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# What about gravity's relativity?

1. Apr 2, 2012

### C. Bernard

Why is not gravity relative like the space-time continuum, since it operates on and is modifying it?
Why is gravity exactly inversely equal to the square root of the distance between objects and not some number like pi or the speed of light?

2. Apr 3, 2012

### TheTechNoir

I don't have time for a proper response right now but I can quickly address one thing.

Why isn't gravity's force inversely equal to the speed of light?
I don't even know what this means. C is a velocity, not a measure of distance. I think you need to be more specific with what units you're using and exactly what you mean.

3. Apr 3, 2012

### C. Bernard

Thank you, Technoir, for your help.
What i meant was: 1) Is gravity absolute or relative like space-time?

2) Why is gravity EXACTLY relative to the square root of the distance, how come it is not a fraction of it, why precisely the square root? It seems too neat to me.
Why not .9999 or 1.0001 or some infinite number like pi?

4. Apr 3, 2012

### zhermes

Gravity is just as relative as space-time is. Keep in mind you need to use general relativity to describe relativistic gravity, instead of newtonian gravity.

That is an excellent question. I don't think there is an entirely definitive answer as of yet. An appealing idea to explain exactly integer exponents (like '2') is by relating them to the exactly integer number of dimensions. For example, the inverse square law would have to do with 2 being 1-less than the number of large spatial dimensions (i.e. because the surface of a 3D region is a 2D surface)....

Last edited: Apr 3, 2012
5. Apr 3, 2012

### clamtrox

That is incorrect. The space-time is very much non-relative. Only space and time as separate concepts are relative.

Do you mean square instead of square root?

You have to understand here that in general relativity, this is not true. In general relativity, gravity is not a force at all. Particles just move as straight as they can, but as space-time is curved, the particle world-lines appear to be bent.

Then the question becomes: why in the limit of weak fields, the apparent force goes down like 1/r². The reason is that gravitational force carriers are conserved. Force of gravity is proportional to the flux of these force carriers, and as they are conserved, the flux at a given radius is proportional to the area of a sphere of that radius.

6. Apr 4, 2012

### Khashishi

Newton formulated the 1/r^2 relationship of gravity because it explained Kepler's laws. It's only approximately true. Note that this is an inverse square relationship, not an inverse square root relationship.

7. Apr 4, 2012

### C. Bernard

To Zhermes, Clamtrox and Khashishi: thanks for your help.

I take note:square not square root.
Still, is it exactly square or approximately square , and if the latter what?
How come magnetic attraction is inversely proportional to the cube of the distance?
If there is a relativistic gravity, is it subject to the same effect as time (such as the so called twins paradox, which could as well be called the two clocks paradox or the two anythings paradox)? In other words does gravity depend on the speed something is moving?
As for the two different explanations received, i am affraid i don't quite understand them; the link between the dimensions of space-time or the volume of a sphere and the force of attaction of gravity on one side and the square of the distance between two objects on the other is not easy to explain in plain english i suppose.
It seems there is no consensus on this matter.

8. Apr 4, 2012

### Staff: Mentor

There is, but remember that most people here are not physics professors. We all have varying amounts of knowledge about a subject, and sometimes one person can simply be incorrect and not know it.

I myself have had difficulty in understanding what you've been asking, as the terminology you used isn't quite correct. Of course this is simply due to lack of knowledge about the subject on both your end and my end. If I understood more I would probably realize exactly what your asking and be able to explain it the correct way.

9. Apr 4, 2012

### Mark M

C. Bernard, there is a definite reason why Newton used the inverse square law for gravity. The inverse square law can be used to describe any force that radiates outward, like gravity. Since the surface area of a sphere is proportional to the square of the radius, as the emitted force gets farther from the source, it is spread out over an area that is increasing in proportion to the square of the distance from the source. Hence, the strength of the force at any unit area is inversely proportional to the square of the distance from the source.

Obviously, we now know that gravity is not a force. But Newton's law provides a very accurate result, and is used for all space missions. Einstein's equations only become relevant in enormous gravitational fields.

There is a consensus on the matter, it is not trivial in any way.

Also, gravity is not relative. Remember Einstein's first postulate of special relativity, the laws of physics are the same for all inertial frames of reference. So, some object approaching the speed of light will not become a black hole, as some people tend to think.

Last edited: Apr 4, 2012
10. Apr 4, 2012

### DaveC426913

many have provided good answers, but I'dd add 2c.

Einsteinian gravity is a field. It is everywhere. It may have different values here and there, but every point in space has a value.

Anything that radiates from a point (such as Newtonian gravity) will have an inverse square relationship. Its value is essentially that of the surface of an expanding sphere. Light does the same thing. If 100 photons leave a point source, the will spread out such that, at twice the distance, they cover 4x the area so their density will be quartered; at 3x the distance, their density will be 1/9th. 4x the distance, their density is 1/16th, etc.

11. Apr 4, 2012

### zhermes

It is exact (as far as anyone can tell), and if it wasn't exact---to my knowledge---the universe wouldn't work anywhere close to the way it does. Kepler's laws, for example, are dependent on it being exactly an inverse square law. Similarly, extended massive objects wouldn't be able to be treated as point masses... etc etc.

It isn't. In general the 'magnetic attraction' depends on the particular geometry, but is based on the same inverse square law as electricity... for two bar magnetic the force is approximately an inverse-square law (if they're end to end); for a general dipole I think its an inverse fourth-power. It all stems from the inverse square law of the electrostatic force.

Mark M gives a good explanation of the surface-area---inverse-square-law connection.

12. Apr 4, 2012

### DaveC426913

Consider:
Take a square n units on a side. Its area is exactly n^2.

This does not seem unusual does it? You don't wonder if it's approximately ^2, or whether there's some strange coincidence at work?

When you relate linear dimensions to area dimensions, you get ^2, exactly , whether it's squares or circles or surfaces of spheres.

13. Apr 6, 2012

### C. Bernard

Many thanks to everybody who has intervened in this discussion, i think i get it.
As for the magnetic attraction being the inverse of the cube of the distance, i found on wikipedia article "magnet", paragraph "fields of a magnet", so it is not entirely reliable.
Still it is amazing than a natural phenomena conforms to a geometrical figure which is a virtual reality more or less invented by man, after all a perfect sphere does not exist in nature.

14. Apr 6, 2012

### zhermes

Yes, the field, in that case, goes like one over the distance-cubed---not the force.

It doesn't matter if the shape is perfect or not---that just changes the constant of proportionality, as long as you're talking about the surface area - it scales like $r^2$, where the exponent is exactly 2. If general relativity teaches us anything, its that the underlying geometry of nature is incredibly decisive in its behavior.

15. Apr 7, 2012

### Nabeshin

I feel compelled to contribute here and say that 1/r^2 is NOT the whole story. In Einstein's general relativity, for one, gravity isn't even thought of in the traditional sense of a force, so this kind of thing isn't really well defined. BUT what you can do is in the limit of slow moving objects that aren't massive compared to black holes, you can see what gravity looks like. It turns out that when you reduce to the case of Newtonian mechanics, you do get 1/r^2, but it actually looks something like
$$F(r)=\frac{\alpha_2}{r^2}+ \frac{\alpha_3}{r^3} + \frac{\alpha_4}{r^4} + \cdots$$

Where the $\alpha$s are some proportionality constants (They're related to v/c, the velocity divided by the speed of light. So when this is small, each next term in the series is much smaller than the last, so is less important). What's important is that the deviation from Newton DOESN'T appear in a difference from two in the lowest order term, but rather the existence of these higher order terms like 1/r^3, 1/r^4 etc.

16. Apr 7, 2012

### Chronos

If you prefer to picture gravity as a force that permeates space, the answer is intuitively simple - its declines with distance squared because the volume of space it encompasses increases by a factor of distance squared.

17. Apr 7, 2012

### DaveC426913

You're looking at it wrong.

When you examine a single factor influencing a phenomenon, the formula describing the effect will be correspondingly simple.

Perfect spheres do exist in nature: the surface where all points are equidistant from a central point is a perfect sphere. So the surface where all points are exactly 1 light second from Earth's centre-of-gravity is a perfect sphere 1 light second in diameter. (This it trivial).

That formula is quite exact - and simple. And it is that principle that we are talking about in this thread.

However, when you examine the reality - you take all other factors into account - for example, there are more and less dense areas on Earth's surface that will cause local gravity over that point to vary. So that means that the surface where all points are under the same force of gravity is not a perfect sphere - it is quite lumpy.

18. Apr 8, 2012

### C. Bernard

Well, i may have been overly optimistic when i wrote i got it.I sure don't get it anymore.
Magnetic fields are function of a volume, but magnetic forces of a surface. But both gravity forces and field are a function of a surface.
It seems gravity is a function of "v"/c. What is v: the speed of gravity? Or is gravity instantaneous? So gravity was relative after all.
Or is v the speed two objects are moving relative to each other?
If r=1, does F=1? If r=0,does F=∞?Does the sum of all the ∂ tend to 1? If so, is F≤1/r^2?

Please somebody, help me.

