text:
# Left adjoints to some inclusions of categories of topological spaces

Let us note $\textbf{Top}_{n}$ for the full subcategory of all $T_n$-spaces $n=0,...,4$. Construct left adoints to the inclusion functors $\textbf{Top}_{n+1}\rightarrow \textbf{Top}_n$ for $n=0,...,3$.

I can "solve" the cases $n=0,1,2$ using the adjoint functor theorem, but $\textbf{Top}_4$ is not as nice a category as the others (it does not have products, or, at least, the inclusion functor does not preserve them) and, therefore, I cannot use the adjoint fucntor theorem. Plus, the phrasing of the problem hints at an explicit construction of these left adjoints.

Hence, my question is how do I construct the left adjoints to these functor explicitly?

NB: This is problem V.9.4. from Mac Lane.

• There's something odd here. If the inclusion functor does not preserve products, it cannot have a left adjoint. – Arnaud D. Jul 26 '17 at 18:41
• I agree that this isn't possible. – Kevin Carlson Jul 26 '17 at 19:34
• You are absolutely right! The question is then just simply wrong, which surprises me of a problem from Mac Lane. – JeCl Jul 26 '17 at 19:42
• For the other ones, the idea is to simply take a quotient, identifying all points which aren't separated enough. You may have to iterate this process to stabilize at a sufficiently separated space. – Kevin Carlson Jul 26 '17 at 19:48
• Is this CWM? Which page? – magma Jul 26 '17 at 21:19

The inclusion $\mathbf{Top}_{4}\to\mathbf{Top}_3$ has no left adjoint, since $\mathbf{Top}_{4}$ is not closed under products and so cannot be a reflective subcategory of $\mathbf{Top}_{3}$.

All of the other left adjoints can be constructed by a transfinite induction where you step by step "throw out" any failure of your space to satisfy the required separation axiom. Let's start with the left adjoint to the inclusion $\mathbf{Top}_1\to\mathbf{Top}_0$. Given a space $X$ and two points $x,y\in X$ such that every open set containing $x$ contains $y$, any map from $X$ to a $T_1$ space must make $x$ and $y$ become equal. So take the equivalence relation generated by saying any two such points $x$ and $y$ are equivalent, and let $X_1$ be the quotient of $X$ by this equivalence relation. Every map from $X$ to a $T_1$ space factors uniquely through the quotient map $X\to X_1$. If $X_1$ is $T_1$, that means $X_1$ is the value of our desired left adjoint to $\mathbf{Top}_1\to\mathbf{Top}_0$ on $X$. If not, we can again take the quotient of $X_1$ by the equivalence relation that identifies all witnesses to its failure to be $T_1$. Call this quotient $X_2$.

We can repeat this process by transfinite induction to get a sequence of smaller and smaller quotients $X_\alpha$ of $X$, taking colimits at limit steps. This process must eventually terminate since $X$ is a set, so we eventually reach a quotient $X_\alpha$ which is $T_1$. Our left adjoint then sends $X$ to this space $X_\alpha$, since every map from $X$ to a $T_1$ space factors uniquely through the quotient map $X\to X_\alpha$.

For the left adjoint to $\mathbf{Top}_2\to\mathbf{Top}_1$, you do the exact same thing, except you identify pairs of points which witness the failure of the $T_2$ axiom instead of the $T_1$ axiom.

The left adjoint to $\mathbf{Top}_3\to\mathbf{Top}_2$ is more complicated, since it may not always take a space $X$ to a quotient. Given a space $X$ with topology $\tau$, let $\tau_1\subseteq\tau$ be the collection of open sets $U$ such that for each $x\in U$, there exist disjoint open sets $V$ and $W$ separating $x$ and $X\setminus U$. Note that if $f:X\to Y$ is any continuous map from $X$ to a $T_3$ space $Y$, then $f^{-1}(U)\in \tau_1$ for every open subset $U\subseteq Y$. Now let $\tau_2\subseteq \tau_1$ be the set of $U\in \tau_1$ such that for each $x\in U$, there exist disjoint $V,W\in \tau_1$ separating $x$ and $X\setminus U$. Again, if $f:X\to Y$ and $Y$ is $T_3$, then $f^{-1}(U)\in \tau_2$ for all open $U\subseteq Y$. We can iterate this by transfinite induction to get a descending sequence $\tau_\alpha$, taking intersections at limit steps. This eventually stabilizes, giving a topology $\tau_\alpha\subseteq\tau$ for that points and closed sets can be separated by open sets. Moreover, every $\tau$-continuous map to a $T_3$ space is also continuous with respect to $\tau_\alpha$.

The set $X$ with the topology $\tau_\alpha$ would then be the value of the left adjoint on $X$, except that points may not be closed in this topology. So we must now apply the construction described in the second and third paragraphs above to take the largest quotient of this topology which is $T_1$. This gives us a new space which we will call $X_1$.

Unfortunately, this space $X_1$ may still not be $T_3$, since taking the quotient may have destroyed the fact that points and closed sets can be separated by open sets. So we repeat the whole process over again with $X_1$, first shrinking the topology to separate points and closed sets, then taking a quotient to make points closed. This gives us a new space $X_2$. We repeat this whole process by transfinite induction, taking colimits at limit steps. At each step, we are taking a smaller quotient set of $X$ and/or a smaller topology, so it eventually stabilizes. Thus we reach a space $X_\alpha$ which is $T_3$, and every map from $X$ to a $T_3$ space factors uniquely through the canonical map $X\to X_\alpha$.

It is instructive to think about how this construction would fail to work if you tried to do it for the $T_4$ axiom instead of $T_3$. The problem is in defining $\tau_1$: a failure of the $T_4$ axiom is detected by a pair of open sets, and there's no way to decide which of the two should not be in $\tau_1$. You might think you can define $\tau_1$ to consist of those $U\in \tau$ such that for all $V\in\tau$ such that $U\cup V=X$, $X\setminus U$ and $X\setminus V$ can be separated by open sets. But this doesn't work, since a map $f:X\to Y$ to a $T_4$ space need not be continuous with respect to $\tau_1$. The problem is in saying "for all $V\in\tau$", since we really want to require this condition only for those $V$ which will be in our eventual $T_4$ topology, but we don't know what that topology is yet.

• What is your reference for the first sentence? I don't get this fact from the nlab page. e.g. $\mathbf{Top}_3$ does not have exponential objects. – Henno Brandsma Jul 27 '17 at 7:12
• At the start of Section 4 of that page, it mentions that a reflective subcategory is always closed under limits that exist in the ambient category. – Eric Wofsey Jul 27 '17 at 7:17
• As a quick proof of this fact, suppose the reflector $F$ exists and you have a diagram in the subcategory with a limit $L$ in the ambient category. Then $FL$ has the exact same universal property as $L$ in the ambient category (it corepresents cones over the diagram, since every such cone factors through the map $L\to FL$). So the map $L\to FL$ is an isomorphism, and so the object $L$ is in the subcategory (at least, up to isomorphism). – Eric Wofsey Jul 27 '17 at 7:20
• OK, I see. So the $\mathbf{Top}_4$ embeddings are always going to be hard, as this category has awful limit properties. CWM only mentions $\mathbf{Top}_0$ being reflective in $\mathbf{Top}$, which is indeed true. – Henno Brandsma Jul 27 '17 at 7:31
• It is exercise 5.9.4 in CWM, and it's refuted by this paper – Henno Brandsma Jul 27 '17 at 7:45

The nlab page has some info and links under the reflection heading where constructions are discussed as well. But it seems, only for $n=0,1,2$. (Kolmogorov quotient, Hausdorff quotient etc.).

Maybe a similar construction to the Hausdorff quotient exists, identifying closed sets that can not be separated, somehow.


text:
# Brain works in sleep

Many mathematicians believe that that their brains continue to do mathematics during sleep. A paper

Kouider et al., Inducing Task-Relevant Responses to Speech in the Sleeping Brain, Current Biology (2014), http://dx.doi.org/10.1016/j.cub.2014.08.016

Proves that brain continues in sleep some mental activities of the day.

