Can I get additional reputation reward from badge like nice question or nice answers?

Question: We have the badge system that encourage users use more features on the site. Does that make sense to give user more reputations after achieving badges like nice questions or nice answers?
Comment: It's not completely clear what you're saying. Which badges in particular are you referring to? Are you suggesting adding a reputation bonus to some badges, or are you suggesting that with certain badges (like Good Question and Good Answer, for example) since the poster has already gained reputation for that answer or question that a badge as well is unnecessary?
Comment: While there's not much issue with discussing these things, note that changes to the system would be network wide, so if your intent is to propose some change this would probably be better placed on the network meta (meta.stackexchange.com). If that's what you're trying to achieve, once your question is polished up it would be possible to migrate it. (Though I expect this has probably already been raised there, so check for duplicates first.)
Comment: The latter. Thanks
Comment: Could you edit your question to be more specific and/or explicit, please?
Comment: Sorry for unclear question, I am new to CV and have a lot of questions in mind. When I ask this question I did not thing though.
Comment: Your title still seems to be asking the former (it appears to be a request for additional reputation for earning a badge) rather than the latter (questioning whether both badges and reputation are necessary)
Comment: @Glen_b Yes, I think myself is not clear what is the question, but your answers explains my confusion 100% in detail in both ways. If you think it is necessary to make it more clear, please feel free to edit my question.
Answer: Reputation and badges are not simply rewards (and indeed, as rewards they're  fairly flimsy when one can earn very real rewards putting the same effort into other activities).
Instead, reputation and badges also (and perhaps more usefully) serve as signals to others, but they signal different things.
At the present time I have a similar reputation to whuber (and I hope he will excuse me mentioning him in my answer), but he has a substantially higher number of "Good Answer" badges than I do (about 5 for every 3 I have, even though I have answered more questions). Good Answer badges are quite rare, so someone holding a lot of them tells you something important about the standard of their answers.
More of the story comes when you look at the tag badges, for example, whuber has a spatial tag badge, spatial statistics being a topic on which he has vastly more knowledge than me. Knowing that he has that tag badge and I don't might help to indicate his relative expertise on the topic.
There are some users on site that have a large amount of reputation that has been mainly gained from asking good questions. That certainly indicates important contributions to the site, but might give you a different view of their answer to some question when compared to a user of similar reputation which has been almost entirely earned on good answers.
Reputation tells our readers only part of the story. Badges tell another part. Both can be useful in giving users a picture of the likely value of what is being said (if it's not otherwise clear). It's no guarantee of course -- anybody might have an off day and give an answer below their usual standard, for example, and ultimately an answer must stand on its own merits -- but such indications can sometimes be useful.


Illustrator: how to save a JPG export preset

Question: Every time I want to export a JPG, I have to set the quality setting, which never happened with previous versions of Illustrator. It's resets to 0 each time, which is unusable for me.
Answer: The 'Save for Web' (CTRL+SHIFT+ALT+S) feature gives much more options, including creating export presets, saving the image at scale, clipping to artboard, etc.
If you prefer using the 'Export' option, normally it should remember the last setting you used (it does on my machine). Try resetting the AI preferences by holding CTRL+SHIFT+ALT (Windows) as you re-start Illustrator.
Comment: Unfortunately, save for web is out of the question, since the final destination for the JPG is Word, and strange things occur when it's inserted into the document (must be resized, sizing to 100% provides an inaccurate representation, etc.).
Resetting the preferences at work requires an administrator, which I am not. Maybe this has been resolved in a patch from Adobe, but knowing them, I would bet against that.
Comment: Try to use Bridge if you have it available. See this [answer](https://graphicdesign.stackexchange.com/a/94585/62949) and the comments below. Apparently only works for CS6 or older, not CC.
Answer: Here is how to fix it (make sure Illustrator isn't running):

In C:\\AppData\Roaming\Adobe\Adobe Illustrator 21 Settings\en_US\x64 there is a file called "Adobe Illustrator Prefs" with no extension.
Make a copy of this file somewhere so that you can get it back if something goes wrong. :)
Open the file with notepad and search for "/JPEGFormat" without quotes.
There are several options there that are associated with the Export dialog box above. In my case, I changed the "\Quality" setting to "8" and VIOLA! it saved the setting so I don't have to change it each time.
Save the file (File>Save, NOT File>Save As) and launch Illustrator to test it out.


How a compression of gas molecules change their velocity therefore their KE?

Question: How can a compression of gas molecules change their velocity therefore their KE? My intuition suggest me that there should be a change of mean molecule path length and so the frequency they hit the container should rise and that should be the cause of greater pressure. Can we compare this with a tunnel with a particle oscillating between the two entrances of the tunnel. When we shrink the tunnel length the particle should hit the entrances more frequently.Let say we move a wall against them... they will gain twice the velocity of the wall as it ca be considered as infinite mass in comparision with the molecules and if we count how many times the molecule intercts with the moving wall during motion of the wall we get the incease of velocity. So greater the frequency of collisions with the wall the greater the change in velocity of the molecule?
Answer: The average kinetic energy of gas particles is proportional to the temperature of the gas. If we compress the gas without changing its temperature, then the average kinetic energy of the gas particles will stay constant. There will be no change in the speed with which the particles collide if you increase the pressure (isothermally).
You may have increased the frequency at which the particles strike the container's walls and each other, which decreases the mean free path, but their kinetic energy will stay the same.
Comment: But, but, but I think the question is about simple compression,  not the isothermal one that requires special pistons etc?
Answer:
So greater the frequency of collisions with the wall the greater the change in velocity of the molecule?

If we're being precise, we'd have to know not only the collision count, but the collision speed, since the KE gain depends on the mass and the inbound speed.
But that renders just fine to the expected form here which is that the increase in KE is equal to the work done, and that is Force across a distance.  And for a gas with a greater number of collisions, you'll have a larger force.


Forgot email associated with my itunes account

Question: Is there a way to find out what email address is used on an itunes account. I have my password but can't remember the email address! Help please!!
Answer: You could have a look in your keychain.
/Applications/Utilities/Keychain Access.app
Select the Login keychain in the top left panel and the All items in the bottom one.
Use the search in the top right corner.
Search for iCloud, MobileMe ,iMessage or Apple ID.
Double click any of the items that come up and see if the jog you memory.
Most of these will bring up the email for your apple id.
Answer: If you bought anything from the store with this account, just select the matching file in iTunes, press Cmd-I to access the info window and go to the first tab (Overview). Your Apple ID should get displayed there.


What is the origin of being in "hot water"?

Question: When a person is suspected of doing something that is wrong, they are said to be in "hot water". Why? Did law breakers long ago have hot water thrown on them for their punishment? Were they dunked in hot water?
Comment: I heard on PBS that in Old England a suspected criminal was forced to put their arm in a cauldron of hot water up to the elbow. If the arm did not heal in a couple of days, they were found guilty.
Answer: The OED's first citation for hot water in the sense "trouble" is:

1537  Lisle Papers (P.R.O.) XI. 100  I can get no conserve dishes, for those that my Lady Fitzwilliam hath came out of Levaunt; howbeit, if they be to be had, I will have of them, or it shall cost me hot water.

Cost me hot water seems to have been a set phrase:

1593  G. Peele Edward I  it shall cost me hot water but thou shalt be King Edward's man
1640  T. Hooker The Soules Humiliation 38  Nay, the Lord knowes that my corruptions have cost me hot water, my heart hath beene exceedingly vexed with them
1654  E. Gayton Pleasant Notes Upon Don Quixote 79  this same search hath not cost me hot water (as they say)

My speculation: maybe hot water was a canonical example of something troublesome to procure (in the days before hot and cold running water)?
Answer: All, I could find was:

This idiom means to be in big trouble or be in an embarrassing situation with someone. It originated in the early 1500s. It may refer to when you're cooking and you spill hot water or hot food. You would be in trouble. Hot water is definitely something you don't want to be in except in a hot bath! An example of how to use this is, "I was in hot water with Ms. Izenberg when I didn't do my homework."

etymonline gives:

hot water
      c.1400, literal; 1530s in figurative sense of "trouble."
Answer: I am coming more and more to the opinion that it is fruitless to argue, or even speculate, about the "origin" of many idioms.
Obviously we can in principle find out when this expression was first recorded (we may or may not be able to do so in fact). But unless this first occurrence was in a context that makes it clear that the speaker was thinking about, say, cooking, then I suggest that we cannot know what image the speaker had in mind.
And even if that person who first used the expression had a particular image in mind, that doesn't mean that the people who heard it and used it themselves had the same image.
To me, the picture that "be in hot water" suggests is the old cartoon image of the explorer captured by "cannibal", tied up in a huge pot of water. But from the dates given by other replies, I suspect that the phrase is much older than that picture. (I may be wrong: perhaps that image goes back to the fifteenth century).
But that is just my picture. You can hear me use the expression, and understand it, even if you have quite a different picture. I suggest that this has always been the case, and that there is not and never has been a specific concrete image which the phrase recalled.
I would also note that both of the specific suggestions reported in other answers
are unconvincing to me, as neither of them accounts for "in hot water".
Edit: the above applies to those idioms which are more or less transparent. There are others which are pretty well opaque (such as "kick the bucket", or "take a rain-check"), and it does make sense to discuss the semantic origin of these. There are no doubt examples which some of us will find transparent and others not.
Comment: Although I'll respectfully disagree about the fruitlessness of speculation, I agree that none of the answers adequately address the "in" part of "in hot water". (I also have a similar mental image as you, while sharing your concern that such an image is unlikely to have been available in the 15th or 16th century.)
Comment: I'll give +1 overall to your answer, since I feel it brings good points to discussion, even though I don't agree with parts. Especially 'You can hear me use the expression, and understand it, even if you have quite a different picture.' - this definitively *does not* work with pure idioms (and I would say this is the reason why pure idioms usually have an associated story - since they have to be taught, either by examples or through story; if through stories then they can be either pseudo-etymologies or real etymologies).
Comment: @Unreason: Agreed. Some idioms are not transparent, and you either have to know their specific origin, or learn them as an unanalysable chunk: for these, it is appropriate to discuss their origin. I'll edit the answer.
Answer: From the book, "A Hog on Ice":

Possibly the allusion was to the ancient way that unwelcome guests were sometimes warded off - by heaving a kettleful of boiling water, when available, upon troublesome intruders.

