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"Requires axiom of choice" vs. "explicitly constructible" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T13:39:01Z http://mathoverflow.net/feeds/question/5329 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5329/requires-axiom-of-choice-vs-explicitly-constructible "Requires axiom of choice" vs. "explicitly constructible" Darsh Ranjan 2009-11-13T09:28:57Z 2009-12-22T05:30:27Z
I think I'm a bit confused about the relationship between some concepts in mathematical logic, namely constructions that require the axiom of choice and "explicit" results.

For example, let's take the existence of well-orderings on $\mathbb{R}$. As we all know after reading this answer by Ori Gurel-Gurevich, this is independent of ZF, so it "requires the axiom of choice." However, the proof of the well-ordering theorem that I (and probably others) have seen using the axiom of choice is nonconstructive: it doesn't produce an explicit well-ordering. By an explicit well-ordering, I simply mean a formal predicate $P(x,y)$ with domain $\mathbb{R}\times\mathbb{R}$ (i. e., a subset of the domain defined by an explicit set-theoretic formula) along with a proof (in ZFC, say, or some natural extension) of the formal sentence "$P$ defines a well-ordering." Does there exist such a $P$, and does that answer relate to the independence result mentioned above?

More generally, we can consider an existential set-theoretic statement $\exists P: F(P)$ where $F$ is some set-theoretic formula. Looking to the previous example, $F(P)$ could be the formal version of "$P$ defines a well-ordering on $\mathbb{R}$." (We would probably begin by rewording that as something like "for all $z\in P$, $z$ is an ordered pair of real numbers, and for all real numbers $x$ and $y$ with $x\neq y$, $((x,y)\in P \vee (y,x)\in P) \wedge \lnot ((x,y)\in P \wedge (y,x)\in P)$, etc.) On the one hand, such a statement may be a theorem of ZF, or it may be independent of ZF but a theorem of ZFC. On the other hand, we can ask whether there is an explicit set-theoretic formula defining a set $P^*$ and a proof that $F(P^*)$ holds. How are these concepts related:

the theoremhood of "$\exists P: F(P)$" in ZF, or its independence from ZF and theoremhood in ZFC;

the existence of an explicit $P^*$ (defined by a formula) with $F(P^*)$ being provable.

Are they related at all?
 http://mathoverflow.net/questions/5329/requires-axiom-of-choice-vs-explicitly-constructible/5356#5356 Answer by Gerald Edgar for "Requires axiom of choice" vs. "explicitly constructible" Gerald Edgar 2009-11-13T14:44:49Z 2009-11-13T14:44:49Z
In Goedel's proof of consistency of AC, we in fact get much more. There is an explicit relation defined, which is (provably in ZF) a well-ordering of a certain subset of the reals. It is consistent (and follows from the axiom V=L) that the subset is all of the reals.
 http://mathoverflow.net/questions/5329/requires-axiom-of-choice-vs-explicitly-constructible/9532#9532 Answer by Ashutosh for "Requires axiom of choice" vs. "explicitly constructible" Ashutosh 2009-12-22T05:30:27Z 2009-12-22T05:30:27Z
Levy has a few interesting things to say on the definability of a set whose existence requires the axiom of choice, see pages 171- 175 Basic Set Theory, Perspectives in Mathematical Logic. In particular, he mentions Feferman's result on the unprovability in ZFC of the existence of definable well ordering of reals.

On your last question: Levy mentions an example (exceedingly trivial) of a formula F(x) for which one does have a definable set satisfying it in ZFC although the existential formula "for some x, F(x)" is unprovable in ZF.

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# Sample Size — Continuous Parameters

#### Exhibit 34.2  90% of the observations fall within a range of ±1.65 standard deviation from the mean.

Consider research programmes such as retail measurement that track item sales in store, consumer panels that track the item purchases in homes or TAM (TV audience measurement) that tracks TV viewership. Each of these platforms track a continuous variable (e.g., sales, purchases or viewership) in a population.

The CLT (central limit theorem) provides the foundation for setting the sample size for the variables measured in these programmes.

As mentioned before, the sample size required to estimate a variable is dependent on population variance, sampling design and the desired level of precision. And the desired level of precision is defined by two statistics — confidence interval or margin of error and confidence level.

If the confidence level is set at 90%, and the margin of error is e, then based on the normal distribution (see Exhibit 34.2):

$$e = 1.65 σ$$

In general:

$$e = Z σ$$

Where:

• e: margin of error
• Z: standardized value associated with the level of confidence. If level of confidence is:
• 90% then Z=1.65
• 95% then Z=1.96
• 99% then Z=2.58

Furthermore, according to the CLT the variance of the sampling distribution (σ2) is a function of the sample size (n), and the universe variance (S2): $$σ^2 = \frac {S^2}{n}$$

Substituting for s 2 :

$$e^2= \frac {Z^2 S^2}{n}$$ $$\mathbf {n=\frac{Z^2S^2}{e^2}=\left(\frac{ZS}{e}\right)^2, \;\; e = \frac {ZS}{\sqrt n}}$$

If the margin of error is expressed as a proportion of parameter value, e' = e/µ, and S' = RSE (relative sampling error) = S/µ, the above formulae may be rewritten as:

$$\mathbf {n=\frac{Z^2S'^2}{e'^2}=\left(\frac{ZS'}{e'}\right)^2, \;\; e' = \frac {ZS'}{\sqrt n}}$$

Example 1: If the standard deviation of the sale of a category in provision shops is 80, and the average sales per store, μ, is about 200, then the required sample size, so that the sales estimates fall within ±6% of its true value with a confidence level of 90% is equal to:

$$n_{prov}=\left(\frac{ZS}{e}\right)^2=\left(\frac{1.65×80}{0.06 × 200}\right)^2=121$$

Take note that in order to half a sampling error we need to quadruple the sampling size. If the sampling error was reduced to 3% (0.03), the required sample size will be 484.

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# How do phase transitions occur in the solid phase? [closed]

For example, how does graphite transform into diamond as pressure increases at a certain (adequate) temperature? I'm not sure if this is a great example to look at the bigger picture, so another example would be pure iron. At atmospheric pressure, as liquid iron cools down it solidifies to $\delta\text{-}\ce{Fe}$, with a bcc crystal structure. Further cooling down brings it to $\gamma\text{-}\ce{Fe}$, which has a fcc. Then, it keeps cooling down and it solidifies to $\alpha\text{-}\ce{Fe}$, again with a bcc structure. I don't really see how do the atoms physically rearrange to form the different crystal structures, neither do I see why would they jump back and forth between these. To add to this, I believe the difference between $\alpha$ and $\delta$ are their paramagnetic properties.

I guess the general question is, how do the atoms in the solid phase rearrange themselves by changes in pressure or temperature? (in contrast to, for example, a liquid-gas phase transition). And of course, why? What drives this change?

• Magnetic properties have nothing to do with all this and are not the cause of difference between $\alpha$ and $\delta$. Indeed, $\alpha$ iron may or may not be ferromagnetic (dependent on temperature), while its crystal structure remains the same. – Ivan Neretin Mar 15 '18 at 8:26
• So they're both bcc, yet different? – ralk912 Mar 15 '18 at 8:30
• In a way, they can be thought of as not different at all. Imagine one phase which is stable until certain temperature, then ceases to be stable, then becomes stable again, then melts. – Ivan Neretin Mar 15 '18 at 8:57
• Then why call them differently? And so, why would this happen? – ralk912 Mar 15 '18 at 9:00
• Because one exists in one range of temperatures, and another exist in a different range, and never the twain shall meet. As to why, that's a very broad question, about as broad as chemistry itself. Solid phase structures are notoriously hard to predict. – Ivan Neretin Mar 15 '18 at 9:47

Phase transitions in solids occur in the same manner as in liquids or gases, through nucleation and growth. Yes, those processes require atomic motion. Yet,atoms diffuse in solids all the time, through various mechanisms. While potentially slower, the thermodynamics of solid phase transitions are easily seen in calorimetry measurements for example.

Nucleation and growth of precipitates is a key part of many engineering alloys. On the other hand, some phase changes (like diamond to graphite) are very hard to make happen at near-human-tolerable temperatures.

The solid phase also sees the martensitic phase transitions, which are second-order diffusionless atomic rearrangements. The name comes from the martensite phase observed in steels. Shape memory alloys are another example of martensitic phases.

A classic materials science text would be 'Phase Transformations in Metals and Alloys", by D.A. Porter and K.E. Easterling. In my opinion, the second edition is vastly preferable to the more recent third edition where the changes mainly seemed to be reintroducing all the errors that were removed in the second edition (sigh).


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# How do you write b^- (4/7) in radical form?

Aug 28, 2017

b^(-(4/7))=color(red)(1/(root(7)(b^4))

#### Explanation:

General observations on exponents:
$\textcolor{w h i t e}{\text{XXX}} {p}^{- q} = \frac{1}{{p}^{q}}$

$\textcolor{w h i t e}{\text{XXX}} {p}^{\frac{r}{s}} = {\left({p}^{r}\right)}^{\frac{1}{s}} = \sqrt[s]{{p}^{r}}$

Base on the above, we have
$\textcolor{w h i t e}{\text{XXX}} b \left(- \frac{4}{7}\right) = \frac{1}{{b}^{\frac{4}{7}}} = \frac{1}{\sqrt[7]{{b}^{4}}}$


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# Category: Networking

If we are connected to the internet via one single ISP, it is most likely that we will have a default route set up to one of their access routers. However, if we want to multi-home, either for reliability or load-balancing reasons, the most straightforward way is to set up BGP peering with our upstream providers. For example, if we connect to three upstream ISPs – A, B and C, there is no point in sending packets destined to one of B’s (or its clients’) IP addresses, to A or C. In order to make informed decisions on where to send packets, we will usually need to receive a full BGP feed so as to see the “whole internet” in terms of routing tables.

In the ideal world, we would see all the IP prefixes that are advertised in the internet, via all of our peers, leaving us to choose the best path, based on the AS path length and local preference. This is not always the case, though. Some routes will be missing in BGP feeds received from our upstream providers. That is – we will have some routes in our routing tables that are advertised only by some of our peers. These can be our other peers’ internal networks (no-advertise/no-export), in which case we don’t really care. If these routes, however, represent actual parts of the internet that should be reachable, this can be a cause for concern.

For example, if we only have two upstream providers and only one of them advertises a route to a certain network and the connection to that ISP is temporarily lost, that network will be unreachable to us.

## The problem

If there are routes missing that should not be, there is little that can be done other than escalating the issue to the problematic ISP. In order to do so, we must first identify the routes that are missing. By the time of writing this article, the global IPv4 routing table has more than 400,000 entries and the number is expected to grow in the future. Therefore, we need a way to work with large collections of routes efficiently. The other problem is that due to different route summarization rules of our peers, routes may be split in different parts by different peers, so we cannot run a simple diff through all of our routing tables.

We would like to have a tool to perform a binary difference operation of two sets of IP address ranges (subnets) in an efficient manner. Ideally, the tool would also be able to perform other set operations, such as  union or intersection, so we can merge several ISP’s advertised routes prior to comparing it with a single ISP’s route table and so on.

What we want to achieve is perhaps best illustrated by an illustration:

## Analysis of the problem

While making a tool like this, we would like to avoid pitfalls, such as using expensive binary trees, even though these may seem at first to be the most versatile and offer most flexibility while working with datasets. Since the IP address space is essentially a binary tree in itself, we can be tempted to model our internal representation of the route table by it.

