text:
# Beamer Ilmenau theme: navigation switching between sections

I'm using the beamer Ilmenau theme. I have two sections (pros and cons) that need to be presented simultaneously during the presentation. Currently the navigation shows the bullets for each subsection (frames) of pros and cons separately. I would prefer if there was a way to switch between the two sections, for instance, start with first dot in the pros section and then jump to the first dot in the cons section and so on with this type of flow. Is this possible?

\documentclass{beamer}

\usetheme{Ilmenau}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

\begin{frame}[t]
\tableofcontents
\end{frame}

\section{Introduction}
\begin{frame}[t]
\frametitle{Welcome}
So this is where we start..
\end{frame}

\begin{frame}[t]
\frametitle{Electric Vehicles}\vspace{4pt}
\begin{enumerate}
\item 1
\item 2
\end{enumerate}
\end{frame}

\section{Pros}
\begin{frame}[t]
\frametitle{Really good}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{Awesome to say the least}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{No doubt}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{One word: Epic}\vspace{4pt}
\end{frame}

\section{Cons}
\begin{frame}[t]
\frametitle{Privacy issues}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{Danger!}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{Keep away}\vspace{4pt}
\end{frame}

\begin{frame}[t]
\frametitle{Are you serious?}\vspace{4pt}
\end{frame}

\section{Conclusion}
\begin{frame}[t]
\frametitle{The verdict...}\vspace{4pt}
\end{frame}

\end{document}

• Welcome to TeX.Stackexchange! You could always rearrange the slides in the finished pdf - either using tex or some other tools. – user36296 Dec 29 '17 at 20:05
• If the majority of your frames is top aligned, maybe its easier to write \documentclass[t]{beamer} – user36296 Dec 29 '17 at 20:06
• Are you aware of "display priority", i.e. \item<2> things? You can basically freely choose how many build-up pages for a complete frame you get and what lands on which. – Oleg Lobachev Dec 29 '17 at 20:36

I think you are looking for something like this:

\documentclass{beamer}
\usepackage{nameref}
\usetheme{Ilmenau}

\begin{document}

\begin{frame}
\titlepage
\end{frame}

\begin{frame}[t]\label{toc}
\tableofcontents
\end{frame}

\section{Introduction}
\begin{frame}[t]\label{welcome}
\frametitle{Welcome}
So this is where we start..
\end{frame}

\begin{frame}[t,label={Vehicles}]
\frametitle{Electric Vehicles}\vspace{4pt}
\begin{enumerate}
\item 1
\item 2
\end{enumerate}
\end{frame}

\section{Pros}
\subsection{Pros 1}
\begin{frame}[t,label={Pros1}]
\frametitle{Really good}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Pros 2}
\begin{frame}[t,label={Pros2}]
\frametitle{Awesome to say the least}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Pros 3}
\begin{frame}[t,label={Pros3}]
\frametitle{No doubt}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Pros 4}

\begin{frame}[t,label={Pros4}]
\frametitle{One word: Epic}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\section{Cons}
\subsection{Cons 1}
\begin{frame}[t,label={Cons1}]
\frametitle{Privacy issues}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Cons 2}
\begin{frame}[t]\label{Cons2}
\frametitle{Danger!}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Cons 3}
\begin{frame}[t]\label{Cons3}
\frametitle{Keep away}\vspace{4pt}

\begin{center}
\end{center}
\end{frame}

\subsection{Cons 4}

\begin{frame}[t]\label{Cons4}
\frametitle{Are you serious?}\vspace{4pt}

\end{frame}

\section{Conclusion}
\begin{frame}[t]
\frametitle{The verdict...}\vspace{4pt}
\end{frame}

\end{document}

• I think the OP uses the standard theme Ilmenau included with the package, but mistyped the name with a lower-case i. This theme has little "navigation dots" called mini frames in the headline, and he wants the active slide marker to jump back and forth between the pros and cons sections. – diabonas Dec 29 '17 at 20:37
• You don't need \usepackage{hyperref} with beamer. – user36296 Dec 29 '17 at 20:42
• OP here.. Thanks a lot for your efforts. Not exactly what I had in mind though. I was looking for sth like a jump from Pros1 to Cons1 then again from Pros2 to Cons2 and so on.. – user151561 Dec 29 '17 at 21:04
• Just fixed... The idea was already the same... the numbers are different – koleygr Dec 29 '17 at 21:12
• Works great now!! just a remark: the hyperlink is too obvious, is there a way around to make it look more pro? thanks :) – user151561 Dec 29 '17 at 21:24


text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=5$
The given equation, $x=2$, is a vertical line. The line parallel to this that passes through $( 5,4 )$ is $x=5$.

format: plain


text:
[jboss-user] [JBoss jBPM] - Re: Parameter userName does not exist as a named parameter

