Can an (pre 3.0) Android program connect to a bluetooth A2DP device?

Question: Can a program that is written using the gingerbread/froyo/eclair (e.g. 2.1, 2.2, 2.3 -> NOT honeycomb) connect to a A2DP bluetooth sink device??  Or, do I need to have the user go to their settings to do so? Any help or advice would be highly appreciated.
Answer: So, after much more research, it seems that it is impossible to programmatically connect to a A2DP device on a pre-3.0 Android device. I am going mark this as the answer but, if someone finds otherwise, please correct me on this as I would really like to do it programmatically.
Answer: Can't you use BluetoothA2dp::connectSink()?  Create a BluetoothA2dp instance, then call connectSink(BluetoothDevice device) with the device you're trying to connect to.
code:
http://www.google.com/codesearch#uX1GffpyOZk/core/java/android/bluetooth/BluetoothA2dp.java&q=BluetoothA2dp&l=112
It seems like this API is not listed in the Android API reference for BluetoothA2dp though:
http://developer.android.com/reference/android/bluetooth/BluetoothA2dp.html
Comment: That class is only available, post honeycomb.  I need it for eclair/froyo devices...
Answer: So, after much more research, I found and verified that it is actually possible to use the hidden API's to connect to a a2dp device via bluetooth. However, it seems that post-honeycomb, that method isn't possible anymore... I will post another question regarding that.


How to tell if an adjective is attributive or predicative [EFL context]?

Question: I've been reviewing my grammar lately and I feel a bit stuck in adjectives. Yeah, I know it's not a big deal of a topic, but I think I need some help here because I'm in the language teacher role. I do understand how attributive and predicative adjectives work, yet I can't find an easier way for learners to differentiate between both types - especially if such differentiation doesn't exist in their mother tongues. And yeah, I know I can just explain the whole thing to them, but what if they come across other adjectives they don't know how to use? I just feel a good dictionary might do the trick, that is, with a tagging system that tells adjectives apart according to that syntactic feature: pointing out if they're attributive, predicative or both. The Oxford Learner's Dictionary online sometimes does this -click here and here for examples-, but it seems to me that a more specialised source might be the answer to this dilemma. Well, I'd like to know what your take on this issue is. Suggestions are welcome. Thanks!
Comment: Practically any adjective can be used either as an attributive or as a predicate. It's dependent on the sentence, not the adjective, so a dictionary won't help. A predicate adjective is essentially the verb of the clause, only it can't get inflected for tense, so it has to go around with an auxiliary _be_ that holds the tense: _She is/was/may have been tired_. An attributive adjective modifies a noun: _the tired woman_. There are a very few adjectives that can't function as both attributive and predicative -- _gala_ is one; you can say _It was a gala party_, but not **The party was gala.*
Comment: If you are expecting to find a rule that says something like  "adjectives ending in 'xx' are attributive, the rest are predicate," then you are in for a disappointment. No, adjectives are not identified by their form/structure, but by their context, as to whether the **occurrence** is attributive or predicate.  In other words, the **occurrence** in context of an adjective, not the adjective by itself, that decides it.
Comment: @JohnLawler — I am interested that you state Gala is an attributive adjective. I, a non-grammarian, would have said "a noun used as an adjective". Seems the same thing. However when I used that term in a comment on [another post](https://english.stackexchange.com/questions/395616/the-use-of-usability-is-confusing#comment937989_395616) regarding the word "usability" tchrist (a mod, if not a grammarian) bit my head off, saying "No, this is NOT a noun used as an adjective; it is a noun used in attribution of another noun." It's not relevant to the question, but if he's right can you explain?
Comment: As I've said many times before, what "part of speech" one calls an individual word in an individual sentence is completely arbitrary and says nothing about the grammar, only about the definitions used. In the case of _gala_, there are facts one could ascertain. One could look up its earliest occurrence and see whether _gala_ appeared first as a noun, or as an adjective. If one cared.
Answer:
I do understand how attributive and predicative adjectives work, yet I can't find an easier way for learners to differentiate between both types - especially if such differentiation doesn't exist in their mother tongues.

I'd like to suggest that maybe a reference grammar might be able to help you here. For instance, the 2002 reference grammar by Huddleston and Pullum et al., The Cambridge Grammar of the English Language (CGEL), discusses this very issue: "Chapter 6 Adjectives and adverbs", part "4 Restricted function for adjectives", on pages 553-562.
CGEL, in "4.1 Attributive-only adjectives",

Adjectives that do not normally occur except as (heads of) attributive modifiers include:

which includes an example list of 30 adjectives (e.g. drunken, future, mere, putative, umpteenth). In this part, they also discuss "Meaning differences between attributive and non-attributive uses".
They also discuss "Potential differences between attributive-only and ordinary attributive adjectives"; where they discuss the four properties that ascriptive attributive adjectives characteristically have, but most attributive-only adjectives lack one or more of them. The properties are: entailment, subset, modifiability, pro-form.
Then they go on and discuss "Some types of attributive-only adjectives". In this part, they say:

Attributive-only adjectives are too numerous and semantically heterogeneous to permit a simple and exhaustive classification. Instead we will here illustrate and comment on some of the most important semantic types.

Some of those types are: degree and quantifying attributives, temporal and locational attributives, associative attributives, process-oriented attributives, modal attributives, particularizing attributives, expressive attributives, hypallage: transferred attributives.
CGEL, in "4.2 Never-attributive adjectives", discusses "Adjectives which can occur predicatively or postpositively, but not attributively". These include "Adjectives formed with the 'a' prefix", and provide an example list of 24 adjectives (e.g. afloat, aglitter, alike, averse).
Also, includes "Adjectives with complements". And also includes "A small set of other adjectives".
And then includes "Postpositive-only adjectives", which are described as:

A handful of adjectives are restricted to postpositive function:

and they provide an example list of 6 adjectives (e.g. flowers galore, the President elect).
Okayee, my fingers are tired. Hopefully I haven't made too many typos. It seems to me that CGEL has a lot of good info in those pages that might help you out (pages 553-562).
Comment: 'Galore' is a quantifier as well as an adjective (and I _don't_ mean in different constructions).
Answer: I've found the Collins Cobuild English Grammar to be very helpful hereabouts. It's not totally comprehensive (what is?), but contrasts three of the four positionings of adjectives that I'm aware of pretty well.
One thing it makes clear is that different positioning sometimes controls semantic difference:

The aeroplane had to land on Runway 7, as the proper runway, Runway 1,
  was damaged.
We veered off the runway proper onto the verge.
The concerned parents / the parents concerned

(Participial adjectives always need watching.)
However, some adjectives modify nouns they're not obviously attached to:

I'd love a quiet pint.
It was a proud day for her parents.

(These are 'transferred epithets' where the noun being modified has to be recovered from the context: quiet occasion / surroundings; the girl's parents.)
Even more peripheral non-semantically-predicative usages are:

A former president / a future king
A mere youth / fake diamonds

where what are traditionally labelled adjectives as they pre-not-really-modify 'their' nouns refer to former or future states, or reference whole groups to which the nouns' referents belong (or not). Most of these adjectives are never used predicatively (These diamonds are fake // *The president is now former / *I'm having a pint that is quiet / *Those youths are mere / *This day is proud for her parents).
................
I've worked up a list (after Elizabeth Coppock's Thesis_5 Adjectives, and with input from Will Oxford_Identity Adjectives:
[In] the preceding sections [I] have argued for the value of a grammatical distinction between functional adjectives and lexical adjectives. To my knowledge, this exact distinction has not previously been made in the literature. However, it is not a new observation that some adjectives lack the full range of prototypical adjectival behaviour ...
The list of "defective adjectives" [is shown in] the rough classification suggested here; it is for descriptive convenience only—the development of a grammatically and semantically adequate classification requires further work.
(8) a. Identity adjectives: same, different, other
<code> b. Ordinal adjectives: next, last, previous, subsequent, preceding, further

 c. Degree adjectives: utter, sheer, outright

 d. Significance & particularity adjectives (predicate-evaluating): main, chief, principal, mere, common

 e. Extent adjectives: entire, whole

 f. Temporal adjectives: future, then [former, past, now? once?] (termed modal II by Coppock)
</code>
I'll risk adding, though there is often semantic content in addition to a non-canonical syntactic role, and predicative use may occur:
<code> g. Frequency etc adjectives: occasional, rare, common, frequent, sporadic, intermittent ...

  g' Other adjectives referring to how assembled referents co-relate: various, assorted, diverse, disparate, diversiform, similar,

 h. Adjectives of potential / lack of (etc) ...: potential, budding, would-be, wannabe, manqué, frustrated [barred sense] ...

  i. Modal and veridical (including privative) adjectives: true, suspected, alleged; (false, fake – can be used attributively)

  j. Adjectives of selection / confirmation / rebuttal (very, exact, correct, wrong)

  k. Event-manner adjectives: eg beautiful in a beautiful dancer, heavy in a heavy smoker

  l. Transferred epithets, usually psychological experience adjectives: eg proud in a proud day, disabled in disabled toilet
</code>
Those listed in h are certainly 'peripheral' (not grading or normally appearing predicatively). Those in g semantically are non-central; they specify the type of occurrence etc of the noun's referent rather than a truly inherent property - a background is implied with dots marking occurrences. With temporal adjectives, lines against a time background (marking eg the past) are implied.
..........
Coppock has a later publication on The Predictability of Predicativity in which she enunciates the rule: 'An adjective is syntactically predicative
if and only if it is semantically predicative'.
Answer: In an unsearchable and potentially ephemeral comment to the original posting, Professor Lawler kindly presented the following answer:

Practically any adjective can be used either as an attributive or as a predicate. It's dependent on the sentence, not the adjective, so a dictionary won't help.

A predicate adjective is essentially the verb of the clause, only it can't get inflected for tense, so it has to go around with an auxiliary be that holds the tense: She is/was/may have been tired.

An attributive adjective modifies a noun: the tired woman.

There are a very few adjectives that can't function as both attributive and predicative — gala is one; you can say:

It was a gala party

but not:

*The party was gala.

