Why is the minimum value of ints, doubles, etc 1 farther from zero than the positive value?

Question: I know it has something to do with 2's complement and adding 1, but I don't really get how you can encode one more number with the same amount of bits when it comes to negative numbers.
Comment: The really nasty part of this is that Abs(MinValue) is negative.
Comment: in java Double.MIN_VALUE is the smallest **positive** value, and the (real) numbers furthest away from zero have equal magnitude (as it has a proper sign bit)
Answer: Think about it in these terms. Take a 2-bit number with a preceding sign:
000 = 0
001 = 1
010 = 2
011 = 3

Now let's have some negatives:
111 = -1
110 = -2
101 = -3

Wait, we also have
100 ...

It has to be negative, because the sign-bit is 1. So, logically, it must be -4.
(Edit: As WorldEngineer rightly points out, not all numbering systems work this way -- but the ones you're asking about do.)
Answer: Because there are not two classes of numbers in the integer range, but three: negative numbers, zero, and positive numbers. Zero has to take up a slot (would be rather impractical not to be able to represent zero...), so either the positive or the negative class has to give up a slot. The fact that it's usually the positive range that has to make that sacrifice is to a certain extent arbitrary, but on the level of bit manipulations there are some things that this decision makes more convenient.
Comment: It's not just bit manipulations.  The set of 32-bit signed numbers is the set of numbers whose binary representation has the same value in all bits after the 31st, and one such number has an infinite set of ones followed by 31 zeroes.  The additive inverse of that number, an infinite string of zeroes followed by a single 1 and 31 zeroes, does not fit the pattern required of signed values.
Answer: There are BASICALLY three ways to represent signed integers in binary: 2's complement, 1's complement, and sign-magnitude.  (Biquinary went the way of the Dodo Bird a long time ago.)
1's complement and sign-magnitude have two zero values, +0 and -0, each with a unique representation.  2's complement only has one zero value, and one representation.
Now, a field of N bits can encode 2^N values.  Subtract one in 2's complement, and you have 2^N-1 = 2^(N-1) + 2^(N-1) + 1.  Since the representation for zero is all zero bits, and a + sign is zero, there will be one more possible nonzero representation with the sign bit set to 1.
This is a very long-winded way of saying 2's complement represents values in the range -(2^(N-1)) .. +(2^(N-1) - 1).
1's complement actually has an advantage over 2's complement if you are doing integer digital signal processing computations.  1's complement operations inherently truncate toward zero.  2's complement truncates toward -infinity.  I learned this one the HARD way...


How can I stop sweaty hands?

Question: Sweaty hands are plaguing my rock climbing and pull up exercises, as well as being annoying at other times. I've read a few (unsourced) ideas that nutrition can be a big factor in sweaty hands. Can I do to stop sweaty palms by changing my diet? If so, how?
Comment: What makes you think your diet has any (measurable) effect? If the claims are unsourced, then they might as well not exist as far as we're concerned. I suggest you rephrase your question to dealing with sweaty hands through non-diet means.
Comment: Rock climber here: even just talking about this makes my palms sweat. If you just started noticing the sweaty hands since rock climbing, I wouldn't worry about it. Most my rock climbing friends have similar issues.
Comment: Have you considered keeping a pouch of chalk on your belt while rock climbing? I've personally found it incredibly useful.
Comment: I do use chalk, but I'm using far more than I should be. My jeans are getting warn out from drying sweaty hands on them. It's bad during climbing, but its happening all the time, and I can't really chalk up in the middle of class.
Comment: I have the same problem, sometimes I'll end up having to re-chalk every couple of moves. You can buy a tube of tite-grip online, which helps, but if you want the serious prescription grade stuff go see a dermatologist.
Answer: Sweaty hands is a common ailment for weight lifters and rock climbers, with equally detrimental effects.  The solution really isn't nutrition related.  It's simply chalk.  Chalk dries the hands, and improves your grip.
If you feel that chalk would be detrimental to the environment, you can go for an "Eco Ball" by Metolius.  Same effects, although I might argue that it is not quite as effective as good old fashioned chalk.

After reading the comments, you also may want to look at this article on diet and hyperhidrosis.  The thing is, you do need to rule out whether you have hyperhidrosis or not.  Take a look on at this article that describes what hyperhidrosis actually is.  It also has tips on food to avoid in addition to other tips.
Also do remember that rock climbing is an activity that causes an adrenaline rush, and is therefore stressful.  Not necessarily in a bad way, but it is a factor to consider in trying to understand the sweaty hands.
Answer: Sweating is generally either: genetics, or high level of stress (or both).
If you have been dealing with sweaty palms for a while, and do not have high levels of stress, it might be genetics. (Some food might make you feel hot like spices and stuff, but i personally do not believe that they cause constant sweating..)
I am not a doctor, but myself I have a very high metabolism and energy levels (genetics) and this makes me get overheated and sweat more than the average person on my forehead and sometimes my underarms.
And during periods of stress it gets worse of course.
The only Solution that kinda fixed the underarms sweat (and am not scaming you or selling you a product, i do not work for this company and I do not live in france) is a European (french) product called etiaxil. They say you can apply it on hands( if you need help in translating let me know).
What i did, i used etiaxil at first 3 times a week following their instructions. And after only one month of usage, i did not have that problem for underarms anymore, even when i am on vacation in a 38 degrees celsius temperature.(too bad they don't have a product that can be used on forehead).
Check it out, and most important , do no be bothered by having sweaty palms because in reality the bodies that sweat more last longer because such bodies keep on cleaning themselves from toxins with sweat.
Answer: Consider putting underarm antiperspirant on the palms of your hands at bedtime. I don't know if underarm antiperspirant is safe or effective on the hands. See this study which studied the use of underarm antiperspirant on the feet.
There exist $20 antiperspirants labeled for use on hands. The inactive ingredients might differ from those of underarm antiperspirant; I haven't researched the matter. The active ingredient is likely to be an aluminum compound which is either identical or related to the aluminum compounds found in underarm antiperspirants. Make sure the product you choose contains an aluminum compound: if it doesn't, it's unlikely to work very well.
Dear all: if you do some PubMed searching to see if underarm antiperspirant works on hands, please edit this answer and point us to what you found. I can approve your suggested edit.

How can I stop sweaty hands?


Im using lubuntu with no dual boot, it has had a fatal crash and needs to be completely removed

Question: So my lubuntu OS v. Willey 14.04 has had a melt down the initial cause was a rewrite of the chown file and has grown from there, I can currently only access the terminal or an unusable desktop, I need to know how to completely remove lubuntu using its terminal as I'm on a PC without deul boot. And than I need to know how to install mint rather via same terminal or via the ISO disk I have preburned, please make answer s vey simple this is my first Linux and I've had it only 3 days I do not own a flash drive and am on a relatively stable wifi network, my first language is English but I live in Mexico, so my region would be that. The PC I'm working with is a 64 bit Intel HP notebook 15" display with Blu-ray ROM and HDMI I'm using a b and g connection. Unfortunately I don't know any of my other specs besides that I have 2 TB ram
Answer: You don't need to remove Lubuntu before installing Linux Mint. The installer can do that for you.
Just boot from the Linux Mint installation disk and start the installation of Linux Mint.
If you get to the part where it asks whether or not you are sure that you want to overwrite the existing data on the disk, answer yes. If the installer detects Lubuntu and suggests that it can keep it and make your system dual-boot, you can select to remove it.
This will get rid of Lubuntu and all of the other data on the disk as well. So make sure you have backed up everything you want to keep.
If your system doesn't boot from the disk, but instead always boots from the hard disk, you have to get into the boot menu of your laptop and change the boot order.
When you turn on on your laptop, it normally will tell you what key or key combination to press to get into the boot menu. It is often Del, F2 or F10, but could be any key or key combination.
Good luck.
Comment: I have tried all of your mentioned with no luck, I've looked into boot order but can only do it from hard disk or from a flash drive, I live in Mexico and flash drives are not readily available or afordible here, I can get into the terminal by pressing shift + f-4 is there a way to do it through this, I've been able to get desktops but nothing else
Comment: If your laptop comes with a DVD/Blu ray player, it should be able to boot from it. According to http://support.hp.com/nz-en/document/c00364979 the optical drive should boot before the hard disk. The only way to install a new OS is to boot into an installation medium. It is not possible from a terminal, because the shell is running of the OS on your disk, which is going to be wiped when you install a new OS. The system will have to go through a reboot before you can install a new OS.
Comment: Thank you I punched my model number into HP's website it's f10 to get into BIOS from there I was able to install from the blu-ray-rom drive. You may mark this as solved
Comment: @user6232162. I can't mark the question as solved, you have to do that :-) If you are happy with my answer, then you can tick it as the one that solved your problem (click on the little green check-mark below the up-vote and down-vote arrows).


How to tell who is connected via vnc to my machine?

Question: How do I check how many vnc connections or different users are currently connected to my machine? Only through wireshark? And how do I kill any other users connected through vnc?
Answer: On Linux,
ss sport = :5900

Would tell you the currently established TCP connections on port 5900.
For anything else, we'd need to know what VNC server you're using as there exist dozens.
If you know the name of the VNC server command,
lsof -ai tcp -c that-command

(as the user running the VNC server or as root) would also tell you the currently established TCP connections handled by that VNC server (in case it's not on port 5900).
Among the methods to close a TCP connection in the general case, there's:

tcpkill (from dsniff package) that fakes the TCP packets that would close a connection.
iptstate (Linux). If you're using a statefull firewall. You can remove the connection from the Linux connection tracker table (x in iptstate), which would usually cause newer packets to be ignored (if the firewall is configured to only accept new connections with TCP SYN). It wouldn't kill the connection, just make it inactive though.
Add firewall rules to reject further packets transmitted/received in that connection (on Linux, iptables with target "REJECT", matching on source and destination TCP ports and IP addresses in OUTPUT and INPUT filter chains)
attach gdb to the running vnc server and do call close(fd) where fd is the file descriptor for the corresponding socket (which you've found out with lsof) followed by detach (it may confuse the vnc server a lot those, calling shutdown instead of close may be safer).
Comment: FYI, I needed to use `sudo` to run `lsof` to get output on my Raspberry Pi (Raspbian/Debian Linux).
Comment: @HeatfanJohn. Yes a user can only see the information from the processes it owns. So you'd need to be root or the user running the vnc server.
Answer: According to this article the following command will show connections to processes containing "VNC command-line":
# sudo netstat -p | grep <name of VNC command-line>

Update
Please note that the netstat command has to run as root
Answer: Assuming your vnc server is Xvnc, try this one-liner:
lsof -Pni | grep Xvnc | grep -v LISTEN

This will list active connections and show you their process PID and IP addresses.
Answer: I was looking for a similar command to figure out what port(s) I had vnc servers listening on and this is what I came up with.
ss -a -t '( dport >= :5900 & dport <= :5999 | sport >= :5900 & sport <= :5999 )'

See a longer explanation of the filter syntax and links to the source in the other answers/comments here.


