text:
# Equivalence test

Equivalence tests are a variation of hypothesis tests used to draw statistical inferences from observed data. In equivalence tests, the null hypothesis is defined as an effect large enough to be deemed interesting, specified by an equivalence bound. The alternative hypothesis is any effect that is less extreme than said equivalence bound. The observed data is statistically compared against the equivalence bounds. If the statistical test indicates the observed data is surprising, assuming that true effects at least as extreme as the equivalence bounds, a Neyman-Pearson approach to statistical inferences can be used to reject effect sizes larger than the equivalence bounds with a pre-specified Type 1 error rate.

Equivalence testing originates from the field of pharmacokinetics.[1] One application is to show that a new drug that is cheaper than available alternatives works just as well as an existing drug. In essence, equivalence tests consist of calculating a confidence interval around an observed effect size, and rejecting effects more extreme than the equivalence bound when the confidence interval does not overlap with the equivalence bound. In two-sided tests an upper and lower equivalence bound is specified. In non-inferiority trials, where the goal is to test the hypothesis that a new treatment is not worse than existing treatments, only a lower equivalence bound is pre-specified.

Mean differences (black squares) and 90% confidence intervals (horizontal lines) with equivalence bounds ΔL = -0.5 and ΔU= 0.5 for four combinations of test results that are statistically equivalent or not and statistically different from zero or not. Pattern A is statistically equivalent, pattern B is statistically different from 0, pattern C is practically insignificant, and pattern D is inconclusive (neither statistically different from 0 nor equivalent).

Equivalence tests can be performed in addition to null-hypothesis significance tests.[2][3][4][5] This might prevent common misinterpretations of p-values larger than the alpha level as support for the absence of a true effect. Furthermore, equivalence tests can identify effects that are statistically significant but practically insignificant, whenever effects are statistically different from zero, but also statistically smaller than any effect size deemed worthwhile (see first Figure).[6]

## TOST procedure

"A very simple equivalence testing approach is the ‘two-one-sided t-tests’ (TOST) procedure.[7] In the TOST procedure an upper (ΔU) and lower (–ΔL) equivalence bound is specified based on the smallest effect size of interest (e.g., a positive or negative difference of d = 0.3). Two composite null hypotheses are tested: H01: Δ ≤ –ΔL and H02: Δ ≥ ΔU. When both these one-sided tests can be statistically rejected, we can conclude that –ΔL < Δ < ΔU, or that the observed effect falls within the equivalence bounds and is statistically smaller than any effect deemed worthwhile, and considered practically equivalent.[8]" [Lakens 2017] Alternatives to the TOST procedure have been developed as well.[9] A recent modification to TOST makes the approach feasible in cases of repeated measures and assessing multiple variables. [10]

## Comparison between t-test and equivalence test

The equivalence test can, for comparison purposes, be induced from the t-test.[11] Considering a t-test at the significance level αt-test achieving a power of 1-βt-test for a relevant effect size dr, both tests lead to the same inference whenever parameters Δ=dr as well as αequiv.-testt-test and βequiv.-testt-test coincide, i.e. the error types (type I and type II) are interchanged between the t-test and the equivalence test. To achieve this for the t-test, either the sample size calculation needs to be carried out correctly, or by adjusting the t-test significance level αt-test, referred to as the so-called revised t-test.[11] Both approaches have difficulties in practice, since sample size planning relies on unverifiable assumptions of the standard deviation ${\textstyle \sigma }$, and the revised t-test yields numerical problems.[11] Preserving the test behaviour, those limitations can be removed by using an equivalence test.

The second Figure allows a visual comparison of the equivalence test and the t-test when the sample size calculation is affected by differences between the a priori standard deviation ${\textstyle \sigma }$ and the sample's standard deviation ${\textstyle {\widehat {\sigma }}}$, which is a common problem. Using an equivalence test instead of a t-test additionally ensures that αequiv.-test is bounded, which the t-test does not do in case that ${\textstyle {\widehat {\sigma }}>\sigma }$ with the type II error growing arbitrary large. On the other hand, having ${\textstyle {\widehat {\sigma }}<\sigma }$ results in the t-test being stricter than the dr specified in the planning, which may randomly penalize the sample source (e.g. a device manufacturer). This makes the equivalence test safer to use.

Chances to pass (a) the t-test and (b) the equivalence test, depending on the actual error 𝜇. For more details, see[11]

• Walker, Esteban; Nowacki, Amy S. (February 2011). "Understanding Equivalence and Noninferiority Testing". Journal of General Internal Medicine. 26 (2): 192–6. doi:10.1007/s11606-010-1513-8. PMC 3019319. PMID 20857339.

## References

1. ^ Hauck, Walter W.; Anderson, Sharon (1984-02-01). "A new statistical procedure for testing equivalence in two-group comparative bioavailability trials". Journal of Pharmacokinetics and Biopharmaceutics. 12 (1): 83–91. doi:10.1007/BF01063612. ISSN 0090-466X. PMID 6747820. S2CID 29838725.
2. ^ Rogers, James L.; Howard, Kenneth I.; Vessey, John T. (1993). "Using significance tests to evaluate equivalence between two experimental groups". Psychological Bulletin. 113 (3): 553–565. doi:10.1037/0033-2909.113.3.553. PMID 8316613.
3. ^ Statistics applied to clinical trials (4th ed.). Springer. ISBN 978-1402095221.
4. ^ Piaggio, Gilda; Elbourne, Diana R.; Altman, Douglas G.; Pocock, Stuart J.; Evans, Stephen J. W.; CONSORT Group, for the (8 March 2006). "Reporting of Noninferiority and Equivalence Randomized Trials" (PDF). JAMA. 295 (10): 1152–60. doi:10.1001/jama.295.10.1152. PMID 16522836.
5. ^ Piantadosi, Steven (28 August 2017). Clinical trials : a methodologic perspective (Third ed.). p. 8.6.2. ISBN 978-1-118-95920-6.
6. ^ Lakens, Daniël (2017-05-05). "Equivalence Tests". Social Psychological and Personality Science. 8 (4): 355–362. doi:10.1177/1948550617697177. PMC 5502906. PMID 28736600.
7. ^ Schuirmann, Donald J. (1987-12-01). "A comparison of the Two One-Sided Tests Procedure and the Power Approach for assessing the equivalence of average bioavailability". Journal of Pharmacokinetics and Biopharmaceutics. 15 (6): 657–680. doi:10.1007/BF01068419. ISSN 0090-466X. PMID 3450848. S2CID 206788664.
8. ^ Seaman, Michael A.; Serlin, Ronald C. (1998). "Equivalence confidence intervals for two-group comparisons of means". Psychological Methods. 3 (4): 403–411. doi:10.1037/1082-989x.3.4.403.
9. ^ Wellek, Stefan (2010). Testing statistical hypotheses of equivalence and noninferiority. Chapman and Hall/CRC. ISBN 978-1439808184.
10. ^ Rose, Evangeline M.; Mathew, Thomas; Coss, Derek A.; Lohr, Bernard; Omland, Kevin E. (2018). "A new statistical method to test equivalence: an application in male and female eastern bluebird song". Animal Behaviour. 145: 77–85. doi:10.1016/j.anbehav.2018.09.004. ISSN 0003-3472. S2CID 53152801.
11. ^ a b c d Siebert, Michael; Ellenberger, David (2019-04-10). "Validation of automatic passenger counting: introducing the t-test-induced equivalence test". Transportation. doi:10.1007/s11116-019-09991-9. ISSN 0049-4488.


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Function for encoding strings in-place

I recently wrote a function that replaces every white space with '%20', just for fun (and sharping my coding skills). The input string is terminated with extra white spaces, with the length that the string should have after encoding takes place. For example:

input = "Not encoded  " //2 white spaces in the end
output = "Not%20encoded"

Although I was able to implement such encoding in-place, I was still not pleased with the result, specially due to the inner loop that shifts the whole string to the right everytime a white space is found. Here is my solution, written in C#:

    public string encode(string str)
{
var array = str.ToCharArray();
for (int i = 0; i < array.Length; i++) {
if (array[i] != ' ') continue;

//switch every char >> 2, from the end of the string to i
for (int j = array.Length - 1; j > i; j--) {
array[j] = array[j - 2];
}

array[i++] = '%';
array[i++] = '2';
array[i] = '0';
}
return new String(array);
}

I'm looking for a solution that operates in O(n), possibly performing a single traversal on the input string. Additional feedback on how to improve performance/readability of my function are always very appreciated :)

• Have you looked at the StringBuilder class? It does away with your 2nd shift-y for loop. – shivsky Oct 21 '13 at 20:58
• I have :) But I was trying to optimize space complexity. That's why I performed the string replacements in-place. – rla4 Oct 21 '13 at 21:01
• How come the string already has exactly the right amount of trailing spaces? Do you have special code for that? Doesn't that perform more allocations? – svick Oct 22 '13 at 0:29
• Maybe my question was not clear enough: this is not a "real world" application nor a "real world" problem. It's just a coding challenge, and the extra trailing spaces is part of the problem specification. – rla4 Oct 22 '13 at 0:39

As strings are immutable you can't really avoid intermediate store of the data as it's being encoded.

As you already know the length of the final string you can avoid the shuffling:

    int spaceCount = 0;
for (int i = str.Length - 1; str[i] == ' '; i--)
{
spaceCount++;
}
if (spaceCount == 0) { return str; }

var array = new char[str.Length];
int idx = 0;
for (int i = 0; i < str.Length - spaceCount; i++)
{
var current = str[i];
if (current  == ' ')
{
array[idx++] = '%';
array[idx++] = '2';
array[idx++] = '0';
}
else
{
array[idx++] = current ;
}
}

var encoded = new string(array);

Space required is 2 * O(n) (array and final string) and time is 2 * O(n) (encoding string into array, building final string)

• Saying that something is 2 * O(n) doesn't make much sense. You can either calculate how man times some specific operations happens (e.g. 2n array accesses) or you can say that it's O(n), but it doesn't make sense to combine the two. That's because 2 * O(n) means exactly the same thing as O(n). – svick Oct 22 '13 at 0:37
• @svick: While in the theoretical world O(c * n) = c * O(n) = O(n) and c as a constant doesn't matter, in the practical world reducing c can sometimes be a valuable optimization. So I like to state this factor. – ChrisWue Oct 22 '13 at 1:29
• Yes, but the factor you're static doesn't say anything unless you explain it. For example, using your counting, 2 * O(n) could easily be faster than O(n). Constants factors are important, but it's very hard to represent them in a way that makes any sense. – svick Oct 22 '13 at 9:35

• copies the whole string twice (1. string.ToCharArray() 2. new string())
• needs n bytes of extra storage (apart from the input and output; n is the size of the output)
• potentially moves the characters in the string a lot (O(n2))
• requires the input to already have free space for the expansion

Solution using StringBuilder:

• copies the whole string twice (1. calls to Append() 2. StringBuilder.ToString())
• needs about n bytes of extra storage
• doesn't need to move characters at all
• doesn't require the input to be in any special format

Based on this, StringBuilder is clearly the better solution.