However, I do know for certain that it originated round about the sixteenth century
There's also this explanation, but I wouldn't place too much on it:

It originated in the early 1500s. It may refer to when you're cooking and you spill hot water or hot food. You would be in trouble.

One thing's for certain, it has been in used since the early 1500s.
Comment: I think that kitchen has pretty good chance to be etymologically related.
Comment: Somehow, I doubt that reason, but please explain @Unreason, why does it seem more plausible?
Comment: Usually idioms come from common phrase or a very impressive, extreme phrases. As eating is a basic need that was present during not only development of particular language, but during development of all languages, there are many idioms related to food and cooking. From top of my head: "The amplifier got *fried*", "I got *burned*", "That story sounds *cooked*", "We are in such a *jam*", etc.. For these reasons I chose the words - "it has pretty good chance to be related", however, I think that actually the burden is on you to justify why you "wouldn't place too much on it".
Answer: My first thought was the "frog in boiling water" metaphor, that if you turn up the heat gradually it won't notice until it's been cooked alive.  Depending on the age of the saying, it might alternatively be referring to those foods which are actually cooked alive (crabs, lobster) for roughly the same meaning, before the metaphor was a set saying.
So the saying someone's "in hot water" would mean the person was in an uncomfortable situation (in trouble) that's only getting worse, so suggesting that the person had better fix it before it's too late - then their goose is cooked (to borrow a different idiom).


How to root Moto G 3rd generation running Marshmallow?

Question: For the past few months I have tried various methods of rooting my Motorola Moto G 3rd gen (Android 6.0.1) and they have all failed. How on Earth do I root my phone?
Answer: Excerpt from my XDA Tutorial on rooting the Moto G 2015 -

This guide covers ALL Moto G 2015 models (including the G Turbo) that can be bootloader unlocked and have TWRP available, it doesn't matter if it is the XT1540, XT1541, XT1543, etc., the method is identical. In fact, this method will work on all 2015 and 2016 series Moto phones running Marshmallow, including the Moto E, X, G3, G4, and Z.
Please read completely and make sure you understand each step and how to do it, if you don't get something search the thread or ask an open question.
tl;dr version for advanced users: Skip to "How to do it" section
I am writing this tutorial because there are still lots of failed root attempts using older "standard" methods that do not work on this device. I also do not like the "one click" root methods, because they can and do fail (KingoRoot will brick a Moto G3, regardless of what it's web page says), and when they do people have no idea how to fix it. The manual way is not difficult, and it teaches you how to work on, fix, and use your device on a level above that of the average smartphone user.
I will only cover the details of rooting, the prerequisites are covered elsewhere in detail and I will link to reliable sources for the information. Specifics of the prerequisites are outside of the scope of this tutorial, but are open for discussion in this thread.
Prerequisites:

Device must have an unlocked bootloader. See Moto - Unlocking the Bootloader for more info. NOTE: Performing this prerequisite will give you all the tools (fastboot) and drivers needed to continue, and wipe your device, meaning erase all your information and programs. Make sure to backup any important data first. This step will also permanently void your warranty once an unlock code has been issued by Moto, it doesn't matter if you even use it, your warranty is done.

You need to have TWRP installed or one-time booted via fastboot. CWM and other recoveries will NOT work at this time. See TWRP for the Moto G 2015 or TWRP for Moto G Turbo.

You need a copy of the latest STABLE SuperSU ZIP from Chainfire's site on the internal storage or SD card of your device.

Reboot and start TWRP recovery, the method you use to do this doesn't matter (boot or flash).

How to do it:
Now, the procedure is the same whether you are trying to root the first time, or you did it the old way just flashing SuperSU and are now not able to boot...
In TWRP, go to Advanced and open the Terminal, in the terminal type this EXACTLY as shown:
echo SYSTEMLESS=true>>/data/.supersu
Now press enter (there is no confirmation returned), then exit and press the Home key. Go to Install and select the SuperSU zip file you downloaded from Prerequisite #3 and swipe to flash it and reboot. No need to clear caches or anything else but you are welcome to if you wish. You can install SuperSU updates normally through the app going forward (as of this posting).
Why do I have to do this???
For whatever reason, the install script for SuperSU does not recognize that this device (like many others) requires a systemless root installation. By creating /data/.supersu in the TWRP recovery environment, the SuperSU install script parses the file and sees "SYSTEMLESS=true" and ignores what it auto-detects and forces a systemless root installation.
Hope this is helpful to someone!
As always, if this is the first time you have booted TWRP or attempted root... BACKUP IN TWRP FIRST!!! Once the system is modified, it cannot be undone (easily) and you will always have a known good starting place if the worst happens.
DISCLAIMER: I am not responsible for anything that happens... Your device, your responsibility, no matter what happens... Although all the information here has been tested and is known to work with no issues except where noted, things occasionally can and do go wrong and we cannot foresee every possible scenario or circumstance.
If you need assistance, we need detailed information about what what's going on and what you have tried to do... Please provide details including:

Device model number and Android version
TWRP version
SuperSU version
Output of 'cat /data/.supersu' from TWRP terminal

And a copy of recovery.log if possible, and any other pertinent information you can think of. Without  specific information there is nothing that I can do to assist you and request for help may be ignored.
Photos added showing what a proper command and flash should look like. Note that in picture 1 the exit command is not needed, you can just back out. In pictures 2 and 3 a proper flash of SuperSU is shown, note that system-less mode is specified and the boot image is patched, this is what should occur. It is normal for it to loop once or twice, but that is it, first boot could take 10 minutes plus.
Comment: acejavelin my Moto G 3rd Gen already boot-loader unlocked. If I 2 to 4 after that will it wipe the existing data from internal memory.
Comment: @Ajay It will not... The unlocking process is what wipes the data.


Brahmā of Brahmā Sütrā

Question: What is the name of the Brahman who has been depicted in the Brahma Sutras?
Answer: सीतानाथ समारम्भां रामानन्दार्य मध्यमाम्।
अस्मदाचार्य पर्यन्तां वन्दे श्रीगुरू परम्पराम् ।।
In the scriptures it is said that one scripture is an explanation of another scripture. Like wise if we go with Brahmā Sütrā, in Vedas it has been already depicted for whom that "Athatö Brahmā Jīggyasā" been called, who is "Ānand Swarōop", etc. So let's start,(Featuring only few sutra which will give you clear insight who is that Brahmā?)

In the begining, Brahman desired Let me be many, so he appeared in
many forms. Shruti says "Bhagavān didn't enjoy (alone), so He desired
a second, and Therefore He transformed himself as a man and woman
locked in embrace, and then he divided this own-self into two equal
parts and became (Pati) Husband and (Patni) wife to sport blissfully
(in his divine Loka)."

(Bṛhadāraṇyaka Upaniṣad 1.4.3)
One thing is clear from here that One thing has become clear from here that if something has been said for one of the eternal couple, then it will automatically be applicable for the other. Because both are the same there is no difference between the two that what vedas says.
Let's see Atharvaveda Shruti Sita Upanishad,

श्रीरामसान्निध्यवशाज्जगदानन्दकारिणी । उत्पत्तिस्थितिसंहारकारिणी
सर्वदेहिनाम् सीता भगवती ज्ञेया मूलप्रकृतिसंज्ञिता प्रणवत्वात्प्रकृरिति
वदन्ति ब्रह्मवादिन इति अथातो ब्रह्मजिज्ञासेति च ॥

Śrī Sītā has the constant company of Lord Śrī Rāma, because of which she is the benefactor of the world. They are the origin, condition and destruction of all living beings. Bhagwati Sita should be known as the original form of nature. because of their promiscuous nature Brahmoists call them 'Prakriti'. In the Brahmā Sütrā, "Athatö Brahmā Jīggyasā", her form has been depicted.
Now the question arises Why only Śrī Sītā Rāma? So it's answer has been further explained in same text.

ओङ्कारात्परतो राम वैखानसपर्वतः ।
तत्पर्वते कर्मज्ञानमयीभिर्बहुशाखा भवन्ति ॥

Beyond 'Om' is the

Vaīkhanās mountain in the form of Rāma

. There are many branches of that mountain in the form of knowledge and action.

एकविंशतिशाखायामृग्वेदः परिकीर्तितः । शतं च नवशाखासु यजुषामेव जन्मनाम्
साम्नः सहस्रशाखाः स्युः पञ्चशाखा अथर्वणः। वैखानसमतस्तस्मिन्नादौ
प्रत्यक्षदर्शनम् स्मर्यते मुनिभिर्नित्यं वैखानसमतः परम्

There are twenty one branches of Rigveda, one hundred and nine branches of Yajurveda, one thousand branches of Samaveda and five branches of Atharvaveda. In the Vedas the Vaïkhanās school has been considered as first direct philosophy (प्रत्यक्ष दर्शन).That's why sages always meditates on Param Vaikhanas Śrī Rāma.
Śrī Sītā has been depicted as the Original form of all Shakti in Vedās as shown below,

सा देवी त्रिविधा भवति शक्त्यासना इच्छाशक्तिः क्रियाशक्तिः
साक्षाच्छक्तिरिति इच्छाशक्तिस्त्रिविधा भवति श्रीभूमिनीलात्मिका
भद्ररूपिणी प्रभावरूपिणी सोमसूर्याग्निरूपा भवति

Atharvavedā Shrüti Sita Upnishad
Śrī Sītā is threefold through Her power, namely the power of desire, the power of action, and the power of knowledge. Taking the incarnation of Śrī Devi, Bhudevi, Neeladevi, the one who benefits everyone with her influence, shines in the form of moon, sun and fire.
That's why in Śrīmad Vālmiki Rāmayan which is itself Incarnation of Vedās clearly says that Śrī Sītā is even Śrī of Bhagwati Śrī.