Double batteries capactior circuit

Question:
Can anyone help me solve this problem?
Comment: seems like a bad question to me, unless the initial charge on the capacitors is specified the rest state is undefined. \$V_{NM}+V_{AB} = 5V \$ is all that can be derived.
Answer: Ok. First things first:
We have four elements in series so we can arrange them however we like and it's all the same.
We have:
-3v+ , C1, +8V- , C2
But it's the same if we have:
+8v-, -3v+, C1, C2
That's equivalent to having one 5v battery (8v - 3v) charging two capacitors in series.
So we have a 5v battery charging a 2uf and a 3uf capacitor.
We will assume capacitors start with no charge. We are looking for the voltage between the two capacitors, let's call it Vo.
We know that both capacitors have the same charge, the same number of q, or electrons attached to its plate since they are in series.
We also know that the voltage of both capacitors must add up to 5v.
We know that for a capacitor q = C*V. this is equal to V = q/C
Let's put all together:
Vc1 = q1/C1
Vc2 = q2/C2
Let's divide this equations:
Vc1/Vc2 = q1/q2 * C2/C1
but we said q1 = q2, so we can cancel them.
Vc1 / Vc2 = C2/C1
That's equation number 1. Equation number 2 is:
Vc1 + Vc2 = 5v.
Now just replace:
Vc1 +C1/C2*Vc1 = 5v
Vc1*(1+2/3) = 5v
Vc1 = 5v/(1+2/3) = 5v/(5/3) = 3v
and Vc1 + Vc2 = 5v --> Vc2 = 2v
Cheers!

Double batteries capactior circuit


Do disposible cameras go bad if you don't develop soon after using?

Question: I've just found an old disposable camera after a bit of a tidy up, no idea what's on it.
On the bottom it says develop by 08/2010 so it's about 6 months overdue.
Will the images still be OK to develop? I.e., does the film degrade after time? I am not looking for prints, just need photos onto disc.
Comment: People already showed you there is the possibility that it is still good. I had the opposite experience after about one year Pictures came out with severe degradation. Mind you it was not kept in controlled conditions just my home which varies temperatures with the seasons and get extremely humid in the summer.
Comment: the cost of developing and scanning this should be pretty low, so why not give it a try? You may find treasures there...
Comment: [No, they go better!](http://content.photojojo.com/photo-projects/disposable-camera-photo-time-capsule/)
Comment: Im so happy to read these post i have like 30 undeveloped cameras from 8 yrs to 3 yrs ago. Had them put away in my boxes in my aunts house ! I was so down thinking i might not get any pics! Can't wait to start developing them! Can walmart help me, or do i need a special lab????
Answer: As long as the camera has been protected from the elements, and stored in moderate conditions (ie, no extreme hot/cold/humidity), I think you should be fine.  Six months isn't that long, and the guidelines take into account that people might not be taking care of the camera.
You might see some degradation in the colors, but probably not much.  It's worth a shot getting it developed, and see what you get!
Answer: Unless you seriously abused this film, there's no problem.  I have taken film of all types--B&W, color slides, color prints--on repeated long trips, including extended ones outdoors in the desert, developed it late--sometimes years late--and noticed no degradation at all.  The only exception was a roll I found last year (Ektacolor or something like that) which had been left almost 20 years in a camera (Nikonos V).  There was noticeable washing out of color around the top and bottom edges of the film, a slight loss of contrast, and a little blurriness overall, but even so the photos are serviceable.  Here's one from the middle of the roll, shown in full exactly as scanned by the processor (converted to JPEG by Picasa for upload here).  Consider it an extreme example of the degradation that happens to all film over time.
Comment: @Evan Good eyes!  You can see those artifacts at the top and bottom.  (They're not whole bands, though: they extend only about 10-20% of the way into the picture from the top and bottom.)  They correspond to the sprocket holes along the edge of the film.  Over the years--even though this is a *waterproof, metal* camera kept in a case or drawer--some photons evidently did get in.  In effect, each sprocket hole acted like a little window of purple light.
Comment: I see about 8 vertical bands; are those present in the original print, or are they artifacts of scanning?
Comment: Very respectable results for 20 years! Like you said, most of the image is still quite good.
Answer: I've seen the results of film that was developed after fifty years of living on a shelf. While it did take some special processing, the results were fabulous.
A few years ago, I found a roll of TMAX that I shot when I was a teenager (going on 20 years ago), and the lab was able to develop it just like it was exposed yesterday.
I've even seen unexposed film that was decades past it's expiration shot and developed perfectly fine.
In essence, unless you're dealing with Kodachrome, there's a good chance that any film you find can be processed.


Accidentally formatted a boot partition

Question: I accidentally formatted boot partition, so now when I boot the computer it just said "no bootable device found".
How can I repair the boot partition?
Comment: use a livcd to create a new one. Install grub.
Comment: in the hope you read this message... if you solve the problem pick the answer that helped you to solve the problem and _upvote_ the all the work you find useful. If you find the solution by yourself, you can post it as your own answer...(and you can still pick another one as best answer). Welcome on SuperUser.
Answer: you could download
supergrub2
put this on a cd or usb then
boot into your system then reinstall grub from your system
Im only guessing your using linux
Answer: use the Ubuntu installation CD to repair the installation to a working condition.
Also try attaching the disk as secondary disk in another machine and try recreating the /boot with files copied from similar OS. For more details check the link below
https://help.ubuntu.com/community/CreateBootPartitionAfterInstall
Answer: Depends on what you mean by "boot" partition. If it's the linux partition that is mounted as /boot then earlier answers involving 'repair', or using your original installation media and looking for a repair option, are probably all you need. I'd recommend trying the original install media first, since it's going to be closest to what you had already, but any grub tool will probably suffice.
OTH if it's the main boot partition for the system and was mounted as / in linux, without a separate /boot partition, it can be a whole lot more work. (If even possible.)
From beginning, if not already done and if you have the hardware available, make an image of the partition using something like dd before doing anything else that can possibly write to that partition.
Depending on the original file system, different tools are available to "unformat" the disk. The one I've seen most promising results from for others is TestDisk.
I can't give testimonial to that, or any other tools, since the list time I formated the wrong disk I was using DOS and had to use DOS debug and lots of hex math to recreate the FAT table by hand. :'( I haven't made that mistake again!
Hopefully you had ext3/4 and the backup superblocks will save the day. And now you have personal experience in the need for backups, and the wisdom of compartmentalized partitions. I'd rather sacrifice some diskspace to extra partitions than my data to errors - human or mechanical.
Best wishes and lots of luck.
Comment: with a volume group name I'm guessing you used LVM during install, and you've entered a realm I have zero experience. Discretion being the better part of valor I must defer to other minds. I suggest you update your Q to give as much detail as possible: what command/process used to format the partitions, what the partition descriptions were (name/size/mount/etc.), and _everything_ you've done since that might have affected the partitions. Also, "no bootable device found" looks like a BIOS message, so there may not be any partitions with the boot flag attached.
Comment: Glad you've found a solution! Not sure if it's possible, but since you're not going to try to fix the problem stated in the Q, maybe you can close the Q so others don't try to help find a solution you cannot try. Enjoy the fresh install, and practice good backups.
Comment: Formatted two partition, an efi partition and an unknown partition.
Comment: The only one partition left is the xubuntu volume group (xubuntu _vg)
Comment: Decided to switch from xubuntu to ubuntu. Xubuntu on laptop is quite a mess IMO. (Doing scaling, Bluetooth, multi monitor, etc.) I've made backup and then restore to ubuntu. It's all good now. Thanks for the help!

Accidentally formatted a boot partition


Snapshot size in lvm

Question: I am trying to understand how to select a proper size for snapshot.
As I know simple snapshot uses space which is needed for covering difference between two logical volumes.
So, what size of snapshot should be if it was made of (for e.g.)30G partition on which all files had been rewritten? Just 30G? or it depends on future changes of partition?
Comment: If you allocate 30GB for a snapshot of a 30GB volume, then you should be able to do literally anything to the volume and not run out of space (except resize it bigger, maybe).
Answer: A snapshot will require as much storage space as changes are made to the contents in the logical volume.
So, let's say that you have a 30GB volume and you take a snapshot. Initially, no space is required. But after a week you see that the snapshot is using 4GB of storage. That means that a subset of the contents in that volume has changed enough to generate 4GB of changes.
If the changes are made over the same subset, the size will remain quite constant. But if changes are mode on different subsets, the size will start varying accordingly.
So, in short terms, the size will depend on the 'changeability' of the volume and the retention time you want to use.
For that example, a 1 month retention policy would require:

4GB for a single snapshot
8GB for 2 snapshots (every other week)
Up to 32GB for 1 weekly snapshot
and so on.
Comment: Not quite.  A snapshot doesn't store multiple versions of changes, so it never needs more space than the original volume.  If the original volume is 30GB, there's only 30GB of data that might possibly ever need to be copied to the snapshot.  (However, I think snapshots have a slight size overhead due to the way they structure the copied data, so you might need to make the snapshot slightly larger than 30GB for that reason.  But it's not related to the rate of change to the original volume.)
Comment: I agree, with you @Wyzard . I was actually thinking about multiple snapshots, and since I was multitasking, I guess I wrote the answer slightly mistaken :) Sorry about the confusion. Editing ;)
Comment: Regarding your edit:  the size overhead in a snapshot has nothing to do with the filesystem (or any aspect of the data) in the base volume.  It's just that the snapshot has to store data copied from the base volume, *plus* information about which storage locations in the snapshot volume correspond to which storage locations in the base volume.

Snapshot size in lvm


What are the major differences between theoretical cryptography and applied (practical) cryptography?

Question: There are a lot of theoreticians who don´t have a clue of the applied aspects involving cryptography.
Are the two disciplines mutually exclusive?
Can you learn one without the other?
Comment: theoretical works on new theory, applied as the name introduces envisions practical deployment trying to solve real world problems solved with real cryptographic protocols. Theoretical cryptography may gave a new, novel theoretical construction with a math-minded logic, without taking care whether this is applicable or efficient, while applied crypto foresees practical aspects
Answer: I cannot tell any obvious feature to differentiate them, but I think you can distinguish theoretical cryptography and applied cryptography from the writing style of papers. In the paper about theoretical cryptography, they usually prove some cryptographic primitive(s) imply another one or construct some schemes or protocols that could be reduced to a standard assumption (e.g., identity-based encryption can be reduced to BDH assumption). Nevertheless, in the paper of applied cryptography, there are usually some implementation results which includes, for example, how to set exact security parameters or message length and so on. Furthermore, theoretical cryptography papers are usually published in conferences like CRYPTO, EUROCRYPT, and TCC, and applied cryptography papers are usually published in conferences like USENIX and ACM CCS.
Comment: That is not really precise. Applied crypto also gives security evidence of the construction with a proof. Block ciphers as AES ( we agree I guess that is purely applied) comes with a security proof
Comment: Sure, that's why I say "includes", because the construction and proof are usually followed by implementation details in applied cryptography papers, but there is no implementation details in theoretical ones.