From the summary of the paper:

using semantic categorization and lexical decision tasks, we studied task-relevant responses triggered by spoken stimuli in the sleeping brain. Awake participants classified words as either animals or objects (experiment 1) or as either words or pseudowords (experiment 2) by pressing a button with their right or left hand, while transitioning toward sleep. The lateralized readiness potential (LRP), an electrophysiological index of response preparation, revealed that task-specific preparatory responses are preserved during sleep. These findings demonstrate that despite the absence of awareness and behavioral responsiveness, sleepers can still extract task relevant information from external stimuli and covertly prepare for appropriate motor responses.

The paper generated a huge response in mass media: BBC, New Scientist, NBC News. It is mentioned in this blog because the study of brain activity  is relevant to mathematics education. A naive question: do our students get enough sleep?

# Misha Gavrilovich: Point-set topology as diagram chasing computations

M. Gavrilovich, Point-set topology as diagram chasing computations, The De Morgan Gazette 5 no. 4 (2014), 23-32.

Abstract:

We observe that some natural mathematical definitions are lifting properties relative to simplest counterexamples, namely the definitions of surjectivity and injectivity of maps, as well as of being connected, separation axioms $$T_0$$ and $$T_1$$ in topology, having dense image, induced (pullback) topology, and every real-valued function being bounded (on a connected domain).

We also offer a couple of brief speculations on cognitive and AI aspects of this observation, particularly that in point-set topology some arguments read as diagram chasing computations with finite preorders.

# Citizen Maths MOOC

An experimental MOOC (Massive Open Online Course) Citizen Maths is launched and the first phase of the course is open for registration. It is free and open for everyone; its motto is Powerful Ideas in Action.

The readers of this blog may like to register for the course, because, as the organisers say,

The success of this first phase of Citizen Maths will depend crucially on the feedback that we obtain. We are particularly keen to get feedback from:

• learners who do the course;
• those with an interest in the learning and teaching of maths, and in the design of online courses.

There is a link to a feedback form on every page of the Citizen Maths web site, and there will be a similar link on every page of the course when it goes live on or around 12 September.

The first pilot stage will run for four weeks and cover the first “powerful idea”: proportion. An admirable choice (a detailed discussion of the role of proportions in elementary mathematics can be found in this paper by Tony Gardiner).

# Matilde Lalin: Attending conferences with small children

[Republished from Terry Tao's Blog -- because of the importance of the issue.-- AB]

[This guest post is authored by Matilde Lalin, an Associate Professor in the Département de mathématiques et de statistique at the Université de Montréal.  I have lightly edited the text, mostly by adding some HTML formatting. -T.]

Mathematicians (and likely other academics!) with small children face some unique challenges when traveling to conferences and workshops. The goal of this post is to reflect on these, and to start a constructive discussion what institutions and event organizers could do to improve the experiences of such participants.

The first necessary step is to recognize that different families have different needs. While it is hard to completely address everybody’s needs, there are some general measures that have a good chance to help most of the people traveling with young children. In this post, I will mostly focus on nursing mothers with infants ($\leq 24$ months old) because that is my personal experience. Many of the suggestions will apply to other cases such as non-nursing babies, children of single parents, children of couples of mathematicians who are interested in attending the same conference, etc..

The mother of a nursing infant that wishes to attend a conference has three options:

# QS World University Ranking 2014, Mathematics

Top 5: Harvard, MIT, Oxford, Berkeley, Cambridge

British Universities in top 50: 3 Oxford. 4-5 Cambridge, 11 Imperial, 25 Warwick, 31 Edinburgh, 34 Bristol, 36 UCL, 50 Manchester

# Maryam Mirzakhani, the first woman to win Fields Medal

From Maryam Mirzakhani’s interview to the Clay Institute:

My older brother was the person who got me interested in science in general. He used to tell me what he learned in school. My first memory of mathematics is probably the time that he told me about the problem of adding numbers from 1 to 100. I think he had read in a popular science journal how Gauss solved this problem. The solution was quite fascinating for me. That was the first time I enjoyed a beautiful solution, though I couldn’t find it myself. [...]

I was very lucky in many ways. The war ended when I finished elementary school; I couldn’t have had the great opportunities that I had if I had been born 10 years earlier. I went to a great high school in Tehran – Farzanegan – and had very good teachers.[...]

Our school was close to a street full of bookstores in Tehran. I remember how walking along this crowded street, and going to the bookstores, was so exciting for us. We couldn’t skim through the books like people usually do here in a bookstore, so we would end up buying a lot of random books. Also, our school principal was a strong-willed woman who was willing to go a long way to provide us with the same opportunities as the boys’ school.

Later, I got involved in Math Olympiads that made me think about harder problems. As a teenager, I enjoyed the challenge. But most importantly, I met many inspiring mathematicians and friends at Sharif University. The more I spent time on mathematics, the more excited I became.

Read the full text of the interview at The Guardian website.

# International Mathematics Competition for University Students 2014

Team results, individual result. British teams, out of 73:

39. University of Warwick

64. UCL

# How to Play like a Mathematician

[Originally posted on Edmund Harriss' blog Maxwell's Demon, this is a transcript of a talk at the Twitter Math Camp 2014, a truly energising event, teacher organised peer professional development. Anyone interested in education, whether parent, academic, teacher or administrator should check it out.]

### One…

Now imagine one dot pop into view.

### Two…

A second dot joins it. Let the two dots flow around each other, rotating and getting closer and further apart.

### Three…

Now a third dot, creating a line or a triangle…

### Four…

Keep on adding, with each addition try to see all the dots, find a shape you like…

### Seventeen…

This is about having fun and playing with math, which often sounds a little:

This is not the holy grail, it is not even a challenge to bring into the classroom. Teachers have too many challenges, sometimes the challenge is just to get through the day without messing up too badly. It is an encouragement to relax and have fun, yet remember that this fun is part of your teaching prep!

Playing with maths can often start with going back, returning to something you know well, and trying something new, testing an idea. If it fails try something a little different, or go back to work out how it went wrong. If it works, can you try everything? Mathematicians can say everything and really mean it! Even then do not settle, go back with your new knowledge and try something new. You might notice once you have started you cannot escape! You can always just stop. This is play not work. Though it might not be relaxing, just as playing a sport is exciting, fun and cathartic but you put effort in.

This is why this can build into your teaching, once you have fun you have a chance to help your students have fun. If they have fun they will put far more effort in than if you have to push them. Also I do not feel that mathematics has a huge number of facts, but isolated they are not that useful, going back and playing with ideas helps build the dense web of connections that really drives understanding.

General strategies are great, but it can be hard to know where to start, I will describe two tools:

• Analogy and the concept of same/different (mathematics is the world’s greatest metaphor!)
• Breaking rules! (yes mathematics is often about creating them, but also about changing them and seeing what happens).

To get further, we need an example, and not one that will lose half the audience just with its title so…

### Counting.

Three dots, are they the same or different? They are in different positions, but are the same shape. We have to be clear what we mean.

Now we take pairs of dots, we can spin them around and pull them apart. We could say they were the same if they can be moved on top of each other. Yet to define that precisely we have to use most of plane geometry. We have not even counted past two and we already need that!

Getting to three the line and the triangle, different in ways that the pair of dots can never be.

Lets change tack, we have been looking at how the same number of dots can be different, what about how different numbers of dots can be the same?

These patterns for four, six and eight have some similar features. How might we describe those precisely so we can identify other ones? Saying that the numbers are all even is an obvious way to do it, but maybe they also share something with this:

Like the earlier examples nine dots drawn like this form a rectangle (specifically a square). Following this definition we can define prime numbers (technically composite numbers!).

Here are another collection of dot patterns that share features, one dimension, two dimension and three dimension, and at this point reality gives up on us. Yet we really went past our page after two, we can use the notions of analogy to push further. We know the next pattern will have sixteen dots. For example we can make this image, with lines to show the structure. Can you find the eight cubes?

With a little work from here we can work out that an nn-dimensional cube has 2n (n-1)-dimensional faces. So we know very little about 172 dimensional space, but we do know that a hypercube in that space has 344 faces! Playing with some of these tricks we can get this:

There is a lot more to discover in this image. If you are interested in getting a version send me an email, I am looking into options.