19. Apr 8, 2012

### Nabeshin

I apologize if I confused the issue any with my response! v is intended to be the relative velocity of the bodies in question, yes. Keep in mind though, that the expansion I wrote is just that: an expansion. To say that gravity is a function if v/c is a little misleading, then. The ACTUAL equations of gravity are those of general relativity, in which you cannot even write F(r)=??? So again, keeping in mind that we're in this expansion framework, not full general relativity, the equation to first order is this:
http://en.wikipedia.org/wiki/Einstein–Infeld–Hoffmann_equations

As you can see, very complicated! My schematic description was merely to illustrate that it's not the full story.

If r=1, F= whatever, nothing special. In general, though, F(0) is undefined, as you'll be dividing by zero.


text:
Dr Ford

• Prior to the relocation of Clan Wars, he used to be able to heal players for ${\displaystyle \left\lfloor {\text{Hitpoints}}\cdot {\frac {1}{5}}\right\rfloor +2}$ hitpoints an unlimited number of times.

format: plain


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# Tag Info

2

Mathias' answer is perfectly suited for the question as asked. A much better, more maintainable way to accomplish the goal, however, is to use a DHCP server with reservations.

3

If your txt file is just a space-separated list of [computername] [ip-address], you can use the Import-Csv cmdlet: $Computers = Import-Csv -Path C:\file.txt -Delimiter " " -Header Name,IP foreach($Computer in $Computers){ # Query$Computer.Name # Assign $Computer.IP as the static address } 0 The solution was to remove the default gateway from the configuration: DefaultGateway = '192.168.0.1' It appears that if there are any configurations in addition to the basic ones (IPAddress, InterfaceAlias, SubnetMask, AddressFamily), DSC will focus on the additional items and consider the basic ones as references. Consider the following: xIPAddress ... 1 So, there is no way to directly set up file permissions in a GPO using Powershell. (Computer/Policies/Windows Settings/Security Settings/File System) However, I was able to work around this by creating a GPO and manually backing it up (one time thing). In respect to the specific answer I was looking for, there are 3 files that need to be edited in the GPO ... 0 Windows is horrific. Freesshd seems to be the way to go for an SSH Daemon for Windows, and to get Powershell to work at all over SSH was the worst experience I have had with computers in a long time. In the end, with a suggestion from my coworker, this is how I solved it: Initially, I set this up as a Windows Task Scheduler job, and it was fine, but my ... 0 Use WinRM to invoke the script with the proper security context remotely SSH doesn't know how to delegate impersonation (nor should it) also use the scheduler built in to windows which can also invoke with the proper credentials. 2 You seem to have two different issues preventing you from doing what you want here - ExecutionPolicy and SysWOW64 File System Redirection. Bypass Execution Policy To bypass the execution policy, do so when launching PowerShell, like so: PowerShell.exe -ExecutionPolicy Bypass -File .\scriptname.ps1 Bypass File System Redirection Since freeSSHd seems ... 3 You should never configure your VM to Static Ip addresses inside the operating system, use azure portal or powershell instead. Configure a Static Internal IP Address for a VM: https://msdn.microsoft.com/en-us/library/azure/dn630228.aspx Basically you are telling the DHCP server to always give the same ip to your VM. If you don't want to use powershell, ... 1 Manual Configuration of network settings is not supported on Azure, which explicitly requires virtual network adapters to use DHCP, and resets this setting when an adapter is replaced (which usually happens each time the VM is shut down and deallocated). However, you can control which private IP address a VM gets by reserving a static IP, and which DNS ... 0 This worked for me to ADD a new manager (still need to know how to subtract a manager):$mgr = get-user gerald foreach ($thing in$things) { $group = Get-DistributionGroup$thing $managers =$group.ManagedBy $newmanagers =$managers+$mgr Set-DistributionGroup -BypassSecurityGroupManagerCheck -ManagedBy$newmanagers -Identity $thing } 1 You have to do in the computer configuration. Go to: Computer Configuration\Policies\Windows Settings\Security Settings\File System Right-click in the empty space in the right pane and choose "Add File". If whatever system you are on doesn't have the folder you want, you can create it in this file browser. After selecting the folder, assign the NTFS ... 0 I have found the answer:$oldpw = "oldpassword" $newpw = "newpassword"$user = $env:username$computer = $env:computername$user = [adsi]"WinNT://$computer/$user" $user.ChangePassword($oldpw, $newpw) This worked for me. Thank you for your replies! 0 I don't have enough points here to change it, but this question looks like a duplicate of this one. 0 I've noticed the problem: The destination user "osman" I was testing this on was a domain admin and apparently, domain admins do not inherit delegation (no idea why, not documented as far as I can see). All other non-admins work just fine! Thanks for all you suggestions. 0 There's no problem with using a self-signed cert for WinRM https, but when connecting to that computer's endpoint you need to specify that the (winRM) client should skip cert checks. You can do this using:$options = New-PSSessionOption -SkipCACheck -SkipCNCheck -SkipRevocationCheck and then set up the session like this: $session = New-PSSession -UseSSL ... 0 See this answer on Stackoverflow (short answer - you cannot directly): http://stackoverflow.com/questions/5544844/how-to-call-a-complex-com-method-from-powershell 1 You can edit the original AD.msc and add functionality as you like. Here are some links: Custom AD console Right Click password reset Right Click Unlock Add additional columns 0 The only way I know of for a user to set their own password is this command. Set-ADAccountPassword -Identity$Name -Reset -NewPassword (ConvertTo-SecureString $Password -AsPlainText -force) -PassThru It will prompt them for their current password or if you leave out NewPassword it will prompt for both. If you are trying to give a helpdesk type person these ... 6 If these systems are members of an Active Directory domain you can use group policy to add a group of admin users to the Local Administrators group. If these systems are members of a domain but the users are local, you can use Restricted Groups to produce the desired effect. If these computers are not members of an AD domain you can use PowerShell to ... 2 If the end result is to replace "David" with "Gerald" as the owner, avoid the Add/Replace syntax altogether: Set-DistributionGroup Sales -ManagedBy "Gerald" If you must do these two operations atomically (if you have cases where you might not want to replace existing values), split it into two distinct operations: Set-DistributionGroup Sales -ManagedBy ... 2 From https://support.microsoft.com/en-us/kb/2019527 : The purpose of configuring WinRM for HTTPS is to encrypt the data being sent across the wire. WinRM HTTPS requires a local computer "Server Authentication" certificate with a CN matching the hostname, that is not expired, revoked, or self-signed to be installed. So you need to install a ... 0$root = 'C:\logs' Get-ChildItem -Path $root | Where-Object {$.name -match "^\d*$" -and$.LastWriteTime -lt (Get-Date).Adddays(-45) } | Remove-Item -Recurse -Force

3

No - "Parallel" does not support a throttle limit I think you are confusing two distinct functionalities in PowerShell Workflows: Parallel (which is a Workflow Foundation keyword) -Parallel (a Workflow-specific parameter to the foreach loop) The Parallel keyword means: "Each of the following statements/sequences can be executed in parallel". So a ...

0

I use these great Subnet Math functions written by Chris Dent. Store your list of addresses in the pure decimal (not dotted decimal) or pure binary values. Sort the list, then retrieve the list, and finally display the list utilizing the ConvertTo-DottedDecimalIP function. Lots of other great subnet functions in his blog.

0

The code is off in a number of ways: If you are in a Windows Domain environment, why don't you just enable Powershell remoting via GPO? Enter-PSSession is a command for interactive remote sessions, what you are looking for is New-PSSession and Invoke-Command with the session parameter CredSSP is only required if you want to access remote systems from the ...

0

PSCredential encrypts the string using that machines private key. The encrypted data has the same level of protection at rest as it does in motion. The level of encryption is dependent upon the certificate being used, so check your environment's certificate template for machine auto enrollment (if you use it). Alternatively you could just look at the ...

2

That thread Tim linked has good info, here is a native way: (Get-Acl -Path "AD:\CN=John Doe,OU=Users,DC=contoso,DC=com" |  select -ExpandProperty Access) |  select ActiveDirectoryRights,IdentityReference With that, you should be able to get that into an Object and compare the user you're working as with -contains. Here are a couple blogs from Microsoft ...

