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# RAID - Information and Tutorial

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I wrote this article for my site, which i decided to do since recently i upped my raid setup from a raid 0 to a raid 10 and figured i could show the concepts and ideas behind the process. So, hopefully i have achieved my goal at making the concepts clear and concise. Enjoy!

# Setting Up Raid

Before we go any further, i must give the standard disclaimer. I can not be held liable for any event that is derived from the use of this article. Nor any damages or loss of data. If you're unsure of anything below, do not attempt it, or thoroughly backup your data before hand.

## What Is Raid?

RAID stands for, Redundant Array of Independent/Inexpensive Disks. Here you're basically taking two or more "physical" disks and forming one or more "logical" disks. The intent is to either increase performance by utilizing two or more drives at once, or to provide some level of redundancy for the protection of data. However, as we will discuss below, no matter what the intention is, some RAID levels offer both performance and redundancy.

## Terms

Strip Width- Here strip width defines the amount of parallel commands that can be executed at once to a stripped raid. Basically, if you have 2 drives, your strip width is 2. That is to say that you can send 2 requests that will be carried out immediately by two different drives at once. The more drives you add the larger your strip width can be, and the more data you can read/write in parallel. In this way you can see why a stripped array of 4 30 gig drives will have better transfer performance than 2 60 gig drives Strip Size- Strip size defines the minimum size a file needs to be to be stripped across drives. If you write a 4 KiB (binary kilobyte) file to a stripped array, with an array size of 32 KiB, that file will only be written to the first drive and will only take up 4 KiB. However, If you have 4 drives in a stripped array with a 32 KiB strip size, then writing a 128 KiB file to the array will result in the file being evenly split into 32 KiB chunks across all drives in the array. Choosing the strip size is an important factor in performance of a stripped array, and the choice varies drastically depending on the use of the array. If you choose a smaller strip size you're going to split files into more chunks/blocks/strips across more drives. This will have the effect of increasing your throughput/transfer performance, but positioning performance (access timing) will decrease as all of the drives will be busy handling a data request already (since the file must be accessed by all drives). If you increase the strip size you're decreasing the amount of drives files will be split across, but this will increase positioning performance as most controllers will then allow another drive that is not in use to fulfill another request while the original request is being completed. Generally there is no rule of thumb to setting the strip size as it depends on many different factors, some of which include the arrays use and the drives being used. Parity- Parity data is error correction data that has been calculated using the actual data being stored. An example of this is a single parity bit added to serial connection transmissions, where 7 bits of actual data is transmitted, with the last bit being the parity bit, then a stop bit/s. Here The parity bit is calculated before the data is sent, after the data is transmitted and received the parity bit is recalculated, if it matches the parity bit transmitted then the data is accepted, if not then the data is resent. The same type of parity information is calculated, but just on a larger scale in raid setups (except for raid 2, which does bit level stripping and bit level error correction, however raid 2 is no longer used). Raid Controller- A raid controller is a controller that handles all raid operation and protocols. There are three types of raid:
Software- All raid operations including logic and calculation is done purely in software. Usually an operating system has support for the less intensive raid levels (1,0). This will generally provide "ok" performance at the cost of a larger cpu utilization due to the entire process being done in software. And as you can probably guess, true software raid does not use any hardware.
Hardware- All raid operations including logic and protocols are handled by a dedicated processor and bios on the hardware controller. The controller may also have an onboard buffer/cache, and maybe even a battery backup for it to keep data from corrupting when utilizing write-back cache mode (in more expensive controllers). A hardware raid controller has to interface with the computer in some way. Here you'll see raid controllers for pci, pci-x, and most recently pci-e, though this list is not exhaustive and there may be raid controllers for other, more obscure interfaces. It is generally a good idea to research the actual throughput of the interfaces before choosing a raid controller.
Pseudo Hardware Controller- Here you have a hardware controller that has its own buffer and bios. However it may be lacking a dedicated processor. This means that all of the protocol and logic of the raid controller is contained on the controller, but the controller uses your processor (cpu) in absence of its own. This generally will lead to a smaller cpu utilization than just software raid, and performance is better than plain software raid, but it will be larger than complete dedicated hardware raid.
Keep in mind that the cheaper you go the cheaper the controller itself will be. While this may be obvious, cheaper controllers offer less advanced performance improving features, and may reach their bottleneck sooner. Depending on how the controller are designed you will generally start to see the controller, or its interface bottleneck. Cheaper controllers tend to do this sooner.

## Drive Interface Choice

For a drive to connect to the computer it needs to use a certain interface, most common choices being SCSI, SAS (basically SCSI over a serial Connection), SATA, and IDE. SCSI/SAS interfaced drives tend to be a bit more expensive, but they generally offer the best performance due to maturity of the interface protocols and large use within the server and mass storage industry. Next we have IDE, whose creation was for cheapness. IDE came into the computer industry as a need for a cheap, economical interface was required to lower the cost of drives and their controllers, and IDE filled that void. Because it was cheaper than other options its popularity grew, even though it was never meant to be an interface for performance. The original IDE interface design had the processor being used for the interface, this mode was called PIO, now there are DMA modes that allow for better performance. But there is still the lingering design problem of single device access per channel. Basically if you have two devices on a single IDE ribbon cable, only one of those devices can communicate with the computer at one time. This makes IDE not suited for raid whatsoever, as benefits of raid come from simutanious access of multiple devices at once. So, i would recommend using only SCSI, its cousin SAS, or SATA for raid. SATA is the successor to IDE, and is a serial based interface. SATA does offer simutanious access to multiple devices and offers better throughput than IDE. Along with this SATA is generally cheaper than SCSI but offers only slightly lesser performance. Lession... Use only SCSI, SAS, or SATA for raid setups.

## RAID Levels

There are many different RAID Levels, including some proprietary ones, that we could discuss. However, for this article i will concentrate on the most common implementations.

### Raid 0

Here you'll find a raid level that actually is not redundant at all. Raid 0 is the simplest of all the levels. Here you have two or more physical drives being combined into one or more logical drives with the recorded data being "stripped" between the drives. The intent with this case is absolute performance with no regard to data safety and redundancy.
With raid 0 you will take two more more drives combine them, then data will be stripped, based on the chosen strip size, between them. This has the effect of spreading out the workload of both writing and reading data from the logical drive, which is composed of multiple physical drives. However, as stated above, how the stripping affects performance, I.E. access times or throughput, is dependent on the strip size used. The stripping in raid 0 will help throughput with smaller strip sizes, and help access times with smart controller if a larger strip size is used.
Raid 0 is generally not used if you have any data you wish to keep safe from data loss. This is because in raid 0, since all data is simply stripped and stripped across multiple drive, if a single drive dies in the array, all data will be lost. Data can not be retrieved if it is in pieces across two or more drives, so if a single drive dies in the array, you're effectively deleting a chunk out of each and every file that was stripped, in a way corrupting the files. However, out of all the raid level's, raid 0 has the best overall read and write performance combination.

### Raid 1

Raid 1 is another simple raid, however it is the first level that offers redundancy. Here you have two disks, that are "mirroring" each other. That is to say, bit per bit, all data that is written to drive A is also written to Drive B. This create a write speed penalty for all data being written to the array, since it must be written to both drives. However, when you read back the data, you will get an improvement. Under most controllers when you request data, it will be read from drive A, and while drive A is busy, drive B can then service an request that may come in the meantime. Thus you get the throughput of a single drive but an increase in access time performance as both drives can service two different read requests at once. In a mirror, one drive of the pair can die and the raid can still function. Drive A could die and drive B would take over, or vice versa. Here one backs up the other in case of failure, so one drive death can be tolerated before risk of data loss.

### Raid 3

Raid 3 offers byte level stripping combined with a dedicated parity drive. Here you have the data stripped, similar to raid 0, but with parity data being calculated and stored on a single dedicated drive. This setup requires a minimum of 3 drives. With raid 3 the read performance is quite similar to raid 0 with a slight hit in performance from byte level stripping. However the write performance suffers considerably due to overhead of calculating parity data and the fact that the single dedicated drive becomes a bottleneck as it must be accessed every time new data is written to the array.
Raid 3 can have a single drive die before any data loss is incurred. Raid 3,4,5 have a 2-1 drive capacity. Meaning the space given by the 3 drives with 1 drive subtracted is the usable space capacity.

### Raid 4

Raid 4 improves upon raid 3 by doing away with byte level stripping and instead does block level stripping and parity data calculation. However writes still suffer from the parity calculation and raid 4 still makes use of a dedicated parity data drive, that is still a bottleneck.
Raid 4 can have a single drive die before any data loss.

### Raid 5

Raid 5 improves by doing the same as raid 4 except that it now disperses the parity data throughout all drives, as opposed to a single dedicated parity drive. This removes the dedicated drive bottleneck. However, writes still suffer from the extra overhead of having to be calculated and written along with the other data.
Raid 5 can have a single drive die without data loss.

### Raid 6

Raid 6 is the same as raid 5 with with an extra drive added to the minimum requirement. Then dual parity data sets are calculated and written. This allows a maximum of two drives to die without data loss. But as you may guess, write performance is even worse than raid 5 as there is now two data sets to be written per write request.

## Nested Raid

You can also have different raid levels nested. Meaning you can mix or combine two raid levels together. The most common being:

### Raid 10 (1+0)

Raid 10 requires a minimum of 4 drives. It will separate the four drives into 2 pairs of 2, each drive in each of those pairs will mirror each other, this forms two logical drives. These logical drives are then combined in a raid 0, forming one logical drive. Here You can have two drives die before any data loss, a drive can die in each mirror pair. By probability this is more redundant compared to raid 01 (0+1).
Raid 10 is excellent with both read and write performance. Raid 10 is considered one of the better raid levels as it offers the same level of redundancy as others but without any parity calculations. Only downside is a 50% usage capacity, meaning out of all the storage space combined from all the drives, only half is usable. However, with the prices of high storage drives getting cheaper it is almost a non-issue.