But there is even better solution: just use string.Replace(). That's likely going to be quite efficient and it's much shorter than any other of the proposed alternatives.

• @downvoter: it would help if you explained what you think is wrong with my answer. – svick Oct 22 '13 at 0:25

Here's how I would do it (sans assertions/error checks):

private static string Encode(string input)
{
var output = new char[input.Length];
var i = 0;
var o = 0;

while (o < output.Length)
{
if (input[i] != ' ')
{
output[o++] = input[i++];
}
else
{
output[o++] = '%';
output[o++] = '2';
output[o++] = '0';
++i;
}
}

return new string(output);
}

It's O(n) in both time and memory.

Using a StringBuilder doesn't gain you much in this instance, because you know exactly how large your output will be (the same length as your input). It's just a wrapper around a char[] anyway, so you'd end up with slightly slower code without any benefit.

format: plain


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Adding second set of parents in an online tree at Ancestry?

6

1

I want to add a second set of parents to someone in my Ancestry.com tree. I'm surprised that I'm unable to find a way to do this. I've searched online and found references to a relationships tab which appears to no longer exist.

There is an + Add Family Member dropdown but it only has options for brother, sister, spouse, and child.

8

1. Go to edit person below their name.
2. Go to Relationships.
3. Click on "Add Alternative Parent".

4

UPDATED VERSION:

1. Go to your family tree
2. Click the person you want to add to
3. Click "Profile"
4. At the top right of the profile it should have a "tools" button and an "Edit" button
5. Click "Edit" and three different buttons should come down from there. One is an "edit relationship" or "add relationship" button.
6. Click the relationship button and it should pull up a screen with the current parents and that is where you can change them to adoptive or step parents or biological parents.
7. To add a new parent click "add alternative father" or "add alternative mother"

Hopefully this helps! We were looking for the new way to do it and nobody had the updated version yet.

1

Since the previous answers were written, Ancestry has added the capability to display two sets of parents on a person's profile page. Crista Cowan demonstrates this in her Barefoot Genealogist Video, September 2019 edition, What's New at Ancestry. The segment starts shortly after 11 minutes into the video.

1. From the Edit drop-down menu (top right of the profile) choose Edit Relationships.
2. This will open an overlay window showing all the relationships for the people in the immediate family (parents, spouses, children) for the person whose profile you are on.
3. Add the new parents by clicking the plus button next to "Choose Alternate Father" or "Choose Alternate Mother". The default setting is "Biological" but there are several other choices on a drop-down menu, including Adopted [sic], Step, and so on.
4. When adding a parent (either the first parent or an alternate) you'll get an option to create a new person, or choose someone who is already in your tree. If the person is already in the tree, start typing the name in the box and the system should prompt you with the person's name and dates.
5. Once you have multiple parents, you'll see an option to chose one as the preferred parent. This will determine which one shows in the pedigree view.
6. The new addition as of September 2019 is that a note will show on the person's profile that there are multiple sets of parents. You can see this in Crista's video around 16:15 and following. Click the down arrow next to the blue message Additional Parent Relationships to see the additional parents.


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# Existence of God

“If God doesn't have a girlfriend or wife, does He need to create a wormhole to have sex?”
“God is mortally wounded”
~ Friedrich Nietzsche on God two minutes before He died
“It is self-evident that this God character has a Messiah Complex. I would advise Him to seek counselling at the earliest possible opportunity.”
~ Sigmund Freud on God

Does God exist? If you have a life then you probably don’t give a rat's ass. However, if you’re bored as hell, then you might want to contemplate this question that will never be answered.

## editOverly Long Section Concerning the Definition of God

To approach the existence of God it is important to understand the meaning of the word “God”. The great dead philosopher Ludwig Wittgenstein spoke on this matter. He said, “God spelled backwards is Dog. One would then assume that God is a dog. However I don’t believe that is the case. Why isn’t God a dog? Because dogs don’t really exist. If God was going to be an animal, he would probably be a cat. Why? Because cats kick ass. Did Egyptians ever worship dogs? No. They worshiped cats. The Norasii worshiped dogs. Who are the Norasii? Exactly. Wikipedia doesn’t even have an article on the Norasii. Wikipedia most definitely has an article on Egypt. Egyptians built the pyramids and the Sphinx and enslaved the Jews! The Norasii didn’t do shit. They just sat around worshiping dogs and not getting articles written about them. Dogs shit on carpets and drool and do what you tell them to do like the little bitches they are. Cats on the other hand are calm and collective and do useful things like lying around and soaking in the sun’s energy while quietly planning to take over the world. God most definitely stands for ‘cat’”.

Another widely accepted theory of god's creation is, A man by the name of Chad Rucker that lives in Richmond Kentuckistan took a massive shit, and then there was god.

However, despite Wittgenstein’s attempts to define God, his strange rant comparing canines to felines drew much criticism mainly due to the undeniable fact that a dog could totally kick a cat’s ass. Therefore, a grand council was held to discuss the definition of God. The council was made up of philosophers, theologians, psychologists, gynecologists, astronomers, and mimes. The meeting was held in Kaliningrad for the sake of Immanuel Kant who didn’t want to leave his hometown. The meeting began with Kant being called forward to speak. After much stifled giggling, (his name sounds a lot like “cunt”) Immanuel spoke in great and confusing detail about human knowledge, the source of transcendental ideas, and a bunch of other crap the caused David Hume to become so frustrated that he killed a mime with a fork.

Everyone ultimately decided to dismiss Kant’s ideas and settle on the fact that God was an old man that looked similar to father time, lived in the clouds, and spoke with an electronically enhanced voice. This seemed to make most people happy until Stanley Milgram decided to deliver electrical shocks to the mimes until they died, which caused everyone to be happy.

## editProof of God’s Existence

Once everyone was agreed on what God was, people who had nothing better to do decided to prove that He (note the masculinity) existed. Some people, mainly Christians, decided to do prove God’s existence illogically (referencing The Bible), while other people, mainly Christians, decided to prove God’s existence logically (referencing themselves). One of these good Christian men (note that the masculinity) was St. Anslem who almost successfully proved God’s existence through his Ontological Argument which is shown below.

Anslem’s Ontological Argument for the Existence of God

1. Think of the greatest thing you can think of.
2. Now think of something even greater.
3. Now think of something even greater than that.
4. Now think of something a little bit greater than the last great thing you thought.
5. Right now you should be thinking about God, if you’re thinking about Mt. Everest go back to 1 and try again. If you’re thinking about Oprah shoot yourself in the head. If you’re thinking about a hot dog stand then stop fucking around because this is serious.
6. The greatest thing in the universe would have to be everything in the universe according to the equation E=mc² where E stands for everything and mc² stands for the greatest thing in the universe.
7. Since the greatest thing is the universe is everything and God is the greatest thing in the universe, God is everything, even food.
8. You eat food, and since you are what you eat, you are God.
9. If you’ve made it this far on the Ontological Proof you are probably thinking. Therefore you probably exist.
10. Since you exist, and you are God, God exists (until you die and then God will be dead).

St. Thomas Aquinas rejected St. Anselm’s argument and instead embraced his own argument mainly because he was an egotistical bastard. St. Thomas Aquinas’s proof that God exists is shown below.

If God doesn’t exist…

1. Then who was the first thing to make something happen?
2. Then who was the first thing to make something happen?
3. Then who was the first thing to make something happen?
4. Then who was the first thing to make something happen?
5. Then who was the first thing to make something happen?
6. Then who was the first thing to make something happen?
7. Then who was the first thing to make something happen?
8. Then who was the first thing to make something happen?
9. Then who was the first thing to make something happen?
10. Then why does everything work?
12. And on a related note… How could have we come from monkeys if there are still monkeys around? Or are we just monkeying around?
However most philosophers and smart dudes (note the masculinity), reject the aforementioned proofs because, as Oscar Wilde put it, “the proofs are mystic, unrealistic, altruistic, dualistic, propagandistic, pantheistic and anachronistic”.

As philanthropists have pointed-out to cult religious leaders time and time again, the greatest contradiction to the postulate of a monotheistic God is the existence of Good.

It is self-evident that good exists. Occasionally, people are destined to do good things to others. Some parts of this universe are a good design, from which life-forms benefit. Thus, not all of this universe we inhabit is definable as the creation of God. This is a contradiction, since God is supposed to be the creator of all that exists.

From reading the Old Temperance, there is little doubt that God was a total badass at that time in history. Not only that, it is obvious that He really, really enjoyed doing all those nasty things to people. One can just imagine Him sniggering with glee as He issues the instruction to Abraham to exterminate the Canaanites, for example. The kind of world-picture we are presented with for this bygone age is a straightforward and reassuring one; that God is just as bad as we are. If not actually worse, assuming that is possible. We, evil beings, it seemed, lived in an evil uninverse created by an evil God, in which we were all hunky-dory so long as we did his evil bidding. And, then some.

Yet, it is intuitively the case that a God of Absolute Evil such as the one portrayed by the Old Temperance would not allow Good to exist, once its existence had come to his attention. Which would happen immediately, since He is postulated to be all-powerful and all-seeing. The moment any Good was seen to emerge in his Realm, He would act to deal with this departure from His own basic nature. The fact that this observedly does not happen (well, mostly it does, but not always) leads us to one of two conclusions:

1. That God is Omnipotent, but not pure Evil.
2. That God is pure Evil, but He is not Omnipotent.

Accepting either of these conclusions defines God as being non-absolute. In which case -as Teal'c would be quick to point out- He is Not a God.

It is self-evident that a badass God would not permit Good to exist. That would be a violation of His basic nature. Yet, this leaves us with the perplexing question of who created good, if God did not? After all, if God created Good, then God cannot be the essence of Evil. Therefore, in this case, God cannot exist. Either that, or someone else sneakily created Good, just for the hell of it. His Son, maybe, just to be rebellious? Who knows.

If that was the case, then oh boy, look what happened to His Son whilst paying us a social-visit.. On the basis of that, I would advise anyone against acting contrary to God's instructions. If he would do that to His own Son, of all people, then... Yeah. Watch out. It is for good reasons that the Kabbalists never write God's name in full, instead using the acronym YHBW.