Will a PVC pipe crack in concrete?

Question: If a plastic PVC pipe is totally covered with cement? Does a minor crack/s in the concrete will also crack PVC pipe or damage it? I mean cracks such as due to very minor seismic movements?
Comment: Covered by or encased in?
Comment: Hello, and welcome to Home Improvement. Interesting question; what size pipe, and what type of cement? And, you should probably [take our tour](//diy.stackexchange.com/tour) so you'll know how best to participate here.
Comment: One approach to minimize problems from minor movement is to wrap the pipes in thin closed-cell foam before pouring concrete around them. "Sill-seal" is a convenient material, available at most building supply places.
Comment: Pipe used for sanitation, 6 inch in diameter, and encased.
Answer: If the pipe is totally encased in concrete small surface cracks would not affect. If the cracks go all the way through, like a total crack in a foundation, it could also damage the pipe. If the concrete has shifted along the crack line then the pipe is going to be damaged. The amount of damage will depend on the type of re- enforcement done when laying the concrete.
What is the pipe used for?
Comment: Pipe used for sanitation, 6 inch in diameter.
Comment: The purpose of asking was that should the pipe be encased with cement or just be buried if there are very minor seismic movements in the area?
Comment: This might depend on location and type of construction but in the residential areas where I have lived the pvc is direct buried with the slab poured on top.
Answer: California building code requires to insulate sewer pvc pipes that will touch concrete if has potential to condensation. This is to prevent pipe cracks. I would recommend insulating since its involves little labor and cost to avoid future problems.

Piping subject to operation at temperatures that will form
  condensation on the exterior of the pipe shall be thermally insulated.

Will a PVC pipe crack in concrete?


Which older SF book pits a space being against a world?

Question: This book would be older than the 50's or 60's, when I read it. It has a planet-destroying entity/creature who tries to destroy a planet. The people on the planet (not earth?) fight back and drive it off. The creature learns to counter the attack and comes back. Then the people learn to rebuff the creature's new attack. This goes back and forth until they are throwing worlds at each other, growing stronger and smarter after each attack. I would like to read it again if still available. Thank You!
Comment: Sounds interesting! Do you recall the creature's motive for destroying planets? Did it eat planets, or was it annoyed by them? Did it have a spaceship, or did it live in space? Was it man-sized, moon-sized, planet-sized, sun-sized, bigger than that? Was it animal, vegetable, mineral, gaseous, or an energy being? Was the book hardcover or paperback, possibly an Ace Double Novel?
Comment: There is a list of planet killers in wikipedia, probably will help - http://en.wikipedia.org/wiki/List_of_planet_killers#Other_literature
Comment: I was a teenager when I read it, and 65 now, so things are a bit hazy...but here is my best try at your questions. I think the creature lived and traveled in space. Don't know motive. Was not after the whole planet at first, but became stronger and larger as it developed during the fighting. I believe it was a paperback. Can not recall the type of life it was.
Answer: This reminds me of "The Searcher", by James H Schmitz, from the collection The Hub: Dangerous territory.  it was first published in 1966.
There is a space entity which attacks, and attempts to destroy a planet.  The reason is the planet finds a sensor the entity needs, and moves it to the planet for research.  there is a series of counterattacks and defenses back and forth between the entity and the people on the planet, and the space entity grows more and more dangerous after each attempt, gaining knowledge and power.  There are attempts to communicate (for various levels of creepiness and threat).  It does not end in them actually throwing planets at each other, but the energy attacks, and the people's rebuffs, are spectacular and dangerous enough to be very memorable.  Does it sound familiar?


Oldest Exercising Equipments

Question: What is the oldest exsistence of physical exercising equipments found in world, through literary evidences or archaeological  excavations?
Answer: The 4th-5th century A.D. "Bikini Girls" mosaic from the Villa del Casale of Piazza Armerina shows female athletes practicing or competing in athletic competition with dumbbells and a big ball that looks like a medicine ball (both in upper-left).

Without knowing the materials used to make that ball, it's hard to tell how heavy it would be. This site goes over the possibilities:

The Romans had several different types of balls for their games --
  hard balls, soft balls, bouncing balls, large balls and small balls.
  They even had a glass ball and a ball of stone the size of a bowling
  ball was found in Pompeii. At least six distinct types of balls can be
  identified . . . The two types of follis most certainly bounced. The
  paganica and harpastum did not. Balls made of wool would have bounced
  poorly. The two balls [below] were found in an Egyptian tomb from the
  Roman period and are about 5" in diameter. Notice the striking
  similarity to the Bikini Girls ball. They are made of linen and hair
  wrapped in string and sewn together. These wouldn't have bounced well
  but were probably used for games like Trigon, harpasta, or Field
  Hockey.
Comment: While it is indeed an interesting answer, 4-5th Century AD? Are there no older ones found?
Comment: @Rohit: It's the oldest solid evidence I've been able to find so far. I followed a few other leads, but they aren't well sourced.

Oldest Exercising Equipments


Difference between video captured with Google Earth and Google Earth Pro

Question: Is there any difference between video which is captured with Google Earth and Google Earth Pro? Maybe some kind of tag showing 'pro'?
Answer: You can't produce movies on the free Google Earth software (at least with my current version on OSX).  There is a Google Earth tag placed on the video from the Pro version, and you have the option to produce videos with a range of resolution. For still images, you can produce a range of different image sizes also.


Can we express $\sin 1^\circ$ in a real closed, not repetitive, radical forms?

Question: Can we express $\sin 1^\circ$ in a real closed, not repetitive radical forms? Any radical forms mean you can use any roots but without constants $\pi$, $e$ or other trigonometry functions.
Comment: [Related question.](http://math.stackexchange.com/questions/94478) I gave a radical expression for $\sin 1^\circ$ in my answer there.
Answer: In principle, yes.
This paper gives a value for $\sin 3^{\circ}$:
$$\sin 3^{\circ} = \frac{1}{4} \sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}} \, .$$
Moreover, we have the triple-angle identity for $\sin$, which I will suggestively write as:
$$
4 \sin^3 \theta-3 \sin \theta + \sin 3\theta=0 \, .
$$
Combining these, you can see that $x=\sin 1^{\circ}$ is a root of the cubic polynomial
$$
4x^3-3x+ \frac{1}{4} \sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}} = 0\, .
$$
You could then use the cubic formula to find a closed-form expression of $\sin 1^\circ$ that uses only radicals.
Two caveats:

When you use the cubic formula on a polynomial with three real roots, the radical expression you get must always involve complex numbers. This will be the case here, since $\sin 121^\circ$ and $\sin 241^\circ$ must also be roots of the same polynomial. So if you want to express $\sin 1^\circ$ in terms of radicals using only real numbers, you're out of luck.
The expression you get will be so horrifically complicated as to be totally useless for any practical or computational purpose.
Comment: In my understanding only two are complex conjugates.
Answer: FWIW, the minimal polynomial of $\sin(\pi/180)$ over the rationals is
$$
281474976710656\,{z}^{48}-3377699720527872\,{z}^{46}+18999560927969280
\,{z}^{44}-66568831992070144\,{z}^{42}+162828875980603392\,{z}^{40}-
295364007592722432\,{z}^{38}+411985976135516160\,{z}^{36}-
452180272956309504\,{z}^{34}+396366279591591936\,{z}^{32}-
280058255978266624\,{z}^{30}+160303703377575936\,{z}^{28}-
74448984852135936\,{z}^{26}+28011510450094080\,{z}^{24}-
8500299631165440\,{z}^{22}+2064791072931840\,{z}^{20}-397107008634880
\,{z}^{18}+59570604933120\,{z}^{16}-6832518856704\,{z}^{14}+
583456329728\,{z}^{12}-35782471680\,{z}^{10}+1497954816\,{z}^{8}-
39625728\,{z}^{6}+579456\,{z}^{4}-3456\,{z}^{2}+1
$$
EDIT: the explicit expression obtained from Micah's cubic is not all that bad:
$$ \frac{-1-\sqrt{3}i}{4} v^{1/3} + \frac{-1+\sqrt{3}i}{4} v^{-1/3}
$$
where $$v=-\frac{1}{4}\sqrt {8-\sqrt {3}-\sqrt {15}-\sqrt {10-2\,\sqrt {5}}}+\frac{i}{4}
\sqrt {8+\sqrt {3}+\sqrt {15}+\sqrt {10-2\,\sqrt {5}}}
$$
Answer: If we're allowed to use complex numbers, then sure. Note that $\sin(1 ^\circ)=\sin(\pi/180)=\frac{e^{i \pi / 180} - e^{-i \pi/180}}{2i}$. We can express $e^{i \pi/180} = (-1)^{1/180}$ and  $e^{-i \pi/180} = -(-1)^{179/180}$. So the result is $\sin(1^\circ) = -\frac{1}{2} (-1)^{1/2}((-1)^{1/180}+(-1)^{179/180})$
Answer: Since $(-1)^{89/180}=\cos(89^\circ)+i\sin(89^\circ)$, we get
$$
\sin(1^\circ)=\cos(89^\circ)=\frac12\left((-1)^{89/180}+(-1)^{-89/180}\right)\tag{1}
$$
However, I think more in keeping with the spirit of the question, if we start with $\sin(6^\circ)=\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}8$ and apply
$$
\sin\left(\frac x2\right)=\sqrt{\frac{1-\sqrt{1-\sin^2(x)}}2}\tag{2}
$$
and
$$
\sin\left(\frac x3\right)=\frac{\sqrt[\Large3]{-\sin(x)+\sqrt{\sin^2(x)-1}}+\sqrt[\Large3]{-\sin(x)-\sqrt{\sin^2(x)-1}}}2\tag{3}
$$
we get $\sin(1^\circ)$. Of course, we still get into the realm of complex numbers when applying $(3)$.
Answer: Well you can construct a regular pentagon and an equilateral triangle inscribed in a circle with ruler and compasses (equivalent to taking square roots) - that gets you angles of $72 ^\circ$ and $60^\circ$ - you can get sin and cos of both angles, so can get sin and cos of the difference $12^\circ$. You can halve that twice to get down to $3^\circ$, and then you need to solve a cubic, which is solvable by radicals.
Answer: Here's a table of exact values of the sine, cosine, tangent, and cotangent of integer multiples of $3^\circ$: http://en.wikipedia.org/wiki/Exact_trigonometric_constants
Answer: This page has closed formulas for $\sin(n^\circ)$ for all $n=1,\dots,90$.  Many of them involved $\sqrt{-1}$, not sure if that is avoidable but somebody in another answer on this page says it is not.
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
I would like to give credit to the person who made me aware of that list, but their post was deleted so I can't recover who it was.  But thank you, whoever you are.
Comment: This page doesn't exist.
Answer: Using the formula
$$
\sin(\pi/n) = \frac 12 i e^{-\pi i/n}-\frac 12 i e^{\pi i/n} = \frac 12 (-1)^{1/2} (-1)^{(n-1)/n}-\frac 12 (-1)^{1/2} (-1)^{1/n},
$$
one can write $\sin(\pi/n)$ using radicals for any positive integer $n$.
Comment: How do you evaluate $(-1)^{1/n}$ as $a+bi$, where $a$ and $b$ are real and are written using radicals?