1

I know this is an old post but I'd like to point out that the ForEach-Object cmdlet doesn't suffer the same issue as using the ForEach key word. So you can pipe the results of DIR to ForEach and just reference the file using $_, like:$backups | ForEach{ Remove-Item _ } You can actually forward the Dir command itself through the pipe and avoid even ... 2 Easiest, native solution according to me. using powershell: Invoke-WebRequest http://your.url/path Save as test.ps1 and just invoke it. Then schedule it using task scheduler. 1 In Powershell 4.0 you can use requires at the top of your script: #Requires -RunAsAdministrator Outputs: The script 'MyScript.ps1' cannot be run because it contains a "#requires" statement for running as Administrator. The current Windows PowerShell session is not running as Administrator. Start Windows PowerShell by using the Run as ... 3 Presumably you're trying to format them this way so that they all show up nicely aligned. The only problem is that 020 isn't the same as 20. Not according to this anyway. Windows for Workgroups with TCP/IP-32 and Windows NT utilities can accept Internet Protocol (IP) addresses comprised of decimal, octal, or hexadecimal numbers. This can cause confusion ... 0 How locked down is your server from an access perspective? Who has access to the server and what can be done if someone does get that password? Most folks could read your script, figure out what the password is used for and ultimately utilize it for malicious purposes. Even if its encrypted. If it was me, I'd look at a few different options: If you ... 2 Install the Windows Update PowerShell Module, then run Get-WUList, it will take a while but will show you the required updates with their size. You can then use Get-WUInstall to download and install them. 2 From Preview Portal: Browse -> Virtual Machines -> Select VM -> All Settings -> Extensions -> Add -> Select Extension -> Create. Using Powershell: https://msdn.microsoft.com/en-us/library/azure/dn606311.aspx Each extension has different pre-requisites so it's not only a matter of installing the extension using powershell commands. Check the above list ... 1 To actually return an object of type Microsoft.ActiveDirectory.Management.ADOrganizationalUnit of a User Object, use this: Get-ADOrganizationalUnit -Identity(($adUser = Get-ADUser -Identity$env:USERNAME).DistinguishedName.SubString($adUser.DistinguishedName.IndexOf("OU="))) The same can be accomplished for a computer: Get-ADOrganizationalUnit ... 0 When executing the script in the ISE or also in the console, everything runs fine. No errors or anything else that would stand out. Runs fine doesn't means there is no errors encountered. Your script probably just ignores them, using -ErrorAction SilentlyContinue parameter with cmdlets. Example: Get-ChildItem -LiteralPath ZZZ:\ This will generate ... 0 What else can I do to free up some space? Try Disk Cleanup first. Here is how to run it if you don't have Desktop Expirience installed, or don't want to install it. To run Disk cleanup: Click Start Click All Programs → Accessories → System Tools → Disk Cleanup Select Drive C from the drop-down menu Click OK 0 Try this. You can add any amount of folders, file types and file counts to the vairable$FoldersToCheck: # File to store log $LogFile = '.\FileCount.log'$FoldersToCheck = @( @{ Path = 'C:\path\to\folder' FileType = '*.cab' FileCount = 13 }, @{ Path = 'C:\path\to\folder\subfolder' FileType = '*.txt' ...

0

When you get an error do $error[0] | fl * -f You get a better stack and it gives you an idea of what was attempted. Looks like you maybe having issues authenticating to a web proxy in order to get to the Azure endpoint. What authentication does your web proxy (i assume its websense) support? I hope the proxy isn't set just for basic auth support. It ... 1 You don't have to be the global admin for an entire SharePoint Online deployment to upload files to a single library, so no, it can certainly be done 4 I think the problem might be with the way you're generating the credential using ConvertTo-SecureString. By default, that cmdlet uses an encryption key specific to the current host (I think) unless you provide an explicit encryption key with the -Key parameter. On the remote side, it needs to decrypt the string using the same encryption key which it ... 1 Find something else to delete first, like IIS logs. 1 unfortunately search-mailbox cmdlet is no longer available. not sure why they removed it. There is no option to recover ONLY recoverable items.. 2 You can easily copy the contents of a file over-the-wire through a PSRemoting session using Invoke-Command and Set-Content:$Session = New-PSSession -ComputerName "remotehost.domain.tld" -Credential (Get-Credential) -UseSsl $FileContents = Get-Content -Path 'C:\path\to\arbitrary.file' Invoke-Command -Session$Session -ScriptBlock { ...

2

Get-Process won't get you anywhere. Assign the Count to a variable and test if the value of that variable is then 13: $cabFileCount = (Get-ChildItem -Filter "*.cab" "C:\path\to\folder").Count Write-Host$cabFileCount if($cabFileCount -eq 13){ # Success! Write-Host "$cabFileCount files found, perfect!" } else { # Failure! Write-Host ...

0

As of Windows 8.1 / Server 2012 R2, there are built-in PowerShell cmdlets for managing firewall rules, and netsh advfirewall has a message that it may be removed in a future version of windows. For this specific case: Get-NetFirewallRule -DisplayGroup "Remote Desktop" | Set-NetFirewallRule -Enabled True You can see the rest with: Get-Help *firewall* ...

0

Here is what works for me, using Powershell v4. PS C:\Scripts> Start-Process cmd -ArgumentList "/c 1.cmd" -WorkingDirectory c:\test4 Where 1.cmd lives in c:\test4

0

I'm not sure what exactly is the process behind the WebClient::DownloadFile method, but it seems to me that it interrupts the ftp connection right before the download transfer completes causing the operation to fail. By using the asynchronous method I managed to successfully download the file. Here's my updated function using the ...

0

Try using -erroraction silentlycontinue

1

This is due to the version of PowerShell you are running. This module runs with 2.0 ONLY. If you are getting the error run 2.0 by running "powershell.exe -version 2.0" Then import the module.

Top 50 recent answers are included


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# Astroparticle Physics - A Joint TeVPA/ IDM Conference

23-28 June 2014
Amsterdam
Europe/Zurich timezone

## HESS J1640$-$465 - an exceptionally luminous TeV gamma-ray SNR

25 Jun 2014, 14:35
15m
Room 4 (Tuschinski Theatre)

### Room 4

#### Tuschinski Theatre

Presentation Gamma-Ray Astrophysics

Stefan Ohm (D)

### Description

HESS J1640$-$465 is among the brightest Galactic TeV gamma-ray sources ever discovered by the High Energy Stereoscopic System (H.E.S.S.). Its likely association with the shell-type supernova remnant (SNR) G338.3$-$0.0 at a distance of $\sim$10 kpc makes it the most luminous Galactic source in the TeV regime. Our recent analysis of follow-up observations with H.E.S.S. reveal a significantly extended TeV morphology with a substantial overlap with the northern part of the SNR shell. Furthermore, the source features a seamless powerlaw spectrum over four orders of magnitude from GeV to TeV energies, with a spectral index of $\Gamma = 2.15\pm 0.10_\mathrm{stat}\pm 0.10_\mathrm{sys}$ and a cut-off energy of $E_c = 7.3^{+2.5}_{-1.8}$ TeV. These new spectral and morphological results suggest that at least part of the TeV emission is likely of hadronic origin where the product of total proton energy and mean target density could be as high as $W_p n_H \sim 4 \times 10^{52}(d/10\mathrm{kpc})^2$ erg cm$^{-3}$.

### Primary authors

H.E.S.S. Collaboration (http://www.mpi-hd.mpg.de/hfm/HESS/) Dr Peter Eger (MPIK, Heidelberg) Stefan Ohm (D)

### Presentation Materials

 HESSJ1640_TeVPA_Ohm_2014.pdf HESSJ1640_TeVPA_Ohm_2014.pptx


text:
Trim to the nearest word under limit

Given a text and the length limit I need to trim the text to the word which is nearest to the limit.

function trim(text, limit) {
if (text.length < limit) return text;

let end = -1;
var i;
for (i = 0; i < limit - 1; i++) {
let current = text.charAt(i);
let next = text.charAt(i + 1);
if (current !== ' ' && next === ' ') {
end = i;
}
}
if (i === text.length - 1) {
end = i;
}

return text.substring(0, end + 1);
}

[0, 2, 4, 6, 8, 10, 12, 50].forEach((limit) => {
console.log(trim('Hello World', limit));
});

O/P:

//

Hello
Hello
Hello
Hello World
Hello World

The current code has a bug when the limit is exactly is exactly the length of a word. For example,

trim('Hello World', 5)

will return an empty string, when it should return "Hello", since it is of length 5, therefore it fits in the length limit.

The issue stems from the loop over i going from 0 to limit - 1 (excluded). The variable end represents the index of the last character to keep in the string, (because the method is returning text.substring(0, end + 1)). Therefore, it could be the case that end is equal to limit - 1, this happens when the limit falls right before a space. For example, when the limit is 5, and the string is 'Hello World', the last character to keep is 'o' at index 4. So the loop should go up to limit (excluded).

Then, there are slight adjustements to make:

• The early-return testing text.length < limit should be text.length <= limit (because otherwise the next character in the loop over i won't exist for the last character of the string);
• The added check if (i === text.length - 1) can safely be removed as a consequence: the only case where i would be equal to text.length - 1 is if limit == text.length (since i goes up to limit - 1 included). But that was already handled with the modified early-return.

There is a simpler and efficient way to tackle this problem, using the built-in lastIndexOf(searchValue, fromIndex). This method starts at the specified fromIndex, and searches backwards the first occurence of the given value. If we tell it to search the first occurence of ' ' searching backwards from limit, we'll exactly have the length of the substring to return.

The only corner case is if limit is greater than the length of the string, in which case the method should simply return the text unchanged (like the current code is doing). It needs to be tested separately; otherwise lastIndexOf would start searching from the end of the string, and the last word would always be cut. This is what it could look like:

function trim(text, limit) {
if (limit >= text.length) {
return text;
}
return text.substring(0, text.lastIndexOf(' ', limit));
}

[0, 2, 4, 5, 6, 8, 10, 11, 12, 50].forEach((limit) => {
console.log(limit + ' -> ' + trim('Hello World', limit));
});

Note that, even if the lastIndexOf call returns -1, because there was no space character found before the limit length, it is still safe to call substring since:

If either argument is less than 0 [...], it is treated as if it were 0.