## editProof of God’s Nonexistence

God’s existence has been tested on a number of occasions. In 1993, Andrew Wiles proved Fermat's Last Theorem. The theorem had never been proven all though it had been hanging around annoying the fuck out of mathematicians for about 357 years. After Andrew Wiles proved Fermat’s Last Theorem, he proceeded to walk to the nearest local Baptist Church, stroll up to the pulpit, and proclaim “where’s your God now bitches?! I just proved Fermat’s Last Theorem! Can your God prove that no nontrivial solutions in integers exist for the equation: $x^n + y^n = z^n$ if $n$ is greater than two? I don’t think so. You know why? Because I did it. God didn’t do shit. What was God doing the last 357 years? Not proving Fermat’s Last Theorem, that’s what he was doing. It only took me seven years! Wow, God couldn’t prove it even though he got a 350 year head start. Well, since I’ve already shown that I’m obviously superior to God, I think I’ll stop here.” Similar tests include the invention of Benham’s Top and the existence of Bruce Lee.

Also, God cannot make a Thai dish so spicy that He Himself could not eat it.

## editOther Shit

### editEpistemology

Epistemology is sort of like studying magic without the luxury of playing Quidditch during your spare time. Epistemologists try to prove God exists by casting two types of spells called a posteriori and a priori. No one really knows the difference between these two spells. However, one of the spells usually ends up describing beliefs while the other one leads to methodology (another form of wizardry that has an extremely boring name and therefore requires that we ignore it completely).

### editIdentifying Attributes

Suppoose you encountered God, would you recognise Him? Here are a few pointers:

Glowing eyes. It is established that God's eyes will invariably glow if you get him sufficiently annoyed at you. As Godliness-tests go, however, this is not a preferred method from a health-and-safety point of view. You may only get to do this once, and may not be able to report-back on the results of the test.

Speaking in a 'phlanged' voice. Although widely reported, it is apparently the case that God only does this for effect. He does not have to speak like that. Therefore, this is not a reliable indicator of Godliness.

### editThe Theory That God Is Actually A Goddess

God is always identified by way of His trademark beard, which is of truly impressive proportions. This, incidentally, tends to rule-out any possibility of God being 'sHe' - Unless of course, the beard happens to be one of those stick-on ones which female Pharaohs such as Nebuchadnezzari the 43rd wore.

Yeah right. Get in the kitchen and fix me a sammich.

## editConclusion

For those without comedic tastes, the so-called experts at Wikipedia have an article about Existence of God.

God might exist or He might not exist if you are an agnostic. If you are a Christian, Muslim, or Jew then He (note the masculinity) does exist. If you are an atheist, He (again note the masculinity) does not exist. And if you are me, then you’re sexist.


text:
# Deriving equations for the “Bianchi-Pinkall torus”

I am trying to work out explicit parametric equations for the "Bianchi-Pinkall flat torus" as depicted in this note, but I seem to have gotten stuck in understanding the descriptions given in that note.

Thus far, I figured that the parametric equations in $\mathbb R^4$ should look like this:

\begin{align*} x_1&=\cos(a+b\sin(2p v))\cos(u+v)\\ x_2&=\cos(a+b\sin(2p v))\sin(u+v)\\ x_3&=\sin(a+b\sin(2p v))\cos(u-v)\\ x_4&=\sin(a+b\sin(2p v))\sin(u-v) \end{align*}

where I changed $aa$ in the note to $a$ here, and respectively $bb$ to $b$, and $ee$ to $p$. I have gotten stuck on the part that says that to obtain the $\mathbb R^3$ embedding, one should stereographically project this torus from the point $(\cos c\pi,0,\sin c\pi,0)$, as well as on the part about varying a parameter $ff$ from $0$ to $2\pi$, as this parameter is nowhere to be found in the parametric equations given.

I am familiar with the usual "north pole" stereographic projection,

$$\left(\frac{x_1}{1-x_4},\frac{x_2}{1-x_4},\frac{x_3}{1-x_4}\right)$$

but I do not know how to generalize this formula to the projection point given in the note.

In particular I wanted to replicate the surface depicted here in another surface plotting program, but I have been unsuccessful in doing so. Thanks for any help.

-

According to this document, $f\!f$ is just another parameter controlling the surface. Your linked PDF says that it "isometrically rotates $S^3$ so that the Hopf circle $v = 0$ is the rotation axis." This circle is the set of points $(e^{iu}\cos a,e^{iu}\sin a)$ for all $u\in[0,2\pi)$. It lies in the subspace with basis $b_1=(\cos a,0,\sin a,0)$ and $b_2=(0,\cos a,0,\sin a)$, and the orthogonal subspace has basis $b_3=(-\sin a,0,\cos a,0)$ and $b_4=(0,-\sin a,0,\cos a)$. To perform the rotation, express the point to be rotated as a linear combination of $b_i$, and then rotate by an angle of $f\!f$ in the $b_3b_4$ plane.
As for projecting relative to an arbitrary point $P$, you just have to rotate the sphere so that $P$ coincides with $(0,0,0,1)$ and then apply the usual stereographic projection formula. For the particularly simple form $P=(\cos c\pi,0,\sin c\pi,0)$, this is essentially a rotation by $-c\pi$ in the $x_1x_3$ plane, namely $$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\mapsto\begin{pmatrix}x_1\cos c\pi+x_3\sin c\pi\\x_2\\x_3\cos c\pi-x_1\sin c\pi\\x_4\end{pmatrix}.$$ Actually, this maps $P$ to $(1,0,0,0)$, so switch $x_1$ and $x_4$ in the projection formula.
By the way, it seems a little too specific to call this sort of surface "the Bianchi-Pinkall torus". A cursory search reveals that what Pinkall proved [1] was that "the inverse image under the Hopf mapping of a simple closed curve on $S^2$ is a flat torus on $S^3$" [2]. So any smooth curve on the familiar 2-sphere gives you a flat torus in the 3-sphere, which you would probably call a Pinkall torus. I haven't been able to find out how Bianchi comes into the picture.
1. Pinkall, "Hopf Tori in $S^3$", Invent. math. 81, 1985.


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# Math Help - Set Theory-functions, need helpp

1. ## Set Theory-functions, need helpp

1) Let P be the set of all U.S. presidents, and let G be the set of all ordered pairs (a,b) in P X P such that b succeeded a in office. Is G the graph of a function? Explain.

2) Prove that for each set X there is a unique function from the empty set to X, regardless of whether or not X is nonempty. Also prove that there are no functions from X to the empty set if X is nonempty.

2. Hello, jenjen!

#1 is a trick question . . .

1) Let $P$ be the set of all U.S. presidents,
and let $G$ be the set of all ordered pairs $(a,b) \in P \times P$
such that $b$ succeeded $a$ in office.
Is $G$ the graph of a function?

It is true that every President had a successor (well, except Dubya).

But Grover Cleveland served two nonconsecutive terms.
. . He was the 22nd President and was succeeded by Benjamin Harrison.
. . He was reelected as 24th President and was succeeded by William McKinley.
Hence, set $G$ contains: (Cleveland, Harrison) and (Cleveland, McKinley).

Therefore, $G$ is not a function.

3. Hey Soroban!!

Thank you so much for the quick reply!

4. Originally Posted by jenjen
]

2) Prove that for each set X there is a unique function from the empty set to X, regardless of whether or not X is nonempty. Also prove that there are no functions from X to the empty set if X is nonempty.
I do not know what definition you are using but there are no funtions between two sets if at least one is empty.

Because the Cartesian product between sets was defined for non-empty sets. Thus, a function can only between two non-empty set because it is a type of Cartesian product.


text:
A simple python package that solves sudokus using genetic algorithms

## Project description

# Genedoku

Solve sudokus of any size using genetic algorithms with an easy approach

## Basic usage

python
from genedoku.Evolution import Evolution
from genedoku.SudokuChromosome import SudokuChromosome

e = Evolution(problem, SudokuChromosome, 50, 20000)
r = e.start()

Where problem contains an array of NxN representing the sudoku and using 0 in the empty spaces.

## Test

bash
python tests/test_genedoku.py [options] < tests/example.txt

Where example.txt it's an input file containing the matrix in the following format:

Each row it's separated by '\n', each element in row per ' ' and the final line should be blank. Empty spaces are 0

## Project details

Uploaded py2


text:
| Home | E-Submission | Sitemap | Editorial Office |
 Korean J Child Stud > Volume 15(1); 1994 > Article
 Korean Journal of Child Studies 1994;15(1): 173-185.
 아동의 읽기에 대한 인식 김영실, 심성경 Children's Perception of Reading Young Sil Kim,Sung Kyoung Sim Abstract The purpose of this study was to investigate childrens perception of reading. The researchers interviewed 120 children aged 5 to 7 (40 for each age group) by using a specially designed interview format. The study of Johns (1973) was referred to as a guide in preparing the format. Responses of the children were classified in accordance with criteria worked out on the basis of the researchers previous studies and analyzed in terms of frequencies and percentiles by age group. χ² test was used to analyze the data. The findings of the study were: (1) The younger the children, the higher the rate of showing no response or of responding only to concrete incidents or things. Older children in general showed higher responses relevant to reading comprehension than younger ones. (2) Older children were more inclined than younger ones to regard the process of reading as a skill-oriented, teacher-based process. (3) Children in these age groups showed little perception of readings objectives of comprehension and communication.
TOOLS
Full text via DOI
 Related articles The Nature of Young Childrens Informational Picturebook Reading  2010 ;31(3) Childrens Play  2009 ;30(6) College Students Perception of Parental Roles and Childrearing attitude  2002 ;23(5) Young Children`s Perceptions and Responses to Negative Emotions  2002 ;23(2) Children's Understanding of Other's Beliefs  1999 ;20(1)


text:
# nLab 2-trivial model structure

Contents

## Model structures

for ∞-groupoids

### for $(\infty,1)$-sheaves / $\infty$-stacks

#### 2-Category theory

2-category theory

Definitions

Transfors between 2-categories

Morphisms in 2-categories

Structures in 2-categories

Limits in 2-categories

Structures on 2-categories

# Contents

## Definition

Every strict 2-category $K$ with finite strict 2-limits and finite strict 2-colimits becomes a model category (or, rather, its underlying 1-category does) in a canonical way, where:

• The weak equivalences are the equivalences.

• The fibrations are the morphisms that are representably isofibrations, i.e. the morphisms $e\to b$ such that $K(x,e)\to K(x,b)$ is an isofibration for all $x\in K$.

• The cofibrations are determined.

We call it the 2-trivial model structure, as it is a 2-categorical analogue of the trivial model structure on any 1-category. It can be said to regard $C$ as an (∞,1)-category with only trivial k-morphisms for $k \geq 3$.

## Properties

• Every object is fibrant and cofibrant.

• By duality, any such category has another model structure, with the same weak equivalences but where the cofibrations are the iso-cofibrations and the fibrations are determined. In $Cat$, the two model structures are the same.