Lets move to the other trick, breaking the rules. Mathematics is made of rules, yet there is not one rule that is not broken somewhere else in mathematics. For example this might make you uncomfortable:

7 + 7 = 2

2 + 1 = 2

If I say instead that seven months after July (the seventh month) is February then the first makes perfect sense. In this case 7+7 is still 14 but 14 is the same as 2, we have modular arithmetic.

That trick will not work for 2 + 1 = 2. Yet in Chemistry two hydrogen molecules combine with an oxygen molecule to create two water molecules. There is an even greater rule, though one that has been enshrined in legend. Yet this image shows what happens when we divide by zero (at the centre)!

(the mathematical trick is to use what is called the Riemann sphere).

In conclusion playing with math can happen with the simplest structures and lead to a variety of thoughts and adventures. No one should be shy of having a go!

Here is a neat animation from my play:

## Notes

I have a list of some other materials to inspire your mathematical play, and there is a whole world of examples in Sue van Hattan’s book Playing with mathematics. That should be available for pre-order soon!

Many other have explored the idea of simple pictorial versions of numbers, often using prime factorisation. With dots and circles, with monsters, or even to make a game. Although my personal favorite are these dots, with their illusion of simplicity.

format: plain


text:
# Bing metrization theorem

In topology, the Bing metrization theorem, named after R. H. Bing, characterizes when a topological space is metrizable.

## Formal statement

The theorem states that a topological space ${\displaystyle X}$ is metrizable if and only if it is regular and T0 and has a σ-discrete basis. A family of sets is called σ-discrete when it is a union of countably many discrete collections, where a family ${\displaystyle {\mathcal {F}}}$ of subsets of a space ${\displaystyle X}$ is called discrete, when every point of ${\displaystyle X}$ has a neighborhood that intersects at most one member of ${\displaystyle {\mathcal {F}}}$.

## History

The theorem was proven by Bing in 1951 and was an independent discovery with the Nagata–Smirnov metrization theorem that was proved independently by both Nagata (1950) and Smirnov (1951). Both theorems are often merged in the Bing-Nagata-Smirnov metrization theorem. It is a common tool to prove other metrization theorems, e.g. the Moore metrization theorem – a collectionwise normal, Moore space is metrizable – is a direct consequence.

## Comparison with other metrization theorems

Unlike the Urysohn's metrization theorem which provides a sufficient condition for metrization, this theorem provides both a necessary and sufficient condition for a topological space to be metrizable.

## References

• "General Topology", Ryszard Engelking, Heldermann Verlag Berlin, 1989. ISBN 3-88538-006-4


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# Box office receipts for independent movies for the first

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Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year.

C) The price of a movie ticket has not increased since last year.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year.

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.
[Reveal] Spoiler: OA
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Re: Box office receipts for independent movies for the [#permalink]

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20 Oct 2012, 12:32
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nitzz wrote:
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year. Not necessarily. If every movie that is going to be released this year bombs badly and generates no receipts the difference wont be significantly higher than 20%.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year. Not necessarily. What if all the movies last year were released in the second half.

C) The price of a movie ticket has not increased since last year. This might infact be even the opposite of what is implied by the passage and is also out of scope.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year. 50 movies were released last year. Lets say box office receipts was 100$, Average is 2$. 20 movies released this year. 20% increase in box office receipts. So 120$. Average is 6$. Answer

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.
Not necessarily. Ticket prices could have been hiked and hence the reason for increased receipts.
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Re: Box office receipts for independent movies for the first [#permalink]

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08 Nov 2012, 01:36
D is the clear winner all the way.

Thank you MacFauz for your explanation
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Re: Box office receipts for independent movies for the first [#permalink]

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13 Dec 2012, 04:20
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I narrowed it down to D and E
Eliminated E as it says number of persons this year is more than last years ....may or may not be true because if a person likes the movies he will watch again hence the receipt will increase but not the number of different people who watched. though different is not mentioned but i assumed so.
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Re: Box office receipts for independent movies for the first [#permalink]

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13 Dec 2012, 06:51
I also thought the same as Archit did... I narrowed down my answer to D and E... Increase in the number of reciepts might be due to people watching those muvies again n again...so statement E is of 'could be true' type....and hence winner ir D all the way...

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Re: Box office receipts for independent movies for the [#permalink]

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13 Dec 2012, 09:11
MacFauz wrote:
nitzz wrote:
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year. Not necessarily. If every movie that is going to be released this year bombs badly and generates no receipts the difference wont be significantly higher than 20%.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year. Not necessarily. What if all the movies last year were released in the second half.

C) The price of a movie ticket has not increased since last year. This might infact be even the opposite of what is implied by the passage and is also out of scope.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year. 50 movies were released last year. Lets say box office receipts was 100$, Average is 2$. 20 movies released this year. 20% increase in box office receipts. So 120$. Average is 6$. Answer

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.
Not necessarily. Ticket prices could have been hiked and hence the reason for increased receipts.

Can you help me in understanding an argument?
How to approach inference question?
I always get confuse after seeing too much of fact.
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Re: Box office receipts for independent movies for the [#permalink]

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13 Dec 2012, 09:29
Aristocrat wrote:
MacFauz wrote:
nitzz wrote:
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year. Not necessarily. If every movie that is going to be released this year bombs badly and generates no receipts the difference wont be significantly higher than 20%.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year. Not necessarily. What if all the movies last year were released in the second half.

C) The price of a movie ticket has not increased since last year. This might infact be even the opposite of what is implied by the passage and is also out of scope.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year. 50 movies were released last year. Lets say box office receipts was 100$, Average is 2$. 20 movies released this year. 20% increase in box office receipts. So 120$. Average is 6$. Answer

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.
Not necessarily. Ticket prices could have been hiked and hence the reason for increased receipts.

Can you help me in understanding an argument?
How to approach inference question?
I always get confuse after seeing too much of fact.

This one is not a typical inference question. It can be called an inference and numbers and percents question. For these types of questions, I find it helpful to break the facts given point by point and based on these facts we can check each answer choice.. Generally on inference questions I like to pick the mildest statement first and check if it is true based on the argument and more often than not, the mildest statement would be the right one. This question in particular can be approached as how one would approach a quant problem.. That is an easier way when lots of numbers and ratios are involved in the argument.
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Re: Box office receipts for independent movies for the [#permalink]

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13 Dec 2012, 23:24
Aristocrat wrote:
MacFauz wrote:
nitzz wrote:
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year. Not necessarily. If every movie that is going to be released this year bombs badly and generates no receipts the difference wont be significantly higher than 20%.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year. Not necessarily. What if all the movies last year were released in the second half.

C) The price of a movie ticket has not increased since last year. This might infact be even the opposite of what is implied by the passage and is also out of scope.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year. 50 movies were released last year. Lets say box office receipts was 100$, Average is 2$. 20 movies released this year. 20% increase in box office receipts. So 120$. Average is 6$. Answer

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.
Not necessarily. Ticket prices could have been hiked and hence the reason for increased receipts.

Can you help me in understanding an argument?
How to approach inference question?
I always get confuse after seeing too much of fact.

Hi,

This is a question that goes slightly beyond the norms of a normal inference question.

Similar questions are asked in the two-part section of IR. We had recently posted a similar question, along with a detailed approach to such questions.

Click on the image to go to the question:

Hope this helps

Thanks,
Chiranjeev
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Re: Box office receipts for independent movies for the first [#permalink]

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10 May 2015, 03:35
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Re: Box office receipts for independent movies for the first [#permalink]

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11 May 2015, 04:29
nitzz wrote:
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year.

C) The price of a movie ticket has not increased since last year.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year.

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.

D is correct
Other options can be proved incorrect by using the info given in the stimulus
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Re: Box office receipts for independent movies for the first [#permalink]

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15 Jun 2015, 22:54
Box office receipts for independent movies for the first half of this year have increased by 20 percent over the total receipts for independent movies for all of last year. Last year, 50 independent movies were released, while so far this year only 20 independent movies have been released. The number of independent movies slated for release in the second half of this year is roughly equal to the number released so far.

If the statements above are true, which of the following must be true?