• If you could help me with reducing off by one errors it would be good, do you have some method to avoid that? because I spend a considerable amount of time handling that. – CodeYogi Jan 7 '17 at 15:02
• @CodeYogi Fair enough. I edited with the issues. – Tunaki Jan 7 '17 at 15:31
• What if I have replicate the same logic in ruby since I find no such method in it? – CodeYogi Jan 8 '17 at 18:21
• @CodeYogi I'm afraid I do not know Ruby at all... From what I searched, perhaps a combination of substring and rindex. – Tunaki Jan 8 '17 at 18:28


text:
# zbMATH — the first resource for mathematics

##### Examples
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Boundedness results for a certain third order nonlinear differential equation. (English) Zbl 1201.34055

The authors consider a class of third order non-linear differential equations of the form

$\stackrel{⃛}{x}+f\left(\stackrel{¨}{x}\right)+g\left(\stackrel{˙}{x}\right)+h\left(x\right)=p\left(t,x,\stackrel{˙}{x},\stackrel{¨}{x}\right),\phantom{\rule{2.em}{0ex}}\left(1\right)$

where $f,g,h\in C\left(ℝ,ℝ\right)$, $p\in C\left({ℝ}^{+}×{ℝ}^{3},ℝ\right)$, ${ℝ}^{+}=\left[0,\infty \right)$. The authors give some new sufficient conditions which guarantee the boundedness and uniform ultimate boundedness of the solutions (1). By defining an appropriate Lyapunov function, they prove their main results.

##### MSC:
 34C11 Qualitative theory of solutions of ODE: growth, boundedness
##### Keywords:
third order nonlinear differential equation
##### References:
 [1] Ademola, T. A.; Ogundiran, M. O.; Arawomo, P. O.; Adesina, O. A.: Stability results for the solutions of a certain third order nonlinear differential equation, Math. sci. Res. J. 12, No. 6, 124134 (2008) · Zbl 1198.34083 [2] Afuwape, A. U.: Further ultimate boundedness results for a third order nonlinear system of differential equations, Boll. un. Mat. ital. C (6) 4, No. 1, 347-361 (1985) · Zbl 0592.34024 [3] A.U. Afuwape, Uniform ultimate boundedness results for some third order nonlinear differential equations, Int Center for Theor. Phys., IC/90/405, November, 1990, pp. 1 – 14. [4] Afuwape, A. U.: Remarks on barbashin – ezeilo problem on third order nonlinear differential equations, J. math. Anal. appl. 317, 613-619 (2006) · Zbl 1099.34039 · doi:10.1016/j.jmaa.2005.05.068 [5] Afuwape, A. U.; Adesina, O. A.: On the bounds for mean-values of solutions to certain third order nonlinear differential equations, Fasc. math. 36, 5-14 (2005) · Zbl 1127.34019 [6] Andres, J.: Boundedness results for solutions of the equation x$·+ax$¨+$g\left(x\right)$x˙+$h\left(x\right)=p\left(t\right)$ without the hypothesis $h\left(x\right)$sgnx$⩾0$ for ∣x∣>R, Atti. accad. Naz. lincei rend. Cl. sci. Fis. mat. Natur. 80, No. 8, 533-539 (1986) [7] Bereketog&caron, H.; Lu; Györi, I.: On the boundedness of solutions of a third order nonlinear differential equation, Dyn. syst. Appl. 6, No. 2, 263-270 (1997) [8] Ezeilo, J. O. C.: A note on a boundedness theorem for some third order differential equations, J. London math. Soc. 36, 439-444 (1961) · Zbl 0104.06501 · doi:10.1112/jlms/s1-36.1.439 [9] Ezeilo, J. O. C.: An elementary proof of a boundedness theorem for a certain third order differential equation, J. London math. Soc. 38, 11-16 (1963) · Zbl 0116.06902 · doi:10.1112/jlms/s1-38.1.11 [10] Ezeilo, J. O. C.: Further results for the solutions of a third order differential equation, Proc. camb. Phil. soc. 59, 111-116 (1963) · Zbl 0115.30501 [11] Ezeilo, J. O. C.: A boundedness theorem for a certain third order differential equation, Proc. London math. Soc. 13, No. 3, 99-124 (1963) · Zbl 0108.09001 · doi:10.1112/plms/s3-13.1.99 [12] Ezeilo, J. O. C.: A generalization of a boundedness theorem for the equation x$·+x$¨+$\varphi 2$(x˙)+$\varphi 3\left(x\right)=\psi$(t,x,x˙,x¨), Atti. accad. Naz. lincei rend. Cl. sci. Fis. mat. Natur. 50, No. 13, 424-431 (1971) [13] Ezeilo, J. O. C.; Tejumola, H. O.: Boundedness theorems for certain third order differential equations, Atti. accad. Naz. lincei rend. Cl. sci. Fis. mat. Natur. 55, 194-201 (1973) · Zbl 0295.34022 [14] Hara, T.: On the uniform ultimate boundedness of solutions of certain third order differential equations, J. math. Anal. appl. 80, 533-544 (1981) · Zbl 0484.34024 · doi:10.1016/0022-247X(81)90122-0 [15] Qian, C.: Asymptotic behavior of a third order nonlinear differential equation, J. math. Anal. appl. 284, No. 1, 191-205 (2003) · Zbl 1054.34078 · doi:10.1016/S0022-247X(03)00302-0 [16] Reissig, R.; Sansone, G.; Conti, R.: Nonlinear differential equations of higher order, (1974) · Zbl 0275.34001 [17] Rouche, N.; Habets, N.; Laloy, M.: Stability theory by Lyapunov’s direct method, Appl. math. Sci. 22 (1977) · Zbl 0364.34022 [18] Swick, K. E.: Asymptotic behavior of the solutions of certain third order differential equations, SIAM J. Appl. 19, No. 1, 96-102 (1970) · Zbl 0212.11403 · doi:10.1137/0119008 [19] Swick, K. E.: Boundedness and stability for a nonlinear third order differential equation, Atti. accad. Naz. lincei rend. Cl. sci. Fis. mat. Natur. 56, No. 20, 859-865 (1974) · Zbl 0326.34062 [20] Tejumola, H. O.: A note on the boundedness of solutions of some nonlinear differential equations of the third order, Ghana J. Of sci. 11, No. 2, 117-118 (1970) [21] Tejumola, H. O.: Periodic boundary value problems for some fifth, forth and third order ordinary differential equations, J. nigerian math. Soc. 25, 37-46 (2006) [22] Tunç, C.: On the asymptotic behavior of solutions of certain third order nonlinear differential equations, J. appl. Math. stoc. Anal. 1, 29-35 (2005) · Zbl 1077.34052 · doi:10.1155/JAMSA.2005.29 [23] Tunç, C.: Boundedness of solutions of a third- order nonlinear differential equation, J. inequal. Pure appl. Math. 6, No. 1, 1-6 (2005) · Zbl 1082.34514 [24] Yoshizawa, T.: Stability theory and existence of periodic solutions and almost periodic solutions, (1975)


text:
# Trapezoid ABCD v2

Trapezoid ABCD has length of bases in ratio 3:10. The area of riangle ACD is 825 dm2. What is the area of trapezoid ABCD?

Correct result:

S =  1072.5 dm2

#### Solution:

$a/c = 3/10 \ \\ S_{ACD}= 825 = ch / 2 \ \\ h = 2S_{ACD}/c \ \\ \ \\ S = \dfrac{a+c}{2} h = \dfrac{a+c}{2} \dfrac{2 S_{ACD}}{c} \ \\ S = (a+c)S_{ACD}/c \ \\ S = S_{ACD}(a/c+1) = 825\cdot (3/10+1) \ \\ S = 1072.5 \ \text{dm}^2$

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text:
# How do you solve and check your solution given -8.4=-6f?

May 5, 2017

I got $f = 1.4$

#### Explanation:

I would multiply both sides by $10$ to get:

$10 \cdot \left(- 8.4\right) = 10 \cdot \left(- 6 f\right)$

$- 84 = - 60 f$

rearrange:

$f = \frac{- 84}{- 60} = \frac{84}{60}$

divide:

So that $f = 1.4$

To check your solution let us substitute this value of $f$ into the original:

$- 8.4 = - 6 \cdot \textcolor{red}{1.4}$

$- 8.4 = - 8.4$ ok


text:
Article

# On the presence of thermal Sunyaev-Zel'dovich induced signal in the first-year WMAP temperature maps

Monthly Notices of the Royal Astronomical Society (Impact Factor: 5.52). 01/2004; 347(2):403-410. DOI: 10.1111/j.1365-2966.2004.07205.x
Source: arXiv

ABSTRACT Using available optical and X-ray catalogues of clusters and
superclusters of galaxies, we build templates of thermal
Sunyaev-Zel'dovich (tSZ) emission as they should be detected by the WMAP
experiment. We compute the cross-correlation of our templates with WMAP
temperature maps, and interpret our results separately for clusters and
for superclusters of galaxies. For clusters of galaxies, we claim
2-5σ detections in our templates built from the BCS (Ebeling et
al.), NORAS (Böhringer et al.) and de Grandi et al. catalogues. In
these templates, the typical cluster temperature decrements in WMAP maps
are around 15-35 μK in the Rayleigh-Jeans range (no beam
deconvolution applied). Several tests probing the possible influence of
foregrounds in our analyses demonstrate that our results are robust
against Galactic contamination. On supercluster scales, we detect a
diffuse component in the V and W WMAP bands, which cannot be generated
by superclusters in our catalogues (Einasto et al.), and which is not
present in the clean map of Tegmark et al. Using this clean map, our
analyses yield, for Einasto's supercluster catalogues, the following
upper limit for the Comptonization parameter associated with
supercluster scales: ySC < 2.18 × 10-8 at
the 95 per cent confidence level.