## Examples

• In model categories built from various kinds of topological spaces, there are often analogous Hurewicz model structures. These are not actually examples of a 2-trivial model structure (for instance, the 2-category of spaces, continuous functions and homotopy classes of homotopies does not have finite limits as a 2-category, or even as a 1-category), but they share a common intuition and can sometimes be obtained as two instances of a more general construction.

## References

Last revised on December 19, 2021 at 23:14:02. See the history of this page for a list of all contributions to it.


text:
# The Unapologetic Mathematician

## Restricting and Inducing Representations

Two of the most interesting constructions involving group representations are restriction and induction. For our discussion of both of them, we let $H\subseteq G$ be a subgroup; it doesn’t have to be normal.

Now, given a representation $\rho:G\to\mathrm{End}(V)$, it’s easy to “restrict” it to just apply to elements of $H$. In other words, we can compose the representing homomorphism $\rho$ with the inclusion $\iota:H\to G$: $\rho\circ\iota:H\to\mathrm{End}(V)$. We write this restricted representation as $\rho\!\!\downarrow^G_H$; if we are focused on the representing space $V$, we can write $V\!\!\downarrow^G_H$; if we pick a basis for $V$ to get a matrix representation $X$ we can write $X\!\!\downarrow^G_H$. Sometimes, if the original group $G$ is clear from the context we omit it. For instance, we may write $V\!\!\downarrow_H$.

It should be clear that restriction is transitive. That is, if $K\subseteq H\subseteq G$ is a chain of subgroups, then the inclusion mapping $\iota_{K,G}K\hookrightarrow G$ is the exactly composition of the inclusion arrows $\iota_{K,H}K\hookrightarrow H$ and $\iota_{H,G}H\hookrightarrow G$. And so we conclude that

\displaystyle\begin{aligned}\rho\!\!\downarrow^G_K&=\rho\circ\iota_{K,G}\\&=\rho\circ\iota_{K,H}\circ\iota_{H,G}\\&=\left(\rho\circ\iota_{K,H}\right)\!\!\downarrow^G_H\\&=\left(\rho\!\!\downarrow^H_K\right)\!\!\downarrow^G_H\end{aligned}

So whether we restrict from $G$ directly to $K$, or we stop restrict from $G$ to $H$ and from there to $K$, we get the same representation in the end.

Induction is a somewhat more mysterious process. If $V$ is a left $H$-module, we want to use it to construct a left $G$-module, which we will write $V\!\!\uparrow_H^G$, or simply $V\!\!\uparrow^G$ if the first group $H$ is clear from the context. To get this representation, we will take the tensor product over $H$ with the group algebra of $G$.

To be more explicit, remember that the group algebra $\mathbb{C}[G]$ carries an action of $G$ on both the left and the right. We leave the left action alone, but we restrict the right action down to $H$. So we have a $G\times H$-module ${}_G\mathbb{C}[G]_H$, and we take the tensor product over $H$ with ${}_HV$. We get the space $V\!\!\uparrow_H^G=\mathbb{C}[G]\otimes_HV$; in the process the tensor product over $H$ “eats up” the right action of $H$ on the $\mathbb{C}[G]$ and the left action of $H$ on $V$. The extra left action of $G$ on $\mathbb{C}[G]$ leaves a residual left action on the tensor product, and this is the left action we seek.

Again, induction is transitive. If $K\subseteq H\subseteq G$ is a chain of subgroups, and if $V$ is a left $K$-module, then

\displaystyle\begin{aligned}\left(V\!\!\uparrow_K^H\right)\!\!\uparrow_H^G&=\mathbb{C}[G]\otimes_H\left(V\!\!\uparrow_K^H\right)\\&=\mathbb{C}[G]\otimes_H\mathbb{C}[H]\otimes_KV\\&\cong\mathbb{C}[G]\otimes_KV\\&=V\!\!\uparrow_K^G\end{aligned}

The key step here is that $\mathbb{C}[G]\otimes_H\mathbb{C}[H]\cong\mathbb{C}[G]$. But if we have any simple tensor $g\otimes h\in\mathbb{C}[G]\otimes_H\mathbb{C}[H]$, we can use the relation that lets us pull elements of $H$ across the tensor product. We get $gh\otimes1\in\mathbb{C}[G]\otimes_H\mathbb{C}[H]$. That is, we can specify any tensor by an element in $\mathbb{C}[G]$ alone.

November 23, 2010 -

1. I’m enjoying every step, John. Have a happy Thanksgiving.

In my daily 2,000 words of new fiction writing (over 300,000 words since the regimen was self-imposed on the advice of Stephen King and Ray Bradbury) I found myself this morning trying to explain why it would be preposterous to discover tensor calculus computations on a 3,750-year-old Babylonian clay tablet.

In this science fiction story, tablets are unburied which have Clebsch–Gordan coefficients and Calabi-Yau manifold tables. The poor NSA officer is trying to explain to the President of the United States.

Comment by Jonathan Vos Post | November 24, 2010 | Reply

2. […] want to work out the matrices of induced representations. Explicitly, if is a left -module of degree , where is a subgroup of , then is a left -module. […]

Pingback by Induced Matrix Representations « The Unapologetic Mathematician | November 25, 2010 | Reply

3. […] of Induced Representations We know how to restrict and induce representations. Now we want to see what this looks like on the level of […]

Pingback by Characters of Induced Representations « The Unapologetic Mathematician | November 29, 2010 | Reply

4. […] Today, we can prove the Frobenius’ reciprocity formula, which relates induced characters to restricted […]

Pingback by (Fake) Frobenius Reciprocity « The Unapologetic Mathematician | November 30, 2010 | Reply

5. […] Before we can prove the full version of Frobenius reciprocity, we need to see that induction and restriction are actually additive […]

Pingback by Induction and Restriction are Additive Functors « The Unapologetic Mathematician | December 1, 2010 | Reply

6. […] Now we come to the real version of Frobenius reciprocity. It takes the form of an adjunction between the functors of induction and restriction: […]

Pingback by (Real) Frobenius Reciprocity « The Unapologetic Mathematician | December 3, 2010 | Reply

7. […] really should see an example of inducing a representation. One example we’ll find extremely useful is when we start with the trivial […]

Pingback by Inducing the Trivial Representation « The Unapologetic Mathematician | December 6, 2010 | Reply

8. […] they’re restricted to , it also tells us how representations of decompose when they’re induced to […]

Pingback by The Branching Rule, Part 3 « The Unapologetic Mathematician | January 31, 2011 | Reply

9. […] we see, the right hand side is obvious, but he left takes a little more work. We consult the definition of induction to […]

Pingback by The Branching Rule, Part 4 « The Unapologetic Mathematician | February 1, 2011 | Reply


text:
# Exponential1D¶

class astropy.modeling.functional_models.Exponential1D(amplitude=1, tau=1, **kwargs)[source]

One dimensional exponential model.

Parameters
amplitudefloat, optional
taufloat, optional

Attributes Summary

 amplitude input_units This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted). param_names tau

Methods Summary

 evaluate(x, amplitude, tau) Evaluate the model on some input variables. fit_deriv(x, amplitude, tau) Derivative with respect to parameters

Attributes Documentation

amplitude = Parameter('amplitude', value=1.0)
input_units

This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted).

Model sub-classes can also use function annotations in evaluate to indicate valid input units, in which case this property should not be overridden since it will return the input units based on the annotations.

param_names = ('amplitude', 'tau')
tau = Parameter('tau', value=1.0)

Methods Documentation

static evaluate(x, amplitude, tau)[source]

Evaluate the model on some input variables.

static fit_deriv(x, amplitude, tau)[source]

Derivative with respect to parameters


text:
Free Version
Moderate

# Match Equivalent Capacitance: Capacitors in Series or Parallel

APPHEM-JWJ16E

A student is working on a circuitry project and wants to create an equivalent capacitance of $700 \mu F$. However, the student's only supply is a bag of $470 \mu F$ capacitors.

How can the student arrange any number of the $470 \mu F$ capacitors so that their equivalent capacitance is as close to $700 \mu F$ as possible?

A

Two capacitors in parallel.

B

Two capacitors in series.

C

Three capacitors in parallel.

D

Two capacitors in series connected to one other in parallel.

E

Two capacitors in parallel connected to one other in series.


text:
# How To Find Period Of Rotation

## How To Find Period Of Rotation?

Calculating Planetary-Rotation Periods

The planetary-rotation period can even be calculated from the velocity-distance-time equation T = D/V where D = distance in this case the circumference (2πR) and V = tangential velocity.Jun 19 2009

## How long is its period of rotation?

The period of rotation of a Solar System object is the length of time it takes that object to spin once around on its axis. For example the Earth takes 24 hours to spin once around its axis. Its period of rotation is 24 hours or a day.

## What is the formula for period?

… each complete oscillation called the period is constant. The formula for the period T of a pendulum is T = 2π Square root of√L/g where L is the length of the pendulum and g is the acceleration due to gravity.

58d 15h 30m

## How do you calculate the period of a wave?

The formula for period is T = 1 / f where “T” is period – the time it takes for one cycle to complete and “f” is frequency. To get period from frequency first convert frequency from Hertz to 1/s. Now divide 1 by the frequency. The result will be time (period) expressed in seconds.

## What is a period in rotational physics?

Period is defined as the amount of time it takes to go around once – the time to cover an angle of radians. Frequency is defined as the rate of rotation or the number of rotations in some unit of time. … These three quantities are related by f = 1 T = ω 2 π .

## How do you calculate period in math?

We may also calculate the period using the formula derived from the basic sine and cosine equations. The period for function y = A sin(Bx + C) and y = A cos(Bx + C) is 2π/|B| radians. Frequency is defined as the number of cycles completed in one second.

## How do you calculate time period and frequency?

In the frequency formula f = 1/T T is the time period. T refers to the time to complete one cycle (in seconds). Time period is inversely proportional to the frequency.

## How long is Saturn’s rotation period?

10 hours 45 minutes 45 seconds
Cassini took readings of the day-length indicator regarded as most reliable the rhythm of natural radio signals from the planet. The results give 10 hours 45 minutes 45 seconds (plus or minus 36 seconds) as the length of time it takes Saturn to complete each rotation.Jun 28 2004

## What is Venus rotation period?

Venus: Planet Profile
Mass (kg) 4.87 x 1024
Rotation period (length of day in Earth days) 243.02 (retrograde)
Revolution period (length of year in Earth days) 224.7
Obliquity (tilt of axis degrees) 178
Orbit inclination (degrees) 3.39

225 days

## How do you find the period in physics circular motion?

The distance around a circle is equivalent to a circumference and calculated as 2•pi•R where R is the radius. The time for one revolution around the circle is referred to as the period and denoted by the symbol T. Thus the average speed of an object in circular motion is given by the expression 2•pi•R / T.