A) The total box office receipts for independent movies this year will be significantly more than 20 percent greater than the receipts for independent movies last year.

B) The number of independent movies released in the first half of this year is equal to the number released in the first half of last year.

C) The price of a movie ticket has not increased since last year.

D) The average receipts for the independent films released during the first half of this year is greater than that of all independent films released last year.

E) The number of people seeing independent movies during the first half of this year is greater than the number who saw independent movies last year.

Pre think step-We can set up the equation as Average Receipts(Total B.O Receipts/Total no.of movies) is greater this year than compared to last year considering the data given.

A-This choice may be true since it has been mentioned in the argument that B.O receipts for IM is greater than 20% in first half of this year as compared to total receipts of last year however,it must not be true since in the second half the trend may go up or go down.Hence incorrect
B-We cannot logically infer this since the number of independent movies released in first half of last year has not been mentioned in the argument.It only states that total 50 movies released last year without giving its breakdown.
C-Price factor has also not been discussed in the argument.It may be true that price has increased this year as compared to last year however we cannot conclusively infer the same.
D-Similar to our pre think step hence correct
E-The number of people seeing the IM has not been stated hence we cannot infer it logically
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Re: Box office receipts for independent movies for the first [#permalink]

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05 Jan 2016, 08:30
daviesj wrote:
I also thought the same as Archit did... I narrowed down my answer to D and E... Increase in the number of reciepts might be due to people watching those muvies again n again...so statement E is of 'could be true' type....and hence winner ir D all the way...

Posted from my mobile device

there may be huge difference in the prices in two year and no of people may vary by only one or by 1000000 making a colud be true option D.
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Re: Box office receipts for independent movies for the first [#permalink]

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20 Jan 2016, 12:23
Could someone explain what "box office receipts" means? Is it a receipts = to the sale of one ticket....or is it a numerical value, as in receipts = amount of \$?
Re: Box office receipts for independent movies for the first   [#permalink] 20 Jan 2016, 12:23
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12 Inference Revision: Box office receipts for independent movies 13 14 Feb 2015, 01:42
16 Box office receipts for independent movies for the first 18 15 Jul 2013, 02:46
10 Holiday receipts the total sales recorded in the fourth 4 30 Apr 2013, 13:04
4 CR: Box office receipts 6 31 Mar 2008, 16:51
Holiday receipts the total sales recorded in the fourth 8 27 Feb 2007, 22:00
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# zbMATH — the first resource for mathematics

Iterative and non-iterative methods for non-linear Volterra integro-differential equations. (English) Zbl 1202.65179
The author considers the problem of numerical solving the initial problem for the equation $A(t)u^{(n)}(t)=f(t,u(t)) + \int_{t_0}^{t}g(s,u(s))ds, \; t_0<t< \infty,$ subject to $$u^{(j)}(t_0)=\alpha_j, \; 0<j\leq(n-1)$$, where $$A(t)$$ are invertible square matrices of the order $$N$$ and $$u^{(j)}$$ denotes the $$j$$-th order derivative of the unknown $$N$$-dimensional function $$u(t)$$. This problem is deep theoretically investigated, and the local existence theorem 1 presented and proved in the paper is presented (with non-essential simplification $$A(t)=I$$) in the text-book by A. B. Vasilieva and A. N. Tikhonov [Integralnye Uravnenia (Russian). Izdat. Moskovskogo Universiteta, Moscow (1986)]. The author presents and discusses a few variants of iterative algorithms of Picard kind, and a series method of solving the initial problem. The latter is the presentation of an approximation of the required solution as a finite functional sum. The first summand is the given initial value $$\alpha$$, and the each following summand is an integral iteration of one or two predecessors. Convergence of the sum to the required solution is proved.

##### MSC:
 65R20 Numerical methods for integral equations 45J05 Integro-ordinary differential equations
Full Text:
##### References:
 [1] Lovitt, W.V., Linear integral equations, (1950), Dover Publications, Inc. New York · JFM 50.0290.05 [2] Delves, L.M.; Mohamed, J.L., Computational methods for integral equations, (1985), Cambridge University Press New York · Zbl 0592.65093 [3] Davis, H.T., Introduction to nonlinear differential and integral equations, (1962), Dover Publications, Inc. Ne York [4] Tricomi, F.G., Integral equations, (1985), Dover Publications, Inc. New York [5] Petrovski, I., Lectures on the theory of integral equations, (1971), MIR Moscow, USSR [6] Krasnov, M.; Kiseliov, A.; Makarenko, G., Integral equations, (1982), MIR Moscow, USSR [7] Porter, D.; Stirling, D.S.G., Integral equations, (1990), Cambridge University Press New York · Zbl 0908.45001 [8] Agarwal, R.P., Boundary value problems for high ordinary differential equations, (1986), World Scientific Singapore · Zbl 0598.65062 [9] Ramos, J.I., Piecewise-quasilinearization techniques for singularly perturbed Volterra integro-differential equations, Applied mathematics and computation, 188, 1221-1233, (2007) · Zbl 1118.65130 [10] Wazwaz, A.-M., A comparison study between the modified decomposition method and the traditional methods for solving nonlinear integral equations, Applied mathematics and computation, 181, 1703-1712, (2006) · Zbl 1105.65128 [11] Hashim, I., Adomian decomposition method for solving BVPs for fourth-order integro-differential equations, Journal of computational and applied mathematics, 193, 658-664, (2006) · Zbl 1093.65122 [12] He, J.-H., Variational iteration method – a kind of non-linear analytical technique: some examples, International journal of nonlinear mechanics, 34, 699-708, (1999) · Zbl 1342.34005 [13] He, J.-H., Some asymptotic methods for strongly nonlinear equations, International journal of modern physics, 20, 1141-1199, (2006) · Zbl 1102.34039 [14] Sweilam, N.H., Fourth-order integro-differential equations using variational iteration method, Computer and mathematics with applications, 54, 1086-1091, (2007) · Zbl 1141.65399 [15] Wang, S.-Q.; He, J.-H., Variational iteration method for solving integro-differential equations, Physics letters A, 367, 188-191, (2007) · Zbl 1209.65152 [16] Shou, D.-H.; He, J.-H., Beyond Adomian: the variational iteration method for solving heat-like and wave-like equations with variable coefficients, Physics letters A, 372, 233-237, (2008) · Zbl 1217.35091 [17] Saberi-Nadjafi, J.; Tamamgar, M., The variational iteration method: a highly promising method for solving the system of integro-differential equations, Computer and mathematics with applications, 56, 346-351, (2008) · Zbl 1155.65399 [18] Adomian, G., Stochastic systems, (1983), Academic Press Inc. New York · Zbl 0504.60067 [19] Adomian, G., Nonlinear stochastic operator equations, (1986), Academic Press Inc. New York · Zbl 0614.35013 [20] Adomian, G., Solving frontier problems of physics: the decomposition method, (1994), Kluwer Boston · Zbl 0802.65122 [21] Ramos, J.I., On the variational iteration method and other iterative techniques for nonlinear differential equations, Applied mathematics and computation, 199, 39-69, (2008) · Zbl 1142.65082 [22] Ramos, J.I., On the Picard lindelof method for nonlinear second-order differential equations, Applied mathematics and computation, 203, 238-242, (2008) · Zbl 1195.65095 [23] Ramos, J.I., A non-iterative derivative-free method for nonlinear ordinary differential equations, Applied mathematics and computation, 203, 672-678, (2008) · Zbl 1157.65414 [24] Keller, H.B., Numerical methods for two-point boundary-value problems, (1992), Dover Publications, Inc. New York · Zbl 0172.19503 [25] Stakgold, I., Boundary value problems of mathematical physics, vols. I & II, (1967), MacMillan New York · Zbl 0158.04801 [26] Stakgold, I., Green’s functions and boundary value problems, (1998), John Wiley & Sons Inc. New York · Zbl 0897.35001 [27] Widder, D.V., Advanced calculus, (1989), Dover Publications, Inc. New York · Zbl 0728.26002 [28] Halmos, P.R., Finite-dimensional vector spaces, (1993), Springer New York · Zbl 0107.01404 [29] Lax, P.D., Functional analysis, (2002), Wiley-Interscience New York · Zbl 1009.47001 [30] Ortega, J.; Rheinbolt, W., Iterative solution of nonlinear equations in several variables, (1970), Academic Press Inc. New York [31] W. Rheinbolt, Methods for solving systems of nonlinear equations, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania, 1974. [32] () [33] Atkinson, K.E., An introduction to numerical analysis, (1989), John Wiley & Sons Inc. New York · Zbl 0718.65001 [34] Evans, D.J., Preconditioning methods: theory and applications, (1983), Routledge New York [35] Y. Saad, Iterative Methods for Sparse Linear Systems, second edition, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania, 2003. · Zbl 1031.65046 [36] Axelsson, O., Iterative solution methods, (1994), Cambridge University Press New York · Zbl 0795.65014 [37] Turkel, E., Preconditioning techniques in computational fluid dynamics, Annual review of fluid dynamics, 31, 385-416, (1999) [38] Yu, Z.-H., Variational iteration method for solving the multi-pantograph delay equation, Physics letters A, 372, 6475-6479, (2008) · Zbl 1225.34024 [39] J.I. Ramos, Piecewise-adaptive decomposition methods, Chaos, Solitons & Fractals, in press. · Zbl 1198.65149 [40] Ramos, J.I., Piecewise homotopy methods for nonlinear ordinary differential equations, Applied mathematics and computation, 198, 92-116, (2008) · Zbl 1137.65048 [41] He, Ji-H., Homotopy perturbation technique, Computer methods in applied mechanics and engineering, 178, 257-262, (1999) · Zbl 0956.70017 [42] He, Ji-H., Homotopy perturbation method: a new nonlinear analytical technique, Applied mathematical computation, 135, 73-79, (2003) · Zbl 1030.34013 [43] He, Ji-H., Addendum: new interpretation of homotopy perturbation method, International journal of modern physics, 20, 2561-2568, (2006) [44] J.I. Ramos, Series approach to the Lane-Emden equation and comparison with the homotopy perturbation method, Chaos, Solitons & Fractals 38 (2008) 400-408. · Zbl 1146.34300 [45] Ramos, J.I., An artificial parameter-decomposition method for nonlinear oscillators: applications to oscillators with odd nonlinearities, Journal of sound and vibration, 307, 312-329, (2007)
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:
# Univariate Logistic Regression¶