### Raid 01 (0+1)

Raid 01 (0+1) requires a minimum of 4 drives. It will separate the four drives into 2 pairs of 2, each of those pairs will strip each other, this forms two logical drives. These logical drives are then combined in a raid 1 (mirroring), forming one logical drive. Here You can have two drives die before any data loss. As with Raid 10, it half's the usable drive space.

## Setting Up a Software Raid- Using Windows

Software raid might a be a solution you're looking for, and you may even be able to do it right now with just your operating system. Windows supports software raid 0 and raid 1. To setup a software raid in Windows Xp follow the steps below:
1.)Right-Click on "MY Computer" and go to "Manage."
2.)Under "Storage" click on "Disk Management"
3.)Here you'll want to right-click on the small square next to your newly installed drives, that are unpartitioned (unless you're doing raid 1, in which case only one needs to be unpartitioned), and say "Convert to Dynamic Disk" for each. Please keep in mind that converting your disks to dynamic disks may cause compatibility issues with some software, like Acronis. And once your disk is dynamic it is risky to convert it back to basic, so if you're doing so, backup your data first.
4.)Right-click the unpartitioned space of any one of the drives, click "New Volume." From here you can make a choice of which raid type you want, and unless you have windows server, raid 1 will be grayed out. This was part of a limitation Microsoft put into the OS, reasons for this are unknown. However, if you're not afraid of hex editors, you can attempt to edit the needed files into being the server versions, this method has been tested to work by me, however because it is a gray legal area i can not provide the files directly myself.
5.)After you have selected which raid level you want, you can then click "Next", add the drives you want to raid, click "Next," then follow the menu from there which lets you choose the standard format options. And you're done!
However, as stated above, you will generally get poorer performance through software raid, along with extra CPU utilization. And under raid 1 you will not be able to boot into windows without a boot disk when using the mirrored volume, due to Window's software raid not writing the needed boot information like the MBR and other files to the drive (Window's software mirror only works on the partition level). But the pro with software raid is hardware independence for raid migration.

## Setting Up Hardware Raid

You should start by researching what kind of raid level you're looking for, this will depend on your budget and use. After you have decided, see what interface you can use. Does your motherboard have a spare pci/pci-e/pci-x? Does the interface you have available meet the bandwidth/throughput you predict will be needed by your raid setup? After you figure this out, look online for what choices you have, I'd recommend shopping at Newegg.com (if in the USA), or Tigerdirect.com, who tend to have new prices. Be careful which controller you get, and avoid cheaper ones if you're looking for performance, generally $50 and above will give results. After you select your controller, you should start looking around for drives. I recommend the Raptor series for their excellent access times. However, the Raptor series, and especially the Velociraptor, may be a too expensive for some. If you can't afford those i would look into the Samsung Spinpoint series, whose high platter data density gives excellent transfer performance for a 7,200 drive. However, I would avoid the 1 terabyte models for now as they're as having quality control issues with that model that make for a lot of DOA's. Only downside are the access times are that of any 7,200 rpm drive. Since access times scale near the same as the performance gain, it is important to watch the access times of the drives you buy. As said earlier, depending on your raid setup, some access time issues may be slightly negated by the controllers operation (another reason to go for a quality controller). For my setup i used an Adaptec 1430SA pci-e 4x raid controller which supports raid 0,1, and 10 (bought on newegg for$104). For my drives, i used 4 150 gig, 10,000 rpm Raptor X's. Power supply is an Enermax Liberty 500 watt modular power supply, processor is a Q6600 quad core, 4 gigs of ddr2 800 ram at 5-5-5-12, and an 8800 gts video card. For this article, all are at default speeds.

Here you will see that i installed my controller card into a spare pci-e slot i had available. I then plugged the 4 SATA cables into the controller, which are then connected to my 4 Raptor X's.
As you may notice, i plugged 2 drives into one cable chain, and two others into another chain. Dispersing out the power to two will help disperse out the initial spike of power required by all drives to start their initial spinning. If you don't disperse the power you may end up with a few drives failing to spin up or dropout during testing/use.
To setup the raid with everything plugged in, turn on your computer, and after it passes the initial POST it will ask you to press ctrl+[letter derived from the manufacturer], in my case it is ctrl+A (for Adaptec). Here you'll have to refer to your controller manual on the specifics of the menu's. But in general most raid controller's bios's will allow you to select which drives are to be a part of the array, then it will ask what array type you want, then the strip size. After you're done setting up the raid you will need to exit the bios and restart the computer.
After you restart you will need to install the drivers for the raid before you can use it. In my case i am installing the raid under an existing windows install that is on a separate drive, so when i start up my computer, Window's saw the raid controller, and i simply pointed the install to the drivers on the disc/downloaded from their site. The same should be done by you. However, if you're wanting to install windows on this raid, you will need to start with a fresh install. To do this create a floppy disk that contains the drivers, backup any data you want saved, pop in the windows install disk, and press "F6" at the beginning of the windows install disk, from here the windows install disk will install the drivers and you are set. We are almost there, but before we can use the raid we need to initialize the disk. Now, if you installed windows on the raid array from the step above, then you can just skip this step, as it is not needed, you only need to initialize the disk if you installed the array on an existing windows install. To do this, right-click on the "My computer" icon and select "Manage," then click on "Disk Management" under "Storage." Here you may get a prompt from the manager, ignore the prompt and close it, then you should see your new raid disk, it will be unpartitioned and uninitialized. Right-click on the square to the left of the unpartitioned space and click "Initialize Disk." Now the disk should be active. From here you just need to right-Click on the unpartitioned space and click "New Volume," from here just follow the steps.
Now your hardware raid should be up and running!

## Testing

Now, i decided to setup my 4 raptors in a 4 drive raid 0 setup for testing, after which i then set them up in raid 10 for more permanent use (I.E. redundancy). I tested these drives in various configurations using different strip sizes to illustrate the principles discussed above.
Here you can see our base test. This is the performance of your average 7,200 rpm SATA drive. You can see its access time by itself is over 13 ms, and its average throughput being only 52.2 MB/s. This is about the performance most people get with only1 drive doing one task, the drive performance suffers even further if being requested data from multiple sources. Here are the drives in raid 0. The raid 0 setup has a strip size of 32 KiB, and i get a very nice and even (too even) transfer rate of 202.5 MB/s average. The access time is 8.1 ms, being reported as doubled compared to the Raptor X's single 4.2 ms time performance. Now, without any further testing we can see that our raid array is already considerably faster than our base generic 7200 rpm drive. Both in sustained throughput and access time. But, can we push performance further? Now, here is the raid 0 with a 16 KiB strip size. We can see that the access time is pretty much the same as before, but look at the max throughput, 326.2 MB/s and an average sustained throughput of 279.6 MB/s! Now that is quite an improvement. As discussed above, using lower strip sizes force the file to be split across more of the drives giving a large increase in throughput when reading and writing as all drives are being collectively used. Next we have our raid 10 results... Just for fun i decided to test the raid 10 array starting out with a 64 KiB strip size. The results are a bit hard to judge, being so sporadic. While i achieve a nice maximum value, the average is a little low, being brought down by the random dips. This is a good lesson as to why you should never go off of just maximum transfer rate values, it is very important to see the overall average transfer rate. So, i then changed the strip size and got, This is the final setup i have come to. It is the raid 10 but with a 16 KiB strip size, as with the raid 0. Here we see a more calm and predictable curve, giving a higher average and minimum.

Hopefully, after reading this you can see the benefits of raid arrays. Just by utilizing this technology, even without spending a chunk of change you can achieve considerable performance improvements in hard drives which is important as it is one of the largest bottlenecks in computers, and affects much more than just load times. From here there is now nothing more to do, other than to go and have fun with your raid! Soon I'll be writing an article on how best to optimize your hard drive, which will allow you to make use of your array even further.
until then, enjoy!
[Edited by - Jarrod1937 on July 11, 2008 11:08:27 PM]

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 This create a write speed penalty for all data being written to the array, since it must be written to both drives

Not necessarily.

First, if you have spare controller bandwidth, because it's writing to two drives at the same time, it can write to both drives in parallel. This is a good reason to put the two drives on different channels.

Second, write-behind RAID 1 controllers will buffer up the additional writes, and then flush them out once there's a lull in activity, thus if you're doing something other than full-on video capture, you may never see the slow-down, even if you put both drives on the same channel.

Personally, I find that RAID 1 is the easiest to set up, and the best trade-off of performance and reliability for me. Hard drives do go bad after a few years, and in the last five years, I've replaced 4 failed drives in 2 different systems, without losing a single bit of data, all because of RAID 1. I do, however, still have a remote back-up that gets taken once weekly, in case the entire computer burns out, is stolen, etc.

And, finally, a spelling nit: it's called "striping" as in "the stars and stripes." "Stripping" is something else, usually found in bars where they've sealed the windows and serve cheap, crappy beer.

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Original post by hplus0603
Quote:
 This create a write speed penalty for all data being written to the array, since it must be written to both drives

Not necessarily.

First, if you have spare controller bandwidth, because it's writing to two drives at the same time, it can write to both drives in parallel. This is a good reason to put the two drives on different channels.

Second, write-behind RAID 1 controllers will buffer up the additional writes, and then flush them out once there's a lull in activity, thus if you're doing something other than full-on video capture, you may never see the slow-down, even if you put both drives on the same channel.

Personally, I find that RAID 1 is the easiest to set up, and the best trade-off of performance and reliability for me. Hard drives do go bad after a few years, and in the last five years, I've replaced 4 failed drives in 2 different systems, without losing a single bit of data, all because of RAID 1. I do, however, still have a remote back-up that gets taken once weekly, in case the entire computer burns out, is stolen, etc.