## What is a complete cycle of a wave?

One cycle of a wave is one complete evolution of its shape until the point that it is ready to repeat itself. … Frequency is the number of complete cycles that a wave completes in a given amount of time. Usually measured in Hertz (Hz) 1 Hz being equal to one complete wave cycle per second.

See also how long does it take to go to the sun

## How do you calculate period and velocity?

In other words frequency and period are reverse of one another. Since the formula for velocity (speed actually since we are looking for the magnitude of velocity) is given by distance divided by time we can simply find the circumference and divide it by the period to get the magnitude of velocity of the object.

## How do you find period given radius and velocity?

So r = d/2. Note also the symbol for the period: T. With this equation given an orbiting object’s speed and the radius of the circle you can calculate the object’s period.

## Where is the period in a sine equation?

If we have a sine function of the form f(x) = Asin(Bx + C) + D then the period of the function is 2π / |B|.

## What is the period of Sin2x?

What is the Period of Sin2x? The period of sin bx is (2π)/b in general. So the period of sin 2x is (2π)/2=π.

## How do you calculate period time?

The formula for frequency is: f (frequency) = 1 / T (period). f = c / λ = wave speed c (m/s) / wavelength λ (m). The formula for time is: T (period) = 1 / f (frequency).

## What is the period of 5 in 5934692?

The digit 5 occurs in Lakhs period. The place of 5 is Ten lakhs.

## What is periods in number system?

When a number is written in standard form each group of digits separated by a comma is called a period . The number 5 913 603 800 has four periods.

116d 18h 0m

## How long does it take for Uranus to rotate?

QUICK FACTS (Data is from NASA Goddard)
Average distance from Sun 1.8 billion miles
Sidereal Rotation 17.24 Earth hours
Length of Day 17.24 Earth hours
Sidereal Revolution 84 Earth years
Diameter at Equator 31 693 miles

## How long is the rotation of Uranus?

Uranus takes about 17 hours to rotate once (a Uranian day) and about 84 Earth years to complete an orbit of the Sun (a Uranian year).

## What is Mars rotation period?

24.6 hours
Orbit and Rotation As Mars orbits the Sun it completes one rotation every 24.6 hours which is very similar to one day on Earth (23.9 hours). Martian days are called sols – short for “solar day.” A year on Mars lasts 669.6 sols which is the same as 687 Earth days.Jul 8 2021

See also what determines the number of phenotypes for a given trait

## Do all planets spin?

The planets all revolve around the sun in the same direction and in virtually the same plane. In addition they all rotate in the same general direction with the exceptions of Venus and Uranus. These differences are believed to stem from collisions that occurred late in the planets’ formation.

## Is the moon’s orbit clockwise or counterclockwise?

As seen from the north side of the moon’s orbital plane the Earth rotates counterclockwise on its rotational axis and the moon revolves counterclockwise around Earth.

## What is the period of revolution of each planet?

Planet Period of Revolution
Earth 365 days
Mars 687 days
Jupiter 11.9 years
Saturn 29.5 years

## How do you calculate the period of revolution of a satellite?

Note: We can also solve this equation by directly using the formula given to find time period of a satellite at a height of[h]from earth surface which is given as [T=2pi sqrt{dfrac{{{left( {{R}_{E}}+h right)}^{3}}}{g{{R}_{E}}^{2}}}].

## How is period related to wavelength?

So in a time of one period the wave has moved a distance of one wavelength. Combining this information with the equation for speed (speed = distance/time) it can be said that the speed of a wave is also the wavelength/period.

## What is the dominant period of a wave?

The dominant wave period (in seconds) is a wave period associated with highest energetic waves at a specific point or area in the total wave spectrum and is always either the swell period or the wind- wave period. Dominant wave period is also known as the “peak” period.

Categories FAQ


text:
# thraxil.org:

## Train in Vain

R.I.P. Joe…

(more as I leave work/get drunk…)

## All Things Punk

OK, so I saw The Distillers this past weekend. What souvenir did I pick up? My best guess is either a cracked or bruised rib. Signs pointing to bruised: it didn’t really hurt until Monday (read: it only hurt when I took a deep breath). Signs pointing to cracked: from Monday onwards, it feels like Doubting Thomas himself has decided to poke around my chest, eschewing his fingers for a hot poker.

Anyhoo, the show was a blast. Anytime the headlining act clears the floor for an all-female slam-dance affair with the phrase, “Vaginas Only” it’s worth the price of admission. Especially when the song (punk rock song, mind you) was all about women’s suffrage. So their detractors will call them a bunch of Hole wannabes. I’ve yet to see their detractors stand up to meself or anyone there. Huff huff.

Punk’s not dead, and with cool chicks like Brodie, it’ll be a long, wonderful, painful death while peeing on the graves of the Backdoor Boys.

## Satan Oscillate my metallic sonatas

OK, so upon reading my last post, it dawned on me that the first line actaully was moderately congruous with the remainder of the message. Hooray for serendipity. In the interests of veracity, here’s the real story…

So I was a few sheets to the wind a few nights past when I sent an e-mail out to a former significant other o’ mine. ‘Twas her birthday and all, and we hadn’t spoken in about a year. That was my Caesar-ish Rubicon quip. The fact that it blended in so wonderfully with my present employer’s ignorance of my skills just made things a tad more ironic.

Thppt.

One of these days, I’m gonna make a stone cold sober report…

## Was ist ist

The die is cast…

In Viagra-land news, their choice to name me as the “document coordinator” has’t bitten them on the heiny as much as even I thought it would. I guess even the likes of yours truly can clean up his organizational act enough to pick up a paycheck at the end of the week.

‘Course it’s still all over but the crying once I’m off the clock. (read: somewhere in this mess that I call a home, maybe I can find some clean clothes…)

## Stray Bullets

So this weekend, while supping upon cheap pizza, I was perusing my “old” copies of Stray Bullets (everyone’s favorite serialized vision of sex, violence, and uh… violence) when I noticed that there was a letter by me in the back… y’know, where all the geeks write in to seem cool.

OK, so first off, I really f-ed up the number of one of the first books. Yup, go Williams.

Second off, it dawned on me that I’d written that letter several months ago in an attempt to seem cordial while trying to russel up a free bumper sticker. Yeah, no bumper sticker came, so I completely forgot about it. Freaked me out when I finally read it yesterday.

Anyhow, now all you comic book nuts can read and say, “I knew that boy when he thought Knox gelatin was a hair product.” Or not. Whatever. Stray Bullets #25… it’s not even that good an issue, but still Stray Bullets, so it kicks butt.

Back to the land of dreams…

## ten animals I slam in a net

Well, no ATR today… but then again, I’ve avoided getting stuck with asinine assignments. ‘Course that’s mainly due to the fact that I’ve physically avoided those responsible for assigning said tasks. >sigh< Ain’t life grand in the Dilbert world.

So instead, I’m listening to the new album by Common Rider – the band fronted by Jesse of Operation Ivy fame. Zen-Punk, and fun for the whole family. Steal your copy today!

## the saga continues

Well, another day has come and gone at the Viagra Factory (or “Pfactory” as the wags down there would put it). I’ve recently stepped into my new shoes as the “Document Coordinator” for our building. Which, as far as I can tell, makes me the archive bitch. For all of those who knew me in college, I’ll admit freely that my organizational abilities have not improved in the time since graduation… as plainly evidenced by the state of my apartment. And now I’m in charge of shipping junk off to be microfilmed, shipping junk off to be filed somewhere in the bowels of Pfizer, and pretty much every other mundane task associated with data storage.

The one saving grace is that I’m in my very own private abode (the archive room) where I can listen to muzak at the volume to which I’m accustomed. So far it’s been mainly up to the Rancid and Dropkick Murphys level of mayhem. Today, tho, I had to go over an ass-load of SOPs coming up from NJ… I’m bringing in the ATR tomorrow, in case I get stuck with that crap again.

Yeah, work is work is work… it is, however, an endless supply of kooks, weirdos, and other misfits who wear their social problems on their sleeves. Oh, and there are some real losers, too.  It’s pretty much an amalgam of “Dilbert” and the place where Dan Akroyd works at the beginning of “Spies Like Us.” I think it was about my second day there when I was roped into making a prank phone call to some poor lady at the credit union.

I think in the meantime, we’re supposed to be testing pharmaceutical products for impurities and/or degredants… but really, it’s like a cGMP deGrassi high.

Now, if only Connecticut sold beer past 8 p.m. …

## Rabbit Rabbit

Hey, it’s October 1st. Thought for the day: Why isn’t the person responsible for that “skaterboy” sone strung up by her entrails? Or at least poked repeatedly with hot forks (y’know, for those of you opposed to the whole capital punishment shebang)

## test

yrs truly, the Rev, is horribly drunk right now… Testing to see if I can post this to me diary…


text:
KAON '99 Conference Proceedings

*** CONTRIBUTIONS ARE PAST DUE! SEND YOUR MANUSCRIPT TO TED QUINN AT TCQUINN@RAINBOW.UCHICAGO.EDU ASAP! ***

We plan to issue full proceedings for our conference. These will be published by The University of Chicago Press. The rather high standards of the Press speak against the volume being a simple assembly of all contributions. The editors will attempt to provide introductory prose for the various sections of the book and to see that the contributions properly cross-reference each other. In the end, this should enhance the value of the proceedings.

It will be necessary to satisfy the Press' guidelines which are somewhat more demanding than those for conference proceedings. The Press has recently published a volume in memory of S. Chandrasekhar, edited by R. M. Wald, which may guide your thinking in this respect. (A copy will be on display at Kaon 99.) We feel this is appropriate in view of the tremendous strides which have been made in the study of kaon decays in the past few years. We are hoping for articles of the quality of that by J. S. Bell and J. Steinberger in the Proceedings of the 1965 Oxford Conference on Elementary Particles, which has been quoted consistently for more than thirty years.

To begin with, the Press will want to send the completed manuscript of the volume out to two peer reviewers, chosen from a list that we will provide. It is possible that revisions of some contributions will be requested, though this will be less likely if all speakers conform to the Guidelines for Contributors noted below. In order to ensure timely publication, we are asking speakers to have LaTeX files with encapsulated PostScript figures in our hands by July 31. In the rare cases of speakers who do not use this format, we would like to receive camera-ready copy by that date. If you wish to arrive in Chicago with a completed draft, all the better, since that will facilitate coordination with other speakers.

A second point which we wish to stress is that the effective integration of contributions requires that all contributors plan to attend as much of the conference as possible, so that contributions can be properly cross-referenced. In the rare cases where that is not possible, we ask you to let us know ahead of time so that we can make arrangements for you to be informed of what other contributors will be providing. We do not want to produce a volume of disjoint chapters but rather an integrated picture of the state of the field at the turn of the millennium. It would be extremely helpful, therefore, if you would contact other speakers in your sessions or on related subjects in other sessions to coordinate your presentations.