वसुधाया हि वसुधां श्रियः श्रीं भर्तृवत्सलाम् । सीतां
सर्वानवद्याङ्गीमरण्ये विजने शुभाम् ।।

(Valmiki Ramayan 6.111.24)
Same is said by Brahmānd Purān,

सीतायाश्च त्रिविधाशाः श्री भूनीलादिभेदतः । श्री भवेद् रुक्मिणी भूः
स्यात् सत्यभामा वृढव्रता ॥ नीलास्याद् राधिका देवी सर्वलोकैक पूजिता ।

In short, Śrī Sītā manifests herself in three form Śrī, Bhu and Nila. By Śrī she becomes Rukmini by Bhu she becomes Satyabhama and by Nila she becomes Radha Rani Sarkār who is worshipped in all Lokas.
One thing has become clear from here that the Siddhant for Śrī SitāRāmā is a Vedic siddhant. That's why Śrī Rāma has been called as, "Rāmō Vīgrāhvāan Dharmā". The Dharmā itself is Śrī Rāma.
So upto here we found that the Brahmā of Brahmā Sütrā none other than Śrī Sītā Rāma. You may apply it to other form Śrī Rāma as well as they are non other than Incarnation of Supreme Brahmān Śrī Rāma as called by Vedā Nāräyan itself,

ब्रह्मादिपञ्चब्रह्माणो यत्र विश्रान्तिमाप्नुयुः । तदखण्डसुखाकारं
रामचन्द्रपदं भजे ॥

(Yajurvedā Shrüti Pānch Brahmõpanishād)
I worship the lotus feet of Lord Śrī Rāmachandrā, the cause of eternal bliss, where the five Brahmas (Shri Shiva, Shri Hāri Vīshnu, Shri Ganesha, Bhagwati Durga and Shri Surya) along with Demi God like Brahma etc. always find rest (ashraya).
Bhagavān Vishṇu also is the manifestation of Bhagavān Śrī Rāma.

ततस्त्वमसि दुर्धर्षात्तस्माद् भावात् सनातनात् ।
रक्षार्थं सर्वभूतानां विष्णुत्वं उपजग्मिवान् ॥

(Śrī Vālmīki Rāmāyaṇa 7.104.9)
"After my prayers in the begining of the creation, Oh Rama ! You forsaken Your durdharṣa (the swarupa which is difficult to be achieved for many) eternal Bhava (the eternal two armed Śrī Rāma form) and assumed the form of Vishṇu for the sake of protection and sustenance of all the beings."
One more thing Śrīmad Valmiki Ramayana is Veda itself as said in Vālmiki Rāmayan and Agastya Samhita,
As said;

वेदवेद्ये परेपुंसि जाते दशरथात्मजे ।
वेदःप्राचेतसादासीत् साक्षाद् रामायणात्मना ॥
तस्माद्रामायणं देवि "वेद एव न संशयः॥

(Śrīmad Valmiki Ramayan and Agastya Samhita)
The one who is the son of Dasarātha is the Vedvedyā i.e. the supreme personality known from the Vedās, on his appearance the Vedas themselves incarnated in the form of Śrī Rāmayana from the mouth of Prāchetas Vālmiki Muni. Lord Shiva himself says, O Goddess! There is no doubt that Śrī Rāmayana is the Veda itself.
Again in Śrīmad Vālmiki Rāmayan,

"वेदोपबृंहणार्याय" तावग्राहयत प्रभुः ॥

In Skandh Puran Uttarkhand also;

रामायणाद काव्यं सर्व "वेदार्थ सम्मतम्"॥

In Mantra Ramayana aslo;

"वेदार्थानभिज्ञे सति न रामायणमपराद्धघति"

So Ramayana's authority (Śrīmad Vālmiki Rāmayan & Śrīmad Rāmcharit Mānas, as Tuslidās ji maharāj is incarnation of Śrī Vālmiki muni as mentioned in scriptures) is same is of Vedas.
Authority of Śrī Rāma has been discussed in Skāndh Puranā uttarkhand in Narad Sanathkumar Dialogue;

ब्रह्मविष्णु महेशाद्या यस्यांशा लोकसाधकाः ।
नमामि देवं चिद्रूपं विशुद्धं परमं भजे ॥

Meaning:~ I bow down to the Sachidānand Śrī Rāma, who fulfills the desired wishes of the whole world, the deities like Brahmā, Vīshnu and Maheshā etc. are only his mere parts. I concentrate my mind on his hymns.
And so Śrī Krishnā also a manifestation of Śrī Rāma,

यो रामः कृष्णतामेत्य सार्वात्म्यं प्राप्य लीलया । अतोषयद्देव्मौनिपटलं
तं नतोऽस्म्यहम्।।

(Atharvavedā Shrüti Krishnōpanishad)
That Śrī Rāma, who transformed himself (incarnated) as Krishnā, attained Sarvaatmakta (सार्वभौमिकता) by his very lilā (divine exploits, pastimes); and thus deities-sages-masses were completely satisfied on this earth. I make prostration before that same Śrī Rāma.
Now coming to some other Sütrā,

आनन्दादयः प्रधानस्य ॥ 3.3.11

Ananda etc. is predominantly religion of Supreme Brahmān.
Śrī Hanumān ji's incarnation Śrīmad Madhawācharya quotes an evidence from the शांडिल्य -शाख of the Vedas to support this derivation of the epithet Rāma in his immortal गिता भाष्य under विभूति - रूपाध्याय 10.36,

आनन्द-रूपत्वात् पूर्णत्वात् लोक-रमणत्वात् च रामः। आनन्द-रूपो
निष्परिमाण एष लोकश्च एतस्मात् रमते तेन रामः।।
इति शांडिल्य-शाखयाम् ।।

Itī Shändilyā Shākhayãm,
The highest Divinity is called by the epithet Rāma, as He is an embodiment-of- pristine pure-happiness (आनन्द-रूपत्वात्); as He is full-of-innumerable auspicious-qualities (पूर्णत्वात्); as He brings about happiness to all (लोक-रमणत्वात्).
In Atharvavedā Rāma Tapniye Upnishad,

रमन्ते योगिनोऽनन्ते सत्यानन्दे चिदात्मनि।
इति रामपदेनासौ परं ब्रह्माभिधीयते।।

In Padmā Purān aslo than same meaning is present,

रमन्ते योगिनो यस्मिन् चिदानन्दे चिदात्मनि ।
इति राम पदेनैतत् परं ब्रह्मभिधीयते ।।

One thing has become cleared from above statements that the Supreme Brahmā who is called as Ānand is none other than Śrī Rāma himself.
Now coming back to point;

आत्मशब्दाच्च ।। 3.3.14 (Brahmā Sütrā)

Some times Aātma word is also used for Supreme Brahmān, for this also Vedās have cleared itself that for whome this word have been used.

जीववाची नमो नाम चात्मा रामेति गीयते।

Atharvavedā Rāma Tapniye Upnishad
'Namah' word refers to the Jeev and word 'Aātma' refers to the Śrī Rāma.
That's why in Śrī MahaRamayana Śrī Rāma has been called as Aātma of all 7 Billion Maha-mantra ,

अंशांशै रामनाम्नश्च त्रयः सिद्धा भवन्ति हि। बीजमोंकारसोऽहंच
सूत्रमुक्तिमिति श्रुतिः ।। इत्यादयो महामन्त्रा वर्तन्ते सप्तकोटयः।
आत्मा तेषां च सर्वेषां रामनामप्रकाशकः।।

(Maha Rāmayana 52/39 )
Moving Forward ⏩,
सर्वत्र प्रसिद्धोपदेशात् ॥ 1.2.1 (Brahmā Sütrā)
In all Vedantā sentences; The famous Par-Brahmā is preached in the form of Upasyadev.
As again Vedās has cleared that all the Siddhant of all Vedantā is exclusively dedicated to Śrī Rāma alone.

समस्तवेदान्तसारसिद्धान्तार्थकलेवरम् ।  विकलेवरकैवल्यं रामचन्द्रपदं
भजे॥

Yajurvedā Shrüti Sarvasara Upanishad
And even all Maha-Vakyās Maha-Vidyā of Vedās too are decided to Śrī Rāma alone,

यन्महावाक्यसिद्धान्तमहाविद्याकलेवरम् । विकलेवरकैवल्यं रामचन्द्रपदं भजे
॥

Atharvavedā Shrüti Mahāvākya Upnishad
Moving Forward ⏩,

अक्षरमम्बरान्तधृतेः॥ 1.3.10
साच प्रशासनात् ॥ 1.3.11 (Brahmā Sütrā)

1.3.10, The word Aksharā is being depicted for Parabrahma Paramatma, because he has been described as holding the whole world up to the sky.

In the incarnation of Vedās Śrīmad Valmiki Ramayan this has been exclusively said for Śrī Rāma,

akSharaM brahma satyaM cha madhye chaante cha raaghava ||

6-117-14
O Śrī Raghavā You are the Aksharā Brahma, the Truth abiding in the middle as well as at the end of the universe.
Śrī Rāma himself is Aākash that's why Brahmā said him following Shlokā, at the time he was ending his lila,

वैष्णवीं तां महातेजो यद्वाऽऽकाशं सनातनम् । त्वं हि लोकगतिर्देव न त्वां
केचित् प्रजानते ॥

Śrīmad Valmiki Ramayan 7.110.10

As Śrī Hāri Vīshnu is incarnation of Śrī Rāma so it was of his will in which form he want to go to his Dhaam.
So from here we can see Śrī Rāma has been called Aākash, even Vedas calls Aākash everything,

अस्य लोकस्य का गतिरित्याकाश इति होवाच सर्वाणि ह वा इमानि
भूतान्याकाशादेव समुत्पद्यन्त आकाशं प्रत्यस्तं यन्त्याकाशो ह्येवैभ्यो
ज्यायानकाशः परायणम्॥

(Chhandogya upanishad 1.9.1)
Śilaka Śālāvatya asked Pravāhaṇa, 'What is the end of this earth?' Pravāhaņa said: 'Aakash, for everything that exists arises from aakash and also goes back into aakash. Aakash is superior to everything. Aakash is the highest goal'.
Don't get confused with 5 element वाला Aākash, even that Aakash has aslo being established by the Glory of Śrī Rāma,
As Śrī Parashara Smriti clearly says;

राम एव परं ब्रह्म परमात्माभिधीयते।
रामात्परतरं नास्ति यत्किचित्स्थूलसूक्ष्मकम्॥
ब्रह्मविष्णुशिवाः सर्वेइंद्रो निर्वरुणो यमः ।
सूर्यश्चंद्रश्च खं भूमिराकाशस्त्वनिलो ह्यपः॥
सर्वे ते रामचंद्रस्य तेजसा संप्रतिष्ठिताः ।
भूतं भव्यं भविष्यच्च सर्वे रामसमुद्भवम्॥

Meaning- Lord Śrī Rāmbhadrā is the Supreme God, whatever is gross and subtle, none of them is beyond Śrī Rāma, that is why Brahma, Vishnu, Shiva all these and Indra, Agni, Varun, Yama, Sun, Moon, Heaven, Earth, sky, wind, water, all these have been established by the glory of Lord Śrī Rāmachandrā and all these three times past, future and present and whatever will be born and has been born in these three times, all that has been born from Śrī Rāma only.
Finally ending here and with a thought that I follow personally, knowing all the shastra of Sanatana Dharma will not take you to Saket but chanting the Holy name of Śrī SitāRāmā will.
Storing every siddhant every concept like a hard disk will make you of Memory Card in next birth
Please Chant SitaRam SitaRam SitaRam
Even Śrī Hanumān has captured the mind and heart of Śrī Rāma by merely chanting his name,
sumir pawansut pawan namu.
apne bas kari rakhay raamu..
Hari Bol Hari Bol
Hail to Awadh dulare Priya Pritam Sarkar.❤❤
Hail to Ānand Bhāshyakār Śrīmad Jagādgurü Ramānandacharya

Brahmā of Brahmā Sütrā


How do opensea royalties work?