I've marked this posting Community Wiki because it is John's answer not my own, and so I deserve no reputation from it.
Comment: @EdwinAshworth That was John's example, but he won't ping. There are plenty of other examples that resist rearrangement. The oddest ones are the attributive sorts that cannot be predicative but still must follow their head. You can have puppies galore but the puppies cannot be galore, nor can you have galore puppies, either.  Purely predicative are those like *asleep* because when the puppies are asleep, they are not asleep puppies: they are only sleeping puppies.
Comment: When you say "verb of the clause" I get confused. In case of a similar sentence, "She is beautiful", the predicate is "is beautiful" and it is certainly not a verb. I think that it is more about whether the predicate of a sentence (i.e. a grammatical component) is a pure verb ("I run") or is a noun+auxiliary verb ("I am hungry"). To be clear, the verb/noun/etc. distinction is based on lexical meaning and part-of-speech while the predicate is a grammatical component that can be composed of various types of words.
Comment: I'm not sure who responds to queries about community wikis. CED, Lexico, Macmillan and Collins (though not AHD, RHK Webster's,  M-W, Wiktionary) list 'a gala party' as an attributive noun usage, which is how I tend to see it. These rarely if ever exist outside premodification. _It is a pickle fork._ // *'He had a pint that was quiet' shows a non-transformable _usage_, and 'former' is an example of a never-predicative adjective (though I'm probably in favour of a reclassification of 'peripheral adjectives' like mere, former, wannabe, alleged).
Comment: 'Galore' is one of the two words I've got to see as being simultaneously in 2 distinct classes: quantifier and adjective.
Answer: Attributive adjectives are used with a noun, mostly before a noun, sometimes after a noun: a little child, an old man, a large forest
Predicative adjectives are used after to be (and similar verbs):
The man is old.
The Latin grammar term attribute is a term for anything you can attach to a noun: articiles and article words (they go into the same position as the articles: possessive adjectives, demonstrative adjectives), adjectives, participles, and relative clauses)
"Predicative use" of an adjective means the adjective is part of the predicate.
Most dictionaries indicate if the use of an adjective is limited to predicative use: The entry of alive of OALD says: adjective, not before a noun. Most adjectives of the type alive, beginning with the prefix a- can only be used as predicative adjectives.
Answer: I agree with the identification of the two types of adjective by position, but can we consider this idea, it's been working for me.
Considering that using an adjective with/for a noun is like passing a judgement, I identify an adjective as predictive when there is an antecedent or experience that bring about the conviction for your judgement. Consider these:

grass-fed beef  tastes better than others.

The examination was unbelievably easy.

The judgement in the above sentences were based on the conviction drawn from experiences of tasting the different types of beef and writing a particular exam. On the other hand, I consider an adjective to be attributive when the judgement is based on instant perception drawn from visible and perceivable attributes/characteristics of a noun. Let's consider these:

he lives in a two storey building.

He goes in a blue Landrover jeep.


How likely is it that we will discover a new species of big cat?

Question: So, I've been reading some articles about these "scandals" where everybody thought that there was a new species of feline discovered, from some weird skull or carcass or whatever. And always, by the end of the article, they would have concluded that it was just a jaguar or an ocelot or not even a feline. But, it got me thinking, and I was wondering - Is it possible (As in, are there more than minuscule chances) that we will discover a new species of big cat, or Panthera species?
Comment: Welcome to BiologySE! Please cite the source from where you heard that. Also, do take a moment to check out the [tour] of our site.
Answer: Extant (alive) extremely small. There just is no place left to hide so to speak. If we do it will most likely be what we thought as one species being split into two along some currently unknown factor, similar to how chimps and bonobos were thought to be the same until fairly recently.
Extinct extremely high, we have discovered such a small fraction of a percentage of the number of species that have ever existed that any group is a candidate for future discovery.
Answer: Actually the last big cat discovered was only 2011 so it's remotely possible that another can be discovered.

Comparisons between the clouded leopards of mainland Asia and the Sunda Islands of Southeast Asia, which include Borneo, Java, Sumatra and Bali, revealed the Sunda clouded leopard was genetically distinct, making it the newest species of big cat known to science.
(Despite their name, clouded leopards are not actually a kind of leopard, but are completely separate from the genus that includes lions, tigers, leopards and jaguars.)

https://m.csmonitor.com/Science/2011/0125/That-new-species-of-big-cat-Turns-out-it-s-actually-two-new-species
Comment: Yeah, that was one of the things that got me thinking for this question.


Why did Rashi create a new letter system for his commentaries?

Question: I had heard that he created his own lettering system so that his words would be easily distinguishable from the actual words of  the Torah. Is this just hearsay or speculation? If so, is there any reason given for why?
Comment: See: http://en.wikipedia.org/wiki/Rashi_script
Comment: So, 1) Rashi didn't create "Rashi script," and 2) it _was_ indeed used by publishers to distinguish between source texts and commentaries? If you'd encapsulate that in an answer, I'll give you the reward.
Comment: I didn't put it in an answer because what you asked for was simply the information from a wikipedia entry. That sort of question has been rejected in the past, so I wasn't sure it was really appropriate to answer in that way.
Answer: It is mentioned in http://en.wikipedia.org/wiki/Rashi_script that it was not Rashi's script (according to he.wikipedia.org, the script is actually sefardic in origin).
According to Marvin Heller (See page 62 there), the printers used a different script to differentiate the commentaries from the main text. (Tosfos also uses "Rashi letters" as well as many other commentaries)
Comment: This answer is almost entirely incorrect. The printers didn't need a new script at all; "Rashi script" was the script first employed in a printed Hebrew text, and was so named because the book in which it was employed was Rashi's commentary on the Torah. That text didn't contain the Torah, only Rashi's commentary - the notion that it was "invented" in order to differentiate the two is a longstanding myth.
Comment: @ShimonbM why didn't they just use block letters?
Answer: He didn't.  According to a Mansucript Preparation class I attended this year, what's known as Rashi script was the font the printer used.  As a side point, the script used in some Judeo-Arabic written manuscripts - particularly the Rambam's handwriting - is very similar to Rashi script.
Comment: Re "the font the printer used": yes, but as you note in the rest of your answer, it was based on handwriting. (So that's not really a "side point".)
Comment: Touche.  But I was talking about the Rambam, NOT Rashi.  And yes, most early printers based their printed text on handwriting.
Answer: also because it  takes less space, you can write more within the same page, meaning cheaper, smaller and more concise books
Comment: This seems to be a duplicate of [this earlier answer](http://judaism.stackexchange.com/a/8849/472).  We try to avoid repeating other answers wholly.  Is there some difference I've not picked up on?  (Also, is this your own reasoning or did somebody say that about Rashi script?)
Answer: I was taught that since ink and particularly parchment was at a premium, it would make sense to fit as much writing as you could onto a sheet, and as such, what we know as Rashi script was born, as it was more compact than block.
Answer: I learned that it also used less ink than the traditional "square" script. In medieval times, when making ink was a long, expensive process using oak galls and other uncommon ingredients, it's easy to understand why a printer preferred using a font that used less ink.


Curacao visa on arrival

Question: Can an Indian passport holder, going from Delhi to Curacao to Panama, not having a USA or Canada or EU visa, get a visa to enter Curacao on arrival?
Comment: is it a single booking and a through ticket? or will you have to clear immigration and check your bag in again?
Comment: ticket will be Panama - Curacao - Jamaica by Copa Air
Answer: You wont need a visa to transit Curacao. It seems you meet the requirements below.

You are transiting through an airport in one of the Caribbean parts of
the Kingdom, hold an onward ticket and meet the following
requirements:
Aruba:    Transit between 7:00 and 23:00 uur
St Maarten: same calendar day
Bonaire, Curaçao, St Eustatius and Saba:  Transit within 48 hours

As long as your transit is within 48 hours and have an onwards ticket you're fine without a visa.
If you're transit is longer you'll need to apply for a short stay visa
More information can be found on the Kingdom of the Netherlands travel website.

Curacao visa on arrival


How to keep pasta hot, yet not sticky in crock pot?

Question: We are having a pasta "bar" for a party that will last several hours.  How do you keep pasta HOT and NOT sticky.  (I have seen several answers for COLD and not sticky, but not for hot.)
Answer: I've not personally tried this with larger quantities, but I have the best success keeping hot pasta non-sticky in smaller quantities (assuming you want to keep pasta and sauce wholly separate) when I do the following three things:

Cook in a large amount of water; this will keep the amount of starch in the water down to a bare minimum.
After draining, rinse off as much of the surface starch as you can; you might need to keep a separate pot of near-boiling water ready for this.
Add a small amount of whatever oil or fat the sauce is based on (olive oil, butter) to the pasta and mix it in.

This is all assuming it is completely unacceptable to sauce the pasta ahead of time.
Good luck!
Comment: Your basic idea will definitely work. We used an oil and butter mix for keeping it not sticky. I would not use a strong flavored oil such as olive oil. Remember to keep the lid closed except when serving  because the outer exposed pasta could dry up somewhat!
Comment: My usual guideline on the oil to use is based on whatever the sauce has - so for (e.g.) an Alfredo sauce, it would definitely be butter and not olive oil.  If the sauce were a tomato and olive oil sauce, however, olive oil would be my choice.