A quick calculation tells us that even if limit ourselves to networks with the maximum prefix length of /24, we would need $$\sum _{n=0}^{24} 2^n = 2^{25}-1 = 33554431$$ nodes or 384 MiB with 12 bytes per node (2×32-bit pointer, 1×32-bit address). This is workable with IPv4, but completely infeasible when dealing with IPv6 routes. We would like to allow for at least the networks with a maximum prefix of /48, yielding:
$$\sum _{n=0}^{48} 2^n = 2^{49}-1 = 562949953421311$$ nodes or 16 PiB of data with 32 bytes per node (2×64-bit pointer, 1×128-bit address). Clearly impossible to hold in RAM by today’s standards.

Fortunately, constructing a large binary tree can be completely avoided when performing this task and can in fact be done in $$O(n)$$ when the input routing tables are already sorted and usually $$O(n \log n)$$ when they are not, depending on the sorting algorithm.

## Algorithm

The algorithm used for comparison is very simple and as stated above runs in linear time. Suppose we compare two routing tables, A and B.

1. Iterate through all the entries in both routing tables and mark the start and the end of each subnet (all-zeros and all-ones addresses of a subnet). Insert both of them into a single array, while noting their type (startA, startB, endA, endB)
2. Sort the above array, ordered only by IP number.
3. Iterate through the array and keep two counters, say $$c_A$$ and $$c_B$$Each time, startX marker is encountered, increase $$c_X$$ and each time endX is encountered, decrease it.
4. If we are in the region, where $$c_A > 0 \wedge c_B = 0$$, we have found a IP range that is in A but not in B.

The algorithm works well even with overlapping subnets, since the counter is simply going to be larger than one in that case. But even though the algorithm is simple, there are a few things to consider:

• If we want to aggregate the missing subnets for clarity, we should perform the action (print out the subnet, …) when counter drops back to zero.
• If we do this, we should be careful with neighbouring subnets, since the counter may drop to zero on the boundary but rises again immediately afterwards.
• We should be careful when dealing with single-address subnets (/32 or /128), since the start and the end point are going to be the same.

The algorithm works for other binary set operations as well. If we want to produce a union of two sets ($$A \cup B$$), we should mark the regions where $$c_A > 0 \vee c_B > 0$$ and for intersections  ($$A \cap B$$), we use $$c_A > 0 \wedge c_B > 0$$ comparison kernel.

## Summing-up

I implemented the above algorithm, based on this Stack Overflow answer, in a tool that can be used to analyse routing tables, called BgpCompare. It reads IP subnets from two input files and then outputs the result of a set operation to standard output. It uses regular expressions to capture IP addresses and prefix lengths, so it can work with a wide variety of different routing platforms’ routing table dumps, just by providing a new regular expression.

BgpCompare is free and open-source software, licensed under LGPL licence. It is written in C++11 and comes with a handy header-only library for IPv4/IPv6 address manipulation (parsing, textual representation, arithmetics, …).

The latest version, with full source code along with Windows and Linux binaries, is available here (1.02 MiB).

Let’s imagine a hypothetical situation. We have a problem with our internet connection, so we call our ISP’s tech support. We wait for a couple of minutes to get a free operator, and then we spend fifteen minutes describing the layout of our local network and then another 20 minutes debugging a complex problem (intermittent packet loss when link is not saturated, for example). Finally, the problem seems to be resolved and the operator concludes the call with: “Should there be any more problems, do not hesitate to contact us again.”

Half an hour later, the problem reappears. We have no other options than to call the ISP’s switchboard again, wait for a free operator, who is invariably going to be someone else, describe the problem once again and go through a debugging session once again, repeating the answers to the same questions and so on.

This is the frustration that web-services “feel” each time you connect to them via a NAT gateway. In a situation like the one above, we would like to be able to call the previous operator directly but because we do not know the operator’s direct phone number, if there is one, we cannot. Even though we were given the permission to do so!

Even if you use public IP addresses and are thereby connected “directly to the internet”, just these 4 rules on a stateful firewall should give you the same degree of protection for the network as a typical consumer-grade NAT implementation. The stateful firewall is called stateful because it maintains some degree of internal state that is used to guide the decisions on what to do with an incoming packet. Let’s make a breakdown of these four rules:

1. First rule will accept/let through any packet that goes from the internal network to the internet. While doing so, it will also remember the connection parameters, such as TCP sequence numbers, source and destination IP-port pair, …
2. Second rule will let trough any packet coming from the internet directed to the internal network IF the connection parameters match one of those that were recorded when packets were going out. Therefore, only if the connection had previously been established from the inside.
3. Third rule will let in any packet that is somehow related to one of the established connections. This rule is not strictly necessary, but it helps with certain applications, such as active-mode FTP that need connections be established from the outside.
4. Fourth rule will drop anything else coming from the internet, such as SYN packets directed towards our internal servers, services running on our workstations, …

Of course, I will never say that these four rules guarantee network security, as they most certainly do not. Network level security and firewalling are complex topics and much work is needed to render the network fully secure. However, this is the configuration used in many networks, residential and corporate, as it allows full client access to the internet, while blocking any servers from within to serve outside clients. It is a quite restrictive default, but exceptions can easily be made by adding more rules to the firewall.

If you deploy NAT in your network, you will most likely have the above rules configured in the firewall. This is one of the reasons why people consider deploying NAT as adding security to the network. The actual reason for this is of the practical matter. If NAT was not combined with a stateful firewall, it could not work, since it has to keep track of the established sessions in order to know where to forward incoming packets. NAT cannot function without a stateful firewall, whereas the stateful firewall can and will function without NAT functionality.

Not using NAT allows us to configure much more fine-grained security at the network perimeter. If we return to the introductory anecdote, we mentioned that we were given the permission to follow-up on the issue. The “phone firewall” in that case could be configured to let in calls from our number to the operator’s direct phone number for a specific period of time, afterwards, the access could be blocked. Because NAT-like mechanism is deployed, we cannot do this, because operator does not even have a direct number.

## Appropriate usage of NAT

One may ask themselves when using NAT is in fact appropriate. Apart from stateful firewalling, which is not really a characteristic of NAT at all but more of a side product, NAT mechanisms accomplish two more important things:

• conservation of public IP address space, and
• hiding the details of the internal network (individual computers’ IP addresses, subnets, …)

NAT was invented primarily to address the first bullet point. Beforehand, people were already using stateful firewalls, but in the years to come, NAT has become so prevalent that many people simply could not imagine running their networks on public IPs. The sad truth is, that due to the IPv4 address exhaustion, public IPs everywhere are simply not an option anymore. We will have to live with NAT for as long as we have IPv4, the only thing we want to avoid are things like double-NAT and similar atrocities that unmanageably complicate the network and prevent certain applications from functioning at all.

That being said, we should be aware that IPv6 offers so much address space that the first bullet point is simply not applicable. We will never have to use masquerading to hide multiple IPv6 addresses behind one single address, because there is such an abundance of it. The second bullet point still lurks though, leaving people who are deploying IPv6 in their networks anxious, since their computers are not “hidden” anymore but rather present themselves with an unique address to the whole internet.

This is a typical case of the security through obscurity fallacy. Knowing the internal telephone number does not help me at all, because I cannot reach the person unless the operator allows me so when I call the switchboard. Apart from that, there are already mechanisms in place that help providing higher anonymity, such as IPv6 Privacy Extensions.

While the classical NAT masquerading that we are familiar with has no place in the IPv6 world, schemes such as NAT66 that provide a one-on-one mapping between public addresses and unique local (ULA) addresses, are marginally effective, especially to facilitate renumbering when a company changes their ISP. Still, deploying any kind of middlebox that changes IP addresses in packets brings about increased complexity, breaks compatibility with certain applications that require Direct connectivity, such as IPsec, while its positive effective are usually negligible.

The verdict would be that while NAT is unavoidable in IPv4 world, we must not fall into temptation and deploy it on IPv6 networks as disadvantages of NAT schemes far outweigh its advantages. IPv6 allows the internet to function as it was originally conceived. As the proverb goes: When all you have is a hammer, everything starts to look like a nail. We should not apply the same logic that we have become used to dealing with IPv4 networks as it will only cause us trouble further down the line.


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# Make mathematica “forget” numbers beyond the precision goal

I need Mathematica to remember only things upto two decimal places.

I'm currently using the unwieldy Floor[x*100]/100 which works but is there a better way to make it use two decimal places (throughout the notebook is okay).

I tried N[x,{2,2}] but although it displays the requisite two decimal places, I know that the kernel still remembers. For instance, if I try it for N[1.234,{2,2}]-1.23 is not 0. It is in fact 0.004. How can I get rid of this?

• Examine FullForm@N[1.234, {2, 2}] and you'll see it's not doing what you think it is doing. Try instead SetAccuracy[1.234, 2] - 1.232. One problem using this approach is that Mathematica tracks relative precision, not a fixed number number of decimal places. Do you want fixed-point arithmetic or is just about output formatting? – Michael E2 Nov 3 '15 at 11:24

You want Round[] for this:

Round[1.234, 1.*^-2] - 1.23
0.

This can be used with $PrePrint if needed. To make your function work "throughout the workbook" you can use: $PrePrint = N @ Floor[#*100]/100 &;

Now

 {1.23999, Pi}

{1.23, 3.14}

Restore $PrePrint to the default: $PrePrint =.


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Infoscience

Book Chapter

# Networked Slepian-Wolf: Theory and Algorithms

In this paper, we consider the minimization of a relevant energy consumption related cost function in the context of sensor networks where correlated sources are generated at various sets of source nodes and have to be transmitted to some set of sink nodes. The cost function we consider is given by the product [rate] × [link weight]. The minimization is achieved by jointly optimizing the transmission structure, which we show consists of a superposition of trees from each of the source nodes to its corresponding sink nodes, and the rate allocation across the source nodes. We show that the overall minimization can be achieved in two concatenated steps. First, the optimal transmission structure has to be found, which in general amounts to finding a Steiner tree and second, the optimal rate allocation has to be obtained by solving a linear programming problem with linear cost weights determined by the given optimal transmission structure. We also prove that, if any arbitrary traffic matrix is allowed, then the problem of finding the optimal transmission structure is NP-complete. For some particular traffic matrix cases, we fully characterize the optimal transmission structures and we also provide a closed-form solution for the optimal rate-allocation. Finally, we analyze the design of decentralized algorithms in order to obtain exactly or approximately the optimal rate allocation, depending on the traffic matrix case. For the particular case of data gathering, we provide experimental results showing a good performance in terms of approximation ratios.


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# Answer to Question #6702 in Mechanics | Relativity for yah

Question #6702
the pressure inside a 1.0cm radius vessel carrying blood with a speed of 0.50m/s is 5200 Newton/ square meter, while the pressure outside is 3200 Newton/ square meter. to what radius must the vessel be reduced so that the outside pressure is greater than that inside, thus making the vessel close at the constriction?
Let's find the area of the vessel:

S = 4πr² = 4π1² = 4π.

We denote the inside pressure force by Fi and outside pressure force by Fo. When the radius is r=1 the values of these forces are:

Fi = S*5200 = 4π5200 = 20800π,
Fo = S*3200 = 4π3200 = 12800π.