Logistic regression is a supervised classification algorithm This is similar to linear regression, but is used when the dependent variable is categorical (despite its name regression). So, it targets the classification problems unlike linear regression for regression problems. Objective in classification is to map the observation (input) to its associated class or label.

Linear regression can be used for classification problems by using a threshold, however using least squares (the sum of the squares of the residuals) criteria is not more apt way for separating the classes for classification.

So, usually for categorical values best try is Logistic Regression, Decision Trees, SVM and Random Forest etc.

Logistic regression predicts the probability of the outcome for a given input unlike linear function which predicts the outcome for a given input.
So, output value modeled in logistic regression is a binary value (0 or 1) instead of a numeric value in linear regression.

Say P is the probability of success of an event. Then probability of failure is 1-P.
By definition, probabilities and proportions cannot exceed 1 or fall below 0. Odds of success are defined as ratio of probability of success over probability of failure.
Odds = P/(1-P)

$$log(\dfrac{P(x)}{1-P(x)}) = \sum_{i=1}^{m} \theta_0 + \theta_i x_i$$ In vector form $$=\theta^T X$$

This model assumes that the log(odds) of an observation can be expressed as a linear function to the input variable.

LHS is logit of P, so the name logistic regression.
RHS is linear, similar to linear regression.

Inverse of the logit function is

$$P(x) = \dfrac{e^z}{1+e^z}$$

$$= \dfrac{1}{1+e^{-z}}$$ where $z = \theta^T X$

RHS is the sigmoid function of Z, which maps the continuous real line to the interval (0,1)

### Flow chart¶

Algorithm for this is very similar to linear regression, only things we need to change is the representation of response(prediction) and the cost function

The response, while in linear regression we use a linear function of X ($\theta^T X$), in logistic regression we use sigmoid function of the linear function ($\theta^T X$).

Hypothesis function:
Linear Regression : $$h_\theta(x) = (\theta^T X)$$

Logistic Regression : $$h_\theta(x) = g(\theta^T X)$$

$g$ is a link function, this transforms the observed responses to the original data.

$$h_\theta(x) = \dfrac{1}{1+e^{-\theta^T X}}$$

This hypothesis function represents the estimated probability that y = 1 for given input x parameterized by θ: $$h_\theta(x) = P(y=1 | x;\theta)$$

Since the hypothesis function is formed by sigmoid function, our cost function is not going to be a convex function, which happened to be for linear regression as our hypothesis function was a linear function. It means, unlike the cost function in linear regression, this cost function in logistic regression will get many local minimum.
So, to make this a convex function, this is transformed using logarithm.

Image source:Coursera, Neural Networks

$$Error(h_\theta(x),y)=\begin{cases} -log(h_\theta(x)), & \text{if y=1}\\ -log(1-h_\theta(x)), & \text{if y=0} \end{cases}$$

Cost function :

Mean of the errors $$J(\theta) = \dfrac{1}{m} \sum_{i=1}^{m}Error(h_\theta(x^{(i)})),y^{(i)})$$

We are trying to minimize the difference between the prediciont and the response

Putting the two functions into one compact function, we get $$J(\theta) = -\dfrac{1}{m} \sum_{i=1}^{m} \big[y^{(i)} log(h_\theta(x^{(i)})) +(1-y^{(i)}) log(1-h_\theta(x^{(i)}))\big]$$

Binomial/Binary/Univariable/Univariate logistic regression - If the dependent variable is a binary variable [True/False, 0/1]

Multinomial logistic regression - If the dependent variable have more than two outcomes.

We'll try for univariate logistic regression.
We use iris data from sklearn datasets for this.
For that we will consider only one feature (sepal length) and a target vector of size 2 (setosa, versicolor)

### Working¶

In [1]:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from sklearn import datasets
from sklearn import linear_model

In [2]:
iris = datasets.load_iris()

In [3]:
iris.feature_names

Out[3]:
['sepal length (cm)',
'sepal width (cm)',
'petal length (cm)',
'petal width (cm)']
In [4]:
iris.target_names

Out[4]:
array(['setosa', 'versicolor', 'virginica'],
dtype='<U10')
In [5]:
X = iris.data[:, 0]

y_bool = iris.target!=2

y = iris.target[y_bool]

X = X[y_bool]

In [6]:
plt.scatter(X, y)
plt.xlabel('Sepal Length ', fontsize=15)

plt.ylabel('0 - setosa, 1 - versicolor ', fontsize=15)
plt.show()

#### Algorithm¶

In [7]:
X = np.c_[np.ones((X.shape[0],1)), X[:]]
y = y.reshape(-1,1)

# Parameters required for Gradient Descent
alpha = 0.1   #learning rate
m = y.size  #no. of samples
np.random.seed(10)
theta = np.random.rand(2)  #initializing theta with some random values
theta = theta.reshape(-1,1)

In [8]:
def gradient_descent(x, y, m, theta,  alpha):
cost_list = []   #to record all cost values to this list
theta_list = []  #to record all theta_0 and theta_1 values to this list
prediction_list = []
run = True
cost_list.append(1e10)    #we append some large value to the cost list
i=0
while run:
Z = np.dot(x, theta)
prediction = 1 / (1 + np.exp(-Z))   #predicted y values
prediction_list.append(prediction)
error = prediction - y
cost = np.sum(-(y * np.log(prediction) + (1 - y) * np.log(1 - prediction))) / m   #  (1/2m)*sum[(error)^2]

cost_list.append(cost)
theta = theta - (alpha * (1/m) * np.dot(x.T, error))   # alpha * (1/m) * sum[error*x]
theta_list.append(theta)
if cost_list[i]-cost_list[i+1] < 1e-9:   #checking if the change in cost function is less than 10^(-9)
run = False

i+=1
cost_list.pop(0)   # Remove the large number we added in the begining
return prediction_list, cost_list, theta_list