How to generate SOAP message from a WCF service?

Question: Is there any way programmatically or any tool by which i can generate SOAP messages from WCF services ?
Answer: What do you mean? WCF uses SOAP to communicate - all WCF messages (except when using the REST based webHttpBinding) ARE SOAP messages.....
Something I think I'm not understanding what you're really looking for....
If you need a tool to build SOAP messages to send to a WCF service - look at the WcfTestClient.exe  in your Visual Studio Common7/IDE directory - it does that. Works with almost any WCF binding.
Or check out SoapUI for a more powerful toolset to examine and test web services in general (works with WCF's basicHttpBinding).
Marc
Answer: Do you mean you want to view the soap messages? if you do, you can try wcf trace viewer.  It comes with VS 2010 developer edition.  If you don't have that, you can try WCF storm, 15 days free trail.
But yeah, like what marc said, if you want to generate soap messages to test your wcf service host, you can use wcftestclient.exe.


Is a programmer liable for copyright infringement in other people's mods to his games?

Question: If a programmer (P) releases a game, and a modder (M) releases a game modification which includes copyrighted content, can P be held liable for the copyright infringement in the mod?
Answer: I'm going to start off by being That Guy and noting that this site is the wrong place for this question.  It's a potentially non-trivial legal issue whose precise answer, under various circumstances, will likely depend on details of local law and legal precedent, as well as on the relative abilities of the lawyers on each side to twist said law and precedents to their advantage and the attitude and preconceptions of whoever happens to be chosen to adjudicate the case.  In short, if you want a real answer, ask a lawyer.
That said, out of the cases mentioned in your question before it was shortened, the M = P case is the easy one if that fact can be proven.  Note that, again depending on jurisdiction, even just reasonable cause for suspicion might be enough to obtain a subpoena or a search warrant to acquire confirming evidence, e.g. from ISP logs (as noted in another answer) or even directly from P's computers, which might contain revealing traces of their activities as M.
The case where M's identity is known (and M ≠ P) is also usually simple for P, since any plaintiff's attention will probably be mostly focused on M, and since P typically has a fairly good defense against any accusations by pointing to M and saying "he did it, I had nothing to do with it!"  That said, as I'll note below, that defense isn't necessarily ironclad, if the plaintiff does decide that they want to go after P as well (e.g. because M is out of their reach, or too poor to pay for any damages even if found guilty, or just because they can).
Now, on to the meat of the answer: If P does get sued, either alongside M or because M could not be identified, it will likely be for contributory infringement.  Basically, the claim would be that P knowingly aided and encouraged M's infringement, and materially benefited from it, by making the game moddable in the first place and continuing to sell it after becoming aware of M's mod's popularity.
It should be noted that, under most circumstances, that would be a rather tenuous claim.  Making games moddable is standard industry practice, and, indeed, it shouldn't be hard to find expert witnesses to testify that making a game unmoddable is often difficult or impossible.  As long as P's original product was a playable and salable game in itself (and did not, for instance, feature blank sprites and textures with "PLACEHOLDER" written on them) it would typically also be hard to argue deliberate intent to encourage infringement.
As for profiting from the sales driven by M's mod, that might depend on just how popular the mod actually became, but in most jurisdictions there should be either case or statute law — such as, notably, the "Betamax case" (Sony v. Universal) in the U.S. — exempting makers of products used by individuals to infringe copyright from contributory liability, as long as the product also has "substantial non-infringing uses" (to quote the Betamax decision).  Indeed, if such an exemption did not exist, what would prevent the plaintiffs from just going directly after, say, Intel and Microsoft for making computers that could run the game that could host the infringing mod?
Of course, the situation could be different if it really could be demonstrated that an overwhelming fraction of all sales of P's game were driven by M's mod, and that P should have been aware of this.  At least, someone could then argue that P should have patched the game (or at least any subsequently sold copies of it) to not run M's infringing mod (and to communicate this to prospective buyers).  Of course, such efforts would likely be ultimately futile (as M could just patch the mod to compensate), but it could be argued that, if P hadn't even made any such effort after explicitly being made aware of M's mod and its popularity (and illegality), that could constitute evidence of willful encouragement.
For an example of similar arguments, see e.g. the recent Megaupload case (which, of course, has not actually proceeded to trial, but at least illustrates the kind of arguments that could be made).
Comment: It does sound rather stupid that a programmer of a tool(moddable game) could get sued for something a user did. It would be like microsoft getting sued if I used word to write a harry potter fan-fic, or autodesk getting sued for me using maya to recreate micky mouse.

Just my 2 cents, not any advice about how the law works.
Answer:
"If M, how ? It's easy for him to avoid any kind of punishment as "the
  mod" is free and therefore there are no business entities and hence M
  can simply leave no information about himself."

It doesn't matter whether it is non-commercial/free or not. Copyright always applies if something is being distributed. M only wouldn't be guilty if he wouldn't have distributed it at all (e.g. wouldn't have uploaded that to that public server), as long as the mod is kept in private it doesn't matter.
It's unprobable that P could get in trouble because he wasn't the one distributing that mod, whether he profited from it or not doesn't matter. In the case of P = M it first would have to be proved that P = M or at least a relation between P and M.
It would have been a different point if the mod was on P's server, because in that case he would have to make sure that CC isn't uploaded there and/or that it isn't he is not responsible for the content.
Answer: Note: This answer is in response to an earlier version of the question, which asked both whether P is liable in general and whether if P = M allows P to profit from copyright infringement"
As I understand it, the primary question is "If a modder breaks the law, is the creator of the base game liable?" The answer to this is generally no. The other answers have covered this pretty well.
However, you make some assumptions (If M, how? It's easy for him to avoid any kind of punishment as "the mod" is free and therefore there are no business entities and hence M can simply leave no information about himself.). I would disagree with these assumptions. As recent events with file-sharers have found out, if you are on the internet, you are not truly anonymous. Your ISP can and will be subpoena'ed for your information. Regardless of whether the modder is anonymous or not, it is undeniable that he is infringing.
This leads us to the final point "In the case when P = M, and we've established that the modder is guilty(but hard to find), does this mean that P can make money without being caught?" You are basically asking, "If P breaks the law, but do not get caught, does that mean he can make a profit off copyright infringement?"
This is a nonquestion. Of course, if you break the law and don't get caught, you can make a profit. This is how crime works. But let us consider what happens when you do get caught (I am not a lawyer, and my answer is based on information such as http://itlaw.wikia.com/wiki/Copyright_infringement#Willful_infringement)

Copyright infringement becomes a felony when it is both for-profit and willful. If M is caught (via subpoenas, investigators, etc.), it becomes obvious that M = P, and the legal argument that he was infringing willfully becomes that much more convincing.

Willful Infringement carries not only a higher limit on the statutory damages you will be sued for (and if you are making a lot of profit, do not expect that the suit will be just for statutory damages) it also raises the level of felony the infringement constitutes.
Answer: Valve does this kind of thing very well.  Like I believe they do, you can basically state, "If someone modifies our game, that is outside our control and therefor not our responsibility"
Comment: I disagree. An EULA is a contract between the user (Valve) and the licensee, the customer. In general, two parties cannot in agreement create obligations for a third party. The EULA has legal consequences for neither the modder nor the author whose copyright that modder infringes.
Comment: A bit pedantic (I think people knew what I meant), but I did remove the EULA reference.
Answer: Hypothetically, if you're both P and M, then this is copyright infringement. If the P side of you makes tons of money with this, the copyright holder might get angry about it. In this case I hope that you were really anonymous while publishing this mod. IANAL, but the simple fact that you're the only one making a profit might be charged against you.
Conlusion: this is a dangerous move. Forewarned is forearmed.
Comment: just a little note: the question is entirely hypothetical..


What does this chord symbol mean?

Question:
I came across this while studying harmony and couldn't figure out what its trying to say. It comes from Bach's Partita in A Minor for Flute, BWV 1013, measure 2.
Answer: Harmonically, this is a V chord in first inversion, with the chordal 9 added.
Ordinarily, this would be written V9 or, more accurately, Vo7,6, or, even more accurately, V6, because 9 chords really weren't a thing in Bach's time.
This chord symbol seems to be mixing notations by indicating it's a V9 chord and that it contains a diminished interval above the lowest pitch.
Of course, it could just be a mistake in the source.
Comment: Interesting, I'd never seen this circle notation used in that way before. So it indicates a diminished interval above the bass, hence a bass of G-sharp with the F above it? To be devil's advocate, could it also mean a potential B in the bass, since the F above that B would also be diminished? I know that opens the can of worms of second-inversion chords, but still...
Comment: @Richardissteppingdown Yeah, neither the circle nor the 9 really make any sense. I can't help but wonder if it's just an error, or maybe some system unique to the source.
Answer: @aaron is almost there.
It is the bass figure indicating (from the bottom note) fully diminished 7th (°) D (V = fifth note) chord (D°) so inverted with the tonic 9th as the bass note (E).
Typically we now call this #vii° in functional harmony instead of the diminished dominant figuration.

The Bach original inidicates there is no accompaniment, fyi.
Is it wrong: probably as the opening is static a-minor without actually going back and forth V-I, a harmonic feature this figure highlights.  The motion is smoother and normally this beat is taken as a substitute-predominant seventh and not part of the cadential dominant, thus extending the cadence length from two beats to two measures.

What does this chord symbol mean?


Which way does the Earth tilt?

Question: On which side does the earth tilt to? I have tried figuring it out myself but I can't seem to find the answer.
Comment: Your question is a bit vague. The North Pole points towards the North star, Polaris. The plane of the equator is (currently) tilted by 23°26′12.0″ relative to the Earth's orbital plane. There's a diagram in my answer [here](https://astronomy.stackexchange.com/a/35118/16685) which may be helpful.
Comment: If we call the ecliptic the "floor" then Earth's north pole tilts towards Orion. See  https://commons.wikimedia.org/wiki/File:AxialTiltObliquity.png or http://slittlefair.staff.shef.ac.uk/teaching/phy115/session2/sunMotion/ecliptic/ecliptic.html **voting to keep open** in this case because I think the OP needs some assistance first, before they can refactor the question sufficiently In this case an answer would do the job better than a bunch of comments, and **closing blocks answers**
Comment: Hi @Jackson, would an answer like "towards the constellation of Orion" make sense to you? Is that the kind of answer you are looking for? Others are having trouble with your question because right now they are probably thinking "Tilted with respect to what?" There are several coordinate systems but none are absolute, so we always need a reference system. Can you think about "Tilted with respect to what?" a bit? Thanks and Welcome to Stack Exchange!
Answer: The equator (Earth's rotation plane) and ecliptic (Earth's solar orbit plane) cross at the vernal equinox ♈, the zero point of both right ascension and ecliptic longitude.
Relative to the ecliptic, the Earth's north pole is tilted toward ecliptic longitude λ=90°, where the Sun appears at the June solstice.
Constellations in that general direction are highest around midnight in December and January: Auriga, Orion, etc.
The Earth's south pole is tilted toward ecliptic longitude λ=270°, where the Sun appears at the December solstice.
Constellations in that part of the sky are highest around midnight in June and July: Sagittarius, Ophiuchus, Hercules, etc.
Even though axial precession changes the celestial poles' orientation relative to the stars in a cycle of about 26,000 years, their ecliptic longitudes with respect to the equinox of date are constant.
The north and south ecliptic poles precess by much smaller amounts and have constant right ascensions α=18h and α=6h respectively, relative to the equinox of date.
Stellarium might help you visualize all this.
In the View window, Markings tab, you can ask it to plot the equator, ecliptic, and celestial and ecliptic poles of date.
Comment: `+1` I wish I could find a *photograph* of an old transparent celestial sphere globe with the ecliptic horizontal and a small earth at the center, tilted. I further wish that it were high-enough resolution that one could see Earth tilting towards Orion. I wish for world peace and for all my SE reputation to change 1:1 for bitcoin, but that's beside the point.

