 views

Consider ...
32 views

### Plotting dispersion relation [on hold]

Why is the second funtion dos not appear in the graph ? ...
45 views

### Issues solving system of differential equations [on hold]

d = NDSolve [{x''[t] == (x [t] + a[t])/(-2), a''[t] == -x''[t]/2, x[0] == 0, x'[0] == 1, a[0] == 0, a'[0] == 0}, {x, y}, {t, 20}] Keeps returning the error "To ...
57 views

### How do I divide an integral into parts and sum the total?

I have an exercise looking like this: Analyze the integral $4\int_{0}^{1} \sqrt{1-x^2} \, \mathrm{d} x$ numerically by deviding the interval ${]0,1[}$ into three equal parts, then summarize the ...
336 views

### Sort with assumptions

I have a list which looks like this list = {0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9]}; and all the $x_i$'s are positive semidefinite (i....
52 views

### Identifying total derivatives

I would like to identify which terms in my expression are a total derivative and which ones are not. For expressions in which all terms are a total derivative, one can simply use integrate to find ...
44 views

### How to define multivariable vector funtion? [on hold]

I would like to define a function of variables having two components as $u(x,y) = (f(x,y),g(x,y))$. I defined such a function as u[x_, y_] := {f[x, y], g[x, y]} ...
44 views

### How to show the answer in terms of another variable

if we have A = 1; B = 44 + 1 output :: B = 45 how we can show B like this output :: B = 44 + A
37 views

### Changing speeds of orbiting bodies with Manipulate

The code below is attempting to simulate the orbits of Earth and Mars along with a Hohmann transfer orbit. Since each one of these bodies orbit at different speeds, I need a way to specify each orbit ...
30 views

### How do I Select when the key of an association is a multi-word string

I have imported some survey responses into a Dataset. I now want to select those rows where the response to "participant type" == "Virtual". I discovered that this ...
31 views

### Keyed Hash for HMAC SHA256

I'm trying to use AWS REST API's (because I can't rely on AWSLink). The sample C# code from Amazon uses this a lot: ...
90 views

### Can pattern assumptions be made reliable?

It is possible to use patterns in assumptions such as f[_]>0. This does not seem to be explained in the documentation but is for instance pointed out in Is it ...
33 views

### Replacing with Elements from a List [duplicate]

I have a fairly complicated expression with nested lists and I want to substitute a given string for each integer. Here's a simple example where ind is my integer-valued data structure and names is ...
287 views

### Derivative of an interpolated function

I am trying to take the derivative of an interpolated function. I am given the function values and the derivatives at some irregular points. Here is my minimal working example to reproduce the error: ...
68 views

### Divide a list into two lists with equal sums

I have a list with L elements. I want to find an index within the list, that divides the list to two separate lists so the sum of elements in each of them are equal....
156 views

### Taking the numerator and the denominator

Is there any way to get the Numerator and the Denominator of an expression split in the same way that Mathematica graphically represents it. Say I have an expression: ...
66 views

### System of coupled differential equations

I am trying to solve the following system of equations, but I get as the output the command itself. ...
26 views

### Monitoring PolynomialReduce/Alternatives for other CAS packages

I have a large generic polynomial That looks like $N = \sum_{i_1,i_2\cdots}c_{i_1,i_2,i_3,\cdots} {x_{1}}^{i_1}{x_{2}}^{i_2}\cdots$ This could have anywhere between 3000-9000 terms with a maximum ...
73 views

### Delete duplicates in tuples

I use the following code to create a list of input parameters: ...
43 views

### How do I uninstall an old version of Mathematica?

My SSD is getting full, and while trying to clean it up, I found one of the largest things on here is an installation of Mathematica 10.3. As this is an outdated version, I would like to remove it, ...
60 views

### Identifying points in the frontier of a set

Let me start with an example. Let $$\mathbf{A}=\begin{bmatrix}3&1\\2&3\\1&5\end{bmatrix},$$ and let $Q=\{\mathbf{q}\vert\mathbf{q}\in\Bbb R^3_+ \land \sum_i^n q_i=1\}$ and \$\alpha=(.5,.5,...


text:
# Finding all contiguous sublists such that each member appears an even number of times

The program needs to find all contiguous sublists of numbers where the number of occurrences of every member in the contiguous sublist is divisible by 2.