RPM Flucating when ac is on

Question: I have a one question i have owner  Tata Indica V2 . my question is when i start car ac rpm flucating up and down and similar pickup also same with rpm. what is the reasion and please give me solution
Comment: So, how much fluctuation : 500 to 1500 rpm? 800 to 850 ? the A/C load is not constant - depends on what load exists and it varies according to the A/C settings.
Comment: @Al_ what was the rpm fluctuation - until the OP clarifies then we are all p*ssing in the wind which was why I only made a comment to elicit more info - which never happened...
Comment: @SolarMike The AC load aspect is not so simple: a reasoning straight as yours might apply for an externally controlled variable displacement compressor, where the AC system is actually able to regulate the refrigerant flow, but internally controlled VDCs actually work on full flow until the evaporator reachs freezing point and then they start destroking, while fixed displacement compressors are always pumping the same amount of refrigerant around and therefore work in cycles. Compressor load is more likely to be affected by amb temp (condensing pressure and therefore discharge pressure).
Comment: @SolarMike Oops, my fault. Misread the date.
Answer: I suspect you are referring to the apparent dip in idle speed when switching the A/C on and subsequent correction.  What actually happens is that the A/C compressor clutch kicks in so that the drivebelt attached to the A/C compressor pulley wheel is then presented with the additional load required to rotate the compressor internals that circulate the refrigerant.
This additional load placed on the engine causes the idle speed to slow down.  What then happens is that the engine management software on the ECU detects that the idle speed has dropped and adjusts the idle control valve / throttle flap position accordingly.  This then causes the RPM to jump back up again and then settle to a comfortable idle speed.
You see the same thing to a lesser extent when turning on other systems on the car such as the heated rear screen.  It's most evident with the A/C compressor as this has is one of the biggest drains on the system in terms of additional equipment.
Comment: I agree with Steve. One additional thought is that cars with smaller engines are more likely to exhibit this behavior.

RPM Flucating when ac is on


Is there an integral that proves $\pi > 333/106$?

Question: The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
Comment: If $f(x)>0$ then $$\int\limits_{a}^{0} f(x)>0$$
Comment: @Affan, you're integrating a fraction of even powers: it can't be negative.
Comment: You can use $\pi = \int_0^1 \frac{4}{1+x^2}$. Now the power series of $\frac{4}{1+x^2}$ is alternating, thus stopping after odd/even numbers of terms gives under and overevaluations. Thus, for all $a< \pi <b$ you can find $m, n$ so that $a< \int_0^1 P_m(x) dx < \pi < \int_0^1 P_n(x) < b$, where $P_n$ denotes a Taylor Polynomial.
Comment: @AndreaAmbu The powers don't have to be even: $x$ and $(1-x)$ are non-negative between $0$ and $1$.
Comment: Hey, how do you know that it is `clearly positive?` is there something about the integral that makes it positive?
Comment: Where do you get that exact number from? (333/106)
Answer:
This integral would do the job:

$$\int_0^1 \frac{x^5(1-x)^6(197+462x^2)}{530(1+x^2)}\:dx= \pi -\frac{333}{106}$$

Also you can refer to S.K. Lucas Integral proofs that $355/113 > \pi$, Gazette, Aust. Math. Soc.  32 (2005), 263-266.

This is the link. (Thanks to lhf for pointing out.)
Comment: How does one come up with that integral, may I ask?
Comment: The link above is broken. The current one is http://educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf
Comment: @ShreevatsaR Very old thread, but for the benefit of future searchers, how one comes up with that integral along with Maple code for doing so is discussed on pages 8-9 of Lucas' writeup.
Answer: Although this is not exactly an answer to the question, it seems sufficiently related to mention: there are some direct generalizations, given on the Wikipedia page about this integral. For instance, $$0 < \frac14\int_0^1\frac{x^8(1-x)^8}{1+x^2}\ dx=\pi -\frac{47171}{15015}$$
In general,
$$\frac1{2^{2n-1}}\int_0^1 x^{4n}(1-x)^{4n}\ dx
<\frac1{2^{2n-2}}\int_0^1\frac{x^{4n}(1-x)^{4n}}{1+x^2}\ dx
<\frac1{2^{2n-2}}\int_0^1 x^{4n}(1-x)^{4n}\ dx$$
which for $n=1$ (the integral in the question) gives slightly better bounds than just $\pi < 22/7$:
$$ \frac{1}{1260} < \frac{22}{7} - \pi < \frac{1}{630}$$
Comment: @anon For this to work, you'd have to show that the rational approximants converged suitably fast.  See Dirichlet's irrationality test.
Comment: @muad: Without checking the asymptotics of the respective numerators and denominators, I'm not sure. It doesn't follow simply from the fact that the left and right expressions go to 0: for instance 1 is rational, but you could still find sequences of rational numbers $x_n, y_n, z_n$ such that $x_n$ and $y_n$ go to 0, but $x_n < 1 - z_n < y_n$ for all $n$.
Comment: Thank you! Doesn't that imply that pi is irrational?
Answer: In the beginning of 2009 I was posting re similar issue at several sites, namely, at sci.math.symbolic, www.math.utexas.edu, etc.
To repeat:  In Paper 1 Lucas found, by brute-force search using Maple programming, several different variants of integral identities which relate each of several first Pi convergents (described in terms of OEIS <oeis.org> sequences as A002485(n)/A002486(n)) to Pi.
Further, in my above-mentioned postings, I conjectured the following identity below, which represents a generalization of Stephen Lucas' experimentally obtained identities between Pi and its convergents:
$$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$
$$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^m(k+(i+k)x^2)\big)/(1+x^2)\; dx$$
where integer n = 0,1,2,3,... serves as index for terms in OEIS A002485(n) and A002486(n), and {i, j, k, l, m} are some integers (to be found experimentally or otherwise), which are probably some functions of n.
The "interesting" (I think) part of my generalization conjecture is that "i" is present in both:
denominator of the coefficient in front of the integral
and in the body of the integral itself
For example for $\frac{22}{7}$
$$\frac{22}{7} - \pi = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,\mathrm{d}x$$
with $n=3, i=-1, j=0, k=1, l=4, m=4$ - with regards to my above suggested generalization.
In Maple notation
i:=-1; j:=0; k:=1; l:=4; m:=4;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
yields 22/7 - Pi
It also works for found by Lucas
http://www.math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf
formula for $\frac{333}{106}$
$$\pi - \frac{333}{106} = \frac{1}{530}\int_{0}^{1}\frac{x^5(1-x)^6(197+462x^2)}{1+x^2}\,\mathrm{d}x$$
with $n=4, i=265, j=1, k=197, l=5, m=6$ -with regards to my above suggested generalization.
In Maple notation
i:=265; j:=1; k:=197; l:=5; m:=6;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
yields Pi - 333/106
And it works for Lucas's formula for $\frac{355}{113}$
$$\frac{355}{113} - \pi = \frac{1}{3164}\int_{0}^{1}\frac{(x^8(1-x)^8(25+816x^2)}{(1+x^2)}$$
with $n=5, i=791, j=2, k=25, l=8, m=8$ -with regards to my above suggested generalization.
In Maple notation
i:=791; j:=2; k:=25; l:=8; m:=8;Int(x^m*(1-x)^l*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
yields 355/113 - Pi
And it works as well for Lucas's formula for $\frac{103993}{33102}$
$$\pi - \frac{103993}{33102} = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(124360+77159x^2)}{1+x^2}\,\mathrm{d}x$$
with $n=6, i= -47201, j=4, k=124360, l=14, m=12$ -with regards to my above suggested generalization.
In Maple notation
i:=-47201; j:=4; k:=124360; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
yields Pi - 103993/33102
And also it works Lucas's formula for $\frac{104348}{33215}$
$$\frac{104348}{33215} - \pi = \frac{1}{38544}\int_{0}^{1}\frac{x^{12}(1-x)^{12}(1349-1060x^2)}{1+x^2}\,\mathrm{d}x$$
with $n=7, i= -2409, j=4, k=1349, l=12, m=12$ - with regards to my above suggested generalization.
In Maple notation
i:=-2409; j:=4; k:=1349; l:=12; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
yields 104348/33215 - Pi
And it works as well for $\frac{618669248999119}{196928538206400}$
which, by the way, is not part of A002485/A002486 OEIS sequences:
$$\frac{618669248999119}{196928538206400} - \pi = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(77159+124360x^2)}{1+x^2}\,\mathrm{d}x$$
with $i= 47201, j=4, k=77159, l=14, m=12$ -with regards to my above suggested generalization.
In Maple notation
i:=47201; j:=4; k:=77159; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1)
618669248999119/196928538206400 - Pi
I do not have computer math resources (Mathematica, Maple, etc.) to
experimentally prove or disprove it for all larger n (but see my comment below).
Best Regards,
Alexander R. Povolotsky
UPDATE #1:
Matt B. in his answer to my question on Mathematics Stack Exchange has provided analytical proof and improved my parametric formula by reducing the number of parameters from 5 to 4
(see Seeking proof for the formula relating Pi with its co seenvergents).
$$(-1)^n (\pi- \frac{p_n}{q_n}) = \int_0^1 \frac{x^{\epsilon+2m'}(1-x)^{2m'}(\alpha + \beta x^2) }{(\alpha - \beta) 2 ^{m'-2} (-1)^{\epsilon}(1+x^2)}dx$$.
Below is the list of parameters in Matt B.'s formula for all cases, covered in Stephen Lucas' publications - the stuff to the right of the arrow sign is the actual Maple code, which one could copy (while in the "edit" mode) and then paste into (let say) Inverse Symbolic Calculator (it accepts Maple code) and run it there.
NB  I replaced for brevity some parameter names used by Matt B: "alpha" by "a", "beta" by "b", "epsilon" by "c", "m' " by "p".
104348/33215 - Pi -> a:=1349;b:=-1060;p:=6;c:=0;Int((x^(c+2p)(1-x)^(2p)(a+bx^2))/((a-b)2^(p-2)((-1)^(c)(1+x^2))),x=0...1)
Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;Int((x^(c+2p)(1-x)^(2p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)(1+x^2))),x=0...1)
355/113 - Pi -> a:=25;b:=816;p:=4;c:=0;Int((x^(c+2p)(1-x)^(2p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)(1+x^2))),x=0...1)
Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;Int((x^(c+2p)(1-x)^(2p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)*(1+x^2))),x=0...1)
22/7 - Pi -> a:=1;b:=0;p:=2;c:=0;Int((x^(c+2p)(1-x)^(2p)(a+bx^2))/((a-b)(2^(p-2))((-1)^(c)(1+x^2))),x=0...1)
Obviously parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what Matt B named as "m'" (and I call it "p") : {2,3,4,4,6,6} ...  Note that when "n" -> infinity, then the integral should come to 0 ...
UPDATE #2:
Analysis of Thomas Baruchel calculations results (see his answer to my question in
Seeking proof for the formula relating Pi with its co seenvergents)
led to observation that in originally supplied five parameter notation (i,j,k,l,m)
j = m/2 - 2
and correspondingly
m=2*(j+2)
This makes the original conjecture to depend on 4 parameters and to look like:
$$ (-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))
=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$
This observation confirms previous Matt B result (see previous update), which also is based on 4 parameters.
Based on his calculations results, Thomas Baruchel also found that even with 4 parameters, this formula yields infinite number of solutions for each n.
Thomas shared with me his calculations results and supplied me with quite a few of valid combinations of i,j,k,l values - so now I have a lot of experimentally found five-tuples {n,i,j,k,l}, which satisfy above parametrization, where n varies in the range from 2 to 26.
Based on this data, of course, it would be nice to find how (if at all) i,j,k,l are inter-related between each other and with "n" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "i" is strongly increasing as "n" is growing from 2 to 26).
Looking at all available {i,j,k,l} solutions (initial ones found by me and those which were found by Thomas Baruchel program) one could observe that in all of them abs(l-j)=2*m where "m" is some positive integer.
If I didn't make a mistake RHS could be reduced (after performing integration) to:
(abs(i)2^j)^(-1)Gamma(2j+5)((k+i)Gamma(l+3)HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2j+l+8)+kGamma(l+1)HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2j+l+6))
May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0, then we have:
Pi = (abs(i)2^j)^(-1)Gamma(2j+5)((k+i)Gamma(l+3)HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2j+l+8)+kGamma(l+1)HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2j+l+6))
Perhaps someone could programmatically check if there are any {i,j,k,l}, which would satisfy above?
Update #3:
Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0
$$\pi = \int_{0}^{1}\frac{4}{1+x^2}\,\mathrm{d}x$$
Should there be an infinite number of such cases?
see Randall's answer at math.stackexchange.com/a/2198869/28343
Update #4
David Trimas looked into the option involving partial solution for my conjecture by setting j=l=0 and deriving formulas for "i" and "k" for j=l=0 condition.
The result is following:
(-1) ^ n * (Pi − A002485(n)/A002486(n)) = ((Abs(i)) * 2 ^ j) ^ (-1) * Int((x ^ l * (1 - x) ^ (2 * (j + 2)) * (k + (i + k) * x ^ 2 ))/(1 + x ^ 2 ), x=0 ...1)
holds true for any n>2 and j=l=0 when
i =(-1)^(n) * 3 * A002486(n)
k = (-1)^(n) * (47 * A002486(n) - 15 * A002485(n)
For example for n=3 where A002485(3)=22 and A002486(3)=7
i=(-1)^337=-21
k=(-1)^3*(477 - 1522)=1
and the quick check via Inverse Symbolic Calculator using Maple (also could be done using Mathematica) provides the confirmation.
i:=-21;j:=0;k:=1;l:=0;(abs(i)2^j)^(-1)int((x^l(1-x)^(2(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1) = 22/7 - Pi
Substitution of
i =(-1)^(n) * 3 * A002486(n)
and
k = (-1)^(n) * (47 * A002486(n) - 15 * A002485(n)
into general conjectured formula with the condition that j=l=0 (under which above formulas for "i" and "k" were derived) confirmed those formulas for "i" and "k" validity by bringing lhs and rhs to be equal.
Though above finding is obviously circular in nature (it doesn't reveal direct dependency of "i" and and "k" on "n" but rather does it via A002485(n) and A002486(n) ) - it is in my view a step forward.
Update #5
As I mentioned above in Update #2, observation of all obtained solutions for my conjectured identity (which represents  generalization of Stephen Lucas' experimentally obtained identities between Pi and its convergents) shows that abs(l-j)=2*m
where "m" is some positive integer.
Going further I make another conjecture re identity which relates Log(2) and its convergents and is analogous to the one which relates Pi and its convergents.
Below is conjectured by me formula (expressed in Maple notations) for relating Log(2) (that is Ln(2)) with ALL of its convergents - those, which are described via A079942(n)/A079943(n) ratio where A079942(n) and A079943(n) are OEIS integer sequences.
(-1)^n*(Log(2) − A079942(n)/A079943(n))=(Abs(i)2^j)^(-1)Int((x^l(1-x)^(2(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)
where integer n>2 serves as the index for the terms in the OEIS A079942(n) and A079943(n) integer sequences, and {i,j,k,l} are some signed integer parameters (which are some implicit functions of "n" and to be found experimentally or otherwise for each value of "n") , and abs(l - j) = 2*m + 1, where "m" is some positive integer.
Comment: One also could check and see that my above generalization formula also applies to identities obtained by Jaume Oliver Lafont, described in the "Following Lucas (2009)"
section in
http://www.oeis.org/wiki/User:Jaume_Oliver_Lafont/Constants#Integrals_involving_convergents_to_Pi
Comment: It would have been nice to have the information contained in the text `104348/33215 - Pi -> a:=....; Int(....)` rather as a table with columns $n; p_n, q_n; a, b, c, p$. The `Int(...)` is always the same (and btw. an inert `int(...)` so you have to use `eval()` to get the result. It should be defined just once as function `chk :=(a,b,c,p)->int(...)` so one can plug in the data from the table and see that it gives $\pm (\pi - p_n/q_n)$. IDK if the answer could be edited in that sense, I will maybe try to do it hoping that this won't give me flames and downvotes etc.
Comment: @Max - go ahead, I don't mind.
Answer: Another solution is given by the integral
$$
0 < \int_0^1 \frac{x^4(1-x)^8}{4(1+x^2)}dx = \pi -\frac{2419}{770} = \pi - \frac{333}{106}-\frac{1}{20405}
$$
This proves the stricter condition
$$
\pi > \frac{333}{106}+\frac{1}{20405}
$$
which implies
$$
\pi > \frac{333}{106}
$$
Similarly, for the fourth convergent (formula (6) http://www.math.ucla.edu/~vsv/resource/general/Lucas.pdf)
$$0<\int_0^1 \frac{x^{10}(1-x)^8}{4(1+x^2)}dx=\frac{3849155}{1225224}-\pi=\frac{355}{113}-\frac{5}{138450312}-\pi$$
Therefore
$$\pi<\frac{355}{113}-\frac{5}{138450312}$$
and
$$\pi<\frac{355}{113}$$
Even for the first convergent
$$0<2\int_0^1 \frac{x(1-x)^2}{(1+x^2)}dx=\pi-3$$
so
$$\pi>3$$
(See https://math.stackexchange.com/a/1618454/134791 for a proof for $3<\pi<4$ that uses this integral)
A series proof that $\pi>\frac{333}{106}$ is given by