In [9]:
prediction_list, cost_list, theta_list = gradient_descent(X, y, m, theta, alpha)
theta = theta_list[-1]

In [10]:
plt.title('Cost Function J', size = 30)
plt.xlabel('No. of iterations', size=20)
plt.ylabel('Cost', size=20)
plt.plot(cost_list)
plt.show()

Weights we got

In [11]:
theta

Out[11]:
array([[-27.53475787],
[  5.08546084]])

We'll plot a decision boundary with the coefficients

We'll take sample data in the training data range

In [12]:
X_test = np.linspace(4, 7, 300)

In [13]:
def sigmoid(x):
return 1 / (1 + np.exp(-x))

In [14]:
loss = sigmoid(X_test*theta[1] + theta[0])

In [15]:
plt.figure(1, figsize=(8, 6))
plt.clf()
plt.plot(X_test, loss, c='C0', label='Hyperplane')
plt.scatter(X[:,1], y, c='C1', label='Training data')
# plt.plot(X_test,X_test*theta[1] + theta[0] )
# plt.axhline(0.5, c='C2',label='0.5 Threshold')
# plt.axvline(5.4147157190635449, c='C2',label='0.5 Threshold')
# plt.axvline
plt.legend()
plt.xlabel('Sepal Length (Input)', fontsize=15)
plt.ylabel("Probability of the output" "\n" "(0 - setosa, 1 - versicolor)", fontsize=15)
plt.show()

We keep this threshold value of 0.5, for the given input if the probability is equal to or above 0.5, we consider it 1 and less than 0.5 we consider it 0

In [16]:
# 0.5 threshold corresponds to
boundary = X_test[np.where(loss >= 0.5)[0][0]]
print(round(boundary,3))

5.415

So, sepal length above 5.415 is categorized as versicolor and less than that is categorized as setosa

In [17]:
plt.figure(1, figsize=(8, 6))
plt.clf()
plt.plot(X_test, loss, c='C0', label='Hyperplane')
plt.scatter(X[:,1], y, c='C1', label='Training data')
plt.axhline(0.5, c='C2',label='0.5 Threshold', linewidth=0.7)
plt.axvline(boundary, c='C3',label='Boundary', linewidth=0.7)
plt.legend()
plt.xlabel('Sepal Length (Input)', fontsize=15)
plt.ylabel("Probability of the output" "\n" "(0 - setosa, 1 - versicolor)", fontsize=15)
plt.show()

## Scikit learn¶

We can check this with sci-kit learn logistic regression module

In [18]:
from sklearn.linear_model import LogisticRegression

In [19]:
lr = LogisticRegression(C=1e100)

Here C from the formal definition from sklearn
Inverse of regularization strength; must be a positive float. Like in support vector machines, smaller values specify stronger regularization.

Since we are not using any regularization here, we need to keep C ($=1/\lambda$) value very high, so the the regularization strength $\lambda \approx 0$

Regularization is used for avoiding overfitting.
We will see more about regularization in another post.

In [20]:
X.shape

Out[20]:
(100, 2)
In [21]:
X[:5]

Out[21]:
array([[ 1. ,  5.1],
[ 1. ,  4.9],
[ 1. ,  4.7],
[ 1. ,  4.6],
[ 1. ,  5. ]])

We added a column of ones for our earlier algorithm, which we don't need here, so taking off that column

In [22]:
lr = lr.fit(X[:,1].reshape(-1,1),y.ravel())

Checking the weights

#### From scikit learn¶

In [23]:
'Theta_0 and Theta_1 are {},{}'.format(round(lr.intercept_[0],3), round(lr.coef_[0,0],3))

Out[23]:
'Theta_0 and Theta_1 are -27.831,5.14'

#### From scratch¶

In [24]:
theta[0,0], theta[1,0]

Out[24]:
(-27.534757871504066, 5.085460835465244)
In [25]:
'Theta_0 and Theta_1 are {},{}'.format(round(theta[0,0],3),round(theta[1,0],3))

Out[25]:
'Theta_0 and Theta_1 are -27.535,5.085'
In [26]:
from IPython.core.display import display, HTML
display(HTML("<style>.container { width:100% !important; }</style>"))


text:
# Variables¶

Variables frequently used in C-API are explained. In interfaces of other languages, the definition may be different, e.g., lattice for Python interface is transposed.

## lattice¶

Attention: In Python interface, lattice is transposed (Crystal structure (cell).)

Basis vectors (in Cartesian)

[ [ a_x, b_x, c_x ],
[ a_y, b_y, c_y ],
[ a_z, b_z, c_z ] ]

Cartesian position $$\mathbf{x}_\mathrm{Cart}$$ is obtained by

$\mathbf{x}_\mathrm{Cart} = \mathrm{L}\cdot\mathbf{x}_\mathrm{frac}$

where $$\mathrm{L}$$ is the basis vectors defined above and is $$\mathrm{L}=(\mathbf{a},\mathbf{b},\mathbf{c})$$, and $$\mathbf{x}_\mathrm{frac}$$ is the atomic position in fractional coordinates given below.

## position¶

Atomic points (in fractional coordinates with respect to basis vectors)

[ [ x1_a, x1_b, x1_c ],
[ x2_a, x2_b, x2_c ],
[ x3_a, x3_b, x3_c ],
...                   ]

## types¶

Atomic species are differenciated by integers. The nubmers are just used to distinguishes atoms, and is not releated to atomic numbers.

[ type_1, type_2, type_3, ... ]

The number of elements is same as the number of atoms.

## rotation¶

Rotation matricies (ingeger) of symmetry operations.

[ [ r_aa, r_ab, r_ac ],
[ r_ba, r_bb, r_bc ],
[ r_ca, r_cb, r_cc ] ]

## translation¶

Translation vectors corresponding to symmetry operations in fractional coordinates.

[ t_a, t_b, t_c ]

## symprec¶

Tolerance of distance between atomic positions and between lengths of lattice vectors to be tolerated in the symmetry finding. The angle distortion between lattice vectors is converted to a length and compared with this distance tolerance. If the explicit angle tolerance is expected, see angle_tolerance.

## angle_tolerance¶

Experimental

Tolerance of angle between lattice vectors in degrees to be tolerated in the symmetry finding. To use angle tolerance, another set of functions are prepared as follows

spgat_get_dataset
spgat_get_symmetry
spgat_get_symmetry_with_collinear_spin
spgat_get_multiplicity
spgat_find_primitive
spgat_get_international
spgat_get_schoenflies
spgat_refine_cell

These functions are called by the same way with an additional argument of const double angle_tolerance in degrees. By specifying a negative value, the behavior becomes the same as usual functions. The default value of angle_tolerance is a negative value.


text:
Chapter 7.1, Problem 24E

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

Chapter
Section

### Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

# Evaluate the integral. ∫ 0 1 ( x 2 + 1 ) e − x d x

To determine

To evaluate: The given integral using the technique of integration by parts.

Explanation

The technique of integration by parts comes in handy when the integrand involves product of two functions. It can be thought of as a rule corresponding to the product rule in differentiation.

Formula used:

The formula for integration by parts for definite integral is given by

abf(x)g(x)dx=f(x)g(x)]ababg(x)f(x)dx

Given:

The integral, 01(x2+1)exdx.

Calculation:

Make the choice for u and dv such that the resulting integration from the formula above is easier to integrate. Let

u=x2+1      dv=exdx

Then, the differentiation of u and antiderivative of dv will be

du=2xdx      v=ex

Using the formula above, the given integration will become

01(x2+1)exdx=(x2+1)(ex)]0101(ex)(2x)dx=(x2+1)(ex)]01+201xexdx …… (1)

Integrate the last term using integration by parts with the following substitution:

u=x     dv=exdx

Then

du=dx

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text:
# Weak convergence in $L^q$ but not in $L^p$, where $1\leq p<q<\infty$?

I've been trying to find a sequence $\{f_n\}$ of functions on $[0,1]$ that converges weakly in $L^q$ but does not converge weakly in $L^p$, where $1\leq p<q<\infty$. I'm stuck and any hints would be greatly appreciated.

(I find this to be tricky because the sequences are defined on $[0,1]$ and the functions cannot spread to infinity. Since we're working on $[0,1]$, we have $L^q\subset L^p$.)

