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Slope of a Line - Maple Help
Slope
Main Concept
The slope of a line through two points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$ is the ratio of the change in the y-coordinates to the change in the x-coordinates: $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$.
The phrase rise over run is often used to describe slope, reflecting the common interpretation of the y-coordinate as representing elevation and the x-coordinate as representing (horizontal) distance. Slope is frequently (but not exclusively) denoted by the letter m.
Note:
• The slope of a line does not depend on which points on the line are chosen to compute it.
• The slope of a vertical line is not defined, since the denominator of the calculation would be 0. In some situations, it is reasonable to represent the slope of a vertical line as infinity ($\infty$).
• The slope of a horizontal line is 0.
Visualizing slope
Click or drag in the plot below to draw a segment of a line. The change in the x-coordinates (the run), the change in the y-coordinates (the rise), and the slope of the line are displayed.
More MathApps
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# Find the Sum of the Following Series: 0.5 + 0.55 + 0.555 + ... to N Terms. - Mathematics
Find the sum of the following series:
0.5 + 0.55 + 0.555 + ... to n terms.
#### Solution
We have,
0.5 + 0.55 + 0.555 + ... n terms
$S_n$ = 5 [0.1 + 0.11+0.111 + ... n terms]
$= \frac{5}{9}\left( 0 . 9 + 0 . 99 + 0 . 999 + . . . +\text { to n terms } \right)$
$= \frac{5}{9}\left\{ \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + . . . \text { n terms } \right\}$
$= \frac{5}{9}\left\{ \left( 1 - \frac{1}{10} \right) + \left( 1 - \frac{1}{100} \right) + \left( 1 - \frac{1}{1000} \right) + . . . \text { n terms } \right\}$
$= \frac{5}{9}\left\{ n - \left( \frac{1}{10} + \frac{1}{{10}^2} + \frac{1}{{10}^3} + . . . \text { n terms } \right) \right\}$
$= \frac{5}{9}\left\{ n - \frac{1}{10}\frac{\left( 1 - \left( \frac{1}{10} \right)^n \right)}{\left( 1 - \frac{1}{10} \right)} \right\}$
$= \frac{5}{9}\left\{ n - \frac{1}{9}\left( 1 - \frac{1}{{10}^n} \right) \right\}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.3 | Q 4.4 | Page 28
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# Time required for deceleration to occur
Given the following, with the correct answer. What type of formula would be applied to this problem and how so?
An object falls freely in a straight line and experiences air resistance proportional to its speed; this means its acceleration is a(t)=−kv(t), where k is a positive constant and v is the object's velocity. The speed of the object decreases from 1300 ft/s to 1200 ft/s over a distance of 1400 ft. Approximate the time required for this deceleration to occur.
I attempted to use the formula (1300^2 - 1200^2) / 1400^2 as suggested but it does not give the correct answer.
Correct Answer is: 1.1206
• It sounds like it belongs on physics.se and we don't like to click through to find a problem, particularly a homework problem with no effort shown. – Ross Millikan Jan 28 '18 at 3:39
• What does "divided by sfts/s" mean? – Mauve Jan 28 '18 at 3:44
• This is a problem related to exponential decay/growth @RossMillikan so it should belong here. – bob Jan 28 '18 at 3:49
• @Useless I fixed the problem. – bob Jan 28 '18 at 3:53
The units of your calculation are sec$^{-2}$ so that formula cannot be right. As a simple approximation, the average speed is about $1250$ ft/sec, so the deceleration time is $1400/1250=1.12$ seconds. For more accuracy, you should solve $s=\frac 12at^2+v_0t$ with your data which will lower the average speed slightly and increase the time slightly.
• Thank you for the response and information, so if I were to set up the given information with this formula $s=\frac 12at^2+v_0t$ would it look something like this? s = 1/2(1250)(1.12) + (1.12) – bob Jan 28 '18 at 5:47
• No You can't plug in $1.12-$ that is what we are trying to solve for. You also didn't square anything for the $\frac 12at^2$ term. We know $s=1400$. The velocity drops by $100$ so we have $at=-100$. Plug those in and solve for $t$. We get $1400=\frac 12 (-\frac {100}t)t^2+1300t$ – Ross Millikan Jan 28 '18 at 12:34
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# What is the requirement for bubble sort to complete in 1 pass?
I am working on a problem where you're given $$n$$ distinct numbers, and you want to find the number of permutations such that it takes bubble sort at most 1 pass to complete.
e.g., if n=3, then the following permutations would only require 1 pass
1 2 3
1 3 2
3 1 2
2 1 3
But
3 2 1
2 3 1
would require more than 1 pass. Apparently the answer is $$2^{n - 1}$$, but I am not sure how to prove this for the general $$n$$ case.
My question is, what are the general constraints for a sequence to allow for it to be sorted with 1 pass of bubble sort?
It is difficult for me to come up with a generic formula to generate the permutations for larger $$n$$.
• Hint: let the position of the maximum element to be $i$. What can you say about elements to the left of $i$ and to the right of $i$? Find the number of such permutations for all possible $i$. You'll get the recurrent formula, for which it would be easy to show the answer by induction.
– user114966
Oct 6 '20 at 2:21
• @Dmitry Doesn't both the elements to the left and right of $i$ have to be less than the element at $i$? Oct 6 '20 at 2:34
• Welcome to COMPUTER SCIENCE @SE. I think that to depend on the termination condition of the bubble sort you argue about: please specify in the question. Oct 6 '20 at 6:38
I'm assuming that a pass of bubble sort on the array $$A[1],\ldots,A[n]$$ proceeds as follows:
• If $$A[1] > A[2]$$ then swap $$A[1]$$ and $$A[2]$$.
• If $$A[2] > A[3]$$ then swap $$A[2]$$ and $$A[3]$$.
• ...
• If $$A[n-1] > A[n]$$ then swap $$A[n-1]$$ and $$A[n]$$.
Bubble sort halts after a pass in which no swaps were made.
As is evident, if bubble sort halts after one pass, then the array must have been sorted.
When does bubble sort halt after two passes? If after the first pass, the array is sorted. In this case, if we know which swaps were made, then we can recover the input permutation. Furthermore, different sets of swaps correspond to different initial permutations. It follows that there are $$2^{n-1}-1$$ permutations that cause bubble sort to halt after exactly two passes. Together with the identity permutation, it follows that there are $$2^{n-1}$$ permutations that cause bubble sort to halt within two passes.
How do these permutations look like? Suppose that the first $$k$$ swaps are made, and then the following swap isn't made. Since the first $$k$$ swaps are made but the following one isn't made, at the end of the pass the first $$k+1$$ elements of the array will be $$A[2],\ldots,A[k+1],A[1]$$, which must correspond to the permutation $$1,\ldots,k+1$$. Hence the original permutation started $$k+1,1,\ldots,k$$.
From here, it's not too difficult to describe all $$2^{n-1}$$ permutations. Starting with the identity permutation, partition it in an arbitrary way, and rotate each part one step to the right. For example, when $$n = 3$$ we get:
• $$1|2|3 \to 1|2|3$$.
• $$1|23 \to 1|32$$.
• $$12|3 \to 21|3$$.
• $$123 \to 312$$.
Each division in the partition corresponds to a step in the pass during which there was no swap.
Here is another way of describing these permutations: they avoid the patterns $$231$$ and $$321$$. This means that for any $$i, it cannot be that $$A[k] < A[i],A[j]$$. To prove this, we show that any permutation above satisfies this constraint, and then prove the converse.
Consider any of the $$2^{n-1}$$ permutations above, and let $$i. Since each element in a part is smaller than each element in a subsequent part, $$A[k] < A[i]$$ is only possible if $$i,j,k$$ are all in the same part. Moreover, $$A[k] < A[j]$$ for indices $$j in the same part only if $$j$$ is the first element in the part, which is impossible since $$j>i$$.
Suppose now that we are given a $$(231,321)$$-avoiding permutation. Let $$i$$ be any index such that $$A[i] > i$$. All elements in $$i,\ldots,A[i]-1$$ must appear before all elements in $$A[i]+1,\ldots,n$$, since otherwise there will be a copy of $$231$$. Furthermore, the former must appear in increasing order, since otherwise there will be a copy of $$321$$. Therefore the part of the array that follows $$A[i]$$ is $$A[i],i,i+1,\ldots,A[i]-1$$.
Imagine now scanning the array in the order $$A[1],A[2],\ldots$$. The first time that $$A[i] \neq i$$, necessarily $$A[i] > i$$ (since we've already seen all smaller elements), and so the following elements are $$i,\ldots,A[i]-1$$. When we continue the scan, again the first time that $$A[i] \neq i$$ we must have $$A[i] > i$$ (since we've already seen all smaller elements), and so on. Therefore the array is of the form above.
• I'm looking at your partition graphic for the $n = 3$ example. From this, I'm inducing that the number of permutation is equal to the number of possible unique partitioning of an $n$ sized array. Is this correct? For $n = 4$, we have the following $$1\ 2 \ 3\ 4 \\ 1 | 2 | 3| 4 \\ 1 \ 2 | 3| 4 \\ 1\ 2 | 3 \ 4 \\ 1 \ 2 \ 3| 4 \\ 1 | 2 \ 3 \ 4 \\ 1 | 2 | 3 \ 4 \\ 1 | 2 \ 3 | 4 \\$$ Oct 6 '20 at 13:05
• Right. It's also easy to prove combinatorially by noting that each of the internal $n-1$ positions could contain or not contain a bar. Oct 6 '20 at 14:11
• Ahhh yes, that was exactly what I was looking for that "each of the internal......could or could not contain a bar." I knew there was an easy way to see this, but I couldn't figure it out. Also btw, do you know if there's a way to see this from a recurrence approach? Specifically, I was initially viewing this problem from the perspective that we have $n$ numbers, and now we add 1 more number, how many additional valid permutations does this additional number provide us with? I couldn't figure out a good way to think about this. Oct 6 '20 at 14:22
• There might be a way to write a recurrence, but a recurrence is not the only way to count things. Oct 6 '20 at 14:23
• To prove $i < j < k$ and $A[k] < A[i], A[j]$, isn't this also clear from the fact that an element in the unsorted array can only move one space to the left max for each pass, so if you have an element in the unsorted array in which there are two elements before it that are larger, then it is not possible to sort this array in one pass? Oct 7 '20 at 22:14
Bubble sort isn’t completed when the array is sorted - it is completed when we know the array is sorted.
Take the case 1-3-2. We have x-y-z and bubblesort figures out in the first pass that x is less than y and y is greater than z. It doesn’t know how x and z are related and therefore doesn’t know whether the array is sorted after one pass or not, therefore bubblesort is not complete at this point
If you change the question to “how many passes until the array is sorted”, that’s a different question. But to answer the question: If the number of array elements is zero or one, then bubble sort completes with zero passes. Otherwise, if either the array is sorted or the array has two elements then bubble sort complets with one pass. Otherwise, two or more passes are needed.
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Munich Personal RePEc Archive
# On uniqueness of moving average representations of heavy-tailed stationary processes
Gouriéroux, Christian and Zakoian, Jean-Michel (2014): On uniqueness of moving average representations of heavy-tailed stationary processes.
Preview
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MPRA_paper_54907.pdf
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## Abstract
We prove the uniqueness of linear i.i.d. representations of heavy-tailed processes whose distribution belongs to the domain of attraction of an $\alpha$-stable law, with $\alpha<2$. This shows the possibility to identify nonparametrically both the sequence of two-sided moving average coefficients and the distribution of the heavy-tailed i.i.d. process.
MPRA is a RePEc service hosted by
the University Library LMU Munich in Germany.
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## Extending Java / RAD Support with NetBeans
In This Posting, I wrote that Java / RAD – ie Swing – Support could be added to the free version of Eclipse.
Well as a follow-up, I just wanted to mention that the Window Builder that this procedure installed, is not extensible. This can be quite limiting.
Instead, a better approach might be to install “Net Beans“, even under Linux. This IDE will support OpenJDK 7. We do need to make sure though, that we do not only have the JRE package installed, but also the JDK (development) package. It is a bit harder to set up than the other one, but has as advantage a Component Palette which is extensible.
After installing Net Beans, the thing to do is to add some sort of standard, built-in Swing component, thereby making the Component Palette visible in the IDE. We have our Swing Designer at that point. Then, under Linux, it is possible to add the Jar File ‘/usr/share/java/swingx.jar‘ to Net Beans, by way of ‘Tools -> Libraries (Global Libraries)’. Then, we can right-click on the Project and click on ‘Properties -> Libraries’, and add the newly-defined Library to the Project.
Next, we can right-click on the component palette, and open the Palette Manager, from which we first create a new Component Category. It will be empty. Then, still within the Palette Manager, we can add components at will From the newly defined Library. It will ask us which Category to add them to, and presumably we would add them to the Category we just created.
And at that point we will see, that we not only have components such as “JPanel”, but that we potentially have a large number of components, whose names begin with “JX”, such as “JXPanel” …
I never really knew, that Net Beans was so powerful, yet free. However, Net Beans is not open source, and to install it, we need to run their SH File as ‘root’. This could present a daunting leap of trust for some people. Also, even though this setup script wants to be run in root mode, it will also need to display a graphical wizard. This requirement has led to a common but minor error message with many users. And so, in user mode, we need to give the command ‘xhost +‘ first, so that the root-mode process can access our X-server, and then after we have installed Net Beans, back in user mode, we give the command ‘xhost -‘ again.
## Getting Java / RAD support within the Open version of Eclipse
One type of question which some of my friends will ask me from time to time, is ‘Whether I am aware of any good IDEs these days’. This question arises from the knowledge, that writing source code just with a text editor, is not much fun, the fact that IDEs exist which enhance the programming experience, and finally, the memory from our distant past, of the fact that there exist IDEs which implement a kind of ‘RAD’ – a Rational Application Development – or a Rapid Application Development platform. This can sometimes conjure up pleasant memories of the days when Sun Microsystems was still in control of Java.
And so I have taken a minor step to explore this subject only today, and come to the following observations.
Java / RAD in the old days, was already based on ‘Swing‘. And Swing is a set of Java packages, which keep two things in sync: A code view, and a design view of our project. It was Swing which would update the code view, if we made changes graphically to our design view, and vice-versa.
Having understood that, the next thing people who use Linux might do – as I did – would be to install a number of Java Swing libraries from our package manager, and then to hope that we have what it takes to start building Java / RAD projects, the way we used to do it with ‘JBuilder‘. JBuilder was a proprietary, paid-for IDE, which back in the days of ‘Borland Inprise‘, possessed the components in a bundled way, not just to be an IDE, but to be an IDE with RAD capabilities.
But inevitably, we run into an impasse. Under Linux and entirely with Open Source software, we are likely to be using “Eclipse” as our IDE, and a version of it, which does not recognize that we have Swing packages installed – in our shared, non-proprietary Java folders. I do know that paid-for versions of Eclipse exist, which again bundle Java / RAD capabilities, but we might want to obtain those, without having to pay much money. So what can be done?
The answer that suits some use of Eclipse, lies in the fact that with any Linux Eclipse install, there exists a user-space install of extensions, similarly to how the Windows flavor of Eclipse, also has such a built-in mechanism to install extensions to this IDE itself. We find that in the Help menu, there is an entry named “Install New Software”. From there, we can specify a URL to download packages from, and we can extend, or possibly confuse, our Eclipse setup. But what we install from here under Linux does not ever use ‘root’, and thus only exists in our user / home directories.
There is a relevant URL which I just learned about, from which we can tell Eclipse to install extensions to itself:
http://dl.google.com/eclipse/inst/d2wbpro/latest/4.2
And when we have typed in the URL, we can see a set of components available, which include 3 Swing components, plus the critically-important “Window Builder“. This Window Builder is the Eclipse extension, that actually maintains the two views. After we have installed all the relevant packages – because we often tell this wizard to install all the dependencies as well – we need to restart Eclipse. We may first have to acknowledge having to install non-signed packages.
Once this is done, under the New Projects list, we will see as the last menu entry, “Other”, referring to a miscellany of non-standard project types. And when we open that view, we will finally get to see our Swing subsection of New objects that we can create.
Doing so opens a specialized wizard, and sets up our Java / RAD project, which will eventually have a code view and a design view.
Now, the logical question does arise, of whether it did any good, to install Swing packages via the Linux package manager. Doing so places them into shared folders using root privileges, folders that are not even specific to one Java compiler used. I have run into situations before, in which Java projects designed by other programmers did not initially recognize, that these shared Java libraries even existed. This has required some workarounds in my past.
Using Eclipse, the fact is that these packages will not show up in the graphical browsers / wizards.
But I think that this is one reason, to install all 4 components offered by the URL above, to make certain object types visible in the IDE wizard. After one has created a basic project with the code and design views, I think the next thing to try, would be just to add the ‘import‘ statements to the code view, that are meant to pull in the Swing packages we think we need, and then to see, whether the design view renders them correctly.
Whether it does so or not, next depends on the separate question, of whether All Swing packages are in fact compatible with the Google Window Builder. And even though Swing should implement a standard interface, depending on what we are trying to do, the answer might still be ‘No’.
It would be necessary to link our project to the desired JAR File explicitly, in a way that states its location in the shared, root File System… In that case, we would click on the IDE Menu, “File -> Import”. And in the window which appears, instead of choosing “Plug-Ins And Fragments”, we could choose “General -> File System”… And under Linux we would choose the folder ‘/usr/share/java‘, so that in the subsequent view, we can select the required JAR Files, knowing that they will be copied.
Then, we would right-click on the project we want to build, and click on Properties, which opens another window, from which we would choose “Java Build Path -> Libraries” …
Dirk
(Edit : ) It is entirely unproven, whether externally-supplied Swing libraries can be made to work properly, with Google Window Builder. What Google has done, is to name the Swing components of its Window Builder, as offering a ‘JPanel‘, a ‘JFrame‘, a ‘JApplet‘ etc., so that the naming of classes will not interfere with generic Swing class names. Hence, if we were to try to use Swing classes provided by the Linux package manager as stated before, first of all those would have dependencies on one or more base libraries, also located in ‘/usr/share/java‘, which would not only require that the ‘import‘ declarations be put into our Java source file, but also that the libraries be listed after the libraries that already belong to Window Builder, within the link order defined in the ‘Java Build Path‘.
The ultimate success we would want to see, would be that the Component Palette belonging to Window Builder, should offer us additional components, other than J-this or J-that. But the mere fact that the naming of classes has been separated, suggests the possibility that these external components will not be recognized by Window Builder at all.
One quick test a user could perform, would take into account the observation that Window Builder source code tends to import one specific package, such as ‘javax.swing.jframe‘. After we have appended the desired libraries to the Java Build Path, base library first, we could try to declare something that generic Swing projects frequently used:
import javax.swing.*;
This declaration should either work, or crash Window Builder.
(Answer : ) Doing so simply causes Window Builder to crash temporarily, until the offending declaration is removed from the code view. That is all.
In the past, the way the email service of my ISP – “Bell Canada” – worked, was that email was outsourced to Microsoft, such that at least my own ‘Sympatico’ email address, was also accessible as Web-mail, via ‘Hotmail’. The way this used to work, was that I had assigned sub-domains for accessing the POP and SMTP servers, which seemed to belong to Bell, but which nevertheless used to access my email with a Hotmail server.
Now, it had happened before that I had received emails from Bell, telling me to make sure I had updated my email server settings, to keep up with improvements that were being promised. But all those emails simply led to a site with Bell, according to which my settings were already correct as they were.
My explanation for that would be, that maybe not all Bell customers were on the same plan before, and that my arrangement with them may have been more of a testing arrangement. In any case I had visited the link each time, to make sure that the server settings recommended there, still corresponded to the server settings I had been using, and they had.
The aspect of this which is pleasantly surprising, was that as of May 3, the service was in fact upgraded – this time – and that it happened without any interruption of the service available to my POP clients, since my POP clients were already configured for the change. Thus, I had not really noticed that the upgrade had in fact taken place until later. I did, however, receive an email from Bell, saying ‘Welcome to the New Service.’
There was however a way in which I was bound to discover the improvement eventually. As long as my email was always hosted on Hotmail servers, a peculiarity of the service was, that one of my POP clients was configured to ‘Leave Email On The Server, For At Most 3 Days’, while all my other POP clients were just configured to ‘Leave Email On The Server’ (with no time limit set). This was one way I had intended to keep all my received email, on POP clients, in sync.
The problem with the earlier Hotmail arrangement was, that their servers would flagrantly disregard the command sent by my own email client, eventually to delete the emails. In fact, Hotmail had announced suspicion of my email client programs in the past, stating that ‘Some program has asked us to delete your emails. But because we do not trust this program, we have put those emails in a special folder for you.’ AFAIK, It could have been true that Hotmail did not trust this mysterious email client of mine, because mine was not a Microsoft client, being a Linux client, and there may have been some way for the Hotmail server to detect that. In fact, I am sure that email clients state a User Agent, the same way Web browsers state a User Agent.
What this meant in practice, was that every few days I would actually need to log in to my Web-mail, and to delete some emails manually, which the setting on that one email client were not sufficient to delete. Hence, it was inevitable that I would be heading back to the Hotmail, Web-mail server eventually, to look at and delete those emails.
Except that more recently, there were no emails in my Hotmail Inbox. My POP clients were still receiving emails, but my Web-mail did not reflect them. In response to this, I had actually changed the setting in the interim, so that none of my POP clients were set to have emails deleted from the server for now.
But it turns out, that the real reason for which my (old) Hotmail Web-mail was no longer showing any received emails, was this real upgrade. The new Web-mail site with Bell, shows me all the emails I have received since May 3.
What this also means, is that the potential now exists, for Bell email servers actually to follow the request of my client, to delete emails in the Inbox, older than 3 days.
Also, I happen to like the new Bell Web-mail layout better than the old Hotmail layout. And, there are No cross-scripted sites which the new Web-mail site links to, which could have been intended as advertisement under Hotmail, but which would trigger the script-blocker on my Web-browsers, that selectively blocks scripts from excessively-abusing domains. There seem to be no linked domains on the new Web-mail site, which my script blocker would want to block.
I guess that maybe, Hotmail was making some additional money on the side, which Bell does not need to make, because I am already paying Bell in full, for my services?
Dirk
## The laptop ‘Klystron’ suspends to RAM half-decently.
One subject which I have written a lot about, was that as soon as I close the lid of the laptop I name ‘Klystron’, it seems to lose its WiFi signal, and that that can get in the way of comfortable use, because to close the lid also helps shield the keyboard of dust etc..
This Linux laptop boots decently fast, and yet is still a hassle to reboot very often. And so I needed to come up with a different way of solving my problem, on a practical level. My solution for now, is to tell the laptop to Suspend To RAM, as soon as I close the lid. That way, the WiFi signal is gone more properly, and when the laptop resumes its session, the scripts that govern this behavior also re-initialize the WiFi chipset and its status on my LAN. This causes less confusion with running Samba servers etc., on my other computers.
There is a bit of terminology, which I do not think that the whole population understands, but which I think that people are simply using differently from how it was used in my past.
It used to be, that under Linux, we had ‘Suspend To RAM’ and ‘Suspend To Disk’. In the Windows world, these terms corresponded to ‘Standby’ and ‘Hibernate’ respectively. Well in the terminology today, they stand for ‘Sleep’ and ‘Hibernate’, borrowing those terms from mobile devices.
There are two types of Suspend working in any case.
In past days of Linux, we could not cause a laptop just to Hibernate. We needed to install special packages and modify the Grand Unified Bootloader, before we could even Suspend To Disk. Suspending To RAM used to be less reliable. Well one development with modern Linux which I welcome, among many, is the fact that Sleep and Hibernate should, in most cases, work out-of-the-box.
I just tried Sleep mode tonight, and it works 90%.
One oddity: When we Resume, on this laptop, the message is displayed on the screen numerous times, of a CPU Error. But after a few seconds of CPU errors, apparently the session is restored without corruption. Given that I have 300 (+) processes, I cannot be 100% sure that the Restore is perfectly without corruption. But I am reasonably sure, with one exception:
The second oddity is of greater relevance. After Waking Up, the clock of the laptop seems to be displaced 2 days and a certain number of hours into the future. This bug has been observed on some other devices, and I needed to add a script to the configuration files as a workaround, which simply sets the system clock back that many days and that many hours, after Waking. Thankfully, I believe that doing so, was as much of a workaround as was needed.
One side-effect of not having done so, before being aware of the problem, was that the ‘KNotify’ alarms for the next two days have also all sounded, so that it will take another two days, before personal organizer – PIM – notifications may sound for me again.
The fact that numerous CPU errors are displayed bothers me not. What that means, is that the way the CPU goes to sleep, and then wakes up, involves power-cycling in ways that do not guarantee the integrity of data throughout. But it would seem that good programming of the kernel does provide data integrity, with the exception of the system clock issue.
But the fact that the hardware is a bit testy when using the Linux version of Sleep, also suggests that maybe this is also the kind of laptop that powers down its VRAM. It is a good thing then, that I disabled the advanced compositing effects, that involve vertex arrays.
There is a side-note on the desktop cube animation I wanted to make.
In general, when raster-rendering a complex scene with models, each model is defined by a vertex array, an index array, one or more texture images etc., and the vertex array stores the model geometry statically, as relative to the coordinate-origin of the model. Then, a model-view-projection matrix is applied – or just a rotation matrix for the normal vectors – to position it with respect to the screen. Moving the models is then a question of the CPU updating the model-view matrix only.
Well when a desktop cube animation is the only model in the scene, as part of compositing, I think that the way in which this is managed differs slightly. I think that what happens here, is that instead of the cube having vertex coordinates of +/- 1 all the time, the model-view matrix is kept as an identity matrix.
Instead, the actual vertex data is rewritten to the vertex array, to reposition the vertices with respect to the view.
Why is this significant? Well, if it was true that Suspending the session To RAM also cut power to the VRAM, it would be useful to know, which types of data stored therein will seem corrupted after a resume, and which will not.
Technically, texture images can also get garbled. But if all it takes is one frame cycle for texture images to get refreshed, the net result is that the displayed desktop will look normal again, by the time the user unlocks it.
Similarly, if the vertex array of the only model is being rewritten by the CPU, doing so will also rewrite the header information in the vertex array, that tells the GPU how many vertices there are, as well as rewriting the normal vectors, as when they are a part of any normal vertex animation, etc.. So anything resulting from the vertex array should still not look corrupted.
But one element which generally does not get rewritten, is the index array. The index array states in its header information, whether the array is a point list, a line list, a triangle list, a line strip, a triangle strip… It then states how many primitives exist, for the GPU to draw. And then it states sets of elements, each of which is a vertex number.
The only theoretical reason fw the CPU would rewrite that, would be if the topology of the model was to change, which is as good as never in practice. And so, if the VRAM gets garbled, what was stored in the index array would get lost – and not refreshed.
And this can lead to the view, of numerous nonsensical triangles on the screen, which many of us have learned to associated with a GPU crash.
Dirk
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Measurement of charged jet cross section in pp collisions at ${\sqrt{s}=5.02}$ TeV
The collaboration
No Journal Information, 2019
Abstract
The cross section of jets reconstructed from charged particles is measured in the transverse momentum range of $5<p_\mathrm{T}<100\ \mathrm{GeV}/c$ in pp collisions at the center-of-mass energy of $\sqrt{s} = 5.02\ \mathrm{TeV}$ with the ALICE detector. The jets are reconstructed using the anti-$k_\mathrm{T}$ algorithm with resolution parameters $R=0.2$, $0.3$, $0.4$, and $0.6$ in the pseudorapidity range $|\eta|< 0.9-R$. The charged jet cross sections are compared with the leading order (LO) and to next-to-leading order (NLO) perturbative Quantum ChromoDynamics (pQCD) calculations. It was found that the NLO calculations agree better with the measurements. The cross section ratios for different resolution parameters were also measured. These ratios increase from low $p_\mathrm{T}$ to high $p_\mathrm{T}$ and saturate at high $p_\mathrm{T}$, indicating that jet collimation is larger at high $p_\mathrm{T}$ than at low $p_\mathrm{T}$. These results provide a precision test of pQCD predictions and serve as a baseline for the measurement in Pb$-$Pb collisions at the same energy to quantify the effects of the hot and dense medium created in heavy-ion collisions at the LHC.
• Jets in pp 5.02 TeV
Figure 3
10.17182/hepdata.91239.v1/t1
Fig. 3: Fully corrected charged jet differential cross sections in pp collisions at $\\sqrt{s}$ = 5.02 TeV. Statistical uncertainties are...
• Cross section ratio with radius
Figure 6
10.17182/hepdata.91239.v1/t2
Fig. 6: Charged jet cross section ratios for $\\sigma$(R = 0.2)/$\\sigma$(R = 0.4) (Red) and $\\sigma$(R = 0.2)/$\\sigma$(R = 0.6)....
• Jets with UE subtraction in pp 5.02 TeV
Figure A1
10.17182/hepdata.91239.v1/t3
Charged jet differential cross sections with UE subtraction in pp collisions at $\sqrt{s}$ = 5.02 TeV. Statistical uncertainties are displayed...
• Jets with leading $p_{T}$ 5 GeV/$c$ in pp 5.02 TeV
Figure A5
10.17182/hepdata.91239.v1/t4
Charged jet differential cross sections without UE subtraction in pp collisions at $\sqrt{s}$ = 5.02 TeV with the leading track...
Loading Data...