Question: Can you please explain or share some useful links about how opensea manages to validate royalties amount and addresses? I was exploring seaport contract for a while and didn't get answers. The simplest case is "fulfillBasicOrder". This function has a lot of validations but I didn't find there how it validates additional recipients so it just makes ERC20 transfers through for loop with address and amount that came from an external source.
I mean I can recreate EIP712 structure that uses opensea, fill additional recipients with needed addresses and sign it without using opensea UI, and similarly send transaction to fulfill basic order. - it is just my thoughts and I firmly believe that I am wrong just didn't find needed file.
Does royalty data have to be validated on behalf of smart contract or smart contract can only validate the signature and then make all needed transfers?
Answer: It seems like your confusion here is mostly about how orders actually work. There is no section of Seaport that handles royalties specifically. The "address and amount that came from an external source" are part of the order, the same as any fee. When ever an order is created using OpenSea's front end, API, or some other marketplace building on top of seaport, you get to choose all the parts of the order to be fulfilled. That signed order is stored off-chain and includes an offer, for example it could be "User Alice sells 1 tubby cat with id 123, Alice receives 0.1 eth, tubby cat owner address receives .01 eth, opensea address receives, .005 eth". When the order is fulfilled each part of it is executed and it's someone meeting the offer. The royalty amount is calculated by the front end, or API, etc. and added as one of the parts of the offer.
Check out the structs section, you might find the OrderComponents comments really helpful: https://github.com/ProjectOpenSea/seaport/blob/main/contracts/lib/ConsiderationStructs.sol
Comment: Thank you so much for your answer! @CupOjoseph. However I have some confusions about the verification moment.
Comment: So I do understand that the order structure is created offchain and signed also offchain. Onchain function like "fulfill order" accepts such arguments - signature, order parameters (royalties data is inculded). These parameters came from some external source and with the help of signature we can verify address that signed our order. My concern is whether order can be compromised, because there is no verification on behalf of seaport. Does valid signature means, that I can believe the data that came from outside? @CupOjoseph
Comment: The entire order, including:
- what NFT/item is sold by seller
- How much is received by seller
- How much is received by collection owner
- How much is received by marketplace
- any other conditions to meet..

are all signed into 1 valid order. Its not added later.
Answer: You can take a look at how royalties work in OpenSea and Seaport overview:
https://support.opensea.io/hc/en-us/articles/1500009575482-How-do-creator-earnings-work-on-OpenSea-
https://docs.opensea.io/docs/10-setting-fees-on-secondary-sales
https://docs.opensea.io/v2.0/reference/seaport-overview

How do opensea royalties work?


How to install VMware Tools in Android x86 VM?

Question: I installed Android-x86 in a VMware virtual machine under Windows 7.
Is there a way to install VMWare tools in this Android-x86 VM to share files between this machine and the host? If yes, how?
Comment: use dropbox or similar, easy than ftp
Answer: VMWare tools are not yet supported for Android, you can have a look on the guest os support matrix on the VMware website.
But you could always make a network share on your host and then get android to connect to it
Answer: You could put your files on a flash drive and then disconnect it from host and connect it to the guest android OS.
Comment: Saved my whole day!
Comment: VM Ware does not support USB forwarding without VMware tools installed. So it is impossible to connect flash drive to android guest.
Answer: for file sharing with your host and guest machine
first try to connect your android vm to internet then download
AirDroid  from google play ,now you can easily download and upload
to your android vmware machine
Comment: Are you affiliated with the developer of this app? This looks like spam.
Answer: If you want to share files between your Windows machine and your Android VM, you could do the following:
On your Windows machine
1. Download TFTP at http://Sourceforge.net/Projects/TFTP-Server
2. Once installed, ensure that the service (Open TFTPServer) is in started (local services in control panel)
3. Copy files to you c:\OpenTFTPServer folder.
In your Android VM
1. Your Android VM should already contain the TFTP application.
2. Launch the Terminal Emulator
3. type in> su root
4. Change directory to a modifiable on> cd /sdcard/download
5. Use TFTP to copy your file from your Windows machine to your Android VM>
tftp -l  -g
6. To verify that the file has been copied type in> ls
Troubleshooting
1> Ensure you're able to ping the IP of the windows machine
Answer: I have a much simpler solution to sharing files between the Windows host and Android-x86 guest. All you need is a network sharing program that supports the Samba (SMB) protocol. I use AndSMB free to download from Play Store and it is very good. You create a new connection that has your host computer name and you can give credentials or connect as Guest. To connect as Guest you could check the Anonymous box but in case this doesn't work, I give user pcname\guest and a blank password, where pcname is your computer's network name. One you open the connection you can see all the shared folder. All you have to do is copy the files you want in a shared folder which you can access from AndSMB! You can also browse all the android folders and upload any file you want to the shared folder, provided you have set read/write access rights to all users (including the Guest account).


Why can't massless particle exceed speed of light?

Question: Why massless particle can't exceed speed of light?
Answer: It's just what it means for a particle to be massless. For a general motion of a particle in the vacuum, the rest mass $m_0$ satisfies
$$m_0^2 c^4 = E^2-p^2 c^2$$
If the left hand side is zero, it follows that
$$E=|p|c$$
which means that the energy-momentum vector is null (light-like). In general, the speed may be calculated from the direction of the energy-momentum vector as $v=pc^2/E$ and it equals $v=c$ if the formula above is satisfied. Massless particles are not only unable to move faster than light; they're unable to move slower than light, too. The massless particles must move exactly by the speed of light.
The general absence of any signals that move faster than light is required by the special theory of relativity; a signal or particle that would be moving faster than light would be equivalent (from a viewpoint of a different inertial system) to a signal or particle that moves backwards in time, which would lead to logical contradictions (killing grandfather before he met your grandmother etc.).
Comment: Dear Jay, there is nothing classical about the argument, about energy, about momentum, or the rest mass. All these are standard quantum mechanical observables and my answer was the up-to-date quantum particle-physics answer. Higher dimensions may affect details of particle physics or not but they do not affect this elementary question.
Comment: I think your answer is only correct under classical thinking... the more modern answer IMHO would be that ' They must move at the velocity of least resistance which is localized to the quantum fluctuations in their system with respect to uncertain energy introduction from higher dimensions. '
Answer: To be absolutely simple.... A particle has got a certain mass and acquires a certain dimension in space..
We CAN'T define a Particle with mass=0.
Comment: Aritro you are new here , but you have to realize that there are many physicists here who know that you can have a particle with zero mass, the most known one is called a photon and demonstrates the so called wave/particle duality. Elementary particles are dimensionless anyway, as far as we have been able to measure.


Has the idea for my story already been taken?

Question: I want to write a story about a place called Kingdom Come, where you can come in but you cannot go (i.e. you are not allowed to leave). Has this idea already been taken? It really sounds like the sort of idea that has already occured to some writer. If so, can I considered a plagarist?
Comment: The Eagles: Hotel California - lyrics  Welcome to the Hotel California / ... / Last thing I remember, I was
Running for the door
I had to find the passage back
To the place I was before
"Relax," said the night man
"We are programmed to receive
You can check out any time you like
But you can never leave!"
Comment: It's been done sufficiently there's an article about it:  [Closed Circle](https://tvtropes.org/pmwiki/pmwiki.php/Main/ClosedCircle)  That said, just because it's been done before doesn't mean it can't be again.  Ideas can't be copyrighted, just put your own unique spin on it.
Answer: The answer to "Has this story idea already been done?" is always yes, regardless of what the idea actually is. Everything's already been done. In your particular case, the one example I can think of immediately is the TV show Once Upon a Time, where anyone who enters the town of Storybrook becomes unable to leave due to mysterious forces (there are probably better examples, that's just the one that came to mind).
Fortunately for you, ideas are not protected by copyright and cannot be "taken"; one person cannot claim a story idea and then prevent anyone else from ever using that same idea. So it doesn't matter one iota whether your idea has already been done or not. As long as your story isn't a blatant rip-off of an existing story, you cannot reasonably be accused of plagiarism (you can still be accused, but the accuser is unlikely to get anywhere). And as long as your story is good, most readers won't mind if it uses story beats they've seen elsewhere.
Comment: You can always be accused of plagiarism.  Down to and including accusations of plagiarizing works that were written after yours.  (I mention this because it can be startling to new writers.)
Answer: All things are used at least once before. All ideas come from inspiration from different books. Some books may even have the same theme but different settings, characters, and most importantly the plot or styles.
A book called Cut Off by Adrianne Finlay is similar to the Hunger game by Suzanne Collins. The School for Good and Evil by Soman Chainani, Harry Potter by J.K. Rowling, and The Iron Trial by Cassandra Clare and Holly Black is all based on the topic of Magic Academies.
So no, you're not doing any plagiarism if you're just trying to collect some ideas for your book. No one is going to sue you for trying to let readers enjoy the book that you made your efforts on.