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• Source
##### Article: The Integrated Sachs-Wolfe Effect in Interacting Dark Energy Models
[Hide abstract]
ABSTRACT: Models with dark energy decaying into dark matter have been proposed in Cosmology to solve the coincidence problem. We study the effect of such coupling on the cosmic microwave background temperature anisotropies. The interaction changes the rate of evolution of the metric potentials, the growth rate of matter density perturbations and modifies the Integrated Sachs-Wolfe component of cosmic microwave background temperature anisotropies, enhancing the effect. Cross-correlation of galaxy catalogs with CMB maps provides a model independent test to constrain the interaction. We particularize our analysis for a specific interacting model and show that galaxy catalogs with median redshifts $z_m=0.1-0.9$ can rule out models with an interaction parameter strength of $c^2\simeq 0.1$ better than 99.95% confidence level. Values of $c^2\le 0.01$ are compatible with the data and may account for the possible discrepancy between the fraction of dark energy derived from WMAP 3yr data and the fraction obtained from the ISW effect. Measuring the fraction of dark energy by these two methods could provide evidence of an interaction.
Physical review D: Particles and fields 02/2008;
• Source
##### Article: Spectral indications of thermal Sunyaev-Zel'dovich effect in ARCHEOPS and WMAP data
[Hide abstract]
ABSTRACT: In this paper, we present a joint cross-correlation analysis of the Archeops CMB maps at 143 and 217 GHz and the WMAP CMB maps at 41, 61 and 94 GHz with sky templates of projected galaxy density constructed from the 2MASS Extended Source catalog. These templates have been divided in patches sorted in decreasing galaxy density with a fixed number of pixels (we considered patches having 32, 64 and 128 pixels) and the cross correlation has been performed independently for each of these patches. We find that the densest patch shows a strong temperature decrement in the $Q$, $V$, $W$ bands of WMAP and in the 143 GHz channel of Archeops, but not in the 217 GHz channel. Furthermore, we find that the spectral behavior of the amplitude of this decrement is compatible with that expected for the non-relativistic thermal Sunyaev-Zel'dovich effect, and is incompatible (at 4.5$\sigma$ level) with the null hypothesis of having only CMB, noise and a dust component (${\propto}\nu^2$) in those pixels. We find that the first 32-pixel sized patch samples the cores of 11 known massive nearby galaxy clusters. Under the assumption that the decrement found in this patch is due entirely to the thermal Sunyaev-Zel'dovich effect, we obtain an average comptonization parameter for those sources of $y = (0.39 \pm 0.07) \times 10^{-4}$ at 13 arcmin angular scales. This value is compatible at 1-$\sigma$ with the expectation, $y = 0.49 \times 10^{-4}$, from a model of the cluster flux number counts based on the standard $\Lambda$-CDM cosmology (Xue & Wu 2001). The observed value of $y$ is slightly diluted when computed for the first patch of 64 and 128 pixels, presumably due to the inclusion of less massive clusters, and the dilution factor inferred is also compatible with the quoted model.
Astronomy and Astrophysics 04/2006; · 5.08 Impact Factor
• Source
##### Article: Constrain intergalactic medium from the SZ effect map
[Hide abstract]
ABSTRACT: In this paper, we try to detect the SZ effect in the 2MASS DWT clusters and less bound objects in order to constrain the warm-hot intergalactic medium distribution on large scales by cross-correlation analysis. The results of both observed WMAP and mock SZ effect map indicate that the hot gas distributes from inside as well as outside of the high density regions of galaxy clusters, which is consistent with the results of both observation and hydro simulation. Therefore, the DWT measurement of the cross-correlation would be a powerful tool to probe the missing of baryons in the Universe. Comment: 9 pages,2 figures. Accepted for publication in Mod. Phys. Lett. A
Modern Physics Letters A 03/2006; · 1.11 Impact Factor


text:
How to compute number of ways one S and one F can be chosen from group of (2F, 2S, 2J, 2R)?

I thought I knew combinations pretty well, until recently I came across this question:

In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist one freshman and one sophomore?

The way I approached this problem was to compute $$\frac{N}{D}$$, where

• $$N =$$ number of ways a sophomore and freshman can be chosen
• $$D$$ = total number of ways students can be chosen from different grades

The I broke it down as follows:

• $$D$$ = total number of ways to choose 2 students - number of ways to choose students from the same grade. So, $$D = C(n=8, k=2) - 4 = 28 - 4= 24$$
• $$N$$ = 1, because there is only one way to choose a sophomore and a freshman.

But this gets me $$\frac{1}{24}$$, which is incorrect. I understand there are other ways of doing this problem, but I'm interested in knowing where I went wrong in my logic/approach of the problem?

Thanks.

You have correctly calculated the denominator. However, there are four ways to select a freshman and a sophomore since there are two ways to select a freshman and two ways to select a sophomore. Hence, the probability that a freshman and a sophomore are selected for the competition is $$\Pr(\text{selected students are a freshman and a sophomore}) = \frac{\binom{2}{1}\binom{2}{1}}{\binom{8}{2} - \binom{4}{1}\binom{2}{2}} = \frac{4}{24} = \frac{1}{6}$$ assuming each individual is equally likely to be chosen.
• Right...I realized this myself too. In fact, there's an easier way to compute the denominator if you consider multiplying these two together: 1) number of ways to choose which two different grades to choose from = $C(4,2) = 6$, and 2) number of ways to choose the 2 individuals from those 2 grades = 4 (as you pointed out). 4*6 = 24. thanks – Sother Aug 26 at 15:53


text:
# [Tex/LaTex] tlmgr: Remote repository is newer than local (2018 < 2019)

babeldebiantexlivetlmgr

I know this question looks like a duplicate but it's not, all the other solutions are not working for me.

I'm using debian 10 buster and I installed texlive with: sudo aptitude install texlive texlive-base, then I tried to compile a document and got this error:

Package babel Error: Unkown option french'. Either you misspelled it or the language definition file french.ldf was not found

Next I tried this:

tlmgr init-usertree

then this:

tlmgr install babel-french

but I got this error:

tlmgr: Remote repository is newer than local (2018 < 2019)
Cross release updates are only supported with update-tlmgr-latest(.sh/.exe) --update

Debian packages a complete TeXLive, so you can install matching packages from the Debian repositories. In your case you should install texlive-lang-french via apt-get (or texlive-full` if you want to avoid this in the future).


text:
Question

# Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.

A
1 : 2
B
2 : 3
C
3 : 2
D
5 : 3

Solution

## The correct option is B 2 : 3$$f_1 = (2m)(acc.)$$                       $$= 2m \times \dfrac{P}{5m} = \dfrac{2}{5} P$$                 $$f_2 = (3m) (acc.)$$                      $$= 3m \times \dfrac{P}{5m} = \dfrac{3}{5} P$$ $$\dfrac{f_1}{f_2} = \dfrac{ 2 P/5}{3P/5} = \dfrac{2}{3}$$ Physics

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text:
# Prove that the series $\sum_{n=1}^{\infty}\int_{n}^{n+1}(\frac{1}{n^z} - \frac{1}{x^z})dx$ converges uniformly on $\mathbb{C}_{\epsilon}$

I was taking my Ph.D Analysis qualifying exam and then they asked me to prove the following exercise.

Let $\epsilon > 0$ and $\mathbb{C}_\epsilon := \{z \in \mathbb{C}, \Re(z) > \epsilon \}$.

(a) $$F(z) = \sum_{n = 1}^{\infty}\frac{1}{n^z}$$ converges uniformly on $\mathbb{C}_{1 + \epsilon}$.

I prove this part easily using the Weierestrass M-Test bounding the terms of this series by a p-series with $p > 1$ and the uniform convergence follows.

(b) Prove that the series $$\sum_{n=1}^{\infty}\int_{n}^{n+1}(\frac{1}{n^z} - \frac{1}{x^z})dx$$ converges uniformly on $\mathbb{C}_{\epsilon}$

(c) Prove that there exists an analytic function $G$ on $\{z \in \mathbb{C}, \Re(z) > 0\}$ such that $$G(z) = F(z) - \frac{1}{z-1}$$ for $\Re(z) > 1$

In part (b) I try to bound the terms as in part (a) but I was not able to show the uniform convergence on $\mathbb{C}_{\epsilon}$. However in my attempt I can show that it was uniformly convergent on $\mathbb{C}_{1+\epsilon}$ but I think it must be a mistake.

For part (c) I have no idea how to approach.

Partial answer to b). I will prove uniform convergence on bounded subsets of $\mathcal C_\epsilon$. I doubt if the convergence is uniform on the whole of $\mathcal C_\epsilon$. Let $f(x)=x^{-z}$. Then $|f(x)-f(n)| \leq |x-z| |z|n^{-\Re z-1}$ $\leq |z| n^{-\Re z-1}$ for x between n and $n+1$. Hence $|\int _n ^{n+1} |f(x)-f(n)|dx \leq |z| n^{-\Re z-1}$ proving uniform convergence of the series in b) for $|z|$ bounded.