The first line of input is N (the number of ai, where N ≤ 353535). The second line of input contains the ai, delimited by spaces (where each ai < 109).

Here is an example of input and what the output should be:

# N is the user input, and a is the user input with
#numbers separated by space and converted to list with .split(" ") but
#we'll look for examles first :
N = 8
a = [3,5,3,5,9,9,5,3]

The output of this input should be 5 because there are 5 contiguous sublists that meet the conditions:

• [3,5,9,9,5,3]
• [3,5,3,5,9,9]
• [5,9,9,5]
• [3,5,3,5]
• [9,9]

Here is my solution:

import time
total = 0
try:
listlength = int(raw_input("Enter the number of list elements >> "))
listofnumbers   = raw_input("Enter the list elements , space separated >>").split(" ")
start = time.clock()
for i in range(listlength,1,-1):
for i1 in range(0,listlength + 1 - i):
numbers = []
countofnumbers=[]
currentlist = listofnumbers[i1:listlength-abs(i-listlength+i1)]
#print "I:",i,",TR nIZ ",tr_niz
for k in currentlist:
if k not in numbers:
numbers.append(k)
countofnumbers.append(currentlist.count(k) if currentlist.count(k)%2==0 else 0)
if 0 not in countofnumbers:
if sum(countofnumbers)%2==0:
total+=1

print total
print "Time: ",time.clock()-start

except ValueError:
print "Data wrong"

My script solves small test cases really well. However, when there are 350350 numbers, It was working all the night, and it didn't turn up with result. How can I fix my algorithm to finish in a reasonable amount of time? Maybe implement multiprocessing because it's using 100% od my core when running , but doesn't use any of 3 other cores available ? And maybe some enhanced algorithm you suggest :)

• Looking at his code it is pretty obvious that by subgroup he means "contiguous sublist," which would mean there's a typo in his explanation, and the third subgroup should be [5, 9, 9, 5]. But other than that it seems a well posed question. – Jaime Mar 30 '16 at 4:48
• I agree with Jaime - it can only be contiguous subsequences, and the [5, 9, 5, 9] thing must be a simple transcription mistake. Otherwise the number of result groups would be higher. A valid question (classic TLE) with a really neat problem. – DarthGizka Mar 30 '16 at 5:10
• I'm sorry for this subgroup, i didn't know how to express in english :) now i know, So the order really matters because you must search input for patterns. But output is as i listed only number of contiguous subsequences. My script does the following . Let's say i have input with 8 numbers, it'll first search first 7, then move one and search the other 7. Then it will search by 6, so it has 3 spaces to move, and so on, up to 2, where it has 7 spaces to move in input. That's how i split it into subsequences and check how many of each number is in that subsequence , if it matches the condition-> – Marko Mackic Mar 30 '16 at 9:38
• It adds one to total.. now it all works where there is small number of items in list, just for comparison, for 100 items it takse 0.14 s, and for 1000 items it takes 159 s, let's round up (100 times loneger) I'd expect 140s and then for 350350 items, it would take a very long period to finish.. – Marko Mackic Mar 30 '16 at 9:55
• EDIT TO the above, corrected it to fist search all the list, because i was skipping it :) so let's say it's 8, it will look if the whole list meets the condition, and then it will do 7,6, ... as explained in comment above. I edited the question with new code ,and edited subgroups to contiguous subsequences ... – Marko Mackic Mar 30 '16 at 10:26

The algorithm can be slimmed by tracking only those numbers for which the number of occurrences in the current window is unbalanced (i.e. odd). Whenever this auxiliary structure is empty, everything is balanced and the current window can be added to the set of solutions. Here's some pseudo code to illustrate the approach:

static void print_balanced_contiguous_subsequences (int[] a)
{
var unbalanced = new HashSet<int>();

for (int i = 0, e = a.Length - 1; i < e; ++i)
{
unbalanced.Clear();

for (int j = i; j <= e; ++j)
{
if (unbalanced.Contains(a[j]))
unbalanced.Remove(a[j]);
else

if (unbalanced.Count == 0) // print the newly found sequence
for (int k = i; k <= j; ++k)
Console.Write("{0}{1}", a[k], k < j ? ' ' : '\n');
}
}
}

However, this algorithm is still quadratic in n (20 ms for 1000 elements, 2 seconds for 10000) and the key idea for a proper solution still eludes me...