What is the real use of "lightningMessageFields" while using LightningMessageChannel

Question: I was thinking - while publishing or subscribing to a LightningMessageChannel, the payload has to be in some way match exactly to the names or fields defined in the lightningMessageFields  XML definition of the channel. Found out my code/components are working without any consideration to the actual names used in the lightningMessageFields.  Whatever I send to the channel( by any name ) I can get back and process in the subscribing component. Am I doing something wrong here or is there any special benefit sticking to the fields defined? if so where can I find details of usage patterns for this specific aspect of LightningMessageChannel fields?
Answer: It is correct and as per design that the LightningMessageChannel fields are not reference-checked and you can add in a message field in the message of your choice. While the channel name is strictly checked!
I think it's left open because some developers (coming from JavaScript background) do not like strict behavior (A message is a mutatable JavaScript Object) while fellow salesforce devs are used to Salesforce Metadata and like to keep it strict and version it.
It's left as it is for now but hopefully, in the future (safe harbor) there is an additional switch to toggle this behavior!
Comment: Thank you so much.  All makes sense now. Pretty much all the examples and samples out there use "message"   as a field in XML as well as in subscriber component and in publisher component which made me think this "message" is somehow magically connected and the same across. I'll rename my subscriber message to messageRx and publisher message to messageTx  (something to that effect) so that whoever maintains my code not that  familiar with channels will not get confused.  Thank you


Solaris 10 ssh auto logout

Question: For some reason, I am getting automatically logged out of one of my shells on one Solaris box but not on the other one.  I don't seem to have any bash variables set to auto logout.  What else should I check?
Answer: the auto logout is happening due to the IdleTimeout variable is set in /etc/ssh/sshd_config
I believe if you have KeepAlive set in your .ssh/config then it should stop (I haven't tested this though, so ymmv).
Answer: I assume you checked for the TMOUT variable in your bash session, correct?
Is there a firewall between your ssh client and the Solaris  server in question?  It might be dropping you after some amount of inactivity.  If so, you could modify the sshd_config on the server and add the KeepAlive and ClientAliveInterval settings.
Answer: I've found that Cisco devices in their default configuration can cause this.  Do you have  different Cisco devices between you and each server?
Answer: Try setting ServerAliveInterval 600 in /etc/ssh/ssh_config on your client machine.
Answer: Check the /etc/default/login (TIMEOUT) and /etc/profile (TMOUT) files.
They have timeout settings for the login shell.

Solaris 10 ssh auto logout


Can't delete printer or add a new one

Question: My printer doesn't work. It appears in the settings but I can't delete it:

If I try to add it again nothings happens.
Any idea how to make it work?
Answer: I had a similar problem. Since eos uses CUPS for printing if you check cupsd (browse to http://localhost:631 is one way) you should see the printers that are installed. You can add/change/delete printers from this interface as well. Once I made some changes (deleted multiple installs of home printer) the standard printer window was updated.

Can't delete printer or add a new one


Create encrypted file container without installing additional applications

Question: How can I create an encrypted file container with the tools already available in Ubuntu, without installing any extra applications? I've tried VeraCrypt and zuluCrypt, but they both require installation.
Comment: I think the best way is to start with a fresh installation, and during the installation, at the partitioning window, select LVM with LUKS encryption 'encrypted disk'. You do this from the Ubuntu live drive (created from the Ubuntu iso file), so you need no extra application program.
Comment: I want a separate file container (or multiple containers) not disk/partition encryption.
Comment: I see. Do you want containers, that you can use as file systems, or would it be OK to use for example an zip archive with encryption (lower security level) or a gpg encrypted tarball (higher security level)?
Answer: I am compiling this as an answer, but a much more comprehensive explanation can be found at this link. I recommend reading through it.
To create a file container encrypted with LUKS/dm-crypt, using cryptsetup as the user-space tool (both available in Ubuntu repositories), follow these steps:

Install cryptsetup:

sudo apt install -y cryptsetup

Create the actual container, containing random data:

sudo dd if=/dev/urandom of=/path/to/your/file.bin bs=1M count=1024
This will create a 1GB (1024 x 1MB) container at the location you specify in the of parameter, adapt the size to your need via the count parameter.

Mount the file container as a loopback device:

sudo losetup -f --show /path/to/your/file.bin
Note the loop device number that is assigned by losetup.

Create the encrypted container:

sudo cryptsetup luksFormat /dev/loopX
where you replace loopX with the actual device number you noted just before. Choose your password when prompted. You can verify that the container has been correctly formatted by LUKS by doing
sudo cryptsetup luksDump /dev/loopX

Map the encrypted container (you can replace c1 by whatever name you want):

sudo cryptsetup luksOpen /dev/loopX c1

Create a filesystem in the mapped container - here I use ext4:

sudo mkfs.ext4 /dev/mapper/c1

Finally, mount the file system anywhere you want:

sudo mount /dev/mapper/c1 /mnt
After that, to use the container you repeat steps 2, 4 and 6. This can easily be automated in a script.
LUKS encryption has many other useful features, such as keyslots, detached headers etc. that I let you discover for yourself.
And remember, the single most important thing when encrypting stuff is CHOOSE A STRONG PASSWORD.


Are free-nuclear division and endomitosis the same?

Question: As far as I understood it, both are cases of karyokinesis, not followed by cytokinesis.
Answer: No. If you google the terms you'll get a lot of sites with definitions. For example:

Nuclear division
Definition
noun
The process by which a nucleus divides, resulting in the segregation of the genome to opposite poles of a dividing cell.

source: http://www.biology-online.org/dictionary/Nuclear_division
Edit:
or

free nuclear division    mitotic division of nuclei without accompanying cytokinesis, i.e. nuclei divide in a common cytoplasm, the cells walls only forming around each later

source: http://ecflora.cavehill.uwi.edu/bio_courses/bl14apl/Gloss.htm
versus

endomitosis
mitosis taking place without dissolution of the nuclear membrane, and not followed by cytoplasmic division, resulting in doubling of the number of chromosomes within the nucleus.

source: http://medical-dictionary.thefreedictionary.com/endomitosis
or a bit more revealing:

Duplicated chromosomes produced by endomitosis exist as discrete units
in a single polyploid nucleus or may be packaged into separate nuclei,
depending on the phase at which mitosis is aborted

source: http://en.wikipedia.org/wiki/Endoreduplication
So as you see by definition nuclear division is part of a bigger process (cell division), and accoriding to the first source karyokinesis is a synonim for nuclear division (karyo = nucleus kinesis = moving, both come form greek language).
Edit:
If you check the definition above, you can see that free-nuclear division is a mitosis without cytokinesis, thus chromosome separation still occurs.
In endomitosis the can end up with a polyploid nucleus, in contrast to the other two aforementioned mechanism where no polyploidy occurs.
Comment: I was talking about free nuclear divisions, not just nuclear division - which is just karyokinesis. Free nuclear divisions happen for instance during formation of the female gametophyte from the megaspore in angiosperms - the nucleus divides but separate cells don't form.
Comment: Oh I see, I missed that part sorry. I will look into that
Comment: I have a doubt though: in case when endomitosis is aborted in between anaphase (i.e. centromere splits but chromatids are not segregated) then in that case I thought nuclear membrane has disappeared in prophase, and now reforms to enclose 4n chromosomes. But one of these definitions says the nuclear membrane does not disintegrate - so how can spindle fibres attach to the chromosomes? Then again, since chromatids don't separate, spindle fibre formation isn't needed and may not occur.
Comment: Errors / failures / abortions in during mitosis results (most of the time) in pausing the process to try to fix the problem and if its not possible than the cell undergoes apoptosis to avoid having cells with abnormal genome. If somehow the cell still gets through (cell) division, then you get abnormal cells with chromosome trisomies, missing chromosomets etc.
Answer: In the cases of pollen grains free nuclear division occurs. entering into the s phase the chromosome no remains as 46 but the DNA doubles ie 4n condition occurs. During the anaphase the sister chromatids separate and reaches both ends. And when telo phase occurs nuclear membrane is surrounded around 2 genetic material. But cytokinesis is blocked there. But when again this cell enters s phase there again the genetic material replicated ie 2 nucleus having 4n condition is formed.this is free nuclear mitosis. These chromosomes doesn't separate in anaphase due to failure in attachment of spindle fibres. So karyokinesis also get stopped resulting cell to 8n condition or polyploidy. This is endomitosis
Comment: Welcome - can you add your sources here so people can background read on your answer?


Confidence interval estimation

Question: Suppose there are two parameters $\theta$ and $\eta$. I have shown that $\theta\in(\eta_{\min},\eta_{\max})$ if statement $A$ holds. Let $\hat{\theta}$ and $\hat{\eta}$ be the MLE's of $\theta$ and $\eta$ respectively. Can I say that if $\hat{\theta}\not\in(\hat{\eta}_{\min},\hat{\eta}_{\max})$, then statement $A$ doesn't hold ?
Answer: Maximum likelihood estimators are random variables because they are functions of the sample, which is a random variable. With a test-like approach, the question would be "What is the probability of $\hat{\theta}\not\in(\hat{\eta}_{\min},\hat{\eta}_{\max})$ assuming that statement $A$ does hold?". If that probability is zero, $\hat{\theta}\not\in(\hat{\eta}_{\min},\hat{\eta}_{\max})$ implies that statement $A$ doesn't hold, but if it's different than zero it could just be used to build a test on $A$ or estimate probability of $A$.
A little more of context could be useful to give a more helpful answer.
Comment: Thanks for the reply. How do I obtain the probability that \hat{\theta}\not\in(\hat{\eta}_{\min},\hat{\eta}_{\max}) assuming statement A holds ? In confidence interval estimation, we usually test whether some population parameter belongs to a confidence interval formed by sample statistics with a certain probability. There is some associated test statistic (involving the parameter and its estimate) which follows a certain distribution that is used to contstruct the confidence interval for the parameter. Here, the context is different.
Comment: @user143423 That's why I suggested giving more context. Knowing what statement $A$ is, what your parameters are and how your confidence interval is computed is needed to compute probabilities.
Comment: The question could also have something to do with properties of MLEs but that is not evident from the question as currently posed.
Answer: Here statement $A$ refers to the specific missing mechanism (NMAR or MAR or MCAR) of a variable in a missing data model for an incomplete contingency table. The parameters $\theta$ and $\eta$ refer to the vectors of fully observed and partially observed cell probabilities respectively in the table. It can be shown theoretically that $\theta\in(\eta_{\min},\eta_{\max})$ always under the model. The intention is to say something about the type of missing mechanism of the variable if the above inclusion doesn't hold for the MLE's of the parameters under the model.