We see that Fi > Fo.

We know that inside pressure force is constant under the volume changes. Let's denote the new radius of the vessel by R.

Fi = Fo
20800π = 3200*4πR² ==> R = sqrt(20800/4/3200) = sqrt(1.625) = 1.2747

So, the vessel must expand a bit, then the outside pressure will grow up.

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# Alien vs Alien Hunter Puzzle

I found this puzzle posted on a student website I frequent, but no one including the poster nor I could solve it. So I'm posting this puzzle here with hope that some of math whiz in here could illuminate on how to approach it:

There is an alien on the surface of a spherical planet. The alien can run at a top speed of $u$. An alien hunter is hunting the alien in his spaceship which can fly at a top speed of $v$. Once the hunter sees the alien, he fires, and the alien dies. Show that the alien will always die if $v > 10u$.

Note that there's no information given about the planet radius nor the altitude at which the spaceship hovers so I'm suspecting that the solution somehow ignores them or they get canceled out somewhere in the calculation.

p/s: again I repeat that I do not know of the solution myself

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The assumption I would make is that the hunter gets to choose its altitude at will. There is a tradeoff between being high so you see a lot of the surface and being low so you can have a high angular velocity. The answer (I don't have it yet) will be a search strategy, which could vary both altitude and heading, that will find the alien. –  Ross Millikan Nov 23 '10 at 4:48
@Ross So I'm assuming the hunter would start off at low attitude in order to get nearer to the alien and then slowly increase the altitude in order to spot the alien faster. The strategy is to find the optimum rate at which the hunter should increase altitude, and then proof that in order to survive that the alien should run at least 10 times faster than the hunter. Am I right? –  Lukman Nov 23 '10 at 5:16
Surely if either $u>0$ or $v>0$ and the planet is finite, the alien will always die - the only question is when. The only exceptions would be using carefully constructed movement plans so that the hunter and alien never see each other. That is, assume that neither knows where the other is until they are within sight of each other - at which point the hunter fires and the alien dies. Given enough time, they're bound to blunder into each other. –  Steve314 Nov 23 '10 at 7:45
@Steve314: Usually here you have to prove you can catch the alien in finite time (though it can be arbitrarily long) even if it knows your search strategy. For example, if u=v, the alien can just stay at the point opposite the hunter. –  Ross Millikan Nov 23 '10 at 14:15
@Ross - understood, but that isn't specified in the question. A strategy that works efficiently even in the worst case (planned or fluke) is valuable - I've suffered enough algorithm theory to understand that. But the question as stated allows a much simpler answer. Besides, in the real world, unpredictability is a normal part of realistic search and evasion strategies for game theory reasons - and as advocates of randomized algorithms will point out, a failure rate in the order of once every thousand universes or so is often more than acceptable. –  Steve314 Nov 25 '10 at 0:10

Let us measure distance in planet radius and time such that the alien can make one radian per unit time on the surface. If I fly along the surface at altitude h, I can see a swath of width $2 \arccos (\frac{1}{1+h})$ and can fly around the equator in $\frac{2 \pi (1+h)}{10}$, so as long as $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{10} \gt 0$ I can make a spiral starting at one pole and push the alien in front of me down to the other pole. At h of about 0.55, this only requires my speed to be about 5.6 instead of 10.

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I'm getting the swath width to be a function of radius $r$ and $h$ as in $r * arccos(\frac{r}{r+h})$ .. am I wrong? Also, I'm expecting to see $\frac{dh}{dt}$ or something ... –  Lukman Nov 23 '10 at 5:51
It looks like the radius is taken to be one unit (or 1) in the above answer. –  Matt Groff Nov 23 '10 at 5:59
I see. So h is relative to the radius then. But one more thing: how did you decide on $h = 0.55$. Is there any calculation involved to come to that value? –  Lukman Nov 23 '10 at 6:38
Yes, I just set r=1. The critical moment comes when I am flying around the equator. I need to prevent the alien from running across my track behind me, so I need to complete the circuit faster than that. To decide on h=0.55, I just varied the speed ratio until $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{v} \gt 0$ was barely satisfied, which I found just by experiment to be h=.55, v=5.6 –  Ross Millikan Nov 23 '10 at 13:48
Slight correction, the optimum altitude is closer to 0.535 –  Ross Millikan Nov 23 '10 at 14:42

There's a similar puzzle here (with solution).

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The goblin puzzle and the alien puzzle kinda like opposite in many ways so I'm having difficulties to adapt the solution. For example, in the goblin puzzle the destination is clear which is the lakeside, but for this alien puzzle there is no limit. The proof needs to consider if the hunter is very far enough the planet surface (well maybe to the point that he can see the entire surface of one hemisphere). And do I have to break the path to two phases too? –  Lukman Nov 23 '10 at 5:30
@~Lukman you could consider flying in from far away, forcing the alien onto one hemisphere, but I did not make use of that. In essence I started at my preferred altitude above the north pole and chased the alien down to the south pole. –  Ross Millikan Nov 23 '10 at 13:49


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# Definition:Distance/Points/Complex Numbers

## Definition

Let $x, y \in \C$ be complex numbers.

Let $\cmod {x - y}$ be the complex modulus of $x - y$.

Then the function $d: \C^2 \to \R$:

$\map d {x, y} = \cmod {x - y}$

is called the distance between $x$ and $y$.


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## Chemistry: The Molecular Nature of Matter and Change 7th Edition

$K_{sp} = (2.8 \times 10^{-11})$
1. Calculate the molar mass $(Ag_2Cr_2O_7)$: 107.87* 2 + 52* 2 + 16* 7 = 431.74g/mol 2. Calculate the number of moles $(Ag_2Cr_2O_7)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{8.3 \times 10^{-3}}{ 431.74}$ $n(moles) = 1.9\times 10^{- 5}$ 3. Find the concentration in mol/L $(Ag_2Cr_2O_7)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 1.9\times 10^{- 5}}{ 0.1}$ $C(mol/L) = 1.9\times 10^{- 4}$ 4. Write the $K_{sp}$ expression: $Ag_2Cr_2O_7(s) \lt -- \gt 2Ag^{+}(aq) + 1Cr_2O{_7}^{2-}(aq)$ $K_{sp} = [Ag^{+}]^ 2[Cr_2O{_7}^{2-}]^ 1$ 5. Determine the ion concentrations: $[Ag^{+}] = [Ag_2Cr_2O_7] * 2 = [1.9 \times 10^{-4}] * 2 = 3.8 \times 10^{-4}$ $[Cr_2O{_7}^{2-}] = [Ag_2Cr_2O_7] * 1 = 1.9 \times 10^{-4}$ 6. Calculate the $K_{sp}$: $K_{sp} = (3.8 \times 10^{-4})^ 2 \times (1.9 \times 10^{-4})^ 1$ $K_{sp} = (1.5 \times 10^{-7}) \times (1.9 \times 10^{-4})$ $K_{sp} = (2.8 \times 10^{-11})$


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# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.2 | Set 1

• Last Updated : 16 Feb, 2022

### Question 1. Form the differential equation of the family of curves represented by y2 = (x – c)3

Solution:

y2 = (x – c)3

On differentiating the given equation w.r.t x,

2y(dy/dx) = 3(x – c)2

(x – c)2 = (2y/3)(dy/dx)

(x – c) = [(2y/3)(dy/dx)]1/2          -(1)

On putting the value of (x – c) in equation (1), we get

y2 = [(2y/3)(dy/dx)]3/2

On squaring both side, we get

y4 = [(2y/3)(dy/dx)]3

y4 = (8y3/27)(dy/dx)3

27y4 = 8y3(dy/dx)3

27y = 8(dy/dx)3

### Question 2. Form the differential equation corresponding to y = emx by eliminating m.

Solution:

y = emx          -(1)

On differentiating the given equation w.r.t x,

dy/dx = memx          -(2)

From eq(1), we get

y = emx

logy = mx

m = (logy/x)

Now, put the value of m in equation(2), we get

x(dy/dx) = ylogy

### (i) y2 = 4ax

Solution:

y2 = 4ax          -(1)

y2/4x = a

On differentiating the given equation w.r.t x,

2y(dy/dx) = 4a          -(2)

Now, put the value of a in equation(2), we get

2y(dy/dx) = 4(y2/4x)

2y(dy/dx) = y2/x

2x(dy/dx) = y

### (ii) y = cx + 2c2 + c3

Solution:

y = cx + 2c2 + c         -(1)

On differentiating the given equation w.r.t x,

dy/dx = c          -(2)

Now, put the value of c in equation(1), we get

y = x(dy/dx) + 2(dy/dx)2 + (dy/dx)3

### (iii) xy = a2

Solution:

xy = a         -(1)

On differentiating the given equation w.r.t x,

x(dy/dx) + y = 0

### (iv) y = ax2 + bx + c

Solution:

y = ax2 + bx + c          -(1)

On differentiating the given equation w.r.t x,

dy/dx = 2ax + b          -(2)

Again differentiating the given equation w.r.t x,

d2y/dx2 = 2a          -(3)

Again, differentiating the given equation w.r.t x, we get

d3y/dx3 = 0

### Question 4. Find the differential equation of the family of curves y = Ae2x + Be-2x where A and B are arbitrary constants.

Solution:

y = Ae2x + Be-2x          -(1)

On differentiating the given equation w.r.t x,

(dy/dx) = 2Ae2x – 2Be-2x          -(2)

Again, differentiating the given equation w.r.t x,

d2y/dx2 = 4Ae2x + 4Be-2x

d2y/dx2 = 4(Ae2x + Be-2x)

d2y/dx2 = 4y

### Question 5. Find the differential equation of the family of curves, x = Aconst + Bsinnt where A and B are arbitrary constants.

Solution:

x = Acosnt + Bsinnt          -(1)

On differentiating the given equation w.r.t x,

dy/dx = -Ansinnt + Bncosnt          -(2)

Again, differentiating the given equation w.r.t x,

d2y/dx2 = -An2cosnt – Bn2sinnt

d2y/dx2 = -n2(Acosnt + Bsinnt)

d2y/dx2 + n2x = 0

### Question 6. Form the differential equation corresponding to y2 = a(b – x2) by eliminating a and b.

Solution:

y2 = a(b – x2

On differentiating the given equation w.r.t x,

2y(dy/dx) = a(0 – 2x)

Again, differentiating the given equation w.r.t x,

x[

### Question 7. Form the differential equation corresponding to y2 – 2ay + x2 = a2 by eliminating a.

Solution:

y2 – 2ay + x2 = a         -(1)

On differentiating the given equation w.r.t x,

2y(dy/dx) – 2a(dy/dx) + 2x = 0

2y(dy/dx) + 2x = 2a(dy/dx)

-(2)

Let us considered dy/dx = y’

On putting the value of ‘a’ in the eq(1), we get

On solving this equation, we get

(x2 – 2y2)y’2 – 4xyy’ – x2 = 0

(x2 – 2y2)(dy/dx)2 – 4xy(dy/dx) – x2 = 0

### Question 8. Form the differential equation corresponding to (x – a)2 + (y – b)2 = r2 by eliminating a and b.

Solution:

(x – a)2 + (y – b)2 = r         -(1)

On differentiating the given equation w.r.t x,

2(x – a) + 2(y – b)(dy/dx) = 0

(x – a) + (y – b)(dy/dx) = 0          -(2)

Again, differentiating the given equation w.r.t x,

1 + (y – b)(d2y/dx2) + (dy/dx)(dy/dx) = 0

-(3)

On putting the value of (y – b) in eq(2),

(x – a)(d2y/dx2) – (dy/dx)3 – (dy/dx) = 0

-(4)

On putting the value of (x – a) and (y – b) in eq(1)

Put (dy/dx) = y’ and d2y/dx2 = y”

y’2(1 + y’2)2 + (1 + y’2)2 = r2y”2

### Question 9. Find the differential equation of all the circles which pass through the origin and whose centres lie on-axis.