And, finally, a spelling nit: it's called "striping" as in "the stars and stripes." "Stripping" is something else, usually found in bars where they've sealed the windows and serve cheap, crappy beer.

You're correct actually, though it depends on how intelligent the controller is. I'll make the corrections later.

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So .. my raid 0 is on the fence and I want to add a raid 1 to it now that i have money... this can be done right? I was told once you hook in the new drives you can go to rebuild as 01 and it will keep your data intact with the new drives?? I'm nervous about it though i dont want to lose my data if possible.

I have an Nvidia sli Mobo (i think 680i or something like that)

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 Original post by AverageJoeSSUSo .. my raid 0 is on the fence and I want to add a raid 1 to it now that i have money... this can be done right? I was told once you hook in the new drives you can go to rebuild as 01 and it will keep your data intact with the new drives?? I'm nervous about it though i dont want to lose my data if possible.I have an Nvidia sli Mobo (i think 680i or something like that)

Lol. It's doable on more advanced controllers. On the cheap little built on software raid controllers most consumer motherboards come with, you'll probably have issues.

Your best bet is to backup your data or create a ghost image, then configure your new raid configuration, and then dump your image back onto the new array.

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Original post by Washu
Quote:
 Original post by AverageJoeSSUSo .. my raid 0 is on the fence and I want to add a raid 1 to it now that i have money... this can be done right? I was told once you hook in the new drives you can go to rebuild as 01 and it will keep your data intact with the new drives?? I'm nervous about it though i dont want to lose my data if possible.I have an Nvidia sli Mobo (i think 680i or something like that)

Lol. It's doable on more advanced controllers. On the cheap little built on software raid controllers most consumer motherboards come with, you'll probably have issues.

Your best bet is to backup your data or create a ghost image, then configure your new raid configuration, and then dump your image back onto the new array.

Ugh... Sucky... Sounds like a safer bet though. Thanks!

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# Triangle congruence worksheet

Given below are the Class 9 Maths Triangle congruence worksheet
(a) Concepts questions
(b) Calculation problems
(c) Multiple choice questions
(e) Fill in the blanks
(f)Subjective Questions

## Subjective Questions

Question 1
Let ABC and PQR are two triangles.

Which cases the triangles will be congruent? Write down the congruence equation
1. $AB =PQ \; , \; BC=QR \; and \; AC=PR$
2. $BC=QR \; , \; \angle B= \angle Q \; , \;\angle C=\angle R$
3. $AB=PQ, BC=QR, \angle C= \angle P$
4. $AB=PQ, BC=QR, \angle B= \angle Q$
5. $AB=PQ \; , \; \angle A= \angle P \;,\; \angle C= \angle R$
6. $AB =PQ \; , \; BC=QR$
7. $\angle B= \angle Q \; , \; \angle C= \angle R$
8. $\angle A= \angle P, \angle B=\angle Q ,\angle C= \angle R$
9. $AB=PR , \angle A= \angle P, \angle B=\angle R$

## True or False statement

Question 2
True or False statement
a. We cannot construct a triangle of side 9,5,4 cm
b. In a Right angle triangle, hyponotuse is the longest side c. centroid is the point of intersection of the median of the triangles
d. A triangle can have two obtuse angles
e. Orthocenter is the point of intersection of altitudes
f. if $ABC \cong PQR$ then AC=PQ
g. if a triangle ABC such that AB > BC , Then $\angle C > \angle A$
h. Mid point of the hypotenuse of right angled triangle is circumcenter

## Multiple choice Questions

Question 4
In a triangle ABC , $\angle A=60^0$ ,\angle B=40^0, which side is the longest
a. AB
b. BC
c. AC

Question 5
In triangle ABC , AB=AC and D is the point inside triangle such that BD=DC as shown in figure

Which of the following is true
a. $\Delta ABD \cong \Delta ACD$
b. $\angle ABD= \angle ACD$
c. $\angle DAC= \angle DAB$
d. All the above

Question 6
An exterior angle of the triangle is $110^0$ , One of the opposite interior angle is $50^0$
What are the other two angles
a. $60^0,70^0$
b. $55^0,55^0$
c. $70^0,50^0$
d. None of the above

Question 7
AD is the median of the triangle . Which of the following is true?

a. $AB+BC+AC > 2AD$
b. $AB +BC > AC$
c. $AB+BC+AC> AD$
d. none of these

Question 8

In the above figure AB || CD ,O is the mid point BC. Which of the following is true
a. $\Delta AOB \cong \Delta DOC$
b. O is the mid point of AD
c. $AB=CD$
d. All the above

Question 9
In an isosceles triangle ΔABC with AB = AC. D is mid point on BC.
Which of the following is true

a. Orthocenter lies on line AD
b. AD is the perpendicular bisector BC
c. Centroid lies on the line AD
d. AD is the bisector of angle A

Question 10
PQR is a right angle triangle in with $P=90^0$ and PQ=PR . What is the value of Q and R
a. $45^0,45^0$
b. $30^0,60^0$
c. $20^0,60^0$

Question 11

In the above quadrilateral ACBD, we have AC=AD and AB bisect the ∠A
Which of the following is true
a. $\Delta ABC \cong \Delta ABD$
b. BC=BD
c. $\angle C= \angle D$
d. None of the above

## Match the column

 Column A Column B Point of intersection of altitude Circumcenter Point of intersection of Median In-centre Point of intersection of angle bisector Orthocenter Point of intersection of perpendicular bisector of sides Centroid

Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Note to our visitors :-

Thanks for visiting our website.

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# Bounded sequence has no convergent subsequence

How can you prove that in the metric space $(\mathbb{R},d)$ where $d(x,y)=|\arctan{x}-\arctan{y}|$ the sequence $(x_n)=n$ is bounded but it has no convergent subsequence ?

Edit 1. Can I say that $(x_n)$ is bounded because $d(n,0)\leq \frac{\pi}{2}$ ?

Edit 2. to see that $x_n$ has no convergent subsequence, I suppose by contradiction that there exists a subsequence $x_{\varphi(n)}$ which converges to $\ell$ it means that $d(x_{\varphi(n)},\ell)\rightarrow 0$ it means that $|\arctan(x_{\varphi(n)})-\arctan(\ell)|\rightarrow 0,$ so $\frac{\pi}{2}-\arctan{\ell} =0$ but $\frac{\pi}{2}-\arctan{\ell}\neq 0$ contradiction.

Thank you.

• i know that $\lim_{x\rightarrow +\infty} \arctan(x)=\frac{\pi}{2}$ – Vrouvrou Jan 3 '15 at 10:20
• Can you find any $\ell\in\mathbb R$ s.t. indeed $\arctan(\ell)=\frac{\pi}{2}$?? If not then there you have the contradiction. – drhab Jan 3 '15 at 11:40
• yes i understand $tag(\pi/2)$ do not exist – Vrouvrou Jan 3 '15 at 12:04
• The existence of a convergent subsequence implies the existence of an $\ell$ with $\arctan(\ell)=\frac{\pi}{2}$. Backwards the non-existence of this $\ell$ implies that no such convergent subsequence exists. This is exactly what had to be proved. So you are ready. – drhab Jan 3 '15 at 12:09
• That's something else. Btw, I did not downvote. Don't worry about that. You have your answers, haven't you? That is more important. Cheers. – drhab Jan 3 '15 at 12:13

$1)$ $x_n$ is bounded as the maximum distance between $x$ and $y$ is $|1-(-1)|=2$ as $-1 \leq |\arctan x|\leq 1$.

$2)$ Suppose that $x_n$ has a convergent subsequence $x_{n_k}$. Suppose further that $\displaystyle \lim_{k \rightarrow \infty}x_{n_k}=x_*$ say.

Then $d(x_{n_k},x_*)=|\arctan n_k-\arctan x_*|> \arctan (x_*+1)-\arctan x_*$ for $n_k > x_*+1$.

Note that $x \mapsto \arctan x : \mathbb{R} \to (-\pi/2,\pi/2)$ is a bijection. The metric on $\mathbb{R}$ which you are using is the pullback of the standard metric on $(-\pi/2,\pi/2)$ through this bijection. In other words, the arctangent function is, by design, an isometric isomorphism from $\mathbb{R}$ in the given metric onto $(-\pi/2,\pi/2)$ in the standard metric.

Thus, it is equivalent to prove that the sequence $y_n = \arctan(n)$ in $(-\pi/2,\pi/2)$ is bounded, but not convergent, for the standard metric. Can you take it from here?

• i want to prove that $x_n=n$ is bounded and have no convergent subsequence not $1/n$ – Vrouvrou Jan 3 '15 at 10:31
• Right. But my point is that is that if $\phi : (X,d) \to (X',d')$ is an isometric isomorphism from one metric space to another, then a sequence $x_n$ in $X$ is bounded and non convergent if and only if the sequence $\phi(x_n)$ in $X'$ is bounded and non convergent. Here "isometric isomorphism" is a synonym for "distance preserving bijection". – Mike F Jan 3 '15 at 10:33
• don't understand – Vrouvrou Jan 3 '15 at 10:34
• but why $\frac1n$ ? see my edit – Vrouvrou Jan 3 '15 at 10:42
• Sorry it was a typo. I'll fix it. Regarding your edit, yes your observation is enough since then $d(n,m) \leq d(n,0) + d(m,0) \leq 2 \frac{\pi}{2}$, so the distance between any two terms is at most $\pi$. I think it sort of misses the point, though, that the whole metric $d$ is actually bounded. Since $\arctan(x)$ and $\arctan(y)$ are both numbers in $(-\pi/2,\pi/2)$, the distance $d(x,y) = |\arctan(x)-\arctan(y)|$ is less than $\pi$ for any $x,y \in \mathbb{R}$. – Mike F Jan 3 '15 at 10:51

Boundedness is simple, by your own comment.