The original problem description might contain a clue, or some restriction that can be leveraged to design a faster solution. It would be nice if it could be quoted verbatim in the question, or at least linked (even if it is not in English).

• OK, so the task is in Serbian, and link is here :staritakprog.dms.rs/takmicenja/2013.3.sio/tekstovi.pdf , it's the first one , I will try your solution, thanks ;) – Marko Mackic Mar 30 '16 at 12:52
• @Marko: Interesting! The last two lines of the task description (about Ai <= 53 in 50% of tests?/testcases? and so on) should be translated and added to your question. The description of the task does not seem to specify a time limit. Is there a global/implied time limit mentioned elsewhere in the document? – DarthGizka Mar 30 '16 at 13:14
• Nope :) just my code didn't pass the hardest case, and that's my problem, i ran for a day for only 350350 numbers, but anyways if your solution is efficient as quoted above it is really acceptible :) – Marko Mackic Mar 30 '16 at 13:18
• But anyways what is the point of A(i) being small, when number of elements is high ? :) Like there are 20 test cases, and in the 4th the list of numbers contains 1000 elements, in 5th 15000(something a bit more) and then goes to 350350 elements :) – Marko Mackic Mar 30 '16 at 13:22
• Yeah , it's working, i'll just post the code in python i wrote, thanks :) – Marko Mackic Mar 30 '16 at 15:03

So @DarthGizka solved it in efficient manner :) Here is the code in python :

import time
try:
total = 0

listofnumbers   = raw_input("Enter the list elements , space separated").split(" ")
start = time.clock()

for i in range(0,len(listofnumbers)-1):
unbalanced = set()
for k in range(i,len(listofnumbers)):
if listofnumbers[k] in unbalanced:
unbalanced.remove(listofnumbers[k])
else:
if len(unbalanced) == 0:
total+=1

print total
print "Time : ",time.clock()-start

except ValueError:
print "Data not ok"

1000 numbers takes 0.45 seconds to complete if anyone has more improvments let me know :) Thanks a lot to @DarthGizka


text:
# MOSFET switching on

I have a doubt. Consider an N-MOSFET: which is the voltage that can switch on it? The voltage between Gate and? Sometimes I read "between Gate and Bulk", sometimes "between Gate and Source", sometimes the ambiguous sentence "Gate voltage".

Then I have another question, related to the previous one. Consider the following pass transistor circuit:

How can the signal at the Gate activate the MOSFET? It must be high with respect to what point (moreover in this circuit Drain and Source are not fixed points)?


text:
# E. Chocolate Bar

### гаралт стандарт гаралт

You have a rectangular chocolate bar consisting of $n × m$ single squares. You want to eat exactlyk$squares, so you may need to break the chocolate bar. In one move you can break any single rectangular piece of chocolate in two rectangular pieces. You can break only by lines between squares: horizontally or vertically. The cost of breaking is equal to square of the break length. For example, if you have a chocolate bar consisting of$2 × 3$unit squares then you can break it horizontally and get two$1 × 3$pieces (the cost of such breaking is$3^{2} = 9$), or you can break it vertically in two ways and get two pieces:$2 × 1$and$2 × 2$(the cost of such breaking is$2^{2} = 4$). For several given values$n$,$m$and$k$find the minimum total cost of breaking. You can eat exactly$k$squares of chocolate if after all operations of breaking there is a set of rectangular pieces of chocolate with the total size equal to$k$squares. The remaining$n*m - k$squares are not necessarily form a single rectangular piece. ## Оролт The first line of the input contains a single integer$t$($1 ≤ t ≤ 40910$) -- the number of values$n$,$m$and$k$to process. Each of the next$t$lines contains three integers$n$,$m$and$k$($1 ≤ n, m ≤ 30, 1 ≤ k ≤ min(n*m, 50)$) -- the dimensions of the chocolate bar and the number of squares you want to eat respectively. ## Гаралт For each$n$,$m$and$k$print the minimum total cost needed to break the chocolate bar, in order to make it possible to eat exactly$k$squares. #### Жишээ тэстүүд ##### Оролт 4 2 2 1 2 2 3 2 2 2 2 2 4  ##### Гаралт 5 5 4 0  ## Тэмдэглэл In the first query of the sample one needs to perform two breaks: * to split$2 × 2$bar into two pieces of$2 × 1$(cost is$2^{2} = 4$), * to split the resulting$2 × 1$into two$1 × 1$pieces (cost is$1^{2} = 1$). In the second query of the sample one wants to eat$3\$ unit squares. One can use exactly the same strategy as in the first query of the sample.
Сэтгэгдлүүдийг ачааллаж байна...