$$\frac{48}{371} \sum_{k=0}^\infty \frac{118720 k^2+762311 k+1409424}{(4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21) (4 k+23)} \\=\pi-\frac{333}{106}$$
Comment: The 1st case in the above answer fits i=-1, j=2, k=1, l=4
for parametric formula

$$ (-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))
=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$

The second case fits parameters: $$ i=-1, j=2, k=1, l=8 $$

The  3rd case fits at least the following 2 combinations

$$ i=-1, j=-1, k=1, l=1 $$

$$ i=-2, j=-1, k=2, l=1 $$

$$ (abs(-1)*(-1)^2)^(-1)*Int((x^1*(1-x)^(2*(-1+2))*(1+(-1+1)*x^2))/(1+x^2),x=0...1)=(Pi -3)/2  $$

$$ (abs(-2)*(-1)^2)^(-1)*Int((x^1*(1-x)^(2*(-1+2))*(2+(-2+2)*x^2))/(1+x^2),x=0...1)=(Pi -3)/2  $$
Comment: Might you have a source for that magnificent series?
Comment: @clathratus , answered in your question https://math.stackexchange.com/questions/3479363/a-magnificent-series-for-pi-333-106
Answer: From the relationship
$$\frac{333}{106}=\frac{377-2·22}{120-2·7}$$
and integrals
$$\pi = \frac{22}{7} - \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx$$
and
$$\pi = \frac{377}{120} - \frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx$$
we obtain
$$\int_0^1 \frac{14x^4(1-x)^4-60x^5(1-x)^6}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x(1-x)^2)}{1+x^2}dx = \int_0^1 \frac{2x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = 106\pi-333$$
Therefore,

$$\frac{1}{53} \int_0^1 \frac{x^4(1-x)^4(7-30x+60x^2-30x^3)}{1+x^2}dx = \pi-\frac{333}{106}$$

The numerator can be shown to be nonnegative for $0\leq x\leq 1$.
In fact, we can get a smaller numerator that is still nonnegative, which leads to a closer approximation to $\pi$

$$\pi=\frac{21991}{7000} +\frac{1}{50} \int_0^1 \frac{x^4 (1 - x)^4 \left(4 - 27 x (1 - x)^2\right)}{\left(1 + x^2\right)} dx$$

Here the numerator has been adjusted to have a double zero in $(0,1)$ while not crossing the axis WA link
The resulting fraction is
$$\frac{21991}{7000} = \frac{22}{7}-\frac{9}{7000}$$
which allows for writing the double inequality
$$\frac{22}{7}-\frac{9}{7000}<\pi<\frac{22}{7}$$
or, equivalently, an upper bound for the error in Archimedes' approximation
$$\frac{22}{7}-\pi<\frac{9}{7000}$$
Answer: Let us consider the polynomial
$$P_n(x):=1-x^2+x^4-x^6+\cdots x^{2n-2}=\frac{x^{2n}+1}{x^2+1}.$$
We have
$$0<\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx<\int_0^1\left(\frac{x^{2n}}{0+1}\right)^2dx=\frac1{4n+1},$$ and the integral can be made as small as desired.
On another hand, the remainder of the long division of $(P_n(x)-1)^2$ by $(x^2+1)^2$ is a binomial $ax^2+b$, with $a,b$ integer. Then
$$\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx=\int_0^1\left(Q(x)+\frac{ax^2+b}{(x^2+1)^2}\right)dx.$$
As
$$\int_0^1\frac{ax^2+b}{(x^2+1)^2}dx=\frac{b-a}4+\frac{b+a}8\pi$$
we can get arbitrarily close rational approximations by a rational integral.