To show it has no convergent subsequences:

Assume it has a convergent subsequence, $(x_{n_k})$, with limit $K$. Then $n_k\ge n=x_n$. Now take $\epsilon=\frac{\frac{\pi}{2}-\arctan{K}}{2}$. Then take $N$ s.t. $\arctan{N}>\arctan{K}+\epsilon$ (possible again by your own comment out the limit of $\arctan$). Then we easily see that for every $n_k\ge N$, $d(x_{n_k},K)>\epsilon$ (because $\arctan$ is increasing). This is a contradiction, so $(x_n)$ cannot have converging subsequences.

• Can you tel me where is the contradiction with the fact that $\frac{\pi}{2}-\arctan{\ell} =0$ – Vrouvrou Jan 3 '15 at 11:30
• You should use what you know of what a limit really is. If there is a limit, then for ALL $\epsilon>0$ you must be able to find a $N$ s.t. bla bla bla. You don't do any of that. You just have to work a little harder to find the contradiction. But basically my answer here is in the same vein as your attempt, but you just did not quite finish it. – user2520938 Jan 3 '15 at 11:36


text:
# Math Help - Series and Sequence/Application (sum of a geometric series) - Supperannuation

1. ## Series and Sequence/Application (sum of a geometric series) - Supperannuation

hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? (p.a. means per annum so yearly) answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.

2. Originally Posted by KavX
hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.
What is p.a.?

3. p.a means per annum
so in other words it means yearly

4. Originally Posted by KavX
p.a means per annum
so in other words it means yearly
Then, "the investment earns 8.2% p.a., paid monthly" means?
Each month invesment is 8.2%/12?

5. see thats were im confused aswell

6. Originally Posted by KavX
hi

on the last question from my maths text book and i've gotten all previous questions right but this last one.

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? (p.a. means per annum so yearly) answer is :$813.16

btw this chapter is based on using the sum of a geometric series formula.
so ive been using this formula

Sn = a[(r^n) - 1]
aaaaaaar - 1

a: being the first amount in the geometric series
r: being the ratio or interest rate
n: being the no. of years or the time period.

unless theres another formula or way of figuring this out because i keep stuffing up something in my working out.

thx all help appreciated.
$20, 20r+20, r(20r+20)+20,\dots$
$20, 20r+20, 20r^{2}+20r+20,\dots, 20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$

$r=1+\frac{0.082}{12}=1.0068333$

$T_{36}$
$=20r^{35}+20r^{34}+20r^{33}+\dots +20r^{2}+20r+1$
$=20(r^{35}+r^{34}+r^{33}+\dots +r^{2}+r+1)$
$=20\left[\frac{r^{36}-1}{r-1}\right]$
$=813.16$

7. kavx,

Here is a lesson you can learn:

Annuity Immediate Accumulated Value

8.2/12 for interest rate, 20 for payment, 36 for number of periods. This is the annuity immediate accumulated value formula.

8. thanx seng

9. Hello, KavX!

25. Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% p.a., paid monthly? Answer is:$813.16

This is an Annuity, which has its own formula: . $A \;=\;D\,\frac{(1+i)^n-1}{i}$

. . where: . $\begin{Bmatrix}A &=& \text{Final amount} \\ D &=& \text{periodic deposite} \\i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}$

We have: . $D = 20,\;i = \frac{0.082}{12},\;n = 36$

Therefore: . $A \;=\;20\,\frac{\left(1 + \frac{0.082}{12}\right)^{36}}{\frac{0.082}{12}} \;=\;813.1609107 \;\approx\;\boxed{\813.16}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The derivation of the Annuity Formula is quite involved,
. . but it does involve a geometric series.

The formula should have been given to you.
Usually, we are not expected to derive it ourselves.

10. never got that formula , havnt even seen it before, anyways thanx for the heads up, i'll keep that formula in mind now.

11. In case anyone is interested, here's the derivation of that formula.

We deposit $D$ dollars each period.
The account pays $i$ percent interest every period.
We make these deposits for $n$ periods.

Reminder: If there are $X$ dollars in the account,
. . . . . . . . after one period, the account has grown to $X(1+i)$ dollars.

We deposit $D$ dollars.

At the end of period 1, our account has $D(1 +i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D(1+i) + D\:=\:D[(1+i) + 1]$ dollars.

At the end of period 2, our account has $D[(1 +i) + 1](1+i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D[(1+i) + 1](1+i) + D \:=\:D\left[(1+1)^2 + (1+i) + 1\right]$ dollars.

At the end of period 3, our account has $D\left[(1 +i)^2 + (1+i) + 1\right](1+i)$ dollars.
We deposit another $D$ dollars,
. . and the balance is: $D\left[(1+i)^3 +(1+1)^2 + (1+i) + 1\right]$ dollars.

. . . and so on . . .

At the end of period $n$, our final balance is:

. . $D\underbrace{\left[(1+i)^n + (1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}} \;=\;A$

The geometric series has the sum: . $\frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$

Therefore: . $\boxed{A \;=\;D\,\frac{(1+i)^n-1}{i}} \quad\hdots\quad \text{ta-}DAA!$

12. In case anyone is interested, here's the derivation of the Amortization Formula.

We take out a loan of $P$ dollars.
We will pay it back in $n$ periodic installments of $M$ dollars each.
They are charging us $i$ percent interest per period.

Reminder: If we owe $X$ dollars,
. . . . . . . . after one period, we will owe $X(1+i)$ dollars.

At time 0, we owe them $P$ dollars

At the end of period 1, we owe them: $P(1 +i)$ dollars.
We pay $M$ dollars, and our balance is: $P(1+i) - M$ dollars.

At the end of period 2, we owe them: $[P(1 +i) + M]\,(1+i)$ dollars.
We pay another $M$ dollars, and our balance is:
. . $[P[(1+i) - M]\,(1+i) - M\;=\;P(1+i)^2 - M(1+i) - M$ dollars.

At the end of period 3, we owe them: $[P(1+i)^2 - M(1+i) - M]\,(1+i)$ dollars.
We pay another $M$ dollars, and our balance is:
. . $[P(1+i)^2 - M(1+i) - M]\,(1+i) - M \;=\;P(1+i)^3 - M(1+i)^2 - M(1+i) - M$ dollars.

. . . and so on . . .

At the end of period $n$, the final balance is zero. .(We're paid up!)

. . $P(1+i)^n - M(1+i)^{n-1} - M(1+i)^{n-2} - \hdots - M(1+i) - M\;=\;0$

$\text{We have: }\;P(1 + i)^n \;=\;M\underbrace{\left[(1+i)^{n-1} + (1+i)^{n-2} + \hdots + (1+i) + 1\right]}_{\text{geometric series}}\;\;{\color{blue}[1]}$

. . The geometric series has the sum: . $\frac{(1+i)^n-1}{(1+i)-1} \;=\;\frac{(1+i)^n-1}{i}$

Substitute into [1]: . $P(1+i)^n \;=\;M\cdot\frac{(1+i)^n-1}{i}$

Therefore: . $\boxed{M \;=\;P\cdot\frac{i(1+i)^n}{(1+i)^n-1}}\quad\hdots\quad \text{There!}$


text:
# Determining the Equilibrium Constant

"An equilibrium constant expression describes the relationship among the concentrations (or partial pressures) of the substances present in a system at equilibrium" (General Chemistry).

For a given chemical reaction:

$aA + bB \rightleftharpoons cC + dD$

the equilibrium constant expression is as follows:

$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

To calculate the equilibrium constant (also known as the dissociation constant), the concentrations of each species in the reaction at equilibrium must be measured. Consider the general acid dissociation equation:

$HA + H_2O \rightleftharpoons A^- + H_3O^+$

Where HA is the acid, H2O is water, A- is the conjugate base of the acid, and H3O+ is the hydronium ion, a protonated water molecule.

Any equilibrium problem given provides some needed information, including a variety of numbers and/or units. Either Ka or Kb, the dissociation constants for the acid or the base in the equation, respectively, will be provided. For this example, Ka = 2.2 x 10-4 and the initial concentration of HA, the acid in solution is 2 mol L-1.

The equilibrium concentrations are calculated using an ICE Table. An ICE Table is a way to keep all your information organized throughout the process. ICE stands for 'Initial', 'Concentration', and 'Equilibrium'.

$HA + H_2O \rightleftharpoons A^- + H_3O^+$

HA + H2O A- + H3O+
I 2 mols/L   n/a   0   0
C -x   n/a   +x   +x
E 2-x   n/a   x   x

In this case,

$K_a = \frac{[A^-][H_3O^+]}{[HA]}$

H2O is ignored in the ICE Table and when calculating the dissociation constant because it is a pure liquid. Any hypothetical solids would also be ignored.

Substituting the values from the "E" row of the table,

$\begin{eqnarray} K_a &=& \frac{(x)(x)}{2-x} \\ &=& \frac{x^2}{2-x} \end{eqnarray}$

A simple way to solve this equation is to first estimate the value for x. Assume x is very small, so the quantity 2-x is essentially 2. Keep in mind that the value for Kis known. The equation simplifies to the following:

$x^2 = 2K_a$

The calculated value for x is very close to 0. It can be plugged into the 2-x term in the original equation to find a more accurate value for x. A few iterations of this process will converge on the correct concentration of H3O+ and, equivalently, A- at equilibrium in this specific case. In this example, the concentration of H3O+ and A- is 0.02086 or 2.09 x 10-2 mol L-1

The relationship between Ka and K b is the following:

$K_W = K_a \times K_b$

Kw is equal to 10-14. If the problem provides a value for Ka but the reaction involves a base,  Kb can be calculated by dividing Kw by Ka. The same holds true if Kb is given for an acid dissociation problem.

### Resources

1. General Chemistry: Principles & Modern Applications 9th edition.

• Gregory Arch


text:
# Find magnetic field in a solenoid for different regions? [closed]

Problem statement:

A steady current I flows along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $\text{P}$ at a distance $r$ from the common axis.