Which way does the Earth tilt?


Calculating the input current that can be taken by an Operational amplifier

Question: Is there a way to  calculate the input current that can be taken by an Operational amplifier, if we know its input Bias and input offset current?
Answer: Offset current is just the difference between bias currents on the two inputs and this can be regarded as fairly constant. For non inverting configurations, input current is bias current plus a dc current proportional to input voltage due to non infinite input resistance and, for higher frequencies, there is a noticeable ac current due to non infinitesimal input capacitance usually a few pico farads.
There are also noise currents if you want to consider those.
Answer: For calculation of these currents (caused by input signals) you need some inputs - either derived from data sheets (with very large tolerances) or from measurements. The latter can be done as follows: After determining the offset voltages and bias currents you can detect the additional output voltage which is created by the current-resistor product at the opamp`s input - of course, with resistive feedback.
Two alternatives: (a) Large resistor in series with the non-inv. input and low-resistive feedback (with gain); (b) Large resistor between the common node of the feedback resistors and the inv. opamp input.


Chitin+Hydrocarbons=Skeleton?

Question: I was wondering, could the use of chitin (assuming it's composed the same as in insect exoskeletons) combined with hydrocarbons feasibly make a working endoskeleton? I know that chitin could be used as a more flexible, power-hungry bone alternative, and in the world of Snaiad, the organisms had an endoskeleton using Hydrocarbons resembling extremely hard wood. I would like to know if a combination of the two could form a working endoskeleton? What I mean by this is, a skeleton made of Chitin reinforced by Hydrocarbons throughout in specific places.
Comment: do you mean replacing bones with chitin bones and muscles? And what do you mean by hydrocarbons, which particular type of thousands for them?
Comment: @MolbOrg Yes, the chitin would be replacing the skeleton of the organism. As for the hydrocarbons, I assume they would be ones found in wood.
Answer: Any relatively hard material could be used in a skeleton.  There will be some advantages and disadvantages depending on what is used, but there is a reason why bone tends to be used for endoskeletons.
Both chitin and wood are more pliable and regenerate better than bone, which is good for an external covering since external coverings are more prone to getting worn away, scratched or damaged.  Bone is not expected to be damaged in an animal's normal life, so more energy could be put into making it strong and supportive.
So a chitin/hydrocarbon blend could be used as a working skeleton.  It probably wouldn't be quite as effective as a bone one, but evolution doesn't have to yield perfect results, just functional ones.

Chitin+Hydrocarbons=Skeleton?


Tricky positive diophantine equation

Question: Find all positive integer solutions to the equation $x^3=y^5+100$.
Notice that $7^3=3^5+100$ is a solution, but I can't find any more.
Thanks for your help!
P.S. I don't know any advanced number theory, just basic olympiad stuff.
Comment: Assume $g=gcd(x, y)$, we can put $x=gt$ and $y=gs$ and get $g^3|100$. The only solution is $g=1$, so $(x, y)=1$ or $x$ and $y$ are co-prime.
Comment: Trivially, $x \equiv y \space (mod 2)$, i.e. $x$ and $y$ are both even or both odd.
Comment: Moreover, x and y are both odd, since if they even putting $x=2a$ and $y=2b$ gives $2a^3=8b^5+25$, which is impossible.
Comment: Brute force to 10,000 failed. Can someone with better computer and computer skills proceed?
Comment: Where have you found this problem?
Comment: The related problem is: https://math.stackexchange.com/questions/4575889/the-diophantine-equation-x324-y5. It asks for all the integral solutions of $x^3+24=y^5$.
Answer: Thanks to Edward we are morally certain that $(x,y) = (7,3)$
is the only solution.  [I extended the search to all odd $y<10^9$ using
the gp command
<code>forstep(y=1,10^9,2, if(ispower(y^5+100,3),print(y)))
</code>
which took about 4 minutes and output only $3$ as expected.]
Also, as a special case of

Siegel's theorem on integral points we know that
there are only finitely many integer solutions.
The original proof is "ineffective", i.e. doesn't yield
an algorithm that can be guaranteed to find all solutions;
more recent work sometimes provides such algorithms, and even ones
that can be carried out in practice, but they're not easy, being
clearly "advanced number theory" as opposed to "basic olympiad stuff".
The existence of one nontrivial solution $(7,3)$ makes it quite unlikely
that an elementary proof is possible; for starters it will never be enough
to just use congruences modulo a few small numbers, because once $(7,3)$
works there will always be infinitely many more surviving candidates.
One thing that makes this just a bit more tractable,
and effectively solvable at least in principle, is that $100$
happens to be a square, and also cube-free, so any solution yields
a pair of solutions $(\pm 10,-x,y)$ to the Diophantine equation
$X^2+Y^3+Z^5=0$ in relatively prime integers.  This lets us build on
previous work in the paper

J. Edwards: A Complete Solution to $X^2 + Y^3 + Z^5 = 0$,
  Journal f. d. reine und angew. Math. (Crelle's Journal)
  571 (2004), 213-236.

which completely solves this equation under the relatively-prime condition.
Edwards finds $27$ identities $X_i(r,s)^2 + Y_i(r,s)^3 + Z_i(r,s)^5 = 0$,
each with $X_i,Y_i,Z_i$ homogeneous polynomials of degree $30$, $20$, $12$
respectively, and shows that every solution of $X^2+Y^3+Z^5=0$
in relatively prime integers is obtained from one of these solutions
by substituting some integers for $r$ and $s$.  Thus any solution with
$X = \pm 10$ is a solution of one of $27$ pairs of
Thue equations
$X_i(r,s) = \pm 10$.  This count is probably exaggerated because
some $i$ might be ruled out by congruence conditions, but at least
one must be possible, again because we know the solution $(X,Y,Z)=(10,-7,3)$.
Unless we're quite lucky, we won't be able to solve all these equations
by elementary means, and since each of the polynomials $X_i(r,s)$
has degree $30$ the more advanced techniques (try $p$-adic methods first,
and the Baker bounds where the $p$-adic analysis does not reach a
complete solutions) could take quite some work to complete.
[added later] The sixth of J.Edwards' identities has
$Z = \sum_{j=0}^{12} a_j r^j s^{12-j}$,
where the coefficients $a_0,\ldots,a_{12}$ are
$$
3,12,-132,0,-1980,-3168,3168,12672,-39600,-10560,-61248,-26112,27072,
$$
and then (as with all these identities) $Y$ is the scaled Hessian
$(Z_{rr} Z_{ss} - Z_{rs}^2)/132^2$ and $X$ is the scaled Jacobian
$(Y_r Z_s - Y_s Z_r) / 240$.  Taking $(r,s)=(0,1)$ recovers the solution
$10^2 - 7^3 + 3^5 = 0$.  So we must at least prove that there are
no other solutions of the degree-$30$ Thue equation $Z_6(r,s) = 10$.
Since $Z_6$ is irreducible and is not positive-definite (the polynomial
$Z_6(1,s)$ has four real roots), even this part of the problem
seems unlikely to have an elementary solution.
Comment: $X^2 + Y^3 + Z^5 = 0$ faintly resembles the relation satisfied by the icosahedral invariant. Can there be any serious connections? Like, for example, I suspect that thing to be isomorphic to some quotient of $X(5)$, so integral points on this curve might be equivalent to finding integral point on some modular equation?
Comment: Yes, there's actually quite a close connection; in each of Edwards'
parametrizations you can make a linear change of variables in $(r:s)$
so that each of $X,Y,Z$ has roots with the same icosahedral symmetry
($12$ vertices, $20$ face centers, and $30$ edge midpoints of the same
regular icosahedron inscribed in the Riemann sphere).  The list of
parametrizations could indeed have been obtained via twists of the
modular curve $X(5)$, though that's not how Edwards did it.
Comment: Ah, nice, thanks professor. Do you know of any references that use the connection with that modular curve to derive the list of parametrizations?
Comment: You're welcome.  No, I don't have a reference for an $X(5)$ approach
to this question, at least not yet :-)  (I've carried out one part of
this analysis already but haven't written it up yet.)
Comment: OK. Will be looking forward for the write-up.
Answer: gcd approach: $x$ and $y$ are coprime, and both not divisible by 2, nor by 5.
Proof. Assume $g$ is the greatest common divisor of $x$ and $y$. we can put $x=gt$ and $y=gs$ to get:
$$\color{red} {g^3t^3}=\color{green}{g^5s^5}+100$$
The expressions $\color{red}{g^3t^3}$ and $\color{green}{g^5s^5}$ are both divisible by $g^3$, so also $g^3|100$. The only cubic number dividing 100 is 1. So $\gcd(x,y)=1$.
Now, let $g'$ be the greatest common divisor of $x$ and 100. From $g'|x$ and $g'|100$ we can infer $g'|x^3-100=y^5$ so $g'|y^5$ and $g'|y$. But if $g'$ divides both $x$ and $y$, he must divide its gcd, hence $g'=1$ ($\gcd(100,x)=1$, or 100 and $x$ are coprime). In similar way we can show that 100 and $y$ are coprime, so $x$ and $y$ are both not divisible by neither 2 nor 5.
In this, $\gcd(a, b)$ means Greatest Common Divisor, the largest integer that divides both $a$ and $b$. If $\gcd(a,b)=1$, we say $a$ and $b$ are coprime.