Solution:

Equation of the circle is (x – a)2 + (y – b)2 = r

Here, a and b are the centre of the circle.

(x – a)2 + (y – b)2 = r         -(1)

When the centre lies on the y-axis, so a = 0

x2 + (y – b)2 = r         -(2)

So, when the circle is passing through origin, so the equation is

02 + b2 = r         -(3)

x2 + (y – b)2 = r2

On squaring both side, we have

x2 + y2 – 2yr + r2 = r2          -(Since  r2 = b2)

2yr = x2 + y2

r = (x2 + y2)(2y)

On differentiating the equation(1) w.r.t x, we get

0 = 4xy + 4y2(dy/dx) – 2x2(dy/dx) – 2y2(dy/dx)

0 = y2(dy/dx) – x2(dy/dx) + 2xy

(x2 – y2)(dy/dx) = 2xy

### Question 10. Find the differential equation of all the circles which pass through the origin and whose centres lie on the x-axis.

Solution:

Equation of the circle is (x – a)2 + (y – b)2 = r2

Here, a and b are the centre of the circle.

When the centre lies on the x-axis, so b = 0

(x – a)2 + y2 = r         -(1)

When the circle is passing through origin, so the equation is

a2 + 02 = r2          -(2)

(x – a)2 + y2 = r2

On squaring both side, we get,

x2 – 2ax + a2 + y2 = r2

x2 + y2 – 2xr + r2 = r2           -(Since  r2 = a2)

2xr = x2 + y

r = (x2 + y2)(2x)          -(3)

On differentiating the equation w.r.t x, we get

0 = 2x2 + 2xy(dy/dx) – x2 – y2

(x2 – y2) + 2xy(dy/dx) = 0

My Personal Notes arrow_drop_up


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## Sunday, May 14, 2017

### Predicting pKa values using PM3 - conformer search using implicit solvation

Disclaimer: These are preliminary results and may contain errors

In a previous post I showed that minimising RDKit-generated conformers generated with MMFF led to slightly worse PM3/COSMO pKa predictions. One reason might be that the MMFF minimisation is done in the gas phase.  Anders Christensen reminded me that TINKER can do MMFF/GBSA minimisations so I used TINKER so minimise 20 and 50 RDKit-generated conformers in used these as initial guesses for the PM3/COSMO energy minimisations (20mmtk and 50mmtk

As you can see there is relatively little difference in the overall performance of Xmm and Xmmtk. The error distribution is a bit tighter for the tk variants and 50mmtk does better at finding structures closer to the "global" minimum, but 50nomm is still the best choice for avoiding >1 pKa errors.

So, either GBSA is not a good substitute for COSMO or MMFF is not a good substitute for PM3.  In light of this recent paper my money is on the latter.  So I'll go with 50nomm for now.

## Saturday, May 13, 2017

### Computing apparent pKa values using QM energies of isodesmic reactions

A few days ago I presented some preliminary pKa results.  Almost all the molecules in the set have more than one ionisable group. Here's how I compute the apparent pKa values

The microscopic pKa values are computed by

\label{eqn:pka}
\mathrm{pK_a}=\mathrm{pK_a^{ref}} + \frac{\Delta G^\circ}{RT\ln (10)}

where $\Delta G^\circ$ denotes the change in standard free energy for the isodesmic reaction

\mathrm{ BH^+ + B_{ref} \rightleftharpoons B + B_{ref}H^+ }

If there are more than one titrateable sites then several protonation states can contribute to the apparent pKa.  Here I follow the approach described by Bochevarov et al. If $\mathrm{D}$ and $\mathrm{P}$ differ by one proton and there are $M$ deprotonated sites and $N$ protonated sites then

\mathrm{K_{app}}  = \frac{([\mathrm{D}_1] + [\mathrm{D}_2] + ... [\mathrm{D}_M])[\mathrm{H}^+]}{[\mathrm{P}_1] + [\mathrm{P}_2] + ... [\mathrm{P}_N]}

which can be rewritten in terms of microscopic pKa values

\mathrm{K_{app}}  = \sum^M_j \frac{1}{\sum^N_i 10^{pK_{ij}} }

The sum contains contains microscopic pKa values for which more than one protonation state is changed. For example, molecule with three ionizable groups $(\mathrm{B_1H^+B_2H^+B_3H^+})$ will have the following microscopic pKa value

\label{eqn:pkapp}
K_{ij} = \frac{ [\mathrm{B_1B_2H^+B_3][H^+]}}{[\mathrm{B_1H^+B_2B_3H^+}]}

in the expression for the apparent pKa value for going from the doubly to the singly protonated state. However, the error in such a pKa value is considerably higher due to less error cancellation and such pKa values are therefore neglected in the sum.

In the Bochevarov et al. implementation the user defines the titrateable group for which the apparent pKa is computed.  However, in my approach all possible protonation states so the assignment of the apparent pKa value to a particular ionizable group is not immediately obvious.  Inspection of Eq $\ref{eqn:pkapp}$ shows that the largest microscopic pKa values will dominate the sum over $N$, while the smallest of these maximum pKa values will dominate the sum over $M$. Thus, the apparent pKa is assigned to the functional group corresponding to the microscopic pKa

\label{eqn:max}
pK^\prime_{ij} = \min_{j}(\{ \max_{i}(\{ pK_{ij} \} ) \} )

In some cases there are several microscopic pKa that contribute almost equally to the respective sums and for these cases the apparent pKa value cannot meaningfully be assigned to a single ionizable sites.  In my approach include all microscopic pKa values that are within 0.6 pKa units of the respective maximum and minimum values defined in Eq $\ref{eqn:max}$.   I choose 0.6 because that is the site-site interaction energy below which two groups titrate simultaneously.

You can find the code I wrote to do this here. It is not cleaned up yet.

## Thursday, May 11, 2017

### Small molecule energy minimisation and conformational search using Tinker

In this post I outlined the (possible) need for conformational analysis using a continuum model. Anders Christensen reminded me that Tinker can do this so I looked in to it. This blogpost and this GitHub repo were particularly helpful.  Here's what I found.

Generating atomic charges
Tinker has MMFF force field parameters except the atomic charges.  These can be generated from an sdf file using sdf2tinkerxyz (which relies on OpenBabel). When you download from SourceForge you get a bin file, which I didn't know what to do with so I downloaded the Linux executable from here instead.

./sdf2tinkerxyz -k default.key < xxx.sdf

creates yyy.xyz (coordinates) and yyy.key (charges), where yyy is the title in the sdf file (which I'll assume = xxx in the following). "default.key" is a file with Tinker keywords that is added to xxx.key. Here's mine

# Solvation Model
SOLVATE GBSA

#Force Field Parameters
PARAMETERS /opt/tinker/tinker/params/mmff.prm

Energy minimisation
To isolate the effect of continuum solvation on the pKa predictions, I want to generate conformers with RDKit and minimise them with MMFF/GBSA using Tinker.  This is done by

/opt/tinker/tinker/bin/minimize xxx.xyz -k xxx.key 0.01 > xxx.tout

"0.01" is the convergence criterium (in kcal/molÅ??). xxx.tout contains the energy information and a xxx.xyz_2 file is generated with the optimized coordinates.  This is in a format that OpenBabel can't handle, so I need to write a converter.  The main challenge is to translate the atom types into names. See the list called lookup in this code.

Conformational Search using SCAN
Tinker also has its own conformational search method. It looks like it's based on eigenvector following but I haven't looked closely at it yet.  This is done by

/opt/tinker/tinker/bin/scan xxx.xyz -k xxx.key 0 10 20 0.00001 > xxx.tout

Here I use the settings used in the DP4 GitHub repo.  "0" is automatic selection of torsional angles, "10" is the number of search directions (modes?), "20" is the energy threshold for local minima,and "0.00001" is the optimisation convergence criterium.  You get a bunch of coordinate files and xxx.tout holds the energy information.

I haven't really played with this option yet. If you have any tips, please leave comment.

## Sunday, May 7, 2017

### Predicting pKa values using PM3 - new code and conformer search

Disclaimer: These are preliminary results and may contain errors

The code I used in my previous pKa prediction paper had some limitations that I have now mostly removed: The setup of protonation states is automated and includes acids like acetic acid and phenol, there is a rudimentary tautomer generations, and the reference molecules are chosen a bit more systematically.

I ran the new code on the same set of molecules and saw a few differences.  Some were expected and due to different references and overall protonation states but most major differences were due to different conformations even though I used the same code to generate conformers. This let me to look at the effect of conformer generation.

The original study used 20 MMFF-minimised conformations ("20mm") generated using RDKit as starting points for PM3/COSMO minimisations.  Now I've tried skipping the MMFF minimisation ("nomm"), 50 conformations with and without MMFF minimisation, and a scheme where I MMFF-minimise 50 or 100 conformations and select conformers that have energies within either 10 or 20 kcal/mol of the lowest energy ("XecutY") for each protonation state/tautomer.

The RMSEs are very similar, but 50nomm has the fewest number of cases where the error in the pKa value (ΔpKa) is greater than 1 pH unit. However, the major outlier for 20mm and 20nomm is Sparteine for which 20mm and 20nomm actually find a conformer that is 4.9 kcal/mol lower than for 50nomm, for the protonated state.

This lead me to look at the energies themselves. The numbers labelled ΔE in the table are computed as follows. For a given molecule I find the lowest energy for the protonated and deprotonated state for each method and identify the method with the lowest energy $E_{min}$. Then I count the number of molecules for which the difference to $E_{min}$ is greater than 1.36 kcal/mol and average that number for protonated and deprotonated to get one number per method.

So there are 4 molecules for which 50nomm doesn't something close to the "global" minimum for either the protonated or deprotonated state, or both. In principle, I should go on and try 100nomm to see if I can "converge", but that's starting to become pretty expensive and the point is trying to develop a practically useful tool.

The 100ecut20 results suggest that the MM gas phase energetics don't represent the PM3/COSMO energetics all that well.  So it would be interesting to try a MM conformational search with an implicit solvent model. RDKit can't do this, but Macrmodel and NAMD should be able to do this.  It also looks like the next release of OpenMM will have this capability.


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# How to express disjunction in Probability Theory?

Joint probability is usually expressed with a comma:
$P(A=a, B=b)$ meaning: The probability of random variable A having value a and random variable B having value b

But what would be the notation of the disjunction? The probability of random variable A having value a or random variable B having value b? $P(A=a ? B=b) = P(A=a) + P(B=b) - P(A=a, B=b)$

I could use the logical 'or' $P(A=a \vee B=b)$ but then shouldnt i also use the logical 'and'?