The correct statement(s) is (are):

• (A) In the region $0 < r < R$, the magnetic field is non-zero

• (B) In the region $R < r < 2R$, the magnetic field is along the common axis.

• (C) In the region $R < r < 2R$, the magnetic field is tangential to the circle of radius r, centered on the axis.

• (D) In the region $r > 2R$, the magnetic field is non-zero.

The given solution is that both (A) and (D) are correct statements. However, is there an alternate argument for (C) is also being correct since the magnetic field in the region $R < r < 2R$ is touches the circle of radius $r$ only at a single point and meets all the requirements of being a tangent?

A detailed argument can be found here:

## closed as off-topic by Michael Seifert, John Duffield, tpg2114♦May 11 at 7:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

• There is obviously only one solenoid, and no question asked... – TZDZ Dec 5 '14 at 9:18
• Yeah I corrected it. It would be great if you could answer it too. – stochastic_zeitgeist Dec 5 '14 at 10:45
• Sorry, what is the question exactly ? – TZDZ Dec 5 '14 at 10:52
• well, in your reasoning, every vector is tangent to any circle... – TZDZ Dec 7 '14 at 10:48
• I'm voting to close this question as off-topic because it's more about the definition of a tangent than about physics. – Michael Seifert May 10 at 12:42


text:
Homological integration

In the mathematical fields of differential geometry and geometric measure theory, homological integration or geometric integration is a method for extending the notion of the integral to manifolds. Rather than functions or differential forms, the integral is defined over currents on a manifold.

The theory is "homological" because currents themselves are defined by duality with differential forms. To wit, the space Dk of k-currents on a manifold M is defined as the dual space, in the sense of distributions, of the space of k-forms Ωk on M. Thus there is a pairing between k-currents T and k-forms α, denoted here by

$\langle T, \alpha\rangle.$

Under this duality pairing, the exterior derivative

$d : \Omega^{k-1} \to \Omega^k$

goes over to a boundary operator

$\partial : D^k \to D^{k-1}$

defined by

$\langle\partial T,\alpha\rangle = \langle T, d\alpha\rangle$

for all α ∈ Ωk. This is a homological rather than cohomological construction.


text:
# How to partition set of items into subsets with similar mean, variance and number of elements - looking for some help

Short version:
Given I have a list of 120 traits (happy, sad, etc.) evaluated on a number of scales ranging from -5 (for example highly negative) to +5 (for example highly positive), how can I group these traits into 6 sets of 20 items each, so that each set has a similar mean and variance of those scales?

A word of explanation: The 120 traits will be judged on whether or not they describe one of 6 social objects (20 traits per object) (using a 1-5 scale). Traits I will be using come from a bigger set (of 300 unique items) where each has been evaluated on measures like: valence, agency, communion and a couple of others.

I'd like to make sure that each social object was judged on traits that overall had a similar mean and (possibly) standard deviation, so that I don't end up, with one object that had very extreme traits (highly negative and highly positive) while others where somewhere in the middle.

R data example:

This is a minimal data example.
Note that in my real data, measures (valence etc) are not necessarily normally distributed and most likely correlated with other measures.

I did some reading and found out my problem might be partitioning related. Here is a R code I made up that does (more or less) what I'd like, but only based on one variable (let's pretented x are ratings of valence).

library(tidyverse)

set.seed(1)
x<- rnorm(120) # normal distribution is just an example
x<- sort(x)
g1<- rep(1:6, 20) # grouping strategy 1
g2<- rep(c(1:6, 6:1), 10) # grouping strategy 2
df<- tibble(x, g1, g2)

df %>%
group_by(g1) %>%
summarise(mean = mean(x), var = var(x))

df %>%
group_by(g2) %>%
summarise(mean = mean(x), var = var(x))

With the above code I can see that strategy #2 yields better results, as means of groups are more similar than with strategy #1. q

# A tibble: 6 x 3
g1       mean       var
<int>      <dbl>     <dbl>
1     1 0.01216328 0.8462127
2     2 0.05113158 0.8146632
3     3 0.08567107 0.7907553
4     4 0.13632257 0.7679591
5     5 0.16613787 0.7915441
6     6 0.20683895 0.8273733

# A tibble: 6 x 3
g2      mean       var
<int>     <dbl>     <dbl>
1     1 0.1078368 0.9842131
2     2 0.1094842 0.8930841
3     3 0.1104103 0.8232504
4     4 0.1115834 0.7368137
5     5 0.1077852 0.7200830
6     6 0.1111654 0.7093138

What I can't figure out is how to work with a couple of ranking variables (above example had only x).

In other words, what would be the best strategy to find the best possible grouping solution for a dataframe like this:

data<- data.frame(
trait_number = seq(1:120),
valence_rating = rnorm(120, mean = 0, sd = 3),
agency_rating = rnorm(120, mean = 0, sd = 3),
communion_rating = rnorm(120, mean = 0, sd = 3),
grouping = NA)

(note: once again rnorm is just an example for sample data. My real data is not normally distributed)

If I can explain anything more, please let me know in comments. Thank you.

• This thing is really bugging me and I would love to offer a bounty if only I could. – blazej Nov 24 '17 at 20:33
• I don't understand what you want to do. Given the data in your R code, you want to divide up the observations into groups? Like, group 1 would be observations with valence around -1, agency around 1, and communion around 0. Group 2 would be observations with valence around 2, agency around 0, and communion around 0. Etc? Like that or something else? – Bill Nov 25 '17 at 18:08
• Thanks for stopping by Bill. No that's (probably) not what I want. I'd like to get 6 sets of 20 elements each, where valence, agency and communion have similar mean and variance between and within those 6 sets. In other words I'd like to get set as similar as possible to each other. – blazej Nov 25 '17 at 18:20
• Unfortunately, this is not something I know about. I googled and found that this is a topic which comes up in cross-validation. The paper I link below suggests an algorithm which is 1) choose an element randomly and assign it to set 1. 2) find the closest 5 elements to the one you chose in step 1 and assign these to sets 2 through 6. 3) keep doing this until you run out of elements or your sets are full. You need to operationalize closest in step 2 with some distance measure. – Bill Nov 25 '17 at 19:05
• researchgate.net/profile/Jose_Moreno-Torres/publication/… – Bill Nov 25 '17 at 19:05

I have a mutlivariate nonparametric regression method that uses clusters. Using the cluster ID's to sort the data, then assigning a group to that sorted data yields the following results with your data df:

require(NNS)

a=NNS.reg(cbind(data$valence_rating,data$agency_rating,data$communion_rating), rep(1,length(data$trait_number)))\$Fitted.xy

setkey(a,NNS.ID)

colnames(a)[1:3]=colnames(data[,2:4])

# Re-run from here with different group innovations
a[,:= (group=rep(1:6, 20))]

a[,{valence_mean=mean(valence_rating);valence_sd=sd(valence_rating)
agency_mean=mean(agency_rating);agency_sd=sd(agency_rating)
communion_mean=mean(communion_rating);communion_sd=sd(communion_rating)
list(valence_mean=valence_mean,valence_sd=valence_sd,
agency_mean=agency_mean,agency_sd=agency_sd,
communion_mean=communion_mean,communion_sd=communion_sd)
}, by=group]

group valence_mean valence_sd agency_mean agency_sd communion_mean communion_sd
1:     1  -0.06839077   2.674746 -0.61403779  3.024703     -0.1573398     2.455025
2:     2  -0.73164854   3.603836 -0.06347979  2.197635      0.5897524     3.638972
3:     3   0.87995369   2.425216 -0.70785458  2.915249     -0.1070705     3.266067
4:     4   0.34903443   2.667286 -0.67996477  3.295273      0.4793330     2.250582
5:     5   0.32653867   3.204289  0.35996665  3.191802      0.5791942     3.786095
6:     6   0.82249408   3.037100 -0.18225506  3.531984     -0.3244441     3.280554

I'm sure it could be better, but hopefully this gets you started and it's applicable to your actual data.


text:
Enlarging regions of the MSSM parameter space for large tan$\beta$ via SUSY decays of the heavy Higgs bosons - Archive ouverte HAL Access content directly
Journal Articles Journal of High Energy Physics Year : 2017

## Enlarging regions of the MSSM parameter space for large tan$\beta$ via SUSY decays of the heavy Higgs bosons

Anibal D. Medina
• Function : Correspondent author
• PersonId : 1013191

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Michael A. Schmidt
• Function : Correspondent author
• PersonId : 1013192

Connectez-vous pour contacter l'auteur

#### Abstract

In the Minimal Supersymmetric Standard Model (MSSM) searches for the heaviest CP-even and CP-odd Higgs $H$, $A$ to tau-lepton pairs severely constrain the parameter region for large values of $\tan\beta$ and light Higgs bosons $H$, $A$. We demonstrate how the experimental constraint can be avoided by new decays to light third-generation sfermions, whose left-right couplings to $H$ can be maximised in regions of large trilinear couplings $A_{b}$, $A_{\tau}$ for sbottoms and staus, or large supersymmetric (SUSY) Higgs mass $\mu$ for stops. Due to the $\tan\beta$-enhancement in the production cross-sections via gluon-fusion and in association with bottom-quark pairs for $H$ and $A$, we find that down-type sfermions, in particular, sbottoms perform a better job in allowing more parameter space than up-type sfermions such as stops, which require much larger values of $\mu$ to compensate for $\tan\beta$. Vacuum stability as well as flavour observables constraints and direct searches for SUSY particles are imposed. We also associate the lightest CP-even Higgs with the observed 125 GeV SM-like Higgs and impose the experimental constraints from the LHC.