For example, with $P_2:=1-x^2+x^4$ we have
$$\frac{(x^4-x^2)^2}{(x^2+1)^2}=x^4-4x^2+8-\frac{12x^2+8}{(x^2+1)^2}$$ so that by integration
$$0<\frac15-\frac43+8-\frac{5\pi}2+1<\frac19$$ or
$$\frac{698}{225}<\pi<\frac{236}{75}.$$
Comment: @JaumeOliverLafont: of course not, you need higher $n$, but there is no limit.
Comment: But $\frac{698}{225} < \frac{333}{106} < \pi $. Does the example answer the question?
Answer: Without restrictions on the integrals, the question is virtually meaningless, because you can take any known bounds on $\pi$ (such as terms of the Gregory series), say $a<\pi<b$ and write
$$\int_0^a dx<\pi<\int_0^bdx.$$
Or if the function must be non-trivial, use any convergent squeezing of $\dfrac4{x^2+1}$ and
$$\int_0^1f(x)\,dx<\int_0^1\frac{4\,dx}{x^2+1}=\pi<\int_0^1g(x)\,dx$$ that you can make as tight as wanted.
Or better, any convergent squeezing of $\pi$, and
$$\int_0^1f(x)\,dx<\pi<\int_0^1g(x)\,dx.$$
In my other answer, I showed constructively that that there are bracketings of $\pi$ with integrals of rational functions as tight as you want, even though they are pretty inefficient.
Answer: A short Maple script shows that among the Dalzell-type integrals, $$\alpha \int_0^1 \frac{x^i (1 - x)^j}{1 + x^2} \,dx,$$ the integrals minimal in the sense that there is no other example $(i', j')$ for which $i' \leq i$ and $j' \leq j$, along with their lower bounds for $\pi$ are:
$$\begin{array}{crl}
\hline
\displaystyle\frac{1}{1024} \int_0^1 \frac{(1 - x)^{24}}{1 + x^2} \,dx & \frac{17\,941\,970\,723}{5\,711\,177\,472}\!\!\!\! &= \frac{333}{106} + \frac{13\,399\,231}{302\,692\,406\,016} \\
\displaystyle\frac{1}{128} \int_0^1 \frac{x (1 - x)^{18}}{1 + x^2} \,dx & \frac{13\,685\,795}{4\,356\,352}\!\!\!\! &= \frac{333}{106} + \frac{14\,527}{230\,886\,656} \\
\displaystyle\frac{1}{16} \int_0^1 \frac{x^2 (1 - x)^{12}}{1 + x^2} \,dx & \frac{2\,264\,177}{720\,720}\!\!\!\! &= \frac{333}{106} + \frac{1501}{38\,198\,160} \\
\displaystyle\frac{1}{4} \int_0^1 \frac{x^4 (1 - x)^8}{1 + x^2} \,dx & \frac{2\,419}{770}\!\!\!\! &= \frac{333}{106} + \frac{1}{20\,405} \\
\displaystyle\frac{1}{2} \int_0^1 \frac{x^7 (1 - x)^6}{1 + x^2} \,dx & \frac{174\,169}{55\,440}\!\!\!\! &= \frac{333}{106} + \frac{197}{2\,938\,320} \\
\displaystyle \int_0^1 \frac{x^{10} (1 - x)^4}{1 + x^2} \,dx & \frac{141\,511}{45\,045}\!\!\!\! &= \frac{333}{106} + \frac{181}{4\,774\,770} \\
2\displaystyle \int_0^1 \frac{x^{29} (1 - x)^2}{1 + x^2} \,dx & \frac{457\,304\,942\,543}{145\,568\,097\,675}\!\!\!\! &= \frac{333}{106} + \frac{147\,383\,783}{15\,430\,218\,353\,550} \\
4\displaystyle \int_0^1 \frac{x^{24\,036}}{1 + x^2} \,dx & \ast & \approx \frac{333}{106} + 1.11071 \cdot 10^{-8} \\
\hline
\end{array}$$
The $\ast$ denotes a rational number that for space reasons cannot be conveniently typeset as a ratio of integers, but its value is $\pi - \Psi(\frac{24039}{4}) + \Psi(\frac{24037}{4})$, where $\Psi(x) = \frac{d}{dx} \log \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the digamma function. The fourth of these integrals (namely, $(i, j) = (4, 8)$) is the example Jaume Oliver Lafont mentions in their answer.
Notice in the above table that $\alpha = 2^{2 - \frac{j}{2}}$. All of these integrals can be used to give tighter bounds by generalizing the observation of Dalzell that ShreevatsaR mentioned in their answer: In general we have
$$2^{2 - \frac{j}{2}} \int_0^1 \frac{x^i (1 - x)^j}{1 + x^2} \,dx \geq 2^{2 - \frac{j}{2}} \cdot \frac{1}{2} \int_0^1 x^i (1 - x)^j \,dx = 2^{1 - \frac{j}{2}} \mathrm{B}(i + 1, j + 1),$$
where $\mathrm{B}$ is the beta function, so for any rational approximant $q < \pi$ derived as above, in fact we have the improve bound $\pi > q + 2^{1 - \frac{j}{2}} \mathrm{B}(i + 1, j + 1)$. Accounting for this observation gives improvements for just the last two integrals in the table, replacing them with the $2\displaystyle \int_0^1 \frac{x^{\color{#bf0000}{17}} (1 - x)^2}{1 + x^2} \,dx$ and $4\displaystyle \int_0^1 \frac{x^{\color{#bf0000}{156}}}{1 + x^2} \,dx$, respectively.

D.P. Dalzell, "On $22/7$." J. London Math. Soc., 19 (75 Part 3): 133$-$134. doi:10.1112/jlms/19.75_part_3.133, MR 0013425, Zbl 0060.15306.
Answer: Here are a few more examples of the type user9413 gave, that is, integrals
$$\alpha \int_0^1 \frac{x^i (1 - x)^j (a x^2 + b) \,dx}{1 + x^2}$$ with $i, j \in \Bbb Z_{\geq 0}$, $a > 0$, $b > 0$, and value $\pi - \frac{333}{106}.$ Some quick Maple computations show that the integrals of this type minimal in the sense that there is no other example $(i',j')$ for which $i' \leq i$ and $j' \leq j$, are
$$\begin{array}{rl}
\hline
\displaystyle \frac{1}{2938320} \!\!\!\!& \displaystyle \int_0^1 \frac{(1 - x)^{12} (1\,501 x^2 + 185\,146) \,dx}{1 + x^2} \\
\displaystyle \frac{1}{1484} \!\!\!\!& \displaystyle \int_0^1 \frac{x^2 (1 - x)^8 (36 x^2 + 407) \,dx}{1 + x^2} \\
\displaystyle \frac{1}{530} \!\!\!\!& \displaystyle \int_0^1 \frac{x^5 (1 - x)^6 (197 x^2 + 462) \,dx}{1 + x^2} \\
\displaystyle \frac{1}{742} \!\!\!\!& \displaystyle \int_0^1 \frac{x^8 (1 - x)^4 (181 x^2 + 923) \,dx}{1 + x^2} \\
\displaystyle \frac{1}{2\,533\,697\,595} \!\!\!\!& \displaystyle \int_0^1 \frac{x^{27} (1 - x)^2 (294767566 x^2 + 5362162756) \,dx}{1 + x^2} \\
\hline
\end{array}$$
The third entry ($(i, j) = (5, 6)$) is the example user9413 gave. The second entry ($(i, j) = (2, 8)$) has the numerator with the smallest possible degree ($10$) among integrals of this type.


Do we know why St Oswin was canonised?

Question: Next session my group will be arriving in Triboar, which as far as I am aware is named in honour of St Oswin Boldfist's defeat of three large boars, according to the Terror at Triboar Dungeon Masters Guild adventure. Oswin was apparently canonised by the cult of Gwaeron Windstrom (I didn't even know saints existed in FR!), and honestly I'm not sure why - presumably not for fighting three CR2 (if we generously assume 'large' = 'giant') pigs. Before I retcon the three boars into a boar-headed hydra or something, am I missing some information somewhere that explains why Oswin's deeds were thought to warrant canonisation, given how rare this appears to be in FR? Official sources only, please.
Comment: Where did you find this information?
Comment: @tardigrade can you tell us where you found the name of this hero (Oswin Boldfist)?
Comment: According to [Kobold Fight Club](https://kobold.club/fight/#/encounter-builder), three giant boars in D&D 5E is a Deadly encounter for one lvl10 PC, and a Medium encounter for one lvl16 PC. Might be pretty heroic after all.
Comment: [this document](http://watermark.dmsguild.com/pdf_previews/173361-sample.pdf) published under [dmsguild.com](http://www.dmsguild.com/) mentions a St Oswin, but not the name Boldfist.
Comment: Checked through the adventure, it does give the name Oswin Boldfist further in. There appears to be a variant of the adventure in which St Cuthbert Stoutwalker is the patron of the town; I am uncertain which came first, but it was mentioned that there aren't many saints in FR; this may have been an expediency of the translation between settings.
Comment: That adventure is not official though, correct?
Comment: I would assume not.
Answer: We do not
Every source I have found on Triboar refers to the exploit that named it as being a "tale" of a "wandering traveler" and makes no attribution to any saint.
In point of fact, I was (at the time of posting this answer) unable to locate any source that mentions the name "Oswin Boldfist" at all, and I had done a fairly thorough search across every edition of the game. I'm very curious whether you found this reference in the DM's Guild adventure "Terror at Triboar."
We could conjecture...
These elements are fact:

Gwaeron Windstrom was deified by Mielikki for his battles with the avatar and servants of Malar.
Gwaeron Windstrom is said to sleep in a stand of trees just west of Triboar.
Malar is god of marauding beasts and monsters.
Mythology has used boars as terrible marauding monsters before (Calydonian Boar, Erymanthian Boar etc.)