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## Thinking Mathematically (6th Edition)
perimeter = $1000$ inches
RECALL: (1) The perimeter of a polygon is equal to the sum of the lengths of all its sides. (2) A square has four equal sides. This means that the three unknown side lengths are all 250 inches. Add the lengths of all the sides to find the perimeter: perimeter = $250+250+250+250= 1000$ inches
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# Concurrent queue with pthread implementation
I'm writing a multi-threaded queue implemented with pthreads. Since I came up with the code based on several tutorials from Internet, I would like to make sure there are no logic errors in my code:
template <typename T>
public:
}
m_queue.clear();
}
void push(T t_data) {
m_queue.push_back(t_data);
}
T front() {
T ret;
while (m_queue.empty()) {
}
ret = m_queue.front();
return ret;
}
void pop() {
if (!m_queue.empty())
m_queue.pop_front();
}
private:
std::deque<T> m_queue;
};
Correctly initialized members.
ThreadQueue() {
}
But you don't check the error codes.
The object is potentially created with invalid members. Use exceptions to gurantee the state of your object.
Destructor does not do anything useful.
~ThreadQueue() {
m_queue.clear();
}
The destructr is supposed to clean resources.
It does not. You should destroy the mutex and condition variable in the destructor (reverse order of creation to be consistent).
Also locking before clearing the queue does not buy you anything. If at the point the destructor is called the object is still visible to other threads then it is going to break anyway. The queue is going to call clear in its own destructor so it is pointless doing it manually.
The call to signal should be inside the lock.
void push(T t_data) {
m_queue.push_back(t_data);
}
Since you should be using RAII for you locking anyway (See below) the signal is going to fall inside the lock stratergy unless you do something complex.
Also you should pass the parameter by const reference rather than value.
Normally front() is going to give you a reference to the front object.
T front() {
T ret;
while (m_queue.empty()) {
}
ret = m_queue.front();
return ret;
}
Returning a copy is probably not what you want (especially for anything interesting). But returning a reference is not really an option either as that opens you to situations where you have race conditions (you have a reference to the top object and another thread pops it (destroying it) just before you call a method).
So you need to provide a method for accessing the object while maintaining the lock (this will probably require the return of a wrapper object that maintains a lock on the queue) or alternatively removing this function.
Other non C++ things that should be addressed:
The mutex should be locked/unlocked using RAII. You are leaving your code open to exception handling problems. If an exception is generated by T which you have no control over then you could leave your queue in an unusable state with the mutex locked and no way to unlock it (because the stack of the locker has been unwound past the unlock because of an exception).
Yes this is correct:
if (!m_queue.empty())
m_queue.pop_front();
But don't get into bad habbits. No using the braces can lead to maintenance problems on C/C++ code because of macros. Best practice dictates that you always use {} for sub statements of if/while/for.
To make the code easy to read when including the code in the interface declaration put the variables at the top. Thus you can read the code in with the context of the variables and knowing their type. If you interface declaration is just an interface (and the code is put in the source file rather than the header) then it is fine to put the private variables at the bottom.
• Maybe it would be clearer to show as to how to use RAII for mutex locking/unlocking with a short example? – dev_nut May 6 '15 at 4:24
Personally, I think you've wrapped too many different responsibilities into the queue class. I think I'd start with a small wrapper for a mutex:
class mutex {
public:
};
Then I'd write something similar for your condvar. Again, you should really call pthread_cond_destroy when you're done with the condvar. A wrapper that automatically called pthread_cond_init in its ctor and pthread_cond_destroy in its dtor makes it much easier to ensure the code works correctly.
Then I'd write a small RAII wrapper for locking and unlocking the mutex:
class lock {
mutex &m;
public:
lock(mutex &m) : m(m) { m.lock(); }
~lock() { m.unlock(); }
};
Then, getting the rest of your code correct becomes rather simpler -- creating and destroying the mutex and condvar happen automatically (fixing a couple of bugs, since your code doesn't currently destroy either as it should). Code using the mutex (for example) becomes simpler as well. For example, your front() simplifies (a little) to something like this:
T front() {
T ret;
lock(m_qmtx);
while (m_queue.empty())
ret = m_queue.front();
return ret;
}
I see (at least) two obvious advantages to this:
1. Simpler code that's easier to ensure is correct.
2. The code is much closer to what you'd likely write in C++11, so anybody familiar with C++11 will find it easier to follow.
One other detail: I think I'd try t come up with better names than m_condv and m_qmtx. At least in my opinion, your current names are next to meaningless (and ugly to boot).
The pthread_cond_signal should be done with the mutex locked. And a broadcast might be preferable to a signal.
I'd put the private variables at the start of the class and braces on the if statements, even though they are of only one line.
• 1.Why should pthread_cond_signal be called inside the lock? 2. broadcast might not be preferable since it does something different than signal does. 3. The order of private members and the braces of if are not logic errors – GuLearn Apr 22 '13 at 18:36
• @user2207811: 1: Because threading is usally more complex than most humans understand why risk deadlock without it. 2: signal are broadcast so that is neither here nor there. (all threads waiting on the cond variable are eligable for being woken on a signal). 3: The use of braces and order of private member's relates to best practices and maintenance. You don't have to accept the advice but don't complain when people tell you need to think about it. – Martin York Apr 25 '13 at 10:10
Your use of the mutex and condition variable seem correct.
It's a queue. Use a std::queue internally and mimic its interface. You're missing a lot of things. Here's the interface of one I wrote, which exactly mimics std::queue...
template<typename T, typename Container = std::deque<T>>
class ConcurrentQueue
{
public:
typedef Container container_type;
typedef typename Container::value_type value_type;
typedef typename Container::size_type size_type;
typedef typename Container::reference reference;
typedef typename Container::const_reference const_reference;
explicit ConcurrentQueue(const Container& c);
explicit ConcurrentQueue(Container&& c = Container());
ConcurrentQueue(const ConcurrentQueue& other);
ConcurrentQueue(ConcurrentQueue&& other);
template<typename Allocator> explicit ConcurrentQueue(const Allocator& a);
template<typename Allocator> ConcurrentQueue(const Container& c, const Allocator& a);
template<typename Allocator> ConcurrentQueue(Container&& c, const Allocator& a);
template<typename Allocator> ConcurrentQueue(const ConcurrentQueue& other, const Allocator& a);
template<typename Allocator> ConcurrentQueue(ConcurrentQueue&& other, const Allocator& a);
ConcurrentQueue& operator=(const ConcurrentQueue& other);
ConcurrentQueue& operator=(ConcurrentQueue&& other);
bool empty() const;
size_type size() const;
void push(const T& item);
void push(T&& item);
template< class... Args >
void emplace(Args&&... args );
T pop();
void swap(ConcurrentQueue& other);
private:
std::queue<T, Container> m_Queue;
/* ... */
};
I'll leave it to you to fill in the implementation. I used the C++ standard library, not pthread. I recommend you do too. Note I also called it ConcurrentQueue. It's a more standard name than ThreadQueue for this class. SharedQueue is also quite popular (needless to say, this is a popular thing to write).
• Use of cond variable is not correct. See point 1 in Williams comments. – Martin York Apr 25 '13 at 9:39
• The interface for threadable containers probably should not look like normal containers. front() returning a reference is a very bad idea. By the time the user calls a function another thread could have destroyed the object with a pop. – Martin York Apr 25 '13 at 10:05
• @LokiAstari ya actually I just added front and back for this. Mine actually only has pop. You're right, front and back should be by value I guess. I don't know about posix, but you don't have to keep the mutex locked when calling notify_one on a C++11cond variable, right? – David Apr 25 '13 at 12:05
• Also note. Returning by value for front() is not really useful if the type is anything useful. If you limit it to POD types then fine. – Martin York Apr 27 '13 at 17:25
• @LokiAstari Ya I agree, I removed it altogether. I'm pretty sure you don't want the mutex locked when you use the condition variable, in C++11. Since it just wraps posix I assume the same must be true there... – David Apr 27 '13 at 20:49
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Metabolomics Methodology Development
In January 2010 the NSF Plant Genome Research Program began funding ($1,918,691 for three years) a project to develop plant metabolomics methodology that extends my previous work using stable isotopic labeling for metabolite identification. The support will enable the purchase of two new mass spectrometers ($498,500 budgeted) that will reside in Alderman Hall and includes a significant mass spectral library database creation effort. Jerry Cohen is a Co-PI on the project; the following is the Project Summary from the funded proposal:
The ability to quantitatively monitor the dynamics of metabolites in an organism is critical to fully understand the complex interactions of biology’s Central Dogma. In part because of the diversity of compounds found in living organisms, especially in plants, the ability to define this critical area in the functional analysis of living plants has been slower to develop than nucleic acid and protein methods. This research addresses impediments to the development of techniques and uses metabolic variations occurring in response to stress conditions as a test system. The present limitations include the rather low numbers of compounds that have typically been identified, relatively insensitive procedures that yield insufficient spatial resolution and the dearth of true quantitative methods. The specific aims are: 1) provide improved sample preparation and derivatization methods that yield structural information to create new spectral library resources for metabolite identification by LC and GC MS, 2) use stable isotope metabolically labeled plant materials for isotope dilution with robust quantification by LC and GC MS and MS/MS. The work will be focused on rice and tomato, but with the expectation that the methods and procedures will be generally applicable to many plant species.
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# What is the integral of ln(sqrt(x))?
May 2, 2015
By part :
$\int \ln \left(\sqrt{x}\right) \mathrm{dx}$
$\mathrm{du} = 1$
$u = x$
$v = \ln \left(\sqrt{x}\right)$
$\mathrm{dv} = \frac{1}{2 x}$
$\left[x \ln \left(\sqrt{x}\right)\right] - \frac{1}{2} \int \mathrm{dx}$
$\left[x \ln \left(\sqrt{x}\right) - \frac{1}{2} x\right]$
don't forget $\ln \left({a}^{b}\right) = b \ln \left(a\right)$
$\left[\frac{1}{2} x \ln \left(x\right) - \frac{1}{2} x\right]$
factorize by $\frac{1}{2} x$ and don't forget the constant !
$\left[\frac{1}{2} x \left(\ln \left(x\right) - 1\right) + C\right]$
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# Computing ζ(3)
I’ve started reading Paul Nahin’s new book “In Pursuit of ζ(3).” The actual title is “In Pursuit of Zeta-3.” I can understand why a publisher would go with such a title, but I assume most people who read this blog are not afraid of Greek letters.
I’ve enjoyed reading several of Nahin’s books, so I was pleased to see he had a new one.
Nahin’s latest book is about attempts to find a closed-form expression for the sum
If you replace the “3” with “s” in the sum above, you have an expression for the Riemann zeta function ζ(s), and so ζ(3) is a convenient way of describing the sum and putting it in context [1]. There are closed-form expressions for ζ(s) when s is an even integer [2] but so far not nobody has found one for s = 3.
The value of ζ(3) is sometimes called Apéry’s constant because in 1978 Roger Apéry was the first to prove that it is an irrational number.
The names “ζ(3)” and “Apéry’s constant” are anachronisms because Euler studied the sum in the 18th century, but Riemann came along in the 19th century and Apéry in the 20th.
## Naive calculation
The most obvious way to compute ζ(3) numerically is to replace the upper limit of the infinite sum with a large number N. This works, but it’s very inefficient.
Let’s calculate the sum with upper limit 100 with a little Python code.
>>> sum(n**-3 for n in range(1, 101))
1.2020074006596781
We can judge the accuracy of this approximation using the implementation of the zeta function in SciPy.
>>> from scipy.special import zeta
>>> zeta(3)
1.2020569031595942
We can predict how accurate our sum will be by estimating the tail of the sum using the integral test as follows.
The integral on the left evaluates to N-2/2 and the integral on the right evaluates to (N+1)-2/2.
This says that if we sum up to N, our error will be between (N+1)-2/2 and N-2/2.
We can estimate from this that if we want d decimal places, we’ll need to sum 10d/2 terms. So to get 16 decimal places, we’d have to sum 100,000,000 terms. This is certainly not what the call to zeta(3) above is doing.
## A better approach
Nahin’s book derives an equation for ζ(3) by Ernst Kummer (1810–1893):
Because the denominator in the sum is a 5th degree polynomial, we get faster convergence than we do from directly evaluating the definition of ζ(3). We could find an interval around the error using the integral test above, and it would show that the error is O(N-4).
Let’s try this out in Python.
>>> f = lambda n: (n**3 * (n**2-1))**-1
>>> 1.25 - sum(f(n) for n in range(2, 101))
1.2020569056101726
>>> _ - zeta(3)
2.4505784068651337e-09
This says we get 8 correct decimal places from summing 100 terms, which is just what we’d expect from our error estimate.
## Doing even better
Kummer’s approach is better than naively computing ζ(3) from its definition, but there are much more efficient methods.
With naive summation up to N, we get
2 log10 N
correct decimals. With Kummer’s sum we get twice as many,
4 log10 N
correct decimals.
But there are methods for which the number of correct decimals is linear in N rather than logarithmic. Wikipedia mentions a method that gives 5.04 N decimal places.
Update: See this post for a sequence of more efficient algorithms.
## Related posts
[1] That’s how the Riemann zeta function is defined for s with real part greater than 1. For the rest of the complex plane, except s = 1, the Riemann zeta function is defined by analytic continuation. The sum is not valid for s with real part less than 1, and so, for example, the sum makes no sense at s = -1. But the analytic continuation of the sum is valid at -1, and equals -1/12 there. This leads to the rigorous justification of the otherwise nonsensical expression
1 + 2 + 3 + … = -1/12.
[2] For s = 2n where n is a positive integer,
For s = 0 and negative integers, we have
Here the B‘s are the Bernoulli numbers, which have closed-form expressions.
## 2 thoughts on “Computing ζ(3)”
1. 1. An Amazon.co.jp-book-reviewer, ”Tarosa” has devised a Japanese-decimal-BASIC algorithm for the zeta values: https://www.amazon.co.jp/-/en/%E9%BB%92%E5%B7%9D-%E4%BF%A1%E9%87%8D/product-reviews/4768704476/ref=cm_cr_dp_d_show_all_btm?ie=UTF8&reviewerType=all_reviews (The second review from the top)
2. Dr. Mikio Sugioka of Shimazu Co. in Kyoto, Japan ( = a colleague of the Nobel Prize winning chemist Koiti Tanaka) calculated the zeta values ζ (2n – 1) for some odd 2n – 1:
http://www5b.biglobe.ne.jp/~sugi_m/page021.htm
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# Functor-Diagram in tikz-cd
I wonder how to draw the typical Functor-Diagram (picture below) in tikz-cd. I know how to draw commutative diagrams, however I don't know how to make the short \mapsto arrows as well as the mapsto arrow in between the vertical arrows. Thanks in advance for any help.
Edit (code that produces errors, namely the ngerman package):
\documentclass{amsart}
%Standardpakete
\usepackage{amsmath}
\usepackage[ngerman]{babel}
\usepackage{tikz-cd}
\begin{document}
$$\begin{tikzcd} \mathsf{C}^{\mathsf{op}} \arrow[rr, "F"] & & \mathsf{D} \\ c \arrow[dd, "f"'] & \mapsto & Fc \\ & \mapsto & \\ c' & \mapsto & Fc' \arrow[uu, "Ff"'] \end{tikzcd}$$
\end{document}
Something like this?
\documentclass[a4paper,12pt]{article}
\usepackage{newtxtext,newtxmath}
\usepackage{mathtools}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
\mathsf{C}^{\mathsf{op}} \arrow[rr, "F"] & & \mathsf{D} \\
c \arrow[dd, "f"'] & \mapsto & Fc \\
& \mapsto & \\
c' & \mapsto & Fc' \arrow[uu, "Ff"']
\end{tikzcd}
\end{document}
Addendum on request of the user. I think that the problem it is the ngerman babel. You can fix it putting,
\usetikzlibrary{babel}
\documentclass{amsart}
%Standardpakete
\usepackage{amsmath}
\usepackage[ngerman]{babel}
\usepackage{tikz-cd}
\usetikzlibrary{babel}
\begin{document}
$\begin{tikzcd} \mathsf{C}^{\mathsf{op}} \arrow[rr, "F"] & & \mathsf{D} \\ c \arrow[dd, "f"'] & \mapsto & Fc \\ & \mapsto & \\ c' & \mapsto & Fc' \arrow[uu, "Ff"'] \end{tikzcd}$
\end{document}
• Yes, exactly what I was looking for, thank you very much! Jun 2 at 11:13
• I am using the documentclass amsart and it appears that your code does not work in that class, do you know why or how to fix it? Jun 2 at 11:19
• @user243350 Here in this community it is necessary to add a minimal working example (a full code of your work). After I have tried with amsart class and the code work correctly. Jun 2 at 11:21
• Ok, I am sorry, I will try to extract the necessary piece of code of my document shortly. Thank you for the advice. Jun 2 at 11:22
• It appears there is nothing wrong with your code when I extract the necessary packages into a seperate argument, so there has to be some error on my side. I will try to fix it, thank you for your efforts! Jun 2 at 11:26
As an alternative to Sebastiano's great answer, it is possible to name the arrows f and Ff and add an arrow between them, then shorten the arrow:
\begin{tikzcd}[column sep=large, row sep=tiny]
\mathsf{C}^{\mathsf{op}} \arrow[r, "F"] & \mathsf{D} \\
c \arrow[d, "f"'{name=f}]\arrow[r,mapsto] & Fc \\[5ex]
c'\arrow[r,mapsto] & Fc' \arrow[u, "Ff"'{name=Ff}]
\arrow[mapsto,from=f, to=Ff,shorten=0.7em]
\end{tikzcd}
Another option could be add an empty arrow name in the inner part of the diagram and add the arrow as before
\begin{tikzcd}[column sep=large, row sep=tiny]
\mathsf{C}^{\mathsf{op}} \arrow[r, "F"] & \mathsf{D} \\
c \arrow[d, "f"', ""{name=f}]\arrow[r,mapsto] & Fc \\[5ex]
c'\arrow[r,mapsto] & Fc' \arrow[u, "Ff"', ""{name=Ff}]
\arrow[mapsto,from=f, to=Ff]
\end{tikzcd}
• Thank you very much to have cited me with....."to Sebastiano's great answer". Jun 2 at 16:00
• Well it does exactly what the OP asked, so it is great :) Jun 2 at 19:18
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# In a random sample of soldiers who fought in the Battle of Preston, 774 soldiers were from the New Model Army, and 226 were from the Royalist Army. Use a 0.05 significance level to test the claim that fewer than one quarter of the soldiers were Royalist.
Critical values: $z 0.005=2.575$,$z 0.01=2.325$, $z 0.025=1.96$, $z 0.05=1.645$, $z 0.1=1.282$ when $d.f=31:t 0.005=2.744$,$t 0.01=2.453$,$t0.025=2.040$,$t0.05=1.696$,$t0.1=1.309$.
This article aims to find that less than one-quarter of soldiers were Royalists given significant value. A critical value is a cutoff value used to mark the beginning of the region within which the test statistic obtained in hypothesis testing is unlikely to fall. In hypothesis testing, critical value is compared with test statistic obtained to determine whether or not the null hypothesis must be rejected. The critical value divides the graph into acceptance and rejection regions for hypothesis testing.
A critical value is a value that is compared to a test statistic in hypothesis testing to determine whether null hypothesis should be rejected or not. If value of the test statistic is less extreme than the critical value, the null hypothesis cannot be rejected. However, if the test statistic is more powerful than the critical value, null hypothesis is rejected, and alternative hypothesis is accepted. In other words, critical value divides the distribution plot into acceptance and rejection regions. If the value of the test statistic falls within the rejection region, then the null hypothesis is rejected. Otherwise, it cannot be dismissed.
Depending on the type of distribution to which the test statistic belongs, there are different formulas for calculating the critical value. A confidence interval or significance level can determine the critical value.
Step 1
It is given that:
$X-226$
$n-774$
Sample projection:
$\hat{p}-\dfrac{x}{n}=\dfrac{226}{774}=0.292$
The researcher claims that less than a quarter of the soldiers were Royalists.
Thus, null and alternative hypotheses are:
$H_{0}=p-0.25$
$H_{1}=p<0.25$
Step 2
The standardized test statistic can be found as:
$Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$
$Z=\dfrac{0.292-0.25}{\sqrt{\dfrac{0.25(1-0.25)}{1200}}}=2.698$
The level of significance, $=0.05$
Using $z-table$, the critical value at the level of significance $0.05$ is $-1.645$.
Since calculated statistic value $Z=2.698>|critical\:value|=|-1.645|$ ,We reject the null hypothesis. Therefore, it was concluded that less than one-quarter of the soldiers were royalists.
## Numerical Result
Since calculated statistic value $Z=2.698>|critical\:value|=|-1.645|$ , we reject the null hypothesis. Therefore, it was concluded that less than one-quarter of the soldiers were Royalists.
## Example
In random sample of soldiers who fought in the Battle of Preston, $784$ soldiers who fought in the Battle of Preston, $784$ soldiers were from New Model Army, $226$ were from New Model Army, and $226$ were from Royalist Army. Use the $0.1$ significance level to test the claim that less than one-quarter of the soldiers were royalists.
Critical values are given by : $z 0.005=2.575$,$z 0.01=2.325$, $z 0.025=1.96$, $z 0.05=1.645$, $z 0.1=1.282$ when $d.f=31:t 0.005=2.744$,$t 0.01=2.453$,$t 0.025=2.040$,$t 0.05=1.696$,$t 0.1=1.309$.
Solution
Step 1
It is given that:
$X-226$
$n-784$
Sample projection:
$\hat{p}-\dfrac{x}{n}=\dfrac{226}{784}=0.288$
The researcher claims that less than a quarter of the soldiers were Royalists.
Thus, null and alternative hypotheses are:
$H_{0}=p-0.25$
$H_{1}=p<0.25$
Step 2
The standardized test statistic can be found as:
$Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$
$Z=\dfrac{0.288-0.25}{\sqrt{\dfrac{0.25(1-0.25)}{1200}}}=3.04$
The level of significance, $=0.1$
Using $z-table$, the critical value at the level of significance $0.1$ is $-1.282$.
Since calculated statistic $Z=3.04>|critical\:value|=|-1.282|$ , we reject the null hypothesis. Therefore, it was concluded that less than one-quarter of the soldiers were Royalists.
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How Would You Make A Quad Based On a Direction in 3D.
Recommended Posts
ankhd 2304
Hello.
I have a direction created from 2 3d locations, and a Size value. With this how could I create a quad or for points is what I would like that are
on the plane facing the direction at Size away from each other. the direction vector would be the centre.
So far I have.
float3 dir = start - target;
float3 perp = Perpendicular(dir);
float3 up = float3(0.0, perp.y, 0.0);//not sure if this is correct
now I'm a bit lost to what to do next to get my 4 points.
Do I do offset perp and up by +- Size like. perp * Size = x point.
Do I need to create a plane that faces the direction
Does dir and the length of dir = plane;
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lwm 2518
Some pseudocode:
dir = normalize(target - start)
up = (0, 1, 0)
halfSize = size / 2
// todo: handle case when (dir == up)
o1 = cross(dir, up) * halfSize
o2 = cross(o1, dir) * halfSize
p1 = start + o1 + o2
p2 = start + o1 - o2
p3 = start - o1 - o2
p4 = start - o1 + o2
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ankhd 2304
Nice work Iwm.
I just did a test with a normalize on the 2 crossproducts it works.
Thank you so much. I'm going to work on the up direction tomorrow oh today it is allready.
What exactly going on with that code...
Time for bed.
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ankhd 2304
Hi all.
My first extrude in a direction.
I have updated my code to acount for the up by using a perpendicular vector.
Thank you GameDev.
here is the updated code to create a 8 location of a cube faceing any direction
//the face vertex points defines our cube
//float3 p1, p2, p3, p4, p5, p6, p7, p8;
float3 vetexdata[8];
//we need to get the direction we are heading
float3 dir = normalize(target - pos);
//create a up vector
float3 up = Perpendicular(dir);//float3(0.0, 1.0, 0.0);
float halfsize = 200;//wdthp * 0.5;
float extrudelength = length(target - pos);
// todo: handle case when (dir == up)
float3 o1 = normalize(cross(dir, up)) * halfsize;
float3 o2 = normalize(cross(o1, dir)) * halfsize;
//this is the first face of the cube starting at our position facing the direction we want to extrude in
vetexdata[0] = pos + o1 + o2;
vetexdata[1] = pos + o1 - o2;
vetexdata[2] = pos - o1 - o2;
vetexdata[3] = pos - o1 + o2;
//the second face at the target location
vetexdata[4] = vetexdata[0] + dir* extrudelength;
vetexdata[5] = vetexdata[1] + dir* extrudelength;
vetexdata[6] = vetexdata[2] + dir* extrudelength;
vetexdata[7] = vetexdata[3] + dir* extrudelength;
//the above defines our cubes face vertices now we need to lay them out
//to form eighter the 4 faces with no end caps or the whole cube with end caps
//the easyest way I think is to make a index list with 3 index for a triangle
/*
uint indices[37]
{
// Front face
0,1,2,
0,2,3,
// Back face
4,6,5,
4,7,6,
// Left face
4,5,1,
4,1,0,
// Right face
3,2,6,
3,6,7,
// Top face
1,5,6,
1,6,2,
// Bottom face
4,0,3,
4,3,7,
};
*/
Image has glow but its a cube facing up
Thanks Again All.
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# How to answer this photoelectric effect question?
#### WMDhamnekar
My attempt:- 1)At 0.68e15 Hz frequency of light,Metal A gives 7.2eV Kinetic Energy(KE).
2)At 1e15 Hz frequency of light, Metal B, gives 6eV Kinetic Energy.
3) At 1.1e15 Hz frequency of light, Metal C gives 5.5eV Kinetic Energy.
4)At1.2e15 Hz frequency of light, Metal D gives 5eV Kinetic energy.
Now $KE_{max}=(\frac{hc}{\lambda}-\phi)$ where $\lambda$=wavelength of light. $\phi$= work function of metal and c= speed of light.
Now $7.2eV=(4.14e-15eVs \times 0.68e15 Hz -\phi)$for Metal A. The wavelength of light is 441 nm. Likewise we can compute wavelength of light incident on Metal B, C and D.
But we don't know the work functions of Metal A,B, C and D.
So, how to answer this question?
Last edited:
#### WMDhamnekar
After performing some computational work, I computed answer to this question which is B.
#### topsquark
Forum Staff
It is more common to express the equation as $$\displaystyle KE_{max} = hf - \phi$$. It makes this a little easier to work with in this case, though there is no real difference.
The work function is the "y" intercept of this equation. So what is the equation for each of the lines?
For example, for metal A, we have two points on the line (I'll call this $$\displaystyle (f, KE_{max} )$$ as an ordered pair.)
(0.64, 0) and (2.4, 7.2) (Of course we can use any two points on the line.) So the equation of the line is $$\displaystyle KE_{max} = hf - 2.723 \text{eV}$$, so the work function is $$\displaystyle \phi = 2.726 \text{eV}$$.
Can you get the other three?
-Dan
Addendum: I see you did some work while I was posting. I get metal B as well, but when I calculated the work function of metal B I get 3.93 eV and when I use the data for the unknown metal I get 4.14 eV, so B is the closest answer. I suppose it could be a rounding error.
Last edited:
#### WMDhamnekar
It is more common to express the equation as $$\displaystyle KE_{max} = hf - \phi$$. It makes this a little easier to work with in this case, though there is no real difference.
The work function is the "y" intercept of this equation. So what is the equation for each of the lines?
For example, for metal A, we have two points on the line (I'll call this $$\displaystyle (f, KE_{max} )$$ as an ordered pair.)
(0.64, 0) and (2.4, 7.2) (Of course we can use any two points on the line.) So the equation of the line is $$\displaystyle KE_{max} = hf - 2.723 \text{eV}$$, so the work function is $$\displaystyle \phi = 2.726 \text{eV}$$.
Can you get the other three?
-Dan
Addendum: I see you did some work while I was posting. I get metal B as well, but when I calculated the work function of metal B I get 3.93 eV and when I use the data for the unknown metal I get 4.14 eV, so B is the closest answer. I suppose it could be a rounding error.
Hello,
I computed the answer as follows:-$KE_{max}+\phi=hf\tag{1}$, where hf is energy of light,h=Planck's constant.
Now $KE_{max},\lambda$ are 2.5eV and 1.87e-7 m respectively. We computed work function$\phi$ of metal B is 4.14 eV. Plugging in these values in (1) we get
$2.5 eV+4.14eV=\frac{6.626e-34 J*s*299792458m/s}{1.87e-7m} \Rightarrow 6.63eV=6.63eV$ rounded off to significant figures.
So, we can conclude that Metal B is the unknown metal we are looking for.