Thu Feb 5 08:00:33 EST 2009

```Not to worry, I figured it out.
I originally noticed that hibernate.cfg.xml was referencing hibernate.identity.hbm.xml, and so I copied hibernate.identity.hbm.xml from jbpm-3.3.1.GA\src\jbpm-jpdl-sources to my WEB-INF/classes directory. Problem is, this hibernate.identity.hbm.xml is empty. I didn't realise that jbpm-identity.jar already had the correctly setup hibernate.identity.hbm.xml file.

I simply removed the empty hibernate.identity.hbm.xml from my classes folder, so the proper one from jbpm-identity.jar is used instead, and problem is gone :-)

View the original post : http://www.jboss.com/index.html?module=bb&op=viewtopic&p=4207329#4207329


text:
## L19: Concurrency and Synchronization

It’s not unusual to find that an application is organized as multiple communicating processes. What’s the advantage of using multiple processes instead of just a single process?

Many applications exhibit concurrency, i.e., some of the required computations can be performed in parallel. For example, video compression algorithms represent each video frame as an array of 8-pixel by 8-pixel macroblocks. Each macroblock is individually compressed by converting the 64 intensity and color values from the spatial domain to the frequency domain and then quantizing and Huffman encoding the frequency coefficients. If you’re using a multi-core processor to do the compression, you can perform the macroblock compressions concurrently.

Applications like video games are naturally divided into the “front-end” user interface and “back-end” simulation and rendering engines. Inputs from the user arrive asynchronously with respect to the simulation and it’s easiest to organize the processing of user events separately from the backend processing.

Processes are an effective way to encapsulate the state and computation for what are logically independent components of an application, which communicate with one another when they need to share information.

These sorts of applications are often data- or event-driven, i.e., the processing required is determined by the data to be processed or the arrival of external events.

How should the processes communicate with each other?

If the processes are running out of the same physical memory, it would be easy to arrange to share memory data by mapping the same physical page into the contexts for both processes. Any data written to that page by one process will be able to be read by the other process.

To make it easier to coordinate the processes’ communicating via shared memory, we’ll see it’s convenient to provide synchronization primitives. Some ISAs include instructions that make it easy to do the required synchronization.

Another approach is to add OS supervisor calls to pass messages from one process to another. Message passing involves more overhead than shared memory, but makes the application programming independent of whether the communicating processes are running on the same physical processor.

In this lecture, we’ll use the classic producer-consumer problem as our example of concurrent processes that need to communicate and synchronize. There are two processes: a producer and a consumer. The producer is running in a loop, which performs some computation to generate information, in this case, a single character C. The consumer is also running a loop, which waits for the next character to arrive from the producer, then performs some computation .

The information passing between the producer and consumer could obviously be much more complicated than a single character. For example, a compiler might produce a sequence of assembly language statements that are passed to the assembler to be converted into the appropriate binary representation. The user interface front-end for a video game might pass a sequence of player actions to the simulation and rendering back-end. In fact, the notion of hooking multiple processes together in a processing pipeline is so useful that the Unix and Linux operating systems provide a PIPE primitive in the operating system that connects the output channel of the upstream process to the input channel of the downstream process.

Let’s look at a timing diagram for the actions of our simple producer/consumer example. We’ll use arrows to indicate when one action happens before another. Inside a single process, e.g., the producer, the order of execution implies a particular ordering in time: the first execution of is followed by the sending of the first character. Then there’s the second execution of , followed by the sending of the second character, and so on. In later examples, we’ll omit the timing arrows between successive statements in the same program.

We see a similar order of execution in the consumer: the first character is received, then the computation is performed for the first time, etc. Inside of each process, the process’ program counter is determining the order in which the computations are performed.

So far, so good - each process is running as expected. However, for the producer/consumer system to function correctly as a whole, we’ll need to introduce some additional constraints on the order of execution. These are called “precedence constraints” and we’ll use this stylized less-than sign to indicate that computation A must precede, i.e., come before, computation B.

In the producer/consumer system we can’t consume data before it’s been produced, a constraint we can formalize as requiring that the ith send operation has to precede the ith receive operation. This timing constraint is shown as the solid red arrow in the timing diagram.

Assuming we’re using, say, a shared memory location to hold the character being transmitted from the producer to the consumer, we need to ensure that the producer doesn’t overwrite the previous character before it’s been read by the consumer. In other words, we require the ith receive to precede the i+1st send. These timing constraints are shown as the dotted red arrows in the timing diagram.

Together these precedence constraints mean that the producer and consumer are tightly coupled in the sense that a character has to be read by the consumer before the next character can be sent by the producer, which might be less than optimal if the and computations take a variable amount of time. So let’s see how we can relax the constraints to allow for more independence between the producer and consumer.

We can relax the execution constraints on the producer and consumer by having them communicate via N-character first-in-first-out (FIFO) buffer. As the producer produces characters it inserts them into the buffer. The consumer reads characters from the buffer in the same order as they were produced. The buffer can hold between 0 and N characters. If the buffer holds 0 characters, it’s empty; if it holds N characters, it’s full. The producer should wait if the buffer is full, the consumer should wait if the buffer is empty.

Using the N-character FIFO buffer relaxes our second overwrite constraint to the requirement that the ith receive must happen before i+Nth send. In other words, the producer can get up to N characters ahead of the consumer.

FIFO buffers are implemented as an N-element character array with two indices: the read index indicates the next character to be read, the write index indicates the next character to be written. We’ll also need a counter to keep track of the number of characters held by the buffer, but that’s been omitted from this diagram. The indices are incremented modulo N, i.e., the next element to be accessed after the N-1st element is the 0th element, hence the name “circular buffer”.

Here’s how it works. The producer runs, using the write index to add the first character to the buffer. The producer can produce additional characters, but must wait once the buffer is full.

The consumer can receive a character anytime the buffer is not empty, using the read index to keep track of the next character to be read. Execution of the producer and consumer can proceed in any order so long as the producer doesn’t write into a full buffer and the consumer doesn’t read from any empty buffer.

Here’s what the code for the producer and consumer might look like. The array and indices for the circular buffer live in shared memory where they can be accessed by both processes. The SEND routine in the producer uses the write index IN to keep track of where to write the next character. Similarly the RCV routine in the consumer uses the read index OUT to keep track of the next character to be read. After each use, each index is incremented modulo N.

The problem with this code is that, as currently written, neither of the two precedence constraints is enforced. The consumer can read from an empty buffer and the producer can overwrite entries when the buffer is full.

We’ll need to modify this code to enforce the constraints and for that we’ll introduce a new programming construct that we’ll use to provide the appropriate inter-process synchronization.

What we’d like to do is to create a single abstraction that can be used to address all our synchronization needs. In the early 1960’s, the Dutch computer scientist Edsger Dijkstra proposed a new abstract data type called the semaphore, which has an integer value greater than or equal to 0. A programmer can declare a semaphore as shown here, specifying its initial value. The semaphore lives in a memory location shared by all the processes that need to synchronize their operation.

The semaphore is accessed with two operations: WAIT and SIGNAL. The WAIT operation will wait until the specified semaphore has a value greater than 0, then it will decrement the semaphore value and return to the calling program. If the semaphore value is 0 when WAIT is called, conceptually execution is suspended until the semaphore value is non-zero. In a simple (inefficient) implementation, the WAIT routine loops, periodically testing the value of the semaphore, proceeding when its value is non-zero.

The SIGNAL operation increments the value of the specified semaphore. If there any processes WAITing on that semaphore, exactly one of them may now proceed. We’ll have to be careful with the implementation of SIGNAL and WAIT to ensure the “exactly one” constraint is satisfied, i.e., that two processes both WAITing on the same semaphore won’t both think they can decrement it and proceed after a SIGNAL.

A semaphore initialized with the value K guarantees that the ith call to SIGNAL will precede (i+K)th call to WAIT. In a moment, we’ll see some concrete examples that will make this clear. Note that in 6.004, we’re ruling out semaphores with negative values.

In the literature, you may see P(s) used in place of WAIT(s) and V(s) used in place of SIGNAL(s). These operation names are derived from the Dutch words for “test” and “increase”.

Let’s see how to use semaphores to implement precedence constraints.

Here are two processes, each running a program with 5 statements. Execution proceeds sequentially within each process, so A1 executes before A2, and so on. But there are no constraints on the order of execution between the processes, so statement B1 in Process B might be executed before or after any of the statements in Process A. Even if A and B are running in a timeshared environment on a single physical processor, execution may switch at any time between processes A and B.

Suppose we wish to impose the constraint that the execution of statement A2 completes before execution of statement B4 begins. The red arrow shows the constraint we want.

Here’s the recipe for implementing this sort of simple precedence constraint using semaphores.

First, declare a semaphore (called “s” in this example) and initialize its value to 0.

Place a call to signal(s) at the start of the arrow. In this example, signal(s) is placed after the statement A2 in process A.

Then place a call to wait(s) at the end of the arrow. In this example, wait(s) is placed before the statement B4 in process B.

With these modifications, process A executes as before, with the signal to semaphore s happening after statement A2 is executed.

Statements B1 through B3 also execute as before, but when the wait(s) is executed, execution of process B is suspended until the signal(s) statement has finished execution. This guarantees that execution of B4 will start only after execution of A2 has completed.

By initializing the semaphore s to 0, we enforced the constraint that the first call to signal(s) had to complete before the first call to wait(s) would succeed.

Another way to think about semaphores is as a management tool for a shared pool of K resources, where K is the initial value of the semaphore. You use the SIGNAL operation to add or return resources to the shared pool. And you use the WAIT operation to allocate a resource for your exclusive use.