Your written contribution should forego some specialized details in favor of a more broadly based exposition. Think of it somewhat more like a colloquium rather than a specialized seminar, and bear in mind that the volume should be one which is likely to be valuable as supplemental material in a graduate physics course as well as to your more advanced colleagues.

We wish to keep the length of the book to 500 pages. With about 1500 minutes of presentations (300 per day), this works out to one page for every three minutes. Although this may seem quite stringent, we are confident that you will manage to meet the goal. Specifically, please keep your contributions to the following length:

 Length of presentation (min.) Pages in volume 10 (panelists) 3 15 5 20 7 25 8 30 10 45 (summary only) 15

With your cooperation, we expect to produce a volume which will not be just another Conference Proceedings but an integrated portrait of the field of kaon decays which will be read by physicists for many years in the future.

Guidelines for Contributors

We will be assembling all of the contributions to the conference into a book, which will be typeset using LaTeX. The most important thing that you can do to assist us in this process is to prepare your contribution in the default LaTeX article style, avoiding the use of any specialized style files or visual formatting commands. The reason is that any such special formatting will have to be undone in order to make your contribution conform to the style of the book.

Figures can be included in your contribution using the standard epsfig package (or the graphics package if you are using LaTeX2e).

Your list of references should be produced with the standard LaTeX commands
(\begin{thebibliography} ... \end{thebibliography})
and referred to in the text with the \cite command.

To estimate the length of your contribution as it will appear in the final book, begin your LaTeX file with the following commands:

\documentstyle[12pt]{article}

For LaTeX2e users, the first line should instead be \documentclass[12pt]{article}


text:
# Why does pdflatex write to toc-file \cyr-commands even with utf8 document encoding?

Why does pdflatex (but not lualatex) write to toc-file \cyr-commands even with utf8 document encoding instead of normal cyrillic letters? How to prevent it? I interested in solution with redefining book.cls commands. I redefined the \addcontentsline{toc}{part}{\unexpanded\expandafter{{#1}}}% but it has no result.

% !TeX program = pdflatex
% !TeX encoding = utf8
% !TeX spellcheck = uk_UA
% !BIB program = bibtex8

\documentclass[12pt]{book}

\usepackage[a4paper]{geometry}

\usepackage[utf8]{inputenc}
\usepackage[T2A,T1]{fontenc}
\usepackage[ukrainian]{babel}

\usepackage{indentfirst}
\usepackage{misccorr}

\makeatletter
\def\@part[#1]#2{%
\thispagestyle{empty}
{\centering
\interlinepenalty \@M
\normalfont
\Huge \sffamily\bfseries #1\par}%
\@endpart
}
\makeatother

\begin{document}
\tableofcontents
\part{Перша}

іва
\part{Друга}
іфавіфав
\end{document}

Toc looks like

\select@language {ukrainian}
\contentsline {part}{{\IeC {\CYRP }\IeC {\cyre }\IeC {\cyrr }\IeC {\cyrsh }\IeC {\cyra }}}{3}
\contentsline {part}{{\IeC {\CYRD }\IeC {\cyrr }\IeC {\cyru }\IeC {\cyrg }\IeC {\cyra }}}{7}

• Note that \unexpanded\expandafter{{#1}} does nothing useful with the \expandafter, which tries to expand {. You can use \unexpanded{\unexpanded{#1}} instead, but I'm not sure why bother: the auxiliary files are not for being human readable. Mar 26 '16 at 18:19

You can try with \detokenize.

For use with hyperref: one needs both of \texorpdfstring and \phantomsection in the patch of \@part.

\documentclass[12pt]{book}

\usepackage[a4paper]{geometry}

\usepackage[T2A,T1]{fontenc}

\usepackage[utf8]{inputenc}

\usepackage[ukrainian]{babel}

\usepackage{indentfirst}
\usepackage{misccorr}

% update for hyperref compatibility

\makeatletter
\def\@part[#1]#2{%
\thispagestyle{empty}%
\phantomsection
{\centering
\interlinepenalty \@M
\normalfont
\Huge \sffamily\bfseries #1\par
}%
\@endpart
}
\makeatother

\usepackage[pdfencoding=auto]{hyperref}

\begin{document}
\tableofcontents
\part{Перша}

іва
\part{Друга}
іфавіфав
\end{document}

Contents of toc file:

\select@language {ukrainian}
\contentsline {part}{Перша}{3}
\contentsline {part}{Друга}{7}

and with hyperref:

\select@language {ukrainian}
\contentsline {part}{Перша}{3}{section*.2}
\contentsline {part}{Друга}{7}{section*.3}

• thanx. But I got problem with hyperref package, the bookmarks now become unreadabl. Mar 26 '16 at 14:44
• @sergiokapone see update.
– user4686
Mar 26 '16 at 15:01
• @sergiokapone there is a problem with hyperlinks, though.
– user4686
Mar 26 '16 at 15:19
• with your updates it works fine Mar 26 '16 at 15:23
• I needed to add \phantomsection to your \@part patch, else the hyperlinks either from bookmarks of from TOC did not work with me.
– user4686
Mar 26 '16 at 15:35

In principle at least different parts of the document can be using different input encodings so the mechanism was designed to always write auxiliary files in the encoding neutral LICR ascii markup.

In practice, especially of you are using UTF-8, a single encoding is used for the whole document, but UTF-8, and UTF-8 support on LaTeX are some years younger than the design of the encoding support for toc files.

• As I understand you, there is no solution due to restrictions of pdaflatex? Mar 26 '16 at 13:08
• @sergiokapone no, you could write out and read back in utf8 but you'd want to disable almost all of the inputenc mechanism which is designed to convert input into the markup form, which can then be used for file writing or different font encodings. So to do a proper job of it it is a bigger change than is worth doing for a toc file which is just an internal file. It is more important to be able to write utf8 when making terminal output (or pdf bookmarks) where tex macro syntax is less than ideal. But your question asked why is it that way, so I addressed the "why" not how to change it. Mar 26 '16 at 13:17


text:
+0

0
39
1

Let a and b be the roots of 2x^2 - x - 2 = 0.  Find the value of 1/a + 1/b.

May 6, 2020

#1
+24992
+1

Let a and b be the roots of $$2x^2 - x - 2 = 0$$.
Find the value of $$\dfrac{1}{a} + \dfrac{1}{b}$$.

$$\text{Let \dfrac{1}{a} and \dfrac{1}{b} be the roots of 2\left(\dfrac{1}{x}\right)^2 - \dfrac{1}{x} - 2 = 0 }$$

$$\begin{array}{|rcll|} \hline \mathbf{2\left(\dfrac{1}{x}\right)^2 - \dfrac{1}{x} - 2} &=& \mathbf{0} \\\\ \dfrac{2}{x^2} - \dfrac{1}{x} - 2 &=& 0 \quad | \quad \cdot x^2 \\\\ 2 - x - 2x^2 &=& 0 \\\\ \mathbf{2x^2+x-2} &=& \mathbf{0} \\\\ x &=& \dfrac{-1\pm \sqrt{1^2-4*2*(-2)} } {2*2} \\\\ x &=& \dfrac{-1\pm \sqrt{17} } {4} \\\\ x_1 = \dfrac{1}{a} &=& \dfrac{-1+ \sqrt{17} } {4} \\\\ x_2 = \dfrac{1}{b} &=& \dfrac{-1- \sqrt{17} } {4} \\\\ x_1+x_2 = \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-1+ \sqrt{17} } {4}+\dfrac{-1- \sqrt{17} } {4} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-1+ \sqrt{17} -1- \sqrt{17}} {4} \\\\ \dfrac{1}{a} + \dfrac{1}{b} &=& \dfrac{-2} {4} \\\\ \mathbf{\dfrac{1}{a} + \dfrac{1}{b}} &=& \mathbf{ -\dfrac{1} {2} } \\ \hline \end{array}$$

May 7, 2020


text:
# Density through Stress-distance graph!

Rod of constant cross-section moves towards right with constant acceleration. Graph of stress and distance from left end is given as in figure. If density of material of rod at cross section $$1$$ is $$9 \frac{g}{cm^3}$$ . Find density at cross section $$2$$ in $$\frac{g}{cm^3}$$.

×


text:
# Class 6 Maths NCERT Solutions for Chapter 14 Practical Geometry Ex – 14.3

## Practical Geometry

Question 1.
Draw any line segment $\overline { PQ }$. Without measuring $\overline { PQ }$, construct a copy of $\overline { PQ }$.

Solution:
Steps of Construction:

Let PQ be the given line segment and l be a given line.

1. Mark a point 0 on the line l.
2. Open out the compasses and adjust so that the steel end is on A and the pencil end is A on B.
3. Transfer the compasses to the line l without disturbing their opening so that the steel end is on 0.
4. With the pencil end make a small stroke on the line l to cut it at the point P.
Then, the fine segment OP so obtained = the given line segment AB.

Question 2.
Given some line segment $\overline { AB }$, whose length you do not know, construct $\overline { PQ }$ such that the length of $\overline { PQ }$ is twice that of $\overline { AB }$.

Solution:
Steps of Construction:

Let AB be the given line segment.Draw any line l and make a point 0 on it.

1. Open out the compasses in such way that the steel end is on A and the pencil end is on B.
2. Transfer the compasses without disturbing their opening to the line l so that the steel end is on O.
3. With the pencil end make a small stroke on -the line l to cut it at the point P.
4. Repeat the steps 3 and 4 with same opening having P as the initial point and Q as the terminal point.
Then, the segment OQ = OP + PQ = AB + AB= 2 AB.

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text:
## LaTeX forum ⇒ Kile ⇒ Automatic LaTeX Equivalents for Spanish Accents

Information and discussion about Kile, an integrated LaTeX environment for Linux KDE
Jueves89
Posts: 2
Joined: Mon Aug 26, 2013 10:17 am

### Automatic LaTeX Equivalents for Spanish Accents

Hi all,

I have been using Kile a lot of time on Debian, but now I have to use it on Windows 7 x64. On Debian, when I type an accent, Kile automatically inserts it as \´a when compiling it appears as "á". But now on Windows when I do the same, Kile inserts directly "á" and I don't like it, I prefer the other method because of compatibility.

I'm using Kile 2.1 with KDE 4.10.2, I have also tried with Kile 3 beta, in both cases with and without Spanish spelling package. I always have the option "Automatically insert the LaTeX equivalent of special characters when typing (accents, etc)" activated.