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Demanding someone to use the "and" is the same as demanding someone to write $2 \times x$ instead of $2x$. –  André Caldas Nov 7 '11 at 16:18

Be aware that usually in probability theory $\wedge$ (logical and) denotes minimum: $$a\wedge b = \min\{a,b\}$$ so for the conjuction and disjunction one usually uses $$\mathsf P(\{A = a\}\text{ and }\{B = b\}) = \mathsf P(\{A = a\}\cap \{B = b\})$$ $$\mathsf P(\{A = a\}\text{ or }\{B = b\}) = \mathsf P(\{A = a\}\cup \{B = b\})$$ thinking about events $\{A = a\}$ and $\{B = b\}$ as sets due to measure-theoretical approach to probability theory. On the other hand, joint events appeared more important for problems of probability theory so that may be the reason why probability of joint events has a shorthand.
For the disjunction I never seen any other shorthand but $\cup$.
@Ivana: that may certainly differ for several fields of math, e.g. I sometimes see $\int$ to denote not an integral but rather a time which system spend in some state. I just can tell you that I've never seen such notation in book on classical PT and that $\wedge$ and $vee$ there usually denote $\min$ and $\max$, while $\cap$ and $\cup$ refers to the set operations. –  Ilya Nov 10 '11 at 16:27


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# Definition:Compact Space/Normed Vector Space

## Definition

Let $\struct {X, \norm {\,\cdot\,} }$ be a normed vector space.

Let $K \subseteq X$.

Then $K$ is compact if and only if every sequence in $K$ has a convergent subsequence with limit $L \in K$.

That is, if:

$\sequence {x_n}_{n \mathop \in \N} :\forall n \in \N : x_n \in K \implies \exists \sequence {x_{n_k} }_{k \mathop \in \N} : \exists L \in K: \displaystyle \lim_{k \mathop \to \infty} x_{n_k} = L$


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# Chemical abundances of two extragalactic young massive clusters

Svea Hernandez, Søren Larsen, Scott Trager, Paul Groot, Lex Kaper

9 Citations (Scopus)

## Abstract

We use integrated-light spectroscopic observations to measure metallicities and chemical abundances for two extragalactic young massive star clusters (NGC1313-379 and NGC1705-1). The spectra were obtained with the X-Shooter spectrograph on the ESO Very Large Telescope. We compute synthetic integrated-light spectra, based on colour-magnitude diagrams for the brightest stars in the clusters from Hubble Space Telescope photometry and theoretical isochrones. Furthermore, we test the uncertainties arising from the use of Colour Magnitude Diagram (CMD) +Isochrone method compared to an Isochrone-Only method. The abundances of the model spectra are iteratively adjusted until the best fit to the observations is obtained. In this work we mainly focus on the optical part of the spectra. We find metallicities of [Fe/H] = $-$0.84 $\pm$ 0.07 and [Fe/H] = $-$0.78 $\pm$ 0.10 for NGC1313-379 and NGC1705-1, respectively. We measure [$\alpha$/Fe]=$+$0.06 $\pm$ 0.11 for NGC1313-379 and a super-solar [$\alpha$/Fe]=$+$0.32 $\pm$ 0.12 for NGC1705-1. The roughly solar [$\alpha$/Fe] ratio in NGC1313-379 resembles those for young stellar populations in the Milky Way (MW) and the Magellanic Clouds, whereas the enhanced [$\alpha$/Fe] ratio in NGC1705-1 is similar to that found for the cluster NGC1569-B by previous studies. Such super-solar [$\alpha$/Fe] ratios are also predicted by chemical evolution models that incorporate the bursty star formation histories of these dwarf galaxies. Furthermore, our $\alpha$-element abundances agree with abundance measurements from H II regions in both galaxies. In general we derive Fe-peak abundances similar to those observed in the MW and Large Magellanic Cloud (LMC) for both young massive clusters. For these elements, however, we recommend higher-resolution observations to improve the Fe-peak abundance measurements.
Original language English A119 24 Astronomy & astrophysics 603 https://doi.org/10.1051/0004-6361/201730550 Published - Jul-2017

## Keywords

• Astrophysics - Astrophysics of Galaxies
• HIGH-RESOLUTION
• STELLAR POPULATIONS
• LARGE-MAGELLANIC-CLOUD
• DWARF SPHEROIDAL GALAXIES
• BLUE COMPACT GALAXIES
• INFRARED SPECTRAL DISTRIBUTIONS
• HIGH-DISPERSION SPECTROSCOPY
• SUPER-STAR CLUSTER
• GLOBULAR-CLUSTERS
• RED SUPERGIANTS


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Algebraic Multigrid for Problems in Linear Elasticity discretized by DG methods

MSc Erwin Karer

May 26, 2009, 5:30 p.m. MZ 005B

In this talk we will focus on linear elasticity problems using a discontinuous Galerkin discretization, which is robust with respect to the Poisson ratio ν in the limit case of incompressible materials, i.e., $ν \rightarrow 1/2$. Those formulations can be rewritten in terms of a sum over face contributions, which is the starting point for our method. First of all we will consider two space dimensions.

In order to construct an eﬃcient preconditioner for the system matrix, we split the set of elements into two independent sets. First, we eliminate the degrees of freedom of one set. Then we apply a transformation of the remaining matrix in order to get a hierarchical representation. This representation is further used to set up a two-level method, were the coarse-grid matrix is obtained by an assembling of local “algebraic edge” matrices.


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# Associations to the word «Surface»

## Wiktionary

SURFACE, noun. The overside or up-side of a flat object such as a table, or of a liquid.
SURFACE, noun. The outside hull of a tangible object.
SURFACE, noun. (figurative) Outward or external appearance.
SURFACE, noun. (mathematics) (geometry) The locus of an equation (especially one with exactly two degrees of freedom) in a more-than-two-dimensional space.
SURFACE, noun. (fortification) That part of the side which is terminated by the flank prolonged, and the angle of the nearest bastion.
SURFACE, verb. (transitive) To provide something with a surface.
SURFACE, verb. (transitive) To apply a surface to something.
SURFACE, verb. (intransitive) To rise to the surface.
SURFACE, verb. (intransitive) To come out of hiding.
SURFACE, verb. (intransitive) For information or facts to become known.
SURFACE, verb. (intransitive) To work a mine near the surface.
SURFACE, verb. (intransitive) To appear or be found.
SURFACE AREA, noun. The total area on the surface of a three-dimensional figure.
SURFACE AREA, noun. (computing) The total resources of a system that are available to the outside world and thus might be attacked.
SURFACE AREAS, noun. Plural of surface area
SURFACE BOUNDARY LAYER, noun. (meteorology) The portion of the atmosphere lying next to the surface of the earth and extending up to between 50 and 100 meters.
SURFACE BOUNDARY LAYERS, noun. Plural of surface boundary layer
SURFACE CAR, noun. (dated) A streetcar (tram, trolley).
SURFACE CARS, noun. Plural of surface car
SURFACE CLUTTER, noun. (engineering) Unwanted radar or sonar wave reflection arising from the land or sea surface.
SURFACE ENERGY, noun. Interfacial energy
SURFACE FINISH, noun. (engineering) The quality of a surface, in terms of its roughness, microscopic pattern (lay) and macroscopic pattern (waviness).
SURFACE FINISHES, noun. Plural of surface finish
SURFACE LAYER, noun. (physics) The layer of molecules at the surface of a liquid that is responsible for surface tension
SURFACE LAYER, noun. (ecology) The layer of surface water that forms a boundary between sea and air
SURFACE MAIL, noun. Mail which is transported by motor vehicle, by railway, or by seacraft, but not by aircraft.
SURFACE NORMAL, noun. (differential geometry) A unit vector at any given point P of a surface M which is perpendicular to the tangent plane TP(M) of M at P. For a surface with parametrization $$\vec x(u,v)$$, it is given by
SURFACE OF REVOLUTION, noun. (mathematics) A surface formed when a given curve is revolved around a given axis. If the resulting surface is a closed one, it also defines a solid of revolution.
SURFACE PLASMON, noun. (physics) A quantized electromagnetic wave trapped at a metallic surface through interaction with the free electrons of the metal.
SURFACE PLASMONS, noun. Plural of surface plasmon
SURFACE QUALITY, noun. A numerical description of the flatness or roughness of a surface, especially an optical or machined surface
SURFACE RUNOFF, noun. Overland flow of excess water (with or without accumulated contaminants) that cannot be absorbed by the ground as infiltration.
SURFACE SHIP, noun. A military (naval) vessel designed for operation on the marine surface, as distinguished from a submarine or other types of vessels.
SURFACE STREET, noun. A street that is not a freeway and has at-grade intersections with other surface streets.
SURFACE TENSION, noun. (physics) the effect on the surface of a liquid that makes it behave as a stretched elastic membrane; it is caused by unbalanced intermolecular forces
SURFACE TENSION, noun. (physics) a measure of this effect
SURFACE WATER, noun. Water collecting on the ground or in a stream, river, lake, sea or ocean, as opposed to groundwater.
SURFACE WEB, proper noun. (computing) (Internet) the portion of the World Wide Web that is indexed by conventional search engines.

## Dictionary definition

SURFACE, noun. The outer boundary of an artifact or a material layer constituting or resembling such a boundary; "there is a special cleaner for these surfaces"; "the cloth had a pattern of red dots on a white surface".
SURFACE, noun. The extended two-dimensional outer boundary of a three-dimensional object; "they skimmed over the surface of the water"; "a brush small enough to clean every dental surface"; "the sun has no distinct surface".
SURFACE, noun. The outermost level of the land or sea; "earthquakes originate far below the surface"; "three quarters of the Earth's surface is covered by water".
SURFACE, noun. A superficial aspect as opposed to the real nature of something; "it was not what it appeared to be on the surface".
SURFACE, noun. Information that has become public; "all the reports were out in the open"; "the facts had been brought to the surface".
SURFACE, noun. A device that provides reactive force when in motion relative to the surrounding air; can lift or control a plane in flight.
SURFACE, verb. Come to the surface.
SURFACE, verb. Put a coat on; cover the surface of; furnish with a surface; "coat the cake with chocolate".
SURFACE, verb. Appear or become visible; make a showing; "She turned up at the funeral"; "I hope the list key is going to surface again".
SURFACE, adjective. On the surface; "surface materials of the moon".