At any given time, the value of the semaphore gives the number of unallocated resources still available in the shared pool.

Note that the WAIT and SIGNAL operations can be in the same process, or they may be in different processes, depending on when the resource is allocated and returned.

We can use semaphores to manage our N-character FIFO buffer. Here we’ve defined a semaphore CHARS and initialized it to 0. The value of CHARS will tell us how many characters are in the buffer.

So SEND does a signal(CHARS) after it has added a character to the buffer, indicating the buffer now contains an additional character.

And RCV does a wait(CHARS) to ensure the buffer has at least one character before reading from the buffer.

Since CHARS was initialized to 0, we’ve enforced the constraint that the ith call to signal(CHARS) precedes the completion of the ith call to wait(CHARS). In other words, RCV can’t consume a character until it has been placed in the buffer by SEND.

Does this mean our producer and consumer are now properly synchronized? Almost! Using the CHARS semaphore, we implemented *one* of the two precedence constraints we identified as being necessary for correct operation. Next we’ll see how to implement the other precedence constraint.

What keeps the producer from putting more than N characters into the N-character buffer? Nothing. Oops, the producer can start to overwrite characters placed in the buffer earlier even though they haven’t yet been read by the consumer. This is called buffer overflow and the sequence of characters transmitted from producer to consumer becomes hopelessly corrupted.

What we’ve guaranteed so far is that the consumer can read a character only after the producer has placed it in the buffer, i.e., the consumer can’t read from an empty buffer.

What we still need to guarantee is that the producer can’t get too far ahead of the consumer. Since the buffer holds at most N characters, the producer can’t send the (i+N)th character until the consumer has read the ith character.

Here we’ve added a second semaphore, SPACES, to manage the number of spaces in the buffer. Initially the buffer is empty, so it has N spaces. The producer must WAIT for a space to be available. When SPACES in non-zero, the WAIT succeeds, decrementing the number of available spaces by one and then the producer fills that space with the next character.

The consumer signals the availability of another space after it reads a character from the buffer.

There’s a nice symmetry here. The producer consumes spaces and produces characters. The consumer consumes characters and produces spaces. Semaphores are used to track the availability of both resources (i.e., characters and spaces), synchronizing the execution of the producer and consumer.

This works great when there is a single producer process and a single consumer process. Next we’ll think about what will happen if we have multiple producers and multiple consumers.

Let’s take a moment to look at a different example. Automated teller machines allow bank customers to perform a variety of transactions: deposits, withdrawals, transfers, etc. Let’s consider what happens when two customers try to withdraw \$50 from the same account at the same time. A portion of the bank’s code for a withdrawal transaction is shown in the upper right. This code is responsible for adjusting the account balance to reflect the amount of the withdrawal. Presumably the check to see if there is sufficient funds has already happened. What’s supposed to happen? Let’s assume that the bank is using a separate process to handle each transaction, so the two withdrawal transactions cause two different processes to be created, each of which will run the Debit code. If each of the calls to Debit run to completion without interruption, we get the desired outcome: the first transaction debits the account by \$50, then the second transaction does the same. The net result is that you and your friend have \$100 and the balance is \$100 less.

So far, so good.

But what if the process for the first transaction is interrupted just after it’s read the balance? The second process subtracts \$50 from the balance, completing that transaction. Now the first process resumes, using the now out-of-date balance it loaded just before being interrupted. The net result is that you and your friend have \$100, but the balance has only been debited by \\$50.

The moral of the story is that we need to be careful when writing code that reads and writes shared data since other processes might modify the data in the middle of our execution. When, say, updating a shared memory location, we’ll need to LD the current value, modify it, then ST the updated value. We would like to ensure that no other processes access the shared location between the start of the LD and the completion of the ST. The LD/modify/ST code sequence is what we call a “critical section.” We need to arrange that other processes attempting to execute the same critical section are delayed until our execution is complete. This constraint is called “mutual exclusion,” i.e., only one process at a time can be executing code in the same critical section.

Once we’ve identified critical sections, we’ll use semaphores to guarantee they execute atomically, i.e., that once execution of the critical section begins, no other process will be able to enter the critical section until the execution is complete. The combination of the semaphore to enforce the mutual exclusion constraint and the critical section of code implement what’s called a “transaction”. A transaction can perform multiple reads and writes of shared data with the guarantee that none of the data will be read or written by other processes while the transaction is in progress.

Here’s the original code to Debit, which we’ll modify by adding a LOCK semaphore. In this case, the resource controlled by the semaphore is the right to run the code in the critical section. By initializing LOCK to 1, we’re saying that at most one process can execute the critical section at a time.

A process running the Debit code WAITs on the LOCK semaphore. If the value of LOCK is 1, the WAIT will decrement value of LOCK to 0 and let the process enter the critical section. This is called acquiring the lock. If the value of LOCK is 0, some other process has acquired the lock and is executing the critical section and our execution is suspended until the LOCK value is non-zero.

When the process completes execution of the critical section, it releases the LOCK with a call to SIGNAL, which will allow other processes to enter the critical section. If there are multiple WAITing processes, only one will be able to acquire the lock, and the others will still have to wait their turn.

Used in this manner, semaphores are implementing a mutual exclusion constraint, i.e., there’s a guarantee that two executions of the critical section cannot overlap. Note that if multiple processes need to execute the critical section, they may run in any order and the only guarantee is that their executions will not overlap.

There are some interesting engineering issues to consider. There’s the question of the granularity of the lock, i.e., what shared data is controlled by the lock? In our bank example, should there be one lock controlling access to the balance for all accounts? That would mean that no one could access any balance while a transaction was in progress. That would mean that transactions accessing different accounts would have to run one after the other even though they’re accessing different data. So one lock for all the balances would introduce unnecessary precedence constraints, greatly slowing the rate at which transactions could be processed.

Since the guarantee we need is that we shouldn’t permit multiple simultaneous transactions on the same account, it would make more sense to have a separate lock for each account, and change the Debit code to acquire the account’s lock before proceeding. That will only delay transactions that truly overlap, an important efficiency consideration for a large system processing many thousands of mostly non-overlapping transactions each second.

Of course, having per-account locks would mean a lot of locks! If that’s a concern, we can adopt a compromise strategy of having locks that protect groups of accounts, e.g., accounts with same last three digits in the account number. That would mean we’d only need 1000 locks, which would allow up to 1000 transactions to happen simultaneously.

The notion of transactions on shared data is so useful that we often use a separate system called a database that provides the desired functionality. Database systems are engineered to provide low-latency access to shared data, providing the appropriate transactional semantics. The design and implementation of databases and transactions is pretty interesting - to follow up, I recommend reading about databases on the web.

Returning to our producer/consumer example, we see that if multiple producers are trying to insert characters into the buffer at the same time, it’s possible that their execution may overlap in a way that causes characters to be overwritten and/or the index to be improperly incremented.

We just saw this bug in the bank example: the producer code contains a critical section of code that accesses the FIFO buffer and we need to ensure that the critical section is executed atomically.

Here we’ve added a third semaphore, called LOCK, to implement the necessary mutual exclusion constraint for the critical section of code that inserts characters into the FIFO buffer. With this modification, the system will now work correctly when there are multiple producer processes.

There’s a similar issue with multiple consumers, so we’ve used the same LOCK to protect the critical section for reading from the buffer in the RCV code.

Using the same LOCK for producers and consumers will work, but does introduce unnecessary precedence constraints since producers and consumers use different indices, i.e., IN for producers and OUT for consumers. To solve this problem we could use two locks: one for producers and one for consumers.

Semaphores are a pretty handy Swiss army knife when it comes to dealing with synchronization issues. When WAIT and SIGNAL appear in different processes, the semaphore ensures the correct execution timing between processes. In our example, we used two semaphores to ensure that consumers can’t read from an empty buffer and that producers can’t write into a full buffer.

We also used semaphores to ensure that execution of critical sections - in our example, updates of the indices IN and OUT - were guaranteed to be atomic. In other words, that the sequence of reads and writes needed to increment a shared index would not be interrupted by another process between the initial read of the index and the final write.

Now let’s figure out how to implement semaphores. They are themselves shared data and implementing the WAIT and SIGNAL operations will require read/modify/write sequences that must be executed as critical sections. Normally we’d use a lock semaphore to implement the mutual exclusion constraint for critical sections. But obviously we can’t use semaphores to implement semaphores! We have what’s called a bootstrapping problem: we need to implement the required functionality from scratch.

Happily, if we’re running on a timeshared processor with an uninterruptible OS kernel, we can use the supervisor call (SVC) mechanism to implement the required functionality.

We can also extend the ISA to include a special test-and-set instruction that will let us implement a simple lock semaphore, which can then be used to protect critical sections that implement more complex semaphore semantics. Single instructions are inherently atomic and, in a multi-core processor, will do what we want if the shared main memory supports both reading the old value and writing a new value to a specific memory location as a single memory access.

There are other, more complex, software-only solutions that rely only on the atomicity of individual reads and writes to implement a simple lock. For example, see “Dekker’s Algorithm” on Wikipedia.

We’ll look in more detail at the first two approaches.

Here are the OS handlers for the WAIT and SIGNAL supervisor calls. Since SVCs are run kernel mode, they can’t be interrupted, so the handler code is naturally executed as a critical section.

Both handlers expect the address of the semaphore location to be passed as an argument in the user’s R0. The WAIT handler checks the semaphore’s value and if it’s non-zero, the value is decremented and the handler resumes execution of the user’s program at the instruction following the WAIT SVC. If the semaphore is 0, the code arranges to re-execute the WAIT SVC when the user program resumes execution and then calls SLEEP to mark the process as inactive until the corresponding WAKEUP call is made.