Wish someone can help me.
Thanks

Tags:

localghost
Site Moderator
Posts: 9206
Joined: Fri Feb 02, 2007 12:06 pm
Location: Braunschweig, Germany
Jueves89 wrote:[…] But now on Windows when I do the same, Kile inserts directly "á" and I don't like it, I prefer the other method because of compatibility. […]

Looks to me like a bug. Try to contact the developers and ask them for help. But perhaps you should abandon this compatibility in favor of typesetting accents as Unicode characters in a modern way. With Kile you can easily search and replace the critical parts.

A Spanish document in which you can use accented characters can easily be set up in the following way.
\documentclass[spanish]{scrartcl}\usepackage[T1]{fontenc}\usepackage{selinput}\SelectInputMappings{   aacute={á},   ntilde={ñ}}\usepackage{babel} \begin{document}   á ñ\end{document}

Finally you could also use Xe(La)TeX or Lua(La)LaTeX to typeset your documents. These typesetting engines support Unicode by default but need other document setups.

Best regards and welcome to the board
Thorsten
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1

Jueves89
Posts: 2
Joined: Mon Aug 26, 2013 10:17 am
I also think it may be a bug. I'll try to contact the developers as you say. If I have any answer I'll tell you.

My preamble looks like is.
\documentclthass[a4paper,12pt]{article}\usepackage[T1]{fontenc}\usepackage[latin1]{inputenc}\usepackage[english,spanish]{babel}\usepackage[   a4paper,   textheight=22.3cm,   headsep=1.0cm,   footskip=13cm,   left=2.5cm,   right=2.5cm,   voffset=0.0cm]{geometry}\usepackage{enumerate}\usepackage{times}\usepackage{calc}\usepackage{graphicx}\usepackage{graphics}\usepackage{soul}\usepackage{tabularx}\usepackage{textcomp}\usepackage[dvips]{epsfig}\usepackage{ae}\usepackage{multirow}\usepackage{array}\usepackage{amssymb}\usepackage{amsmath}\usepackage{fancybox}\usepackage{boxedminipage}\usepackage{longtable}%\usepackage{ltablex}\usepackage{fancyhdr}\usepackage{rotating}\usepackage{subfigure}\usepackage{color}\usepackage{multicol} \providecommand{\providehyphenmins}[2]{}

When compiling I always have good results. Accents and ñ always appear. It doesn't matter if I insert it as \'a or á.

Return to “Kile”

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text:
# Table for $$A_5(9,d;k)$$

d\k 2 3 4
4 0.0 0.0459644629343 0.04856345689
6 0.0 0.00575501054655
8 0.0


text:
# Fourier Transform of $f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$

I was trying to take the FT of

$$f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$$

This is just the shifting rule applied to the FT of

$$g(x) = \exp(-\pi ax^{2})$$

which is given by

$$\hat g(k) = \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{-k^{2}}{4\pi a} \bigg)$$

Hence, we should get

$$\hat f(k) = \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{- (k – 2\pi b)^{2}}{4\pi a} \bigg)$$

However, when trying to derive this result I got stuck and if someone could help me I would really appreciate it.

We have that

$$\hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx$$

Differentiating wrt to $k$ and integrating by parts, we find

\begin{align} \hat f'(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) (-ix) \exp(-ikx) dx \\ &= \frac{i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} -x \exp(-\pi ax^{2}) \exp(i(2 \pi b – k)x) dx \\ \end{align}

with

\begin{align} u &= \exp(i(2 \pi b – k)x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{\exp(-\pi ax^{2})}{2\pi a} \\ u’ &= i(2 \pi b – k)\exp(i(2 \pi b – k)x) \ \ \ \ \ v’ = -x \exp(-\pi ax^{2}) \end{align}

Hence we get

\begin{align} \hat f'(k) = \frac{i}{\sqrt{2\pi}} \bigg[\frac{\exp(-\pi ax^{2}) \exp(i(2 \pi b – k)x)}{2\pi a}\Biggr \rvert_{-\infty}^{\infty} \\ – \bigg(\frac{i(2 \pi b – k)}{2\pi a} \bigg) \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ \end{align}

noting that the evaluated expression equals $0$. Therefore

\begin{align} \hat f'(k) &= \frac{i}{\sqrt{2\pi}} \bigg[- \bigg(\frac{i(2 \pi b – k)}{2\pi a} \bigg) \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ &= \frac{2\pi b – k}{2\pi a} \bigg[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ &= \frac{2\pi b – k}{2\pi a} \hat f(k) \end{align}

Solving, we find

\begin{align} \hat f(k) &= \hat f(0) \exp \bigg(\frac{4\pi bk – k^{2}}{4\pi a} \bigg) \\ &= \hat f(0) \exp \bigg(\frac{- (k – 2\pi b)^{2} + 4\pi^{2}b^{2}}{4\pi a} \bigg) \end{align}

The only way I can see to get the correct result is if

\begin{align} \hat f(0) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) dx \\ &= \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{-4\pi^{2}b^{2}}{4\pi a} \bigg) \end{align}

and the only way I thought I may be able to solve the integral for $\hat f(0)$ is to change to polar coordinates.

So, does anyone have suggestions on how to solve the $\hat f(0)$ integral?

#### Solutions Collecting From Web of "Fourier Transform of $f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$"
$$\exp \left(ax^2 +bx\right) = \exp \left(\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2 -\frac{b^2}{4a} \right),$$ for given constants $a$ and $b$.
If your goal to find the Fourier transform of the function $f(x) = e^{-\pi ax^{2} + 2\pi ibx}$ then here is an easier approach
$$F(k) = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + 2\pi ibx}e^{-ikx}dx = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + (2\pi b-k)ix}dx.$$


text:
# Statistical models in terms of families of random variables

A statistical model is a function $P : \Theta \to \Delta(X)$, where $\Theta$ is a parameter space, and $\Delta(X)$ is the set of probability measures on a state space $X$.

Suppose that $\Theta$ and $X$ are topological spaces, and equip $\Delta(X)$ with the topology of weak convergence of measures. Suppose that the function $P : \Theta \to \Delta(X)$ is continuous, defining a topological statistical model.

An $X$-valued random variable is a measurable function $x : \Omega \to X$ on some abstract probability space $(\Omega, \mathcal F, \mathbb P)$. The push-forward measure $x_* \mathbb P := \mathbb P \circ x^{-1}$ is called the law or marginal of $x$.

If $P : \Theta \to \Delta(X)$ defines a statistical model, can we find an abstract probability space $(\Omega, \mathcal F, \mathbb P)$ and a family of $X$-valued random variables $\{x_\theta\}$ so that $(x_\theta)_* \mathbb P = P_\theta$ for each $\theta \in \Theta$?

Moreover, if $P$ defines a topological statistical model, what regularity properties does this imply about the higher-order function $\theta \mapsto x_\theta$? If so, can the space $L = L(\Omega,X)$ of $X$-valued random variables be equipped with a natural topology so that the function $x : \Theta \to L$ is continuous?

• For the nontopological case, you can simply take $\Omega=X^\Theta$, $\mathcal{F}$ the product $\sigma$-algebry, $\mathbb{P}=\otimes_{\theta\in\Theta}P_\theta$ and $x_\theta$ the projection onto the $\theta$'s factor. – Michael Greinecker Nov 16 '13 at 8:51
• I'm kind of lost: as Michael Greinecker recalled, you can always build a huge probability space in which all your variables live, whatever $\Theta$ is, thanks to the tensor product construction. Then, maybe the "natural" topology you look for just the convergence in distribution of random variables ? In this case, that $x:\Theta\rightarrow L$ is continuous is by definition equivalent to $P:\Theta\rightarrow \Delta(X)$ continuous. But maybe I misunderstood the question. – Adrien Hardy Nov 16 '13 at 9:58
• The tricky part is regularity, then. This is the whole point of Kolmogorov's continuity theorem: the product construction does not guarantee regularity. I have to think about @John Dawkins' answer a little more. – Tom LaGatta Nov 16 '13 at 14:38

D. Blackwell and L. Dubins [An extension of Skorohod's almost sure representation theorem, Proc. Amer. Math. Soc., vol. 89} (1983) 691–692] have shown that is $X$ is a reasonable space then there is a map $P\to f_P$ of $\Delta(X)$ into $L$ (based on a well-chosen probability space $(\Omega,{\mathcal F}, {\mathbb P})$) such that, for each $P\in\Delta(X)$, (i) the push forward of ${\mathbb P}$ by $f_P$ is $P$ and (ii) $Q\mapsto f_Q(\omega)$ is continuous at $P$, for a.e. $\omega\in\Omega$.


text:
# American Institute of Mathematical Sciences

December  2019, 11(4): 447-485. doi: 10.3934/jgm.2019023

## Networks of coadjoint orbits: From geometric to statistical mechanics

 Department of Mathematics, Imperial College, London SW7 2AZ, UK

* Corresponding author: So Takao

To Darryl Holm, for his 70th birthday

Received  April 2018 Revised  May 2019 Published  November 2019

Fund Project: AA acknowledges EPSRC funding through award EP/N014529/1 via the EPSRC Centre for Mathematics of Precision Healthcare. ST acknowledges funding through Schrödinger scholarship scheme

A class of network models with symmetry group $G$ that evolve as a Lie-Poisson system is derived from the framework of geometric mechanics, which generalises the classical Heisenberg model studied in statistical mechanics. We considered two ways of coupling the spins: one via the momentum and the other via the position and studied in details the equilibrium solutions and their corresponding nonlinear stability properties using the energy-Casimir method. We then took the example $G = SO(3)$ and saw that the momentum-coupled system reduces to the classical Heisenberg model with massive spins and the position-coupled case reduces to a new system that has a broken symmetry group $SO(3)/SO(2)$ similar to the heavy top. In the latter system, we numerically observed an interesting synchronisation-like phenomenon for a certain class of initial conditions. Adding a type of noise and dissipation that preserves the coadjoint orbit of the network model, we found that the invariant measure is given by the Gibbs measure, from which the notion of temperature is defined. We then observed a surprising 'triple-humped' phase transition in the heavy top-like lattice model, where the spins switched from one equilibrium position to another before losing magnetisation as we increased the temperature. This work is only a first step towards connecting geometric mechanics with statistical mechanics and several interesting problems are open for further investigation.