#### Domains

Physics [physics]

### Dates and versions

cea-01565662 , version 1 (20-07-2017)
cea-01565662 , version 2 (07-10-2022)

### Identifiers

• HAL Id : cea-01565662 , version 2
• ARXIV :

### Cite

Anibal D. Medina, Michael A. Schmidt. Enlarging regions of the MSSM parameter space for large tan$\beta$ via SUSY decays of the heavy Higgs bosons. Journal of High Energy Physics, 2017, 2017, pp.95. ⟨cea-01565662v2⟩

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# How to implement a neural network Intermezzo 1

http://peterroelants.github.io/posts/neural_network_implementation_intermezzo01/

This page is part of a 5 (+2) parts tutorial on how to implement a simple neural network model. You can find the links to the rest of the tutorial here:

## Logistic classification function

This intermezzo will cover:

If we want to do classification with neural networks we want to output a probability distribution over the classes from the output targets t

. For the classification of 2 classes t=1 or t=0

we can use the logistic function used in logistic regression . For multiclass classification there exists an extension of this logistic function called the softmax function which is used in multinomial logistic regression . The following section will explain the logistic function and how to optimize it, the next intermezzo will explain the softmax function and how to derive it.

In [1]:

## Logistic function

The goal is to predict the target class t

from an input z. The probability P(t=1|z) that input z is classified as class t=1 is represented by the output y of the logistic function computed as y=σ(z). σ is the logistic function and is defined as:

σ(z)=11+ez

This logistic function, implemented below as logistic(z) , maps the input z

to an output between 0 and 1

as is illustrated in the figure below.

We can write the probabilities that the class is t=1

or t=0 given input z

as:

P(t=1|z)P(t=0|z)=σ(z)=11+ez=1σ(z)=ez1+ez

Note that input z

to the logistic function corresponds to the log odds ratio of P(t=1|z) over P(t=0|z)

.

logP(t=1|z)P(t=0|z)=log11+ezez1+ez=log1ez=log(1)log(ez)=z

This means that the logg odds ratio log(P(t=1|z)/P(t=0|z))

changes linearly with z. And if z=xw as in neural networks, this means that the logg odds ratio changes linearly with the parameters w and input samples x

.

In [2]:

In [3]:

### Derivative of the logistic function

Since neural networks typically use gradient based opimization techniques such as gradient descent it is important to define the derivative of the output y

of the logistic function with respect to its input z. y/z

can be calculated as:

yz=σ(z)z=11+ezz=1(1+ez)2ez1=11+ezez1+ez

And since 1σ(z))=11/(1+ez)=ez/(1+ez)

this can be rewritten as:

yz=11+ezez1+ez=σ(z)(1σ(z))=y(1y)

This derivative is implemented as logistic_derivative(z) and is plotted below.

In [4]:

In [5]:

### Cross-entropy cost function for the logistic function

The output of the model y=σ(z)

can be interpreted as a probability y that input z belongs to one class (t=1), or probability 1y that z belongs to the other class (t=0) in a two class classification problem. We note this down as: P(t=1|z)=σ(z)=y

.

The neural network model will be optimized by maximizing the likelihood that a given set of parameters θ

of the model can result in a prediction of the correct class of each input sample. The parameters θ transform each input sample i into an input to the logistic function zi

. The likelihood maximization can be written as:

argmaxθ(θ|t,z)=argmaxθi=1n(θ|ti,zi)

The likelihood (θ|t,z)

can be rewritten as the joint probability of generating t and z given the parameters θ: P(t,z|θ). Since P(A,B)=P(A|B)P(B)

this can be written as:

P(t,z|θ)=P(t|z,θ)P(z|θ)

Since we are not interested in the probability of z

we can reduce this to: (θ|t,z)=P(t|z,θ)=ni=1P(ti|zi,θ). Since ti is a Bernoulli variable , and the probability P(t|z)=y is fixed for a given θ

we can rewrite this as:

P(t|z)=i=1nP(ti=1|zi)ti(1P(ti=1|zi))1ti=i=1nytii(1yi)1ti

Since the logarithmic function is a monotone increasing function we can optimize the log-likelihood function argmaxθlog(θ|t,z)

. This maximum will be the same as the maximum from the regular likelihood function. The log-likelihood function can be written as:

log(θ|t,z)=logi=1nytii(1yi)1ti=i=1ntilog(yi)+(1ti)log(1yi)

Minimizing the negative of this function (minimizing the negative log likelihood) corresponds to maximizing the likelihood. This error function ξ(t,y)

is typically known as the cross-entropy error function (also known as log-loss):

ξ(t,y)=log(θ|t,z)=i=1n[tilog(yi)+(1ti)log(1yi)]=i=1n[tilog(σ(z)+(1ti)log(1σ(z))]

This function looks complicated but besides the previous derivation there are a couple of intuitions why this function is used as a cost function for logistic regression. First of all it can be rewritten as:

ξ(ti,yi)={log(yi)log(1yi)if ti=1if ti=0

Which in the case of ti=1

is 0 if yi=1 (log(1)=0) and goes to infinity as yi0 (limy0log(y)=+). The reverse effect is happening if ti=0.
So what we end up with is a cost function that is 0 if the probability to predict the correct class is 1 and goes to infinity as the probability to predict the correct class goes to 0

.

Notice that the cost function ξ(t,y)

is equal to the negative log probability that z is classified as its correct class:
log(P(t=1|z))=log(y),
log(P(t=0|z))=log(1y)

.

By minimizing the negative log probability, we will maximize the log probability. And since t

can only be 0 or 1, we can write ξ(t,y) as:

ξ(t,y)=tlog(y)(1t)log(1y)

Which will give ξ(t,y)=ni=1[tilog(yi)+(1ti)log(1yi)]

if we sum over all n

samples.

Another reason to use the cross-entropy function is that in simple logistic regression this results in a convex cost function, of which the global minimum will be easy to find. Note that this is not necessarily the case anymore in multilayer neural networks.

#### Derivative of the cross-entropy cost function for the logistic function

The derivative ξ/y

of the cost function with respect to its input can be calculated as:

ξy=(tlog(y)(1t)log(1y))y=(tlog(y))y+((1t)log(1y))y=ty+1t1y=yty(1y)

This derivative will give a nice formula if it is used to calculate the derivative of the cost function with respect to the inputs of the classifier ξ/z

since the derivative of the logistic function is y/z=y(1y)

:

ξz=yzξy=y(1y)yty(1y)=yt

This post at peterroelants.github.io is generated from an IPython notebook file. Link to the full IPython notebook file

# How to implement a neural network Part 2

http://peterroelants.github.io/posts/neural_network_implementation_part02/

This page is part of a 5 (+2) parts tutorial on how to implement a simple neural network model. You can find the links to the rest of the tutorial here:

## Logistic regression (classification)

This part will cover:

While the previous tutorial described a very simple one-input-one-output linear regression model, this tutorial will describe a 2-class classification neural network with two input dimensions. This model is known in statistics as the logistic regression model. This network can be represented graphically as:

The notebook starts out with importing the libraries we need:

In [1]:

## Define the class distributions

In this example the target classes t

will be generated from 2 class distributions: blue (t=1) and red (t=0). Samples from both classes are sampled from their respective distributions. These samples are plotted in the figure below. Note that X is a N×2 matrix of individual input samples xi, and that t is a corresponding N×1 vector of target values ti

.

In [2]:

In [3]:

## Logistic function and cross-entropy cost function

### Logistic function

The goal is to predict the target class t

from the input values x. The network is defined as having an input x=[x1,x2] which gets transformed by the weights w=[w1,w2] to generate the probability that sample x belongs to class t=1. This probability P(t=1|x,w) is represented by the output y of the network computed as y=σ(xwT). σ is the logistic function and is defined as:

σ(z)=11+ez

This logistic function and its derivative are explained in detail in intermezzo 1 of this tutorial. The logistic function is implemented below by the logistic(z) method.

### Cross-entropy cost function

The cost function used to optimize the classification is the cross-entropy error function . And is defined for sample i

as:

ξ(ti,yi)=tilog(yi)(1ti)log(1yi)

Which will give ξ(t,y)=ni=1[tilog(yi)+(1ti)log(1yi)]

if we sum over all N

samples.

The explanation and derivative of this cost function are given in detail in intermezzo 1 of this tutorial. The cost function is implemented below by the cost(y, t) method, and its output with respect to the parameters w

over all samples x

is plotted in the figure below.

The neural network output is implemented by the nn(x, w) method, and the neural network prediction by the nn_predict(x,w) method.

In [4]:

In [5]:

## Gradient descent optimization of the cost function

The gradient descent algorithm works by taking the derivative of the cost function ξ

with respect to the parameters, and updates the parameters in the direction of the negative gradient .

The parameters w

are updated by taking steps proportional to the negative of the gradient: w(k+1)=w(k)Δw(k+1). Δw is defined as: Δw=μξw with μ

the learning rate.

ξi/w

, for each sample i

is computed as follows:

ξiw=ziwyiziξiyi

Where yi=σ(zi)

is the output of the logistic neuron, and zi=xiwT

the input to the logistic neuron.

• can be calculated as:
ξiyi=yitiyi(1yi)
• can be calculated as:
yizi=yi(1yi)
• zi/w
• can be calculated as:
zw=(xw)w=x

Bringing this together we can write:

ξiw=ziwyiziξiyi=xyi(1yi)yitiyi(1yi)=x(yiti)

Notice how this gradient is the same (negating the constant factor) as the gradient of the squared error regression.

So the full update function Δwj

for each weight will become

Δwj=μξiwj=μxj(yiti)

In the batch processing, we just add up all the gradients for each sample:

Δwj=μi=1Nxij(yiti)

To start out the gradient descent algorithm, you typically start with picking the initial parameters at random and start updating these parameters according to the delta rule with Δw

until convergence.

is implemented by the gradient(w, x, t) function. Δw

is computed by the delta_w(w_k, x, t, learning_rate) .