The SIGNAL handler is simpler: it increments the semaphore value and calls WAKEUP to mark as active any processes that were WAITing for this particular semaphore.

Eventually the round-robin scheduler will select a process that was WAITing and it will be able to decrement the semaphore and proceed. Note that the code makes no provision for fairness, i.e., there’s no guarantee that a WAITing process will eventually succeed in finding the semaphore non-zero. The scheduler has a specific order in which it runs processes, so the next-in-sequence WAITing process will always get the semaphore even if there are later-in-sequence processes that have been WAITing longer. If fairness is desired, WAIT could maintain a queue of waiting processes and use the queue to determine which process is next in line, independent of scheduling order.

Many ISAs support an instruction like the TEST-and-CLEAR instruction shown here. The TCLR instruction reads the current value of a memory location and then sets it to zero, all as a single operation. It’s like a LD except that it zeros the memory location after reading its value.

To implement TCLR, the memory needs to support read-and-clear operations, as well as normal reads and writes.

The assembly code at the bottom of the slide shows how to use TCLR to implement a simple lock. The program uses TCLR to access the value of the lock semaphore. If the returned value in RC is zero, then some other process has the lock and the program loops to try TCLR again. If the returned value is non-zero, the lock has been acquired and execution of the critical section can proceed. In this case, TCLR has also set the lock to zero, so that other processes will be prevented from entering the critical section.

When the critical section has finished executing, a ST instruction is used to set the semaphore to a non-zero value.

If the necessary synchronization requires acquiring more than one lock, there are some special considerations that need to be taken into account. For example, the code below implements the transfer of funds from one bank account to another. The code assumes there is a separate semaphore lock for each account and since it needs to adjust the balance of two accounts, it acquires the lock for each account.

Consider what happens if two customers try simultaneous transfers between their two accounts. The top customer will try to acquire the locks for accounts 6005 and 6004. The bottom customer tries to acquire the same locks, but in the opposite order. Once a customer has acquired both locks, the transfer code will complete, releasing the locks.

But what happens if the top customer acquires his first lock (for account 6005) and the bottom customer simultaneously acquires his first lock (for account 6004). So far, so good, but now each customer will be not be successful in acquiring their second lock, since those locks are already held by the other customer!

This situation is called a “deadlock” or “deadly embrace” because there is no way execution for either process will resume - both will wait indefinitely to acquire a lock that will never be available.

Obviously, synchronization involving multiple resources requires a bit more thought.

The problem of deadlock is elegantly illustrated by the Dining Philosophers problem. Here there are, say, 5 philosophers waiting to eat. Each requires two chopsticks in order to proceed, and there are 5 chopsticks on the table.

The philosophers follow a simple algorithm. First they pick up the chopstick on their left, then the chopstick on their right. When they have both chopsticks they eat until they’re done, at which point they return both chopsticks to the table, perhaps enabling one of their neighbors to pick them up and begin eating. Again, we see the basic setup of needing two (or more) resources before the task can complete.

Hopefully you can see the problem that may arise...

If all philosophers pick up the chopstick on their left, then all the chopsticks have been acquired, and none of the philosophers will be able to acquire their second chopstick and eat. Another deadlock!

Here are the conditions required for a deadlock:

1. Mutual exclusion, where a particular resource can only be acquired by one process at a time.

2. Hold-and-wait, where a process holds allocated resources while waiting to acquire the next resource.

3. No preemption, where a resource cannot be removed from the process which acquired it. Resources are only released after the process has completed its transaction.

4. Circular wait, where resources needed by one process are held by another, and vice versa.

How can we solve the problem of deadlocks when acquiring multiple resources? Either we avoid the problem to begin with, or we detect that deadlock has occurred and implement a recovery strategy. Both techniques are used in practice.

In the Dining Philosophers problem, deadlock can be avoided with a small modification to the algorithm. We start by assigning a unique number to each chopstick to establish a global ordering of all the resources, then rewrite the code to acquire resources using the global ordering to determine which resource to acquire first, which second, and so on.

With the chopsticks numbered, the philosophers pick up the lowest-numbered chopstick from either their left or right. Then they pick up the other, higher-numbered chopstick, eat, and then return the chopsticks to the table.

How does this avoid deadlock? Deadlock happens when all the chopsticks have been picked up but no philosopher can eat. If all the chopsticks have been been picked up, that means some philosopher has picked up the highest-numbered chopstick and so must have earlier picked up the lower-numbered chopstick on his other side. So that philosopher can eat then return both chopsticks to the table, breaking the hold-and-wait cycle.

So if all the processes in the system can agree upon a global ordering for the resources they require, then acquire them in order, there will be no possibility of a deadlock caused by a hold-and-wait cycle.

A global ordering is easy to arrange in our banking code for the transfer transaction. We’ll modify the code to first acquire the lock for the lower-numbered account, then acquire the lock for the higher-numbered account. Now, both customers will first try to acquire the lock for the 6004 account. The customer that succeeds then can acquire the lock for the 6005 account and complete the transaction. The key to deadlock avoidance was that customers contented for the lock for the *first* resource they both needed - acquiring that lock ensured they would be able to acquire the remainder of the shared resources without fear that they would already be allocated to another process in a way that could cause a hold-and-wait cycle.

Establishing and using a global order for shared resources is possible when we can modify all processes to cooperate. Avoiding deadlock without changing the processes is a harder problem. For example, at the operating system level, it would be possible to modify the WAIT SVC to detect circular wait and terminate one of the WAITing processes, releasing its resources and breaking the deadlock.

The other strategy we mentioned was detection and recovery. Database systems detect when there’s been an external access to the shared data used by a particular transaction, which causes the database to abort the transaction. When issuing a transaction to a database, the programmer specifies what should happen if the transaction is aborted, e.g., she can specify that the transaction be retried. The database remembers all the changes to shared data that happen during a transaction and only changes the master copy of the shared data when it is sure that the transaction will not be aborted, at which point the changes are committed to the database.

In summary, we saw that organizing an application as communicating processes is often a convenient way to go. We used semaphores to synchronize the execution of the different processes, providing guarantees that certain precedence constraints would be met, even between statements in different processes.

We also introduced the notion of critical code sections and mutual exclusion constraints that guaranteed that a code sequence would be executed without interruption by another process. We saw that semaphores could also be used to implement those mutual exclusion constraints.

Finally we discussed the problem of deadlock that can occur when multiple processes must acquire multiple shared resources, and we proposed several solutions based on a global ordering of resources or the ability to restart a transaction.

Synchronization primitives play a key role in the world of “big data” where there are vast amounts of shared data, or when trying to coordinate the execution of thousands of processes in the cloud. Understanding synchronization issues and their solutions is a key skill when writing most modern applications.


text:
# SPM Additional Mathematics 2019, Paper 2 (Question 2)

Question 2:
Express 2n + 2 – 2n + 1 + 2n – 1 in the form p(2n – 1), where p is a constant.
Hence, solve the equation 8(2n + 2 – 2n + 1 + 2n – 1) = $5\left({2}^{{n}^{2}}\right).$
[6 marks]

Solution:

$\begin{array}{l}{2}^{n+2}-{2}^{n+1}+{2}^{n-1}\\ =\left({2}^{n}×{2}^{2}\right)-\left({2}^{n}×{2}^{1}\right)+\left({2}^{n}÷{2}^{1}\right)\\ ={2}^{n}\left(4-2+\frac{1}{2}\right)\\ ={2}^{n}\left(\frac{5}{2}\right)\\ =5\left(\frac{{2}^{n}}{2}\right)\\ =5\left({2}^{n-1}\right)\\ \therefore p=5\\ \end{array}$


text:
# nLab Leonard S. Charlap

Selected writings

## Selected writings

category: people

Created on May 10, 2022 at 12:19:49. See the history of this page for a list of all contributions to it.


text:
# Algorithms¶

## aif360.algorithms.preprocessing¶

 algorithms.preprocessing.DisparateImpactRemover([…]) Disparate impact remover is a preprocessing technique that edits feature values increase group fairness while preserving rank-ordering within groups [1]_. algorithms.preprocessing.LFR(…[, k, Ax, …]) Learning fair representations is a pre-processing technique that finds a latent representation which encodes the data well but obfuscates information about protected attributes [2]_. algorithms.preprocessing.OptimPreproc(…[, …]) Optimized preprocessing is a preprocessing technique that learns a probabilistic transformation that edits the features and labels in the data with group fairness, individual distortion, and data fidelity constraints and objectives [3]_. algorithms.preprocessing.Reweighing(…) Reweighing is a preprocessing technique that Weights the examples in each (group, label) combination differently to ensure fairness before classification [4]_.

## aif360.algorithms.inprocessing¶

 algorithms.inprocessing.AdversarialDebiasing(…) Adversarial debiasing is an in-processing technique that learns a classifier to maximize prediction accuracy and simultaneously reduce an adversary’s ability to determine the protected attribute from the predictions [5]_. algorithms.inprocessing.ARTClassifier(…) Wraps an instance of an art.classifiers.Classifier to extend Transformer. algorithms.inprocessing.GerryFairClassifier([…]) Model is an algorithm for learning classifiers that are fair with respect to rich subgroups. algorithms.inprocessing.MetaFairClassifier([…]) The meta algorithm here takes the fairness metric as part of the input and returns a classifier optimized w.r.t. algorithms.inprocessing.PrejudiceRemover([…]) Prejudice remover is an in-processing technique that adds a discrimination-aware regularization term to the learning objective [6]_.

## aif360.algorithms.postprocessing¶

 algorithms.postprocessing.CalibratedEqOddsPostprocessing(…) Calibrated equalized odds postprocessing is a post-processing technique that optimizes over calibrated classifier score outputs to find probabilities with which to change output labels with an equalized odds objective [7]_. algorithms.postprocessing.EqOddsPostprocessing(…) Equalized odds postprocessing is a post-processing technique that solves a linear program to find probabilities with which to change output labels to optimize equalized odds [8]_ [9]_. algorithms.postprocessing.RejectOptionClassification(…) Reject option classification is a postprocessing technique that gives favorable outcomes to unpriviliged groups and unfavorable outcomes to priviliged groups in a confidence band around the decision boundary with the highest uncertainty [10]_.

## aif360.algorithms¶

 algorithms.Transformer(**kwargs) Abstract base class for transformers.


text:
1. ## functions

can someone help me out with this question?