Citation: Alexis Arnaudon, So Takao. Networks of coadjoint orbits: From geometric to statistical mechanics. Journal of Geometric Mechanics, 2019, 11 (4) : 447-485. doi: 10.3934/jgm.2019023
##### References:

show all references

##### References:
In this figure, we plot the maximum real part of the eigenvalues $\lambda_S$ of the linearised system (58) corresponding to equilibrium solutions ${\boldsymbol \Pi}_e$ (eigenvectors of $\mathbb L$) with energy $\lambda_e$ (eigenvalue of $\mathbb L$ corresponding to eigenvector ${\boldsymbol \Pi}_e$). The blue crosses indicate the maximum real eigenvalues that are non-zero, corresponding to unstable solutions, and the red dots indicate those that are zero, corresponding to linearly stable solutions. The purple dots indicate the algebraic multiplicity $n$ of $\lambda_e$, whose values are to be read from the right axis. As expected, the equilibria corresponding to the highest and lowest energy are stable in both cases
In this figure, we display four non-trivial examples of linearly stable equilibrium solutions of the $20\times 20$ rigid body lattice. All the solutions are parallel to one of the axes (either $\Pi_1$ or $\Pi_3$) and have an 'argyle' pattern of aligned spins (2(a), 2(d)) or anti-aligned spins (2(b), 2(c)). The former pattern corresponds to configurations with low energy (negative values of $\lambda$), and the latter corresponds to configurations with high energy (positive values of $\lambda$). All four solutions given here correspond to the red dots in figure 1
This figure shows the possible $(\lambda_1, \lambda_2)$-pairs that one can have from solving equation (86). Fixing $\lambda_1$, the possible values of $\lambda_2$ that solve (86) are the eigenvalues of the graph Laplacian $\mathbb L(\lambda_1)$. We see that the eigenvalues $\lambda_2$ in the left panel 3(a) collapse as $\lambda_1 \rightarrow 0$, which does not occur in the right panel 3(b)
The two panels in this figure display the maximum, real part of the eigenvalues $\lambda_S$ of the linearised system (98) around the equilibrium configuration ${\boldsymbol \Gamma}_e$ (eigenvectors of $\mathbb L(\lambda_1)$) plotted against $\lambda_2$ (minus the eigenvalue of $\mathbb L(\lambda_1)$ corresponding to ${\boldsymbol \Gamma}_e$) for $\lambda_1 = 0.5$. Again, the blue crosses correspond to unstable equilibria, the red dots correspond to linearly stable equilibria and the purple dots correspond to the multiplicity of $\lambda_2$, whose values are to be read from the right axis. As expected, the configuration corresponding to the highest $\lambda_2$ (or, the lowest eigenvalue of $\mathbb L(\lambda_1)$) is stable in both cases
This figure shows the time evolution of the averaged spins (5(a), 5(c), 5(e)) and energy (5(b), 5(d), 5(f)) in a $20\times 20$ heavy top lattice, starting from different initial conditions, with dissipation (dashed lines) or without dissipation (bold lines). The blue, orange and green lines in the left panel correspond to the $\Gamma_1$, $\Gamma_2$ and $\Gamma_3$-components of the spins averaged over the lattice respectively and the same colours in the right panel correspond to the total, kinetic and potential energy. We observe metastability of the $\Gamma_1$ and $\Gamma_2$ axes in panels 5(c) and 5(a) in the dissipative case, and a partial synchronisation around the $\Gamma_3$-axis in the non-dissipative case in 5(a), 5(c) and 5(e)
The two panels in this figure show the temperature phase transition of the rigid body lattice for two different cases. The blue dots correspond to averaged magnetisation computed via direct simulation of the full stochastic equation (47) and the black line corresponds to the mean field approximation (103). Both cases exhibit a second order phase transition, characterised by a sudden loss of magnetisation (magnitude of the order parameter) at the critical temperature
The four panels in this figure show phase transitions in the heavy top lattice. In figure 7(a), the blue dots represent the $\Gamma_3$-components of the averaged position $\left\langle {\bf{\Gamma }} \right\rangle$ and in 7(b)-7(d), the blue, green and red dots represent the $\Gamma_1$, $\Gamma_2$ and $\Gamma_3$-components of the averaged position respectively. The black line in all four figures correspond to the mean field approximation (106). The top left panel 7(a) shows a phase transition similar to the rigid body phase transition, and the other panels 7(b)-7(d) display a 'triple-humped' phase transition. The corresponding critical temperatures are indicated by the vertical dashed lines. In 7(b)-7(d), we also varied the Casimirs (where $c_1 = {\boldsymbol \Pi} \cdot {\boldsymbol \Gamma}$ and $c_1 = \|{\boldsymbol \Gamma}\|^2$) to show that the three critical temperatures depend on their values, and may even disappear for large enough $c_1/c_2$. Due to the high dimension and non-compactness of the phase space, the mean field approximation requires intensive computation to be precise, explaining the visible errors, especially at low temperatures
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2018 Impact Factor: 0.525


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### Feynman Teaching Award (w00t!)

My division chair stopped by my office a couple weeks ago and urged me to be present at the next faculty meeting, during which there would be a discussion on online learning. Given that i served on the ad hoc committee addressing this issue, I guessed the request made sense. However, since I wasn't the chair of the committee, and given that there was no real reason for my chair to know I was on this nonstandard committee, the in-persion request was a little odd.

The campus-wide faculty meeting went well, with Prof. Djorgovski (Astro) among those who talked about the success and challenges of online learning. Then came the next action item: announcing the Feynman Teaching Award. To my surprise:

Planetary Astronomer Wins Feynman Prize for Excellence in Teaching

PASADENA, Calif.—John A. Johnson, assistant professor of planetary astronomy at the California Institute of Technology (Caltech), has been awarded the Richard P. Feynman Prize for Excellence in Teaching.

Johnson was recognized for his dedication, passion, and innovation in teaching as well as his ability to inspire his students.

The Feynman Prize was established "to honor annually a professor who demonstrates, in the broadest sense, unusual ability, creativity, and innovation in undergraduate and graduate classroom or laboratory teaching." Caltech faculty members, students, postdoctoral scholars, staff, and alumni may nominate a faculty member for the prize. A committee appointed by the provost then selects the winner.

According to the official citation, Johnson "immediately emerged as exceptional—a 'true outlier,' in the words of a committee member."

"Richard Feynman's writing inspired me to pursue physics and astronomy," Johnson says. "It is an amazing honor to have my name in any way associated with his."

Johnson was lauded for his creative teaching methods, in which he eschews traditional lectures and problem sets and instead has students work on problems in small groups. At various times, he has required students to explain what they were learning in a class blog, forbidden discussion of grades, emailed YouTube videos that illustrate the day's material, and brought in guest lecturers to discuss the course material and provide career advice.

"My goal is to help the students take ownership of their learning by guiding them rather than lecturing them," explains Johnson, who says he learned his teaching philosophy from physicist Ronald Bieniek at the Missouri University of Science and Technology. "I'm very pleased to hear that my students feel I accomplished this goal, and that we all had such an enjoyable time in the process."

In a nomination letter, one student wrote that Johnson "rocked the boat in the astronomy department, challenging our conceptions of how astronomy, and the sciences in general, are taught." Another student wrote, "Classroom experiences that are intellectually engaging, practical, and entertaining are incredibly rare. Through his teaching style, attention to detail, and unique course structure, Professor Johnson provides just such an experience."

Many students cited Johnson's "life-changing" influence beyond academics. One called him "a remarkable teacher who can not only enlighten students in the classroom but also sculpt their spirits for their future careers." A graduate student said, "He reminded me…why I wanted to be a scientist in the first place."

Johnson, whose research focuses on searching for planets around stars other than our sun, earned his BS in physics in 1999 from the University of Missouri-Rolla (now the Missouri University of Science and Technology) and his PhD in astrophysics from the University of California, Berkeley, in 2007. After completing a postdoctoral fellowship at the University of Hawaii Institute for Astronomy, he joined Caltech's faculty in 2009. In 2012, he won the Newton Lacy Pierce Prize from the American Astronomical Society, an Alfred P. Sloan Fellowship, a David and Lucile Packard Fellowship, and a Lyman Spitzer Lectureship from Princeton. Johnson says, however, that of all the awards he has received this past year, he's most proud of the Feynman Prize.

The previous four winners of the Feynman Prize are Paul Asimow, professor of geology and geochemistry; J. Morgan Kousser, professor of history and social science; Dennis Dougherty, the George Grant Hoag Professor of Chemistry; and Jehoshua (Shuki) Bruck, the Gordon and Betty Moore Professor of Computation and Neural Systems and Electrical Engineering. The first Feynman Prize was awarded for the 1993-1994 academic year. Nominations for next year's prize will be solicited in the fall.

Written by Marcus Woo

### On the Height of J.J. Barea

Dallas Mavericks point guard J.J. Barea standing between two very tall people (from: Picassa user photoasisphoto).

Congrats to the Dallas Mavericks, who beat the Miami Heat tonight in game six to win the NBA championship.

Okay, with that out of the way, just how tall is the busy-footed Maverick point guard J.J. Barea? He's listed as 6-foot on NBA.com, but no one, not even the sports casters, believes that he can possibly be that tall. He looks like a super-fast Hobbit out there. But could that just be relative scaling, with him standing next to a bunch of extremely tall people? People on Yahoo! Answers think so---I know because I've been Google searching "J.J. Barea Height" for the past 15 minutes.

So I decided to find a photo and settle the issue once and for all.

I then used the basketball as my metric. Wikipedia states that an NBA basketball is 29.5 inches in circumfe…

### Finding Blissful Clarity by Tuning Out

It's been a minute since I've posted here. My last post was back in April, so it has actually been something like 193,000 minutes, but I like how the kids say "it's been a minute," so I'll stick with that.
As I've said before, I use this space to work out the truths in my life. Writing is a valuable way of taking the non-linear jumble of thoughts in my head and linearizing them by putting them down on the page. In short, writing helps me figure things out. However, logical thinking is not the only way of knowing the world. Another way is to recognize, listen to, and trust one's emotions. Yes, emotions are important for figuring things out.
Back in April, when I last posted here, my emotions were largely characterized by fear, sadness, anger, frustration, confusion and despair. I say largely, because this is what I was feeling on large scales; the world outside of my immediate influence. On smaller scales, where my wife, children and friends reside, I…

### The Force is strong with this one...

Last night we were reviewing multiplication tables with Owen. The family fired off doublets of numbers and Owen confidently multiplied away. In the middle of the review Owen stopped and said, "I noticed something. 2 times 2 is 4. If you subtract 1 it's 3. That's equal to taking 2 and adding 1, and then taking 2 and subtracting 1, and multiplying. So 1 times 3 is 2 times 2 minus 1."

I have to admit, that I didn't quite get it at first. I asked him to repeat with another number and he did with six: "6 times 6 is 36. 36 minus 1 is 35. That's the same as 6-1 times 6+1, which is 35."

Ummmmm....wait. Huh? Lemme see...oh. OH! WOW! Owen figured out

x^2 - 1 = (x - 1) (x +1)

So $6 \times 8 = 7 \times 7 - 1 = (7-1) (7+1) = 48$. That's actually pretty handy!

You can see it in the image above. Look at the elements perpendicular to the diagonal. There's 48 bracketing 49, 35 bracketing 36, etc... After a bit more thought we…