text:
Decide for which positive values on a that makes the series converge

For which positive values on a does the series converge?:

$$\sum _{n=1}^{\infty}na^{\ln(n)}$$

I have tried to rewrite the expression, but that gives me nothing.

Anyone got a clue?

-

I will guess that your $i$ should be an $n$. If $a$ is positive, we may write it as $e^x$ for some $x$. Then $$na^{\ln(n)}=n(e^x)^{\ln(n)}=ne^{\ln(n^x)}=n(n^x)=n^{x+1}.$$ Now you've got yourself a $p$-series.
Yep, changed that. But why can you change a to $$e^x$$ ? –  Curtain Dec 10 '12 at 22:04
@JulianAssange $x = \ln(a)$ –  Tom Oldfield Dec 10 '12 at 22:25


text:
# Real rank zero of group $C^*$-algebras

The concept of real rank zero of a $C^*$-algebra is introduced as non-commutative analogue of dimension ( topological dimension ). For example, it shown (by Brown-Pedersen) such that, if $X$ is a compact Hausdorff space, then $RR$ $C(X)=dim(X)$.

Let $A$ be a $C^*$-algebra. we define $RR(A)=0$ iff every self-adjoint element in $A$ can be approximate by invertible, self-adjoint element in $A$. for example every Von-Neumman algebra has real rank zero.

So, on the other hand since $C^*(\mathbb T)=C(\mathbb Z)$ and $C^*(\mathbb Z)=C(\mathbb T)$, then $RR$ $C^*(\mathbb T)=0$ but
$RR$ $C^*(\mathbb Z)$$\not=$$0$ .

Question: Which locally compact groups $G$ satisfy $RR$ $C^*(G)=0$?

NOTE: If $G$ is locally compact group with fixed Harr measure, let full group $C^*$-algebra $C^*(G)$ be the completion of the convolution algebra $L^1(G)$ with respect to the norm $\lVert f\rVert_{C^*(G)}$=$\sup$$\{\lVert\phi(f)\rVert\}$, where the supremum is taken over all $*$-representation $\phi$ of $L^1(G)$ as an algebra of bounded linear operators in a Hilbert space.

• BTW, I find this a much better question than your older one about "ideal structure", since you have a more limited and hence more focused question. – Yemon Choi Feb 14 '16 at 19:11

Here is a partial answer, and a pointer towards some literature that should be relevant.$\newcommand{\Cst}{{\rm C}^*}$
Doing some searching online brings up an old result of Kaniuth (Proc. AMS, 1993), which implies that if $G$ is a connected non-compact group then its full $\Cst$-algebra has real rank strictly greater than zero. The proof relies heavily on structure theory for Lie groups, but the structure theory used seems to be fairly standard material from that area.
Theorem 2 of Kaniuth's paper gives some examples of nilpotent locally compact groups whose $\Cst$-algebras have real rank zero.
Note that the $c_0$-direct sum of finite-dimensional matrix algebras always has real rank zero (this is a nice exercise in understanding the definition) and hence $\Cst(G)$ has real rank zero for every compact group $G$.


text:
## Basic College Mathematics (10th Edition)

There are 100cm/m so to find meters we must divide number of centimeters by 100 1. $40\div100=0.4$


text:
# zbMATH — the first resource for mathematics

Fixed point theory in cone metric spaces obtained via the scalarization method. (English) Zbl 1197.54055
Summary: Motivated by the scalarization method in vector optimization theory, we take a new approach to fixed point theory on cone metric spaces. By using our method, we prove some fixed point theorems and several common fixed point theorems on cone metric spaces in which the cone need not be normal. Our results improve and generalize many well-known results from the literature.