Euler's and Fermat's theorems approach:
From Euler's theorem we can get some modular equation (thank to Edward for the results mod primes):

mod 2: As I proved above, both $x$ and $y$ are $\equiv 1$
$x^3\equiv x\equiv y+1 \pmod 3$
$x^3\equiv x\equiv y \pmod 4$
$x^3\equiv y \pmod 5$
$x^3\equiv x\equiv y+4 \pmod 6$
$x^3\equiv y+4 \pmod 8$
$x^3\equiv y \pmod{10}$
$x^3\equiv y+4 \pmod{12}$

By the Chinese remainder theorem, we can infer some more modular equations from that list. From 2., 4. and 6. we know that:

$x^3=y+4+8k_1$
$x^3=y+5k_2$
$x^3=y+1+3k_3$

This match into one formula: $x^3 = y+n$ where n is equal to 4 mod 8, 0 mod 5 and 1 mod 3. This has a unique solution mod 120:
$$x^3\equiv y+100 \pmod{120}$$
Similar way brings the formula for $x$:
$$x\equiv y+4 \pmod{12}$$

Brute force approach:
I run this command in python (since I don't have Matmatica nor Matlab):
<code>[(x,y) for (x,y) in itertools.product(xrange(10), xrange(10)) if gcd(x,100)==1 and gcd(y,100)==1 and (x**3)%10==(y**5)%10]
</code>
And got that the only solutions modulo 10 are: $(1, 1), (3, 7), (7, 3), (9, 9)$
Similar command brought me the only 40 solutions modulo 100:
<code>[(x,y) for (x,y) in itertools.product(xrange(100), xrange(100)) if gcd(x,100)==1 and gcd(y,100)==1 and (x**3)%100==(y**5)%100]
</code>
$[(1, 1), (1, 21), (1, 41), (1, 61), (1, 81), (7, 3), (7, 23), (7, 43), (7, 63), (7, 83), (43, 7), (43, 27), (43, 47), (43, 67), (43, 87), (49, 9), (49, 29), (49, 49), (49, 69), (49, 89), (51, 11), (51, 31), (51, 51), (51, 71), (51, 91), (57, 13), (57, 33), (57, 53), (57, 73), (57, 93), (93, 17), (93, 37), (93, 57), (93, 77), (93, 97), (99, 19), (99, 39), (99, 59), (99, 79), (99, 99)]$

EDIT: Divisors of x and y with brute-force:if $q|y$, then $q|y^5=x^3-100$ and $x^3\equiv100\pmod q$. Similar way shows that if $p|x$ then $y^5\equiv -100 \pmod p$. So I run some Python command to detect what $q$ and $p$ cannot be:
<code>>>> qs=[q for q in xrange(1, 1000) if not [x for x in xrange(q) if x**3%q==100%q]]
>>> qs=[q for q in qs if not [x for x in qs if x!=q and q%x==0]]
</code>
Gave me that $y$ is not divisible by any number of:
<code>[7, 8, 13, 19, 31, 43, 61, 67, 97, 109, 125, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 277, 283, 307, 313, 337, 367, 373, 379, 397, 409, 433, 439, 487, 499, 523, 541, 571, 577, 601, 619, 631, 709, 727, 757, 769, 787, 811, 823, 853, 877, 883, 919, 937, 991]
</code>
By similar commands I got:
<code>>>> ps=[p for p in xrange(1, 1000) if not [y for y in xrange(q) if y**5%p==(-100)%p]]
>>> ps=[p for p in ps if not [x for x in ps if x!=p and p%x==0]]
</code>
So $x$ is not divisible by any of:
<code>[8, 31, 41, 61, 71, 125, 131, 151, 181, 191, 211, 241, 271, 311, 331, 401, 421, 431, 461, 491, 541, 571, 601, 631, 661, 691, 701, 751, 761, 811, 821, 881, 911, 941, 971, 991]
</code>
I asked for a proper way to determine possible values of $p$ and $q$. I'll update if found better way.
As in this answer, $x^k\equiv n \pmod p$ iff $n^{\frac{p-1}{k_p}}\equiv 1 \pmod p$, where $k_p = \gcd(k,p-1)$.
In our case, we have $x^3\equiv 100 \pmod q$ iff $100^{\frac{q-1}{k_q}}\equiv 1 \pmod q$. If $k_q=(q-1, 3)=1$, the equation becomes $100^{q-1}\equiv 1 \pmod q$, which is true for every prime $q$ by Fermat's little theorem.
So suppose we have $k_q=(q-1, 3)=3$, or $3|q-1$. Now the equation becomes $100^{\frac{q-1}3}\equiv 1 \pmod q$ or $ord_q(100)|\frac{q-1}3$. Closer form seem impossible.
Answer: Interesting question.  Here's a complete (EDIT: and wrong!) answer that shows that $(x=7, y=3)$ is the only solution.
First, I cheated and wrote a small computer program that evaluated all integers up to one million.  Your solution $(x=7, y=3)$ was the only one found.
Second, we can observe that $x^3 - y^5 = 100$ so it's clear that $x^3 > 100$ and therefore $x \ge 5$.
Third, since $x^3 - y^5 > 0$, it's clear that $x^3 > y^5$ so $x > y$.
Now we  get into a bit of number theory.  Fermat's Little Theorum says that $a^p \equiv a\:\pmod p$ so we can consider the equation $\mod 3$ as follows:
$$\begin{eqnarray}
x^3 &\equiv& y^5 + 100 \pmod{3} \\
x &\equiv&y^5+100 \pmod{3}\\
x &\equiv&y^5+1 \pmod{3}\\
x &\equiv&y^{5 \bmod \phi(3)} + 1 \pmod{3}\\
x &\equiv&y^{5 \bmod 2} +1 \pmod{3}\\
x &\equiv&y + 1 \pmod{3}
\end{eqnarray}$$
In this, the $\phi(n)$ function is Euler's totient function which is equal to $n-1$ when $n$ is a prime number.  We can do a similar operation considering a rearranged version of the original equation $\mod 5$:
$$\begin{eqnarray}
x^3-100 &\equiv& y^5\pmod{5}\\
x^3-100 &\equiv& y\pmod{5}\\
x^3 &\equiv& y\pmod{5}
\end{eqnarray}$$
Summarizing, we can assert that $x\ge 5$, $x>y$, $x\equiv y+1\pmod{3}$ and $x^3 \equiv y\pmod{5}$ for solutions to this equation.
This should probably allow us to conclude that $(x=7, y=3)$ is the only solution, but at the moment, I lack the insight to bring us to that conclusion. Perhaps this will inspire someone else to fill in that blank.
Edit:
Inspiration has come at last!  As LeeNeverGup has pointed out, we know that $x$ and $y$ are both the same parity (both odd or both even) because the equation $\mod 2$ is $x \equiv y$. We can rule out both even by substituting $x=2a$ and $y=2b$ yielding $2^3a^3=2^5b^5+100 \implies 2a^3=8b^5+25 \implies 0\equiv 0+1 \pmod{2}$.  Since that's not true, both $x$ and $y$ must be odd numbers.
We also know from the $\mod 3$ equation above the $x\equiv y+1 \pmod{3}$. Put another way, $x-y =3n+1$ where $n \in \mathbb{N}$ (the set of natural numbers).  Since $x$ and $y$ are both odd numbers, we can substitute $x=2a+1$ and $y=2b+1$ (with $a,b \in \mathbb{N}$).  This gives us $2a+1-2b-1=3n+1$.  Examining this in $\mod 2$ yields $0+1-0-1\equiv n+1 \pmod{2}$ so clearly $n\equiv 1\pmod{2}$. Again we do a substitution to show this fact and we get $x-y=3(2c+1)+1; c\in\mathbb{N}$ or $x-y=6c+4$ and $y = x-6c-4$.
Substituting this into the cubic equation, we get $x^3-x+6c+4=5m$, so again looking at this equation $\mod 2$, we can see that $m \equiv 0\pmod{2}$.  Once more we substitute to reflect this and get $x^3-x+6c+4=10f; f \in\mathbb{N}$.
We know that $(x=7,y=3)$ is a solution and we know that $x\ge 5$ and that $x \equiv 1 \pmod{2}$ (that is, $x$ is odd). If there is another solution, it must be of the form $x=2d+9$ where $d\in\mathbb{N}$.  Given this, we can rewrite the cubic equation above:
$$\begin{eqnarray}
(2d+9)^3 -2d-9+6c+4&=&10f\\
(2d+9)^3 -2d+6c-5&=&10f\\
8d^3+108d^2+484d+6c+724&=&10f
\end{eqnarray}$$
If we solve for $d$ (omitting a lot of messy algebra) we get three solutions, two of which involve imaginary numbers, and therefore can be ruled out because $d \in \mathbb{N}$.  That leaves us with
$$d=\left(\frac{\sqrt{675f^2+\left(-810c-540\right)f+243c^2+324c+107}}{8\times 3^{\frac{3}{2}}} - \frac{-5f+3c+2}{8}\right)
 ^{\frac{1}{3}}+\\
{\frac{1}{12\left(\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{8\times 3^\frac{3}{2}}-\frac{-5
 f+3c+2}{8}\right)^\frac{1}{3}}} -\frac{9}{2}$$
Substituting $$g=\left(\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{8\times 3^\frac{3}{2}}-\frac{-5
 f+3c+2}{8}\right)^\frac{1}{3}$$
If we multiply both sides by 2, we get $$ 2d = 2g+\frac{1}{6g}-9$$.  Since $2d$ and $9$ are integers, $2g+\frac{1}{6g}$ also must be an integer.  So $\frac{12g+1}{6g} = \frac{2g + \frac{1}{6}}{g}$ must be an integer (specifically an odd integer) and so $g$ must be an odd multiple of $\frac{1}{6}$. That assertion is not correct!  Therefore the rest of the proof is also erroneous and should be ignored.  Thanks to @barto for pointing this out in a comment, and apologies to all for the error.
In other words $6g \in \mathbb{Z}$ and $6g \equiv 1\pmod{2}$. We can return to our definition of $g$ and substitute: $$6g=2h+1=6\left(\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{8\times 3^\frac{3}{2}}-\frac{-5
 f+3c+2}{8}\right)^\frac{1}{3}$$ where $h \in \mathbb{N}$.
$$2h+1=3\left(\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{3^\frac{3}{2}}+5
 f-3c-2\right)^\frac{1}{3}$$
Since $2h+1$ is an odd integer, $\left(\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{3^\frac{3}{2}}+5
 f-3c-2\right)^\frac{1}{3}$ must also be an odd integer.  So $\frac{\sqrt{675f^2+\left(-810c-
 540\right)f+243c^2+324c+107}}{3^\frac{3}{2}}+5
 f-3c-2$ must be an odd cubic integer
This implies that $\sqrt{\frac{675f^2-810cf-540f+243c^2+324c+107}{27}}$ is an integer, and so is $\frac{675f^2-810cf-540f+243c^2+324c+107}{27}$. So $675f^2-810cf-540f+243c^2+324c+107 \equiv 0 \pmod{27}$.  However, each of the factors except 107 is divisible by 27 so we get $26 \equiv 0 \pmod{27}$ which is a contradiction.
Therefore $(x=7, y=3)$ is the only solution.
Comment: Ah, I see you already mentioned this in a comment.  $2^3 m^3 = 2^5 n^5 + 100 \implies 2m^3 + 8n^5 + 25 \implies 0 \equiv 0 + 1 \pmod{2}$ which is a contradiction.
Comment: In you last solution there must be a mistake somewhere. (You didn't take in the fifth power, so it is very unlikely to be true.) You concluded that the equation $x^3-x+6c+4=10f$ has no solutions, but sure it does. Take for example $x=11$, $c=1$ and $f=133$. Unless I'm missing some important detail...
Comment: @barto: You're right and there's an error.  I asserted that because $2d = 2g+\frac{1}{6g}-9$, $g$ must be an odd integral multiple of $\frac{1}{6}$ but that's not correct. Solving for $g$ in terms of $d$ in that equation gives: $g = \pm \frac{\sqrt{3}\sqrt{12d^2+108d+239}-6d-27}{12}$, so the rest of the "proof" after that is faulty.
Comment: Start by modulo 2: $x$ and $y$ are both odd.
Comment: Also, using Euler's theorem, $x\equiv y \space (mod\space 4)$ and $x\equiv y + 4 \space (mod\space 6)$
Comment: If we start with $\mod 2$, it seems to me that we get $x\equiv y\pmod{2}$, but that only says that $x$ and $y$ are either both odd or both even.  We need to add additional information to conclude that they're both odd unless I'm missing something.
Comment: The option of both even lead to contrafiction after subtituating x by 2m and y by 2n
Comment: $2^3 m^3 = 2^5 n^5 + 100 \implies 0 \equiv 0 \pmod{2}$ doesn't seem to lead to contradiction.  Can you elucidate?
Comment: $2^3 m^3 = 2^5 n^5 + 100 \implies 0 \equiv 4 \pmod{8}$
Comment: @barto: Because $x^3\equiv y\pmod{5}$ we can equivalently say that $x^3-y=5m$ where $m\in\mathbb{N}$.  We don't have a fifth power because $a^5\equiv a\pmod{5}$.
Comment: @barto: I think you're right that something is wrong, but I don't see what, unless it's my derivation of the cube root.  Your $x=11 = 2d + 9 \implies d=1$.  Using my cube root equation gives $g=\left(\frac{181+\sqrt{359}}{8\times 3^\frac{3}{2}}+\frac{165}{2}\right)^\frac{1}{3}$ and $d=g+\frac{1}{12g}-\frac{9}{2}$.  Something's awry.
Answer: Again not a complete solution, just some new insights...
The fact that you already know a solution makes it interesting. It allows us to rewrite the equation as
$$x^3-7^3=y^5-3^5,$$
or
$$(x-7)(x^2+7x+49)=(y-3)(y^4+3y^3+9y^2+27y+81).$$
Note that every prime divisor of $\frac{x^3-7^3}{x-7}$ is either $3$ or congruent to $1$ modulo $3$. Similarly, every prime divisor of $\frac{y^5-3^5}{y-3}$ is either $5$ or congruent to $1$ modulo $5$. (This is all due to a more general result involving cyclotomic polynomials of primes.)
The factor that 'makes' the RHS large is $y^4+3y^3+9y^2+27y+81$. The fact that this can only have primes $\equiv0,1\pmod5$ suggests that there will not be many solutions as $x^3-7^3$ might have a lot of prime divisors that have to divide $y-3$. Finding a close upper bound on the primes dividing $x^3-7^3$ that are $\equiv0,1\pmod5$ possibly will exclude further solutions.
I'll start off considering an easy case.  If $x-7\mid\frac{y^5-3^5}{y-3}$ then either $x\equiv2$ or $x\equiv3\pmod5$. In any case $x^2+7x+49$ is not congruent to $0$ or $1$ modulo $5$, so it should have some divisor $>1$ in common with $y-3$. We want this divisor to be big in order to obtain a contradiction.
Perhaps finding an upper bound on $\gcd(x-7,y-3)$ might help. Certainly the $\gcd$ should divide $3x-7y$ but I can't really make anything interesting of this.