## Wise words

Words - so innocent and powerless as they are, as standing in a dictionary, how potent for good and evil they become in the hands of one who knows how to combine them.
Nathaniel Hawthorne


text:
# Non-parametric test

A statistical test of a hypothesis $H_0$: $\theta\in\Theta_0\subset\Theta$ against the alternative $H_1$: $\theta\in\Theta_1=\Theta\setminus\Theta_0$ when at least one of the two parameter sets $\Theta_0$ and $\Theta_1$ is not topologically equivalent to a subset of a Euclidean space. Apart from this definition there is also another, wider one, according to which a statistical test is called non-parametric if the statistical inferences obtained using it do not depend on the particular null-hypothesis probability distribution of the observable random variables on the basis of which one wants to test $H_0$ against $H_1$. In this case, instead of the term "non-parametric test" one speaks frequently of a "distribution-free statistical testdistribution-free test" . The Kolmogorov test is a classic example of a non-parametric test. See also Non-parametric methods in statistics; Kolmogorov–Smirnov test.


text:
# Homework Help: Fourier series convergence test

1. Sep 10, 2011

### JamesGoh

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

A function f(x) is given as follows

f(x) = 0, , -pi <= x <= pi/2
f(x) = x -pi/2 , pi/2 < x <= pi

determine if it's fourier series (given below)

$F(x)=\pi/16 + (1/\pi)\sum=[ (1/n^{2})(cos(n\pi) - cos(n\pi/2))cos(nx) - (1/(2n^{2}))( n\pi cos(n\pi) + 2sin(n\pi/2) )sin(nx) ]$

for n= 1 to infinity

converges to it for the case of x = pi/2

2. Relevant equations

F(x) = 0.5*[f(x+) + f(x-) ]

3. The attempt at a solution

see pdf attachment

#### Attached Files:

• ###### q37a.pdf
File size:
493.6 KB
Views:
63
2. Sep 12, 2011

### vela

Staff Emeritus
Your work is kind of hard to follow, but it looks like you incorrectly got rid of the cosine terms. You have

$$F(x)=\frac{\pi}{16} + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\cos n\pi -\cos \frac{n\pi}{2}}{n^2} \cos nx - \frac{1}{\pi}\sum_{n=1}^\infty \frac{n\pi \cos n\pi + 2\sin \frac{n\pi}{2}}{2n^2}\sin nx$$

You can simplify it a bit by noting that $\cos n\pi = (-1)^n$. Next, when $x=\pi/2$, the values of cos nx and sin nx depend on whether n is odd or even. Use those facts to get the series for $F(\pi/2)$.

Last edited: Sep 12, 2011


text:
## LSU Historical Dissertations and Theses

1991

Dissertation

#### Degree Name

Doctor of Philosophy (PhD)

Chemistry

Robert V. Nauman

#### Abstract

The spectroscopic characteristics of neutral, protonated and deprotonated triphenylformazans were investigated by means of visible absorption and $\sp1$H NMR spectroscopies. Studies of the effect of solvents and substituents on the absorption spectra of the neutral and ionic formazans were conducted. CNDO/S calculations for a series of formazans were performed to assign the electronic transitions observed in the absorption spectra of triphenylformazan. The formazan system was found to be highly delocalized and not tautomeric as had been reported. The visible transition of the U (trans-syn) form of triphenylformazan was found to arise from orbitals localized on the formazan system. The near UV adsorption of triphenylformazan has significant charge transfer character. The spectroscopic trends displayed by neutral and ionic triphenylformazans were found to be similar to those of the $\beta$-diketones. The ionic forms of triphenylformazan derivatives were not photochromic. The anionic forms of the formazans were found to exist in equilibrium mixtures of U and S (cis-syn) isomers that absorb at different wavelengths in the visible region. The equilibrium is affected by the nature of the solvent and by the presence of substituents on the carbon phenyl ring. The cationic forms of the derivatives were found to exist almost exclusively in the U geometry. The site of protonation is believed to be one of the nitrogen atoms which has phenyl substituents.

262

COinS


text:
# Starting from rest a body slides down a $45^{\circ}$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between body and plane is
$(a)\;0.33 \quad (b)\;0.75 \quad (c)\;0.25 \quad (d)\;0.80 N$


text:
# Find the missing side, round to the nearest tenth.2058х

Find the missing side, round to the nearest tenth.
20
58
х

## This Post Has 2 Comments

1. isabelperez063 says:

=20.5

Step-by-step explanation:

1. use the Pythagorean theorem: a^2+ b^2=c^2  (arms+legs squared=hypotenuse)

2. plug in the numbers to the equation:

14^2+15^2=c

225+196=c^2

421=C^2

the square root of 421 is 20.5 which is the hypotenuse of the triangle

2. Killion2022 says:

20.5

Step-by-step explanation:


text:
1. ## Lagrangian Dynamics.

AB is a straight frictionless wire fixed at a point A on the vertical axis OA with constant velocity w(omega). A bead of mass m is constrained to move on the wire.
Set up the Lagrangian's equation.

I know that the lagrangian's equation is L=T-V where V is the potential energy and T is the kinetic Energy.
I was taking as generalised coordinates w but i don't know how to proceed and get the kinetic and potential energy.


text:
# zbMATH — the first resource for mathematics

Ideal convergence in 2-normed spaces. (English) Zbl 1134.46302
Summary: We introduce and investigate $$\mathcal I$$-convergence in 2-normed spaces, and also define and examine some new sequence spaces with a 2-norm.

##### MSC:
 46A70 Saks spaces and their duals (strict topologies, mixed topologies, two-norm spaces, co-Saks spaces, etc.) 40A05 Convergence and divergence of series and sequences
Full Text:


text:
# What is a Homotopy between $L_\infty$-algebra morphisms II

I would like to continue on question I asking, what is a homotopy between Lie infinity algebras, since I'm not satisfied in two directions:

1.) The naive approach to define a homotopy would be ('naive' in my opinion of course) the following:

Let $(L,(l)_{k\in\mathbb{N}})$ and $(M,(d)_{k\in\mathbb{N}})$ be two Lie infinity algebras, let $f_\infty,g_\infty:L \to M$ be two morphism (in the most general sense) and let $(C(L),Q_L)$ and $(C(M),Q_M)$ be the appropriate differential graded coalgebras with induced morphism $F,G:C(L)\to C(M)$.

Then a homotopy between $F$ and $G$ is a degree $+1$ map

$H:C(L)\to C(M)$ such that $F-G = HQ_L \pm Q_MH$

let signs and additional structure of $H$ (linear, coalgebra, ...) aside for a moment.

It is strange, however, that I never saw this approach in the literature. Is this definition of homotopy equivalent to the previously mentioned approaches in question I?

Now for what is more important:

2.) The homotopy theory of Lie infinity algebras as given by Urs Schreiber in question I is obtained by 'transferring' the homotopy theory of differential graded Lie algebras 'along' an adjunction

$F\dashv G$

($F:L_\infty Alg \to DGLA$ and $G:DGLA \to L_\infty Alg$)

Now the question that really irritates me for quite some time is: how can we be sure that we get the correct homotopy theory of Lie infinity algebras by transferring its 'shadow' in the category of DG Lie algebra back along the previous mentioned adjunction? To me it looks like we can not rule out that there is a more general definition of weak equivalences in the category of Lie infinity algebras, which just project under $F$ onto those we already know.

Sorry if the second question is vague.

Edit; I used the 'Lie algebra cohomology tag' since the homotopy theory of Lie infinity algebras affects Lie algebra cohomology, too.

1) Any choice of path (or cylinder) objects will give the same homotopy classes, using the model category structure. Unless M is abelian, these path objects are harder to construct than path objects for chain complexes (exercise). You can't just ignore the coalgebra structure.

2) Hinich's definition of weak equivalences of dg coalgebras was reflected from DGLAs, but for L-infinity algebras, these are equivalent to tangent quasi-isomorphisms. For general dg coalgebras, intrinsic and natural characterisations of weak equivalences exist, but are more involved. Note that these are stronger (not weaker, as the question seems to hope) than dg coalgebra quasi-isomorphisms.

• Thanks for the answer and welcome to mathoverflow, Jon. I need to think about it. – Mark.Neuhaus Jul 25 '14 at 22:09

First of all, you might want to look at the MO question How to define the equivalence of Maurer-Cartan elements in an $L_\infty$-algebra?, since of course this is effectively what you need (morphisms can be described as MC elements in a particular convolution $L_\infty$-algebra, I suspect you know that).

Let me also say that some sort of approach to homotopy of the kind you are asking for is discussed by Martin Markl in this paper, and in this paper it is checked that the sort of definition proposed by Martin can be obtained from a Sullivan-type approach to homotopies (outlined in an answer to your first question) using appropriate homotopy transfer theorem for homotopy cooperads.

It is worth remarking, in particular, that the definition of a "derivation homotopy" suggested in another answer to this question does not work: the crucial difference between $A_\infty$ and $L_\infty$ is that $A_\infty$ is a nonsymmetric operad, and therefore instead of the Sullivan's algebra of differential forms on the 1-simplex we can use the dg algebras of chains on the 1-simplex (which is noncommutative) - this gives the "derivation homotopy".

I did not know where to put this answer, whether here or in this thread. Please, consider it as a remark on a possible definition of $L_\infty$-homotopies. I would like to summarize some considerations on $L_\infty$-algebra homotopies using as prototype the homotopies of $A_\infty$ morphisms.

With $A$ and $B$ we denote $\mathbb Z$-graded vector spaces on a ground field of characteristic zero (for simplicity). In what follows, a graded vector space will be also called "graded object".

1. A little remark: the chain complexes case

If $A$ resp. $B$ are chain complexes, with differentials $d_A$, resp. $d_B$, and $f,g: A\rightarrow B$ are (chain complexes) morphisms then a homotopy $H: A\rightarrow B$ between them is a degree $-1$ morphism of graded objects s.t. $f-g = d_BH+Hd_A$. No extra condition on $H$ is needed, as $d_B (f-g) = (f-g) d_A$ is automatically satisfied.

1. $L_\infty$ case

Let $(A,D_A)$ resp. $(B,D_B)$ by $L_\infty$-algebras; with $D_A$, resp. $D_B$ we denote the squaring to zero, $+1$ degree coderivations on the free graded co-commutative coalgebras over $A$, resp. $B$. These latter are denoted by $(C(A),\Delta_A)$ resp. $(C(B),\Delta_B)$: the $\Delta$'s are the coproducts.

Let $F,G$ be $L_\infty$-morphisms from $A$ to $B$, i.e. differential graded coalgebra maps $F,G:(C(A),D_A,\Delta_A)\rightarrow (C(B),D_B,\Delta_B)$ satisfying the known compatibility conditions w.r.t. coderivations and coproducts.

Definition I An $(F,G)$-derivation $R$ of degree $k$ is a graded object map $R:C(A)\rightarrow C(B)$ of degree $k$ s.t.
$$\Delta_B H = (F\otimes H+ H\otimes G)\Delta_A.~~ (**)$$

We arrive at the main definition of this answer, i.e.

Definition II A homotopy $H$ between the $L_\infty$-algebras morphisms $F$ and $G$ is an $(F,G)$-derivation of degree $-1$ s.t. $F-G = D_BH+HD_A$.

The questions are: why do we impose the $(F,G)$-coderivation relation (**) ? Which consequences does it have? The (inevitably partial) answers to these questions are my motivation for such definition of $L_\infty$-homotopies. In summary:

• As $H$ is an $(F,G)$-coderivation, then it satisfies a lifting property and it is uniquely determined by its components $H_n: A^{\otimes n}\rightarrow B$. $H$ is lifted with the well-known formulae $H=\sum F_{\bullet}^{\otimes\bullet}\otimes H_{\bullet}\otimes G_{\bullet}^{\otimes\bullet}$.

• $D_BH+HD_A$ is an $(F,G)$-coderivation ($D_A$ and $D_B$ are $(1,1)$-coderivations) as well; this implies that the relation $F-G = D_BH+HD_A$ can be expressed with the familiar tower of $L_\infty$-relations found in the literature.

• $G + D_BH+HD_A$ and $F-D_BH+HD_A$ really cocommute with coproducts, as $L_\infty$-algebra morphisms should.

In summary, I believe a definition of homotopy of $L_\infty$-algebras could be the one given in definition II, above. As said in the introduction, similar constructions take place in the realm of $A_\infty$-algebras: in this setting one uses as graded coalgebras the reduced tensor coalgebras. A good reference can be the first chapter in Hasegawa's PhD Thesis.