Confidence interval estimation


Random Emission material in Array

Question: I've created a glowing sphere (basically a visible light source) and made a grid with an array. How do I randomize and animate the strength of the emission so none of the lights are the same brightness and the values change independently?
Answer: Blender doesn't give a lot of ways to distinguish between different elements of an array.  One method that it does provide is UV:

I've created a UV map, then shrunk all vertices onto a single point.  When I tell the array modifier to offset UV for each instance, this creates a unique number for each element that I can use as the "seed" for a noise texture lookup.  The UVs of each vertex of the first element, shown, are 0.5,0.5; the UVs of each vertex of the second element are 1.5, 0.5; etc.  Because these numbers are the same for each individual element, the strength is constant over that element.  Because it varies from element to element, we get a different random output for each element.
Here, I'm also doing some manipulation of the noise texture's output to make it more obvious that I'm getting unique values for each element.  The RGB curves and math/multiply are not necessary for the technique.

Random Emission material in Array


Truncate with a twist

Question: I've been browsing this site for two days now with no avail to my problem.
I have a SharePoint 2013 on premise site with Designer Workflows only. This site for reasons I cannot go into is very limited and doesn't have functionality as you would with some other premise situation. Also note that there is no switching, there is no moving, it is what it is.
So what I'm working with is a word document is created on Premise. This document is sent out to ANOTHER SharePoint 2013 on Premise site that has Nintex and Designer Workflow options.
From Site B I use nintex to convert DOC to PDF and now I need it to email to the user that created it. When moving the file it comes over with just the column "name". The document is labeled one of the following to prevent over writing.  "emailES" "emailCO" "emailPU" "emailSO" "emailSC".
Using this I tried to make a column that would truncate the last 2 characters and leave the email in place. So I need it to go from firstname.lastname@example.com to firstname.lastname@example.com. From there it's easy I simply use the name as the email to send to. But I can't wrap my head around removing the last two and leaving email in place.
I'm sure this is incredibly easy, but maybe this week isn't my week. Any suggestions?
Answer: If you create a new calculated column to include the following, I believe it will give you the string that you're hoping to preserve:
=LEFT(Name,LEN(Name)-2)
Cheers!

Truncate with a twist


Fastest engine for Mass Updates (only updates done on table)

Question: I have a table of 2 columns and 60 million rows. I want to update the table by incrementing one field by 1 on various different rows (This is the only type of operation that will ever be done on the table). The updates that occur happen all at once everytime, so I am usually updating 30-40 million out of the 60 million rows once a day. This table is only used by one user so I am not worried about multiple users trying to connect to the table at the same time or anything like that. Which engine (and parameters) would be fastest to perform just updates and nothing else on (well there will be one select statement once in a while, but I am not concerned about the performance of select statements)? I am currently using MyISAM as I am not worried about ACID, but it currently is quite slow for mass updates. Essentially I am trying to find the fastest way to do MASS Updates.
Comment: Shouldn't the middle sentence in your question read: "This table is only used by one user so I am *not* worried about multiple users trying to connect to the table at the same time or anything like that." ?
Comment: Yes, you are correct. I have just changed that, thanks!
Answer: Your question is somewhat like a question I answered back on Sep 06, 2013 : mySQL for analytics and updating same row
In that answer, I recommended the following

Adding a third column to the table to represent a slot
Select an arbitrary value for the number of slots
Making a composite PRIMARY KEY

hour boundary
slot number

Doing an INSERT with ON DUPLICATE KEY UPDATE to increment the counter

Since you are doing 30-40 million updates, you may need to choose a very high number of slots. In your case, I would go with 250,000 slots. The high the number of slots, the lower the incidence of collision (trying to update the same slot).
As far as the choice of storage engine, you cannot stay with MyISAM. Why ? MyISAM will do a full table lock for every update. Doing 40 million updates require 40 million table locks. My posted answer works well with InnoDB's row level locking. Even if you implement the slot paradigm, MyISAM will still do 40 million table locks with 40 million updates.
Definitely, you must go with InnoDB. You also need to tune InnoDB.

Larger Buffer Pool (increasing innodb_buffer_pool_size)
Larger Log Buffer (increasing innodb_log_buffer_size to 128M)
More Write I/O Threads (setting innodb_write_io_threads to 8 or more)

If you are using MySQL 5.1, upgrade to MySQL 5.6.
Give it a Try !!!
UPDATE 2014-09-15 00:21 EDT
I learned the concept of slots for counters from page 140 of the Book

My post (dated Sep 06, 2013) is the first post in the DBA StackExchange to mention this book for your specific issue.
 Someone else mentioned this technique one month later in this same forum (counter table in mysql dates Oct 18, 2013)
I just tried googling for this concept search for mysql counter slot
I found some posts:

Baron Schwartz, one of the authors of the book, has a URL to that very reference
Avoid locks when storing counters in MySQL

The main point I am making is that this concept is not new. You can implement it.
Give it a Try !!!
Comment: Thanks for the reply! I am unaware of the concept of slots. So if I have a table with |colA|colB|slot| what value goes into the slot rows? Could you possibly give a small example of how this would look like?
Comment: Sorry, I just read your other post. I don't understand what the slots do? I am trying to find the concept on google, but can't find anything.


window.location.pathname alternative in Lightning Experience

Question: In a Salesforce Community's knowledge article detail page, I am using window.location.pathname in lightning component controller to retrieve the page url (I want to later pass the url to the Apex controller to get the article url name). This is working in Salesforce classic but not in Lightning Experience. Is there any alternative?
I have read about pageReference and sforce.one navigation methods but I don't want to navigate, just to extract the url of the current community page.
Answer: You can try the lightning:isUrlAddressable interface on your custom lightning component.
More info here: lightning:isUrlAddressable


Is it possible to create an instance of an object in Java without calling the constructor?

Question: I'm trying to fix a bug in one of my programs which I think might be due to Hibernate figuring out how to instantiate an instance of an object without calling its default (or any other) constructor.
Comment: You should possibly post the original problem rather than a supposed solution.
Comment: Hibernate uses the default empty constructor of entities. You're on the wrong track. Post your problem instead.
Answer: Actually, yes, it is possible to bypass the constructor when you instantiate an object, if you use objenesis to instantiate the object for you. It does bytecode manipulations to achieve this.
Deserializing an object will also bypass the constructor.
It isn't possible to do this using reflection.
Answer: Just to complete the picture: using method clone to create a new object bypasses constructors as well.
Comment: Provided that clone() is correctly implemented, of course. It is possible to override clone to act as a factory method not using super.clone(). (While super.clone() should be used - not the constructor)
Answer: Yes, deserializing an object does not invoke its constructor.
That said, I don't see why hibernate would serialize an object, unless perhaps in the second level cache.
Answer: Deserialization involves creating objects without invoking a constructor.
It's possible (at least with the Sun/Oracle JDK) to do this programmatically. This was covered in "Creating Objects Without Calling Constructors", an edition of The Java Specialists' Newsletter. It involves using classes from the sun.* packages so isn't portable.


Ubuntu 18.10 doesn't recognize my laptop as laptop

Question: I have Lenovo 330-15ich with i5 and gtx1050. But there's one problem. It seems like Ubuntu doesn't recognize my machine as a laptop. I don't have any battery indicator on the top panel, upower -d doesn't show any battery too. What's more, in settings there aren`t almost any options in power configuration. Also laptop doesn't suspend when I close the lid, despite I've chosen this option in Gnome Tweaks. I tried many suggestions from the internet, tried adding acpi=force or acpi_osi=Linux to my Grub configuration file without any results. So I really don't know what to do. Any help would be appreciated)
Comment: Run the command `laptop-detect -v` command (provided by the *laptop-detect* package). If the answer is that your system is NOT recognized as a laptop, then please file a bug report against the 'linux' package. Developers cannot fix what they do not know about.
Comment: `laptop-detect -v` returns `We're a notebook (chassis_type is 10)`
Answer: It would appear this laptop has many strange hardware IDs that cause the Linux kernel to not understand what hardware it's using:

Lenovo IdeaPad 330 touchpad not working

On most vanilla Ubuntu systems for 18.10 you'll be using a kernel of 4.15 or older, and some are claiming that 4.18+ contain the changes to the Linux kernel to have these hardware IDs detected (note, it's not that Linux is missing the functionality to interface with the hardware, simply that the way the hardware is detected isn't matching various identifiers)
You can try using the Ubuntu Kernel Update Utility called ukuu to upgrade your kernel to 4.20:
sudo apt-add-repository -y ppa:teejee2008/ppa
sudo apt-get update
sudo apt-get install ukuu

The run ukuu and select 4.20 to install.
More info on github.com/teejee2008/ukuu
Comment: I've successfully installed kernel 4.20, but, unfortunately, the result is the same(
Answer: I didn't found any answer and tried to install elementary OS, which actually runs on kernel 4.15. Everything was great, except one - my laptop didn't suspend on lid close. So I tried to install newer kernel and battery was no more recognized.
So, the answer is to install old kernel (for me the newest possible was 4.16.18). Unfortunately I didn't manage to resolve issue with suspending, looking for solution now. Also I reported it as Linux kernel bug.


How often should I calculate the position of a bullet?