Tricky positive diophantine equation


Should I edit a question or ask a new one?

Question: I Want to know the difference between display managers and found this question, but it does not include SDDM. Should I edit that question to include SDDM or ask as a separate one?
Comment: I see you asked your question, it is a good idea to link it here so people know http://askubuntu.com/questions/860924/what-are-the-current-display-managers-and-how-to-choose-the-correct-one-what-ar
Answer: IMO you should ask a new question. That one is old and very specific.
You should write the question to include comparisons of ALL current display managers, then we would have a good and detailed up to date comparison.
Also, a lot has change, so the current answer is likely dated anyway.
As this would be a very useful question, I would be willing to give a bounty to anyone who wants to write a grate answer for us ;)
Answer: You shouldn't do either. Big-list type questions aren't well-suited to the Stack Exchange format. They're a pain to keep updated. New entries don't get anywhere near as many views as older ones, so quality control on them is more difficult.
Also, other big list questions (like What Application Indicators are available?) have this notice on them:

This question exists because it has historical significance, but it is not considered a good, on-topic question for this site, so please do not use it as evidence that you can ask similar questions here. While you are encouraged to help maintain its answers, please understand that "big list" questions are not generally allowed on Ask Ubuntu and will be closed per the help center.

Also, the question that you asked is also soliciting opinions:

Why should I choose one over others?

I'd VTC just for that, even if it wasn't too broad.
Comment: @Dan it was the case for the question OP [actually posted at first](http://askubuntu.com/posts/860924/revisions): "I want to know what is a display manager and differences between them. Why should I choose one over others?" That's a classic big-list question.
Comment: Thank you, then I will find it myself. Thank you
Comment: I simplified the question. now you can answer it.
Comment: @muru Although I agree with the big list argument. I don't agree about it being the case for this question.


Could you use the detonation of a Hydrogen bomb or atomic bomb to start a nuclear fusion reactor

Question: Would this lead to a reactor that produced more energy than you put in?
Comment: Possible duplicate of [What stops us from creating a nuclear fusion reactor as we already have the hydrogen bomb working on the same principle of fusion?](https://physics.stackexchange.com/q/330349/)
Comment: No I'm saying could we use the h bomb as the input energy for a nuclear fusion reactor
Comment: How do you stop the bomb from blowing up your reactor? How would you refuel it? But anyway, starting the fusion reaction isn't the hardest part. The problem is sustaining and controlling it.
Answer: This has actually been proposed! Project PACER at Los Alamos explored the possibility of detonating hydrogen bombs in underground cavities (e.g. in salt domes) to produce "geothermal" energy. Exploding a 2 kton device once every 20 minutes in a steel lined underground cavity using molten salt droplets to cushion the effects of the explosive would produce about 1000 MWe of power. This is a fusion reactor since most energy comes from the fusion reaction, started by a nuclear detonation.
Comment: The fact that someone *proposes* an idea does not make it in any way sensible.  How many active projects to *build* such a device are under way ?  A 2kT explosion *every twenty minutes* ?  Doesn't that sound just the tiniest bit crackpot to you ?
Comment: @StephenG - Peer reviewed publications. Check the calculations, they made sense to me. I would expect it to be very impractical, but the question was never if using bombs to power reactors was a *good* solution to energy production, just if it was possible.
Comment: No argument about the source you link to (good find, BTW), however peer reviewed, while a better gauge for quality material, does not make an idea sensible.  The OP asked about power production and that, to me, means practical.  Strictly speaking the OP asked if more energy is produced than you put in and, again strictly speaking, the answer to that is *no*.  It only makes sense as a question if you're considering the practicality of it all (IMO).
Comment: @StephenG - More energy DOES come out. Strictly speaking, the answer is absolutely *yes*. What are you talking about?
Comment: And as to "practical", well you only know that if you try. Some chunk of LLNL was working on this under Plowshares to the point they carried out actual UGX's to test it. I don't know why you would call this crackpot. A good portion of LLNL also worked on mirror fusion, it didn't work in the end either, are you saying Post is a crackpot too?
Answer: I think this confuses what is hard about getting fusion reactors to work.
I assume you are thinking of using a thermonuclear device to start a fusion reactor rather than to use one or more as a fusion reactor.  In other words I assume you are not thinking of harnessing the energy released from one or more such devices as in the various mad schemes people involved with nuclear weapons thought up to harness them for peaceful power generation.  That approach is addressed in another answer.
The problem with getting a fusion reactor to work is not getting fusion started (anyone can do that): it is keeping the stuff you are fusing stable and contained for long enough that you can get more power out than you put in.  There are various approaches to doing this of which we know of only one really successful one: that approach – gravitational confinement – unfortunately doesn't scale down to scales we can use directly.
Given that the problem is not starting fusion but containing it and keeping it stable, it seems unlikely that detonating a thermonuclear weapon inside your fusion reactor is going to help!


How to troubleshoot Application install during SCCM 2012 OS deployment?

Question: When installing an Application (as opposed to a Package/Program) during an OS deployment task sequence the only thing that seems to get written to the smsts.log if an install fails is that the application install failed, (normally with the seemingly unhelpful generic TS error code 0x80004005).
But in SCCM 2007 OSD, if you pressed F8 while the task sequence failed massage was onscreen, the smsts.log used to list the exact command line used to start an install, as well the return code and/or first few lines of any output from the installer as it exited.
It doesn't seem to do this in 2012. The closest equivalent to that in 2012 for Applications seems to be the AppEnforce.log, but none of the App*.log log files seem to exist during an OSD task sequence, they only appear later when task sequence has successfully completed (or if they do exist during the TS, I haven't managed to find them in the normal C:\_SMSTaskSequence\Logs or C:\Windows\CCM\Logs locations).
This is using SCCM 2012 SP1 installing a Windows 7 SP1 32-bit OS and related apps, if that makes any difference.
Answer: AppEnforce.log does exist during OSD, it depends what part of OSD you are running in when the error occurs. As you are installing applications I will assume you have already installed the client as part of the task sequence.
In this case your logs will be in %SystemRoot%\CCM\Logs, I have just forced a task sequence to fail installing an application and you can find it there. Failing that let the machine reboot after the failure and look in the normal location and you will find the logs in the normal place.
If the logs do not appear at all then this would suggest the application inforcement has not even started when the failure is thrown. In this case look for any errors in smsts.log for the step you are running. You will see an exit code, usually 24 (download failed) in the log file for the application install step.
Comment: Thanks a lot! I had VBSscript that did had no "On Error Resume Next" setup and cause the application to time out. By checking the Appenforce.log I got it fixed.