##### MSC:
 54H25 Fixed-point and coincidence theorems (topological aspects) 65J15 Numerical solutions to equations with nonlinear operators 47H10 Fixed-point theorems
Full Text:
##### References:
 [1] Huang, L.G.; Zhang, X., Cone metric spaces and fixed point theorems of contractive mappings, J. math. anal. appl., 332, 1468-1476, (2007) · Zbl 1118.54022 [2] Chen, G.Y.; Huang, X.X.; Hou, S.H., General ekeland’s variational principle for set-valued mappings, J. optim. theory appl., 106, 1, 151-164, (2000) · Zbl 1042.90036 [3] Abbas, M.; Jungck, G., Common fixed point results for noncommuting mappings without continuity in cone metric spaces, J. math. anal. appl., 341, 416-420, (2008) · Zbl 1147.54022 [4] Abbas, M.; Rhoades, B.E., Fixed and periodic point results in cone metric spaces, Appl. math. lett., 22, 511-515, (2009) · Zbl 1167.54014 [5] Arshad, M.; Azam, A.; Beg, I., Common fixed points of two maps in cone metric spaces, Rend. circ. mat. Palermo, 57, 433-441, (2008) · Zbl 1197.54056 [6] Das, K.M.; Naik, K.V., Common fixed point theorems for commuting maps on a metric space, Proc. amer. math. soc., 77, 3, 369-373, (1979) · Zbl 0418.54025 [7] Di Bari, C.; Vetro, P., $$\varphi$$-pairs and common fixed points in cone metric spaces, Rend circ. mat. Palermo, 57, 279-285, (2008) · Zbl 1164.54031 [8] Di Bari, C.; Vetro, P., Weakly $$\varphi$$-pairs and common fixed points in cone metric spaces, Rend circ. mat. Palermo, 58, 125-132, (2009) · Zbl 1197.54060 [9] Ilić, D.; Rakočević, V., Quasi-contraction on a cone metric space, Appl. math. lett., 22, 5, 728-731, (2009) · Zbl 1179.54060 [10] P. Raja, M. Vaezpour, Some extensions of Banach’s contraction principle in complete cone metric spaces, Fixed Point Theory Appl., Vol. 2008, 11 pages, Article ID 768294. · Zbl 1148.54339 [11] Wardowski, D., Endpoints and fixed points of set-valued contractions in cone metric spaces, Nonlinear anal., 71, 512-516, (2009) · Zbl 1169.54023 [12] Rezapour, Sh.; Hamlbarani, R., Some notes on the paper “cone metric spaces and fixed point theorems of contractive mappings”, J. math. anal. appl., 345, 719-724, (2008) · Zbl 1145.54045 [13] Arshad, M.; Azam, A.; Verto, P., Some common fixed point results in cone metric spaces, Fixed point theory appl., 2009, (2009), Article ID 493965, 11 Pages [14] Jungck, G.; Radenović, S.; Radojević, S.; Rakočević, V., Common fixed point theorems for weakly compatible pairs on cone metric spaces, Fixed point theory appl., 2009, (2009), Article ID 643840 · Zbl 1190.54032 [15] Ilić, D.; Rakočević, V., Common fixed points for maps on cone metric space, J. math. anal. appl., 341, 876-882, (2008) · Zbl 1156.54023 [16] Geraghty, M.A., On contractive mappings, Proc. amer. math. soc., 40, 604-608, (1973) · Zbl 0245.54027 [17] Jameson, G., () [18] Jeyakumar, V.; Oettli, W.; Natividad, M., A solvability theorem or a class of quasiconvex mappings with applications to optimization, J. math. anal. appl., 179, 2, 537-546, (1993) · Zbl 0791.46002 [19] Ćirić, Lj.B., A generalization of banach’s contraction principle, Proc. amer. math. soc., 45, 267-273, (1975) · Zbl 0291.54056 [20] Aamri, M.; El Moutawakil, D., Some new fixed point theorems under strict contractive conditions, J. math. anal. appl., 270, 181-188, (2002) · Zbl 1008.54030 [21] Imdad, M.; Ali, Javid, Jungck’s common fixed point theorem and E. A. property, Acta math. appl. sin., engl. ser., 24, 1, 87-94, (2008) · Zbl 1158.54021
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
## Federico Ardila ; Florian Block - Universal Polynomials for Severi Degrees of Toric Surfaces