Question: In game's like Battlefield or Arma, how many times per second is the position of a bullet calculated?
Comment: Specifically, Battlefield servers used to run at a fixed 30hz tick-rate. Recently, the ability to increase the tick-rate up to 120hz was added, although most of the servers running enhanced tick rates run at only 60hz.
Comment: Also, keep in mind that an individual client may be "updating bullet locations" at 1000fps (smaller and more frequent updates) but the server is only notified of those changes TickRate times per second. If a bullet moves in a straight line and is on one side of you in frame X and on the opposite side of you in frame X+1, you know you'll need to check for collision.
Comment: Also, also, there is very little reason to calculate a new bullet position every frame; you only need to integrate the bullet's accumulated flight time. Given the starting point, firing direction, and elapsed flight time, the current location can be calculated. Instead of re-drawing the bullet model at every point along its' trajectory, the entire path is often summarized visually with a brief 'vapor trail' effect. It takes many frames for the actual position of a bullet fired from across the map to matter to a player; it only matters for the last inch.
Answer: You should use projectiles only for comparatively slow "bullets" like grenades from a grenade launcher or RPGs.
For very fast bullets like bullets fired from a gun, A ray cast (or a ray-world intersection test) is used instead. A ray is fired from the barrel of the gun (origin of the ray) in the direction the character is aiming (direction of the ray). If the ray hits something, the hit is registered to the code handling bullet hits after a delay which is proportionate to the distance of the hit point from the ray origin.
If you take the average muzzle velocity to be say 1 km/s, the time delay can be calculated using the simple time = distance / speed formula. In most cases, the distance will not be more than 100 meters (unless using sniper rifles). If the delay is more than say 0.1 seconds, you could check in the code handling bullet hits whether the object hit (or its bounding box) still contains the hit point. If it does, the bullet has hit that object.
You could do extensive research and find the muzzle velocity of each gun you have in the game and use it for each gun separately, or you could use an average figure.


What is an example of 3 or 4 word compound nouns?

Question: I see a ton of examples of compound nouns which are two words combined into one, scattered across the web when you search "compound noun list", such as "blackboard", "aircraft", or "underworld". But there are fewer examples of 2-word compound nouns which remain separated (not even joined by a hyphen), such as town square, or "bus stop". But the question is, can you have long "chains" of compound nouns than just 2, without hyphens (for example, merry-go-round isn't one)? Are there any 3 or 4 (or more!) word sequences which can be made into compound nouns in English? If not, why not? I am looking for regular nouns, not proper names if that helps. Bonus, if there are some, where is a list of a bunch of them?
Comment: Are there really only a few 2 word compound nouns? Birthday cake. Replace cake with card, party, wish, drink, dinner, kiss… Replace birthday with anniversary, retirement, holiday. Soccer ball. Replace ball with game, field, team, stadium, referee…Replace soccer with any other sport (don't use ball if it is in the name, but other nouns work e.g. baseball bat)…hamburger bun, ice cream cone (ooo that's three)…
Comment: Some words involving a compound word and the ending man/person work. For example highwayman (a type of robber that travelled by horse), longshoreman (one employed to load/unload ships), etc. If you look for words ending in man (https://www.thefreedictionary.com/words-that-end-in-man), you can find a few.
Comment: Thanks @Damila, those are great examples and ways to think of them! Hadn't considered that.
Comment: Sure. As a game I thought of many in my current surroundings. And here is a bonus 4-word: ice cream cone holder.
Comment: The simple answer is no. "Town square" and "bus stop" are not compound words but syntactic constructions consisting of noun+noun (i.e. modifier + head). Compounds, by contrast, are always single words consisting of two (or occasionally more) smaller bases. I suggest you obtain a copy of a good book on morphology.
Answer: It's easy to find such expressions with the right tool. In COCA, you can search for NOUN NOUN NOUN NOUN to find strings of four nouns. The most popular is "radio talk show host", with more than 500 hits in the corpus. Looking at strings of 5 nouns, I found "health care cost growth benchmark" and even the six noun string "motor vehicle license revenue anticipation certificates", both of which come from legalese. While many examples come from technical contexts like that, there are others that don't, such as "video game designer job description".
(A lot of the other hits have proper nouns or are garbage, but there are many other good examples.)
There's not really a limit to how many nouns can be strung together like this, except that after a while it looks ridiculous and becomes hard to understand.
Comment: Great! But it looks like it's paywalled? No wait the search link works, it's just that if I search it doesn't work, I'll have to learn some more about COCA then. Thanks!
Comment: @LancePollard You have to make an account. Even with how much I use it, I haven't needed to pay (but they constantly bug me about being a free user).


dimension of smooth connected manifold

Question: Let $M^n$ be a smooth, connected $n$-manifold. By definition, this means that every point $p\in M$ has a neighborhood $U$ that is homeomorphic to some open subset of $\mathbb{R}^n$. Now let us assume we only know that for some point $p\in M$ we know that it is in a neighborhood homeomorphic to $\mathbb R^n$ and for the other points we just know that they have a neighborhood that is homeomorphic to $\mathbb R^k$ for some $k\in \mathbb N$. I think that we can show using path-connectedness that $k$ is always equal to $n$ but I don't know how exactly.
Comment: Would that be guaranteed by the transition maps? (they are bijective homeomorphisms after all.)
Comment: I think you should be clearer about your assumptions on $M$.
Comment: How much do you know about the rigorous definition of a smooth manifold? The transitions maps are the composites given by $\alpha (\beta^{-1}(x))$ where $x$ is an element of $\mathbb{R}^k$. They lift such an $x$ into the manifold, then project it back down to $\mathbb{R}^k$ via a different coordinate chart (and they are defined on their overlap in $M$)
Comment: @JustinBenfield I have understood your point just a few seconds before your second comment and therefore deleted my comment, I'm sorry for that. Your argument shows that dimension is well defined on chain connected charts. However all connected spaces are chain connected w.r.t. open coverings, hence this is sufficient. Thanks!
Answer: Let $\sim$ be the equivalence relation which states that $x \sim y$ if there exists charts around them which go to the same dimension (they are not necessarily on a same chart. This only states that each one has a chart around them that both have the same dimension). This is clearly an equivalence relation, and each class is clearly open. By connectedness, we must have only one class. By invariance of domain (or by looking at the derivatives if you are assuming smoothness), there can be no two charts around the same point which go to different dimensions. Hence, the charts only go to a fixed dimension.

dimension of smooth connected manifold


LGPL open source license

Question: I have recently become curious about open source licenses. I have a question about the LGPL. I created a Java web project and built it as a WAR package. Can I add LGPL jars to this WAR file?
Comment: AFAIK as long as you use the LGPL API without making any modifications to the library, your work won't be considered as derivative work, and thus you can release it under any terms. There are other requirements particularly when you redistribute the library with your code so if you plan to distribute your code, you should make sure you follow the terms of the actual license.
Answer: Warning: I am not a lawyer.  If you need legally sound advice, always ask a lawyer.  In this case, ask a lawyer who understands open source licensing.

The following is a summary:

If you don't distribute the WAR file to anyone else, then LGPL places no restrictions on what you do with the library.
If you do distribute the WAR file to someone else, you have two options:

Release your source code under a license that is compatible with the GPL or LGPL, OR ...
Use the LGPL'ed libraries in your code without modifying the libraries, AND without linking or incorporating the libraries into your code in a way that prevents the recipient from replacing them with a modified copy of the libraries.

(Assuming that the library in question is Java, option 2 means that you have to use the library JAR file as-is.  Don't add your own customizations / fixes, don't change class or package names (etc), don't add classes, and don't merge it with your own JAR files.)
For more information, please refer to the GNU Lesser Public License page.
Comment: If customization is needed, he can do it, but then must release modified library under LGPL. Noone else probubly will not find it usefull, other versions of original library will not be possible to use, but other version of his modified library (eg. user repairing bug in it) will be usable and that is what he is using.

LGPL open source license


Ratio vs division

Question: I remember reading somewhere that in ancient times they were not treating a ratio like a division as we do.
I was wondering is there a subtle distinction between the concept of the ratio and the idea of division that made them cautious about this?
Comment: in ancient times they did mostly straight edge compass geometry style questions. In explicitly geometry with ratios allowing for similar objects it is less clear with division
Comment: But why they were not willing to use division for a ratio?
Comment: they also had less of a formal definition of a number, and more of a given unit length.  in short ratios were fat easier to understand without the existence of numbers. I.e. this stick when compared with this stick is similar to this needle when compared to this needle.  it was difficult to prove that this was always devision at first glance where division is the inverse operator of multiplication. Also ratios don't allow for no stick i.e. an invalid comparison. much easier then visualizing the same for 0 and division.
Comment: See this [post](http://hsm.stackexchange.com/questions/2/irrationality-of-the-square-root-of-2) for some comments on the ancient Greek math concepts of *number* and *ratio*.
Answer: They did not work so much with either ratio or division as with proportion.  This was denoted $a:b::c:d$. For example, altitudes $a$ and $b$ will be in the same proportion as the corresponding triangles with the same base. Today we would write this as $\frac{a}{b}=\frac{c}{d}$ but for the Greeks the only numbers were natural and proportion was a relation among actual entities rather than abstract numbers.

Ratio vs division


Prove $\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\frac{\pi^2}9$

Question: I am in the middle of proving that
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac{\pi^2}{18}$$
And I have reduced the series to
$$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
But this integral is giving me issues. I broke up the integral
$$\int_0^1\frac{\log(t^2-t+1)}{t^2-t}\mathrm dt=\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt$$
I preformed the substitution $t-1=u$ on the first integral, then split it up:
$$\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt=\int_{-1}^0\frac{\log(2u+i\sqrt3+1)}u\mathrm du+\int_{-1}^0\frac{\log(2u-i\sqrt3+1)}u\mathrm du-2\log2\int_{-1}^0\frac{\mathrm du}u$$
But the last term diverges, but I don't know what I did wrong. In any case, I would be surprised if there wasn't an easier way to go about this. Any suggestions? Thanks.
Comment: The correct substitution is $t=1-u$. Then the two integrals become the same
Comment: @Zacky first we note that
$$\frac1{k^2{2k\choose k}}=\frac{\Gamma^2(k)}{2k\Gamma(2k)}$$
We then use the definition of the beta function to find
$$\frac{\Gamma^2(k)}{2k\Gamma(2k)}=\frac1{2k}\int_0^1 t^{k-1}(1-t)^{k-1}\mathrm dt$$
Finally, using the Taylor series
$$\frac{\log(t+1)}t=\sum_{k\geq1}\frac{(-1)^{k-1}t^{k-1}}k$$ we interchange the sum and integral signs to arrive at $$\sum_{k\geq1}\frac1{k^2{2k\choose k}}=\frac12\int_0^1 \frac{\log(t^2-t+1)}{t^2-t}\mathrm dt$$
Comment: The other two integrals, with the $i\sqrt3$'s, diverge as well.  You're essentially winding up with an expression of the form $\infty+\infty-2\infty$.
Comment: Sorry to bump this, but can you tell me how did you arrive at that integral from the sum?
Comment: An equivalent problem is discussed at page 31 of [my notes](https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view).
Answer: I start from
$$
\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt .
$$
In the first integral, substitute $t=1-u$. Then
$$
\int_0^1\frac{\log(t^2-t+1)}{t-1}\mathrm dt
=-\int_0^1\frac{\log(u^2-u+1)}{u}\mathrm du .
$$
So you get
$$
\sum_{k\geq1}\frac1{k^2{2k\choose k}}
= -\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt .
$$
ADDENDUM
After some sleep, I managed to compute the integral with the help of polylogarithm.
For $n\in\mathbb R$, define
$$
\mathrm{Li}_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n}.
$$
Some simple facts.