Rate of production of enzymes in human

Question: In the metabolic pathways, each of the reaction needs an enzyme(protein) in order to occur successfully. Is there any quantitative measure that can be associated to this enzyme? Like synthesis rate of enzyme, degradation rate, mRNA transcription rate, enzyme activity, amount of enzyme present per unit measure of protein in the cell and if there is rate of generation of the enzyme's product, the rate of production of this enzyme in the human body, etc? I wanted to know the list of features that can quantify enzyme activity in the cell.
Comment: Your question is very unclear. There are plenty of quantitative measures when looking at a particular protein - synthesis rate vs. degradation rate, mRNA transcription rate, enzyme activity, amount of enzyme present per unit measure of protein in the cell, rate of generation of the enzyme's product, and on and on and on. Please [edit] your question and explain exactly what you're doing and what kind of data you're looking for. This is definitely an [XY problem](http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem).
Comment: @MattDMo edited the question. Is it clearer now?
Comment: This is very broad. Measures of abundance and synthesis rate of enzymes (and other proteins) can be found in proteomics work such as the paper mentioned by @tsttst below. Measurement of enzyme activity is completely different, and is often studied using isotope-labeled metabolites in a technique known as metabolic flux analysis, see for example http://msb.embopress.org/content/2/1/62.abstract  Both are large, active research fields.
Answer:
Your question already answers a fundamental point: namely that there are several distinct quantitative measures that can be associated with enzymes (the list could also be extended, and you will discover more in biochemical textbooks...).
If experimentally or theoretically accessible, one additional important quantity is the formation of product over time divided by biomass ( e.g.: Jacob et Monod 1961)
Comment: I wanted to know all those points and how they are calculated...
Comment: Note that the more direct quantitative measures usually require enzyme-specific experimental assays (and measuring the activity within cells usually is even one step up - and something that has only been achieved for very few enzymes). If you want to look at quantitative measures, which can be measured well for most enzymes (and quantities that can be estimated reasonably well by models based on those measures), you could find them (for some cell lines and growth conditions) in Schwannäusser et al. 2011 http://www.nature.com/nature/journal/v473/n7347/full/nature10098.html

Rate of production of enzymes in human


Data moving from RDBMS to Hadoop, using SQOOP and FLUME

Question: I am in the process of learning Hadoop and stuck with few concepts on moving data from Relational database to Hadoop and vice versa.
I have transferred files from MySQL to HDFS using SQOOP import queries. The files I transferred were structured datasets and not any server log data. I recently read that we usually use flume for moving log files into Hadoop,
My question is:
1. Can we use SQOOP as well for moving log files?
2. If yes, which of SQOOP or FLUME is more preferred for log files and why?
Answer: 1) Sqoop can be used to transfer data between any rdbms and hdfs. To use scoop the data has to be structured usually specified by schema of database from where data is being imported or exported.Log files are not always structured,depending on source and type of log so sqoop is not used for moving log files.
2)Flume can collect, aggregate data from many different kinds of customizable data sources. It gives more flexibility in controlling what specific events to capture and use in user defined work flow before storing in say hdfs.
I hope it clarified difference between sqoop and flume.
Comment: Thanks for such an explanation!
Answer: SQOOP is designed to transfer data from RDMS to HDFS whereas FLUME is for moving large amounts of log data.
Both are different and specialized for different purposes.
Like
You can use SQOOP to import data via JDBC ( which you can not do in FLUME ),
and
You can use FLUME to say something like "I want to tail 200 lines of log file from this server".
Read more about FLUME here
http://flume.apache.org/
Answer: SQOOP not only transfers data from RDBMS but also from NOSql databases like MongoDB. You can directly transfer data to HDFS or Hive.
Transferring data to Hive you need not have to create table beforehand.. It takes the scheme from database itself.
Flume is used to fetch log data or streaming data


UIPopoverController with UINavigationController change anchor point?

Question: I have a custom UIPopoverController that has a view with UINavigationController in it. The UINavigationController is rigged so it allows dragging of the whole UIPopoverController around by touching the navigation bar. (draggable popover).
Once the UINavigationController pushes a new view, the size of the popover changes in an animated fashion, and location of the popover returns to the anchor point where it was first shown, before being dragged with my custom navigation bar. (also animated) Is there a way to change the anchor point of the UIPopoverController while it is presented, so that it does not animate back to it's initial anchor location after a push to the UINavigationController.
Answer: I'm unsure of what a 'draggable popover' is, but for your issues on the size of the popover and its anchor location changing, here are my views,
use the setPopoverContentSize: method of the UIPopover to set the size of the popover before pushing the new view within the navigation controller.
the presentPopoverFromRect:CGRect inView:UIView permittedtDirections:UIPopoverArrowDirection animated:BOOL to set the location where the popover's arrow points to and in which direction.


Bootrec /fixboot/mbr wont work

Question: Okay I decided to mess around with fedora 22 and dual installed it with my windows 8.1 laptop (has uefi) and for some reason something went wrong with it. After deleting Linux partition and resizing c:\ I go to fix mbr and boot with recovery, but here is the issue... It won't even work. It's still there as if I didn't do anything, asking me if I want to boot in fedora 22, fedora recovery or windows boot manager.
Answer: UEFI booting does not use Master or Partition Boot Records.
UEFI boot entries listed are from NVRAM.
UEFI firmware "setup" should present under "boot options" the functionality to list boot entries, to reorder boot entries, to create new boot entry, to delete boot entry.

Bootrec /fixboot/mbr wont work


Weird icon on my icon tray

Question: Does anyone know what this is? It started to show up after I installed Dropbox I think, and nothing is showing up when I tried to click it. It doesn't change the behavior of my desktop though and the Dropbox folder works fine too. I just want to know what this weird icon is. Thank you!
Comment: Possible duplicate of [Xubuntu Dropbox Icon Fail](http://askubuntu.com/questions/732816/xubuntu-dropbox-icon-fail)
Answer: That is dropbox doing it's recent failing thing. Run this:
dropbox stop && dbus-launch dropbox start
I had the same problem. It's more of a workaround, but it works. It helps to create a custom command to just run that real quick. Create an alias in /home/<user>/.bashrc:
alias db='dropbox stop && dbus dropbox-start`
Or, I wrote a script with that in it, then did
sudo install db.sh /path/to/db.sh/commandname
where commandname runs db.sh from the terminal.
Comment: Hey this was on another question, looks like a permanent workaround to hide this behavior: **The command line command works great but only works for that session. To make the fix permanent, one should, after running this command, go to the Dropbox indicator > Preferences... > uncheck Start Dropbox on system startup. Then add dbus-launch dropbox start to Session and Startup > Application Autostart. This works perfectly for me on Xubuntu 16.04 LTS. – Ads20000**
Comment: Thank you!  That was driving me crazy as mine was doing that too!  =)
Comment: Sure thing. Yeah, I don't know what exactly is going on with it yet, but there are no options shown on right click, right? Just a blank menu... but this works for now, for me, and good, for you too! Cheers!
Comment: That is exactly what mine was doing too.  No options when right clicking at all.  But I needed my dropbox for my Plex Media Server database backup I run, and it wasn't backing up with it failing.  So, this works.  =)
Comment: I guess that makes this question a duplicate sort of, but here's that thread: http://askubuntu.com/questions/732967/dropbox-icon-is-not-working-xubuntu-14-04-lts-64

Weird icon on my icon tray


Can structural equation models be used to derive clinical formulae?

Question: Structural equation models (sem) are used to model latent variables. Renal function is a latent variable measured by serum creatinine levels (with measurement errors) expressed by many different clinical formulae derived from linear regression models. Inulin clearance is a gold standard for renal function. However, hundreds of articles had studied the relative "accuracies" of these formulae as compared to surrogate "gold standards". Can sem be used to derive a clinical formula from serum creatinine and inulin clearance for the estimation of renal function?
Comment: I consider myself to be reasonably good in structural equation modeling, but I cannot really see how to answer this question. What is a clinical formula? There are things absolutely trivial to you as a clinician or a biostatistician, but this is essentially a different language, and you need to provide a better translation.
Comment: For example, glomerular filtration rate (GFR) is a measurement of renal function. A clinical formula for GFR is: GFR (mL/min/1.73 m$^2$) = 175 x (serum creatinine level)$^{-1.154}$ x (age)$^{-0.203}$ x (0.742 if female). This formula was derived from the anti-log of a linear regression of log-transformed X and Y variables. Because both inulin clearance (the gold standard of renal function) and serum creatinine levels are measurements with errors and inulin clearance is not usually measured clinically. Can SEM (which accounts for measurement errors) be used to derive a clinical formula for GFR?
Comment: OK, that clarifies things a bit: my understanding is that in the above formula, everything has been measured somehow with an equipment that provides a numeric reading (GFR, serum creatinine), or is an observable patient characteristic (age, gender). For the model that you are contemplating, please describe what are the variables that you have, how you obtain them, and what you think the causal relations between the variables are (which function causes which rate to go up or down).
Comment: Inulin clearance is theoretically a "gold standard" for GFR if it is not measured with error, although it is measured with error in practice and is usually not measured clinically. Therefore, true GFR is not known and is a "latent variable". GFR is higher in men, blacks, youth, obesity or tall person. GFR is lower in diseases (e.g. diabetes mellitus and hypertension) and drugs (e.g analgesics), etc. Lower GFR non-linearly increases serum creatinine level. All clinical formulae for GFR uses gender, age and creatinine, although these variables account only for 60-80% of variance.
Answer: Yes SEM can do that. You enter the measured and latent variables into the model, specify their relationships, and then you will get quite a lot of output. This output will include a structural equation (looks like a regression equation, with coefficients, standard errors, etc) and an R^2 result. SEM also allows you to specify other relationships such as allowing covariance between serum creatinine and inulin.
Answer: If you can build a regression model for something, it means to me that this is a measurable quantity. While a linear regression model is a special case of the general SEM, the greater strength of SEM is, arguably, being able to accommodate latent variables and measurement error in predictors. If you want to build a structural equation model for the latent variable BLAH (renal function), you need to have a study where several different ways to measure it have been undertaken (serum creatinine levels, inulin clearance). If, in turn, these variables are obtained from "clinical formulae" (another regression model, as far as I understand), that messes up the model quite a bit, and you need to figure out exactly which of the variables affect the measurement process, and which one affects the underlying latent variable (without knowing anything about the biochemisty of whatever it is that you are interested in, I would dare suggesting that age and gender affect the latent variable, rather than the measurement process).  Ideally, you would want your equations linking the latent variables with the measured variables to be linear in parameters, so you would need to apply the typical transformations, such as logs. The standard language of structural equation models are path diagrams; mainstream statisticians tend to stare at them with little understanding of what's going on, but social scientists have found them to be very handy in explaining the relations between variables. I think this paper explains it quite well, although I don't know how closely it follows the biomedical language (it is written by psychometricians): http://www.citeulike.org/user/ctacmo/article/2663951.
Comment: Don't ever mention PLS in my presence if you want me to respond again :). What you have then is a MIMIC (multiple indicators, multiple causes) model, and that's what you should look up. You should be able to set up your this SEM model in any decent SEM software. To identify your latent variable, you need to have at least two "reflective" indicators of it (e.g., measured GFR and creatinine level). While psychology uses "formative" and "reflective" indicators terminology which I find awkward, sociologists use "causal" and "effect" indicators terminology, so you might want to look these up, too.
Comment: Gender, age, race, body weight, body height, diseases and drugs are formative indicators whereas serum creatinine level is a reflective indicator for GFR. Should partial least squares (PLS) be used instead of SEM in this hybrid case of both formative and reflective indicators?