f(xy)=f(x)f(y), f(0) does not equal 0. Find the value of f(2003).

2. ## Re: functions

Note that $f(0) = f(0)f(0)$. Since $f(0) \neq 0$, $f(0) = 1$.

Hence, for any k, $f(0k) = f(k)f(0) \Rightarrow 1 = f(k)(1) \Rightarrow f(k) = 1$ for all k. Therefore, $f(2003) = 1$.


text:
# What the humidity metric is hiding?

So I have been spending time in Moscow Russia which essentially records very similar climatic humidity to my home city of London. However in Moscow my throat is permanently dry, I need to consume more water, many people have humidifiers in there bedrooms and the government spray the roads with water to prevent citizens choking on the dry dusty air.

Clearly the climate is 'drier' (or another more accurate technical adjective). What metric will show this difference between Moscow and London ?

And why is the humidity metric the same ?

• What you say is very odd. When I lived in London I was taking lessons in Russian language, and my language teacher ( a Russian) said he felt the cold much more in London than in Moscow, although Moscow was colder according to the thermometer. He said the reason for this was that the London cold was a damp cold, whereas in Moscow the air was dry. From what you say, he was right, so I don't see how London and Moscow can have similar humidity, especially as England has a marine climate and Russia doesn't. – Michael Walsby Oct 31 '19 at 13:24
• Is this a difference between relative and absolute humidity? Not sure how it feels factors in, but if Moscow is colder then the same relative humidity will be a lower absolute humidity. – winwaed Oct 31 '19 at 13:28

Weather services rarely state the absolute humidity because it is not easy to determine. Instead, they state the relative humidity.

Absolute humidity is the mass of water vapor divided by the mass of dry air in a volume of air at a given temperature. The hotter the air is, the more water it can contain. Absolute humidity is expressed as grams of moisture per cubic meter of air ($$g/m^3$$).

Relative humidity is the ratio of the current absolute humidity to the highest possible absolute humidity (which depends on the current air temperature). A reading of 100 percent relative humidity means that the air is totally saturated with water vapor and cannot hold any more, creating the possibility of rain.

A better measure of how atmospheric moisture will affect people and the environment is the dew point.

The dew point is, in short, the point at which dew droplets form on objects like grass – in other words, when a relative humidity of 100 percent is achieved.

If you want a real judge of just how "dry" or "humid" it will feel outside, look at the dew point instead of the RH. The higher the dew point, the muggier it will feel.

• thanks, this is very interesting – Iain Watt Nov 4 '19 at 10:55


text:
## 20070429

### You weary giants of Roombas and broomsticks

Today, I was reading "I am a strange loop", and while immersed in a story about virtual presence, I realized that it would be really cool if you could build a telepresence robot out of, say, a Roomba, a broomstick, and a Macbook.

Some clever guys obviously beat me to it. Check out these. For those who don't want to bother clicking the link: it's a telepresence robot that looks like... well, not all that different from a Macbook on a broomstick on a Roomba. Note that they plan to sell these at "between $1800 and$3000" in "2008"; by my estimate, the lower end of that might be less than the cost of the parts you'd need to make one.

And here's another fun question: what is the interaction of telepresence and immigration laws? If I live in one country, but my job is "in" another, where am I legally employed? This question first became very real to me when I lived in the US as a student and wondered what the legal consequences would be if I, through the primitive telepresence technologies of e-mail, telephone and ssh, were to do a little free-lance work for a Dutch company in Holland, paid in Euros on my Dutch bank account, with the Dutch company possibly never even realizing I was in California? (I never did it, but an American I know here does do the exact opposite.)

And why shouldn't he? Given ever improved telepresence technologies (I am really starting to like that buzzword, even though it's probably already gone out of style), immigration laws start seeming not only backwards and selfish, but positively hilarious. And that last thing is a good thing, because in the end, the only way you can really fight The Man is to poke fun at Him...

I have more to say about the political implications of this, but meanwhile, you can ponder whether John Perry Barlow was onto something when he wrote this (emphasis mine):

Governments of the Industrial World, you weary giants of flesh and steel, I come from Cyberspace, the new home of Mind. On behalf of the future, I ask you of the past to leave us alone. You are not welcome among us. You have no sovereignty where we gather.

We have no elected government, nor are we likely to have one, so I address you with no greater authority than that with which liberty itself always speaks. I declare the global social space we are building to be naturally independent of the tyrannies you seek to impose on us. You have no moral right to rule us nor do you possess any methods of enforcement we have true reason to fear.

Governments derive their just powers from the consent of the governed. You have neither solicited nor received ours. We did not invite you. You do not know us, nor do you know our world. Cyberspace does not lie within your borders. Do not think that you can build it, as though it were a public construction project. You cannot. It is an act of nature and it grows itself through our collective actions.


text:
Hc Verma I Solutions for Class 12 Science Physics Chapter 1 Introduction To Physics are provided here with simple step-by-step explanations. These solutions for Introduction To Physics are extremely popular among Class 12 Science students for Physics Introduction To Physics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma I Book of Class 12 Science Physics Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 12 Science Physics are prepared by experts and are 100% accurate.

Question 1:

The metre is defined as the distance travelled by light in second. Why didn't people choose some easier number such as second? Why not 1 second?

The speed of light in vacuum is 299,792,458 m/s.
Then time taken by light to cover a distance of 1 metre in vacuum = s
Hence, the metre is defined as the distance travelled by light in s.
As 300,000,000 m/s is an approximate speed of light in vacuum, it cannot be used to define the metre.

The distance travelled by light in one second is 299,792,458 m. This is a large quantity and cannot be used as a base unit. So, the metre is not defined in terms of second.

Question 2:

What are the dimensions of:
(a) volume of a cube of edge a,
(b) volume of a sphere of radius a,
(c) the ratio of the volume of a cube of edge a to the volume of a sphere of radius a?

(a) Volume of a cube of edge a, $V=a×a×a$
i.e., [V] = L $×$ L $×$ L = L3

(b) Volume of a sphere of radius a, $V=\frac{4}{3}\pi \left(a{\right)}^{3}$
i.e., [V] = L $×$ L $×$ L = L3

(c) The ratio of the volume of the cube to the volume of the sphere is a dimensionless quantity.

Question 3:

Suppose you are told that the linear size of everything in the universe has been doubled overnight. Can you test this statement by measuring sizes with a metre stick? Can you test it by using the fact that the speed of light is a universal constant and has not changed? What will happen if all the clocks in the universe also start running at half the speed?

The validity of this statement cannot be tested by measuring sizes with a metre stick, because the size of the metre stick has also got doubled overnight.
Yes, it can be verified by using the fact that speed of light is a universal constant and has not changed.
If the linear size of everything in the universe is doubled and all the clocks in the universe starts running at half the speed, then we cannot test the validity of this statement by any method.

Question 4:

If all the terms in an equation have same units, is it necessary that they have same dimensions? If all the terms in an equation have same dimensions, is it necessary that they have same units?

Yes, if all the terms in an equation have the same units, it is necessary that they have the same dimension.

No, if all the terms in an equation have the same dimensions, it is not necessary that they have the same unit. It is because two quantities with different units can have the same dimension, but two quantities with different dimensions cannot have the same unit. For example angular frequency and frequency, both have the dimensions $\left[{T}^{-1}\right]$ but units of angular frequency is rad/s and frequency is Hertz.Another example is energy per unit volume and pressure.Both have the dimensions of but units of pressure is N/m2 and that of energy per unit volume is J/m3

Question 5:

If two quantities have same dimensions, do they represent same physical content?

No, even if two quantities have the same dimensions, they may represent different physical contents.
Example: Torque and energy have the same dimension, but they represent different physical contents.

Question 6:

It is desirable that the standards of units be easily available, invariable, indestructible and easily reproducible. If we use foot of a person as a standard unit of length, which of the above features are present and which are not?

If we use the foot of a person as a standard unit of length,features that  will not be present are  variability, destructibility and  reproducible nature and the feature that will be present is the availability of  the foot of a person to measure any length.

Question 7:

Suggest a way to measure:
(a) the thickness of a sheet of paper,
(b) the distance between the sun and the moon.

(a) The thickness of a sheet of paper can roughly be determined by measuring the height of a stack of paper.
Example: Let us consider a stack of 100 sheets of paper. We will use a ruler to measure its height. In order to determine the thickness of a sheet of paper, we will divide the height of the stack with the number of sheets (i.e., 100).

(b) The distance between the Sun and the Moon can be measured by using Pythagoras theorem when the Earth makes an angle of 90âˆ˜ with the Sun and the Moon. We already know the distances from the Sun to the Earth and from the Earth to the Moon. However, these distances keep on changing due to the revolution of the Moon around the Earth and the revolution of the Earth around the Sun.

Question 1:

Which of the following sets cannot enter into the list of fundamental quantities in any system of units?
(a) length, mass and velocity,
(b) length, time and velocity,
(c) mass, time and velocity,
(d) length, time and mass.

(b) length, time and velocity

We define length and time separately as it is not possible to define velocity without using these quantities. This means that one fundamental quantity depends on the other. So, these quantities cannot be listed as fundamental quantities in any system of units.

Question 2:

A physical quantity is measured and the result is expressed as nu where u is the unit used and n is the numerical value. If the result is expressed in various units then
(a)

(b)

(c)

(d) .

(d)

The larger the unit used to express the physical quantity, the lesser will be the numerical value.

Example:
1 kg of sugar can be expressed as 1000 g or 10000 mg of sugar.
Here, g (gram) is the larger quantity as compared to mg (milligram), but the numerical value used with gram is lesser than the numerical value used with milligram.

Question 3:

Suppose a quantity x can be dimensionally represented in terms of M, L and T, that is, . The quantity mass
(a) can always be dimensionally represented in terms of L, T and x,
(b) can never be dimensionally represented in terms of L, T and x,
(c) may be represented in terms of L, T and x if a = 0,
(d) may be represented in terms of L, T and x if $a\ne 0$.

(d) may be represented in terms of L, T and x if $a\ne 0$

If a = 0, then we cannot represent mass dimensionally in terms of L, T and x, otherwise it can be represented in terms of L, T and x.

Question 4:

A dimensionless quantity
(a) never has a unit,
(b) always has a unit,
(c) may have a unit,
(d) does not exist.

(a) may have a unit

Dimensionless quantities may have units.

Question 5:

A unitless quantity
(a) never has a non-zero dimension,
(b) always has a non-zero dimension,
(c) may have a non-zero dimension,
(d) does not exist.

(a) never has a non-zero dimension

A unitless quantity never has a non-zero dimension.