So we can speculate some hypothetical origin story, like

Perhaps at some point during his time as a mortal (or afterwards, why not), Gwaeron Windstrom was weakened from battle with Malar's forces and needed to rest. Oswin Boldfist discovered a trio of dire boars, empowered and enraged by Malar's touch, hunting for the weakened Windstrom in order to kill him. Boldfist heroically pursued the boars, leaping onto one of them and turning its goring tusks into the flank of a second. The smell of blood caused the third to whip around in frenzy, and Boldfist stood fast against its charge. When Windstrom woke, he found the daring mortal lying atop the three slain boars and recognized the great service that had been done him.

But given the tone of the adventure...
The adventure Terror at Triboar suggests that "St. Oswin" is more of a local hero than an actual saint, the founder of a small town with a curious tale for a bit of color. Patrons at the tavern mutter about how they're tired of hearing the story of Oswin, which - as you've noted - is a bit sparse on relevance and heroics. It would be just as plausible, working from this source, that Oswin was never actually sainted and was given the title as a courtesy of sorts in recognition of his local status.
Sources Checked
City of Splendors: Waterdeep, Forgotten Realms Campaign Guide, Forgotten Realms Campaign Setting, In Volo's Wake, Powers and Pantheons, Princes of the Apocalypse, Storm King's Thunder, Sword Coast Adventurer's Guide, The North: Guide to the Savage Frontier, Terror at Triboar, The Savage Frontier, Volo's Guide to the North, Waterdeep and the North
Comment: I was checking every book about the north released during TSR's era and found nothing. I don't think the reason exists. But i am also curious about the source of that name (Boldfist).
Comment: Sorry, I was offline for a while - yes, it was from Terror at Triboar, which I'm aware is not an official source - I'd just assumed the name was originally taken from a canon source somewhere. Entirely my fault.
Comment: Nothing to apologize for, it was a valid question. Just wish it were easier to conclusively prove a negative.

Do we know why St Oswin was canonised?


Is there any shloka from any Upanishad or other scriptural text that may depict the content of passage of Brih .Up 3:8:9?

Question: Br Up passage  3:8:9 ( Atasya vaa aksharasya - -pitro nvayattah) --the content of this passage is so revealing. It's not only inferential rather experiential also on the contemplation of The Universal self. I get ecstatic while contemplating on this passage  ! Shall feel blessed if some brother help find me slokas from Upnishads Or other scriptural text ( in the form of poetry) that may depict the content of above referred passage of Br Up. Pl Bless me  !
Comment: Thanks Brother  !
Comment: Mundaka Upanishad, 2nd chapter. Also Katha II.ii.15, Svetasvatara VI.14. Gita 15.6 and Gita 15.12 and Gita 15.17-18
Comment: 'That which is infinite is immortal, and that which is finite is mortal.'- Chandogya Upanishad 7.24.1 'That which is infinite is the source of happiness. There is no happiness in the finite. Happiness is only in the infinite.' The real ecstasy or joy in Maya comes from the infinite Atman aka Sat Chida Ananda but an animal can never understand this and continues to fight for food, mate and shelter but a yogi restrains from Maya, while man is an intermediate stage. Thats why modern hindu monks aka Swami(knower of Self/Swayam) use the word Ananda in their names like Vivekananda, Yogananda etc.,
Comment: Can you clarify or specify bit what you didn't understand or what you're looking for in regards to the mentioned verse?
Comment: @ Pandya. Grateful. Pl have a relook at my jigynasa that has been put  through the question. For me it is simple to memorise shloka and sweet to chant. The deep import of the passage of Br Upnishad  as referred to in my question has deeply influenced me from within. I am in fact looking for some SHLOKA in some Upnishad Or other scriptural text which may have reflection of the content or import of the above referred passage of Br Upnishad  . Beg your pardon for the trouble.
Answer: I think the commentary of Adi Shankaracharya could help you.

The Śruti, by attempting to negate various attributes of the
  Immutable, has indicated Its existence. Yet, anticipating the popular
  misconception about It, it adduces an inferential evidence in favour
  of Its existence: Under the mighty rule of this Immutable, the Brahman
  that has been known to be within all, immediate and direct—the self
  that is devoid of all attributes such as hunger, O Gārgī, the sun and
  moon, which are like two lamps giving light to all beings at day and
  night respectively, are held in their positions, as a kingdom remains
  unbroken and orderly under the mighty rule of a king. They must have
  been created for the purpose of giving light by a Universal Ruler who
  knows of what use they will be to all, for they serve the common good
  of all beings by giving light, as we see in the case of an ordinary
  lamp.[4] Therefore That exists which has made the sun and moon and
  compels them, although they are powerful and independent, to rise and
  set, increase and decrease, according to fixed place, time and
  causes.[5] Thus there exists their mighty Ruler, the Immutable, as the
  lamp has its maker and regulator. Under the mighty rule of this
  Immutable, O Gārgī, heaven and earth maintain their positions,
  although they are by nature subject to disruption because of having
  parts, inclined to fall owing to their weight, liable to separate,
  being a compound, and are independent, being each presided over by a
  conscious deity identifying itself with it. It is this Immutable which
  is like a boundary wall that preserves the distinctions among
  thing—keeps all things within their limits; hence the sun and moon do
  not transgress the mighty rule of this Immutable. Therefore Its
  existence is proved. The unfailing sign of this is the fact that
  heaven and earth obey a fixed order; this would be impossible were
  there not a conscious, transcendent Ruler. Witness the Mantra, 'Who
  has made heaven powerful and the earth firm' (Ṛ. X. cxxi. 5).
Under the mighty rule of this Immutable, O Gārgī, moments, Muhūrtas,
  etc.—all these divisions of time, which count all things past, present
  and future that are subject to birth—are held in their respective
  places. As in life an accountant appointed by his master carefully
  calculates all items of income and expenditure, so are these divisions
  of time controlled by their master, the Immutable. Similarly some
  rivers, such as the Ganges, flow eastward from the White Mountains,
  the Himalayas, for instance, and they, notwithstanding their power to
  do otherwise,[6] keep to their original courses; this too indicates a
  Ruler. Others flowing westward, such as the Indus, continue in that
  direction, and still others keep to their respective courses, do not
  deviate from the courses they have taken; this is another indication.
Moreover, even learned men praise those that give gold etc., even at a
  personal sacrifice. Now the conjunction and disjunction of gifts,
  their donors and their recipients are seen to take place before our
  eyes in this very life. But the subsequent recombination (of the donor
  and the fruit of his gift) is a matter we do not directly see. Still
  people praise the charitable, for they observe on other evidence that
  those that give are rewarded. This would be impossible were there no
  Ruler who, knowing the various results of actions, brought about this
  union of the giver and the reward, for the act of giving obviously
  perishes then and there. Therefore there must be someone who connects
  the givers with the results of their charity.
Objection: Cannot the extraordinary result of an action (Apūrva) serve
  this purpose?
Reply: No, for there is nothing to prove its existence
Objection: Does not the same objection apply to the Ruler too?
Reply: No. for it is an established fact that the Śrutis seek to posit
  His existence. We have already (p. 53) said that the Sruits aim at
  delineating the Reality. Besides, the implication on which the theory
  of the extraordinary result depends is out of place, for the fruition
  can be otherwise accounted for. We observe that the reward of service
  is obtained from the person served; and as service is an act. and
  sacrifices, gifts, offering oblations in the fire, etc., are just as
  much acts, it stands to reason that the reward for their performance
  should come from those in whose honour they are performed, viz. God
  and so forth. Since we can explain the obtaining of rewards without
  sacrificing the directly observed inherent power of acts, it is
  improper to sacrifice that power. Moreover, it involves a superfluity
  of assumptions. We must assume either God or the extraordinary result.
  Now we observe that it is the very nature of an act of service that it
  is rewarded by the person served, not by the extraordinary result; and
  no one has ever actually experienced this result. So (in your view) we
  have to assume that the extraordinary result, which nobody has ever
  observed, exists; that it has the power to confer rewards; and that
  having this power, it does in addition confer them. On our side,
  however, we have to assume only the existence of the person served,
  viz. God, but neither His power to confer rewards nor His exercise of
  it, for we actually observe that the person served rewards the
  service. The grounds for inferring His existence have already been
  shown in the text: 'Heaven and earth maintain their positions,' etc.
  (this text). Likewise the gods, although they are so powerful, depend
  on the sacrificer for their livelihood—for such means of subsistence
  as the porridge and cakes.. That in spite of their ability to live
  otherwise they have taken to this humiliating course of life, is
  possible only because of the mighty rule of the Lord. Similarly the
  Manes depend for their subsistence on independent offerings. The rest
  is to be explained as before

.
Comment: @ Hare krishna - if you found his answer helpful, you should upvote it. Motivates the responder to keep contributing here:)
Comment: Grateful for showing interest for my concern. I do have the bhasya of sri shankar. In fact the content of the above passage is so enchanting that i did wish to memorise the same in shloka form ,had that been available even in some resembling form in some other scriptural text--and keep that chanting  . Anyway my deep gratitude!  My sincere request to blessed brothers to do help me.


Is being angry a prerequisite for sparring?