dmtcs:3089 - Discrete Mathematics & Theoretical Computer Science, January 1, 2012, DMTCS Proceedings vol. AR, 24th International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2012) - https://doi.org/10.46298/dmtcs.3089
Universal Polynomials for Severi Degrees of Toric Surfaces

Authors: Federico Ardila ; Florian Block

The Severi variety parametrizes plane curves of degree $d$ with $\delta$ nodes. Its degree is called the Severi degree. For large enough $d$, the Severi degrees coincide with the Gromov-Witten invariants of $\mathbb{CP}^2$. Fomin and Mikhalkin (2009) proved the 1995 conjecture that for fixed $\delta$, Severi degrees are eventually polynomial in $d$. In this paper, we study the Severi varieties corresponding to a large family of toric surfaces. We prove the analogous result that the Severi degrees are eventually polynomial as a function of the multidegree. More surprisingly, we show that the Severi degrees are also eventually polynomial "as a function of the surface". Our strategy is to use tropical geometry to express Severi degrees in terms of Brugallé and Mikhalkin's floor diagrams, and study those combinatorial objects in detail. An important ingredient in the proof is the polynomiality of the discrete volume of a variable facet-unimodular polytope.

Volume: DMTCS Proceedings vol. AR, 24th International Conference on Formal Power Series and Algebraic Combinatorics (FPSAC 2012)
Section: Proceedings
Published on: January 1, 2012
Imported on: January 31, 2017
Keywords: Enumerative geometry, toric surfaces, Gromov-Witten theory, Severi degrees, node polynomials,[INFO.INFO-DM] Computer Science [cs]/Discrete Mathematics [cs.DM]


text: Quaternions are used in computer graphics...there are quite a number of questions on this, some I have answered, but got tired of typing the same thing over and over. The short version is, you write a 3-D point as a pure quaternion $v,$ (zero real part). Then, given a unit quaternion $\xi,$ a rotated pure quaternion comes from $\xi v \bar{\xi}.$ I'm not sure these give much help with contest problems. –  Will Jagy Jul 28 '12 at 21:36


text:
# Bounds for Integer Expressed in Base k

## Theorem

Let $n \in \Z$ be an integer.

Let $k \in \Z$ such that $k \ge 2$.

Let $n$ be expressed in base $k$ notation:

$n = \displaystyle \sum_{j \mathop = 1}^s a_j k^j$

where each of the $a_j$ are such that $a_j \in \set {0, 1, \ldots, k - 1}$.

Then:

$0 \le n < k^{s + 1}$

## Proof

As none of the coefficients $a_j$ in $\displaystyle \sum_{j \mathop = 1}^s a_j k^j$ is (strictly) negative, the summation itself likewise cannot be negative

Thus:

$0 \le n$

The equality is satisfied when $a_j = 0$ for all $j$.

We then have:

 $\displaystyle n$ $=$ $\displaystyle \sum_{j \mathop = 1}^s a_j k^j$ $\displaystyle$ $\le$ $\displaystyle \paren {k - 1} \sum_{j \mathop = 1}^s k^j$ as $a_j \le k - 1$ for all $j$ $\displaystyle$ $=$ $\displaystyle \paren {k - 1} \dfrac {k^{s + 1} - 1} {k - 1}$ Sum of Geometric Sequence $\displaystyle$ $=$ $\displaystyle k^{s + 1} - 1$ $\displaystyle$ $<$ $\displaystyle k^{s + 1}$

$\blacksquare$