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# Sorry another Beamer question
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## Sorry another Beamer question
Hi all, Thanks for the help about my previous question which came down to incompatibility issues between beamer and graphics (graphicx) which a one-line fix of code at the start of the document can solve (for anyone with the same problem the line to insert at the start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother and thanks very much to Paul Thompson for passing this fix along to me) Emboldened, I thought I'd try out a file with PDF graphs called in it. Needless to say, it did not go well! The following call within the latex file \includegraphics[ natheight=10.6913in, natwidth=14.0472in, height=1.2035in, width=1.5733in] {C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf} Led to the following error .... [17] [18] [19] [20] ! Undefined control sequence. ...@base \Gin@ext image\GPT@AttrShort \ifx \GPT@print \ltx@empty... l.758 \end{frame} % ? Hit return and it compiles but with the graph simply missing from the output. The same latex file compiles fine (with the graphs included) on other well-known tex platforms such as scientific word and tex writer, so is this another conflict between beamer and graphics? Many thanks. Best wishes, Rob Taylor ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
Sorry, the problem yesterday was a conflict between geometry and beamer, not graphics and beamer ... so probably unrelated to today's problem .... > -----Original Message----- > From: Taylor, Robert [mailto:[hidden email]] > Sent: 21 September 2016 12:59 > To: [hidden email] > Subject: [MiKTeX] Sorry another Beamer question > > Hi all, > > Thanks for the help about my previous question which came down to > incompatibility issues between beamer and graphics (graphicx) which a one- > line fix of code at the start of the document can solve (for anyone with the > same problem the line to insert at the start of the latex file is > \makeatletter\let\ifGm@compatii\relax\makeatother > and thanks very much to Paul Thompson for passing this fix along to me) > > Emboldened, I thought I'd try out a file with PDF graphs called in it. Needless > to say, it did not go well! The following call within the latex file > > \includegraphics[ > natheight=10.6913in, natwidth=14.0472in, height=1.2035in, width=1.5733in] > {C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf} > > Led to the following error > > .... [17] [18] [19] [20] > ! Undefined control sequence. > ...@base \Gin@ext image\GPT@AttrShort > \ifx \GPT@print \ltx@empty... > l.758 \end{frame} > % > ? > > Hit return and it compiles but with the graph simply missing from the output. > The same latex file compiles fine (with the graphs included) on other well- > known tex platforms such as scientific word and tex writer, so is this another > conflict between beamer and graphics? > > Many thanks. > > Best wishes, Rob Taylor > ------------------------------------------------------------------------------ > _______________________________________________ > Q: How can I leave the mailing list? > A: See http://docs.miktex.org/faq/support.html#leavingml------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Taylor, Robert I think I solved this myself amazingly enough (!), by finding this on the web: "With natwidth and natheight you are trying to specify a bounding box which is not supported by the pdftex driver: the driver pdftex.def does not support the manual specification of the bounding box, because it is not needed. If the user wants to have a portion of the image, there are options viewport and trim to specify this wish in a cleaner way. Option bb is treated as viewport with a warning." In case this helps anyone else. Cheers, Rob > -----Original Message----- > From: Taylor, Robert [mailto:[hidden email]] > Sent: 21 September 2016 12:59 > To: [hidden email] > Subject: [MiKTeX] Sorry another Beamer question > > Hi all, > > Thanks for the help about my previous question which came down to > incompatibility issues between beamer and graphics (graphicx) which a one- > line fix of code at the start of the document can solve (for anyone with the > same problem the line to insert at the start of the latex file is > \makeatletter\let\ifGm@compatii\relax\makeatother > and thanks very much to Paul Thompson for passing this fix along to me) > > Emboldened, I thought I'd try out a file with PDF graphs called in it. Needless > to say, it did not go well! The following call within the latex file > > \includegraphics[ > natheight=10.6913in, natwidth=14.0472in, height=1.2035in, width=1.5733in] > {C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf} > > Led to the following error > > .... [17] [18] [19] [20] > ! Undefined control sequence. > ...@base \Gin@ext image\GPT@AttrShort > \ifx \GPT@print \ltx@empty... > l.758 \end{frame} > % > ? > > Hit return and it compiles but with the graph simply missing from the output. > The same latex file compiles fine (with the graphs included) on other well- > known tex platforms such as scientific word and tex writer, so is this another > conflict between beamer and graphics? > > Many thanks. > > Best wishes, Rob Taylor > ------------------------------------------------------------------------------ > _______________________________________________ > Q: How can I leave the mailing list? > A: See http://docs.miktex.org/faq/support.html#leavingml------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
Now, there you go. You have figgered out how to hack. That's the issue - it's not how to do things when they work correctly (The LaTeX Companion covers much of this, but does not include my class in its coverage), but how to solve problems. -----Original Message----- From: Taylor, Robert [mailto:[hidden email]] Sent: Wednesday, September 21, 2016 7:26 AM To: A place for MiKTeX users to discuss MiKTeX related questions. Subject: Re: [MiKTeX] Sorry another Beamer question I think I solved this myself amazingly enough (!), by finding this on the web: "With natwidth and natheight you are trying to specify a bounding box which is not supported by the pdftex driver: the driver pdftex.def does not support the manual specification of the bounding box, because it is not needed. If the user wants to have a portion of the image, there are options viewport and trim to specify this wish in a cleaner way. Option bb is treated as viewport with a warning." In case this helps anyone else. Cheers, Rob > -----Original Message----- > From: Taylor, Robert [mailto:[hidden email]] > Sent: 21 September 2016 12:59 > To: [hidden email] > Subject: [MiKTeX] Sorry another Beamer question > > Hi all, > > Thanks for the help about my previous question which came down to > incompatibility issues between beamer and graphics (graphicx) which a > one- line fix of code at the start of the document can solve (for > anyone with the same problem the line to insert at the start of the > latex file is \makeatletter\let\ifGm@compatii\relax\makeatother > and thanks very much to Paul Thompson for passing this fix along to > me) > > Emboldened, I thought I'd try out a file with PDF graphs called in it. > Needless to say, it did not go well! The following call within the > latex file > > \includegraphics[ > natheight=10.6913in, natwidth=14.0472in, height=1.2035in, > width=1.5733in] {C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf} > > Led to the following error > > .... [17] [18] [19] [20] C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf> > ! Undefined control sequence. > ...@base \Gin@ext image\GPT@AttrShort > \ifx \GPT@print \ltx@empty... > l.758 \end{frame} > % > ? > > Hit return and it compiles but with the graph simply missing from the output. > The same latex file compiles fine (with the graphs included) on other > well- known tex platforms such as scientific word and tex writer, so > is this another conflict between beamer and graphics? > > Many thanks. > > Best wishes, Rob Taylor > ---------------------------------------------------------------------- > -------- _______________________________________________ > Q: How can I leave the mailing list? > A: See http://docs.miktex.org/faq/support.html#leavingml------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml----------------------------------------------------------------------- Confidentiality Notice: This e-mail message, including any attachments, is for the sole use of the intended recipient(s) and may contain privileged and confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply e-mail and destroy all copies of the original message. ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Taylor, Robert On 21/09/2016 12:58, Taylor, Robert wrote: > Hi all, > > Thanks for the help about my previous question which came down to incompatibility issues between beamer and graphics (graphicx) which a one-line fix of code at the start of the document can solve (for anyone with the same problem the line to insert at the start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother > and thanks very much to Paul Thompson for passing this fix along to me) > > Emboldened, I thought I'd try out a file with PDF graphs called in it. Needless to say, it did not go well! The following call within the latex file > > \includegraphics[ > natheight=10.6913in, natwidth=14.0472in, height=1.2035in, width=1.5733in] > {C:/Users/rtaylor/Dropbox/tex/graphics/1a__1.pdf} > > Led to the following error > > .... [17] [18] [19] [20] > ! Undefined control sequence. > ...@base \Gin@ext image\GPT@AttrShort > \ifx \GPT@print \ltx@empty... > l.758 \end{frame} > % > ? > > Hit return and it compiles but with the graph simply missing from the output. The same latex file compiles fine (with the graphs included) on other well-known tex platforms such as scientific word and tex writer, so is this another conflict between beamer and graphics? > > Many thanks. > > Best wishes, Rob Taylor Please send a full example: beamer works fine in general with graphicx (as you'd expect, I guess, for a class for making visual presentation material!). Joseph ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Taylor, Robert Am Wed, 21 Sep 2016 11:58:37 +0000 schrieb Taylor, Robert: > Thanks for the help about my previous question which came down to > incompatibility issues between beamer and graphics (graphicx) > which a one-line fix of code at the start of the document can > solve (for anyone with the same problem the line to insert at the > start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother Sorry but this is a hack and not a solution. If you want to use it fine, but you shouldn't spread it around. So correct way to solve the problem is to make a minimal example and to show its log-file so that one can analyze what is wrong with your system. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Taylor, Robert Am Wed, 21 Sep 2016 12:26:24 +0000 schrieb Taylor, Robert: > I think I solved this myself amazingly enough (!), by finding this on the web: > "With natwidth and natheight you are trying to specify a bounding > box which is not supported by the pdftex driver: the driver > pdftex.def does not support the manual specification of the > bounding box, because it is not needed. If the user wants to have > a portion of the image, there are options viewport and trim to > specify this wish in a cleaner way. Option bb is treated as > viewport with a warning." While viewport is the correct option, natwidth/natheight imho shouldn't give an error but like bb issue only a warning. I reported the problem to the LaTeX team and I think David will add some code to avoid the error soon. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Ulrike Fischer-2 Ulrike: I copied that hack from an email with your name on it. Paul A. Thompson -----Original Message----- From: Ulrike Fischer [mailto:[hidden email]] Sent: Wednesday, September 21, 2016 9:56 AM To: [hidden email] Subject: Re: [MiKTeX] Sorry another Beamer question Am Wed, 21 Sep 2016 11:58:37 +0000 schrieb Taylor, Robert: > Thanks for the help about my previous question which came down to > incompatibility issues between beamer and graphics (graphicx) which a > one-line fix of code at the start of the document can solve (for > anyone with the same problem the line to insert at the > start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother Sorry but this is a hack and not a solution. If you want to use it fine, but you shouldn't spread it around. So correct way to solve the problem is to make a minimal example and to show its log-file so that one can analyze what is wrong with your system. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml----------------------------------------------------------------------- Confidentiality Notice: This e-mail message, including any attachments, is for the sole use of the intended recipient(s) and may contain privileged and confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply e-mail and destroy all copies of the original message. ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
Am Wed, 21 Sep 2016 16:12:24 +0000 schrieb Thompson,Paul: > >> Thanks for the help about my previous question which came down to > >> incompatibility issues between beamer and graphics (graphicx) which a > >> one-line fix of code at the start of the document can solve (for > >> anyone with the same problem the line to insert at the > >> start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother > > Sorry but this is a hack and not a solution. > I copied that hack from an email with your name on it. That was from 2010. And the context was: ========================= Update beamer. Or put \makeatletter\let\ifGm@compatii\relax\makeatother before \documentclass. ========================= So it should be clear that with a current beamer it should no longer be necessary. Such measures are sometimes needed for a short time until some update has reached the user. But after 6 years you should move them to the bin. Actually you should never rely on such old code without have checked quite carefully if the context for which the code was written is still valid. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
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## Re: Sorry another Beamer question
In reply to this post by Ulrike Fischer-2 All very evangelical but the updated and latest Beamer files are exactly what was causing the problem .... > On 21 Sep 2016, at 17:34, Ulrike Fischer <[hidden email]> wrote: > > Am Wed, 21 Sep 2016 16:12:24 +0000 schrieb Thompson,Paul: > >>>> Thanks for the help about my previous question which came down to >>>> incompatibility issues between beamer and graphics (graphicx) which a >>>> one-line fix of code at the start of the document can solve (for >>>> anyone with the same problem the line to insert at the >>>> start of the latex file is \makeatletter\let\ifGm@compatii\relax\makeatother > >>> Sorry but this is a hack and not a solution. > >> I copied that hack from an email with your name on it. > > That was from 2010. And the context was: > > ========================= > Update beamer. Or put > > \makeatletter\let\ifGm@compatii\relax\makeatother > > before \documentclass. > ========================= > > So it should be clear that with a current beamer it should no longer > be necessary. Such measures are sometimes needed for a short time > until some update has reached the user. But after 6 years you should > move them to the bin. > > Actually you should never rely on such old code without have checked > quite carefully if the context for which the code was written is > still valid. > > > -- > Ulrike Fischer > http://www.troubleshooting-tex.de/> > > ------------------------------------------------------------------------------ > _______________________________________________ > Q: How can I leave the mailing list? > A: See http://docs.miktex.org/faq/support.html#leavingml------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
Am Wed, 21 Sep 2016 17:31:14 +0000 schrieb Taylor, Robert: > All very evangelical Well as the code you are using is from me you should expect that I know what it is about and that I have every right to claim that it is not a good idea to use it today. > but the updated and latest Beamer files are exactly what was > causing the problem .... I do have a miktex updated today and this here works fine: \documentclass{beamer} \usetheme{CambridgeUS} \begin{document} \begin{frame}{abc} blub \end{frame} \end{document} A remark to a comment you sent me privatly: > Of course I would rather something functioned properly than use a > hack but the hack has worked and for someone whose aim is to > produce some slides to present at a conference that matters WAY > more than the evangelical beauty of a properly functioning latex > system!! As I already wrote: what you are doing in your own code is up to you. But when you start to spread around wrong code and claim this code is a solution you are misleading other people and I think this is unfair to them. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
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## Re: Sorry another Beamer question
On 9/21/2016 3:27 PM, Thompson,Paul wrote: > Don't dump on people for being resourceful. I have never seen Ulrike "dump" on anyone. Unimpeachable patience; endless helpfulness ... these I have seen. fwiw, Alan ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
OK, withdraw "dump" replace with "thoughtful explanation" -----Original Message----- From: Alan Isaac [mailto:[hidden email]] Sent: Wednesday, September 21, 2016 2:34 PM To: A place for MiKTeX users to discuss MiKTeX related questions. Subject: Re: [MiKTeX] Sorry another Beamer question On 9/21/2016 3:27 PM, Thompson,Paul wrote: > Don't dump on people for being resourceful. I have never seen Ulrike "dump" on anyone. Unimpeachable patience; endless helpfulness ... these I have seen. fwiw, Alan ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml----------------------------------------------------------------------- Confidentiality Notice: This e-mail message, including any attachments, is for the sole use of the intended recipient(s) and may contain privileged and confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply e-mail and destroy all copies of the original message. ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by Ulrike Fischer-2 On Wed, Sep 21, 2016 at 08:18:23PM +0200, Ulrike Fischer wrote: > Am Wed, 21 Sep 2016 17:31:14 +0000 schrieb Taylor, Robert: > > > All very evangelical > > Well as the code you are using is from me you should expect that I > know what it is about and that I have every right to claim that it > is not a good idea to use it today. > > > but the updated and latest Beamer files are exactly what was > > causing the problem .... > > I do have a miktex updated today and this here works fine: > > \documentclass{beamer} > \usetheme{CambridgeUS} > > \begin{document} > \begin{frame}{abc} > blub > \end{frame} > \end{document} Please check the logfile exactly which files are being read. Maybe latex reads some outdated version of a file which does not come from the current MiKTeX. -- Siep Kroonenberg ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
> I do have a miktex updated today and this here works fine: > > \documentclass{beamer} > \usetheme{CambridgeUS} > > \begin{document} > \begin{frame}{abc} > blub > \end{frame} > \end{document} Please check the logfile exactly which files are being read. Maybe latex reads some outdated version of a file which does not come from the current MiKTeX. -- Siep Kroonenberg ----------------------------------- Thanks for that reminder. This is again an important tip that inexperienced users do not know about. They think that MikTeX knows which file to pick. Of course, it usually does. But that can be bunged up in a number of ways, and looking for duplicate files is a key step in some difficult cases. I know that I have that problem, in that I tend to stash outdated versions of stuff on the local tree. ----------------------------------------------------------------------- Confidentiality Notice: This e-mail message, including any attachments, is for the sole use of the intended recipient(s) and may contain privileged and confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply e-mail and destroy all copies of the original message. ------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
In reply to this post by AlanIsaac Am Wed, 21 Sep 2016 15:34:06 -0400 schrieb Alan Isaac: >> Don't dump on people for being resourceful. > > > I have never seen Ulrike "dump" on anyone. > Unimpeachable patience; endless helpfulness Then you haven't seen much of me ;-) I sometimes have quite clear views and try to get them around. And in this case I think it is very important not to confuse work-arounds and hacks with real solutions. I don't have any problems with peoples with some time pressure using a hack -- I'm doing is sometimes too. But it is dangerous to think of such a work-around as a solution as it means that you won't tackle the real problem. The problem my code from 2010 tried to solve should be gone long ago and if Robert (or someone else) still have it something is wrong with their texsystem. -- Ulrike Fischer http://www.troubleshooting-tex.de/------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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## Re: Sorry another Beamer question
A tex system I installed yesterday using the basic installer on the CTAN site and updated everything as you are meant to ..... > On 21 Sep 2016, at 21:06, Ulrike Fischer <[hidden email]> wrote: > > Am Wed, 21 Sep 2016 15:34:06 -0400 schrieb Alan Isaac: > >>> Don't dump on people for being resourceful. >> >> >> I have never seen Ulrike "dump" on anyone. >> Unimpeachable patience; endless helpfulness > > Then you haven't seen much of me ;-) > > I sometimes have quite clear views and try to get them around. > > And in this case I think it is very important not to confuse > work-arounds and hacks with real solutions. > > I don't have any problems with peoples with some time pressure using > a hack -- I'm doing is sometimes too. But it is dangerous to think > of such a work-around as a solution as it means that you won't > tackle the real problem. The problem my code from 2010 tried to > solve should be gone long ago and if Robert (or someone else) still > have it something is wrong with their texsystem. > > -- > Ulrike Fischer > http://www.troubleshooting-tex.de/> > > ------------------------------------------------------------------------------ > _______________________________________________ > Q: How can I leave the mailing list? > A: See http://docs.miktex.org/faq/support.html#leavingml------------------------------------------------------------------------------ _______________________________________________ Q: How can I leave the mailing list? A: See http://docs.miktex.org/faq/support.html#leavingml
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# Endomorphism algebra of the tensor product of modules
Let $k$ be a commutative ring, and let $M$ and $N$ be $k$-modules. Let $\mathrm{End}(M) = \mathrm{Hom}_k (M,M)$ be the endomorphism algebra.
Is it true that $\mathrm{End}(M) \otimes \mathrm{End}(N) \cong \mathrm{End}(M \otimes N)$?
I saw this referenced on a blog but I can't find a proof of it. I tried to show that $\mathrm{End}(M \otimes N)$ satisfies the universal property of the tensor product of $\mathrm{End}(M)$ and $\mathrm{End}(N)$ but to no avail.
Thanks for any help.
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Well, it's true when $k$ is a field and $M$ and $N$ are finite-dimensional: this is basically the Kronecker product construction. I don't immediately see how this generalises. – Zhen Lin Sep 17 '12 at 13:06
Seems you need for this at least one of the modules $M,N$ to be a finitely projective $k$-module. See Bourbaki, Algebra, chapter II, $\S$ 4.4, proposition 4.
EDIT. But if you just want a map
$$\mathrm{End} (M) \otimes \mathrm{End}(N) \longrightarrow \mathrm{End}(M\otimes N)$$
Just send $\phi \otimes \psi \in \mathrm{End} (M) \otimes \mathrm{End}(N)$ to $\phi \otimes \psi \in \mathrm{End}(M\otimes N)$. :-D
I mean: the "second" $\phi\otimes \psi$ is
$$(\phi\otimes\psi)(m\otimes n) = \phi(m)\otimes \psi(n) \ .$$
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Thank you. The source I was reading must have been incorrect then. Luckily I only needed a map $\mathrm{End}(M) \otimes \mathrm{End}(N) \to \mathrm{End}(M \otimes N)$. – Paul Slevin Sep 17 '12 at 13:15
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I already specified the monoid part of these monoidal preorders before stating the first puzzle. The puzzles ask what's the preorder part: that is, when does one complex count as \$$\le \$$ another. This depends on what reactions we allow.
Yes, we can think of a complex
$a \text{H} + b \text{O} + c \text{H}_2\text{O}$
as a multisubset of the set \$$\\{\textrm{H}, \textrm{O}, \textrm{H}_2\textrm{O} \\} \$$. Alternatively, and perhaps less stressfully, we can think of it as an ordered triple \$$(a,b,c) \$$ of natural numbers. The monoid operation is just the usual addition of these triples.
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Trigonometry! #151
Geometry Level 3
The number of solutions to the equation $\sin 2x+\cos 2x+\sin x+\cos x+1=0$ in $$[0,2\pi]$$ is
This problem is part of the set Trigonometry.
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# Indiscrete Space is Irreducible
## Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Then $T$ is irreducible.
## Proof
There is only one non-empty open set in $T$.
So there can be no two open sets in $T$ which are disjoint.
Hence (trivially) $T$ is irreducible.
$\blacksquare$
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How to solve integral equation $x(t)-\int_{0}^{1}[\cos (t) \sec (s) x(s)]ds=\sinh (t), 0\leq t\leq 1.$
I was thinking about the problem that was as follows:
The integral equation $x(t)-\displaystyle \int_{0}^{1}[\cos (t) \sec (s) x(s)]ds=\sinh (t), 0\leq t\leq 1,$ has
(a)no solution,
(b)a unique solution,
(c)more than one but finitely many solution,
(d)infinitely many solutions.
My attempts: From the given equation,we get $$x(t)=\cos(t)\int_{0}^{1}\sec(s)x(s)ds + \sinh(t)=C \cos(t)+\sinh(t)$$ where $C=\int_{0}^{1}\sec(s)x(s)ds$ and so $$C=\int_{0}^{1}\sec(s)[\cos(s)C +\sinh(s)]ds=C\int_{0}^{1}ds+\int_{0}^{1}\sec(s)\sinh(s)ds$$ and hence we get, $\int_{0}^{1}\sec(s)\sinh(s)ds=0.$
$$x(t)=\cos t\int_0^1x(s)\sec s\,ds + \sinh t=C\cos t+\sinh t$$ where $C=\int_0^1x(s)\sec s\,ds$ and so $$C\cos t=\int_0^1[C\cos s+\sinh s]\sec s\,ds=C\int_0^1ds+\int_0^1\sec s\sinh s\,ds=C+k,$$ where $k=\int_0^1\sec s\sinh s\,ds\approx0.739$, so since no choice of $C$ will make this equation true for all $t$, there is no solution.
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# How do you find the limit lim_(x->2^+)sqrt(2-x) ?
$x \setminus \to {2}^{+}$ means that $x$ is approaching 2 from the right, that is, $x > 2$. But if $x > 2$ then $2 - x < 0$ so that $2 - x$ is not in the domain of the square root function...So $\setminus {\lim}_{x \setminus \to {2}^{+}} \setminus \sqrt{2 - x}$ does not exist , while from the left:
$\setminus {\lim}_{x \setminus \to {2}^{-}} \setminus \sqrt{2 - x} = \sqrt{0} = 0$
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## 19.9 Embedding abelian categories
In this section we show that an abelian category embeds in the category of abelian sheaves on a site having enough points. The site will be the one described in the following lemma.
Lemma 19.9.1. Let $\mathcal{A}$ be an abelian category. Let
$\text{Cov} = \{ \{ f : V \to U\} \mid f\text{ is surjective}\} .$
Then $(\mathcal{A}, \text{Cov})$ is a site, see Sites, Definition 7.6.2.
Proof. Note that $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is a set by our conventions about categories. An isomorphism is a surjective morphism. The composition of surjective morphisms is surjective. And the base change of a surjective morphism in $\mathcal{A}$ is surjective, see Homology, Lemma 12.5.14. $\square$
Let $\mathcal{A}$ be a pre-additive category. In this case the Yoneda embedding $\mathcal{A} \to \textit{PSh}(\mathcal{A})$, $X \mapsto h_ X$ factors through a functor $\mathcal{A} \to \textit{PAb}(\mathcal{A})$.
Lemma 19.9.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C} = (\mathcal{A}, \text{Cov})$ be the site defined in Lemma 19.9.1. Then $X \mapsto h_ X$ defines a fully faithful, exact functor
$\mathcal{A} \longrightarrow \textit{Ab}(\mathcal{C}).$
Moreover, the site $\mathcal{C}$ has enough points.
Proof. Suppose that $f : V \to U$ is a surjective morphism of $\mathcal{A}$. Let $K = \mathop{\mathrm{Ker}}(f)$. Recall that $V \times _ U V = \mathop{\mathrm{Ker}}((f, -f) : V \oplus V \to U)$, see Homology, Example 12.5.6. In particular there exists an injection $K \oplus K \to V \times _ U V$. Let $p, q : V \times _ U V \to V$ be the two projection morphisms. Note that $p - q : V \times _ U V \to V$ is a morphism such that $f \circ (p - q) = 0$. Hence $p - q$ factors through $K \to V$. Let us denote this morphism by $c : V \times _ U V \to K$. And since the composition $K \oplus K \to V \times _ U V \to K$ is surjective, we conclude that $c$ is surjective. It follows that
$V \times _ U V \xrightarrow {p - q} V \to U \to 0$
is an exact sequence of $\mathcal{A}$. Hence for an object $X$ of $\mathcal{A}$ the sequence
$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(V, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(V \times _ U V, X)$
is an exact sequence of abelian groups, see Homology, Lemma 12.5.8. This means that $h_ X$ satisfies the sheaf condition on $\mathcal{C}$.
The functor is fully faithful by Categories, Lemma 4.3.5. The functor is a left exact functor between abelian categories by Homology, Lemma 12.5.8. To show that it is right exact, let $X \to Y$ be a surjective morphism of $\mathcal{A}$. Let $U$ be an object of $\mathcal{A}$, and let $s \in h_ Y(U) = \mathop{Mor}\nolimits _\mathcal {A}(U, Y)$ be a section of $h_ Y$ over $U$. By Homology, Lemma 12.5.14 the projection $U \times _ Y X \to U$ is surjective. Hence $\{ V = U \times _ Y X \to U\}$ is a covering of $U$ such that $s|_ V$ lifts to a section of $h_ X$. This proves that $h_ X \to h_ Y$ is a surjection of abelian sheaves, see Sites, Lemma 7.11.2.
The site $\mathcal{C}$ has enough points by Sites, Proposition 7.39.3. $\square$
Remark 19.9.3. The Freyd-Mitchell embedding theorem says there exists a fully faithful exact functor from any abelian category $\mathcal{A}$ to the category of modules over a ring. Lemma 19.9.2 is not quite as strong. But the result is suitable for the Stacks project as we have to understand sheaves of abelian groups on sites in detail anyway. Moreover, “diagram chasing” works in the category of abelian sheaves on $\mathcal{C}$, for example by working with sections over objects, or by working on the level of stalks using that $\mathcal{C}$ has enough points. To see how to deduce the Freyd-Mitchell embedding theorem from Lemma 19.9.2 see Remark 19.9.5.
Remark 19.9.4. If $\mathcal{A}$ is a “big” abelian category, i.e., if $\mathcal{A}$ has a class of objects, then Lemma 19.9.2 does not work. In this case, given any set of objects $E \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ there exists an abelian full subcategory $\mathcal{A}' \subset \mathcal{A}$ such that $\mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$ is a set and $E \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}')$. Then one can apply Lemma 19.9.2 to $\mathcal{A}'$. One can use this to prove that results depending on a diagram chase hold in $\mathcal{A}$.
Remark 19.9.5. Let $\mathcal{C}$ be a site. Note that $\textit{Ab}(\mathcal{C})$ has enough injectives, see Theorem 19.7.4. (In the case that $\mathcal{C}$ has enough points this is straightforward because $p_*I$ is an injective sheaf if $I$ is an injective $\mathbf{Z}$-module and $p$ is a point.) Also, $\textit{Ab}(\mathcal{C})$ has a cogenerator (details omitted). Hence Lemma 19.9.2 proves that we have a fully faithful, exact embedding $\mathcal{A} \to \mathcal{B}$ where $\mathcal{B}$ has a cogenerator and enough injectives. We can apply this to $\mathcal{A}^{opp}$ and we get a fully faithful exact functor $i : \mathcal{A} \to \mathcal{D} = \mathcal{B}^{opp}$ where $\mathcal{D}$ has enough projectives and a generator. Hence $\mathcal{D}$ has a projective generator $P$. Set $R = \mathop{Mor}\nolimits _\mathcal {D}(P, P)$. Then
$\mathcal{A} \longrightarrow \text{Mod}_ R, \quad X \longmapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(P, X).$
One can check this is a fully faithful, exact functor. In other words, one retrieves the Freyd-Mitchell theorem mentioned in Remark 19.9.3 above.
Remark 19.9.6. The arguments proving Lemmas 19.9.1 and 19.9.2 work also for exact categories, see [Appendix A, Buhler] and [1.1.4, BBD]. We quickly review this here and we add more details if we ever need it in the Stacks project.
Let $\mathcal{A}$ be an additive category. A kernel-cokernel pair is a pair $(i, p)$ of morphisms of $\mathcal{A}$ with $i : A \to B$, $p : B \to C$ such that $i$ is the kernel of $p$ and $p$ is the cokernel of $i$. Given a set $\mathcal{E}$ of kernel-cokernel pairs we say $i : A \to B$ is an admissible monomorphism if $(i, p) \in \mathcal{E}$ for some morphism $p$. Similarly we say a morphism $p : B \to C$ is an admissible epimorphism if $(i, p) \in \mathcal{E}$ for some morphism $i$. The pair $(\mathcal{A}, \mathcal{E})$ is said to be an exact category if the following axioms hold
1. $\mathcal{E}$ is closed under isomorphisms of kernel-cokernel pairs,
2. for any object $A$ the morphism $1_ A$ is both an admissible epimorphism and an admissible monomorphism,
3. admissible monomorphisms are stable under composition,
4. admissible epimorphisms are stable under composition,
5. the push-out of an admissible monomorphism $i : A \to B$ via any morphism $A \to A'$ exist and the induced morphism $i' : A' \to B'$ is an admissible monomorphism, and
6. the base change of an admissible epimorphism $p : B \to C$ via any morphism $C' \to C$ exist and the induced morphism $p' : B' \to C'$ is an admissible epimorphism.