In [6]:

Gradient descent is run on the example inputs X

and targets t for 10 iterations. The first 3 iterations are shown in the figure below. The blue dots represent the weight parameter values w(k) at iteration k

.

In [7]:

In [8]:

## Visualization of the trained classifier

The resulting decision boundary of running gradient descent on the example inputs X

and targets t

is shown in the figure below. The background color refers to the classification decision of the trained classifier. Note that since this decision plane is linear that not all examples can be classified correctly. Two blue dots will be misclassified as red, and four red spots will be misclassified as blue.

Note that the decision boundary goes through the point (0,0)

since we don’t have a bias parameter on the logistic output unit.

In [9]:

This post at peterroelants.github.io is generated from an IPython notebook file. Link to the full IPython notebook file

The next page is about How to implement a neural network Intermezzo 2

Pages: 1 2 3


text:
## Math and MEMS: More Exponentials

Yesterday, Allyson Hartzell, the Director of Failure Analysis, Test, and Reliability at Pixtronix, Inc., visited Girls’ Angle to tell us about how she uses math in her work.

She works with Microelectricomechanical Systems, or MEMS. MEMS are tiny machines. They are so small that a hundred of them could fit across the width of a human hair. Even though these machines are so tiny, they can be very complex. Allyson showed us pictures of MEMS that contained several working gears. She even showed us a steam engine. According to Allyson, if you have ever used a Wii remote, you have relied on a MEMS, because it is a MEMS that enables the Wii to know how you’re moving the Wii remote around.

Last Wednesday, I blogged about exponential growth. At one point in her presentation, Allyson showed us this graph:

D.M. Tanner, et al. (2000) 'MEMS Reliability: Infrastructure, Test Structures, Experiments, and Failure Modes'. Sandia Report SAND2000-0091, p. 78 (Courtesy Sandia National Laboratories, Radiation and Reliability Physics Dept., http://www.mems.sandia.gov).

This graph gives a perfect example of why understanding exponential growth is so useful. It shows the number of revolutions that a tiny gear can make before one can expect it to break down and no longer work. If you look at the vertical axis, the equally spaced marks are labeled by powers of 10, starting at $10^3$ and ending at $10^{10}$. Having grasped exponential growth, we know that $10^{10}$ is much bigger than $10^3$. Each time you go up a mark along the vertical axis, you are looking at an increase in the longevity of the gear not by a little bit, but by a whole factor of 10. So, for example, one thing this graph shows is that about 70% of the gears will last 1,000,000 revolutions, and of the gears that survive 1,000,000 revolutions, half will last not just a little bit longer, but 100 times longer (assuming that the number of revolutions per unit time is constant).

You might be wondering, “Why bother with all these exponentials? Why not just redo this graph using axes that are labeled so that evenly spaced marks represents evenly spaced numbers?” The best way to answer this question is to go ahead and try to convert the graph to one that uses such axes. You will find that it is practically impossible to do. Don’t take my word for it. Try it yourself! If you succeed, please send us your graph; we’d love to see it!


text:
# Physics motion question!?

## Two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at $20 m {s}^{-} 1$ and the second at $28 m {s}^{-} 1$. If the second car is 6km behind the first car, how much time will it need to catch up with the first car?

May 20, 2018

$t = 12.5 \text{minutes}$

#### Explanation:

Speeds are relative. The rotation of the Earth's surface at the equator goes at about 460 m/s to accomplish its full revolution every 24 hours. Check out this article from Scientific American. If you were in the lead car, the speed, relative to you, of the 2nd car would be $28 m {s}^{-} 1 - 20 m {s}^{-} 1 = 8 m {s}^{-} 1$. That is the speed of that car's approach.

The formula of speed, or velocity if we know the direction, is $v = \frac{d}{t}$

(We tend not to use s for speed, perhaps because s is already used for seconds, and some times distance.)

Solving that formula for time

$t = \frac{d}{v}$

We can use that formula to solve our problem, but first, there is a problem with units coming up. The distance is in km and our speed is in ms^-1. So it will be best to convert the 6 km to meters.

$6 \cancel{k m} \cdot \frac{1000 m}{1 \cancel{k m}} = 6000 m$

Now we can find the time for the 2nd car to catch you.

$t = \frac{6000 \cancel{m}}{8 \frac{\cancel{m}}{s}} = 750 s$

The question does not say what units it wants the answer in, but just for fun ...

$750 \cancel{s} \cdot \frac{1 \min}{60 \cancel{s}} = 12.5 \min$

I hope this helps,
Steve


text:
# Prove that exist $n$ consecutive numbers each one with at least one prime factor repeted $t$ times

"Given $n,t \in \mathbb{N}$ prove that exist $n$ consecutive natural numbers with each one of them having at least one prime factor repeted $t$ times."

Well, at a first glance I couldn't do anything, so I found this:

How do I prove that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite?

And my point is, with the numbers $i+(n+1)!$ with $1<i\leq n+1$, I get $n$ consecutive composite numbers. And if I do $i+((n+1)!)^t$ will I get what I was asked for?

• Here $0 \not \in \mathbb{N}$ – Robson Nov 17 '17 at 17:59
• This approach will not work. Just try it for a small case like $n=1, t=2$. You don’t even get one consecutive value with the desired property. – Erick Wong Nov 17 '17 at 18:02
• Use the Chinese Remainder theorem with a suitable set of prime powers – Joffan Nov 17 '17 at 18:02

$$\left\{\begin{array}{rcl}m&\equiv&0\pmod{2^k}\\ m+1&\equiv&0\pmod{3^k}\\ \ldots &\equiv &\ldots \\ m+n-1 &\equiv &0\pmod{p_n^k}\end{array}\right.$$ has a solution $m\leq (2\cdot 3\cdot\ldots\cdot p_n)^k\approx e^{kn}$ which is associated with an interval $[m,m+n-1]$ of integers fulfilling the wanted constraint.


text:
# Can we conclude structural break from the following results?

The following figure shows the log-likelihood of individual observations for different subsets of a data set - the horizontal line shows the average log-likelihood:

So, the first subset includes all the data points observed up to t=7, the second all data points up to t=33, and so forth. For each subset, I estimate a parameter $\theta$, and then I calculate the likelihood of each observation given this parameter estimate. This is what the figure shows.

It can be seen that on average the likelihood of the observed data (shown as a horizontal line) tends to decrease as the number of observations grows. This pattern can be confirmed in the following plot, which shows how the average likelihood of observations changes as the number of observations included grows, for different data sets:

In the light of this, would you agree that these results suggest that the $\theta$ parameter is not equal for all $t$? If it was equal, I would expect the same average likelihood regardless of the number of observations included.

-
Could you explain why you are carrying out this analysis? Are you looking for outliers? Changepoints? (If so, what kind of change?) Something else? –  whuber Nov 24 '12 at 23:29
Nevermind, the analysis is wrong. –  Ernest A Nov 26 '12 at 1:04


text:
# Optimization with polynomial objective and constraints

Let us be given the following optimization problem.

\begin{aligned} & \text{minimize} && u \cdot x^2 + v \cdot y^2 + w \cdot z^2, \\ & \text{subject to} && x + y + z = 1, \\ & {} && 0 \leq x < y < z \leq 1,\\ & {} && u \cdot x +2 v \cdot y \leq 3, \\ & {} && 3 v \cdot y + 2 w \cdot z \leq 1, \\ & {} && 0 \leq u \leq 1, \\ & {} && 0 \leq v \leq 1, \\ & {} && 0 \leq w \leq 1. \end{aligned}

All $x,y,z,u,v,w$ are variables to be optimized. In general, we have a set of polynomial constraints and the objective function to be minimized is also a polynomial. This problem seems to be done by geometric programming. However, in geometric programming, only monomial equality constraints are allowed. But here, we have the equality constraint $x+y+z = 1$.

So what should be a possible way to solve this kind of problems?

• You could replace each $z$ by $1-x-y$ if you could figure the right restriction so that $0 \le x<y<z \le 1$ would be satisfied. Then there would no longer be equality constraints. – coffeemath May 24 '13 at 12:36


text:
# If the depth of water is 8.64 cm on a field over a base period of 10 days, then the duty is

This question was previously asked in
RSMSSB JE Civil Diploma Oct 2016 Official Paper
View all RSMSSB JE Papers >
1. 10 hectares per cum/s
2. 100 hectares per cum/s
3. 864 hectares per cum/s
4. 1000 hectares per cum/s

Option 4 : 1000 hectares per cum/s

## Detailed Solution

The relation between Duty (D) in ha/cumecs , Delta in meter and Base period (B) in days is given by the following relation:

$$D = \frac{{8.64\;B}}{\Delta }$$

D = $$\frac{{8.64\; \times \;10}}{{8.64\; \times \;{{10}^{ - 2}}}}$$

∴ Hence, Duty = 1000 ha/cumecs

Important point:

Duty means how much area can be irrigated by canal discharging one m3/s water. For Ex. Duty of 10 ha/cumecs means 1 cumecs discharge can irrigate 10 hectare area.


text:
# What is the inverse of Y 3x?

## What is the inverse of Y 3x?

f-1
Answer: The Inverse of y = 3x is f-1(x) = 1/3x.

What is the formula in inverse?

An inverse variation can be represented by the equation xy=k or y=kx . That is, y varies inversely as x if there is some nonzero constant k such that, xy=k or y=kx where x≠0,y≠0 . Suppose y varies inversely as x such that xy=3 or y=3x . That graph of this equation shown.

### How do you find the inverse of a function?

Given the function f(x) we want to find the inverse function, f−1(x) f − 1 ( x ) . First, replace f(x) with y . This is done to make the rest of the process easier. Replace every x with a y and replace every y with an x .

What is the inverse of y 3x 4?

The inverse function of 3x – 4 is (x+4)/3.

#### Is the inverse of y 3x 1 a function?

This is a linear relation and therefore a function (all linear relations are functions).

Which equation is the inverse of Y 2×2 8 quizlet?