Question: My sensei says while sparring you should express your inner tiger. My understanding is that I should be savage and angry. But I don't like being angry! Is it possible to be good at sparring without anger impulses?
Comment: Tigers are fierce, but they are not *angry* when they hunt and kill, any more than you are angry when you go to the market and get your dinner.
Comment: Why not simply ask the sensei himself, what the actual meaning is?
Comment: It's 'sabum', not 'sensei' for TKD. A nice, simple question with some (IMHO) good answers!
Comment: Did they specifically mention anger? "Express your inner tiger" doesn't (necessarily) imply anger, nor savagery.
Answer: Emotions in general are detrimental to your sparring.
Usually when I have seen this said to a student it is not because we want you to get angry, but that you are sparring as if you are afraid. If you are afraid of striking your opponent you become SLOW and hesitant. Remember here that you should only ever spar willingly (never spar if you don't want to) and also that your partner is in the same position.
Hitting your partner (or at least trying in earnest to do so) is actually beneficial to them, if you don't hit them (by reacting calmly to the openings they present) then they will have a false sense of security/confidence going into other bouts and wind up getting hit more.
Sparring 'Angry' is usually an attempt by your opponent to intimidate you into losing confidence and not trying hard. Many people will try this tactic, learn to deal with it and keep your calm as they are using that instead of skill to try to win. Once you have dealt with people sparring this way a couple of times you will learn to cherish new opponents that try it as they are easy to pick apart and usually don't have any other tactic to fall back on.
As @Kristina Lex says - reacting to the situation and putting a hand or foot to any target that presents itself is key - and much easier if you aren't angry/afraid/flustered - leave emotions at the door of the Dojang or on the pads/bags.
Comment: Remember, anger is a weapon only to your opponent.  Forgetting this is a good way to get a boot to the head!
Answer: What works for your sensei might not work from you. He is teaching you, what had worked for him. For some, they use anger for the aggressiveness it brings. Anger dominates certain opponents. And in the case of draw, the more aggressive fighter that attacks more (whether contact or not) is usually declared the winner.
But anger does not work for everyone. It results in your judgment being clouded, tunnel vision, rash decisions leading to mistakes, disqualifications due to fouls, not to mention raising your blood pressure and heart rate for the wrong reasons.
I prefer teaching students to fight with a cool head, but maintaining an 'afraid-to-lose' mentality. As in, never let the other guy score more points than you. And reactive fighting. Seeing an opening results in automatic launching of a kick to that area.
Comment: "express your inner tiger" is about being angry, it about being confident and aggressive (vs afraid and passive).
Comment: I wish I could accept more than one answer! Thanks again =)
Answer: While tigers can be considered savage in some sense, as in "uncivilized" or "fierce", they certainly don't get angry. When they are attacking, they have a goal in mind ("kill this prey to satisfy my hunger") and they use their instinct and physical might to reach that goal, there is no emotion. When they are defending themselves, they again use their instinct and physical abilities to save their lives, whether it's fighting back or running away.
I think that is what your sensei may have meant.
Answer: Anger is definitely not a prerequisite for sparring, specially since angry people tend to neglect their technique, but I think that you might have misinterpreted what your sensei told you.
A tiger hunts with everything it has, full concentration, full physical force and deadly precision, since they hunt completely alone.
Keeping this in mind and the fact that you yourself said that you don't like being angry, he might have ment that you should try to be less timid in sparring and go for it with more "power".
It's about focus, not anger.
Answer: No, anger is not a prerequisite for sparring.
There is more than one school of thought for what your mental state should be to spar/fight. This list is not meant to be exhaustive:

Fight without emotion. Buddhist or Daoist training for a state where thought is stopped (mushin) also means that your emotions do not disturb your mental state and create thoughts. You can act decisively and without hesitation if you are mentally unencumbered.
Harness emotion for strength. This sounds like what your instructor is suggesting. Many athletes use music to prime themselves before competition to improve performance. A snarling animal can be intimidating and cause an opponent to think twice before engaging.
Crazy. Examples are Viking berserkers or Moro rebels in the Philippines. This mental state is for killing without worrying about getting killed. The US Army switched to using higher caliber (larger) bullets with more "stopping power" to fight the Moro rebels because they would keep fighting even after being shot.

There is a range of possibilities.


Human somatic cells have 22 pairs of homologous chromosomes?

Question: I am studying for my Bio class and I am stuck with questions that I came up with during this chapter of Meiosis. I understand that human somatic cell is made up of 22 pairs of autosomes and 1 pair of sex chromosome. I am confused about what made up each pair of autosomes? is that 1 chromosome from the father and 1 chromosome from mother made up 1 pair of autosomes in somatic cells? and therefore, since each pair of autosomes consisted of 1 chromosome from dad and 1 from mom, we can say human somatic cells consisted of 22 pairs of homologous chromosomes (not to mention the sex of the human), right?
Comment: @PolypipeWrangler 22.5?!  You can't have half a homologous pair.  But I get what you're trying to say and it's true... There are 22 homologous pairs and half the time there is a homologous XX pair.  The other half the time there's a heterologous XY pair, though even portions of those chromosomes are homologous...
Comment: Got it! thank you so much for the clarification!
Answer: Answer: *Yes  , human somatic cells have 22 pairs of homologous chromosomes
The human genome consists of 46 (2n*(22+1) ) chromosomes.
Most of our cells are somatic and diploid (2n), they have two homolgous chromosoms (called homologs, one "father" & "mother") these cells multiply by mitosis.
Reproductiv cells or gonosomal cells, (gametes; Sperm and eggcell(ovum)) are the product of meiosis ( a part of gametogenesis ) . These cells are haploid (1n), they only have one set of chromosomes (1n * (22+1)) which is a combination of "fater&mother". When 2 gametes  combine a diploid zygote is formed.
In both cases (Mei- and Mitosis) chromosome (pair (2n)oder single(1n)) number 1 -22 are called autosome , while
Chormosome number 23 is an allosome (sex chromosome), either Y (male) or X (female).
Since the gametes are haploid (1n) they have only 1 single allosome, egg cell always X , and  sperm cell either X or Y.
If the allosome (the 23 chormosome pair) of an human is hemizygot ( X–Y ), he is genetically male. Is the allosome pair X-X
,homozygot the person is genetically female.
Somatic cells can have therefore 22  pairs of homologous chromosomes the 22 autosomes. The allosome pair of chromosomes is not considert homologous but either "homozygot / homogametic"[XX] or "hemizygot/ hemigametic"[XY].


Is there a difference between foreign and native accents?

Question: As the title says, are there any linguistic differences between accents acquired from birth/childhood and accents ued by adult language learners who speak the language fluently, but still with a pronounced accent? Can native speakers identify which accents are native and which are developed by adult learners?
Comment: I know a few high level ESL speakers who are very articulate in English, but all of them have traces of the accent of their first language. It might not be obvious all the time but it will show up here and there. Some people have an accent that is so slight that many native speakers don't even notice until some linguistic slip tips them off and they start listening for it. I guess if a person had a real talent for mimicry, and learnt a second language to a very high level, they might eventually be able to pull off a native accent - but that situation is going to be extremely rare.
Comment: @Minty from my understanding, pretty much only children and younger teenagers can pick up another (native) accent perfectly. My question though is rather if there are a difference in the nature of native accents and accents that is influenced by the speakers native tounge.
Comment: Don't forget - there are also **suprasegmental** features, although it's not quite clear how broad your understanding of "accent" is. The fundamental difference is, naturally, L1 transfer - in the case of L2 phonetics.
Answer: An "accent", which is not a technical linguistic concept, generally refers to some deviation from a social norm, and unless you have e.g. the speech of the Queen as your guide, knowing exactly what constitutes that norm can be very difficult. In the context of American speech, "southern accents" are not the standard, nor are northern dialects (Chicago, Minnesota, Buffalo). Given that, I think your question has to come down to asking whether – given a target social standard of pronunciation – do people growing up in that environment acquiring that dialect as their first language pronounce things differently from adults who learn the same dialect as a native speaker?
Obviously in the case of L2 acquisition there is a vast continuum in how well L2 speakers acquire local norms of pronunciation, so you would want to only consider those speakers who have achieved very high phonetic fluency in the target language. I know a guy who migrated to the US from Israel as an adult, and he has an impeccable New York accent, to the point that other New Yorkers (not linguists) have commented on how he sounds just like a native speaker. I also know people who "have foreign accents" (Japanese, Spanish, Norwegian), but these are native-foreign accents, that is, they are the result of a local modification of pronunciation norms, arising from there being very many speakers of the dialect / foreign language and a high degree of bilingualism. Although most if not all second-generation Japanese are native speakers of English, their dialect is/was detectable.
I conclude that native speakers can make educated guesses as to whether a person is a native speaker vs. L2 speaker, but not always. If you are not familiar with the target dialect, your odds of guessing right are decreased.
Comment: I don't think you will find many linguists who agree that an accent is "some deviation from a perceived norm". That would imply that people whose speech is close to that norm have no accent.
Comment: This linguist would say that "accent" is just another term for "phonological idiolect" (though some people include local lexical items like _pop_ and _soda_ in "accent"). That is, the way somebody sounds that one remarks as different from one's own (naturally accent-free) pronunciation. This is true for everybody; they all speak the correct, standard way, and hear any recognizable difference as "having an accent". The question is whether the "accent" identifies the speaker as being out-group or in-group with regard to some important social group.
Comment: As for the difference between foreign and local, that's basically the difference between native and non-native. Non-native accents usually have some of the characteristics of the native language, whatever it was, but that's not true of regional or social accents. George Mason University has a [Speech Accent Archive](http://accent.gmu.edu/), with both local native and various non-native speakers saying the same English paragraph; each recording  is transcribed in IPA. See whether you can distinguish.
Comment: Accent not being a linguistic concept (it's a popular one) I wouldn't expect linguists to agree. That's why non-linguists say things like "he speaks without an accent", and why linguists talk of dialects.


Reduced special fiber implies reduced generic fiber for a projective morphism?