Given such a structure let $\mathcal{C} = (\mathcal{A}, \text{Cov})$ where coverings (i.e., elements of $\text{Cov}$) are given by admissible epimorphisms. The axioms listed above immediately imply that this is a site. Consider the functor
$F : \mathcal{A} \longrightarrow \textit{Ab}(\mathcal{C}), \quad X \longmapsto h_ X$
exactly as in Lemma 19.9.2. It turns out that this functor is fully faithful, exact, and reflects exactness. Moreover, any extension of objects in the essential image of $F$ is in the essential image of $F$.
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# Samuel Coskey: Introduction to continuous logic
Wednesday, March 4 from 3 to 4pm
Room: Math 124
Speaker: Samuel Coskey (BSU)
Title: An introduction to continuous logic
Abstract: Continous logic is a proper generalization of first order logic where the usual binary truth values are replaced by the unit interval $[0,1]$. The models for this logic are metric structures, which are metric spaces together with continuous functions and $[0,1]$-valued relations. Just as ordinary logic has typical applications in discrete math, continuous logic has applications in analysis. In this talk we will introduce just the basic concepts and theory of continuous logic.
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## Precalculus (6th Edition) Blitzer
In a right angled triangle, the acute angles are complements. Thus, $\sin (90^{\circ}-\theta) =\cos \theta$ So, the sine of one acute angle is equal to the cosine of the other acute angle.
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2013-07-14, 16:34 #34 chris2be8 Sep 2009 22×3×167 Posts To clarify my last post, I'm now sieving 84^131-1 (the last of the 3 I listed). I meant I'll queue the first of the two firejuggler found for processing. Sorry for the confusion. Chris PS. Has anyone worked on 39^158+1 (the middle one of the 3 I listed)? wombatman found a poly for 37+148+1 (the first one I listed). PPS. And many thanks again.
2013-07-14, 16:46 #35 firejuggler Apr 2010 Over the rainbow 32×281 Posts Will work on the fourth number of post 85 once i'm done proccessing the hits on the C176 Last fiddled with by firejuggler on 2013-07-14 at 17:03
2013-07-14, 16:55 #36 VBCurtis "Curtis" Feb 2005 Riverside, CA 52×11×17 Posts Chris- See post #123 for the poly for the middle number. -Curtis
2013-07-14, 21:39 #37 VBCurtis "Curtis" Feb 2005 Riverside, CA 124316 Posts I am doing poly search for the C152 from Aliquot sequence 3408:1287. I'm using my usual settings from this thread: Default stage1 norm, -nps stage 2 norm roughly default/18 (1.5e20 in this case). I get my usual 100kb of -nps output in 12 hrs, but -npr produces ZERO polys, even when I loosen the min e value lower than default (3.5e-12 instead of default 3.63e-12). Are there numbers where the polys found are just terrible compared to expectations? I just loosened the min value to 3.0e-12, and I get plenty of poly output- so it's not a software error, far as I can tell. Coeffs for these tests are in the 3-6M range, which seems pretty standard for a C152. Edit: Should I experiment with stage-2 norm for the npr step? Last fiddled with by VBCurtis on 2013-07-14 at 21:52
2013-07-14, 23:09 #38 fivemack (loop (#_fork)) Feb 2006 Cambridge, England 2×3,191 Posts I think there are still some infelicities in the table of norms and min-E; I'm getting ludicrously high yield around 160 digits and, as you mention, very low yields at C152. For 9096168730717325471174345357545088389800010197483699618393510880224063584308788722339816387794620862849936796290458644418085736533343831602295288484811 (151 digits) I ended up with (at least in October 2011, parameters may be changed now) with 20k polynomials in msieve.dat.p for each 1000-A5 range. But running 3936090042216234613709067994268312375269653484027874112960552153626006376667321970650341587753839767751476856325227586616119340003347991887464416435517 (151 digits) last week I got no more than ten polynomials in some such ranges. For 4202701474560599145864668379214403877602720577806362075074321031589277391358279336093140444841799054199486202393045071085680275027063857193327313335405064090859 I got 3-5k polynomials for each 1000-A5 range
2013-07-14, 23:09 #39 jasonp Tribal Bullet Oct 2004 67168 Posts We've sometimes seen input sizes where the chosen bounds are too conservative, in that too many hits get found, but almost never the opposite. Also, the numerical value of the skew that you find doesn't have very much to do with sieving quality, so it isn't a big deal if you find a poly with large skew. It usually is just an indication that your search is in an early phase, where you get a polynomial with a very good root score that requires really stretching the skew to achieve.
2013-07-15, 00:22 #40 firejuggler Apr 2010 Over the rainbow 32×281 Posts nothing better in the 400-420M range than what I already posted for the C176
2013-07-15, 02:21 #41 firejuggler Apr 2010 Over the rainbow 9E116 Posts for (421^101-1)/((421-1)*3637*52859291287277*15527015834461272375419*384360771211140230121323*3103491858106402597710257788494888754189303) I have a flare Code: R0: -3827996057819835127917748724463 R1: 131539370518073 A0: -602715843958817538679066753523087553532960 A1: 45781605633077304430392767759952648 A2: 1068428398276002038958183140 A3: -48864994613567679099 A4: -571757632456 A5: 9180 Mon Jul 15 02:40:31 2013 skew 46132459.25, size 3.041e-015, alpha -8.806, combined = 2.126e-012 rroots = 5 skew is horrible; found a second hit @ Code: R0: -1766358190766735704168317054222 R1: 112645411133573 A0: 493231366602757377598300196387454965 A1: 2761322107950780473345799391567 A2: -16280562559665314935321596 A3: -18743805941490767137 A4: 7812064738909 A5: 438840 skew 1418693.51, size 2.267e-015, alpha -5.886, combined = 1.806e-012 rroots = 5 Skew is better but score, not so much I will extend search upto a leading coef of 3M
2013-07-15, 03:05 #42 wombatman I moo ablest echo power! May 2013 1,741 Posts It's not better than what Curtis got, but I'll include my best so far for posterity and comparison of skews and whatnot: Code: R0: -2985364064788479756004867693430704 R1: 338252966469030421 A0: -850411162568663002775199166333595393602815 A1: 912278472619088502716551535935935096 A2: 30604247888096781733208334160 A3: -36949595121636590255444 A4: -2832475140989925 A5: 100568520 skew 8586463.91, size 2.728e-017, alpha -8.174, combined = 1.281e-013 rroots = 3
2013-07-15, 04:26 #43 VBCurtis "Curtis" Feb 2005 Riverside, CA 111038 Posts Each 5-digit increase in size takes roughly double the effort to sieve- and should have a corresponding increase in poly-select time. This C176, at roughly 20 digits larger than our other searches, deserves ~15x more effort than we have given the other numbers. I've been putting ~3 days into the 155-157s in this thread, which may be why I'm often getting better polys. I will be putting at least a week into my part of the C176 search, more if I/we don't find a "lucky" poly. If the three of us put a week each into it, we have a good chance to find a poly worth sieving with. For this search, I think that's 1.45e-13 at minimum.
2013-07-15, 09:45 #44 firejuggler Apr 2010 Over the rainbow 47418 Posts (421^101-1)/((421-1)*3637*52859291287277*15527015834461272375419*384360771211140230121323*3103491858106402597710257788494888754189303) again, Code: R0: -1379023836637462618160987168271 R1: 117578686639117 A0: -36460967635057249759007362058565617216 A1: 82932889219472424950843244760688 A2: 66710061781474557308819682 A3: -21316592155418774201 A4: -9644497418536 A5: 1513008 skew 2662773.43, size 2.688e-015, alpha -7.134, combined = 2.021e-012 rroots = 5 I will resume my search on the C176
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## Pat Blythe – SING! Toronto, Jeff Jones Trio and …Music!
Posted in music, Opinion, Review with tags , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , on June 8, 2019 by segarini
I am still in awe of what happens every time I step out the door whether it’s for a walk, head to a club, meet a friend for a meal, hop in the vehicle and drive….somewhere, photograph a festival, have a random conversation with someone you just met…..something totally unexpected invariably transpires, something I was totally unprepared for. Whether I’m learning it or experiencing it at least it’s consistent. Standing back and studying the broader picture I find there are a hundred million large and teeny tiny connections, that, as the universe unfolds and does its thing, links us all together. Whether it’s a person, a concert, a band, a type of music, an experience, a friend of a friend, we are all our own “six degrees of separation”. I find it all rather fascinating……and I’m still in awe.
## JAIMIE VERNON – THE FERGUS CONVERGENCE
Posted in Opinion with tags , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , on May 9, 2015 by segarini
I first met Fergus Hambleton in person while I was performing with my trio, Graye featuring Todd Miller and Lawrie Ingles, at a small club on Yonge Street in Toronto called Rocky Raccoons. It was a mid-week jam night and we were going to do a few songs when I looked up and Fergus was sitting in the front row taking in what can only be described as a “loose” performance of one of my original songs.
## Pat Blythe: Yesiree! CMW 2015 Continues….
Posted in Opinion with tags , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , on May 6, 2015 by segarini
Yessirreee….it’s that time once again. My dawgs are barking, my body’s aching and wondering what the hell I’m doing to it. Ahhhh….CMW. It’s been such a busy four days and the party hasn’t even started yet.
## Roxanne Tellier: CMW 2015 … And So It Begins
Posted in Opinion with tags , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , on May 3, 2015 by segarini
Finally, after a long, cold winter, I’m coming up for airto a bright spring promisinga musical overload. Perfect weather, indeed, for enjoying Fergus Hambleton at Hugh’s Room on Wednesday, where he showcased his sunny signature sound and new CD “Written on the Wind. “
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## FANDOM
25,433 Pages
The ammo press is a machine that can be used to create ammunition upon completion of the quest Free Labor in the Fallout 3 add-on The Pitt. Access is given by either Wernher or Lord Ashur.
## CharacteristicsEdit
Located overlooking The Hole, the ammo press consists of a terminal used to operate the machine and a materials box at the bottom of the stairs where ammunition or scrap metal is placed. The amount of ammunition received depends on how much ammunition/scrap metal is inserted into the machine.
Any variety of conventional ammunition can be placed in the materials box, from 5mm ammunition to .44 Magnum. Once placed in the materials box, the ammunition inside can be melted down and re-shaped by the ammo press into any other variety of conventional ammunition by the terminal.
The ammo press takes unwanted ammunition from the early game (10mm or .32 caliber, for example) and converts it to more desirable ammunition. It also makes ammunition not commonly used (for example, 5mm rounds not used by a Small Guns character) somewhat more useful than simply selling it. Unfortunately, the press is unable to create any energy weapon ammunition, only conventional ammunition.
The ammo press, combined with The Outcast Collection Agent quest, means one can build up a considerable amount of any type of conventional ammunition by simply farming Enclave outposts for weapons and power armor, and trading the hardware in for 5.56mm ammunition. The 5.56mm ammunition can then be recalibrated in the ammo press into new ammunition. Combined with the perk, Scrounger, it makes making ammo much easier.
## CraftingEdit
In order to make ammunition, unused ammunition or scrap metal must be loaded into the materials bin. The bin is a rectangular box with a large X located at the front base of the press that is closest to the Ammunition Mill terminal. Once loaded, return to the computer to select the desired ammunition. After selecting one's ammunition of choice, return to the bin and retrieve the new ammunition.
### FormulaEdit
Ammunition Type Press Value Press/Caps
10mm round 2 2
.308 caliber 15 5
.32 caliber round 5 5
.44 Magnum 15 7.5
5.56mm 3 3
5mm 1 1
Shotgun shell 10 5
Scrap metal 20 --
The ammo press calculates the quantity produced as an integer based on the value of the materials entered into the press, divided by the value of the goods required:
$\frac{nv}{p}=\Z t$
n= Number of units placed into hopper
v= Value of individual unit of type that was placed into hopper
p= Value of individual unit of same type as intended product
t= Total number of intended product capable of being made
In a scenario in which multiple types of items are placed into bin at same time use the following formula: $\frac{n_1 v + ... + n_x v}{p}=\Z t$
Example: 2 types of items placed in bin would use:
$\frac{(n_1v) + (n_2v)}{p}=\Z t$
To calculate how much ammunition of a given type the Press will produce, use the table to the right. For every kind of ammunition or scrap metal in the hopper, multiply its quantity by its value in the table, and add the resulting values up to get the total value of metal in the hopper. Then divide by the value of the type of ammunition you are pressing, rounding down; this is the number of cartridges the press will make.
Example
Pressing scrap metal into 5.56mm rounds.
$\frac{10*20}{3}=66$
• If you have 10 units of scrap metal in the hopper, its value is 10 × 20 = 200 units.
$10 * 20 = 200$
• Pressing this into 5.56mm rounds gives 200 / 3 = 66.7 cartridges, which rounds down to 66. So pressing 10 scrap metal into 5.56mm rounds will make 66 cartridges.
$\frac{200}{3}=66$
If you wanted to make .44 rounds instead, you could turn the same 10 scrap metal into thirteen .44 rounds, meaning 15 5mm rounds would be transformed into one .44 round and vice versa, however 14 5mm rounds will provide 0 .44 rounds resulting in a complete loss of 14 cartridges.
$\frac {200}{15}=13$
Replenish Depleted Ammunition - See the above example. Use debris ammunition to accumulate desired ammunition type:
• First convert undesired ammunition into 5mm ammunition.
• Then, keeping the Press Value in mind, go ahead with conversion. In the above example, convert 10 scrap into 200 5mm rounds.
• Now, we know .44 costs 15. So, use a multiple of 15 to maximize production. i.e., Press 195 (15*13=195) 5mm rounds to get 13 .44 and have remaining 5 5mm rounds as spare. If you had used scrap to press directly, those five 5mm rounds would have been lost!
## NotesEdit
• Any material may be loaded into the bin, but only scrap metal and conventional ammunition will be converted into new ammunition, the rest will simply remain in the bin unused.
• The 30-3 Rule - To prevent loss of ammo through conversion one can follow this easy to remember rule of thumb. Only insert ammo in multiples of 30 and scrap metal in multiples of 3. Doing so won't waste any ammo as the press value's lowest common multiple across all ammo types is 30 (with scrap metal the press value's LCM is 60 which requires 3 scrap metal). Following this rule will allow you to change any ammo type into any other type without fear of wasting any potential rounds and is easier than remembering the entire value table.
## BugsEdit
pc xbox360 When creating any type of ammo, you will return to another page in the terminal and the action of pressing the last round will still be there granting you the possibility to make infinite amount of rounds. [verified]
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## The country that used to exist
What the hell is history, anyway?
History is just a bunch of stuff that happened. Mostly to people now dead. We owe these people nothing. They’re dead, after all. Sometimes we have some scraps of paper they scribbled on. Sometimes we don’t.
I was reading Tacitus’ Annals the other week (not for any good reason; I was just somewhere where there was a copy of Tacitus) and I was rather looking forward to the story of Caligula. (Who Tacitus quite confusingly calls “Caius Caesar”—as if there was some shortage of Romans by this name.) Suddenly, though, there was a gap. Tacitus did write about Caligula. But no one knows what he said. That book of the Annals is lost.
We have most of the Annals, though, including the opening, and here’s how it goes:
Rome at the beginning was ruled by kings. Freedom and the consulship were established by Lucius Brutus. Dictatorships were held for a temporary crisis. The power of the decemvirs did not last beyond two years, nor was the consular jurisdiction of the military tribunes of long duration. The despotisms of Cinna and Sulla were brief; the rule of Pompeius and of Crassus soon yielded before Caesar; the arms of Lepidus and Antonius before Augustus; who, when the world was wearied by civil strife, subjected it to empire under the title of “Prince.” But the successes and reverses of the old Roman people have been recorded by famous historians; and fine intellects were not wanting to describe the times of Augustus, till growing sycophancy scared them away. The histories of Tiberius, Caius, Claudius, and Nero, while they were in power, were falsified through terror, and after their death were written under the irritation of a recent hatred. Hence my purpose is to relate a few facts about Augustus—more particularly his last acts, then the reign of Tiberius, and all which follows, without either bitterness or partiality, from any motives to which I am far removed.
“Without either bitterness or partiality.” Sine ira et studio. I doubt these words were idly chosen, and we kids these days could do a lot worse than to imitate them. No one gives a rat’s ass about Tiberius now. But I see no shortage of either ira or studio.
Anyway, it’s not too hard to lose a book of Tacitus. Nor is it any great tragedy. They actually made a documentary about Caligula. So obviously we know something.
It’s slightly more difficult to lose a country. It happened, though, and it was not Dalmatia or Dacia, Rhaetia or Bithynia. Here’s a picture of it:
(Source.) The country was called Rhodesia, and it is no longer found on any map. Earth’s crust being relatively stable, the place where it was still exists. But you wouldn’t want to go there.
The story of Rhodesia is in a strange category we might call “living history.” Rhodesia is dead, but there are still a few Rhodesians kicking around, and some of them even have blogs. Such as this fellow—whose opinions I’m sure many UR readers might contest.
Because that innocuous little motto, sine ira et studio, acquires a slight edge when we come to the “living history” category. When Tacitus wrote about, say, Vespasian, he was writing of events of a roughly similar age. You can be sure there was plenty of both ira and studio.
History is not a list of facts. We all agree: in 1973, when I was born (nowhere near Rhodesia—I have no personal connection at all to the place) there was a country by that name. In 2007, as I write, there is not. These are facts. But they are not history. If Tacitus had merely compiled lists of facts, no one would have copied his books, and we would never even have heard of any Tacitus.
The genius of Tacitus, and his classical peers—Thucydides, for example, is even more readable—is that they actually tell us what happened. Not just the facts, but the story. And perhaps we have no choice but to believe them, but we do. Their voices are credible.
So what the hell happened to Rhodesia? Why was there a Rhodesia, and why isn’t there a Rhodesia anymore? What’s the story?
This is a rather large question. It involves the lives of millions of people. It’s not quite as large a question as “what the hell happened to the Roman Empire.” But at least no one involved in the fall of the Roman Empire is still alive and liable to send me angry email.
Before we try to answer this question, you might enjoy a few minutes immersing yourself in Rhodesian memorabilia, large piles of which can be found at rhodesia.nl. For example, this is quite a typical story. This and this (more cerebral), or this and this (more lurid) represent the Rhodesian Ministry of Information’s view of the conflict. Rhodesia after its unilateral declaration of independence in 1965 became quite isolated from the Western world, and as a result even the graphic design of Rhodesian publications can strike us as quite unusual—see, for example, this large PDF, which is the final issue of Cheetah, the magazine of the Rhodesian Light Infantry.
If you are willing to take a minute for this 8MB download, perhaps you will see why I find the story of Rhodesia so fascinating. It is simply an introduction to an alien world. It might as well have been printed on Alpha Centauri, except that it seems to be in English. Aside from the language barrier, I am quite sure I have more in common with Tacitus himself than with any of the odd bipeds who adorn the pages of Cheetah. Perhaps a modern German has rather the same reaction to documents from the Third Reich. (The last two pages of the PDF, which are quite artistic, may be the most unusual—don’t miss the selection of past Cheetah covers.)
Of course, this does nothing to answer the question. Perhaps first we should get the official story, which we can hear from our good old friend Samantha Power. As Power wrote, in an excellent 2003 article called How To Kill a Country:
Although Zimbabwe is as broken as any country on the planet, it offers a testament not to some inherent African inability to govern but to a minority rule as oppressive and inconsiderate of the welfare of citizens as its ignominious white predecessor.
I suspect the readers of Cheetah might take some issue with the word “ignominious.” But, of course, the winners write the history books. Or at least they get to teach at Harvard. And the losers are, well—losers.
In any case, Power is no fan of ZANU-PF, and she closes with this same point:
For all their differences, Mugabe and Ian Smith share a basic misconception about power: they both fail to realize that a government cannot survive indefinitely when it advances the political and economic desires of the few at the expense of the many. When I asked Smith whether he would stop leaving his front door open now that starving Zimbabweans are prowling the city, he replied, “I’m not going to change now.” The same, alas, is most likely true of Robert Mugabe.
See, this is how bad Robert Mugabe is: he is just as bad as Ian Smith.
(Who, almost unbelievably, is still alive—though shortly after this article was published, Smith’s home was stolen and he moved, like so many of his compatriots, to South Africa. Which presumably will survive long enough for him to at least die of old age in his own bed. You can buy Smith’s memoirs here. Be warned, he’s not a very good writer.)
Of course the idea that starving Rhodesians, whatever their skin tone, were ever “prowling” Salisbury, is rather surprising. And quite untrue. As the Rhodesian government put it in 1979 (bear in mind that the Rhodesian dollar was pegged 1:1 to the US dollar):
A United States State Department Agency report says that in the decade 1965–1975, Rhodesian economic growth outstripped that of almost all its neighbours; her Gross National Product rose by almost 80 per cent. and the per capita income by 26 per cent. The 1974 per capita income of $422 compared very favourably with that of$348 in Botswana, $148 in Zaire and$126 in Tanzania.
In the purely material sphere earnings have increased phenomenally: in 1958 the 652,000 workers in the industrial sector earned an average per capita income of $169, ranging from a high of$285 in the finance and insurance fields to a low of $104 for an agricultural worker; in 1965 the figures were 656,000 workers, average income$250, and by 1975, 982,000 workers had a mean per capita income of $692, with those employed in the educational, transport and communications and finance, insurance and real estate spheres, earning an average of$981, while the least paid employees—agricultural workers, domestic servants and miners and quarry workers—earned \$336 on the average. [...] Black private enterprise is growing, particularly in the fields of transport, distribution and agriculture: black bus-owners manage highly successful businesses which provide transport between the urban areas and the Tribal Trust Lands; black retailers own and run about 2,500 stores, varying in size and sophistication; and blacks own and manage over 8,000 commercial farms.
In the professional field, there are many successful black doctors, lawyers, teachers, accountants, etc. In the private sector of business and commerce, there are significant examples of appropriate responsibilities offered to, and successfully discharged by, qualified blacks and many large companies employ blacks in management or trainee-management positions.
The Civil Service employs increasing numbers of blacks in established posts, especially in the fields of education, health and social welfare.
I know it seems odd, and I do not have a link handy, but one of the mainstays of the early postwar case for decolonialization was actually that European government was artificially retarding the development of black Africa, confining native populations to traditional pre-industrial economic, social and political structures, and preventing them from moving forward to a modern, centrally-planned socialist economy. Hopefully if you are a regular UR reader, I have established some credibility and you are willing to take me on trust when I say that people actually did believe this. Otherwise I’ll have to dig up a link or something.
Of course this proposition is no longer uttered as such. But it no longer needs to be uttered, because so few remember Africa before decolonialization, and those who do don’t exactly teach at Harvard. So it has passed into history, and fossilized in a way. It can be assumed and included in arguments without any actual thought.
In fact, what Power really means is that a government “cannot survive indefinitely” if it defies the wishes of the international community. I.e., of, well—Power. Here we have the ultimate cause of Rhodesia’s destruction. I believe this is a clear and undisputed fact.
The proximate cause of Rhodesia’s demise is also clear. Rhodesia was throttled by carnations, stabbed in the back by South Africa, and finished off by Jimmy Carter.
The carnations were dropped into rifles and used to overthrow the Estado Novo of Portugal, which then abandoned Moçambique, Portuguese territory for the last 500 years, which in 1975 reverted to the international community—specifically its Cyrillic arm.
The Carnation Revolution left South Africa as the last ally of landlocked, blockaded Rhodesia. As everyone in the world can surely now agree, the National Party regime in South Africa were swine, and they thought they could appease the international community by selling out the Rhodesians—for example, cutting off their fuel supply. While this did them no good at all, it essentially forced the Rhodesian government to surrender.
As this article describes, Henry Kissinger approved—or at least led the Rhodesians to believe he’d approved—a plan under which Rhodesia would adopt a universal-suffrage voting system. Rhodesia, unlike South Africa, never had an apartheid system or race-specific elections; the entire cause of the dispute was that the Rhodesian electoral system, following Cecil Rhodes’ dictum of “equal rights for all civilized men,” had a property qualification for elections.
However, this first plan for majority rule was designed to elect the moderate black leader Abel Muzorewa. It was boycotted by the guerrilla leaders Mugabe and Nkomo, and (worst of all) it specifically preserved the white Rhodesian civil service. In other words, at least for the time being, it was essentially cosmetic. Carter, and his advisor on African affairs Andrew Young, refused to accept the results of Kissinger’s diplomacy and forced a new election, which was won (quite violently) by Mugabe.
Of course presumably Young would disagree with this perception. As one Rhodesian Ministry of Information pamphlet put it:
This weekend, in the worst atrocity committed against white civilians in the history of Rhodesia’s six-year war, terrorists of Robert Mugabe’s Zimbabwe National Liberation Army hacked and battered to death almost the entire white staff and their families at the Elim Pentacostal Mission in the Eastern border mountains.
Mr. Young is asked: “Does Mr. Mugabe strike you as a violent man?”
He replies: “Not at all, he’s a very gentle man. In fact, one of the ironies of the whole struggle is that I can’t imagine Joshua Nkomo, or Robert Mugabe, ever pulling the trigger on a gun to kill anyone. I doubt that they ever have… The violent people are Smith’s people and hopefully they won’t be around for the new Zimbabwe.”
This weekend, when local and international journalists arrived at the scene of the massacre 15 km from Umtali and less than 7 km from the Mozambique border, the mutilated and blood-stained bodies of three men, four women and five children—including a three-week-old baby—were lying as they had been found that morning.
Mr. Young is asked how he gets on with Mr. Mugabe.
He replies: “I find that I am fascinated by his intelligence, by his dedication. The only thing that frustrates me about Robert Mugabe is that he is so damned incorruptible…. The problem is he was educated by the Jesuits, and when you get the combination of a Jesuit and a Marxist kind of philosophy merging in one person, you’ve got a hell of a guy to deal with.” So we have the ultimate cause and the proximate cause. But do we really know the story?
All these facts are easy to assemble into the Rhodesian point of view. From this perspective, what happened is that Rhodesia, a white British nation which extended the benefits of peace, law and economic development to the Shona and Matabele nations which shared its idyllic little corner of the Dark Continent, was smashed by International Pinko Communism, which destroyed the country by handing it over to the vicious terrorist gangs its soldiers had been bravely resisting. The subsequent debacle of Mugabe’s Zimbabwe was as predictable as it was tragic, and if God-fearing American and British citizens had subscribed to Cheetah and read more of those Ministry of Information pamphlets, the whole thing could have been avoided.
Do I believe this? Well, in a certain sense, I believe it. I certainly think it’s a perspective that is worth appreciating, albeit in slightly less caricatured terms, as long as you also know the official story. I’d definitely pay good money to see Samantha Power debate “The Last of the Rhodesians” on the subject.
However, I don’t think either of these narratives is quite how Tacitus would put it. And nor is the combination—truly ira et studio—completely satisfying.
Perhaps if there is a unifying thread to the story, it is in the life of Garfield Todd, eulogized by the Grauniad here and described from a liberal Rhodesian perspective here.
Because there was such a thing as a liberal Rhodesian perspective. Rhodesia was a country, not a cartoon, and it had politics of its own. And these politics, I think, are relevant to more than Rhodesia (which after all is dead). They are a kind of microcosm of postwar “Anglosphere” politics as a whole, in which each faction had its strange Rhodesian reflection.
Todd, basically, was the Universalist to end all Universalists. He was not just crypto-Christian—he was Christian, a representative of the powerful missionary system. He devoted himself, with typical Universalist noblesse oblige, to the uplift and education of Africans.
And he was, in fact, Robert Mugabe’s boss. Mugabe got his start as a schoolteacher in the “Dadaya New Zealand Churches of Christ Mission School,” where Todd was the principal. Small wonder that in 1972 Todd, the former PM of Rhodesia, was confined under house arrest for having worked underground to support Mugabe’s guerrillas.
His daughter Judith Todd, who among other efforts underwent a hunger strike against Smith’s government, recently revealed that during the 1983 Matabeleland massacres—in which tens of thousands of Mugabe’s tribal opponents were murdered—she was raped by an officer of ZANU’s North Korean-trained security forces.
At the time her father was a senator of Zimbabwe, appointed by his former protege Mugabe, whose position he had done so much to obtain—only three years previously. And Garfield Todd remained in this role for two more years.
What would Tacitus make of this? Frankly, I’m not sure he could even process it. The whole affair is just not in his philosophy. Wittgenstein said that if a lion could talk, we would not understand him. Probably Garfield Todd, simply by virtue of being born in 1908, had some Latin. But I’m not sure Tacitus would find him any more intelligible than Wittgenstein’s lion.
The social relationship between Todd and Smith is particularly fascinating. Travelers to Rhodesia in its later years often remarked that it seemed like a piece of prewar Britain, even 19th-century Britain, that had somehow fossilized and failed to develop.