Proof of c): let $H(z)= \sum _1 ^{\infty} \int _n ^{n+1} x^{-z} dx+\frac 1 {1-z}$. Since $\int _n ^{n+1} x^{-z} dx=\frac {(n+1)^{-z} -n^{-z}} {1-z}$, The N-th partial sum becomes $\frac {(N+1)^{-z} -1} {1-z}$ which converges to $- \frac 1 {1-z}$. It follows that $H(z)=0$ in $\mathcal C_\epsilon$. By b) $F(z)-H(z)- \frac 1 {1-z}$ is analytic in Re (z) >1.

• Thank you for this. Really appreciated. – Richard Clare Jan 13 '18 at 2:20


text:
Percent Composition Calculator + Online Solver With Free Steps

The online Percent Composition Calculator is a calculator that finds the percentage composition of a chemical compound.

The Percent Composition Calculator is an easy-to-use calculator that helps chemists and scientists quickly determine the percentage composition of a chemical compound.

What Is a Percent Composition Calculator?

A Percent Composition Calculator is an online calculator that allows you to determine the percentage composition of a chemical compound.

The Percent Composition Calculator takes only a single input: the compound’s chemical formula. The calculator then displays a range of details about the percentage composition of the compound.

How To Use the Percent Composition Calculator?

To use the Percent Composition Calculator, you input the compound’s chemical formula into the calculator and click the “Submit” button.

The detailed instructions for using the Percent Composition Calculator are given below:

Step 1

To use the Percent Composition Calculator, you need first enter the compound’s chemical formula in its respective box.

Step 2

After entering the compound’s chemical formula into the Percent Composition Calculator, you click the “Submit” button. The calculator will open a new window and display the chemical compound’s detailed properties and percentage composition.

How Does a Percent Composition Calculator Work?

The Percent Composition Calculator works by taking the compound’s chemical formula and a ratio between the elements in the mixture. The ratio is then multiplied by 100 to give us a percentage.

The following formula can be used to calculate the percentage composition of a compound:

$\% C_{E}= \frac{g^{E}}{g^{T}} \times 100$

What Is a Percent Composition?

The percent composition of any given compound is the ratio of the quantity of each element contained in the compound to the total amount of individual components that are present in the compound multiplied by 100. In this case, the quantity is expressed in grams of the elements contained in the solution.

The percent composition of any chemical is an expression of its composition in terms of all the elements present. The chemical analysis reveals the importance of this composition calculation.

The following formula can represent the percentage composition:

$\% C_{E}= \frac{g^{E}}{g^{T}} \times 100$

The percent composition of an element E is represented by percent CE. This is the value we will compute. The numerator on the right side represents the total quantity of element E present in the compound. The denominator, however, represents the total amount of all the elements in the mixture.

What Is Mass Percent Composition?

The mass percent composition represents the concentration of an element in a compound or a component in a combination. This word refers to the overall percentage by mass of each element present in a compound.

It is vital to remember that the mass percent composition can be calculated by dividing the mass of a component by the total mass of the combination. This proportion is then multiplied by 100. It is also known as the mass percent $(\frac{w}{w})\%$.

This formula is useful for determining the smallest entire number of moles and the relative number of atoms in each element of a compound. Chemists can derive the real molecular formula using the empirical formula. This formula specifies the number of atoms in the chemical.

Solved Examples

The Percent Composition Calculator instantly allows you to find the percentage composition of a chemical compound.

The following examples are solved using the Percent Composition Calculator:

Example 1

While experimenting, a chemist comes across the following strong acid:

$H_{2}SO_{4}$

The chemist needs to find the percentage composition of the strong acid to complete his experiment. Use the Percent Composition Calculator to find the percent composition of the strong acid.

Solution

Using the Percent Composition Calculator, you can easily determine the percentage composition of the chemical compound. First, we input the compound’s chemical formula into the Percent Composition Calculator; the compound’s chemical formula is $H_{2}SO_{4}$.

Once we have input the compound’s chemical formula into the Percent Composition Calculator, we click the “Submit” button. The calculator quickly finds the percentage composition of the compound along with its properties in a new window.

The following results are taken from the Percent Composition Calculator:

Sulphuric Acid

Chemical Names and Formulas

$\text{Formula} = H_{2}SO_{4}$

$\text{Hill \ Formula} = H_{2}O_{4}S$

Name = Sulphuric Acid

Figure 1

Properties

Here are some basic properties of sulphuric acid determined using this calculator:

Molar mass = 98.07 g/mol

Phase = liquid (at STP)

Melting Point = $10.371^{\circ}$C

Boiling Point = $279.6^{\circ}$C

Density = 1.8305 g/$cm^{3}$

Solubility in water = very soluble

Liquid Properties (at STP)

Here are the liquid properties of Sulphuric acid at standard temperature and pressure:

Density = 1.8305 g/$cm^{3}$

Vapor Pressure = 6 x $10^{-5}$ mmHg

Dynamic Viscosity = 0.021 Pas (at $25^{\circ}$C)

Surface Tension = 0.0735 N/m

Refractive Index = 1.41827

Thermodynamic Properties

Thermodynamic properties are given below:

Specific heat capacity cp (liquid) = 1.416 J/(gK)

Specific free energy of formation $\Delta_{f}G^{\circ}$ (liquid) = -7.036 kJ/g

Specific heat of formation $\Delta_{f} H^{ \circ}$ (aqueous) = -9.269 kJ/g

Specific heat of formation $\Delta_{f} H^{ \circ}$ (liquid) = -8.3 kJ/g}

Specific entropy $S^{\circ}$ (aqueous) = 0.2039 J/(gK)

Specific entropy $S^{\circ}$ (liquid) = 1.601 J/(gK)

Specific heat of vaporization = 0.57 KJ/J

Specific heat of fusion = 0.1092 KJ/g

Chemical Identifiers

Every chemical has certain associated numbers. For Sulphuric acid various identifiers are listed below:

CAS number = 7664-93-9

Beilstein number = 2037554

PubChem CID Number = 1118

PubChem SID Number = 24859176

SMILES identifier = OS(=O)(=O)O

Toxicity Property

The following properties identify the smell and odor of the chemical substance:

Odor = odorless

Threshold limit value = 0.05 ppmv

Ion Equivalent

The ionic composition is given below:

$H^{+}$ (hydrogen cation) = 2

$SO_{4}^{2-}$ (sulfate anion) = 1

Example 2

A scientist needs to find the percentage composition of hydrochloric acid HCl. Using the Percent Composition Calculator, find the percent composition of hydrochloric acid.

Solution

You can simply determine the percentage composition of a chemical substance using the Percent Composition Calculator. First, we enter the compound’s chemical formula into the Percent Composition Calculator; the compound’s chemical formula is HCl.

We click the “Submit” button after entering the compound’s chemical formula into the Percent Composition Calculator. The calculator instantly determines the compound’s percentage composition and attributes in a new window.

The following results are extracted from the Percent Composition Calculator:

Input Interpretation

The input entered in the calculator is interpreted in the following manner:

Hydrogen chloride

Chemical Name and Formula

The chemical name and formula of the acid are given below:

Formula = HCl

Name = hydrogen chloride

Figure 2

Basic Properties

Molar mass = 36.46 g/mol

Phase = gas (at STP)

Melting Point = $-114.17^{\circ}C$

Boiling Point = $-85^{\circ}C$

Density = 0.00149 g/$cm^{3}$ (at $25^{\circ}$C

Solubility in water = miscible

Gas Properties (at STP)

Gaseous properties of HCL at standard pressure and temperature conditions are given below:

Density = 0.00149 g/$cm^{3}$ (at $25^{\circ}$C

Vapor density = 1.3 (relative to air )

Molar Volume = 24470 $cm^{3}$/mol}

Thermodynamic Properties

Listed below are some of the thermodynamic properties of HCL:

Specific heat capacity cp (gas) = 0.7982 J/(gK)

Specific free energy of formation $\Delta_{f}G^{\circ}$ (gas) = -2.614 kJ/g

Specific heat of formation $\Delta_{f} H^{ \circ}$ (gas) = -2.532 kJ/g}

Specific entropy $S^{\circ}$ (gas) = 5.129 J/(gK)

Specific heat of vaporization = 0.414 \KJ/J

Specific heat of fusion = 0.05 KJ/g}

Critical Temperature = 325K

Critical Pressure = 8.27 MPa

Toxicity Property

Following are some of the toxicity properties of HCL:

Short term exposure limit = 7 mg/$m^{3}$

Ion Equivalent

The ion equivalence of HCL is given below:

$H^{+}$  (hydrogen cation) = 1

$Cl^{-}$(chloride anion) = 1

All Images/graphs are made using GeoGebra.


text:
×

# RAINBOWA python code wrong output

here is my python code:

q="1234567"

k=[]

# function to omit test cases like 2 1 3 4 5 6 7 6 5 4 3 1 2

def is_increasing(n):

for i in range(int(len(n)/2)):

if n[i]>n[i+1]:

return False

return True

# function to check given array a rainbow array

def is_rainbow(n):

for i in range(len(n)):

if n[i]==n[len(n)-i-1] and str(n[i]) in q and is_increasing(n):

continue

else:

return False

return True

x=int(input())

for i in range(x):

s1=int(input())

s=list(map(int,input().split()))

if (is_rainbow(s)):

k.append(1)

else:

k.append(0)

for l in k:

if l==1:

print("yes")

else:

print("no")

can you please tell me whats wrong in my code.