Question 6:

.

The value of n is
(a) 0
(b) −1
(c) 1
(d) none of these.
You may use dimensional analysis to solve the problem.

(a) 0

[ax] = [x2]
⇒ [a] = [x]    ...(1)

Dimension of LHS = Dimension of RHS

Question 1:

The dimensions ML−1 T−2 may correspond to
(a) work done by a force
(b) linear momentum
(c) pressure
(d) energy per unit volume.

(c) pressure
(d) energy per unit volume

[Work done] = [ML2T−2]
[Linear momentum] = [MLT−1]
[Pressure] = [ML−1 T−2]
[Energy per unit volume] = [ML−1 T−2]

From the above, we can see that pressure and energy per unit volume have the same dimension, i.e., ML−1 T−2.

Question 2:

Choose the correct statements(s):
(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(c) A dimensionally incorrect equation may be correct.
(d) A dimensionally incorrect equation may be incorrect.

(a) A dimensionally correct equation may be correct.
(b) A dimensionally correct equation may be incorrect.
(d) A dimensionally incorrect equation may be incorrect.

It is not possible that a dimensionally incorrect equation is correct. All the other situations are possible.

Question 3:

Choose the correct statements(s):
(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.
(d) The dimension of a derived quantity is never zero in any base quantity.

The statements which are correct are:
(a) All quantities may be represented dimensionally in terms of the base quantities.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities.
(c) The dimensions of a base quantity in other base quantities is always zero.

Statement (d) is not correct because A derived quantity can exist which is dimensionless for example fine structure constant which is given by

Question 1:

Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure.

(a) Linear momentum = mv
Here, [m] = [M] and [v] = [LT−1]
∴ Dimension of linear momentum, [mv] = [MLT−1]

(b) Frequency = $\frac{1}{\mathrm{Time}}$
∴ Dimension of frequence = $\left[\frac{1}{T}\right]=\left[{\mathrm{M}}^{0}{\mathrm{L}}^{0}{\mathrm{T}}^{-1}\right]$

(c) Pressure = $\frac{\mathrm{Force}}{\mathrm{Area}}$

Question 2:

Find the dimensions of
(a) angular speed ω,
(b) angular acceleration α,
(c) torque $\tau$ and
(d) moment of interia I.
Some of the equations involving these quantities are
.
The symbols have standard meanings.

(a) Dimensions of angular speed, $\mathrm{\omega }=\frac{\mathrm{\theta }}{t}=\left[{\mathrm{M}}^{0}{\mathrm{L}}^{0}{\mathrm{T}}^{-1}\right]$

(b) Angular acceleration,
Here, ω = [M0L0T−1] and t = [T]
So, dimensions of angular acceleration = [M0L0T−2]

(c) Torque, τ =Frsin$\theta$
Here, F = [MLT−2] and r = [L]
So, dimensions of torque = [ML2T−2]

(d) Moment of inertia = mr2
Here, m = [M] and r2 = [L2]
So, dimensions of moment of inertia = [ML2T0]

Question 3:

Find the dimensions of
(a) electric field E.
(b) magnetic field B and
(c) magnetic permeability ${\mu }_{0}$.
The relevant equation are

where F is force, q is charge, v is speed, I is current, and a is distance.

(a) Electric field is defined as electric force per unit charge.
i.e., $\mathrm{E}=\frac{F}{q}$

(b) Magnetic field, $B=\frac{\mathit{F}}{\mathit{q}\mathit{v}}$

(c) Magnetic permeability, ${\mathrm{\mu }}_{0}=\frac{B×2\pi r}{I}$

Question 4:

Find the dimensions of
(a) electric dipole moment p and
(b) magnetic dipole moment M.
The defining equations are p = q.d and M = IA;
where d is distance, A is area, q is charge and I is current.

(a) Electric dipole moment, P = q.(2d)
Here, [q] = [AT] and d = [L]
∴ Dimension of electric dipole moment = [LTA]

(b) Magnetic dipole moment, M = IA
Here, A = [L2]
∴ Dimension of magnetic dipole moment = [L2A]

Question 5:

Find the dimensions of Planck's constant h from the equation E = hv where E is the energy and v is the frequency.

E = hv, where E is the energy and v is the frequency

Question 6:

Find the dimensions of
(a) the specific heat capacity c,
(b) the coefficient of linear expansion α and
(c) the gas constant R.
Some of the equations involving these quantities are
and $PV=nRT$.

(a) Specific heat capacity, $C=\frac{Q}{m∆T}$

(b) Coefficient of linear expansion,$\alpha =\frac{{L}_{1}-{L}_{0}}{{L}_{0}∆T}$
So, $\left[\alpha \right]=\frac{\left[\mathrm{L}\right]}{\left[\mathrm{LK}\right]}=\left[{\mathrm{K}}^{-1}\right]$

(c) Gas constant, $\mathrm{R}=\frac{\mathrm{PV}}{n\mathrm{T}}$

Question 7:

Taking force, length and time to be the fundamental quantities, find the dimensions of
(a) density,
(b) pressure,
(c) momentum and
(d) energy.

Let F be the dimension of force.

(a) Density$=\frac{m}{v}=\frac{\left(\mathrm{force}/\mathrm{acceleration}\right)}{\mathrm{volume}}$

(b) Pressure$=\frac{\mathrm{force}}{\mathrm{area}}\phantom{\rule{0ex}{0ex}}$

(c) Momentum = mv = (force/acceleration) × velocity

(d) Energy$=\frac{1}{2}m{v}^{2}=\frac{\mathrm{force}}{\mathrm{acceleration}}×{\left(\mathrm{velocity}\right)}^{2}$

Question 8:

Suppose the acceleration due to gravity at a place is 10 m/s2. Find its value if cm/(minute)2.

Acceleration due to gravity, g = 10 $m/{s}^{2}$

g = 10 m/s2 = 10 × 100 cm $×\frac{1}{{\left(\frac{1}{60}\mathrm{min}\right)}^{2}}$
g = 1000 × 3600 cm/min2 = 36 × 105 cm/min2

Question 9:

The average speed of a snail is $0·020$ miles/ hour and that of a leopard is 70 miles/ hour. Convert these speeds in SI units.

1 mi = 1.6 km
1 km = 1000 m

For the snail, average speed = 0.02 mi/h =

For the leopard, average speed = 70 mi/h =

Question 10:

The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data : Specific gravity of mercury = $13·6$, Density of at Calcutta. Pressure $=h\rho g$ in usual symbols.

Height, h = 75 cm = 0.75 m
Density of mercury = 13600 kg/m3
g = 9.8 m/s2

In SI units, pressure = hρg = 0.75 × 13600 × 9.8 = 10 × 104 N/m2 (approximately)

In CGS units, pressure = 10 × 104 N/m2

Question 11:

Express the power of a 100 watt bulb in CGS unit.

In SI unit, watt = joule/s
In CGS unit, 1 joule = 107 erg
So, 100 watt = 100 joule/s

Question 12:

The normal duration of I.Sc. Physics practical period in Indian colleges is 100 minutes. Express this period in microcenturies. 1 microcentury = 106 × 100 years. How many microcenturies did you sleep yesterday?

1 microcentury = 10−6 × 100 years = 10−4 × 365 × 24 × 60 minutes

Suppose, I slept x minutes yesterday.
x min = 0.019x microcenturies

Question 13:

The surface tension of water is 72 dyne/cm. Convert it in SI unit.

1 dyne = 10−5 N
1 cm = 10  −2 m

Question 14:

The kinetic energy K of a rotating body depends on its moment of inertia I and its angular speed ω. Assuming the relation to be $K=k{I}^{a}{\omega }^{B}$ where k is a dimensionless constant, find a and b. Moment of inertia of a sphere about its diameter is $\frac{2}{5}M{r}^{2}$

Kinetic energy of a rotating body is K = kI aωb.

Dimensions of the quantities are [K] = [ML2T−2], [I] = [ML2] and [ω] = [T−1].

Now, dimension of the right side are [I]a = [ML2]a and [ω]b = [T−1]b.

According to the principal of homogeneity of dimension, we have:
[ML2T−2] = [ML2]a [T−1]b

Equating the dimensions of both sides, we get:
2 = 2a
a = 1
And,
−2 = −b
b = 2

Question 15:

Theory of relativity reveals that mass can be converted into energy. The energy E so obtained is proportional to certain powers of mass m and the speed c of light. Guess a relation among the quantities using the method of dimensions.

According to the theory of relativity, E α macb
⇒ E = kmacb, where k = proportionality constant

Dimension of the left side, [E] = [ML2T−2]

Dimension of the right side, [macb]= [M]a [LT−1]b

Equating the dimensions of both sides, we get:
[ML2T−2] = [M]a [LT−1]b
a = 1, b = 2
E = kmc2

Question 16:

Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm's law, product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for R and V are ${\mathrm{ML}}^{2}{\mathrm{I}}^{-2}{\mathrm{T}}^{-3}$ and ${\mathrm{ML}}^{2}{\mathrm{T}}^{-3}{\mathrm{I}}^{-1}$ respectively.

Dimensional formula of resistance, [R] = [ML2A−2T−3]    ...(1)
Dimensional formula of potential difference, [V] = [ML2A−1T−3]    ...(2)
Dimensional formula of current,  I = [A]

Dividing (2) by (1), we get:
$\frac{\left[V\right]}{\left[R\right]}=\frac{\left[{\mathrm{ML}}^{2}{\mathrm{A}}^{-1}{\mathrm{T}}^{-3}\right]}{\left[{\mathrm{ML}}^{2}{\mathrm{A}}^{-2}{\mathrm{T}}^{-3}\right]}=\left[\mathrm{A}\right]$
⇒ V = IR

Question 17:

The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

Frequency, f $\propto$LaFbmc
f = kLaFbmc   ...(1)
Dimension of [f] = [T−1]

Dimension of the right side components:
[L] = [L]
[F] = [MLT−2]
[m] = [ML−1]

Writing equation (1) in dimensional form, we get:
[T−1] = [L]a [MLT−2]b [ML−1]c
[M0L0T−1] = [Mb + c La + bc T−2b]

Equating the dimensions of both sides, we get:
b + c = 0               ....(i)
a + bc = 0      ....(ii)
−2b = −1              ....(iii)
Solving equations (i), (ii) and (iii), we get:

∴ Frequency, f = kL−1F1/2m−1/2

Question 18:

Test if the following equations are dimensionally correct:
(a)

(b) $v=\sqrt{\frac{P}{\rho }},$

(c)

(d)
where h = height, S = surface tension, $\rho$ = density, P = pressure, V = volume, $\eta =$ coefficient of viscosity, v = frequency and I = moment of interia.

(a)
Height, [h] = [L]
Surface Tension, $\left[\mathit{S}\right]=\frac{\left[F\right]}{\left[L\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]}=\left[{\mathrm{MT}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[\mathrm{I}\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^{0}\right]$
Radius, [r] = [L], [g]= [LT−2]
Now,

Since the dimensions of both sides are the same, the equation is dimensionally correct.

(b) $\nu =\sqrt{\left(\frac{\mathit{P}}{\mathit{\rho }}\right)}$
Velcocity, [ν] = [LT−1]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
Density, $\left[\rho \right]=\frac{\left[M\right]}{\left[V\right]}=\left[{\mathrm{ML}}^{-3}{\mathrm{T}}^{0}\right]$
Now, $\sqrt{\frac{\mathit{P}}{\mathit{\rho }}}={\left[\frac{\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{ML}}^{-3}\right]}\right]}^{\frac{1}{2}}={\left[{\mathrm{L}}^{2}{\mathrm{T}}^{-2}\right]}^{1/2}=\left[{\mathrm{LT}}^{-1}\right]$
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(c)
Volume, [V] = [L3]
Pressure, $P=\frac{\left[F\right]}{\left[A\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-2}\right]$
[r]= [L] and [t] = [T]
Coefficient of viscosity,
$\left[\eta \right]=\frac{\left[F\right]}{6\mathrm{\pi }\left[\mathrm{r}\right]\left[\mathrm{v}\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[\mathrm{L}\right]\left[{\mathrm{LT}}^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$

Now,
Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

(d) $\mathrm{\nu }=\frac{1}{2\pi }\sqrt{\left(\frac{mgl}{I}\right)}$
Frequency, ν = [T−1]

Since the dimensions of both sides of the equation are the same, the equation is dimensionally correct.

Question 19:

Let x and a stand for distance. Is $\int \frac{dx}{\sqrt{{a}^{2}-{x}^{2}}}=\frac{1}{a}{\mathrm{sin}}^{-1}\frac{a}{x}$ dimensionally correct?

Dimension of the left side of the equation= $\left[\int \frac{dx}{\sqrt{\left({a}^{2}-{x}^{2}\right)}}\right]=\left[\int \frac{\mathrm{L}}{\sqrt{\left({\mathrm{L}}^{2}-{\mathrm{L}}^{2}\right)}}\right]=\left[{\mathrm{L}}^{0}\right]$


text:
# What are the mean and standard deviation of the probability density function given by (p(x))/k=x^3-49x for  x in [0,7], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Jan 27, 2016

Integrate p(x) over $\left[0 , 7\right]$, set equal to 1, then solve for k ...

#### Explanation:

$k = \frac{- 4}{2401}$

Next, using k above, integrate $x f \left(x\right)$ over $\left[0 , 7\right]$ to find $E \left(X\right)$

$E \left(X\right) = \frac{56}{15}$

Now, using k above, integrate ${x}^{2} f \left(x\right)$ over $\left[0 , 7\right]$ to find $E \left({X}^{2}\right)$

$E \left({X}^{2}\right) = \frac{49}{3}$

Find the Variance:

${\sigma}^{2} = E \left({X}^{2}\right) = {\left[E \left(X\right)\right]}^{2} = \frac{49}{3} - {\left(\frac{56}{15}\right)}^{2} = \frac{539}{225}$

Finally, standard deviation $\sigma = \sqrt{{\sigma}^{2}} = \frac{7 \sqrt{11}}{15}$

If you really need to know the mean and standard deviation in terms of k , then simply divide these parameters by $k = \frac{- 4}{2401}$, then multiply each by the variable $k$.

hope that helped!

Note: Used the Solver feature of my TI-84 to find all these values:)


text:
# Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$

Physics – Nuclear Physics – Nuclear Experiment

Scientific paper

[ 0.00 ] – not rated yet Voters 0   Comments 0

## Details Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$ Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$

30 pages, 17 figures, 4 tables. Final version to appear in Phys. Rev. C

Scientific paper

10.1103/PhysRevC.85.045203