A figure like Ian Smith could never have achieved any political prominence in prewar Britain, because his origins were decidedly middle-class. Smith was an RAF officer in WWII, but his accent was no doubt anything but Old Etonian. Much as Todd was the ultimate missionary, Smith was the ultimate settler—and the battle between missionaries and settlers, which ended of course in the victory of the former, was the great conflict of colonialism.
Rhodesia can be seen as a sort of British alternate present. Perhaps something like a Rhodesian Britain could have existed in a Europe where WWII never happened, and middle-class right-wing movements were not considered “populist” and intrinsically dangerous. Today’s British National Party, for example, is very reminiscent of the Rhodesian Front in its general cultural associations. (Note that, like the Rhodesian sites I’ve linked to, the BNP also has the worst possible taste in Web design. This is very petty-bourgeois.)
Smith’s Rhodesian Front was a profoundly Optimate-Vaisya concoction: basically lower middle-class, with a light spattering of degenerate aristocrats. Given the BDH-OV conflict, that Smith’s Rhodesia would fight a war with the Universalist “world community,” or with the likes of the über-Universalist Garfield Todd, is utterly unsurprising.
Remember that this is not just a political conflict, but a religious war. Universalism and traditionalist Revelationism are the two leading clades of Christianity in the world today. If Revelationism had any chance of extirpating Universalism, it would surely seize it, so one can’t really be too shocked and offended that the project is proceeding in the opposite direction.
The United States is the only country in which any political force with even the slightest resemblance to the Rhodesian Front is tolerated as a legitimate party. For all of Europe and Britain, the analogy to National Socialism—which of course was another Vaisya party—is simply too close to home. Today’s Europeans simply cannot understand why Republicans, at least populist Republicans, are allowed to exist within the American political system. They look at it rather the way you’d look at someone who kept a pet leopard in his closet.
Fortunately for Universalists, the political wind is certainly in their sails. It is increasingly difficult for “populist” Republicans to survive in the American political system, simply because the press they get is so bad. The likes of Tom Tancredo and Duncan Hunter could never get themselves appointed as first-tier presidential candidates, whereas John McCain, Mitt Romney and Rudolph Giuliani—all of whom have proved themselves acceptable to the Universalist press—can. Even George W. Bush outdistanced the field on the basis of his “compassionate conservatism,” and (as some commenters here have observed)—neoconservatism itself contains a powerful dose of Universalism. Until a new outbreak of Middle American populism creates a new McCarthy or Reagan, the Republicans are headed back toward “post-partisan” territory, where like Willkie, Eisenhower or Rockefeller they will again serve as the Polygon’s faithful lapdogs, deeply sorry that they ever even thought of digging a hole in the sofa.
In other words, my prediction for the future: more progress.
There is only one slight problem with this picture. The problem is that, basically, Ian Smith and his cronies were right. And basically, Garfield Todd and his friends were wrong.
Universalist Christianity is profoundly adaptive when it comes to outcompeting its opponents. It’s very difficult for a Christian to argue against Universalism, because it is manifestly so pure and Christlike. Even in a time and place like Rhodesia in the ’60s—when Universalism was blatantly suicidal—it was by no means easy for Smith’s insurgent party to brush the Establishment aside. The Rhodesian Front could win the votes of Rhodesian farmers who didn’t want to be murdered in their beds. It had no chance of convincing the world.
However, Universalism also has enemies. Smith never really managed to hate the likes of Todd—he thought they were misguided. I was brought up a Universalist, so obviously I have the same feeling, only more so. It was not Smith, but Mugabe, who proved the real enemy of the Todds.
And being an enemy, he saw no shame at all in feigning Christian sentiments to manipulate his enemies. This is the permanent strategy of the militarist kleptocracies against the Universalist democracies. Negotiate with us, we are really moderate, we would not hurt a fly. True, we are armed to the teeth, but we only care about social justice, all we want is our rights. We, in fact, are budding Universalists, and if you call off your dogs, disband your commandos and accept our just and reasonable demands, we too will learn to turn the other cheek.
If you are any sort of a Christian, this song cannot fail to charm your heart a little. And all the more if you are a Universalist, which after all is a very fanatical sort of a Christian. A Universalist by definition believes that the world is progressing toward a future in which everyone will be a Universalist. When he sees evidence of this, he wants to believe it.
When I think about Universalist support for Third World “liberation movements”—hardly a single one of which turned out to be anything but evil and corrupt, and certainly none of which came even close to providing better government than its colonial predecessor—of course it is easy for me, rejecting Universalism, to sneer.
But this is just evidence of the same mentality. Imagine if, say, the Congo, or Saudi Arabia, or Samoa, was taken over tomorrow by a movement which claimed to be libertarian. Possibly even by the formalist corporate-capitalist state of my dreams. I have trouble imagining this, of course. But if it did happen, presumably my first inclination would be to support it. And presumably I’d have a bit of trouble realizing that these acolytes of my preferred future were, in fact, a gang of killers and thieves.
When we read history sine ira et studio, we cannot hate the Garfield Todds. We can describe them as stupid, hubristic, fanatical and destructive, of course, and so of course they were—at least in my opinion. But any effort to rid human society of stupidity, hubris, fanaticism or destructiveness is itself so hubristic that it simply redoubles the mistake.
If certain people had thought clearly and acted effectively, perhaps Rhodesia would have survived. They didn’t, and it didn’t. The only lesson we can draw from this is that it’s important to think clearly and act effectively. The past is dead; we owe it nothing.
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This post will be directed towards those newer in investing, with an explanation of drawdowns–in my opinion, a simple and highly important risk statistic.
Would you invest in this?
As it turns out, millions of people do, and did. That is the S&P 500, from 2000 through 2012, more colloquially referred to as “the stock market”. Plenty of people around the world invest in it, and for a risk to reward payoff that is very bad, in my opinion. This is an investment that, in ten years, lost half of its value–twice!
At its simplest, an investment–placing your money in an asset like a stock, a savings account, and so on, instead of spending it, has two things you need to look at.
First, what’s your reward? If you open up a bank CD, you might be fortunate to get 3%. If you invest it in the stock market, you might get 8% per year (on average) if you held it for 20 years. In other words, you stow away $100 on January 1st, and you might come back and find$108 in your account on December 31st. This is often called the compound annualized growth rate (CAGR)–meaning that if you have $100 one year, earn 8%, you have 108, and then earn 8% on that, and so on. The second thing to look at is the risk. What can you lose? The simplest answer to this is “the maximum drawdown”. If this sounds complicated, it simply means “the biggest loss”. So, if you had$100 one month, $120 next month, and$90 the month after that, your maximum drawdown (that is, your maximum loss) would be 1 – 90/120 = 25%.
When you put the reward and risk together, you can create a ratio, to see how your rewards and risks line up. This is called a Calmar ratio, and you get it by dividing your CAGR by your maximum drawdown. The Calmar Ratio is a ratio that I interpret as “for every dollar you lose in your investment’s worst performance, how many dollars can you make back in a year?” For my own investments, I prefer this number to be at least 1, and know of a strategy for which that number is above 2 since 2011, or higher than 3 if simulated back to 2008.
Most stocks don’t even have a Calmar ratio of 1, which means that on average, an investment makes more than it can possibly lose in a year. Even Amazon, the company whose stock made Jeff Bezos now the richest man in the world, only has a Calmar Ratio of less than 2/5, with a maximum loss of more than 90% in the dot-com crash. The S&P 500, again, “the stock market”, since 1993, has a Calmar Ratio of around 1/6. That is, the worst losses can take *years* to make back.
A lot of wealth advisers like to say that they recommend a large holding of stocks for young people. In my opinion, whether you’re young or old, losing half of everything hurts, and there are much better ways to make money than to simply buy and hold a collection of stocks.
****
For those with coding skills, one way to gauge just how good or bad an investment is, is this:
An investment has a history–that is, in January, it made 3%, in February, it lost 2%, in March, it made 5%, and so on. By shuffling that history around, so that say, January loses 2%, February makes 5%, and March makes 3%, you can create an alternate history of the investment. It will start and end in the same place, but the journey will be different. For investments that have existed for a few years, it is possible to create many different histories, and compare the Calmar ratio of the original investment to its shuffled “alternate histories”. Ideally, you want the investment to be ranked among the highest possible ways to have made the money it did.
To put it simply: would you rather fall one inch a thousand times, or fall a thousand inches once? Well, the first one is no different than jumping rope. The second one will kill you.
Here is some code I wrote in R (if you don’t code in R, don’t worry) to see just how the S&P 500 (the stock market) did compared to how it could have done.
require(downloader)
require(quantmod)
require(PerformanceAnalytics)
require(TTR)
require(Quandl)
require(data.table)
SPY <- Quandl("EOD/SPY", start_date="1990-01-01", type = "xts")
SPYrets <- na.omit(Return.calculate(SPY$Adj_Close)) spySims <- list() set.seed(123) for(i in 1:999) { simulatedSpy <- xts(sample(coredata(SPYrets), size = length(SPYrets), replace = FALSE), order.by=index(SPYrets)) colnames(simulatedSpy) <- paste("sampleSPY", i, sep="_") spySims[[i]] <- simulatedSpy } spySims <- do.call(cbind, spySims) spySims <- cbind(spySims, SPYrets) colnames(spySims)[1000] <- "Original SPY" dailyReturnCAGR <- function(rets) { return(prod(1+rets)^(252/length(rets))-1) } rets <- sapply(spySims, dailyReturnCAGR) drawdowns <- maxDrawdown(spySims) calmars <- rets/drawdowns ranks <- rank(calmars) plot(density(as.numeric(calmars)), main = 'Calmars of reshuffled SPY, realized reality in red') abline(v=as.numeric(calmars[1000]), col = 'red') This is the resulting plot: That red line is the actual performance of the S&P 500 compared to what could have been. And of the 1000 different simulations, only 91 did worse than what happened in reality. This means that the stock market isn’t a particularly good investment, and that you can do much better using tactical asset allocation strategies. **** One site I’m affiliated with, is AllocateSmartly. It is a cheap investment subscription service ($30 a month) that compiles a collection of asset allocation strategies that perform better than many wealth advisers. When you combine some of those strategies, the performance is better still. To put it into perspective, one model strategy I’ve come up with has this performance:
In this case, the compound annualized growth rate is nearly double that of the maximum loss. For those interested in something a bit more aggressive, this strategy ensemble uses some fairly conservative strategies in its approach.
****
In conclusion, when considering how to invest your money, keep in mind both the reward, and the risk. One very simple and important way to understand risk is how much an investment can possibly lose, from its highest, to its lowest value following that peak. When you combine the reward and the risk, you can get a ratio that tells you about how much you can stand to make for every dollar lost in an investment’s worst performance.
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# Consider a circle with area 81π . An arc on this circle is such that i
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Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 01:44
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If angle ABC is 40 degrees (see figure), and the area of the circle is $$81\pi$$, how long is arc AXC? (CB is a diameter of the circle)
A. $$\frac{\pi}{2}$$
B. $$\pi$$
C. $$2\pi$$
D. $$4\pi$$
E. $$8\pi$$
Kudos for a correct solution.
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 03:04
1
1
Concept: Length of an arc =[ø/360]*2*pi*r --------------- (1)
ø - is the central angle say COA, where O is the center of the circle!
Now central ∟COA = 2*∟CAB = 80 degrees ---------------- (2)
Area of circle = π*r*r = 81*π
Therefore r*r = 81
r = 9 -------------- (3)
From 1, 2 & 3
Length of arc AXC = (80/360)*2*π*9 = 4π
Hence D !!
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 03:57
Bunuel wrote:
If angle ABC is 40 degrees (see figure), and the area of the circle is $$81\pi$$, how long is arc AXC? (CB is a diameter of the circle)
A. $$\frac{\pi}{2}$$
B. $$\pi$$
C. $$2\pi$$
D. $$4\pi$$
E. $$8\pi$$
Kudos for a correct solution.
Attachment:
2015-07-02_1243.png
Property-1: Length of the Arc = (Angle/360)*2*π*r
So we need to calculate the angle made by Arc AXC at the centre of the Circle
Property-2: Angle made at the centre by any arc is always twice the angle made at the circumference by the same arc
i.e. Since Arc AXC makes an angle of 40 degrees at the circumference so
i.e. Arc AXC will make an angle of 2*40 = 80 degrees at the Centre of the circle
i.e. Length of the Arc AXC = (80/360)*2*π*9 [Area = πr^2 = 81π i.e. Radius = 8]
i.e. Length of the Arc AXC = 4π
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 04:14
1
IMO D is correct
Length of arc AXC= angle subtended at centre/360 * 2 *pi*radius
Angle at centre=2 * angle at circumference=80
Area=pi* r^2=81pi
Length of arc= 80/360 * 2* pi* 9= 4pi
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 04:32
1
Bunuel wrote:
If angle ABC is 40 degrees (see figure), and the area of the circle is $$81\pi$$, how long is arc AXC? (CB is a diameter of the circle)
A. $$\frac{\pi}{2}$$
B. $$\pi$$
C. $$2\pi$$
D. $$4\pi$$
E. $$8\pi$$
Kudos for a correct solution.
Attachment:
2015-07-02_1243.png
GIven area of the circle = 81*$$\pi$$
Thus, $$\pi$$$$r^2$$ = 81*$$\pi$$
r=9
NOw, as any angle subtended by an arc on the circumference is half the angle subtended by the same arc at the center of the circle, thus the angle subtended by arc AXC at the center of the circle is 2*40 deg.= 80 deg.
For the complete circumference, 2$$\pi$$ or 360 is subtended by 2$$\pi$$ r of the circumference. 80 deg will thus subtend = 80/360 * 2$$\pi$$ r = 4$$\pi$$. thus D is the answer.
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 06:13
So in other words the central angle is twice the size of an arc angle?
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 07:06
2
noTh1ng wrote:
So in other words the central angle is twice the size of an arc angle?
Yes, the angle made by the arc is half the angle made by the SAME arc at the center.
In the attached figure, the same arc is minor arc AB.
Attachments
Central Angle Theorem.jpg [ 15.13 KiB | Viewed 4126 times ]
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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02 Jul 2015, 09:41
1
Area =81 pi
R=9
Arch AXC forms the angle at the center =twice the angle formed at circumference= 2*40 =80 degree
Length of Arc AXC=Angle formed at center/360 * 2 pi * R
80/360*2.Pi.9=4.pi
Ans is D
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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03 Jul 2015, 09:10
does this mean ... whenever we are calculating length of arc we need to know angle subtended by it at centre ... not at circumference ..
Right ?
I calculated it as 2*pi
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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03 Jul 2015, 09:23
2
does this mean ... whenever we are calculating length of arc we need to know angle subtended by it at centre ... not at circumference ..
Right ?
I calculated it as 2*pi
Yes, the angle always has to be the one that the arc subtends at the center of the circle.
the direct formula for calculating the length of any arc of a circle = l = $$r*\theta$$ , where $$\theta$$ is the angle measure at the center of the circle in RADIANS and not in degrees and r is the radius of the circle.
$$2\pi$$ radians equal 360 degrees.
In the question above, 80 degrees will be equal to $$(80/360) * 2\pi$$ radians. Radius, r = 9.
Thus the length of the minor arc = $$r*\theta$$ = $$9*(80/360) * 2\pi$$ = $$4\pi$$
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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06 Jul 2015, 04:34
Bunuel wrote:
If angle ABC is 40 degrees (see figure), and the area of the circle is $$81\pi$$, how long is arc AXC? (CB is a diameter of the circle)
A. $$\frac{\pi}{2}$$
B. $$\pi$$
C. $$2\pi$$
D. $$4\pi$$
E. $$8\pi$$
Kudos for a correct solution.
Attachment:
2015-07-02_1243.png
MANHATTAN GMAT OFFICIAL SOLUTION:
If the area of the circle is $$81\pi$$, then the radius of the circle is 9 ($$A = \pi r^2$$). Therefore, the total circumference of the circle is $$18\pi$$ ($$C = 2\pi r$$). Angle ABC, an inscribed angle of 40°, corresponds to a central angle of 80°. Thus, arc AXC is equal to 80/360 = 2/9 of the total circumference:
$$\frac{2}{9}*18\pi = 4\pi$$.
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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25 Mar 2016, 23:01
If angle ABC is 40 degrees (see figure), and the area of the circle is 81π, how long is arc AXC? (CB is a diameter of the circle)
pi*r^2 =81 pi
r=9
Angle at the center of the arc can be found as shown in fig...
for 80degree arc
Total perimeter*80/360
2pi*r*80/360
4pi
Attachments
Circle.png [ 27.99 KiB | Viewed 3368 times ]
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Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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01 Apr 2016, 20:15
Chiragjordan wrote:
Consider a circle with area 81π.An arc on this circle is such that it makes a 40 degree inscribed angle with a point P on the circle. What is the length of this arc?
[A] 2π
[C] 8π
[D] 12π
[E] Cannot be determined
Hi,
[b]A GOOD Q.. +1kudos
Difficulty level should 600-700...
1) first, since we are talking of arc length, lets change the Area to Perimeter..
$$A=81π=π*r^2$$
so r=9 and P= 18π
2) Next would be to find the angle that this arc makes at center..
Since the ARC makes 40 degree angle at a point on the circumference, ARC will make 2*40 at the center..
3) Finally lets correlate the angle and perimeter to find length of ARC..
360 degree makes the perimeter or 18π..
so 1 degree will make $$\frac{18π}{360}$$, and
2*40 or 80 degree will make$$\frac{18π}{360} * 80 = 4π$$
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink]
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Re: Consider a circle with area 81π . An arc on this circle is such that i [#permalink] 09 Mar 2019, 14:55
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# Gibbs Sampler, Bivariate Normal, Subchain
In Example 7.1 of "Introducing Monte Carlo Methods with R", the authors write
$(X,Y)\sim N\Bigg((0,0),\begin{pmatrix}1 &\rho \\ \rho & 1\end{pmatrix}\Bigg)$
Then,
1. Given $x_t$,
2. $Y_{t+1}\mid x_t \sim N(\rho x_t, 1-\rho^2)$
3. $X_{t+1}\mid y_{t+1} \sim N(\rho y_{t+1}, 1-\rho^2)$
Finally, they say: The subchain $(X_t)_t$ satisfies $$X_{t+1} \mid X_t = x_t\sim N(\rho ^2 x_t,1-\rho^4)$$.
I'm at a loss for how to show this. I did
\begin{align} f(x_{t+1}\mid x_{t}) &= \int_\mathbb{R} f(x_{t+1}, y_{t+1} \mid x_t)dy_{t+1}\\ &=\frac{1}{2\pi(1-\rho^2)}\int_{\mathbb{R}}exp\Big[ \frac{-1}{2(1-\rho^2)}\big((y_{t+1} - \rho x_t)^2+ (x_{t+1}-\rho y_{t+1})^2\big) \Big] dy_{t+1} \end{align} where I have assumed $X_{t+1}$ is conditionally independent of $X_t$ given $Y_{t+1}$.
But I just don't know where to go from there. Any help would be appreciated.
Sorry if this is unclear in our book, but when \begin{align*} Y_{t+1}\mid X_t=x_t &\sim \mathrm{N}(\rho x_t, 1-\rho^2)\\ X_{t+1}\mid Y_{t-1}=y_{t+1},X_t=x_t &\sim \mathrm{N}(\rho y_{t+1}, 1-\rho^2) \end{align*} one gets $$X_{t+1}|X_t=x_t \sim \mathrm{N}(\rho \times \rho x_t,1-\rho^2+\rho^2(1-\rho^2))$$ as $$\mathbb{E}[X_{t+1}|X_t]=\underbrace{\mathbb{E}[\mathbb{E}[X_{t+1}|Y_{t+1}]|X_t]}_{\mathbb{E}[\rho Y_{t+1}|X_t]=\rho^2 X_t}$$ and $$\text{var}(X_{t+1}|X_t)=\underbrace{\mathbb{E}[\text{var}(X_{t+1}|Y_{t+1})|X_t)}_{\mathbb{E}[1-\rho^2|X_t]=1-\rho^2}+\underbrace{\text{var}(\mathbb{E}[X_{t+1}|Y_{t+1}]|X_t)}_{\rho^2\text{var}(Y_{t+1}|X_t)=\rho^2(1-\rho^2)}$$ If one wants to solve by the integral in the question, \begin{align*} &\exp\left\{\frac{-1}{2(1-\rho^2)}\big((y - \rho x_t)^2+ (x_{t+1}-\rho y)^2\big)\right\}\\ &\quad = \exp\left\{\frac{-1}{2(1-\rho^2)}\big(y^2(1+\rho^2)-2y(\rho x_t+\rho x_{t+1})+\rho^2 x_t^2+x_{t+1}^2\big)\right\}\\ &\quad = \exp\left\{\frac{-(1+\rho^2)}{2(1-\rho^2)}\big(y^2-2\rho (x_t+x_{t+1})y/(1+\rho^2)+\rho^2(x_t+x_{t+1})^2/(1+\rho^2)^2\big)\right\}\\ &\quad\times\exp\left\{\frac{-1}{2(1-\rho^2)}\big(\rho^2 x_t^2+x_{t+1}^2-\rho^2(x_t+x_{t+1})^2/(1+\rho^2)\big)\right\}\\ \end{align*} which makes the first exponential term a perfect normal density in $y$ and the second term can be simplified into \begin{align*} &\exp\left\{\frac{-1}{2(1-\rho^2)}\big(\rho^2 x_t^2+x_{t+1}^2-\rho^2(x_t+x_{t+1})^2/(1+\rho^2)\big)\right\}\\ &\quad=\exp\left\{\frac{-1}{2(1-\rho^2)(1+\rho^2)}\big( (1+\rho^2)[\rho^2 x_t^2+x_{t+1}^2]-\rho^2(x_t+x_{t+1})^2\big)\right\}\\ &\quad=\exp\left\{\frac{-1}{2(1-\rho^4}\big( x_{t+1}^2[1+\rho^2-\rho^2]-2x_tx_{t+1} \rho^2+x_t^2\rho^2[1+\rho^2-\rho^2]\big)\right\}\\&\quad=\exp\left\{\frac{-1}{2(1-\rho^4)}\big( x_{t+1}-\rho^2 x_t\big)^2\right\}\\\end{align*} which leads to the same result (!)
• Indeed, I also used the fact that $(X_{t+1},Y_{t+1})$ given $X_t$ being normal implies that $X_{t+1}$ given $X_t$ is also normal. – Xi'an Mar 17 '17 at 6:58
Perhaps this view helps: \begin{align} Y_{t + 1} &= \rho x_t + \sqrt{1 - \rho^2} N_t \\ X_{t + 1} &= \rho Y_{t + 1} + \sqrt{1 - \rho^2} N_{t + 1} \\ &= \rho^2 x_t + \rho \sqrt{1 - \rho^2} N_t + \sqrt{1 - \rho^2} N_{t + 1}, \end{align} where $N_t$ and $N_{t + 1}$ are standard normal distributed. It is then easy to see that $$\mathbb{E}[X_{t + 1} | x_t] = \rho^2 x_t + 0 + 0$$ and $$\mathbb{V}[X_{t + 1} \mid x_t] = \rho^2(1 - \rho^2) + (1 - \rho^2) = (1 + \rho^2)(1 - \rho^2) = 1 - \rho^4,$$ where we have used that the variance of the sum is the sum of the variances for uncorrelated variables and the third binomial formula.
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# Poincaré inequality for $W_0^{1,\infty}$
In the book A first course in Sobolev spaces by Leoni, the following Poincaré inequality for $W_0^{1,p}(\Omega)$ is stated:
Suppose $\Omega\subset \mathbb{R}^n$ has finite width (lies between two parallel hyperplanes) and $p\in [1,\infty)$. Then for all $u\in W^{1,p}_0(\Omega)$, $$\|u\|_{L^p} \leq C \|\nabla u\|_{L^p}$$
My question is: Does the Poincaré inequality above still hold for $p=\infty$? If yes, how to prove it? And if no, what is the counterexample?
Thank you.
• Is $W_0^{1,p}(\Omega)$ here the closure of $C_c^{\infty}(\Omega)$ under the Sobolev norm? – detnvvp Oct 16 '13 at 4:08
• yes that's right. – digiboy1 Oct 16 '13 at 4:16
• I think the answer would be false if $\Omega$ is not bounded. Let $\Omega =\mathbb R$ and $u$ be piecewise linear such that $u(0)=1$, $u(-r) = u(r) = 0$. Then $||u|| = 1$ but $||Du|| = 1/r$. – user99914 Oct 16 '13 at 4:30
• What if I assume that $\Omega$ is bounded? – digiboy1 Oct 16 '13 at 4:58
• The assumption that $\Omega$ has finite width implies that $\Omega$ is bounded in dimension $1$. – Tomás Oct 16 '13 at 12:15
By the fundamental theorem of calculus we have that, if $u\in C_c^\infty(\mathbb{R}^n)$ and $x_0$ is such that $u(x_0)=0$ then
$$u(x)=\int_0^1 \nabla u(tx+(1-t)x_0)\cdot (x-x_0)dt$$
from which we get
$$|u(x)|\leq \| \nabla u \|_\infty |x-x_0|$$
Now if $\Omega$ has finite width and $u\in C_c^\infty(\Omega)$, then for every $x\in \Omega$ there is an $x_0 \notin \Omega$ such that $|x-x_0| \leq D$ (where $D$ is the distance between the two parallel hyperplanes bounding $\Omega$). We conclude that
$$\| u\|_{\infty} \leq D \| \nabla u\|_{\infty}.$$
• How do you get around the fact that $C^{\infty}_{c}(\Omega)$ isn't dense in $W^{1,\infty}_{c}(\Omega)$? – fourierwho Apr 13 '18 at 2:36
• @fourierwho: We can treat functions in $W^{1,\infty}_c(\mathbb{R}^n)$ directly since these are Lipschitz and the argument above still holds. – Jose27 Apr 13 '18 at 2:52
• We need to assume the boundary of $U$ is sufficiently regular (e.g. $C^{1}$). Otherwise, $W^{1,\infty}(U)$ functions aren't necessarily Lipschitz. – fourierwho Apr 13 '18 at 3:00
• @fourierwho: We're working with $W^{1,\infty}_0$ which has no such problems (although maybe I'm using a nonstandard definition of this space?). – Jose27 Apr 13 '18 at 3:03
• That's a fair point. Can one prove that $W^{1,\infty}_{0}$ functions are Lipschitz regardless of the boundary regularity? I don't know the answer. I was only thinking of the fact that $W^{1,\infty}$ doesn't imply Lipschitz. – fourierwho Apr 13 '18 at 3:08
$W^{1,\infty}$ is indeed larger than the bounded Lipschitz class. For example, if the domain has an internal cusp, say $U$ is the planar domain given by $U=\{(x,y)\in\mathbb{R}^2\, :\, |y|>e^{-1/x^2}\text{ when }x>0\}$, consider the function $f$ given by $f(x,y)$ is the inner distance (with respect to $U$) between $(1/2, e^{-4})$ and $(x,y)$, we have that $f$ is (at least locally) in $W^{1,\infty}(U)$ but fails to be Lipschitz. Functions in $W^{1,\infty}$ are locally Lipschitz though.
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LHC Seminar
# Time-dependent CP violation in Bs0 decays at LHCb
## by Dr Manuel Tobias Schiller (University of Glasgow (GB))
Europe/Zurich
Video only (CERN)
### Video only
#### CERN
Description
CP violation was first observed more than five decades ago in particles called neutral K mesons, and has since been observed in other types of particle – including in neutral B0 mesons in 2001 by the Babar and Belle experiments and in D0 mesons by LHCb.
Recently the LHCb experiment observed time-dependent CP violation in decays of Bs0 mesons for the first time. In this seminar the recent results on B decays to two light hadrons (pions and kaons), and of the decays of Bs0->DsKpipi will be presented. Both these types of Bs0 decays are sensitive to the CKM angle gamma, that quantifies the amount of CP violation in the Standard Model.
Organized by
Michelangelo Mangano, Monica Pepe-Altarelli and Guillaume Unal.
Webcast
There is a live webcast for this event
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# Chapter 7 - Quiz (Sections 7.1-7.3) - Page 325: 8
(a) SAS is given in the question. (b) We should use the law of cosines to begin solving the triangle.
#### Work Step by Step
(a) The lengths of two sides are given and the angle between them is given. Therefore, SAS is given in the question. (b) If we tried to use the law of sines to begin solving the triangle, we would be missing two pieces of information in the equation. So this method would not work. We should use the law of cosines to begin solving the triangle. We should use the law of cosines to find the length of the side $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# A Running coupling of the Higgs potential
1. Sep 27, 2016
### spaghetti3451
Why is the coupling constant of the Higgs potential not a constant, but is a running constant?
In other words, $\lambda (h) \sim \lambda_{0} - \frac{y_{t}}{16\pi^{2}}\text{ln}\big(\frac{h^{2}}{m^{2}}\big) + \frac{\lambda_{0}^{2}}{16\pi^{2}}\text{ln}\big(\frac{h^{2}}{m^{2}}+\cdots$,
where $\lambda_{0}$ is the bare coupling and $y_{t}=\frac{m_{t}}{VEV}$, $m_{t}$ is the mass of the top quark and VEV is the HIggs vacuum expectation value?
2. Sep 27, 2016
### vanhees71
Look for "renormalization group equation".
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## limits velocity function- 800 meters
hello can u please show the work because I will have to be able to solve these on the test. thanks!!
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# All finite path connected CW complexes are homotopy equivalent to a CW complex with only one 0-cell.