11
accept rate: 0%

 0 Where are you calling the "is increasing" function? Also, what if your array is like 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1 Above is not a rainbow array as it has an element more than 7, i.e. 8. answered 09 Sep '17, 14:58 15.2k●1●18●59 accept rate: 18%
 0 Yep, @vijju123 is right. answered 09 Sep '17, 19:23 11●2 accept rate: 0%
 0 Here is my solution in python ! what i essentially did was checked for palindrome and increment the checking factor by one starting with zero and also note the middle element has to be 7 and rest all cases are false answered 09 Sep '17, 23:35 94●4 accept rate: 0%
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text:
# Key terms and concepts for sections 14.7-14.8

• The curl of a vector field. Mnemonic: $\nabla\times\mathbf F$.
• Stokes' theorem:

$\oint_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r} = \iint_S\mathrm{curl}\,\mathbf{F}\cdot\mathbf{n}\,\mathrm d\sigma$

• Gradient fields have zero curl, and curls have zero divergence:

$\mathrm{curl}\,\mathrm{grad}\, f = 0 \quad\quad\quad\quad \mathrm{div}\,\mathrm{curl}\,\mathbf{F} = 0$


text:
# Fluid Statics

## Introduction

1. The branch of physics which deals with study of fluid at rest is called Hydrostatics.$\\$“Hydro”, meaning water/water like fluid$\\$“Statics”, meaning at rest$\\$

2. A fluid is a substance that flows from one point to another such as liquid and gas.

## Density

1. The ratio of mass of a body to its volume is called density.$\\$Mathematically, Density(ρ)=Mass(m)/Volume(v) $\to$ $ρ = \frac{m}{v}$

2. When two liquids of equal volume are mixed whose densities are $d_1$ and $d_2$ :$\\$Density of mixture =$\frac{d_1+d_2}{2}$

3. When two liquids of equal mass are mixed whose densities are $d_1$ and $d_2$ :$\\$Density of mixture =$(\frac{2.d_1.d_2}{d1+d2})$

## Relative Density/Specific Gravity

1. Relative density is different from density. It’s the comparison of a substance’s density with density of pure water at 4 degree celsius.$\\$Ways to calculate relative density/specific gravity:$\\$

Specific Gravity$\\$$\frac{weight- of- the- body}{weight- of- same- volume -of- water- at- 4 \degree C.} \\$=$\frac{density- of- the- body}{density -of- water- at -4 \degree C}$$\\$$\textbf {NOTE:}$ density of water at 4 degree celsius is 1gm/cc(CGS unit)$\\$Specific Gravity =density of body(only when CGS unit is chosen)$\\$

2. Some other deviations of Specific Gravity:$\\$

Sp.Gr=$(\frac{wt. -of -object -in -air}{loss -of -wt. -in -water})$ x sp. gr of water at room temperature$\\$Sp.Gr(for a given liquid)=$\frac{wt.- of- given -volume- of- liquid}{wt.- of- equal- volume- of- water}$x sp. gr of water at room temperature.

PRESSURE:

## Laws of liquid pressure

1. The pressure of a liquid is directly proportional to its density.$\\$$P \propto d$

2. The pressure of a liquid is directly proportional to its depth from the free surface.$\\$$P \propto h$

3. Pressure at a point in a liquid is the same in all directions.

4. Pressure of a liquid is independent of the shape of the vessel.

## ARCHIMEDES’ PRINCIPLE

1. The principle states that , ‘When a body is partially or completely immersed in liquid then loss of weight of body is equal to weight of the displaced liquid.$\\$Mathematically, Loss of weight = weight of liquid displaced

2. $\textbf{Upthrust:} \\$An upward force acting on a body when it is partially or completely immersed.

3. According to Archimedes’ principle,$\\$" Upthrust = wt. of liquid displaced"$\\$"Upthrust = wt. of body in air - wt. of body in water"

4. Mathematically, Upthrust(U)$\\$= d.g.v$\\$(d= density of the liquid in which the body is)$\\$(v= volume of body inside the liquid)

## Pascal’s Law

1. This is the law of transmission of liquid pressure.

2. The law states that pressure applied to an enclosed liquid(i.e liquid within a container like system), pressure is equally transmitted to every portion of the vessel/container.$\\$We get the relation$\\$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$ (where $F_1$ and $F_2$ are forces and $A_1$ and $A_2$ are area of each opening)

## Application of Pascal’s Law

1. $\\$Smaller force is magnified to a larger force. Force applied at the smaller opening magnifies to a larger magnitude at bigger opening, thus helping to lift heavier loads easily.

2. As pressure on both ends are equal(From Pascal’s law)$\\$$F_2=\frac{A_2}{A_1} \times F_1$( as $A_1$ is smaller and $A_2$ is larger $F_2$ is a bigger quantity )

3. $\textbf{Hydraulic Brakes:}$

4. $\textbf{Hydraulic Cranes:}$

## Principle of Floatation

1. Considering a body of density $d$ is placed in a liquid of density $L$ then,$\\$

$\to$ $d>L$ i.e wt. of body> upthrust , the body sinks$\\$$\to$ $d=L$ i.e wt. of body= upthrust , the body just floats on the surface$\\$$\to$ $d i.e wt. of body< upthrust , the body floats with certain part of it inside the liquid and certain part of it outside,$\\$fraction outside $= 1-\frac{d}{l}$

## Equilibrium in Floating Bodies

1. For a body such as a ship to keep floating, the Metacenter(MC) of the body should be above its Center of Gravity(CG), else it sinks.$\\$

$\textbf{Metacenter:}$ Point of intersection of the line from CG and the line from Center of Buoyancy(CB).$\\$$\textbf{Center of Buoyancy:}$ The geometric center of the submerged portion of the body.$\\$

2. Some Important points:

$\to$ Liquid obeys volumetric conservation at all times under normal conditions.$\\$

$\to$ If a body just floating in a liquid is pressed down and released it will sink.$\\$

$\to$ Force of buoyancy depends on the mass of liquid displaced.$\\$

$\to$ Apparent weight of a floating body is 0.$\\$

$\to$ When an ice block floating in water melts the water level remains constant.$\\$

$\to$ If ice containing metal floating in water melts, water level falls.$\\$

$\to$ Even though g(acceleration due to gravity) varies, the barometric height remains constant.$\\$

$\to$ If P is atmospheric pressure total pressure at depth h is$\\$

        P’= P+ d.g.h

Reactions


text:
I think this is because I havent given the auth argument, on my local I usually need use to start the container with this kind of cmd:

docker container create --name doccano


text:
# Show that T is a linear transformation

Consider the function $T: P_1 \rightarrow P_1$ defined by $T(at+b) = (3a+b)t + 4a +2b$.
Show that T is a linear transformation.

Solution: From the definition of a linear transformation we must show that the function satisfy the two properties of a linear transformation(vector addition and scalar multiplication)
This is what I have:
$T((at+b)+(a_1t+b_1)) = T((a+a_1)t + (b+b_1)) = T(at+b)+T(a_1t+b_1)$

But I don't think this shows that the function holds for vector addition.

-
Does your proof still work if I define $T(at+b)=1$? –  wj32 Nov 24 '12 at 9:38

Be careful, your proof does not prove much! We have to show that $T((at+b)+(a_1t+b_1)) = T(at+b)+T(a_1t+b_1)$ and $T( \alpha (at+b))= \alpha T(at+b)$ for any $\alpha$ in the field (over which the polynomial are defined). $T((at+b)+(a_1t+b_1))=T((a+a_1)t+(b+b_1))=(3(a+a_1)+(b+b_1))t+4(a+a_1)+2(b+b_1)=(3a+3a_1+b+b_1)t+4a+4a_1+2b+2b_1=(3a+b)t+2a+2b+(3a_1+b_1)t+4a_1+2b_1=T(at+b)+T(a_1t+b_1)$

and also we have: $T( \alpha (at+b))=T(\alpha at +\alpha b)=(3(\alpha a)+\alpha b)t+4(\alpha a)+2(\alpha b)=\alpha(3a+b)t+\alpha (4a)+\alpha (2b)=\alpha T(at+b)$

(Of course, I used the commutative, associative and distributive laws for a field).

So $T$ is indeed a linear transformation.