The functions of the example below are not supported in $[0,1]$.

## 1 Answer

Sorry for overlooking the requirement that the functions are supported on $[0,1]$. If this condition is added, then the statement is true.

Straight from the definition, $f_n$ converges weakly to $f\in L^q$ iff for every $g\in L^{q^*}$,

$$\int f_ng\to \int fg,$$

where $1/q+1/q^*=1$. Since $p<q$, $p^*>q^*$, so $L^{p^*}([0,1])\subset L^{q^*}([0,1])$. Then the above convergence holds for every $g\in L^{p^*}$. This implies $f_n$ converges weakly to $f$ in $L^p$. As a sanity check, we see that $f$ is in $L^p$ because $L^q([0,1])\subset L^p([0,1])$.

More succinctly, this is because continuous linear maps between Banach spaces are also weakly continuous (See Theorem 1.1 of Chapter 6 of Conway, A Course in Functional Analysis).

Just for reference, the result is false if the functions are supported on $\mathbb R$.

Let $r\in(p,q)$ and $f_n=n^{-1/r}$ on $[0,n]$ and $f_n=0$ elsewhere. Then

$$\|f_n\|_{q}=(n\cdot n^{-q/r})^{1/q}=n^{1/q-1/r}\to 0$$

as $n\to\infty$, so $f_n\to0$ in $L^q$. On the other hand, we replace $q$ by $p$ in the above inequality to get

$$\|f_n\|_{p}=n^{1/p-1/r}\to\infty$$

as $n\to\infty$. By the uniform boundedness principle (among other approaches), $f_n$ does not converge weakly in $L^p$.

• The example seems actually impossible: Let p*, q* be the conjugate exponents of p, q resp. Since [0,1] contains sets of measure at most 1, if p<q (i.e., p*>q*) then L^q([0,1]) is a subset of L^p([0,1]) and L^{p^*}([0,1]) is a subset of L^{q^*}([0,1]). Therefore, if f_n are in L^q, hence also in L^p, then weak convergence in L^q implies weak convergence in L^p. – Aubrey Mar 10 '15 at 2:33
• Thank you! Perhaps you should just change the phrase "the statement is true" to "the statement is false". – Aubrey Mar 10 '15 at 2:35


text:
# naginterfaces.library.tsa.multi_spectrum_lag¶

naginterfaces.library.tsa.multi_spectrum_lag(nxy, mtxy, pxy, iw, mw, ish, ic, cxy, cyx, kc, l, xg=None, yg=None)[source]

multi_spectrum_lag calculates the smoothed sample cross spectrum of a bivariate time series using one of four lag windows: rectangular, Bartlett, Tukey or Parzen.

For full information please refer to the NAG Library document for g13cc

https://www.nag.com/numeric/nl/nagdoc_27.3/flhtml/g13/g13ccf.html

Parameters
nxyint

, the length of the time series and .

mtxyint

If cross-covariances are to be calculated by the function (), must specify whether the data is to be initially mean or trend corrected.

For no correction.

For mean correction.

For trend correction.

If cross-covariances are supplied , is not used.

pxyfloat

If cross-covariances are to be calculated by the function (), must specify the proportion of the data (totalled over both ends) to be initially tapered by the split cosine bell taper. A value of implies no tapering.

If cross-covariances are supplied , is not used.

iwint

The choice of lag window.

Rectangular.

Bartlett.

Tukey.

Parzen.

mwint

, the ‘cut-off’ point of the lag window, relative to any alignment shift that has been applied. Windowed cross-covariances at lags or less, and at lags or greater are zero.

ishint

, the alignment shift between the and series. If leads , the shift is positive.

icint

Indicates whether cross-covariances are to be calculated in the function or supplied in the call to the function.

Cross-covariances are to be calculated.

Cross-covariances are to be supplied.

cxyfloat, array-like, shape

If , must contain the cross-covariances between values in the series and earlier values in time in the series, for lags from to .

If , need not be set.

cyxfloat, array-like, shape

If , must contain the cross-covariances between values in the series and later values in time in the series, for lags from to .

If , need not be set.

kcint

If , must specify the order of the fast Fourier transform (FFT) used to calculate the cross-covariances.

If , that is if covariances are supplied, is not used.

lint

, the frequency division of the spectral estimates as . Therefore, it is also the order of the FFT used to construct the sample spectrum from the cross-covariances.

xgNone or float, array-like, shape , optional

Note: the required length for this argument is determined as follows: if : ; otherwise: .

If the cross-covariances are to be calculated, then must contain the data points of the series. If covariances are supplied, need not be set.

ygNone or float, array-like, shape , optional

Note: the required length for this argument is determined as follows: if : ; otherwise: .

If cross-covariances are to be calculated, must contain the data points of the series. If covariances are supplied, need not be set.

Returns
cxyfloat, ndarray, shape

If , will contain the calculated cross-covariances.

If , the contents of will be unchanged.

cyxfloat, ndarray, shape

If , will contain the calculated cross-covariances.

If , the contents of will be unchanged.

xgfloat, ndarray, shape

Contains the real parts of the complex spectral estimates in elements to , and to contain . The series leads the series.

ygfloat, ndarray, shape

Contains the imaginary parts of the complex spectral estimates in elements to , and to contain . The series leads the series.

ngint

The number, , of complex spectral estimates, whose separate parts are held in and .

Raises
NagValueError
(errno )

On entry, , and .

Constraint: if , .

(errno )

On entry, , and .

Constraint: if , .

(errno )

On entry, .

Constraint: if , .

(errno )

On entry, .

Constraint: if , .

(errno )

On entry, .

Constraint: if then , or .

(errno )

On entry, and .

Constraint: .

(errno )

On entry, , and .

Constraint: .

(errno )

On entry, , and .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, and .

Constraint: .

(errno )

On entry, .

Constraint: , , or .

(errno )

On entry, .

Constraint: .

(errno )

On entry, , and .

Constraint: if , .

(errno )

On entry, and .

Constraint: .

Notes

The smoothed sample cross spectrum is a complex valued function of frequency , , defined by its real part or co-spectrum

and imaginary part or quadrature spectrum

where , for , is the smoothing lag window as defined in the description of uni_spectrum_lag(). The alignment shift is recommended to be chosen as the lag at which the cross-covariances peak, so as to minimize bias.

The results are calculated for frequency values

where denotes the integer part.

The cross-covariances may be supplied by you, or constructed from supplied series ; as

this convolution being carried out using the finite Fourier transform.

The supplied series may be mean and trend corrected and tapered before calculation of the cross-covariances, in exactly the manner described in uni_spectrum_lag() for univariate spectrum estimation. The results are corrected for any bias due to tapering.

The bandwidth associated with the estimates is not returned. It will normally already have been calculated in previous calls of uni_spectrum_lag() for estimating the univariate spectra of and .

References

Bloomfield, P, 1976, Fourier Analysis of Time Series: An Introduction, Wiley

Jenkins, G M and Watts, D G, 1968, Spectral Analysis and its Applications, Holden–Day


text:
# American Institute of Mathematical Sciences

2015, 22: 87-91. doi: 10.3934/era.2015.22.87

## A Besicovitch cylindrical transformation with Hölder function

 1 Lomonosov Moscow State University, Russian Federation

Received  April 2015 Revised  August 2015 Published  October 2015

For any $\gamma\in(0,1)$ and any $\varepsilon>0$ we construct a cylindrical cascade over some irrational circle rotation with a $\gamma$-Hölder function such that the Besicovitch condition holds and the Hausdorff dimension of the set of points in the circle having discrete orbits is more than $1-\gamma-\varepsilon$. This result gives the answers to some questions of K. Frączek and M. Lemańczyk [1].
Citation: Andrey Kochergin. A Besicovitch cylindrical transformation with Hölder function. Electronic Research Announcements, 2015, 22: 87-91. doi: 10.3934/era.2015.22.87
##### References:
 [1] K. Frączek and M. Lemańczyk, On the Hausdorff dimension of the set of closed orbits for a cylindrical transformation, Nonlinearity, 23 (2010), 2393-2422. doi: 10.1088/0951-7715/23/10/003.  Google Scholar [2] A. S. Besicovitch, A problem on topological transformations of the plane. II, Proc. Cambridge Philos. Soc., 47 (1951), 38-45. doi: 10.1017/S0305004100026347.  Google Scholar [3] W. H. Gottschalk and G. A. Hedlund, Topological Dynamics, Amer. Math. Soc. Colloq. Publ., Vol. 36, Amer. Math. Soc., Providence, RI, 1955.  Google Scholar [4] E. Dymek, Transitive cylinder flows whose set of discrete points is of full Hausdorff dimension, arXiv:1303.3099v1, 2013. Google Scholar [5] A. Kochergin, A mixing special flow over a circle rotation with almost Lipschitz function, Sbornik: Mathematics, 193 (2002), 359-385. doi: 10.1070/SM2002v193n03ABEH000636.  Google Scholar [6] K. Falconer, Fractal Geometry. Mathematical Foundations and Applications, Second edition, John Wiley & Sons, Inc., Hoboken, NJ, 2003. doi: 10.1002/0470013850.  Google Scholar