Statement: if $$X$$ is a finite path connected CW-complex, then $$X$$ is homotopy equivalent to a CW-complex with only one $$0$$-cell.
Thoughts: we can use the theorem that for a contractable subspace $$A \subseteq X$$ such that $$(X, A)$$ has the homotopy extension property, the projection $$X \to X/A$$ is a homotopy equivalence. I want to apply this for each $$1$$-cell that does not have its ends glued together. This is ok because then it is a contractable, there are only finitely many, so eventually this procedure will stop and $$(X, A)$$ is a relative CW-complex, so it has HEP.
We will be left with only $$1$$-cells that have their ends glued together. I'd like to conclude from path connectedness that there is only one $$0$$-cell left, but could not manage this.
Consider the $$1$$-skeleton $$X^1$$ of $$X$$. Since $$X$$ is path connected, the cellular approximation theorem shows that $$X^1$$ is path connected. We have finitely many $$0$$-cells $$\{x_i\}$$, $$i = 1,\ldots,m$$. Let us show how to reduce the number of $$0$$-cells by $$1$$ (if $$m > 1$$). There exists a path from $$x_m$$ to $$x_1$$. Hence there must exist a closed $$1$$-cell $$e^1$$ attached to $$\{x_m\}$$ and some $$\{x_i\}$$ for $$i < m$$. Clearly $$e^1$$ is a contractible subcomplex of $$X$$ (with $$0$$-skeleton $$\{x_m,x_i\}$$) , thus the quotient map $$X \to X' = X/e^1$$ is a homotopy equivalence. The space $$X'$$ is a path connected CW-complex. The number of $$0$$-cells is $$m-1$$. Thus, proceeding inductively, we get the desired CW-complex with one $$0$$-cell.
In fact $$X^1$$ is a multigraph, i.e. a graph which is permitted to have multiple edges between two vertices (note that edges can be self-loops connecting a vertex to itself). It is well-known that each connected multigraph has a spanning tree which contains all vertices. Recall that a tree is a contractible subgraph. See for example Hatcher's "Algebraic Topology" Proposition 1A.1. So let $$T$$ be a spanning tree of $$X^1$$. It is a subcomplex of $$X$$, thus the quotient map $$X \to X/T$$ is a homotopy equivalence. The CW-complex $$X/T$$ has only one $$0$$-cell.
If $$X$$ is a CW-complex, then the inclusion $$i:X^1\to X$$ of the $$1$$-skeleton induces a bijection $$\pi_0(X^1)\to \pi_0(X)$$. Indeed, $$X$$ is obtained from $$X^1$$ by attaching cells of degree $$2$$ and higher, which do not affect $$\pi_0$$ (this follows from cellular approximation, but you can see it really concretely: if $$n\geq 2$$ then $$S^{n-1}$$ is connected so each new cell just gets attached to one of the connected components you already had).
In particular, if $$X$$ is connected, then its $$1$$-skeleton $$X^1$$ is connected. If all edges in $$X^1$$ start and end at the same vertex, then it is clear that each vertex together with all its edges forms a connected component of $$X^1$$, so if $$X^1$$ is connected it can only have one vertex.
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# A question regarding a particular space
I have a question regarding the notation of a space that I came across in a book. What is the definition of $C^0(X, Y)$, where $X$ and $Y$ and normed spaces?
-
What is the book? – Philip Brooker Jul 29 '12 at 11:21
Maybe it denotes the set of all the continuous mappings from $X$ to $Y$. – Paul Jul 29 '12 at 11:31
It's the standard notation (in French, at least!) for the set of continuous functions from $X$ to $Y$. It's often written simply as $\mathcal{C}(X,Y)$.
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## Cryptology ePrint Archive: Report 2014/823
On the Oblivious Transfer Capacity of Generalized Erasure Channels against Malicious Adversaries
Rafael Dowsley and Anderson C. A. Nascimento
Abstract: Noisy channels are a powerful resource for cryptography as they can be used to obtain information-theoretically secure key agreement, commitment and oblivious transfer protocols, among others. Oblivious transfer (OT) is a fundamental primitive since it is complete for secure multi-party computation, and the OT capacity characterizes how efficiently a channel can be used for obtaining string oblivious transfer. Ahlswede and Csiszár (\emph{ISIT'07}) presented upper and lower bounds on the OT capacity of generalized erasure channels (GEC) against passive adversaries. In the case of GEC with erasure probability at least 1/2, the upper and lower bounds match and therefore the OT capacity was determined. It was later proved by Pinto et al. (\emph{IEEE Trans. Inf. Theory 57(8)}) that in this case there is also a protocol against malicious adversaries achieving the same lower bound, and hence the OT capacity is identical for passive and malicious adversaries. In the case of GEC with erasure probability smaller than 1/2, the known lower bound against passive adversaries that was established by Ahlswede and Csiszár does not match their upper bound and it was unknown whether this OT rate could be achieved against malicious adversaries as well. In this work we show that there is a protocol against malicious adversaries achieving the same OT rate that was obtained against passive adversaries.
In order to obtain our results we introduce a novel use of interactive hashing that is suitable for dealing with the case of low erasure probability ($p^* <1/2$).
Category / Keywords: Oblivious transfer, generalized erasure channel, oblivious transfer capacity, malicious adversaries, information-theoretic security.
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#### Transmit Power Minimization for MIMO Systems of Exponential Average BER with Fixed Outage Probability
##### Dian-Wu Yue, Yichuang Sun
This paper is concerned with a wireless system operating in MIMO fading channels with channel state information being known at both transmitter and receiver. By spatiotemporal subchannel selection and power control, it aims to minimize the average transmit power (ATP) of the MIMO system while achieving an exponential type of average bit error rate (BER) for each data stream. Under the constraints of a given fixed individual outage probability (OP) and average BER for each subchannel, based on a traditional upper bound and a dynamic upper bound of Q function, two closed-form ATP expressions are derived, respectively, and they correspond to two different power allocation schemes. Numerical results are provided to validate the theoretical analysis, and show that the power allocation scheme with the dynamic upper bound can achieve more power savings than the one with the traditional upper bound.
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# Capital Costs
If you were offered $1,000 today or the same amount in ten years’ time, which would you take? Most of us would realise that, because of inflation, the value of the money would decrease over the period so that the$ 1,000 would not buy as much in ten years’ time as it does today. Seen from another perspective, if you take the money today, you can put it in the bank and earn interest, so that the actual amount increases over the period. This simple example illustrates the need for us to treat for capital and operating expenditure differently, and this forms the content of this unit.
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# 1962 AHSME Problems
## Problem 1
The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:
$\textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8}$
## Problem 2
The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}$ (Error compiling LaTeX. ! Missing } inserted.) is equal to:
$\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$
## Problem 3
The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$
## Problem 4
If $8^x = 32$, then x equals:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4}$
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# What is the full EM + matter lagrangian?
1. Mar 14, 2008
### pellman
This is a followup to the old thread What are the FULL classical electrodynamic equations? which never really provided a satisfactory answer.
I have decided to phrase it perhaps in a more straightforward manner. Given that we have the EM field, or the equivalent potential field, and charged matter. What is the lagrangian for this system?
I will be happy if given in terms of point charges or charge density. But it has to include both the action of the field on the charges and the fact that the sources of the field are those same charges. Writing down the lagrangian for an EM field with fixed sources or point particles influenced by a fixed field is easy and in every textbook.
Has anyone ever even seen what I am asking for here?
For an example of what I am talking about, here is the lagrangian density for spin 0 charges + EM field from quantum field theory. (This is just conceptual. Signs and constants might be wrong. H-bar and c are set to 1.)
$$(i\partial_{\mu}\phi^{\dag}-eA_{\mu})(i\partial^{\mu}\phi-eA^{\mu}) + m\phi^{\dag}\phi -\frac{1}{16}F^{\alpha\beta}F_{\alpha\beta}$$
I am looking for the classical analog to this lagrangian.
Last edited: Mar 14, 2008
2. Mar 14, 2008
### kdv
$$-m \int d\tau \sqrt{- g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu} + \int dx^4 j^\mu A_\mu - \frac{1}{4} \int d^4 x F^{\alpha \beta} F_{\alpha \beta}$$
with the current being
$$q \int d\tau ~ \delta^4(x-x(\tau)) \frac{dx^\mu (\tau)}{d\tau}$$
(I'm not sure of all the signs)
Is that what you were looking for?
3. Mar 14, 2008
### pellman
Yep. That looks like it. But let me ponder it a bit. Might be a couple days. Please check back then so that I can ask you half a dozen questions.
But, really. Thanks!
Last edited: Mar 14, 2008
4. Mar 14, 2008
### kdv
You're very welcome!
5. Mar 14, 2008
### kdv
Looking at the other thread I noticed that the Lagrangian had already been given! (by two people)
6. Mar 16, 2008
### pellman
kdv, I haven't much chance for this yet. But I can clarify a couple things first, please. What you have provided is the action not the lagrangian or lagrangian density itself, right?
So the interaction term written out fully looks like
$$q \int dx'^4 \int d\tau ~ \delta^4(x'-x(\tau)) \frac{dx^\mu (\tau)}{d\tau} A_\mu(x')$$
correct? I put in a prime on the x as variable of integration because its double use as particle position confused me, esp in the delta function. Am I ok so far?
I'm trying to get to the point where I could understand how to write the Euler-Lagrange equations from this action.
Last edited: Mar 16, 2008
7. Mar 16, 2008
### kdv
yes
This is correct.
Note that the Dirac delta can be used to do the four-dimensional integral which leaves only a one-dimensional integral over the path of the particle. This si what you see in some books: the interaction part of the action is an integral over the path of the particle, not a four-dimensional integral.
8. Mar 17, 2008
### pellman
I don't know if I would want to do that. It gives the correct value for the action of course, but wouldn't it affect being able to make a variation in the system and looking for the extremum of the action integral?
The main problem is rather general: how to deal with a system consisting of both discrete particles and continuous fields from a calculus of variations approach? My curiosity is aroused here. I'm sure this situation must arise in other contexts and has been dealt with. I'm going to dig around in my books.
The other approach is to use continuous charge distributions. Then the interaction is $$\int dx^4 j^\mu A_\mu$$, period. We would just have to replace the free matter term for point particles with one in terms of j.
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• # question_answer In a first-order reaction, the reacting substance has half-life period of 10 min. What fraction of the substance will be left after an hour the reaction has occurred? A) 1/6 of initial concentration B) 1/64 of initial concentration C) 1/12 of initial concentration D) 1/32 of initial concentration
${{t}_{1/2}}=10;$ Number of half-life $=\frac{60}{10}=6=$ half-life ${{C}_{t}}=\frac{{{C}_{0}}}{{{\left( 2 \right)}^{n}}}=\frac{{{C}_{0}}}{{{\left( 2 \right)}^{6}}}=\left( \frac{{{C}_{0}}}{64} \right)$
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How Many Sides Does A Circle Have? And the explanation isâ¦. We draw this as a circle and not a beyond-plank limit tiny sides shape because it's a bloody circle, stop being a cunt. Theyâre simple but so difficult. Category : Misc; Contributor : Riddle Admin; Rating: 3.25; Print Riddle; Can you solve this riddle? Or Right Click and Copy the Address Below to Share on Forums. one. Guess the answer for this riddle and select from the given options How many sides does a circle have? 1 decade ago. McConnell: Big stimulus not needed based on jobs report. We definitely wouldnât expect people to tackle this one before their morning coffee, as itâs easy to quickly declare the answer as one. Show Answer Hide Answer . All you have to do for the riddle is simple ask your friends, family, or post onto social media: âHow many sides does a circle have?â, and then wait for their response. ANOTHER RIDDLE: âYou enter a bedroom there are 34 peopleâ riddle explained. 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# Motor Rpm Formula
4 x 1540 x 0. When you're looking at the tachometer on a vehicle's dashboard, you are looking at the current rpm of. 9AKK105481 EN 05-2011 Application summary Speed 4-pole 2-pole Fixed below 3000/3600 rpm ++ Gearbox often needed Gearbox always needed 3000/3600 rpm Gearbox always needed + Gearbox not needed. The Kv rating of a brushless motor is the ratio of the motor's unloaded RPM to the peak (not RMS) voltage on the wires connected to the coils (the back EMF). Step 1: Select transmission from the drop down list or, if it's not in the list and you know the gear ratios of your transmission, enter them below. Discover LiftMaster replacement parts for garage door openers. The rotor rpm based on 320 gpm = 0. The minimum suggested RPM will be 600; the maximum, 900. Formula E motors: 17,500rpm, 26kg and instant torque The McLaren powertrain that will drive the Spark-Renault Formula E racing car weighs just 26kg, yet delivers near-instant torque and revs to 17,500rpm, states technical journal The Engineer. airflow rate of 10,000 CFM. Again, moving the same torque curve to the right another 1500 RPM (blue, 587 lb-ft torque peak at 6000 RPM) causes the power to peak at about 696 HP at 6500 RPM Using the black curves as an example, note that the engine produces 500 HP at both 4500 and 5400 RPM, which means the engine can do the same amount of work per unit time (power) at 4500. However it does not tell you anything about the motor, that we would care about in terms of performance. Adjust the cells to calculate the max motor RPM required. Los motores automotrices operan normalmente en un rango de entre 600 y 6. CL-0820-17-11T coreless motor for Latrax Alias. per hp At 1200 rpm, a motor develops a 4. They asked me if I could increase the HP and I know that I have to have the right formula. When people say an engine has "lots of low-end torque," what they mean is that the peak torque occurs at a fairly low rpm value, like 2,000 or 3,000 rpm. If done correctly, a small home will have a “quality” feel because you’ll invest thought, and sweat, and money into all the details that matter to YOU. The first step in calculating static thrust is determining the power transmitted by the motors to the propellers in terms of rpm. But if you still want to use RPM formula, then diameter value is the actual diameter of work piece. To change the output frequency to 50 Hz for the same generator configuration, the engine speed needs to be reduced to 3,000 rpm. Simply enter the values and click "Calculate". DC Motor Winding Circuit Connection Diagrams. If additional advice is needed please contact Douglas Manufacturing with your specific need or problem. Used 450 HP Horizontal Electric Motor (Westinghouse) Used Westinghouse 450 HP electric motor with following features: • 3557 RPM • 4160 Volts • 56 Amps • 3 Phase • 60 Hz • 1. 2L Supercharged offers 550 HP and torque that is not found on any other engine in its class. This calculator will provide the speed of the vehicle based on the transmission gear ratio, the engine RPM, the tire height, and the ring gear and pinion gear ratio in the rear end axle differential housing. keyway, tapped 7/16-20 in. Electric motors are not internal combustion motors. Even more revealing, at peak power RPM (table line 19) the Formula One engine MPS (table line 23) is 25. Using the example of a motor that operates at 55 hertz, the result would be 120 x 55 = 6600. Motor RPM = 1500 Motor Displacement x Motor RPM ÷ 231 = 3. RPM = Motor KV Racing * Battery Voltage. Drum roll please…here’s the forumula: Hz x 60 seconds / 6 (wave cycles per revolution) = RPM Simplified the formula for me is Hz x 10 = RPM. • The 3-phase set of currents, each of equal magnitude and with a phase. A small transmission in a mixer has a 12-tooth gear on the motor shaft driving a 72-tooth gear on the mixer blade. RPM = F x 120/ P. Gallons per minute of supply x 231 RPM = Cubic inches of motor displacement Gallons per minute x 231 Motor cubic inches = RPM RPM x cubic inches of motor displacement. For example, 2 or 4. Receptacles (qty. Flow Rate = Rate of flow into motor in Gallons Per Minute (GPM) Motor Displacement = Cubic Inches per Revolution. Of note that it doesn't mean peak torque was at max rpm, it was likely at half that, maybe less. 4 x 1540 x 0. Calculating the RPM resulting from a motor and speed reducer assembly requires only basic mathematical knowledge. Estimating engine power. Gallons per Minute. Assume the following: Rpm= 1755, ms= 8", ds= 14", gbr= 25. All powertrains carry over unchanged from the 2020 cars. Calculation Backup: the formula used for Critical Speed is: N c =76. This calculator will help you estimate the potential of an engine. Rotation Reversible ; Duty. The torque produced by an induction motor is a function of the shaft power and the shaft speed where the torque reduces with speed for constant power. There are slip for motor, so the RPM not exactly 1500, but 1440. Pipe and Fitting Friction Loss Estimates. The smaller pulley will always wear the fastest as it is turning more revolutions per minute and it has a smaller circumference for the belt to spread the wear over. To figure the speed the belt is moving you need to know the diameter of the contact wheel, multiply that times 3. , driver of the No. Propeller Selection. e gen rating / impedance. formulas may be helpful for quick analysis of limited applications. Define the terms in the general centrifugal fan formula and rearrange them to use the formula to solve for the higher air flow: CFM 1/CFM 2 = RPM 1/RPM 2 x (D1/D2)^3 (cubed). They asked me if I could increase the HP and I know that I have to have the right formula. Use for drive wheel on robots. What is the function of an electric motor in terms of electric. Note: Using a higher thrust motor will not give you a higher top speed. RPM is related to voltage, that motor is rated for 12vdc then it should rotate at 6100 rpm with 12v applied if it has been rated correctly. The formula to convert rpm to $\omega$ is: $\omega = rpm \cdot \frac{\pi }{{30}}$, in your case that's $\omega = 200 \cdot \frac{\pi }{{30}} = 20. 2094: 80 Revolutions Per Minute to Radians Per Second = 8. At a motor speed of 100 rps (6000 rpm), the required pulse frequency is more than 5 MHz. Each propeller blade is a rotating airfoil which produces lift and drag, and because of a (complex helical) trailing vortex system has an induced upwash and an induced downwash. RPM of motor = 60Hz x 120_ No. Compound Wound Motor – The DC motor which has both the parallel and series connection of the field winding is known as the compound wound rotor. Enter any 3 known values to calculate the 4th If you know any 3 values (Pulley sizes or RPM) and need to calculate the 4th, enter the 3 known values and hit Calculate to find the missing value. Una hélice cuyo paso sea igual al diámetro no tiene nada de especial ni es necesariamente la mejor. In order to use the equation, four pieces of information are required: 1. Or, change input variables (engine RPM, axle gear ratio and tire height) to determine vehicle speed in each forward gear. Using the motor from generator kit is the best way to do it. Pipe and Fitting Friction Loss Estimates. Formula One currently uses 1. Basically, if RPM is too low, the vehicle won’t move. Not to be used in place of real testing. There are four shoes. Each propeller blade is a rotating airfoil which produces lift and drag, and because of a (complex helical) trailing vortex system has an induced upwash and an induced downwash. Operating motor speed would be roughly equivalent to engine speed, with a range of 13,000 to 21,000 rpm (engine speed was limited to 18,000 rpm during the 2009 F1 season). For example, suppose that a motor rotates at 65 hertz, which means that it completes 65 revolutions per second. World Motor Sport Council. †For the formula, please see "Straight-line Speed and Motor Rotation Speed" on Page 11. These formulas calculate the brake horse power and total power. Thus divide the resistance(l-l) by 2 to get the phase resistance. These values can be inserted into the formula to calculate the speed: N = 120 x 60 / 4 N = 1800 (rpm). Answer: The Cell-plate (45 cell) x 0. Formula Plus; Resources. 1416 (SFM) 12 Revolutions Per Minute: The rotating velocity of the machine spindle. 43 Richard Petty Motorsports Chevrolet will also have to start at the rear of the field in the. where n is the rotation in revolutions-per-minute (rpm) (to be consistent with the Maxon datasheet below) and k n is called the speed constant, the inverse of which is often called the electrical constant, which depends on the motor design. The speed at which engagement begins is 3/4th of the running speed. The definition of the Kv value is how many RPM a motor outputs per every volt. speed of 2000 RPM, and a ratio of. Electric motor basics. Twice in the race, Wallace was hit with pit road speeding. Our selection of Airplane Engines is quality built and designed for maximum fun. In the case of these direct drive tools, such as handheld circular saws (not worm driven), table saws and radial arm saws, this will be the RPM which the blade is operating at. Speed 45 RPM; Voltage 24 DC; Amperage 0. And is this even accounting for resistive heat loss yet?. La velocidad será 10 HP * 5252 / 21 lb-pie = 2500 RPM. RPM = Motor KV Racing * Battery Voltage. Motor RPM = 1500 Motor Displacement x Motor RPM ÷ 231 = 3. I have an industrial customer, that bought a machine tool (grinder) built in 1941. For motors that reverse, jog, etc. However it does not tell you anything about the motor, that we would care about in terms of performance. Stall torque: 160 g*cm @12V Maximum speed: 29,000 rpm The new 29k motor is exactly the same as the MX10 except for two things: 1. Suppose you have a motor that turns 1200 RPM (revolutions per minute), and you need to turn something at 500 RPM. Find out what the motors recommended RPM range is. 60Hz is the most common frequency in the US. 0L: 200 225: 5300-5800 5500-6000: 1. Four pole designs are the most common pole configuration for AC induction motors and are typically. The formula above will result in a very accurate RPM vs. 55, so if RPM = Volts x Kv and Nm = Amps x 8. For your convenience, below is auto formula to get the value of PRM, just need to enter the kv rating. 4-7/8" wheel with o-ring on edge to friction drive equipment. 7721 Toll-Free: 800. Our Briggs World Formula engines are #1! Extraordinary Machining. Motor Top Speed RPM Calculator. These calculators are designed to give the approximate engine RPM (revolutions per minute) based on data entered. Suppose your motor vehicles speedometer not working and you need to know your vehicle speed for the given engine speed for a certain gear ratio. 4248: 4 Revolutions Per Minute to Radians Per Second = 0. “It is better to make torque at high rpm than at low rpm, because you can take advantage of gearing. For example a motor that in the template shows 1800 rpm @ 60 Hz, means that with an electrical signal of 60 Hz will turn at 1800 RPM. 0L: 200/225: 5000-6000: 1. To calculate synchronous speed of an induction motor, apply this formula: rpmsyn = 120 x f Nprpmsyn = synchronous speed (in rpm)f = supply frequency in (cycles/sec)Np = number of motor polesExample: What is the synchronous speed of a. So, for example, if you have a 4 pole motor on 60 Hz then RPM=60*(2/4)*60=1800 rpm. and r 2 = 7. 1 inches of water (25 Pa). 230V/single phase. TO OBTAIN HAVING FORMULA Belt Speed in Feet per Minute Diameter (D) of Pulley in Inches & Rev. Good torque and power up to 6200 rpm. Again, moving the same torque curve to the right another 1500 RPM (blue, 587 lb-ft torque peak at 6000 RPM) causes the power to peak at about 696 HP at 6500 RPM Using the black curves as an example, note that the engine produces 500 HP at both 4500 and 5400 RPM, which means the engine can do the same amount of work per unit time (power) at 4500. 686 hp (at 4000 rpm ). We are proud to be delivering a bespoke Hybrid System for the British Touring Car Championship, technology that promises to redefine the series for a new era. Once you have the RPM, you can calculate speed using the following formula: Multiply the RPM by the propeller pitch (e. In this unprecedented time, B2B content marketing strategist Megan Thudium presents the top three natural ways that companies can ‘get human’ by thinking like people connecting to other people; and why it starts with listening, continues with empathy and ends with being meaningful to what those individuals need to hear when they need to hear it…. • MotorEFF = Motor drive efficienc y usually 80-95% Tip Speed Ts = 3. Gear Ratio Speed Calculator By David Nguyen Calculate theoretical speed for a car in each of its gear ratios. The pedals and the Formula 1-style steering wheel can be adjusted as necessary. Using the above formula – Ferrari F40 Horsepower = (425 lb. Note: Hertz is a metric unit of frequency. If you're seeing this message, it means we're having trouble loading external resources on our website. A = number of parallel paths. The Kv rating of a brushless motor is the ratio of the motor's unloaded RPM to the peak (not RMS) voltage on the wires connected to the coils (the back EMF). Using the example of a motor that operates at 55 hertz, the result would be 120 x 55 = 6600. Automatic BEAM PRESS machines, ranging from 25 up to 250 ton maximum cutting force. 205 lb = 35. Bugatti Veyron is one of the cars with the highest torque figures. This is not the case!! if you require a quality motor that follows spec this one is not for you. Enter Cutter Diameter and Surface Speed to Calculate the R. 2618= slat speed in Feet Per Minute. 5 amp per hp At 230 volts, a single-phase motor draws 5 amp per hp. La velocidad será 10 HP * 5252 / 21 lb-pie = 2500 RPM. 63 CID, SAE 10 Ports, 2576 PSI, 470 RPM, 2584 Torque. Some time we required to check the RPM of the motor while creating projects. RPM can affect car performance in a variety of ways. To figure feet per minute just divide by 12. The Kv rating of a brushless motor is the ratio of the motor's unloaded RPM to the peak (not RMS) voltage on the wires connected to the coils (the back EMF). Z = total no. More than a few Westech customers walked over to the dyno room when. Chief BMPH Motor: 2-Bolt A, 23. That is what you can get out of the engine at wide open throttle at that rpm. It also supports Google Docs, DropBox, OneDrive, etc. • On large generators and motors, brushless exciters are used. Good torque and power up to 6200 rpm. Download WPS Office for Linux. WERA Motorcycle Roadracing, Inc is one of the oldest and largest national sanctioning bodies conducting motorcycle races at road courses across the United States. The following formula will work for any GM, Ford or Chrysler vehicle and is very useful when there is no established baseline from which to work from such as replacing a transmission in a vehicle that was purchased without a drive line. Conceived in 2012, the championship was intended by the FIA to serve as an R&D platform for the electric vehicle and promote interest in EVs and sustainability. Edelbrock is the most respected name in performance. ft x 4000)/5252 = 323. the maximum number of spectators they can accommodate. Enter any 3 known values to calculate the 4th If you know any 3 values (Pulley sizes or RPM) and need to calculate the 4th, enter the 3 known values and hit Calculate to find the missing value. Nice shopping on ' Coleman 1/6HP 230V 42 Frame 1050 RPM Furnace Blower Motor S1-1468-120P '. Vicprop, marine propeller specialists. Say you want to know what size pulley to use on that old planer to get 3450 RPM, you know the motor RPM is 1725 and the pulley on the motor is 6". 2 HOURS AGO. Variable Speed Drives) What They Are, How They Work Application Information Adjustable Speed Drives - Application Information DC Drives - Principles of Operation DC Drive Types DC. One hertz is equal to 60 rpm, since there are 60 seconds in a minute. The minimum torque (zero) occurs at maximum rpm, reached when the motor is not under a load, and is thus called free rpm. Motor RPM = 1500 Motor Displacement x Motor RPM ÷ 231 = 3. The sculptured racing seats create an unmistakable association with motor racing. An initial concern, particularly given the high motor speed, was the effect of iron losses in the stator core, which is typically an AC motor’s heaviest component. Example: The output speed of a motor with 25 GPM flow rate and. Motor racing-Binotto admits to questioning his role at suffering Ferrari. Take the result of the first part of the equation and divide by the number of poles in the motor. Practica realizando el cálculo de las RPM. If rate r is the same as speed s, r = s = d/t. = 2" (just for sake of argument) 2000 rpm, use a 2" driven pulley (1/1) 1000 rpm, use a 4" driven pulley (1/2). • The gauge of the wire is a function of your operating voltage and length. Seadoo Oem Black Handle Grip Kit 295500110. 6L: 150/175: 4500-5500: 1. electricity moves (oscillates) at 60-waves per second "Poles" are the positive and negative electromagnetic fields in the motor. That is why the shaft and frame of a 900 rpm motor are usually larger than those of an 1800 rpm motor of the same HP. rpm = piston speed in fpm x 6 / stroke in inches Formulas for brake horsepower. The propeller revolution rate is the engine rpm divided by the gear ratio. (RPM) S =. The Formula One World Championship raced at the Indianapolis Motor Speedway from 2000-2007 in the United States Grand Prix at Indianapolis. 55, so if RPM = Volts x Kv and Nm = Amps x 8. where the torque is measured in foot-pounds. Adding Rheostat in Stator Circuit. = 2" (just for sake of argument) 2000 rpm, use a 2" driven pulley (1/1) 1000 rpm, use a 4" driven pulley (1/2). In this method of speed control of three phase induction motor rheostat is added in the stator circuit due to this voltage gets dropped. per hp At 575 volts, a 3-phase motor draws 1 amp per hp At 460 volts, a 3-phase motor draws 1. Note that when an ac motor is loaded, the spinning magnetic field in the stator does not change speed. RPM returns in 2020 with a fresh new look. 5lt overhead camshaft unit using performance components developed by Cosworth Racing. how to Formula For Pulley Size And Motor Rpm for the 1 last update 2019/11/12 1. A way to figure the RPM without using the chart is to simply divide 6,000 (or 9,000) by the diameter of the piece in inches. These formulas calculate the brake horse power and total power. For example, if you know the motor speed is 1778 rpm and the fan speed is 944 rpm the multiplication factor would be: 944/1778 =. For example, if your small pulley is 80mm diameter, and spins at 1000 RPM, and you need to find the second pulley size to spin it at 400 RPM, Enter Pulley1 80, Pulley 1 RPM 1000, Pulley 2 RPM 400. The average velocity of gas particles is found using the root mean square velocity formula: μ rms = (3RT/M) ½ μ rms = root mean square velocity in m/sec R = ideal gas constant = 8. ) Transmission Gear ratio (i. 2618 x D x RPM Shaft Speed in Rev. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. Electric Motor Pulley Rpm Calculator Formula What Makes The Perfect Woodworking Plan? 21 Jul 2020 (☑ Step-By-Step Ideas) | Electric Motor Pulley Rpm Calculator Formula Get Access To Plans!!. It can be shown that the synchronous speed of a motor is determined by the following formula: where n s is the (synchronous) speed of the rotor (in rpm), f is the frequency of the AC supply (in Hz) and p is the number of magnetic poles per phase. Name of Person Title / Representative / Manager 123-456-7890 [email protected] With a dc motor, use a dc voltmeter to measure the. Series Motor – In this motor the field winding is connected in series with the armature of the motor. electricity moves (oscillates) at 60-waves per second "Poles" are the positive and negative electromagnetic fields in the motor. Output RPM formula = Input RPM times number of driving gear teeth divided by number of driven gear teeth. 0L: 200 225: 5300-5800 5500-6000: 1. 3/4 HP Motor 24 Volt 270 RPM SKU. RPM Output/RPM Input = Multiplication Factor On a belt-driven fan the two primary speeds required are the RPM of the Motor and the RPM of the fan. For a linear DC motor the "back emf constant" in units of volts/rad/sec is equivalent to the torque constant of the motor in units of N-m/Amp. Back in the 1990s, F1 was still running V-12s. The synchronous speed for an electric induction motor is determined by. horsepower = rpm x torque / 5252. Determining compound gear ratios (multiple stages). Typically for engines and motors, power and available torque are provided as curves on performance data sheets as a plot of BHP, Torque, and fuel consumption as a function of RPM. How can we calculate the rotational speed of an induction motor? A: Generally, the speed (N) of an induction motor is expressed by the following formula. Physics Formulas Physics Calculators: To link to this RPM to Linear Velocity Calculator page, copy the following code to your site: More Topics. The synchronous speed can be adjusted by changing the line frequency and number of poles of an AC motor. From the rating plate we can see that the min -1 (revolutions per minute) is 930 to 950 at 50 Hz. The Formula One World Championship raced at the Indianapolis Motor Speedway from 2000-2007 in the United States Grand Prix at Indianapolis. where the torque is measured in foot-pounds. f = frequency (Hz). Every revolution of the stepper motor is divided into a discrete number of steps, in many cases 200 steps, and the motor must be sent a separate pulse for each step. ) motors with three Holley Model 2300 2 bbl. f RPM = 60 × f Hz. History of Motor. This can be generalised as RPM = 120 x f / poles. bhp loss = elevation in feet / 1000 x 0. Maximum Rotation Speed Maximum Rotation Speed ≤ Rated Rotation Speed of a motor †Try to get as close to the motor's rated rotations as possible. Using the motor from generator kit is the best way to do it. Example: How much torque does a motor develop at 1,200 psi, 1500 rpm, with 10 gpm input? Formula: GPM x PSI x 36. This propeller rpm times the prop pitch determines the theoretical distance that the boat should have moved in one minute, which can be converted to a theoretical boat speed in miles per hour. torque = 5252 x horsepower / rpm. 50Hz is the most common frequency outside the US. η m = Mechanical shaft speed. Nice shopping on ' Coleman 1/6HP 230V 42 Frame 1050 RPM Furnace Blower Motor S1-1468-120P '. Extrapolation formula is (12/14. 77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information. Master Follower Motor Controls. The formula above will result in a very accurate RPM vs. That data consists of the ring gear and pinion gear ratio, the height of the tires (diameter from ground to top of the tire while mounted on the vehicle) and the transmission final gear ratio. Since induction motors slip the full-load RPM will be somewhat less than the calculated speed. The pedals and the Formula 1-style steering wheel can be adjusted as necessary. 61 Km, Tc=(Tf-32)*5/9. What is the mixer blade speed?. 85 x 60 = 4. In this case the more the merrier! dBA. Revolutions/Minute. The rated speed is the speed of the rotor. Darrell Wallace Jr. Some time we required to check the RPM of the motor while creating projects. 6:1 reduction gear). If you're seeing this message, it means we're having trouble loading external resources on our website. The general rule is - large to small gear means 'multiply' the velocity ratio by the rpm of the first gear. What is the mixer blade speed?. “It is better to make torque at high rpm than at low rpm, because you can take advantage of gearing. τ = (I * V * E *60) / (rpm * 2π) Connect the motor to the load. 77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information. 5 amp per hp At 230 volts, a single-phase motor draws 5 amp per hp. 5 m/sec, while the Cup MPS is a stunning 27. 76 seeds = 7. = 2" (just for sake of argument) 2000 rpm, use a 2" driven pulley (1/1) 1000 rpm, use a 4" driven pulley (1/2). They asked me if I could increase the HP and I know that I have to have the right formula. However, simple gears with only 5 teeth tend to run a bit rough, so your best bet is to make (or obtain) gears with 10 and 24 teeth. To find your vehicle’s RPM enter your Ring and Pinion Gear Ratio, Tire Height and Speed and press "Solve". How to convert hertz to revolutions per minute [Hz to RPM]: f RPM = 60 × f Hz. The formula for torque at any given rpm is: T = T s - (N T s ÷ N f) where T is the torque at the given rpm N, T s is the stall torque, and N f is the free rpm. F1 competed on a 13-turn, 2. This directly translates in to unloaded RPM for any specific application when the Battery Voltage is known. DC controls adjust speed by varying the voltage sent to the motor (this differs from AC motor controls which adjust the line frequency to the motor). the ratio formula is pretty easy, the ratio of the pulley diameters is the opposite of the ratio of the shaft rpm. The motor on there now is a 5hp, 480V 3 phase 1725 RPM the drive. 395 RPM = 17. I will use my 6/1 genset as an example. If the motor in the above example had four poles, the solution would be 6600 / 4. RPM returns in 2020 with a fresh new look. rpm = piston speed in fpm x 6 / stroke in inches Formulas for brake horsepower. Download WPS Office for Linux. To calculate synchronous speed of an induction motor, apply this formula: rpmsyn = 120 x f Nprpmsyn = synchronous speed (in rpm)f = supply frequency in (cycles/sec)Np = number of motor polesExample: What is the synchronous speed of a. Manual transmission recommended. 4 hertz to RPM = 240 RPM. To figure feet per minute just divide by 12. , RPM 5699 x 6 = 34194). BTCC Hybrid 2022 The future of motorsport requires innovative solutions, cutting-edge technology, and out-of-the-box thinking. Developed by the South African Caudwell Racing team, the engine is a 4-stroke, 3. 2L Supercharged offers 550 HP and torque that is not found on any other engine in its class. 0% Other Gases WET AIR = Same as dry air plus water vapor SPECIFIC DENSITY = 1_____ Specific Volume SPECIFIC DENSITY OF AIR = __1__ =. using the RPM, first calculate angular velocity using the formula (2*pi*RPM)/60. Revolutions/Minute. Variable Speed Drives) What They Are, How They Work Application Information Adjustable Speed Drives - Application Information DC Drives - Principles of Operation DC Drive Types DC. Los motores automotrices operan normalmente en un rango de entre 600 y 6. Feature image credit: Oriental Motor USA Corp. Operating motor speed would be roughly equivalent to engine speed, with a range of 13,000 to 21,000 rpm (engine speed was limited to 18,000 rpm during the 2009 F1 season). Estimate Electric Motor & Prop Combo Read the GUIDELINES to help you choose your Plane's Power System - Last updated: January, 15 - 2018. “It is better to make torque at high rpm than at low rpm, because you can take advantage of gearing. Gauge the existing propeller and determine what pitch it is. This is referred to as an armature-controlled motor. You divide the figure in RPM by 60, multiply by 2π and then multiply by the radius of the circle. T = (100 hp) 63025 / (1000 rpm) = 6303 (lb f in). 6991 HR/Personnel Fax: 765-751-9507. 82 x cutting speed)/diameter. Our competitors want you to believe that mixer sizing requires impossible calculations and a pinch of black magic. 00 ounces. Rotary speed = 220 rpm. Note: Using a higher thrust motor will not give you a higher top speed. The formula is simply expressed as: [(Balance Grade) x (Weight) / (Rated RPM)] = Max allowable imbalance per plane in units of force (oz-in). Calculates average speed to ride a fixed distance in a given time. (2) In SI units, the motor torque and back emf constants are equal, that is, ; therefore, we will use to represent both the motor torque constant and the back emf constant. That means most Camrys get a 2. ) Transmission Gear ratio (i. When the rotor magnetic field approaches the stator one, the torque is reduced. 0-liter unit capable of revving past 15,000 rpm. Formula One currently uses 1. Thanks to the big bore and short stroke, the 339-inch motor produced 607 hp at 7,900 rpm and 466 lb-ft of torque at 6,200 rpm. 5kW You also need to spec the right power cable • PVC flat-jacketed wire is the most common. 141592(pi) to get the circumference of the wheel. RPM of motor = 60Hz x 120_ No. For your convenience, below is auto formula to get the value of PRM, just need to enter the kv rating. CL-0820-17-11T coreless motor for Latrax Alias. 0 Service Factor • WI Enclosure • H586J Frame • HS Type • Weight 2,950 lbs • Dims 54"Lx66"Wx48"H. 5) where, N c is the critical speed,in revolutions per minute,. The term, “Service Factor” for an electric motor, is defined as: “a multiplier. 90 =$5,172. World Motor Sport Council. Formula: Speed = Distance ÷ Time. Here's how to calculate the proper RPM of your machines for a given material and diameter. formulas may be helpful for quick analysis of limited applications. using the RPM, first calculate angular velocity using the formula (2*pi*RPM)/60. Multiply the motor's frequency by the constant 120. DC-motors, direct current motors. At any point in the assembly’s rotation, the sum of all of the forces are […]. So the question would be, is there any formula that would give me the output HP at the output of the gear? I hope you understood my not very well put question. 6 litre four-stroke turbocharged 90 degree V6 double-overhead camshaft (DOHC) reciprocating engines. The speed of the motor is measured in rpm Rpm means revolutions of shaft per minute Ex:- if the motor is running at 2000 rpm that means it's shaft making 2000 revolutions per minute Note:-the speed can also be measured in rps revolutions per second. Example: The output speed of a motor with 25 GPM flow rate and. Average Speed Calculator. 732 x Eff x PF)/746. An equation expresses a mathematical relationship between the quantities present in that equation. motor itself. Four pole designs are the most common pole configuration for AC induction motors and are typically. 147 x 320 = 47 rpm. Among all the Mercedes activity at the 2017 Geneva Motor Show, Project One would rev its slightly altered 1. 76 seeds = 7. no need to enter anything in '6th' gear for a car with only 5 speeds). 3L: 200/225/ 225FHL/250: 5500-6000: 1. The RPM (Revolutions Per Minute) defines how fast a fan spins. The following formula will work for any GM, Ford or Chrysler vehicle and is very useful when there is no established baseline from which to work from such as replacing a transmission in a vehicle that was purchased without a drive line. Basically, if RPM is too low, the vehicle won’t move. The formula above will result in a very accurate RPM vs. There are four shoes. 45 RPM 24 VDC INLINE GEARMOTOR New, Molon model CHM-2435-1. The compound wound motor is further categorised into short-shunt and long-shunt motor. Adjust the cells to calculate the max motor RPM required. See our Technical Resource Center. Using the above formula of: Gas consumption per hour= horsepower/15, how much gasoline would an engine burn with the following horse-power and given time of operation? a 40 ph for 4 hours? A 25 hp for 3 hours? A 90 hp for 6 hours? A 6 hp for 9 hours? Write the formula for your family's outboard motor gas consumption. Because AC motors face loads—such as centrifugal pumps, fans and blowers—their torque requirement vary with the square or cube of the speed. The factor WK 2 (inertia) is the weight of a rotating object multiplied by the square of the radius of gyration K. Enter the Motor RPM: OR Enter the Tire Diameter (inches) OR Enter the Speed (MPH) OR Enter the Differential Ratio: OR Home | About. How you can do that , let’s see. 3600 RPM Output voltage (volts) 120/240 Phase 1 Product Height 8 in. Brake or shaft horsepower is…. So the gear ratio is S / (R+S) Constraints on number of teeth and planets If you want the planetary gears to be evenly spaced, and all be engaging the next tooth at the same time, then both your sun and your ring gear needs to be evenly divisible by the number of planets. If your vehicle has been running roughly or losing power, there may be a lack of pressure in one or more cylinders. DC Motor. com has compiled empirical data used to calculate power [1] , and the formula used for their datasheet is given in Equation 1. The speed at which engagement begins is 3/4th of the running speed. 8 hertz to RPM = 480 RPM. Using this formula, you can see that a four-pole motor operating on the bench under no-load conditions runs at 1,800 rpm (7,200 ÷ 4 poles). Example: The output speed of a motor with 25 GPM flow rate and. Por ejemplo, encuentra las RPM si el motor de un automóvil de 10 HP puede aplicar un torque de 21 lb-pie a un eje que hace girar las ruedas del vehículo automotor. electricity moves (oscillates) at 60-waves per second "Poles" are the positive and negative electromagnetic fields in the motor. In this unprecedented time, B2B content marketing strategist Megan Thudium presents the top three natural ways that companies can ‘get human’ by thinking like people connecting to other people; and why it starts with listening, continues with empathy and ends with being meaningful to what those individuals need to hear when they need to hear it…. Every revolution of the stepper motor is divided into a discrete number of steps, in many cases 200 steps, and the motor must be sent a separate pulse for each step. Then the convertion will be: 60 Hz = 1800 RPM 1 Hz = 30 RPM And those values are the ones that an inverter or drive uses for control the motor speed. WPS Office suite is fully compatible with Microsoft Presentation, Word, Spreadsheets or PDF files. of Poles 1800 RPM Motor – slippage makes it about 1750 3600 RPM Motor – slippage makes it about 3450 DRY AIR = 78. Multiply this number (2) by the rpm (120). ) What horsepower is it producing when it operates in this manner? Using the formula above: Horsepower = Torque x RPM / 5252 = 40 ft-lb x 6,000 RPM / 5252. Check the circle next to the item you are solving for and enter the remain three items in the spaces provided. 3 hertz to RPM = 180 RPM. p = number of motor poles. By using our formula to determine the horsepower on Mars, we know that if the torque (measured in ft-lb) of our motor is 30, and the rpm on Mars is 900, the horsepower would be about 5. Many power tools are direct drive, meaning the blade mounts directly to the motor shaft. What is the mixer blade speed?. 35:1 ratio is used as stage 2 of the reduction, input 11T and 16T for stage 1, and 1T and 2. 5, while 3600 RPM motors are typically balanced to G1. How to convert hertz to revolutions per minute [Hz to RPM]: f RPM = 60 × f Hz. Performing a conversion from RPM to speed in a linear direction involves two steps: first convert the RPM to a standard angular velocity, and then use the formula v = ωr to convert to linear velocity. 0 kg and diameter 76. Whereas a 2 hp 3600 RPM motor will offer 2. When people say an engine has "lots of low-end torque," what they mean is that the peak torque occurs at a fairly low rpm value, like 2,000 or 3,000 rpm. 0-liter unit capable of revving past 15,000 rpm. That data consists of the ring gear and pinion gear ratio, the height of the tires (diameter from ground to top of the tire while mounted on the vehicle) and the transmission final gear ratio. This figure can be used to determine the most efficient size and type of fan for a particular application. The basic equation used to calculate the proper RPM for machining is as follows: RPM = (cutting speed x 12)/(pi x diameter) Where the cutting speed is in feet/min and the diameter is in inches. (algebra required). Or, change input variables (engine RPM, axle gear ratio and tire height) to determine vehicle speed in each forward gear. 3145 (kg·m 2 /sec 2 )/K·mol T = absolute temperature in Kelvin M = mass of a mole of the gas in kilograms. 82 x cutting speed)/diameter. The Motor Output HP Available column and the Motor Torque column is what you need to be looking at. A small transmission in a mixer has a 12-tooth gear on the motor shaft driving a 72-tooth gear on the mixer blade. Starting Torque Formula for Induction Motor The starting torque of a motor is that torque which is produce by it at start. Calculation Backup: the formula used for Critical Speed is: N c =76. You will often see this expressed in a brochure or a review in a magazine as "320 HP @ 6500 rpm, 290 lb-ft torque @ 5000 rpm" (the figures for the 1999 Shelby Series 1). Because 1 horsepower is equal to 746 watts, the formula above will include a division by 746 to complete the conversion. If the motor in the above example had four poles, the solution would be 6600 / 4. RPM, which stands for revolutions per minute, is the amount of times the shaft of a DC motor completes a full spin cycle per minute. It withstands the intense mechanical action of continuous, high RPM operation to deliver superior viscosity protection. Motor are designed for 60 hz use have synchronous speeds of 3600,1800, 1200, 900, 720, 600, 514, and 450 rpm. It also supports Google Docs, DropBox, OneDrive, etc. Determine the RPM The best way to determine the RPM of a motor/propeller is to use a tachometer while running up the motor on the ground or on a workbench. Receptacles (qty. Seadoo Oem Black Handle Grip Kit 295500110. My guess of between 265 to 275 hp at 4150 rpm was pretty much spot on. At a motor speed of 100 rps (6000 rpm), the required pulse frequency is more than 5 MHz. If done correctly, a small home will have a “quality” feel because you’ll invest thought, and sweat, and money into all the details that matter to YOU. Roll your pointer to the left of the table to see typical values. Revolutions/Minute. 205 lb = 35. () Q p d D D p d RPM Q NS S 4 1 4 3 / / = = Generally, efficiency increases and fan size decreases as specific speed increases. There is no other monster truck like E-Revo. τ = (I * V * E *60) / (rpm * 2π) Connect the motor to the load. 2 HOURS AGO. This can be found on the nameplate of the motor. Drilling operations are those in which a cutting tool with sharp teeth, such as a twist drill, rotates and feeds into the workpiece axially, forming a hole with a diameter equal to that of the tool. Los motores automotrices operan normalmente en un rango de entre 600 y 6. Poles = number of poles of your motor. RPM = a/360 * fz * 60 RPM = Revolutions per minute. (diameter of wheel x pi) x rpm of motor = inches per. Pin = Power In of the Motor expressed in Watts Vin = Voltage Into the Motor Iin = Current Into the Motor Io = Noload Current of the Motor, derived from running a motor WOT without a prop at varying voltages. These formulas calculate the brake horse power and total power. INERTIA: Inertia is a measure of a body’s resistance to changes in velocity. Roller Chain Dimensions in Inches. Maximum Rotation Speed Maximum Rotation Speed ≤ Rated Rotation Speed of a motor †Try to get as close to the motor's rated rotations as possible. Seed spacing down the row then is: 125 cm / 17. My guess of between 265 to 275 hp at 4150 rpm was pretty much spot on. HP=Tn/63025. τ * rpm * 2π / 60 = I * V * E. The higher this number, the faster the fan will spin. The Season 1 motor provided by McLaren had 17500 rpm and 130 Nm of peak torque. calculate rcf (x g) from speed (rpm) Enter RPM: x g. The results will then be: Automatic transmission in Drive; Automatic transmission in Overdrive. Therefore # of poles = 120 x f / RPM. Slip is proportional to torque. Pump Input. The propeller revolution rate is the engine rpm divided by the gear ratio. • MotorEFF = Motor drive efficienc y usually 80-95% Tip Speed Ts = 3. If a car has less than 6 speeds, only enter numbers for the ratios that apply (i. Typically for engines and motors, power and available torque are provided as curves on performance data sheets as a plot of BHP, Torque, and fuel consumption as a function of RPM. Inventory Review and Quantity Breakpoint Model Tutorials. 2094: 80 Revolutions Per Minute to Radians Per Second = 8. The motor is assumed to be a typical 3-phase induction motor. Multiply this number (2) by the rpm (120). it S-Can motor, 29,000 rpm V12/4 (FC130) - 160 g*cm torque, supplied without wires and pinion. Back in the 1990s, F1 was still running V-12s. Series Motor – In this motor the field winding is connected in series with the armature of the motor. 50Hz is the most common frequency outside the US. The formula used to determine the speed of an AC motor uses the frequency of the current and the number of poles per phase and the slip in the case of an induction motor. To determine whether pressure is escaping from the engine, you need to check the compression in the cylinders with a compression gauge, which measures the amount of pressure that the piston exerts on the fuel/air mixture before the spark plug fires the mixture. Because the 2. 77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information. Nice shopping on ' Coleman 1/6HP 230V 42 Frame 1050 RPM Furnace Blower Motor S1-1468-120P '. The frequency of the power applied to an AC motor determines the motor speed, based on the equation: N = 120 x ƒ / p where N = speed (rpm). This propeller rpm times the prop pitch determines the theoretical distance that the boat should have moved in one minute, which can be converted to a theoretical boat speed in miles per hour. 3/4 HP Motor 24 Volt 270 RPM SKU. Find out what RPM the prop turns. The compound wound motor is further categorised into short-shunt and long-shunt motor. 5 hp at 900 rpm. That is what you can get out of the engine at wide open throttle at that rpm. The Motor Output HP Available column and the Motor Torque column is what you need to be looking at. Calculating the RPM resulting from a motor and speed reducer assembly requires only basic mathematical knowledge. Permanent magnet motor. There is no other monster truck like E-Revo. It is usually given in RPM (Revolutions per minute) on an electric motor nameplate. In the year 1821 British scientist Michael Faraday explained the conversion of electrical energy into mechanical energy by placing a current carrying conductor in a magnetic field which resulted in the rotation of the conductor due to the torque produced by the mutual action of electrical current and field. User Manuals; Safety Data Sheets - SDS 3/4 HP Motor 24 Volt 270 RPM Be the first to review this product. 1 rpm = 2π/60 rad/s. 00 ₹ 3,095. speed of 2000 RPM, and a ratio of. It is claimed to be able to run as high as 8,500 rpm in race trim. 1750 RPM ÷ 1000 RPM × 3. Loading Unsubscribe from theINVERSEside? Cancel Unsubscribe. Define the terms in the general centrifugal fan formula and rearrange them to use the formula to solve for the higher air flow: CFM 1/CFM 2 = RPM 1/RPM 2 x (D1/D2)^3 (cubed). Formula : HP = (E x I x Eff) / 746 Where , HP = Horsepower E = Voltage I = Current Eff = Efficiency DC Motor Horsepower calculation is made easier here using this electrical calculator. Even more revealing, at peak power RPM (table line 19) the Formula One engine MPS (table line 23) is 25. 75 cubic inches of displacement is: Speed (RPM) = (231 x 25) / 2. with r 1 = 1. You can use the equivalent formula d = rt which means. Output RPM formula = Input RPM times number of driving gear teeth divided by number of driven gear teeth. To find the output speed of a hydraulic motor, use the hydaulic motor speed formula: Output RPM = (231 x Flow Rate) / Motor Displacement. Motor racing-Formula One statistics for the Spanish Grand Prix Aug 12 (Reuters) - Some statistics for Sunday's Spanish Formula One Grand Prix at Barcelona's Circuit de Catalunya, the sixth race of. Countersink screw pilot bits. With a dc motor, use a dc voltmeter to measure the. (diameter of wheel x pi) x rpm of motor = inches per. 7867 Customer Service Fax: 765. Note: Hertz is a metric unit of frequency. RPM (n) = motor speed in revolutions per minute (RPM) ns = synchronous speed in revolutions per minute (RPM) Rotor Poles (P) = number of poles: Hertz (f) = frequency in cycles per second (CPS) T = torque in pound-feet: EFF = efficiency as a decimal: PF = power factor as a decimal: HP = horsepower. Determine your engine’s RPM based on the transmission gear ratio, tire height, MPH and the ratio of your ring gear and pinion gear. The compound wound motor is further categorised into short-shunt and long-shunt motor. A 115/230 volt electric motor that. The formula for torque at any given rpm is: T = T s - (N T s ÷ N f) where T is the torque at the given rpm N, T s is the stall torque, and N f is the free rpm. It embodies the highest level of creativity in design and engineering that unleashed a stunning array of new technology, new innovation, and unheard of speed that reset every benchmark for racing monster truck performance. Say you want to know what size pulley to use on that old planer to get 3450 RPM, you know the motor RPM is 1725 and the pulley on the motor is 6". τ * rpm * 2π / 60 = I * V * E. What is the mixer blade speed?. \$100 Promotion. Knurling is the precess to produce diamond shaped impression on the surface of a component to make its surface rough for easy grip. One example of a 2 hp 1800 RPM motor will offer 5. The RPM is restricted by frictional losses in the engine and the tendancy of the engine to explode if you rotate it too fast. Calculate your conversion factor to convert hertz to rpm. The rotor will constantly be turning to align its magnetic field with that of the stator field. 75 cubic inches of displacement is: Speed (RPM) = (231 x 25) / 2. 5 amp per hp At 230 volts, a single-phase motor draws 5 amp per hp. distance = speed x time. The calculator below can be used to obtain an idea of how a given ratio will effect engine RPM. Calculating RPM for a three phase induction motor is relatively simple… ACThree Phase Induction MotorRPMis determined by the formula: RPM = (120 * Frequency) / # of poles in the motor; Since the number of poles of athree phaseinduction motoris established when it is manufactured, the only way to change the speed of the motor is to change. In electrical engineering, the synchronous speed is the motor velocity where the speed of stator's rotating magnetic field and rotor are same to produce the required torque. 00 ounces. 262 x RPM (SFM) Cutting Speed: The velocity of the workpiece as it passes the insert measured in surface feet per minute. 75 cubic inch motor turn with 6 gpm input? GPM = 6 Motor Displacement = 0. Other articles where Brake horsepower is discussed: horsepower: …turbine, or motor is termed brake horsepower or shaft horsepower, depending on what kind of instrument is used to measure it. In the year 1821 British scientist Michael Faraday explained the conversion of electrical energy into mechanical energy by placing a current carrying conductor in a magnetic field which resulted in the rotation of the conductor due to the torque produced by the mutual action of electrical current and field. For a 50-Hz system, the formula would be: 50 x 60 x 2 = 6,000 no-load rpm ÷ number of poles. Thanks to the big bore and short stroke, the 339-inch motor produced 607 hp at 7,900 rpm and 466 lb-ft of torque at 6,200 rpm. 1 out of 5 stars 15 ₹ 3,095. 85 x 60 = 4. Formula One currently uses 1. Photo by: Richard Petty Motorsports “From a business perspective, the addition of DoorDash expands our resources, enabling us to build the best team and in turn grow our organization at a pace that will enhance our ability to win races," Brian Moffitt, CEO of RPM, said. Note: The rotational speed data for the tools relates to their use under load. Calculate Horsepower from Full Load Amps. Suppose your motor vehicles speedometer not working and you need to know your vehicle speed for the given engine speed for a certain gear ratio. Horsepower=Torque Output (inch pounds) x RPM/63025. DC controls adjust speed by varying the voltage sent to the motor (this differs from AC motor controls which adjust the line frequency to the motor). I hope this 3 phase motor current calculation formula (how to calculate 3 phase motor current) help you. Using this formula, you can see that a four-pole motor operating on the bench under no-load conditions runs at 1,800 rpm (7,200 ÷ 4 poles). USEFUL FORMULAS FOR COMPRESSED AIR COMPRESSOR RPM = motor pulley pitch diameter X motor rpm compressor pulley pitch diameter MOTOR PULLEY pitch diameter = compressor pulley pitch diameter X compressor rpm. 262 x RPM (SFM) Cutting Speed: The velocity of the workpiece as it passes the insert measured in surface feet per minute. 205 lb = 35. Perhaps the greatest sounding in the field was Ferrari's 412, equipped with a 3. motor itself. 5, while 3600 RPM motors are typically balanced to G1. Perhaps the greatest sounding in the field was Ferrari's 412, equipped with a 3. Calculate the approximate V-belt length using the following formula:. motor pulley rpm formula poleshow to motor pulley rpm formula poles for By building smaller, you may be able to afford upgrades you might not have considered otherwise. By Deanna Sclar. T = (100 hp) 63025 / (1000 rpm) = 6303 (lb f in). Los motores automotrices operan normalmente en un rango de entre 600 y 6. It can be used for measuring engine rpm and by making this bracket measuring starter motor rpm as stated in the repair manuals. then from this value of ang. 6-10 hp, IP Rating: IP55 at Rs 24000/piece in Ahmedabad, Gujarat. Knurling is the precess to produce diamond shaped impression on the surface of a component to make its surface rough for easy grip. 5 amp per hp At 230 volts, a single-phase motor draws 5 amp per hp. b) Calculate the current draw from the battery to perform this operation. Solution: Rotor speed = η m = 1200 r/min. How many revolutions per minute in a hertz: If f Hz = 1 then f RPM = 60 × 1 = 60 RPM. Fluid Motor Power. 91 CFM in. 3 hertz to RPM = 180 RPM. 5, while 3600 RPM motors are typically balanced to G1. To find the pitch diameter of the motor pulley use the following formula:. 77 ÷ RPM = torque in inch pounds This company assumes no liability for errors in data nor in safe and/or satisfactory operation of equipment designed from this information. The formula for calculating vehicle speed is: Car or automobile speed in consistent unit,.
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## Thurston, Selberg, and Random Polynomials, Part II.
What is the probability that the largest root of a polynomial is real?