Question: Let $f:X \to Y$ be a projective morphism between irreducible Noetherian schemes. If a fiber over a closed point of $Y$ is reduced is the generic fiber reduced?
Comment: You may be interested to know that if $f:X \rightarrow Y$ is a proper flat map between noetherian schemes then the locus of $y \in Y$ such that $X_y$ is *geometrically* reduced is always open. This is EGA IV$_3$, 12.2.4 (which provides a wealth of openness results for properties of *geometric* fibers).  In practice,  it is properties of geometric fibers (rather than actual fibers) that define reasonable loci in the base (e.g., geometric irreducibility works well, whereas mere irreducibility isn't even a constructible condition on the base in general).
Answer: Let's impose a flatness hypothesis (whose necessity is explained in Jason Starr's answer). The answer is still negative, but to explain the context for the counterexample it is instructive to first record some necessary features of any counterexample, so we know where to look.
In view of my above comment about geometric fibers, any search for a counterexample will necessarily have to involve a special fiber that is reduced yet not geometrically reduced.  In other words, any counterexample will have to involve an imperfect residue field at the special point in the base.
Moreover, the answer is affirmative if the base $Y$ is Dedekind.  Indeed, we can assume in such a case that the base $Y$ is Spec($R$) for a dvr $R$ with uniformizer $t$, so $X$ is a proper flat $R$-scheme having reduced special fiber $X_0$.  If $N$ is the coherent sheaf of nilpotents inside $O_X$ then for any section $f \in N(U)$ over an open $U \subset X$ we know that $f_0 = f|_{U_0}$ vanishes since $X_0$ is reduced, so $f$ is a section of $tO_X$. In other words, $N \subset tO_X$. By $R$-flatness of $X$, it is clear (check!) that $N \cap tO_X = tN$, so $N \subset tN$.  Hence, by Nakayama's Lemma along points of the special fiber, it follows that $N$ has vanishing stalks along the special fiber $X_0 \subset X$.  The closed support of the coherent $N$ inside the $R$-proper $X$ is therefore disjoint from the special fiber. But every non-empty closed set in $X$ must meet $X_0$ due to $R$-properness of $X$ and locality of $R$.  Thus, the support is empty, which is to say $N = 0$.  Then $X$ is reduced, so its localization given by the generic fiber over the integral base $Y$ is also reduced.  That settles the question affirmatively when $Y$ is Dedekind (and $X$ is any proper flat $Y$-scheme).
In view of the preceding observations, an "optimal" counterexample should involve a local base $Y = {\rm{Spec}}(R)$ where $R$ is a non-Dedekind 1-dimensional local noetherian domain having imperfect residue field with characteristic $p > 0$.  And in fact there are counterexamples over such $R$, as we now construct.  This shows that the EGA result with reducedness for geometric fibers is essentially "best possible".  We will give counterexamples in equicharacteristic $p$. Maybe someone else can address the story with generic characteristic 0. (SEE THE END FOR SUCH AN EXAMPLE)
The idea is to make such an $R$ that is an order in a dvr of characteristic $p$ whose fraction field contains a certain $p$th root but for which $R$ does not contain that $p$th root.  Let $k$ be a field of characteristic $p > 0$ and let $A = k(t)[x]_{(x)}$, a dvr with uniformizer $x$ and residue field $k(t)$ that is obviously not perfect.  Let $F = {\rm{Frac}}(A) = k(t,x)$ and let $A' = A[T]/(T^p - t) = k(T)[x]_{(x)}$, so the dvr $A'$ with uniformizer $x$ is the integral closure of $A$ in $F' = F[T]/(T^p - t) = F(t^{1/p})$. Note that $A \rightarrow A'$ is an example of a finite extension of discrete valuation ring whose ramification degree is 1 but residual extension is not separable (so it is not "unramified").
The residue field of $A'$ is $k(T)$ in which the element $t$ from the residue field of $A$ has image $T^p$ that is a $p$th power, but we can "fix" that (or rather, "ruin" it) by considering the order
$R = A + xA'$.  Concretely, $R$ is the preimage of $k(t)$ under the reduction map $A' \twoheadrightarrow k(T)$.
Clearly $R$ is a 1-dimensional local noetherian domain whose fraction field is $F'$, residue field is $k(t)$, and normalization is $A'$. Note that the element $T = t^{1/p}$ in the fraction field $F'$ of $R$ does not lie in $R$.
Inside $\mathbf{P}^2_A$ with homogeneous coordinates $[U,V,W]$, consider the closed subscheme $Z$ defined by $U^p + t V^p$. This is $A$-flat since $A$ is Dedekind.  Its special and generic fibers are the "same" projective scheme over the residue field $k(t)$ and fraction field $k(t,x)$ respectively, and as such these fibers are both reduced (as $t$ is not a $p$th power in either the residue field or fraction field of $A$, each of characteristic $p$) and are also both geometrically irreducible. The base change
$X = Z_R$ is certainly $R$-flat (since $Z$ is $A$-flat) and projective with special fiber that is the same as that of $Z$ since $R$ and $A$ have the same residue field by design.  In particular, $X_0$ is reduced (though not geometrically so!). The generic fiber $X_{F'} = Z_{F'}$ is non-reduced since $t$ is a $p$th power in $F'$.  Note however that the $R$-flat $X$ has irreducible generic fiber, so $X$ is also irreducible.   Voila.

EDIT:
Here is an even better example in the same spirit, but with a twist on the idea so that it works in generic characteristic 0. It will use an order in
$\mathbf{Z}[T]_{(p)}$ as I had been hoping.  This example was pointed out to me by somebody "offline".
Let $A = \mathbf{Z}[t]_{(p)}$, a dvr with uniformizer $p$, residue field $\mathbf{F}_p(t)$, and fraction field $F = \mathbf{Q}(t)$. Let $A'= \mathbf{Z}[T]_{(p)}$ made into an $A$-algebra via $t \mapsto T^p$, so its fraction field $F'$ is $F(t^{1/p})$.
Let $R = A + pA'$, a 1-dimensional local noetherian domain with residue field $\mathbf{F}_p(t)$ and fraction field $F'$.
Consider the polynomial $f(U,V) = (U + T V)^p \in U^p + t V^p + pA'[U,V]
\subset R[U,V]$.
Obviously $f$ is a $p$th power over $F' = \mathbf{Q}(T)$, but it is not a (unit multiple of a) $p$th power over $R$ since the reduction of $f$ over the residue field of $R$ is $U^p + t V^p \in \mathbf{F}_p(t)[U,V]$, which is even irreducible.
Let $X \subset \mathbf{P}^1_R$ be defined by the vanishing of the homogeneous $f$.
(The equicharacteristic-$p$ example above ought to have been considered inside $\mathbf{P}^1$ just as well; I mistakenly thought being in $\mathbf{P}^2$ would provide better irreducibility properties, but those hold already with $\mathbf{P}^1$.)
The $R$-scheme is $R$-flat because each of $U^p$ and $V^p$ have coefficients that are units in $R$ (ensuring that after dehomogenization, the coordinate ring of the affine open chart of $X$ is a finite free $R$-module). In fact, we see that $X$ is finite flat over $R$, with degree $p$.
The special fiber of $X$ is the zero scheme of $U^p + t V^p$ in $\mathbf{P}^1$ over $\mathbf{F}_p(t)$, and this is visibly reduced.  On the other hand, the generic fiber is $(U+TV)^p=0$ in $\mathbf{P}^1_{F'}$, and this is visibly not reduced (but is irreducible, and likewise $X$ is irreducible; in fact, $X_{\rm{red}} = {\rm{Spec}}(R)$!).
Comment: this is a very nice answer !
Comment: OK, thanks to a suggestions from someone offline, an example has now been given with generic characteristic zero. I have put it at the bottom of the answer.
Comment: This is a truly great answer.
Comment: Thanks, Olivier.  I hope someone can figure out if there is a counterexample with generic characteristic 0, such as over an order in $\mathbf{Z}[t]_{(p)}$ or something like that; it would be even nicer. :)
Comment: It is a pleasure to read your answers.
Answer: That is not true if you do not add some sort of flatness hypothesis.  For a quick counterexample, let $Y$ be $\mathbb{A}^2$ with coordinates $(s,t)$.  Let $\mathbb{P}^2$ have homogeneous coordinates $[U,V,W]$.  Let $X$ be the closed subscheme of $\mathbb{A}^2\times \mathbb{P}^2$ with defining ideal $\langle W(tU-sV), (tU-sV)^2 \rangle$.  Set-theoretically, $X$ equals the Cartier divisor $Z(tU-sV)$, which is irreducible.  The generic fiber is non-reduced: it is the line in $\mathbb{P}^2$ with equation $tU-sV=0$, but with an embedded point at the point $Z(tU-sV,W)$.  However, the scheme-theoretic fiber over the origin in $\mathbb{A}^2$ is reduced: it is all of $\mathbb{P}^2$.
If you add a flatness hypothesis, this should follow from Auslander's Theorem, cf. EGA IV.6.11.
Comment: @Starr: The "problem" with applying Auslander's theorem is that it does not address fibral aspects, only properties of local rings on the "total space".
Comment: @Marguax:  I was assuming that the OP was asking about geometric fibers.  In this case, I think Auslander's Theorem does apply.  I can base change to a DVR whose generic point maps to the generic point of $Y$ and whose special point maps to the given closed point.  So assume that $Y$ is regular.  Then, if the closed fiber is reduced, or more generally $S_d$, then also the domain is $S_d$.  Now apply Auslander's theorem.  I would need to double-check in EGA, but I have a feeling this is roughly how the openness result is proved in EGA as well.
Comment: @Starr: Ah, I didn't realize you had in mind just geometric fibers.  Your proposed argument (and sense of how it probably goes in EGA) sounds reasonable to me.


Chemical characteristics of taste

Question: I know this is probably a quirky question for this forum, but here goes. I'm getting into home brewing, and today I stumbled a cross the website for American Society for Brewing Chemists (ASBC) http://www.asbcnet.org/checksample/beer_analysis.htm. The page I just linked to is for the beer analysis service offered by ASBC. For the analysis, they test 32 different properties of the beer. Does anyone know if there is a known way to quantify or approximate the taste of a beer just with these 32 properties? Again, sorry if this is out of place for this forum.
Answer: There are a lot chemical compounds that contribute to the overall taste/aroma/etc of a beer let alone the proportions of each said compounds. Unless those 32 tests are incredibly comprehensive, I am guessing it could probably at only guess at beer category. The yeast make up around 600 chemical compounds alone. Most of those compounds are barely perceivable by taste/aroma so if any of the proportions move around, the flavors could drastically change.
Comment: Thanks for the Answer. I love craft beer, and I'm taking a graduate level machine learning course at my university. I'm just curious as to if anybody has ever tried to map quantifiable beer properties to say Beer Advocate ratings. Conceptually, you could predict the rating of a beer without needing to taste it. This assumes, however, that there is a way to roughly quantify beer taste by doing chemical analysis.

Chemical characteristics of taste