• I tried that definition, but could not arrive anywhere, becauase the equations seems to be non computable. – Mark.Neuhaus Nov 16 '14 at 18:10
• Sorry to hear that; could you please provide us with more details on your particular application? Which equations are you dealing with? – Avitus Nov 16 '14 at 21:03
• The difference from $A_\infty$ is very crucial (symmetric operad vs nonsymmetric), unfortunately, and this definition would be problematic. See e.g. Remark 3 in arxiv.org/pdf/1208.4695v2.pdf – Vladimir Dotsenko Nov 16 '14 at 21:34


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# Exponentiation when the exponent is irrational

I am just curious about what inference we can draw when we calculate something like $$\text{base}^\text{exponent}$$ where base = rational or irrational number and exponent = irrational number

-
As far as I can tell, this is one of the problems by Hilbert, and is quite difficult to answer. However, Gelfond-Schneider tells us that, if the base is algebraic, then it is always transcendental. –  awllower Feb 6 '13 at 10:51
@awllower, do you have a reference to which of Hilbert's problem this is related to? –  Ittay Weiss Feb 6 '13 at 10:54
@IttayWeiss In the link. –  awllower Feb 6 '13 at 10:55
Hilbert's seventh is not quite what is being asked here. –  Ittay Weiss Feb 6 '13 at 10:57
Yes, I know, so I did not post it as an answer. And I shall reformulate the comment as : related to one of the problems of Hilbert. –  awllower Feb 6 '13 at 11:02

An example I have always liked of $$\textbf{irrational}^{\textbf{irrational}} = \textbf{rational}$$ is the following: $$2 = \sqrt{2}^2= (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}.$$ So either $\alpha = \sqrt{2}^{\sqrt{2}}$ is rational, or $\alpha$ is irrational, and then $\alpha^{\sqrt{2}}$ is rational.

PS @IttayWeiss in his post has a much more precise statement. This has the advantage of being elementary.

-
it also has the advantage of being a really nice example of proving a statement of the form $P\vee Q$ without actually proving either $P$ or $Q$. –  Ittay Weiss Feb 6 '13 at 11:10

$2^2$ is rational while $2^{1/2}$ is irrational. Similarly, $\sqrt 2^2$ is rational while $\sqrt 2^{\sqrt 2}$ is irrational (though it is not so easily proved), so that pretty much settles all cases. Much more can be said when the base is $e$. The Lindemann-Weierstrass Theorem asserts that $e^a$ where $a$ is a non-zero algebraic number is a transcendental number.

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By the theorem of GFelfond and Schneider, then, one can easily conclude that $\sqrt2 ^{\sqrt2}$ is even transcendental. –  awllower Feb 6 '13 at 10:54
It was Gelfond. Apology here. –  awllower Feb 6 '13 at 11:02
@IttayWeiss- How can 1 prove that √2^√2 is an irrational number???? –  KAVISH Feb 7 '13 at 16:53
it's not that easy @KAVISH. The Gelfond-Schneider theorem implies that it is in fact transcendental. –  Ittay Weiss Feb 7 '13 at 19:20


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# What is the enthalpy of hydrogenation of propene?

Jan 16, 2017

Normally, we might locate the enthalpy of formation for each substance, but since that for ${\text{C"_3"H}}_{6}$ is not in my textbook, we can use enthalpies of bond formation and bond breaking instead.

Draw out the Lewis structures:

Then you should notice that the reactants overall contain these bonds that are broken:

• $6 \times \text{C"-"H}$ single bonds (${\text{CH}}_{3}$, $\text{CH}$, ${\text{CH}}_{2}$)
• $1 \times \text{C"-"C}$ single bonds
• $1 \times \text{C"="C}$ double bond
• $1 \times \text{H"-"H}$ single bond

and these bonds are made in propane:

• $8 \times \text{C"-"H}$ single bonds (${\text{CH}}_{3}$, ${\text{CH}}_{2}$, ${\text{CH}}_{2}$)
• $2 \times \text{C"-"C}$ single bonds

The relevant bond enthalpies are therefore:

• DeltaH_("C"-"H") = "413 kJ/mol"
• DeltaH_("C"-"C") = "348 kJ/mol"
• DeltaH_("C"="C") = "614 kJ/mol"
• DeltaH_("H"-"H") = "436 kJ/mol"

The enthalpy of reaction can be calculated by keeping track of which bonds were broken and which were made in the reaction. So, from the formula:

color(blue)(DeltaH_"rxn") ~~ sum_R n_RDeltaH_"break" - sum_P n_PDeltaH_"form"

= [6DeltaH_("C"-"H") + 1DeltaH_("C"-"C") + 1DeltaH_("C"="C") + 1DeltaH_("H"-"H")] "kJ" - [8DeltaH_("C"-"H") + 2DeltaH_("C"-"C")] "kJ"

$= \left[\left(\text{6 mol")("413 kJ/mol") + ("1 mol")("348 kJ/mol") + ("1 mol")("614 kJ/mol") + ("1 mol")("436 kJ/mol")] - [("8 mol")("413 kJ/mol") + ("2 mol")("348 kJ/mol}\right)\right]$

$= \textcolor{b l u e}{- \text{124 kJ/mol}}$


text:
# Cartesian coordinates vectors

1. Feb 10, 2017

### The black vegetable

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

2. Relevant equations

3. The attempt at a solution

the answer given is the same but without the negative sign, I don't understand because the crossproduct of unit vectors
when using a Cartesian coordinates of the directions given by the right-hand rule? Is the positive z direction pointing out of the page if X and Y are as follows

apologies if this is in the wrong section, thanks for any help in advance

2. Feb 10, 2017

### BvU

Yes. $\hat {k} = \hat {\bf \imath } \times\hat {\bf \jmath}$ , so z points towards you, out of the screen.

Corkscrew rule I call it. Turn $\hat{\bf \imath }$ over the smallest angle towards $\hat {k}$. Corkscrew will go in the minus y direction : $$\hat \imath \times\hat k = -\hat \jmath$$

3. Feb 10, 2017

### TSny

The problem doesn't specify which of the two wires the charge -q is a distance b from. Does the answer change if you pick the other wire to be a distance b from -q?

4. Feb 10, 2017

### The black vegetable

Thank you both your answers, TSny I thought the same, closer to the top wire the combined magnetic field would be in the opposite direction . Maybe the question is just not very good


text:
Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 4

True

Work Step by Step

From 0$^{\circ}$ to 90$^{\circ}$, as the angle increases, so does the $tan$ of the angle. Therefore: $\tan$28$^{\circ}$ $\leq$ $\tan$40$^{\circ}$ is true.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# How can I combine \seq_use:Nn and \seq_map_inline:Nn?

The following 'naïve' solution throws the error:

ERROR: Misplaced \omit.

--- TeX said ---
\multispan ->\omit
\@multispan
l.13   }
\\
--- HELP ---
From the .log file...

I expect to see \omit only after tab marks or the \cr of
an alignment. Proceed, and I'll ignore this case.

\documentclass{article}
\usepackage{expl3}
\ExplSyntaxOn

\begin{document}

\seq_set_split:Nnn \l_tmpa_seq { , } { 1 , 2 , 3 }

\begin{tabular}{*{\seq_count:N \l_tmpa_seq}{c}}
hellow & world \\
\seq_map_inline:Nn \l_tmpa_seq {
\multicolumn{1}{|c|}{#1} &
} \\
goodbye & world
\end{tabular}
\end{document}

where I'd like to come up with a table like:

  hello    world
|   1    |   2   | 3 |
goodbye   world

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Isn't this caused (as the help tells you) by an unequal number of items in each row? You have 2 cells in the first row, and then write a sequence of three cells to the second. Perhaps add an & to the first and last row? – hugovdberg Apr 3 '14 at 18:34
@hugovdberg The error message is clear, but you actually don't need an equal number of items in each row :) as long as you don't make more columns than you specify, you're ok. The question is more to get the result I'm going for, not necessarily to fix the code I have. I'm fairly certain my approach will turn out to be utterly fruitless :( – Sean Allred Apr 3 '14 at 18:35
and what about a newline after the \seq_map_inline and an extra & after the third item? You have 3 items, each with a & so specifying 4 cells in that row actually, while a count of the sequence returns 3. Perhaps a conditional if items < count insert &? – hugovdberg Apr 3 '14 at 18:37
@hugovdberg Ah yes, that's a typo :) see my edit – Sean Allred Apr 3 '14 at 18:39
It's a problem of expandability. \seq_map_inline:Nn is not expandable. You can use \seq_map_function:NN after defining a (non-protected) function which does \multicolumn{...}{...}{#1}. – Bruno Le Floch Apr 3 '14 at 21:02

You have to use an expandable function, because any unexpandable token makes \multicolumn illegal after it.

\documentclass{article}
\usepackage{expl3}
\ExplSyntaxOn

\begin{document}

\seq_set_split:Nnn \l_tmpa_seq { , } { 1 , 2 , 3 }
\seq_pop_left:NN \l_tmpa_seq \l_tmpa_tl
{
& \multicolumn{1}{c|}{#1}
}

\begin{tabular}{*{\int_eval:n { \seq_count:N \l_tmpa_seq + 1 }}{c}}
hellow & world & some \\
\multicolumn{1}{|c|}{\l_tmpa_tl}
goodbye & world & again
\end{tabular}
\end{document}

Note that the first item should be detached because it needs a different treatment.

-


text:
## What Is the Money Flow Index (MFI)?

The Money Flow Index (MFI) is a technical oscillator that uses price and volume data for identifying overbought or oversold signals in an asset. It can also be used to spot divergences which warn of a trend change in price. The oscillator moves between 0 and 100.

Unlike conventional oscillators such as the Relative Strength Index (RSI), the Money Flow Index incorporates both price and volume data, as opposed to just price. For this reason, some analysts call MFI the volume-weighted RSI.

### Key Takeaways

• The Money Flow Index (MFI) is a technical indicator that generates overbought or oversold signals using both prices and volume data.
• An MFI reading above 80 is considered overbought and an MFI reading below 20 is considered oversold, although levels of 90 and 10 are also used as thresholds.
• A divergence between the indicator and price is noteworthy. For example, if the indicator is rising while the price is falling or flat, the price could start rising.

## The Formulas for the Money Flow Index Are:

\begin{aligned} &\text{Money Flow Index}=100-\frac{100}{1+\text{Money Flow Ratio}}\\ &\textbf{where:}\\ &\text{Money Flow Ratio}=\frac{\text{14 Period Positive Money Flow}}{\text{14 Period Negative Money Flow}}\\ &\text{Raw Money Flow}=\text{Typical Price * Volume}\\ &\text{Typical Price}=\frac{\text{High + Low + Close}}{3}\\ \end{aligned}

When the price advances from one period to the next Raw Money Flow is positive and it is added to Positive Money Flow. When Raw Money Flow is negative because the price dropped that period, it is added to Negative Money Flow.

## How to Calculate the Money Flow Index

There are several steps for calculating the Money Flow Index. If doing it by hand, using a spreadsheet is recommended.