Answer: √[(x + 8) / 2] is the inverse of y = 2×2 – 8.

## How do you find the inverse property?

To find the multiplicative inverse of a number, all you have to do is find the reciprocal of the number. If you have the number 99, the reciprocal is 1/99. This is also the multiplicative inverse because when you multiply 99 and 1/99, you get 1 as a result.

What is the inverse square law in simple terms?

Specifically, an inverse square law says that intensity equals the inverse of the square of the distance from the source. For example, the radiation exposure from a point source (with no shielding) gets smaller the farther away it is. If the source is 2x as far away, it’s 1/4 as much exposure.

### How do you tell if a graph is an inverse function?

Let f be a function. If any horizontal line intersects the graph of f more than once, then f does not have an inverse. If no horizontal line intersects the graph of f more than once, then f does have an inverse. The property of having an inverse is very important in mathematics, and it has a name.

What is the inverse of 7?

For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1!

#### What is the inverse of 2X 7?

Reversing a Function If the original function is f(x)=2x-7, the order for the function is to multiply the x by 2 and then subtract 7. The inverse reverses this, so it adds 7 to the y and then divides by 2. So, the inverse of f(x)=2X-7 is f^-1(y)=(y+7)/2.


text:
m
• E

F Nous contacter

0

# Documents  Canzani, Yaiza | enregistrements trouvés : 2

O

P Q

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## Understanding the growth of Laplace eigenfunctions (part 1 of 2) Canzani, Yaiza | CIRM H

Post-edited

Research talks;Partial Differential Equations;Geometry

In this talk we will discuss a new geodesic beam approach to understanding eigenfunction concentration. We characterize the features that cause an eigenfunction to saturate the standard supremum bounds in terms of the distribution of $L^{2}$ mass along geodesic tubes emanating from a point. We also show that the phenomena behind extreme supremum norm growth is identical to that underlying extreme growth of eigenfunctions when averaged along submanifolds. Using the description of concentration, we obtain quantitative improvements on the known bounds in a wide variety of settings. In this talk we will discuss a new geodesic beam approach to understanding eigenfunction concentration. We characterize the features that cause an eigenfunction to saturate the standard supremum bounds in terms of the distribution of $L^{2}$ mass along geodesic tubes emanating from a point. We also show that the phenomena behind extreme supremum norm growth is identical to that underlying extreme growth of eigenfunctions when averaged along ...

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## Probabilistic methods in geometry, topology and spectral theory:CRM workshop on probabilistic methods in spectral geometry PDEMontréal # August 22-26, 2016and on probabilistic methods in topologyMontréal # November 14-18, 2016 Canzani, Yaiza ; Chen, Linan ; Jakobson, Dmitry | American Mathematical Society;Centre De Recherches Mathematiques 2019

Congrès

- viii; 197 p.
ISBN 978-1-4704-4145-6

Contemporary mathematics , 0739

Localisation : Collection 1er étage

théorie spectrale # probabilités # analyse géométrique asymptotique # chaos quantique # géométrie des métriques aléatoires # gravité quantique # espace de triangulation # statistique topologique # noeud aléatoire

#### Filtrer

##### Codes MSC

Titres de périodiques et e-books électroniques (Depuis le CIRM)

Ressources Electroniques

Books & Print journals

Recherche avancée

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text:
Jaynator495 one year ago The Sprauger Bug. Below is a detailed explanation of how its caused, what causes it, how to recreate it, and a recreation of it.

1. Jaynator495

Its caused by guests, See they dont have an image, i mean the do... But it cannot be registered under the "smallAvatarSrc" binding which is required for posts, so when a guest is viewing a post and your respond, this triggers a OS javascript event that REQUIRES the smallAvatarSrc binding to load the viewers of the post, as a result it substitutes whats there with the template, causing the Error! HOW HAVE I NOT FOUND THIS SOONER??!?!?!??

2. AlexandervonHumboldt2

sprauger? interesting

3. Jaynator495

This bug has been around for the longest time... AND I FINALLY FIGURED IT OUT!

4. Elsa213

Do the guest see Sprauger? o:

5. Jaynator495

Nope, they cant respond to posts!

6. Elsa213

Niether can Sprauger? e.e

7. Jaynator495

sprauger isnt really there, its just the template LOL

8. Elsa213

lemme see e.e

9. Jaynator495

-_- how praytell am i going to show you?

10. Elsa213
11. Jaynator495

http://prntscr.com/8pm9d3 YOU DID THIS!

12. Elsa213

i still saw Spraguer cx

13. Jaynator495

well atleast we proved what i just said xD

14. Elsa213

cx

15. Jaynator495

This is the proof that backs up my statement

16. Elsa213

I don't understand that, Jaynator. e.e

17. Jaynator495

look for the details such as "smallAvatarSrc" "is not defined" etc

18. Elsa213
19. Jaynator495

yes, its trying to find the guests avatar, but it returns nothing, so the function cannot be made, and as a result it cannot understand what should be there, and comprimises by showing the template :P

20. Elsa213

Try to make an extension to get rid of adds e.e

21. Elsa213

22. Jaynator495

Not happening, incredibly easy, but not happening. I'm NOT making a Ad-blocker, this harms OpenStudy's and other sites revenue

23. Elsa213

awe ;^;

24. anonymous

He should make an Ad Blocker xD Everybody would love it!

25. Jaynator495

IM NOT MAKING AN AD BLOCKER! Plus it would litterally take me 5 minutes to make it -_- its just removing the tag that google uses for ads, then looking into other ad providers and deleting that tag to LOL

26. Jaynator495
27. anonymous

I don't get it.. ._. I am sorry i'm not a genius like you! |dw:1444447145208:dw| My drawing skills aren't even good on here! Wait doesn't google chrome already have an Ad blocker?

28. Jaynator495

many LOL

29. anonymous

Yeah, ha silly me. I can just download one from the store.

30. AlexandervonHumboldt2

interesting http://openstudy.com/users/spraguer#

31. anonymous

Who is that?^

32. TheSmartOne

lol

33. Jaynator495

crud did i misspell it?

34. nincompoop

rubbish

35. AlexandervonHumboldt2

Chris Sprague is a co-founder and Chief Evangelist of OpenStudy.

36. Jaynator495

Wait i DID mispell it? crud... son of a gun i did! LOL

37. Jaynator495

._.

38. Jaynator495

woah thomaster, stop coming while im tagging you before i tag you, its kinda creepy how you always do this XD

39. pooja195

Hey! Look sprauger is on your post!! :D

40. Jaynator495

its just chibi XD

41. pooja195

#SARCASM

42. Jaynator495

Well i just increased the speed exponentially when you have markup disabled XD now to work on increasing the speed of the actual markup :P

43. Jaynator495

thomaster you have been watching this post all day o_O

44. Jaynator495

my god thomaster your still viewing it

45. Jaynator495

@jagr2713

46. Jaynator495

@jagr2713 come down here :P

47. jagr2713

lol

48. Jaynator495

see em now :P

49. Jaynator495

Jagr? :P

50. Jaynator495

\bold[*UMG thomaster finally stopped viewing it!?!?!?! LOL*]

51. jagr2713

oh emojis

52. Jaynator495

\italic[~what did you think i said?\bold[* ._. nope thomasters still here xD*]~]

53. jagr2713

lol

54. Jaynator495

Im finally making ALL the codes server side, as i got bored with the baitfile, this also means a lot of codes will be added :P

55. Jaynator495

ANNNND done... .xD

56. confluxepic

$$\color{blue}{\text{Originally Posted by}}$$ @Jaynator495 Its caused by guests, See they dont have an image, i mean the do... But it cannot be registered under the "smallAvatarSrc" binding which is required for posts, so when a guest is viewing a post and your respond, this triggers a OS javascript event that REQUIRES the smallAvatarSrc binding to load the viewers of the post, as a result it substitutes whats there with the template, causing the Error! HOW HAVE I NOT FOUND THIS SOONER??!?!?!?? $$\color{blue}{\text{End of Quote}}$$ No no no no. Spraguer is trying to fool you.

57. Jaynator495

Not true... We found it and replicated it around 30 times by the time you posted this conflux xD

58. confluxepic

It's a joke.


text:
Intermediate Algebra (6th Edition)

$2x^{2}$ - 32 = (2x + 8)(x - 4) Check by multiplying factors with FOIL: 2x(x) + 2x(-4) + 8(x) + 8(-4) = 2$x^{2}$ - 8x + 8x - 32 = 2$x^{2}$ - 32


text:
user34241 (imported from SE)
%I

\documentclass[ aspectratio = 1610 ]{beamer}

\usepackage{amsthm}

\usepackage{amsmath}

\usetheme{Frankfurt}

\makeatletter

\setbeamertemplate{theorem begin}{%

\begin{\inserttheoremblockenv}%

}
\setbeamertemplate{theorem end}

{\end{\inserttheoremblockenv}}

\makeatother

\begin{document}

\begin{frame}

\begin{definition}[ Is it possible to produce this heading without the white bar below?]

\end{definition}

\end{frame}

\end{document}
samcarter
Removing an empty body from a rounded beamer block is not trivial, however you can simply reverse roles and use a block with empty title:

\documentclass[ aspectratio = 1610 ]{beamer}

\usetheme{Frankfurt}

\makeatletter

\setbeamertemplate{theorem begin}{%
\begin{\inserttheoremblockenv}%
}
\setbeamertemplate{theorem end}
{\end{\inserttheoremblockenv}}

\makeatother

\begin{document}

\begin{frame}

\begin{definition}[ Is it possible to produce this heading without the white bar below?]
\end{definition}

{
\setbeamercolor{block body}{bg=structure.fg!75!black,fg=white}
\begin{block}{}
\inserttheoremheadfont Is it possible to produce this heading without the white bar below?
\end{block}
}

\end{frame}

\end{document}

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