-
Thank You so much for the explanation and for editing my question. –  Diana Martinez Nov 25 '12 at 1:45


text: A spherical lens or mirror surface has a center of curvature located either along or decentered from the system local optical axis. {\displaystyle r=0} 0 The radius of curvature of a convex mirror used for rearview on a car is 4.00 m. If the location of the bus is 6 meters from this mirror, find the position of the image formed. °ÆA!°Š"°Š#� $� %° °Å°S�Ä z [2], This article is about optical applications. K Optical surfaces with non-spherical profiles, such as the surfaces of aspheric lenses, also have a radius of curvature. A surface that converges a wavefront is taken a positive e.g. from the axis. K Sometimes this image is real (you can form it on a surface such as a piece of paper) and sometimes virtual (it cannot be formed on a surface) Curvature and the sign convention The radius of curvature (R) of a surface is taken as a positive number if it increases the curvature of the waves – i.e. 2 Looking at the diagram you can see that waves with no curvature have been converged to a point a distance f (the focal length) from the lens. α The vertex of the lens surface is located on the local optical axis. ĞÏà¡±á > şÿ y { şÿÿÿ x ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿì¥Á 7 ğ¿ bjbjUU .. 7| 7| ñ ÿÿ ÿÿ ÿÿ l H H H H , ¼ ¼ ¼ 8 ô 4 , *› ì H H " j j j -� -� -� …š ‡š ‡š ‡š ‡š ‡š ‡š$ œ 6� h «š 9 -� W� Ö -� -� -� «š K’ H H j j íy äš K’ K’ K’ -� L H x j j …š K’ -� …š K’ : K’ …š À X …š j °ıŞ“ñNÊ, � ¼ y� Ğ …š …š úš 0 *› …š �� I� �� …š K’ , , H H H H Ù Lenses Lenses affect the light that passes through them making an image of the object from which the light waves come. Waves and curved surfaces When light waves fall on a curved surface that surface changes the curvature of the wave. Sign conventions 4 • Light travels from left to right • A radius of curvature is positive if the surface is convex towards the left • Longitudinal distances are positive if pointing to the right • Lateral distances are positive if pointing up • Ray angles are positive if the ray direction is obtained by rotating the Solution: Given: The radius of curvature (R)= +4.00 m. Object distance(u) = -6.00 m. Image distance(v) = ? is the conic constant, as measured at the vertex (where and Formula used: $$f=\frac{R}{2}$$ $$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$ converges the waves and negative if it decreases the curvature of the ways – i.e. For each R, the convention is such that we make R > 0 if the light hits the curved surface before the center of curvature, and R < 0 if the opposite is true.


text:
# Which of the following is a linear polynomial?

Question:

Which of the following is a linear polynomial?

(a) $x+x^{2}$

(b) $x+1$

(c) $5 x^{2}-x+3$

(d) $x+\frac{1}{x}$

Solution:

(b) $x+1$

Clearly, $x+1$ is a polynomial of degree 1 . So, it is a linear polynomial.

Leave a comment


text:
# On eigenvalues of complex-orthogonal matrices

Suppose that $A$ is a complex matrix satisfying $A^TA = I$ (so $A$ is the entrywise transpose, not the conjugate transpose). What can be said about the eigenvalues of $A$, if $A$ is "complex-orthogonal" in this sense?

Of course, for any eigenpair $(\lambda,x)$, we have $$x^Tx = x^TA^TAx = (Ax)^TAx = \lambda^2 (x^Tx)$$ which allows us to conclude that $\lambda^2 = 1$... so long as $x^Tx \neq 0$. Can anything else be said? Does the case in which $A$ has real entries allow us to conclude that $|\lambda| = 1$?

• For the first question, what about $x^*x=x^*A^*Ax=(Ax)^*Ax=\overline{\lambda}\lambda x^*x=|\lambda|^2x^*x$, so $|\lambda|^2=1$? May 21 '17 at 12:25
• @GerryMyerson $A^*$ is the adjoint of $A$ relative to the Hermitian inner product, i.e. $$\langle x, y \rangle = y^*x = \sum_{j=1}^n x_j \overline{y_j}$$ the point is to avoid this. May 21 '17 at 12:46
• For any of you interested in the (now gone) first part of this question, I've asked it separately over here. May 21 '17 at 13:38
• $A^*$ is the matrix you get from $A$ by flipping it across the main diagonal and taking the complex conjugate. You can do both those actions without ever having heard of inner products. May 21 '17 at 23:08
• @GerryMyerson sure, I guess you could. It would a pretty unintuitive thing to do, though. Why, for instance, should we expect that $(AB)^* = B^*A^*$? May 22 '17 at 0:41

$$A=\pmatrix{\frac{a+a^{-1}}2&i\frac{a-a^{-1}}2\\ -i\frac{a-a^{-1}}2&\frac{a+a^{-1}}2}$$ has $A^tA=I$ and has $a$ and $a^{-1}$ as eigenvalues.

As an example $a=2$ gives $$A=\pmatrix{\frac54&\frac34i\\-\frac34i&\frac54}$$ and $$A^tA=\pmatrix{\frac54&-\frac34i\\\frac34i&\frac54} \pmatrix{\frac54&\frac34i\\-\frac34i&\frac54}=\pmatrix{1&0\\0&1}.$$

• That answers part of the second half of my question, so thanks for that. Still, I'm looking for something more comprehensive. May 21 '17 at 13:01
• @Omnomnomnom For even $n$ the quadratic form $z_1^2+\cdots+z_n^2$ is equivalent to $w_1w_2+w_3w_4+\cdots$. Is this a useful step towards your "comprehensive" vision? May 21 '17 at 13:07
• About your other comment, I suppose that depends on what exactly an "equivalence" of quadratic forms refers to in this case May 21 '17 at 13:15
• @Omnomnomnom For equivalence of quadratic forms see Wikipedia en.wikipedia.org/wiki/Quadratic_form May 21 '17 at 13:30
• You're right about the matrix... it was quick to check by hand after all. I'm surprised that W|A made a mistake there; I don't think I have a typo. May 21 '17 at 13:33

At the risk of stating the obvious, as it hasn't been explicitly stated here, I add that:

1. Each eigenvalue is either (a) one of a pair of eigenvalues $$(\lambda, \lambda^{-1})$$, or (b) $$\lambda = \pm 1$$.
2. The product of all the eigenvalues satisfies $$\prod_i \lambda_i = \pm 1$$.

Both of these claim are straightforward to show:

1. Let $$A x = \lambda x$$ define the left eigenvector $$x$$, then by taking the transpose it follows that $$x^T A = x^T \lambda^{-1}$$, i.e. $$x^T$$ is a right eigenvector with eigenvalue $$\lambda^{-1}$$. If (a) $$x^T x = 0$$ then these are distinct eigenvalues and $$\lambda$$ may take any value, if (b) $$x^T x \neq 0$$ then they are the same eigenvalue, and we require $$\lambda = \lambda^{-1}$$, i.e. $$\lambda = \pm 1$$.
2. Note $$1 = \det(I) = \det(A A^T) = \det(A)\det(A^T)= \det(A)^2$$. This implies $$\det(A) = \prod_i \lambda_i = \pm 1$$.

This has been thoroughly studied in the paper "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices" by Horn and Merino (1999) and also in Olga Ruff's master thesis "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices: characterisation and examples" (2007). In particular, theorem 1.2.3 (pp. 31-32) of Ruff's thesis states that

An $$n\times n$$ complex matrix is similar to a complex orthogonal matrix if and only if its Jordan Canonical Form can be expressed as a direct sum of matrices of only the following three types:

(a) $$J_k(\lambda)\oplus J_k(\lambda^{-1})$$ for $$\lambda\in\mathbb C\setminus\{0\}$$ and any $$k$$,

(b) $$J_k(1)$$ for any odd $$k$$ and

(c) $$J_k(-1)$$ for any odd $$k$$.

In particular, every nonzero complex number is an eigenvalue of some complex orthogonal matrix, and for each complex orthogonal matrix, all eigenvalues $$\ne\pm1$$ must occur in reciprocal pairs.

• Interestingly Ruff and Horn and Merino seem inequivalent. Horn and Merino further explicitly exclude the possibility of $J_k(\pm1) \oplus J_k(\pm 1)$ for odd $k$ (theorem 1), whereas Ruff (quoted here) does not. Mar 3 at 15:30
• @ComptonScattering $J_k(1)\oplus J_k(1)$ for odd $k$ is allowed in Horn and Merino. It is the direct sum of two matrices of type (d). The authors just state the same theorem differently. It might be clearer to say that even-sized Jordan blocks for $\lambda=\pm1$ must occur an even number of times but odd-sized Jordan blocks for $\lambda=\pm1$ can occur any number of times. Mar 3 at 15:45
• Yes you are correct. Mar 3 at 15:48

I have some simple ideas:

Let $A$ be an orthogonal matrix $i.e.$ $A^TA=AA^T=I$, from the very definition of the orthogonal matrix, $A$ and $A^T$ are both inverses of each other. So the definition says that their eigenvalues are reciprocals. $i.e.$ if $a$ $\in \mathbb{C}$ is an eigenvalue of $A$ then $\frac{1}{a}$ is an eigenvalue of $A^T$. But we know that the eigenvalues of $A$ and $A^T$ are same. So $a=\frac{1}{a}$ implies $a^2=1$. If we take $a=a_1+ia_2$ then $a_1^2-a_2^2=1$, $a_1a_2=0$ which implies modulus of $a$ is $1$.

• Noting that $A$ has the same eigenvalues as $A^{-1}$ is not enough to deduce that any eigenvalues satisfy $a = (1/a)$. For instance, note that $$A = \pmatrix{2&0\\0&1/2}$$ has the same eigenvalues as its inverse. May 21 '17 at 13:25
• ohh... I see. I made a mistake doing $a=\frac{1}{a}$. thank you May 21 '17 at 13:29

To clarify the result from the existing answer: we can get a complex-orthogonal matrix with eigenvalues $a,a^{-1}$ with $$A = \frac 12 \pmatrix{a+a^{-1} & i(a - a^{-1})\\ -i(a - a^{-1}) & a + a^{-1}} = \pmatrix{i&-i\\1&1} \pmatrix{a\\&a^{-1}}\pmatrix{i&-i\\1&1}^{-1}$$