1. Calculate the Typical Price for each of the last 14 periods.
2. For each period, mark whether the typical price was higher or lower than the prior period. This will tell you whether Raw Money Flow is positive or negative.
3. Calculate Raw Money Flow by multiplying the Typical Price by Volume for that period. Use negative or positive numbers depending on whether the period was up or down (see step above).
4. Calculate the Money Flow Ratio by adding up all the positive money flows over the last 14 periods and dividing it by the negative money flows for the last 14 periods.
5. Calculate the Money Flow Index (MFI) using the ratio found in step four.
6. Continue doing the calculations as each new period ends, using only the last 14 periods of data.

## What Does the Money Flow Index Tell You?

One of the primary ways to use the Money Flow Index is when there is a divergence. A divergence is when the oscillator is moving in the opposite direction of price. This is a signal of a potential reversal in the prevailing price trend.

For example, a very high Money Flow Index that begins to fall below a reading of 80 while the underlying security continues to climb is a price reversal signal to the downside. Conversely, a very low MFI reading that climbs above a reading of 20 while the underlying security continues to sell off is a price reversal signal to the upside.

Traders also watch for larger divergences using multiple waves in the price and MFI. For example, a stock peaks at $10, pulls back to$8, and then rallies to $12. The price has made two successive highs, at$10 and $12. If MFI makes a lower higher when the price reaches$12, the indicator is not confirming the new high. This could foreshadow a decline in price.

The overbought and oversold levels are also used to signal possible trading opportunities. Moves below 10 and above 90 are rare. Traders watch for the MFI to move back above 10 to signal a long trade, and to drop below 90 to signal a short trade.

Other moves out of overbought or oversold territory can also be useful. For example, when an asset is in an uptrend, a drop below 20 (or even 30) and then a rally back above it could indicate a pullback is over and the price uptrend is resuming. The same goes for a downtrend. A short-term rally could push the MFI up to 70 or 80, but when it drops back below that could be the time to enter a short trade in preparation for another drop.

## The Difference Between the Money Flow Index and the Relative Strength Index (RSI)

The MFI and RSI are very closely related. The main difference is that MFI incorporates volume, while the RSI does not. Proponents of volume analysis believe it is a leading indicator. Therefore, they also believe that MFI will provide signals, and warn of possible reversals, in a more timely fashion than the RSI. One indicator is not better than the other, they are simply incorporating different elements and will, therefore, provide signals at different times.

## Limitations of the Money Flow Index

The MFI is capable of producing false signals. This is when the indicator does something that indicates a good trading opportunity is present, but then the price doesn't move as expected resulting in a losing trade. A divergence may not result in a price reversal, for instance.

The indicator may also fail to warn of something important. For example, while a divergence may result in a price reversing some of the time, divergence won't be present for all price reversals. Because of this, it is recommended that traders use other forms of analysis and risk control and not rely exclusively on one indicator.


text:
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## “Top Ten Physicists” on Radio Wales

The evening before last (Tuesday the 1st of March) I was on the programme Science Cafe on BBC Radio Wales talking about 10 Physicists Who Transformed Our Understanding of the Universe, the book I co-wrote with Brian Clegg. I was interviewed by Adam Walton, who presents this weekly science programme. We discussed how the book came about and then ran through the 10 physicists in the list. You can listen to the programme via the BBC iPlayer by following this link (it will be available on the iPlayer until the end of March; it will also be re-broadcast on Radio Wales on Sunday 6th March at 6:30am GMT)

I discussed our book 10 Physicists Who Transformed Our Understanding of Reality on BBC Radio Wales’ weekly science programme Science Cafe. The book was co-written with Brian Clegg

As you can see from the screen capture above, the ten in the list (it is not our list, it was taken from The Observer newspaper) are

1. Isaac Newton
2. Niels Bohr
3. Galileo Galilei
4. Albert Einstein
5. James Clerk Maxwell
7. Marie Curie
8. Richard Feynman
9. Ernest Rutherford
10. Paul Dirac

Thirty minutes was not enough to talk in any detail about any of the ten, but I hope it gave the listeners a nice taster of the fascinating characters whom we write about in each chapter. The book has been available since early December. You can order a copy by following this link if you are in the UK, and this link if you are in the US.

At the BBC studios in Cardiff recording this interview for Science Cafe

I am currently in Namibia (see my blog from Tuesday), and recorded the interview in the few days between getting back from a 3-week trip to the USA and leaving to give talks on a cruise in South America. The broadcast of the interview was delayed a little as Science Cafe had recently done a programme about antimatter, and as Paul Dirac is one of the physicists in the list the producers decided to leave a few weeks pass before its broadcast.

Last night (Wednesday the 2nd of March) I gave a talk about ‘the oldest light in the Universe’ (the Cosmic Microwave Background – the subject of another of my books) to the Swakopmund Scientific Society here in Swakopmund (Namibia). I will give a summary of that talk next week when I am back in Cardiff.

You can also find out more about 10 Physicists Who Transformed Our Understanding of the Universe by following this link.

As I mentioned in this blog here, a few months ago I contributed some articles to a book called 30-Second Einstein, which will be published by Ivy Press in the not too distant future. One of the articles I wrote for the book was on Indian mathematical physicist Satyendra Bose. It is after Bose that ‘bosons’ are named (as in ‘the Higgs boson’), and also terms like ‘Bose-Einstein statistics’ and ‘Bose-Einstein condensate’. So, who was Satyendra Bose, and why is his name attached to these things?

Satyendra Bose was an Indian mathematical physicist after whom the ‘boson’ and Bose-Einstein statistics are named

Satyendra Bose was born in Calcutta, India, in 1894. He studied applied mathematics at Presidency College, Calcutta, obtaining a BSc in 1913 and an MSc in 1915. On both occasions, he graduated top of his class. In 1919, he made the first English translation of Einstein’s general theory of relativity, and by 1921 he had moved to Dhaka (in present-day Bangladesh) to become Reader (one step below full professor) in in the department of Physics.

It was whilst in Dhaka, in 1924, that he came up with the theory of how to count indistinguishable particles, such as photons (light particles). He showed that such particles follow statistics which are different from particles which can be distinguished. All his attempts to get his paper published failed, so in an act of some desperation he sent it to Einstein. The great man recognised the importance of Bose’s work immediately, translated it into German and got it published in Zeitschrift für Physik, one of the premier  physics journals of the day.

Because of Einstein’s part in getting the theory published, we now know of this way of counting indistinguishable particles as Bose-Einstein statistics. We also name particles which obey this kind of statistics bosons; examples are the photon, the W and Z-particles (which mediate the weak nuclear force), and the most famous boson, the Higgs boson (responsible for mediating the property of mass via the Higgs field).

With the imminent partition of India when it was gaining independence from Britain, Bose returned to his native Calcutta where he spent the rest of his career. He died in 1974 at the age of 80.

You can read more about Satyendra Bose, Bose-Einstein statistics and Bose-Einstein condensates in 30-second Einstein, out soon from Ivy Press.

## 30-second Einstein

Although I will restate it on Thursday, I am not going to be able to post the last part of my series about the derivation of Planck’s radiation law this week as planned. It is taking me too much time, and so I will post it instead on Thursday the 17th. It is nearly finished; honest!

One of the reasons I don’t have as much time to write it as I would wish is that I am writing articles for a book called ’30-second Einstein’, which Ivy Press will be publishing next year (2016). The book is part of their series of ’30-second + subject’, and I have already contributed to their ’30-second physics’ volume, which is due out in March of 2016. For more information about the articles I wrote for ’30-second Physics’, follow this link.

Albert Einstein, before his hair went white and crazy 🙂

As with ’30-second Physics’; ’30-second Einstein’ is being edited by Brian Clegg, and I am writing 16 of the 50-odd articles which will appear in the book. There are about five or six of us contributing to the book.

The articles I am writing are

• Bose-Einstein statistics
• Biography – Satyendra Bose
• Photolectric quanta
• Biography – Max Planck
• Dreaming about light (floating alongside a sunbeam)
• On the electrodynamics of moving bodies
• Simultaneity
• Length, time and mass
• Spacetime
• Biography – Hermann Minkowski
• $E=mc^{2}$
• Heavy clocks
• Gravitational lenses
• The cosmological constant
• Expanding universes

The concept of the books is quite interesting; each topic needs to be explained in about 300 words, which believe me is quite a challenge. This is sufficiently few words that every one counts, and usually I find my initial draft(s) are 450 words or more, so then I have to edit them down making sure I don’t lose the explanation in the process.

Anyway, look out for this book in the next year or so, as soon as it is available I will post a link to it on this blog. In the meantime, check out the other books in the ’30-second’ series (there is a link to the series on this page), I haven’t contributed to any of them but many of them look interesting (and I have downloaded a few and enjoyed them greatly).

## To catch a comet – Rosetta public lecture

Last night (Monday the 24th of August) I went to a public lecture about the Rosetta mission
at the National Museum of Wales in Cardiff. The lecture was given by Mark McCaughrean, who is senior science advisor at the European Space Agency (ESA) and, if I’m correct, also either heads up or is very senior in their public outreach efforts. It was one of the best public lectures I’ve ever attended, and in writing that statement I am trying to figure out how many public lectures I have actually attended. In addition to having given probably over 100 public lectures myself, I have probably attended some 150-200 public lectures given by others in the last 40-odd years.

The opening slide of Mark McCaughrean public lecture about the Rosetta mission at the National Museum of Wales

In addition to learning a lot about the Rosetta mission (I will blog about some of what I learnt next week), the lecture got me thinking about what makes a good public lecture. I have also been thinking about this the last few days because my book on the Cosmic Microwave Background has been reviewed by Physics World (the magazine of the Institute of Physics), and that review will apparently appear in their October magazine. But, the reviewer shared with me some of her observations about the book, and one point she raised is that she felt I was inconsistent in my level of explanations in the book. What she meant was that there are some things I explain so that complete novices can follow my arguments, but other things where more of a physics/astronomy background would be necessary to follow that I am saying.

This is a valid point, and it shows the quandary I was in when trying to decide at what level to pitch the book. My primary audience was that I hoped the book will be used by undergraduates in the Disunited Kingdom and graduate students in the United States as a background text to any course they may be doing on the early Universe. But, in the back of my mind, I also had the interested lay-reader in mind, which is why I explained some things at a level for them. What I probably ended up doing was falling between two stools, and that is not always good in communicating science to the public.

Last night’s lecture by Mark did a wonderful job, as it seemed to me that he was able to keep it at a level that (hopefully) everyone could understand, but at the same time there was some specialist information in there for professional astronomers to give them (and me) the impression that we too had learnt something. This is a difficult tightrope to walk, but Mark did it very well.

Audience participation time – the audience had to jump 4cm in the air to simulate the acceleration felt by Philae when landing on comet 67/P

This is what I try to do in my own public lectures, but I doubt I do it as well as Mark did last night. Whether I’m talking to school groups, astronomical societies, on the radio or TV or lecturing on a cruise, I always try to make sure that I don’t lose any of my audience in the first three quarters or so of the lecture by keeping things as simple as possible. At the same time, I always try to make sure that there is some information in the lecture (maybe some 25% of it) which will be news to even a professional in the field, as even in a public lecture you may have professionals in the audience. This was the case, for example, in lectures I gave on the cruise I did in South America in March – one of my regular attendees had worked at NASA JPL and he and I would have long chats after each lecture where he would quiz me further, or impart some information that I did not know about.

Last night, Mark had a perfect mixture of videos, cartoons, animations, humour and exciting information, and it was all delivered in a relaxed and humorous way. As I say, one of the best public lectures I have ever attended.