$\mathrm{Li}_1(z)=-\log(1-z)$.
$\mathrm{Li}_2(-1)=\sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}$.
$z\frac{d}{dz} \mathrm{Li}_n(z) = \mathrm{Li}_{n-1}(z)$.
$\mathrm{Li}_n(z) = \int_0^z \frac{\mathrm{Li}_{n-1}(s)}{s}\,ds$.

Then
$$
\begin{split}
-\int_0^1\frac{\log(t^2-t+1)}t\mathrm dt
&= -\int_0^1\frac{\log(1+t^3)-\log(1+t)}t\mathrm dt \\
&= \frac13\int_0^1\frac{\mathrm{Li}_1(-t^3)}{t^3}3t^2\mathrm dt
  - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= \frac13 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt
  - \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= -\frac23 \int_0^1\frac{\mathrm{Li}_1(-t)}{t}\mathrm dt \\
&= -\frac23 \int_0^{-1}\frac{\mathrm{Li}_1(t)}{t}\mathrm dt \\
&= -\frac23 \mathrm{Li}_2(-1)
= -\frac23 \left(-\frac{\pi^2}{12}\right)
= \frac{\pi^2}{18}.
\end{split}
$$
Comment: Oh well, I don't have the slightest idea right now :D
Comment: But at least you have only one integral and not two!
Comment: can I have a teeny tiny hint on how to start with this remaining integral?
Comment: Thanks for this addendum, it's pretty neat.
Answer: Let's multiply by $1$ the integral found in @Federico's answer.
$$\int_0^1\frac{\log(1-x+x^2)}x dx =\int_0^1\frac{\ln(1+x^3)-\ln(1+x)}{x}dx$$
$$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$
$$\sum_{n=1}^\infty \frac1{n^2{2n\choose n}}
=\frac23 \int_0^1 \frac{\ln(1+x)}{x}dx=\frac23\sum_{n=1}^\infty \int_0^1\frac{(-1)^{n-1}x^{n-1}}{n}dx$$$$=\frac23\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}=\frac23\cdot\frac{\pi^2}{12}=\frac{\pi^2}{18}$$
Above I used: $\ \displaystyle{\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}}$
Comment: $\frac{1+x}{1+x}$
Comment: Perfect! thank you!
Comment: What was the $1$ you multiplied by in the first step?
Answer: Another Approach is to employ Feynman's Trick:
Let
$$I(x) = \int_{0}^{1} \frac{\ln\left| x^2\left(t^2 - t\right) + 1\right|}{t^2 - t}\:dt$$
Note $I = I(1)$ and $I(0) = 0$
Thus
\begin{align}
I'(x) &= \int_{0}^{1} \frac{2x\left(t^2 - t\right)}{\left(x^2\left(t^2 - t\right) + 1\right)\left( t^2 - t\right)}\:dt = \frac{2}{x}\int_{0}^{1} \frac{1}{\left(t - \frac{1}{2}\right)^2 + \frac{4 - x^2}{4x^2}}\:dt\\
&= \frac{4}{x}\int_{0}^{\frac{1}{2}} \frac{1}{t^2 + \frac{4 - x^2}{4x^2}}\:dt = \frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right)
\end{align}
We now integrate to solve $I(x)$
$$I(x) = \int\frac{8}{\sqrt{4 - x^2}}\arctan\left(\frac{x}{\sqrt{4 -x^2}} \right)  \:dx = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2 + C $$
Where $C$ is a constant of integration. As $I(0) = 0$ we find $C = 0$ and so:
$$I(x) = 4\left[\arctan\left( \frac{x}{\sqrt{4 - x^2}}\right) \right]^2$$
And finally
$$ I = I(1) = 4\left[\arctan\left( \frac{1}{\sqrt{3}}\right) \right]^2 = \frac{\pi^2}{9}$$
Comment: I'm glad you enjoyed it. I've been on a Feynman 'kick' lately and have started to develop pattern matching skills. One element that may seem odd on the surface is the use of $x^2$ rather than just $x$. If you are interested, try both. You will find the $x$ creates an incredibly nasty indefinite integral to solve at the end and yet $x^2$ makes the process incredibly easy.
Comment: @clathratus - Please refer to my page. I've almost exclusively being posting questions/answers in this area. In some parts I combine both Feynman's Trick and Laplace Transforms. I've also used multiple introduced parameters to solve systems. I would love to have a 'Feynman Trick' mate (/buddy for the US Folk) so please contact me at any time. I want to put together a simple LaTeX document with examples of this method and would love contributors.
Comment: This is a really cool approach! In retrospect, seeing the $$\frac{\log(t^2-t+1)}{t^2-t}$$ really begs for the parameterization which you introduced. Great job.
Comment: I've done quite a bit with series, but not much with the 'Feynman' trick. Any resources that you've found especially helpful in learning about when and how to use it?
Comment: How can I contact you? I'm afraid that I may not have much to offer with the Feynman trick, but I do have more experience with generalized series, factorials/products, and I have been recently investigating special functions and finding closed forms, which may be useful. We could combine forces and do some cool stuff. By the way, you may contact me via the email address email@example.com
Comment: @clathratus - I will email you :-)
Answer: Firstly observe that, for $x$ real,
\begin{align}(1-x)^2-(1-x)+1&=(1-2x+x^2)-1+x+1\\
&=x^2-x+1
\end{align}
\begin{align}J&=\int_0^1 \frac{\ln(x^2-x+1)}{x(x-1)}\,dx\\
&=-\int_0^1 \frac{\ln(x^2-x+1)}{1-x}\,dx-\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=1-x$,
\begin{align}J&=-2\int_0^1 \frac{\ln(x^2-x+1)}{x}\,dx\\
&=-2\int_0^1 \frac{\ln\left(\frac{x^3+1}{x+1}\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{\ln\left(x^3+1\right)}{x}\,dx\\
&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-2\int_0^1 \frac{x^2\ln\left(x^3+1\right)}{x^3}\,dx\\
\end{align}
In the latter integral perform the change of variable $y=x^3$,
\begin{align}J&=2\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx-\frac{2}{3}\int_0^1
\frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(x+1\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{\ln\left(1-x^2\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=\frac{4}{3}\int_0^1 \frac{x\ln\left(1-x^2\right)}{x^2}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx
\end{align}
In the first integral perform the change of variable $y=x^2$,
\begin{align}J&=\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx-\frac{4}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln\left(1-x\right)}{x}\,dx\\
&=-\frac{2}{3}\Big[\ln x\ln(1-x)\Big]_0^1 -\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=-\frac{2}{3}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{2}{3}\zeta(2)\\
&=\frac{2}{3}\times \frac{\pi^2}{6}\\
&=\boxed{\frac{\pi^2}{9}}
\end{align}
Comment: I really appreciate the detailed and aesthetically pleasing nature of this response, not to mention the elegance. Thanks for this great answer.
Comment: Thanks, but after reading all the anwers i'm afraid i have only summed up computations already written by other people (with some details ok)
Answer: Let $I(a)=\int_0^1\frac{\ln(4\sin^2a \ (t^2-t)+1)}{t^2-t}dt$. Then
$$I'(a)= \int_0^1 \frac{2\cot a }{t^2-t+\frac14\csc^2a}dt
=8a
$$
and
$$\int_0^1\frac{\ln(t^2-t+1)}{t^2-t}dt=I(\frac\pi6)=\int_0^{\pi/6}8a\ da=\frac{\pi^2}9
$$
Answer: $$\begin{align*}
I &= \int_0^1 \frac{\log(t^2-t+1)}{t^2-t} \, dt \\
&= \int_0^1 \frac{\log(t^2-t+1)}{t-1} \, dt - \int_0^1 \frac{\log(t^2-t+1)}t \, dt \tag1 \\
&= \int_0^1 \frac{\log((1-t)^2-(1-t)+1)}{(1-t)-1} \, dt - \int_0^1 \frac{\log(t^2-t+1)}t \, dt \tag2 \\
&= -2 \int_0^1 \frac{\log(t^2-t+1)}t \, dt \\
&= 2 \int_0^1 \frac{2t-1}{t^2-t+1} \log(t) \, dt \tag3 \\
&= 2 \int_0^1 \frac{2t^2+t-1}{t^3+1} \log(t) \, dt \tag4 \\
&= 2 \sum_{n=0}^\infty (-1)^n \int_0^1 \left(2t^{3n+2}+t^{3n+1}-t^{3n}\right) \log(t) \, dt \tag5 \\
&= -2 \sum_{n=0}^\infty (-1)^n \left(\frac2{(3n+3)^2} + \frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) \tag3 \\
&= -\frac49 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} + 2 \sum_{n=0}^\infty \frac{(-1)^n}{(3n+1)^2} - 2 \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} \\
&= -\frac49 \cdot\frac{\pi^2}{12} + \frac1{18} \left(\psi^{(1)}\left(\frac16\right) - \psi^{(1)}\left(\frac46\right) - \psi^{(1)}\left(\frac26\right) + \psi^{(1)}\left(\frac56\right)\right) \tag6 \\
&= \boxed{\frac{\pi^2}9} \tag7
\end{align*}$$

$(1)$ : partial fractions
$(2)$ : substitute $t\mapsto1-t$ in the first integral
$(3)$ : integrate by parts
$(4)$ : introduce a factor of $t+1$
$(5)$ : exploit the series expansion of $\dfrac1{1-t}$
$(6)$ : the first sum is well-known; $\psi^{(1)}$ denotes the trigamma function
$(7)$ : trigamma reflection formula,

$$\psi^{(1)}(1-z) + \psi^{(1)}(z) = \pi^2 \csc^2(\pi z)$$