Precise measurements of the proton electromagnetic form factor ratio $R = \mu_p G_E^p/G_M^p$ using the polarization transfer method at Jefferson Lab have revolutionized the understanding of nucleon structure by revealing the strong decrease of $R$ with momentum transfer $Q^2$ for $Q^2 \gtrsim 1$ GeV$^2$, in strong disagreement with previous extractions of $R$ from cross section measurements. In particular, the polarization transfer results have exposed the limits of applicability of the one-photon-exchange approximation and highlighted the role of quark orbital angular momentum in the nucleon structure. The GEp-II experiment in Jefferson Lab's Hall A measured $R$ at four $Q^2$ values in the range 3.5 GeV$^2 \le Q^2 \le 5.6$ GeV$^2$. A possible discrepancy between the originally published GEp-II results and more recent measurements at higher $Q^2$ motivated a new analysis of the GEp-II data. This article presents the final results of the GEp-II experiment, including details of the new analysis, an expanded description of the apparatus and an overview of theoretical progress since the original publication. The key result of the final analysis is a systematic increase in the results for $R$, improving the consistency of the polarization transfer data in the high-$Q^2$ region. This increase is the result of an improved selection of elastic events which largely removes the systematic effect of the inelastic contamination, underestimated by the original analysis.

No associations

LandOfFree

## Say what you really think

Search LandOfFree.com for scientists and scientific papers. Rate them and share your experience with other people.

## Rating

Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$ does not yet have a rating. At this time, there are no reviews or comments for this scientific paper.

If you have personal experience with Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$, we encourage you to share that experience with our LandOfFree.com community. Your opinion is very important and Final Analysis of Proton Form Factor Ratio Data at $\mathbf{Q^2 = 4.0}$, 4.8 and 5.6 GeV$\mathbf{^2}$ will most certainly appreciate the feedback.

Profile ID: LFWR-SCP-O-426013

All data on this website is collected from public sources. Our data reflects the most accurate information available at the time of publication.


text:
# How to download the whole BLAST nt database into a specific folder?

I have successfully downloaded the whole nt BLAST database into the current folder using:

wget -b "ftp://ftp.ncbi.nlm.nih.gov/blast/db/nt.??.tar.gz"

However, I would like to download it in a specific folder, e.g. output/

I have tried:

wget -b "ftp://ftp.ncbi.nlm.nih.gov/blast/db/nt.??.tar.gz"  -O output/

but it does not work. Nothing happens. My 'wget-log' is empty.

I could move to that folder and download it there but I would like to remain in the same working directory.

C.

-O is to specify the name of the output file. You want the -P option.
wget -b "ftp://ftp.ncbi.nlm.nih.gov/blast/db/nt.??.tar.gz" -P output/


text:
How to evaluate $\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$?

$$\int \limits_{-\infty}^{\infty}\frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx$$

-
It might be worth noting that the range for $\sin x$ is $-1\leq \sin x \leq 1$. Therefore $|1−\sin x|$ is always positive and so you can remove the absolute value in the denominator. –  E.O. Feb 16 '12 at 10:31
This is the integral on $(0,2\pi)$ of the same function without the absolute values, times $2/(1-\mathrm e^{-2\pi})$. –  Did Feb 16 '12 at 11:21
To see what Didier Piau means you might divide the interval into intervals of length $2\pi$ and use periodicity of the denominator together with the fact that $\exp(-x-2n\pi)= \exp(-2n\pi)\exp(-x)$ (giving a geometrical series). –  AD. Feb 16 '12 at 12:17
@DidierPiau I see what you mean, but not if it helps.. –  AD. Feb 16 '12 at 12:18
Is this homework? For what class? Are you supposed to know complex analysis? –  bgins Feb 16 '12 at 14:14

Here are my thoughts so far -- not an answer, just a possible start.

Observing that $\sin x\le1$, we can easily remove the absolute values: $$I=\int_{-\infty}^{\infty} \frac{e^{-|x|}}{|1-\sin x|^{\frac{1}{4}}} \,dx =2\int_{0}^{\infty} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx.$$ A major challenge of solving this integral with a substitution is that the denominator is periodic, with poles at $x=(2n+\frac12)\pi$. For example, with $z=\tan\frac{x}{2}$, we'd have $dz=\frac12\sec^2\frac{x}{2}dx$ $=$ $\frac12(1+z^2)dx$ or $dx=\frac{2}{1+z^2}dz$ and $x=\frac{2z}{1+z^2}$ which would yield $$I=2\int_{0}^{?} \frac{e^{-2\tan^{-1}z}}{(1-\sin x)^{\frac{1}{4}}} \,dx,$$ but there would be no way to transform the upper limits integration! However, this problem can be surmounted by integrating separately over each period of $\sin x$ to obtain \eqalign{ I &= 2\sum_{n=0}^{\infty} \int_{2\pi n}^{2\pi(n+1)} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx \cr &= 2\sum_{n=0}^{\infty} e^{-2n\pi} \int_{0}^{2\pi} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx \cr &= \frac2{1-e^{-2\pi}} \int_{0}^{2\pi} \frac{e^{-x}}{(1-\sin x)^{\frac{1}{4}}} \,dx. } This is an improper integral, but it exists because $1-\sin x$ as $x$ approaches $\frac\pi 2$ behaves the same way that $1-\cos x$ does as $x$ approaches $0$, namely like $x^2$, and $\int_0^1\frac{dx}{x^p}$ exists for $0<p<1.$

At this point we have at least a few options to pursue. We could:

• use a Taylor series expansion for $(1-\sin x)^{-\frac14}=\sum a_nx^n$ and integrate each term $\sum a_nx^ne^{-x}$

• carry out a substitution like the one above, or like $x=t^2$ to get $$I=\frac2{1-e^{-2\pi}} \int_0^\sqrt{2\pi}\frac{2te^{-t^2}dt}{(1-\sin t^2)^{\frac14}} %\quad\text{or}\quad %I=2\int_0^\infty\frac{2te^{-t^2}dt}{(1-\sin t^2)^{\frac14}}$$

• use $\int e^{-x}(1+\sin x)^{\frac14}(\cosh ix)^{-\frac12}\,dx$ and possibly end up integrating in the complex plane.

There is presumably a way since sage reports $3.2916690469253642$, but I haven't checked how it's getting that.

-
This is what Mathematica gives for the integral that you got stuck $$-\left(\frac{4}{25}-\frac{28 i}{25}\right) \left(\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)-(1+i) \, _2F_1\left(\frac{3}{4}+i,1;\frac{1}{4}+i;-i\right)\right)+\left(\frac{4}{25}-\frac {28 i}{25}\right) e^{-2 \pi } \left(\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)-(1+i) \, _2F_1\left(\frac{3}{4}+i,1;\frac{1}{4}+i;-i\right)\right)+\frac{2^{3/4} e^{-\pi /2} \sqrt{\pi } \Gamma \left(\frac{1}{4}+i\right)}{\Gamma \left(\frac{3}{4}+i\right)}$$ –  userNaN Feb 16 '12 at 13:23
I noticed that (1) the Taylor series for $(\frac{1-\cos x}{2})^{-\frac14}/\sqrt x$ converges very nicely and (2) that $x=t^2$ might take the integral slightly closer in your direction. What is $F_1$? –  bgins Feb 16 '12 at 13:36
Looks like Barnes integral: en.wikipedia.org/wiki/Hypergeometric_function#Barnes_integral. –  bgins Feb 16 '12 at 14:12
Oh, I forgot to simplify the result $$\frac{1}{425 \Gamma \left(\frac{3}{4}+i\right)}\Biggl(e^{-2 \pi } \left((176+432 i) \left(e^{2 \pi }-1\right) \left(-1+\, _2F_1\left(-\frac{1}{4}+i,1;\frac{1}{4}+i;-i\right)\right) \Gamma \left(\frac{7}{4}+i\right)+425\ 2^{3/4} e^{3 \pi /2} \sqrt{\pi } \Gamma \left(\frac{1}{4}+i\right)\right)\Biggr)$$ –  userNaN Feb 16 '12 at 14:22


text:
## 2012-08-29

### Fundamental Rule of Exponents

For a basic algebra class, given the rudimentary order-of-operations that looks like this:
1. Parentheses
2. Exponents & Radicals
3. Multiplication & Division
4. Addition & Subtraction
Then we have:

The Fundamental Rule of Exponents: Operations on same-base powers shift one place down in the order of operations.

Cases:
(1) Exponents will multiply powers, i.e., (am)n = am∙n. Example: (x6)2 = x12.
(2) Radicals will divide powers, i.e., n√am = am/n. Example: 3√x15 = x5.
(3) Multiplying will add powers, i.e., am∙an = am+n. Example: x4∙x7 = x11.
(4) Division will subtract powers, i.e., am/an = am−n. Example: x9/x2 = x7.

We've discussed this before, but I just recently decided to apply the name shown here to the pattern. It doesn't show up on a Google search yet, so I think it's fair-game to do so. Cheers!


text:
# Homework Help: 2 questions regarding circular motion and the law of gravity

1. Jan 2, 2007

### Koolaid

hello i am new to the forums. But I have been working on a packet of problems and I got them all done but i have been stuck on these two for a good hour. I have tried to search to internet for possible equations i could use but can't find any that will work or maybe i am just over looking them so i decided to post here and see what some other peoples thoughts are. But I am looking for some guidance on what equations to use.

1. The problem statement, all variables and given/known data

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4000miles). At this location, the acceleration due to gravity is what factor times the value of g at the Earth's surface?

-thoughts
i think im ight be using

F = GMm/r@2

where G is the universal gravitational constant.

G = 6.67*10-11 N-m2/kg2

and F = mg

2. The problem statement, all variables and given/known data

What angular velocity (in revolutions/second) is needed for a centrifuge to produce an acceleration of 1000g at a radius arm of 15cm?

-thoughts

none at the moment

Last edited: Jan 2, 2007
2. Jan 2, 2007

### neutrino

Hi, koolaid. Welcome to PF!

Here are a couple of hints:

#1: Usually, how would you go about finding g at a point at some distance from an object?

#2 What kind of acceleration does the object undergo?

3. Jan 2, 2007

### Koolaid

well for #1 i updated my post

for #2 wouldn't be tangetal acceleration so [tangetal acceleration = radius * angular acceleration] (a=r*a) right?
or would it be centripetal acceleration i am not sure [ centripetal acceleration = tangental velocity squared / radius] (a=v^2/r)

Last edited: Jan 2, 2007
4. Jan 2, 2007

### neutrino

Well, what you've provided is the force on an object of mass 'm' (or 'M') due to the other. The acceleration would be GM/r2 (assuming M is the Earth's mass). You have been provided the distance (from the centre) as twice the Earth's radius. Use that to find that to find the acceleration as a multiple of g (at the surface of the Earth).

Since it has not been mentioned that the speed is changing, it mustrefer to the centripetal accelaration. Write it down in terms of the angular velocity.

5. Jan 2, 2007

### Koolaid

#1. still don't know g or m cause you only have G = 6.67E-11 and r = 4000mi = 6.437E6 m then multiply by 2 = 1.2874E7 m .

#2. a=rw [ centripetal acceleration = radius * angular velocity]

Last edited: Jan 2, 2007
6. Jan 2, 2007

### neutrino

You need not know the values. Moreover, m is not needed since the answer requires you to find only the acceleration due to gravity.

g(r) = GM/r^2

r = 2R , where R is Earth's radius. Therefore,

g(r) = GM/(2R)^2
...
Can you do it from here?
...

The magnitude of centripetal acceleration = w^2*r

7. Jan 2, 2007

### Koolaid

i am still lost.. i am just not really making the "connection" for both

8. Jan 2, 2007

### HallsofIvy

You "need" practically none of the values you have given because the question asks "what factor times the value of g at the Earth's surface". That is, the acceleration at twice the distance from the center of the earth is what number times the acceleration at the surface of the earth? You are really only asked how gravitational accelaration depends on distance from the center of the earth. You don't need to know G, you don't need to know the mass of the earth, you don't need to know the radius of the earth. And, although you must use it, you don't even need to know g! (g, the acceleration due to gravity at the earth's surface is 9.81 m/s2.)

(And be careful not to confuse "G" with "g"!)

You should know the formula: $a= R \omega^2$ where a is the accelleration, R is the radius arm and $\omega$ is the angular velocity (in radians/second- there are $2\pi$ radians in one revolution).

Last edited by a moderator: Jan 2, 2007
9. Jan 2, 2007

### Koolaid

ok i got the first question all done.

but for question 2. i am not sure how to incorporate the 1000 g into the problem.

the radius also needs to be in meters os its .15 meters

10. Jan 2, 2007

### Hootenanny

Staff Emeritus
1000g means one thousand times the acceleration due to gravity [on earth], i.e. $1000 \times 9.81$. So, to rephrase the question, what is the angular velocity required such that the centripetal acceleration is 9810m/s2?

11. Jan 2, 2007

### Koolaid

wow those were simple now that I think about it lol… well now i understand them.. thank you to everyone that helped


text:
# Revision history [back]

Hello, @curios_mind!

I think you found an interesting problem. My appreciation is that $a=-5/8$ is the correct answer (up to complex representation). Let me explain...