show all references

##### References:
 [1] K. Frączek and M. Lemańczyk, On the Hausdorff dimension of the set of closed orbits for a cylindrical transformation, Nonlinearity, 23 (2010), 2393-2422. doi: 10.1088/0951-7715/23/10/003.  Google Scholar [2] A. S. Besicovitch, A problem on topological transformations of the plane. II, Proc. Cambridge Philos. Soc., 47 (1951), 38-45. doi: 10.1017/S0305004100026347.  Google Scholar [3] W. H. Gottschalk and G. A. Hedlund, Topological Dynamics, Amer. Math. Soc. Colloq. Publ., Vol. 36, Amer. Math. Soc., Providence, RI, 1955.  Google Scholar [4] E. Dymek, Transitive cylinder flows whose set of discrete points is of full Hausdorff dimension, arXiv:1303.3099v1, 2013. Google Scholar [5] A. Kochergin, A mixing special flow over a circle rotation with almost Lipschitz function, Sbornik: Mathematics, 193 (2002), 359-385. doi: 10.1070/SM2002v193n03ABEH000636.  Google Scholar [6] K. Falconer, Fractal Geometry. Mathematical Foundations and Applications, Second edition, John Wiley & Sons, Inc., Hoboken, NJ, 2003. doi: 10.1002/0470013850.  Google Scholar
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1. Find the minimum value of $\displaystyle { \frac{ ( x + \frac{1}{x} )^6 - ( x^6 + \frac{1}{x^6}) - 2}{(x+\frac{1}{x})^3 + (x^3 + \frac{1}{x^3} )} }$ and $x \in \mathbb{R}$ and $x > 0$
SOLUTION: here
2. Given that P and Q are points on the sides AB and AC respectively of $\Delta ABC$. The perpendiculars to the sides AB and AC at P and Q respectively meet at D, an interior point of $\Delta ABC$. If M is the midpoint of BC, prove that PM = QM if and only if $\angle BDP = \angle CDQ$.
3. Let $N = 2^5 + 2^{5^2} + 2^{5^3} + ... + 2^{5^{2013}}$. Written in the usual decimal form, find the last two digits of the number N.
SOLUTION:here
4. Two circles $\Sigma_1$ and $\Sigma_2$ having centers at $C_1$ and $C_2$ intersect at A and B. Let P be a point on the segment AB and let $AP \neq PB$. The line through P perpendicular to $C_1 P$ meets $\Sigma_1$ at C and D. The line through P perpendicular to $C_2P$ meets $\Sigma_2$ at E and F. prove that C,D, E and F form a rectangle.
SOLUTION: here
5. Solve the equation $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers x, y.
SOLUTION: here
6. From the list of natural numbers 1, 2, 3, … suppose we remove all multiples of 7, all multiples of 11 and all multiples of 13.
1. At which position in the resulting list does the number 1002 appear?
2. What number occurs in the position 3600?
SOLUTION: here


text:
Thread: triangulation of a surface

1. 2 manifolds

Anybody have any interesting problems/info on 2-manifolds? Namely regarding their triangulation?

2. clarify

My friend! I think you should clarify more your problem in order to help you better , Shouldn't you ?


text:
# Real analysis Rudin vs abbott/kolmogorov

I've been thinking of buying books from amazon to self-study analysis. I find a lot of positive/negative reviews about either Rudin (too difficult), or Abbott (too basic) plus Kolmogorov book: Kolmogorov book. It is for self study.

I am not a mathematics major, I am computer science student, who deals with a lot of probability in machine learning. I would like to study analysis (or rather measure theory but as I understand it is part of analysis) to apply to this particular field, and to exercise to write simple proofs, and understand articles dealing with epsilons and sets (i can do basic ones from strang linear algebra and multivariable calculus courses, and geometry at school) and I would like to understand probabilistic articles on Wiki that deal with real analysis.

So can someone familiar with both options (Rudin vs abbott + Kolmogorov) advise which one is better in this regard?

Or maybe there is some other book I don't know of - but everywhere I go it is always Rudin.

This question is different from similar ones here because field of application of real analysis is gonna be probability.

PS. Thank you for recommendations here. I decided to go with Abbott for the beginning and then continue with Folland/Rudin, where the second one I found at my university library. Maybe will take a year/1.5 all in all :)

• People still buy books? Impressive! – Bettybel Jul 5 '17 at 17:59
• Abbott doesn't seem to have measure theory. Kolmogorov's is rather old, and it doesn't really have measure theory either. Its last chapter talks about Stieltjes integrals, but that's it. – Bettybel Jul 5 '17 at 18:11
• – Bettybel Jul 5 '17 at 18:15
• try a book that covers some of probability theory, by example the book of Folland. – Masacroso Jul 5 '17 at 18:17

This is probably better as a comment, but too lengthy.

Abbott is a junior level analysis text. Baby Rudin is definitely a senior level analysis text, designed for a two-semester course. Rudin is terse and makes you work for it. I felt the book was too difficult when I was going through the course; but in hindsight, it was appropriate (which means I grew from the course!).

Senior analysis has the burden of building one's mathematical maturity for graduate studies. When I took senior analysis, we were expected to put in 20 hours each week. There were several grad students from other departments (at least for the first semester), that were taking this course to improve their writing skills for publishing research papers. From this perspective, Rudin was appropriate.

Measure theory is usually taught as a graduate course. It doesn't take much work to define a measure and derive basic properties; only basic set theory and basic algebra are involved ($\sigma$-fields are basically free algebras). Once you start defining the Lebesgue integral or talking about modes of convergence, having a stronger analysis background will be helpful. Graduate level probability theory classes usually introduce the requisite measure theory upfront (from what I have heard). So if your goal is probability, I would try and find a graduate probability theory course website or text to help guide you.

I would personally suggest Rudin for the rudiments of analysis. You will definitely want to cover series and sequences, point-set topology, continuity, differentiation, and series and sequences of functions. It may be safe to skip the derivative as a linear operator and the Riemann-Stieltjes integral (though these are always good things to know, especially thinking about the derivative as a linear operator, if you are doing machine learning and stochastic optimization). I would avoid Baby Rudin for measure, as I have not heard good things about his exposition.

Trench is another alternative to Rudin which I have heard recommended. I am not familiar with this text personally.

If you find Rudin too terse, you can always pick up Abbott.

Bartle's The Elements of Integration and Lebesgue Measure is a friendly text for measure theory, though it does assume some maturity. Again, measure is really a graduate level topic.

Also, if you are still a student, it might be worth sitting in on an analysis class at your institution. Having someone with intuition and classmates with whom to converse will definitely be helpful for learning the material.

• I used Abbott as a junior and we only covered the first half of the book - up to sequences of functions. I recommend it to people starting out, but something like Wade or Royden (used at my current school for junior analysis) can help beef one up before moving to say, Carothers or Baby Rudin. – Sean Roberson Jul 5 '17 at 18:20
• Well, attending Uni is not an option though im a student :) I study in a non-english speaking country, and they offer only graduate analysis courses in mathematics, which are too advanced for me (though im CS graduate student) :) – John Aug 1 '17 at 13:05