Observe the numerator of your expression $f$ can be written as $$\sqrt[3]{2}\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)=\sqrt[3]{2}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)$$ which is nothing more than one of the possible solutions of the equation $x^3=-2$, or alternatively, this is one of the possible values of $\sqrt[3]{-2}$ in the complex plain.

Now, consider the denominator for $a=-5/8$: $$\left(-2\left(\frac{-5}{8}\right)-1\right)^{1/3}=\sqrt[3]{\frac{1}{4}}=\sqrt[3]{\frac{1}{2^2}}$$

So, what $f$ is for $a=-5/8$ turns out to be $$\frac{\sqrt[3]{-2}}{\sqrt[3]{\frac{1}{2^2}}}=\sqrt[3]{-2}\sqrt[3]{2^2}=\sqrt[3]{-2^3}=-2$$ So the original equation is satisfied.

If you want to be more formal, $$\frac{\sqrt[3]{-2}}{\sqrt[3]{\frac{1}{2^2}}}=\sqrt[3]{-2}\sqrt[3]{4}=\sqrt[3]{2}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)\sqrt[3]{4}\bigl(\cos(\pi/3)+i\sin(\pi/3)\bigr)=\sqrt[3]{8}(\cos^2(\pi/3)+\sin^2(\pi/3))=\sqrt[3]{8}=2$$

Of course, as you can see, this depends on you choosing the correct representation of complex numbers like $\sqrt[3]{-4}$. If you choose wrongly, for example, $\sqrt[3]{-4}=\sqrt[3]{4}\bigl(\cos(\pi/3)-i\sin(\pi/3)\bigr)$, you won't get the answer, but other possible value for $f$ in the complex plain. That is what happened when you subs $a=-5/8$: Sage didn't choose the correct complex representations of some cube roots.

In summary, the problem is that you have a complex equation with ambiguous numbers like cube roots. Sage and Maxima did what they could, which is to give you a correct answer, up to complex representation. In this case, it is up to you to decide if this is what you wanted or not. If this is not the solution you expected, or you want an alternative solution, I recommend you to rewrite the cube roots explicitly as one of their possible complex number values.

I hope this helps!


text:
# itemps

0th

Percentile

##### Functions to plot summary information about the sampled inverse temperatures, tree heights, etc., stored in the traces of a "tgp"-class object

Functions for making barplots summarizing the progress of importance tempering. The itemps.barplot function can be used to make a histogram of the inverse temperatures visited in the trans-temporal Markov chain. The hist2bar function is useful for making a histogram of integer-valued samples (e.g., tree heights) encountered in one or several Markov chains

Keywords
hplot
##### Usage
itemps.barplot(obj, main = NULL, xlab = "itemps",
ylab = "counts", plot.it = TRUE, ...)
hist2bar(x)
##### Arguments
obj

"tgp"-class object

main

Main plot label to be augmented by itemps.barplot

xlab

Label for the x-axis

ylab

Label for the y-axis

plot.it

whether to plot the barplot in addition to returning the data.frame for later use in a barplot call

other arguments passed to barplot if plot.it = TRUE

x

matrix of integers whose columns are treated as different realizations of similar processes producing where each row represents a sample (e.g., tree height) under that process

##### Details

itemps.barplot specifically works with the \$trace field of a "tgp"-class object. An error will be produced if this field is NULL, i.e., if the b* function used the create the object was not run with the argument trace=TRUE

The hist2bar function can be used on any integer (or discrete) valued matrix. The columns are interpreted as different realizations of similar processes for comparison with one another via a histogram.

The histogram is obtained with the barplot command used with the argument beside=TRUE. See the examples section of default.itemps

##### Value

Both functions return a data.frame that can be used within the barplot function with argument beside=TRUE

##### References

Gramacy, R.B., Samworth, R.J., and King, R. (2007) Importance Tempering. ArXiv article 0707.4242 https://arxiv.org/abs/0707.4242

https://bobby.gramacy.com/r_packages/tgp/

default.itemps, vignette(tgp2), barplot


text:
Prove that the countable union of sets of cardinality $c$ again has cardinality $c$. [duplicate]

Where $c = 2^{\aleph_0}$.

I have no idea how to tackle this problem. Any help is greatly appreciated!

marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 6 '17 at 11:37

Let $\{ X_n \mid n < \omega \}$ be a countable family of sets of cardinality $2^{\aleph_0}$, and suppose (without loss of generality) that they're pairwise disjoint. For each $n$, you can inject $X_n$ into the interval $(n,n+1) \subseteq \mathbb{R}$. Piecing together these injections yields an injection $\bigcup_n X_n \to \mathbb{R}$.

For each $n < \omega$, the interval $(n,n+1)$ has cardinality $2^{\aleph_0}$. Proving this is an easy exercise: first biject any open interval with $\mathbb{R}$, then use a linear ($\Rightarrow$ bijective) function to scale/translate it to $(n,n+1)$.
Since $|X_n|=2^{\aleph_0}$ for each $n$, there is a bijection $f_n : X_n \to (n,n+1)$, which we can view as an injection $f_n : X_n \to \mathbb{R}$ whose image is equal to $(n,n+1)$. The reason we choose this interval is so that, when we build a function $\bigcup_n X_n \to \mathbb{R}$, distinct sets $X_n$ are embedded into disjoint subsets of $\mathbb{R}$.
Since the sets $X_n$ are pairwise disjoint, there is a unique function $f : \bigcup_n X_n \to \mathbb{R}$ such that $f(x)=f_n(x)$ for each $x \in X_n$. Moreover, $f$ is injective, since its restrictions to each $X_n$ are injective and the images of each $X_n$ are pairwise disjoint.
Since $\bigcup_n X_n$ injects into $\mathbb{R}$, we see that $\left| \bigcup_n X_n \right| \le 2^{\aleph_0}$.


text:
## 17Calculus Limits - Solve By Substitution

##### 17Calculus

Evaluating Limits Using The Technique of Substitution

### Finite Limits, Infinite Limits, Limits At Infinity . . . Terminology Explained

The use of the terms finite limits, infinite limits and limits at infinity are used differently in various books and your instructor may have their own idea of what they mean. In this panel, we will try to break down the cases and explain the various ways these terms can be used as well as how we use them here at 17calculus.

When we talk about limits, we are looking at the $$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$. The various terms apply to the description of $$c$$ and $$L$$ and are shown in the table below. The confusion lies with the terms finite limits and infinite limits. They can mean two different things.

$$\displaystyle{ \lim_{x \to c}{f(x)} = L }$$

when

term(s) used

$$c$$ is finite

limits approaching a finite value or finite limits

$$c$$ is infinite $$\pm \infty$$

limits at infinity or infinite limits

$$L$$ is finite

finite limits

$$L$$ is infinite $$\pm \infty$$

infinite limits

You can see where the confusion lies. The terms finite limits and infinite limits are used to mean two different things, referring to either $$c$$ or $$L$$. It is possible to have $$c = \infty$$ and $$L$$ be finite. So is this an infinite limit or a finite limit? It depends if you are talking about $$c$$ or $$L$$.

How 17calculus Uses These Terms
The pages on this site are constructed based on what $$c$$ is, i.e. we use the terms finite limits and infinite limits based on the value of $$c$$ only ( using the first two rows of the table above and ignoring the last two ). This seems to be the best way since, when we are given a problem, we can't tell what $$L$$ is until we finish the problem, and therefore we are unable to determine what type of problem we have and know what techniques to use until we are done with the problem.

Important: Make sure and check with your instructor to see how they use these terms.

This is the first and most important technique for evaluating limits. It is very simple but you need to understand when you can use it.

Technique 1 - Substitution

We know from the concept of continuity, that if a function, $$f(x)$$, is continuous at the point $$x=c$$, then $$\displaystyle{ \lim_{x \to c}{[f(x)]} = f(c) }$$. So this tells us, that if the function is continuous at the point at which we are taking the limit, then we can just substitute that point into the function to determine the limit and we are done.

We can also substitute the limit value into the function to help us determine what to do next, even if the function is not continuous at the point. Here is what substitution tells us.

Case 1: After substitution, if we get an indeterminate form, we need to use one (or more) of the techniques discussed on other pages (factoring, rationalizing, using trig identities or, if you know derivatives, L'Hôpital's Rule).

Case 2: After substitution, if we get a fraction like $$c/0$$ where $$c$$ is any non-zero finite number this indicates a possible asymptote, so we need to look at what is happening on either side of $$x=c$$. There are 4 possibilities. (These graphs look at $$c=0$$. However, the discussion holds regardless of the value of $$c$$.)

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = +\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

The limit exists and $$\displaystyle{ \lim_{x \to c}{[f(x)]} = -\infty }$$.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = +\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = -\infty }$$

The limit does not exist.

$$\displaystyle{ \lim_{x \to c^-}{[f(x)]} = -\infty }$$ and $$\displaystyle{ \lim_{x \to c^+}{[f(x)]} = +\infty }$$

The limit does not exist.

Note - On this site (and in many textbooks), we say that when $$\displaystyle{ \lim_{x \to c}{[f(x)]} = \infty }$$, the limit exists and the limit is infinity (similarly for negative infinity). Some textbooks and instructors say that the limit does not exist. Check with your instructor to see what they require.

Now let's use this technique on some practice problems.

Practice

Use the technique of substitution to solve these limits, giving your answers in exact, simplified form.

$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 3}{\left[\frac{5x^2-8x-13}{x^2-5}\right]}}$$

Solution

 You always want to try direct substitution first. $$\displaystyle{\lim_{x \rightarrow 3} \left[ \frac{5x^2-8x-13}{x^2-5} \right] }$$ $$\displaystyle{\frac{5(9)-8(3)-13}{9-5} }$$ $$\displaystyle{\frac{45-24-13}{4} }$$ $$\displaystyle{\frac{8}{4} = 2 }$$ In this case, the result is determinate. So the limit is 2.

$$\displaystyle{\lim_{x\rightarrow 3}\left[\frac{5x^2-8x-13}{x^2-5}\right]=2}$$

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to1}{\frac{3x+5}{x+4}}}$$

Solution

### PatrickJMT - 22 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{\lim_{x\to0}{(x^2+x-6)}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to0}{(x^2+x-6)}}$$

Solution

### PatrickJMT - 23 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\to14}{(x+\sqrt{x+11})}}$$

Solution

### PatrickJMT - 24 video solution

video by PatrickJMT

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} }$$

Solution

### Krista King Math - 1921 video solution

video by Krista King Math

$$\displaystyle{ \lim_{x \to 3}{(x^3-4x^2+2x+5)} = 2 }$$

Log in to rate this practice problem and to see it's current rating.

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}$$

Problem Statement

Evaluate this limit using substitution.
$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}}$$

Solution

Direct substitution yields $$\displaystyle{ \lim_{x \rightarrow 0}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]} = \frac{-3}{0}}$$. So the answer is $$+\infty$$, $$-\infty$$ or DNE (Does Not Exist). How do you know which one (without a graph)?
We need to determine the behavior of the function on either side of $$x=0$$. On the left side of zero when $$x < 0$$ and is very close to zero, the numerator is negative. We know this since x is very, very small, so $$x^4$$ is positive but very small and $$5x$$ is very small but negative. Both terms will be dominated by -3, so the numerator is negative. In the denominator, we need to look at the square root; $$x^2$$ is always positive and adding a positive number to 4 means that the number under the square root is always greater than 4, so the square root is always larger than 2. This makes the denominator negative. So we have a negative numerator and a negative denominator, so the limit $$\displaystyle{ \lim_{x \rightarrow 0^-}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$
We use similar logic when $$x > 0$$, to get $$\displaystyle{ \lim_{x \rightarrow 0^+}{\left[ \frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}= +\infty }$$

Since the limit from the right is equal to the limit from the left, the original limit is also $$+\infty$$.

$$\displaystyle{\lim_{x\rightarrow 0}{\left[\frac{x^4+5x-3}{2-\sqrt{x^2+4}}\right]}=+\infty}$$

Log in to rate this practice problem and to see it's current rating.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.