Naturally enough, this depends on how one models a random polynomial. If we take polynomials of degree N which are constrained to have all of their roots to be of absolute value at most one (with respect to the normalized Lebesgue measure on $\mathbf{R}^N$), then, as mentioned last time, the probability that the largest root is real is either $1/N$ in odd degree $N$ and $1/(N-1)$ in even degree. A priori, this seems surprisingly small. However, the roots of such polynomials are accumulating on the unit circle, and it’s easier for complex roots to be near the unit circle than real roots. So let’s instead consider the Kac model of polynomials $f(x)$, where the coefficients are chosen to be independent normals with mean zero. If you ask for the probability that the root whose absolute value is closest to one is real, then I suspect that the answer will be approximately $1/N$. However, what about the largest root? The first observation is that the expected number of real roots is $2/\pi \log N$, so a the most naïve guess is that the probability that the largest root is real is approximately $(2/\pi N) \log N$. If you like, you can pause here and guess whether you think this is too high, too low, or about right.
A useful observation is that, instead of considering the largest root, we can consider the smallest root. This is because the map $a_k \rightarrow a_{N-k}$ is measure preserving and inverts the roots. On the other hand, the behavior of random Kac polynomials in large degree inside the unit circle starts to approximate the behavior of random power series
$f(x) = a_0 + a_1 x + a_2 x^2 + \ldots$
where the $a_i$ are all normally distributed with mean zero and standard deviation one. It’s easy to see that $f(x)$ will have radius of convergence $1$ with probability one. So we might instead consider what the probability is that the smallest root of a random power series is real. However, in this case, it is quite elementary to see that this probability $P_{\infty}$ is strictly between zero and one. Quite explicitly, consider the subspace of power series such that the following inequality holds:
$\displaystyle{|a_0| + \frac{1}{2} |a_1 - 2| + \frac{1}{2^2} |a_2| + \frac{1}{2^3} |a_3| + \ldots < 1}.$
This region has positive measure (easy exercise). On the other hand, for all such power series, one can apply Rouché's theorem for the contour $|2x| = 1$ to see that $f(x)$ and $2x$ have the same number of zeroes inside this disc, and hence $f(x)$ has exactly one root of absolute value less than $1/2$. By the reflection principle, this root is real. It follows that the probability that the smallest root of $f(x)$ is real is positive. Equally, one can consider the region:
$\displaystyle{|a_0 - 1| + \frac{1}{2} |a_1| + \frac{1}{2^2} |a_2 - 8| + \frac{1}{2^3} |a_3| + \ldots < 1},$
and by applying Rouché along $|2x| = 1$ and comparing with $1 + 8 x^2$, the corresponding $f(x)$ will have exactly two roots inside this ball, and from the inequalities above it follows that neither of them will be real, and hence $P_{\infty} < 1$.
The same argument shows that if $P_N$ is the probability that the smallest (or largest) root of a Kac polynomial is real, then there are uniform (independent of $N$) estimates $0 < a < P_N < b < 1$ for all $N$. Naturally enough, one should expect that $P_N$ converges to $P_{\infty}$. This is true, and the rough idea is to show that, with probability approaching one (as $N \rightarrow \infty$), one can apply Rouché’s theorem to deduce that the smallest root of $f(x)$ is real if and only if the smallest roots of its truncation $f_N(x)$ is real. The key idea here is that, for the truncation $f_N(x)$, most of the roots of $f_N(x)$ will be uniformly distributed along the unit circle, and so the contribution of the relevant factor $\prod |x - \alpha|$ to $f_N(x)$ will not be too small. Hence one can usually apply Rouché along the contour $|x| = \beta$ as long as there are no roots of $f_N(x)$ of absolute value too close to $\beta$.
The computations above also allow one to give effective gaps between $P_{\infty}$ and either zero or one (by estimating the measure of the corresponding regions as translates of $|a_i| \le 1/2^{i+1}$), although these estimates are not so sharp. Namely, the probability that the smallest root of a random power series is real is at least 0.256% and at most 99.999999999999917%. Some numerical data suggests, however, that the probability that the largest root of a random Kac polynomial (of large degree) will be real is approximately 52%. I have some undergraduates working with me this summer, and one of their projects will be to see if they can prove that the probability is really strictly larger than 50%, or at least to find a good an estimate as they can.
One may ask what happens for other ensembles of polynomials. One natural class to consider is the so-called binomial polynomials, where the $a_i$ are now normal with mean zero and variance $\displaystyle{\binom{n}{i}}$. Here the previous argument doesn't (a priori) work. On the other hand, as Boris Hanin (a Steve Zelditch student from Northwestern who is leaving for a postdoc at MIT next year) pointed out to me, it actually does: to fix it, one should scale all the roots of the relevant polynomials by $\sqrt{N}$, and then there really is a limit distribution as $N \rightarrow \infty$, given by power series
$f(x) = \displaystyle{\frac{a_0}{0!} + \frac{a_1 x}{\sqrt{1!}} + \frac{a_2 x^2}{\sqrt{2!}} + \ldots}$
where the normalized $a_i$ are normals with standard deviation one. Note that these power series have an infinite radius of convergence with probability one. The probability that the smallest root is real will once again be strictly between $0$ and $1$. In order to prove convergence of $P_N$ (by applying Rouché’s theorem), one needs to know that the relevant 2-point correlation functions behave reasonably enough; I’m hoping to get Boris to work out and write down the details here. Numerically, the limit probability in this case is somewhere around 62%.
Gap Probabilities:
I speculated last time on some conjectural relationship between the space of real monic polynomials $\Omega_N$ all of whose roots are at most one, and the space of random Kac polynomials of degree $N$ as $N$ goes to infinity. But now I wanted to point out a more direct an elementary relationship between ensembles of random real polynomials and our space $\Omega_N$. A gap probability is the probability that the eigenvalues/roots of some ensemble avoid some region of the corresponding parameter space. Let’s compute this for a very large gap. That is, let’s compute the probability that a random polynomial has all of its roots less than $T$ as $T \rightarrow 0$.
Let's consider the Kac model of random polynomials
$f(x) = a_0 x^N + a_1 x^{N-1} + \ldots + a_N$
where the $a_i$ are chosen independently from a normal distribution with mean zero and standard deviation one. Hence we are asking: what is the probability that all the roots of $f(x)$ have absolute value at most $T$? This is simply the integral
$\displaystyle{\left(\frac{1}{\sqrt{2 \pi}}\right)^{N+1} \int_{P \Omega_N(T)} e^{-|x|^2/2} dx}$
where $P \Omega_N(T)$ is the space of polynomials (not necessarily monic) all of whose roots are at most $T$. There is a map
$\Omega_N(T) \times [-\infty,\infty] \rightarrow P \Omega_N$
given by $(\lambda,P) \mapsto \lambda P$ with Jacobian $|\lambda|^{N+1}$. Hence we can write our quantity as an integral over $\Omega_N(T)$, which turns out to be
$\displaystyle{\left(\frac{1}{\sqrt{2 \pi}}\right)^{N+1} \int^{\infty}_{-\infty} \int_{ \Omega_N(T)} |\lambda|^{N+1} e^{-(\lambda^2 - a^2_1 \lambda^2 - \ldots - a^2_N \lambda^2)/2} dx}.$
We can now compute the integral over $\lambda$ directly, and then scaling $\Omega_N(T)$ to $\Omega_N$ in the usual way, we find that the probability that all the roots have absolute value $T$ is
$\displaystyle{ T^{N(N+1)/2} \left(\frac{2}{\pi}\right)^{(N+1)/2} \Gamma(N/2+1) \int_{\Omega_N} \frac{dV}{(1 + T^2 a^2_1 + T^4 a^2_2 + \ldots + T^{2N} a^2_{N})^{N/2+1}}}$
Now suppose that $T \rightarrow 0$. Then the integral over $\Omega_N$ converges to the volume of $\Omega_N$, and we obtain an exact asymptotic that all the roots are (highly) concentrated at zero. In fact, one can do this computation with any probability measure $\mu$ which decays sufficiently at infinity.
Curiously enough, we can also ask (in the setting of random polynomials subject to some reasonable measure $\mu$ for each $i$) what the probability is that a random polynomial has $R$ real roots contingent on all the roots of that polynomial being less than $T$. It turns out that, as $T \rightarrow 0$, the answer in this case is simply the ratio of the volume of $\Omega_{R,S}$ to $\Omega_N$ (with $R+2S=N$). This answer does not depend at all on $\mu$. The explanation for this is that, having subjected the polynomials to the constraint that all the roots have absolute value at most $T$ for small $T$, one is restricting to some tiny region where the measure is constant, and so it is converging to a scaled version of Lebesgue measure.
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# rgb2ntsc
Convert RGB color values to NTSC color space
## Syntax
`yiqmap = rgb2ntsc(rgbmap)YIQ = rgb2ntsc(RGB)`
## Description
`yiqmap = rgb2ntsc(rgbmap)` converts the `m`-by-3 RGB values in `rgbmap` to NTSC color space. `yiqmap` is an `m`-by-3 matrix that contains the NTSC luminance (Y) and chrominance (I and Q) color components as columns that are equivalent to the colors in the RGB colormap.
`YIQ = rgb2ntsc(RGB)` converts the truecolor image `RGB` to the equivalent NTSC image `YIQ`.
## Class Support
`RGB` can be of class `uint8`, `uint16`, `int16`, `single`, or `double`. `RGBMAP` can be `double`. The output is `double`.
collapse all
### Tips
In the NTSC color space, the luminance is the grayscale signal used to display pictures on monochrome (black and white) televisions. The other components carry the hue and saturation information.
`rgb2ntsc` defines the NTSC components using
$\left[\begin{array}{c}Y\\ I\\ Q\end{array}\right]=\left[\begin{array}{rrr}\hfill 0.299& \hfill 0.587& \hfill 0.114\\ \hfill 0.596& \hfill -0.274& \hfill -0.322\\ \hfill 0.211& \hfill -0.523& \hfill 0.312\end{array}\right]\left[\begin{array}{c}R\\ G\\ B\end{array}\right]$
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Question-and-Answer Resource for the Building Energy Modeling Community
Get started with the Help page
# Bug in OS SDK 2.9.0 for OS:ElectricLoadCenter:Storage:Converter?
I am trying to model a PV system with battery, where the battery can also be charged from the grid when the PV system is down. So I have an OpenStudio measure that creates, among many others, the following objects.
OS:ElectricLoadCenter:Storage:Converter,
{6f4c1fc9-72d9-4366-a914-b7d95575a797}, !- Handle
PV Charger, !- Name
{4e3c2f1a-8bc8-4717-ba4e-74c8240df8c0}, !- Availability Schedule Name
FunctionOfPower, !- Power Conversion Efficiency Method
, !- Simple Fixed Efficiency
8000, !- Design Maximum Continuous Input Power {W}
{12e72227-f689-4cef-8e33-a646d0d9d28b}, !- Efficiency Function of Power Curve Name
40, !- Ancillary Power Consumed In Standby {W}
, !- Zone Name
OS:Curve:Linear,
{12e72227-f689-4cef-8e33-a646d0d9d28b}, !- Handle
PV Charger Efficiency, !- Name
1, !- Coefficient1 Constant
0, !- Coefficient2 x
0, !- Minimum Value of x
1, !- Maximum Value of x
0, !- Minimum Curve Output
1, !- Maximum Curve Output
Dimensionless, !- Input Unit Type for X
Dimensionless; !- Output Unit Type
When this is run in OpenStudio, the following error appears in eplusout.err:
Program Version,EnergyPlus, Version 9.2.0-921312fa1d, YMD=2020.12.26 19:58,
** Severe ** <root>[ElectricLoadCenter:Storage:Converter][PV Charger][simple_fixed_efficiency] - "8000" - Expected number less than or equal to 1.000000
** Fatal ** Errors occurred on processing input file. Preceding condition(s) cause termination.
...Summary of Errors that led to program termination:
..... Reference severe error count=1
..... Last severe error=<root>[ElectricLoadCenter:Storage:Converter][PV Charger][simple_fixed_efficiency] - "8000" -
Expected number less than or equal to 1.000000
************* Warning: Node connection errors not checked - most system input has not been read (see previous warning).
************* Fatal error -- final processing. Program exited before simulations began. See previous error messages.
************* EnergyPlus Warmup Error Summary. During Warmup: 0 Warning; 0 Severe Errors.
************* EnergyPlus Sizing Error Summary. During Sizing: 0 Warning; 0 Severe Errors.
************* EnergyPlus Terminated--Fatal Error Detected. 0 Warning; 1 Severe Errors; Elapsed Time=00hr 00min 0.27sec
I do not understand why EnergyPlus looking for simple_fixed_efficiency when the OS:ElectricLoadCenter:Storage:Converter object is calling for FunctionOfPower? Plus, why is Design Maximum Continuous Input Power {W} taken for Simple Fixed Efficiency in the first place?
Sorry to be bothering y'all with an old version of SDK, but paying a subscription for SketchUp just to be able to move past 2.9.0 is a tall order for an energy modeler.
Thank you
edit retag close merge delete
What does the ElectricLoadCenter:Storage:Converter object look like after it's translated to EnergyPlus for simulation? That's a good way to determine whether the issue is with OpenStudio or EnergyPlus.
( 2020-12-28 10:25:45 -0500 )edit
Matthew you were correct, as usual. The IDF object has the efficiency and power entries switched! Here is a clip:
FunctionOfPower, !- Power Conversion Efficiency Method
8000, !- Simple Fixed Efficiency
, !- Design Maximum Continuous Input Power {W}
PV Charger Efficiency, !- Efficiency Function of Power Curve Name
Thanks.
( 2020-12-29 11:11:02 -0500 )edit
( 2020-12-29 11:12:28 -0500 )edit
Sort by » oldest newest most voted
### Cause
Looks like a bug due to a copy/paste typo (see below) where the code to ForwardTranslate the Design Maximum Continuous Input Power field is actually setting the Simple Fixed Efficiency field and using the same variable optD.
https://github.com/NREL/OpenStudio/bl...
// SimpleFixedEfficiency, optD
optD = modelObject.simpleFixedEfficiency();
if (optD) {
}
// designMaximumContinuousInputPower, optD
optD = modelObject.designMaximumContinuousInputPower();
if (optD) {
}
I opened an issue in the OpenStudio SDK repo here.
https://github.com/NREL/OpenStudio/is...
### Solution
As a workaround you could write a simple EnergyPlus measure to fix the IDF after it's been translated by OpenStudio, which would look something like this.
# get all objects
# OR
# get one object, e.g. if there's only one in the model
# set Simple Fixed Efficiency field to 0 rather than what's entered in the Design Maximum Continuous Input Power field
more
1
Thanks Matthew, that sounds like a brilliant workaround. I'll see if I can play with that a little. I am curious, though, as I am not familiar with the IDF methods so much:
In your last line, why is the design_maximum_continuous_input_power on the left of an equal sign? It seems to me this might be more appropriate for a getDouble rather than a setDouble?
( 2020-12-29 11:18:51 -0500 )edit
And if so, might I actually need the following?
... so that I correct both efficiency and power?
( 2020-12-29 11:20:42 -0500 )edit
@mattkoch I updated the solution code per your comment. I was originally going to create a variable for it for clarity. Your two lines of code would work too if you need the Simple Fixed Efficiency to be non-zero.
( 2020-12-29 11:22:05 -0500 )edit
@mattkoch I opened an issue for this (see link above) and mentioned you there, not sure if your GitHub username is also mattkoch.
( 2020-12-29 13:14:50 -0500 )edit
Thanks Matthew, makes me think now that adding the entire PV system (arrays, inverter, load center) plus battery and charger may be better done directly in the IDF file, rather than the OSM file, using EnergyPlus measures rather than OpenStudio measures, especially since adding life cycle cost via OSM also seems to be impossible?
( 2020-12-29 21:59:28 -0500 )edit
Following the advice from MatthewSteen, the following snippet in a simple EnergyPlus measure at least makes the model run to completion now. I have not had a chance to check the plausibility of the results. That's an entirely different effort.
pv_chargers = workspace.getObjectsByType('ElectricLoadCenter:Storage:Converter'.to_IddObjectType)
runner.registerFinalCondition("The model started with #{pv_chargers.size} PV Chargers.")
pv_charger = pv_chargers.first
runner.registerInfo("Initial PV Charger #{pv_charger}")
simple_fixed_efficiency = 1.0
design_maximum_continuous_input_power = pv_charger.getDouble(3)
pv_charger.setDouble(3, simple_fixed_efficiency)
pv_charger.setDouble(4, design_maximum_continuous_input_power.to_f)
runner.registerInfo("Final PV Charger #{pv_charger}")
runner.registerFinalCondition("The model finished with #{pv_chargers.size} PV Chargers.")
Note, EnergyPlus does not like zero efficiencies.
more
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Generation function of recurrence sequence
I'm trying to find generation function of recurrence sequence
ClearAll[c];
c[1] := 0;
c[2] := 1;
c[n_] := (2 + 2*(n - 2)*c[n - 2] + (n - 2) (n - 1) c[n - 1])/(n (n - 1));
Table[c[i], {i, 1, 15}]
FindGeneratingFunction[%, x]
but Mathematica just output
FindGeneratingFunction[{0, 1, 2/3, 5/6, 4/5, 37/45, 52/63, 349/420, 338/405,
873/14175, 14554/17325, 157567/187110, 466498/552825,
11994551/14189175, 41582906/49116375}, x]
How I can find GeneratingFunction?
-
Let y[x]=Sum[c[n]x^n,{n,0,Infinity}] be the generation function of c[n]. Then y[x] satisfys a 2rd order ODE (1-x)y''[x]-2xy'[x]-2/(1-x)==0. Try to find solution with DSolve and the result is
y[x] -> (1 + C[1] E^(-2x))/(1 - x) + C[2] - (2 C[1] ExpIntegralEi[2 - 2 x])/E^2
We have initial conditions y'[0]=c[1]=1, y''[0]=2c[2]=2*1=2. Thus,the undetermined coefficents(upper case "C") C[1] and C[2] will be obtained immediately by adding them to the DSolve function above.
However Mathematica returns
"DSolve::bvnul:For some branches of the general solution, the given boundary
conditions lead to an empty solution."
That error massage means C[1] or C[2] can't be determined. This may be due to the value of c[0], c[1] and c[2]. Turn to the recurrence relation, then we get c[2]=(2+0*c[0]+0*c[1])/2=1.
That is to say, c[1] and c[0] can be any value!! No wonder that we can't find a proper solution to that ODE.
By the way, you will get a result of c[n] (a long long expression involved n) by inputing following command
RSolve[{c[n] == (2 + 2*(n - 2)*c[n - 2] + (n - 2) (n - 1) c[n - 1])/(n (n - 1)),
c[1] == 0, c[2] == 1}, c[n], n]
-
Thank you! I found it already math.stackexchange.com/questions/539078/… – Филипп Цветков Oct 25 '13 at 11:07
I realise that it is not the answer for your question about generating function but it may help a little.
I will abuse the fact that you do not ask why. I have no experience with this area in Mathematica. But using documentation I was able to find a solution.
That's what is the best in Mathematica!
For your list, I've added an iterator so it will fit FindSequenceFunction
list = Table[{i, c[i]}, {i, 1, 15}]
f = FindSequenceFunction[list]
DifferenceRoot[
Function[{y, n},
{-2 - 2*n*y[n] - n*(1 + n)*y[1 + n] + (1 + n)*(2 + n)*y[2 + n] == 0,
y[1] == 0, y[2] == 1}]]
Array[f, 15] == Array[c, 15]
True
-
I didn't notice you hit 10k, congrats! – rcollyer Oct 25 '13 at 14:13
@rcollyer I don't know why I have not seen your comment earlier :/ Thanks! :) – Kuba Mar 12 '14 at 9:50
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## Calculating inflation rate macroeconomics
Inflation rate is the percentage increase in general level of prices over a period. It represents the rate at which the purchasing power of money has eroded over a period. Central banks and governments keep track of inflation rate and change monetary and fiscal policies accordingly. HOSP 2207 (Economics) Learning Centre Macroeconomics: GDP, GDP Deflator, CPI, & Inflation Macroeconomics is the big picture view of an economy. Microeconomics looks at the market for a specific good, like cell phones or bicycles, but macroeconomics deals with ALL goods and services produced in an economy and the AVERAGE PRICE LEVEL of those goods. Based on the data in Table 1, what’s the unemployment rate in 2016? In this example, the unemployment rate can be calculated as 7.7 million unemployed people divided by 159.1 million people in the labor force, which works out to an 4.8% rate of unemployment. Read on to walk through the steps of calculating this percentage.
Dr. Econ discusses the Consumer Price Index (CPI) and what it comprises. The maintenance of price stability—avoiding high inflation rates or deflation over Unfortunately, rigorous estimates of such an “op- timal inflation rate” have not been available in the economics literature. This article provides estimates of the Inflation calculator helps you determine the inflation rate basing on the change of prices. Check out 12 similar macroeconomics calculators Read on to learn what is inflation, how to calculate the inflation rate, and how it influences your The Inflation Calculator uses monthly consumer price index (CPI) data from 1914 to the Average Annual Rate of Inflation (%) / Decline in the Value of Money:. Download Save. Aplia Calculating inflation using a simple price index. Course: Introduction to Macroeconomics (ECO1102). You are currently viewing a preview . The inflation rate is the percentage increase in the average price level of goods and services over a period of time, usually one year. You can calculate the inflation
## For example, if a loan has a 12 percent interest rate and the inflation rate is 8 percent, then the real return on that loan is 4 percent. In calculating the real interest
3 Feb 2014 First, the Bureau of Labor Statistics operates under a veil of secrecy. The raw data used to calculate the CPI is not available to the public. When I Annual inflation rate in the US eased to 2.3% in February of 2020 from 2.5% in January which was the highest rate since October of 2018. Figures came slightly Economists calculate the rate of inflation by examining data from the consumer price index (CPI), provided by the Bureau of Labor Statistics (BLS). The CPI is a tool that economic observers use to track inflation. It represents the average change in prices over time for all components of an economy. An inflation rate can be computed for any price index using the general equation for percentage changes between two years, whether in the context of inflation or in any other calculation: $\displaystyle\frac{(\text{Level in new year}-\text{Level in previous year})}{\text{Level in previous year}}=\text{Percentage change}$ Plug your variables into the formula to calculate inflation. The formula for inflation is a ratio of the later CPI minus the earlier CPI over the earlier CPI. After you divide the difference between the 2 CPIs by the earlier CPI, multiply the result by 100 to find the rate of inflation. The Formula for Calculating Inflation. The formula for calculating the Inflation Rate using the Consumer Price Index (CPI) is relatively simple. Every month the Bureau of Labor Statistics (BLS) surveys thousands of prices all over the country and generates the CPI or (Consumer Price Index).
### Inflation rate calculator solving for inflation given consumer price index of this year and CPI of last year.
Intermediate Macroeconomics. Price Indexes, Inflation and Interest Rates. Winter 2000. Last updated: January 4, 2000. Note: These notes are preliminary and The rate of inflation can be calculated by taking the percentage rate of change in the price index for a given period of time. The formula used for calculating inflation Inflation Calculator. Inflation is usually Labor Statistics. This first calculator uses CPI data to show how things have been going. Annualized Inflation Rate: % 12 Jul 2018 Now if you're new to the world of economics and eCommerce, some of these You calculate inflation by looking at the percentage increase or Dr. Econ discusses the Consumer Price Index (CPI) and what it comprises. The maintenance of price stability—avoiding high inflation rates or deflation over Unfortunately, rigorous estimates of such an “op- timal inflation rate” have not been available in the economics literature. This article provides estimates of the Inflation calculator helps you determine the inflation rate basing on the change of prices. Check out 12 similar macroeconomics calculators Read on to learn what is inflation, how to calculate the inflation rate, and how it influences your
### 3 Feb 2014 First, the Bureau of Labor Statistics operates under a veil of secrecy. The raw data used to calculate the CPI is not available to the public. When I
Unfortunately, rigorous estimates of such an “op- timal inflation rate” have not been available in the economics literature. This article provides estimates of the Inflation calculator helps you determine the inflation rate basing on the change of prices. Check out 12 similar macroeconomics calculators Read on to learn what is inflation, how to calculate the inflation rate, and how it influences your The Inflation Calculator uses monthly consumer price index (CPI) data from 1914 to the Average Annual Rate of Inflation (%) / Decline in the Value of Money:. Download Save. Aplia Calculating inflation using a simple price index. Course: Introduction to Macroeconomics (ECO1102). You are currently viewing a preview . The inflation rate is the percentage increase in the average price level of goods and services over a period of time, usually one year. You can calculate the inflation
## The inflation rate is the percentage increase in the average price level of goods and services over a period of time, usually one year. You can calculate the inflation
Download Save. Aplia Calculating inflation using a simple price index. Course: Introduction to Macroeconomics (ECO1102). You are currently viewing a preview . The inflation rate is the percentage increase in the average price level of goods and services over a period of time, usually one year. You can calculate the inflation
Inflation is the rate at which the general level of prices for goods and services is rising and, consequently, the purchasing power of currency is falling. Central banks attempt to limit inflation
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# Launch your favorite apps with useful aliases from the Run dialog
Ever since Windows 95, Windows has had a cool feature known as app paths. It allows end users to create their own commands to run anything. Throughout its long history, this little known feature never gained much popularity, probably because it was initially designed for developers to prevent them from adding their apps to the system path variable. Even in Windows 8.1, this feature still exists without any changes, and is still secretly hidden from the eyes of the average Windows user. In this article, we will look at what app paths are and how you can create your own aliases to tremendously boost your working efficiency.
You can see app paths in action right after you first logon to Windows. Just press the Win + R keys together on your keyboard, type mplayer2.exe into the Run dialog or Start Menu search box and press Enter. Or type mplayer2 into the Start Menu/Start screen's search box and press Enter. You will see that Windows Media Player opens.
Press Win+R to show the Run dialog
But wait, isn't the executable file of Windows Media Player named wmplayer.exe? Moreover, wmplayer.exe is not located in C:\Windows\ or C:\Windows\System32, from any location in the system path variable from where it can easily be located by the OS. It's located in C:\Program Files (x86)\Windows Media Player\wmplayer.exe and yet running mplayer2 started it!
You might wonder how Windows is able to locate and run Media Player using the mplayer2.exe command.
It is possible due to app paths. The Windows Shell uses them to find and run the appropriate executable file associated with the current alias.
Technically every alias is just a subkey of the App Paths registry branch at:
HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\App Paths
App paths
Every subkey of the App Paths key is an alias which looks like someapp.exe. The full path to the target executable file is specified in the default value of this subkey. The value to the target EXE can also have arguments or switches.
Let's look at our example of mplayer2.exe. From the screenshot below you can see that it points to the C:\Program Files (x86)\Windows Media Player\wmplayer.exe, so it will be launched every time the user or some app requests the mplayer2 application.
App paths - mplayer2
Also, there is another alias, wmplayer.exe which points to the same file but it allows you to avoid using the full path. So you can see multiple different aliases, mplayer2.exe and wmplayer.exe, can both launch C:\Program Files (x86)\Windows Media Player\wmplayer.exe.
Aliases which are stored in the HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\App Paths key are system-wide aliases, which all user accounts on that PC can use. Additionally, starting with Windows 7, it is possible to have per-user aliases, which can be specified in the following key:
HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\App Paths
Per-user aliases are only accessible to the specific user who has them defined in his registry.
By default, Windows has several system-wide aliases and no per-user aliases.
You can use this app paths feature and create custom aliases to launch apps faster. By creating shorter aliases, you can use your Run dialog or the Start Menu's search box as your application launcher.
For example, you can create an alias named ie.exe for the following file:
C:\Program Files (x86)\Internet Explorer\iexplore.exe
Using this alias, you will be able to launch Internet Explorer quickly by typing ie into the Run dialog or the Start Menu search box.
Unfortunately, Windows does not provide any GUI to manage app paths. To create an alias, you must use the Registry Editor to create a subkey under 'App Paths' key and set the full path to the target executable file manually. It is not convenient to use the Registry Editor every time you want to manage app paths.
I decided to create a tool with a simple user interface to control per-user and system-wide aliases. My portable, Win+R Alias Manager allows you to create aliases for any application and use your Start Menu or the Run dialog as the application launcher.
Win+R Alias Manager
Using Win+R Alias Manager, you can create aliases for every app you have on your PC, whether it's an installed app or a portable app.
Win+R Alias Manager allows you to create, edit and delete aliases for both user and system applications (i.e. Firefox, Internet Explorer, Skype, Notepad etc).
The application allows you to create a per-user alias by default, but you can change this with a simple checkbox:
Win+R Alias Manager options
The alias and file path fields are required, the checkboxes are optional.
The 'System-wide alias (otherwise per-user)' checkbox will allow you to control which aliases are accessible to all users on the system. The last option will add the path containing the application executable file to its local %PATH% environment variable. Most applications do not require this option to be enabled, use it only if you are sure that it needs to be enabled for that particular software.
If you use Windows 7 or Windows 8, I recommend you to use per-user aliases. Windows XP and Windows Vista do not support per-user aliases so you are limited to using only system-wide aliases.
## 4 thoughts on “Launch your favorite apps with useful aliases from the Run dialog”
1. zydrius32
A small mistake in the sentence in the last screenshot after 2nd checkbox. It should be: ‘You should set this option if you know that…’
Like or Dislike: 0 0
2. Surfin
Nice tool ! Thanks…
I have a question about the Run dialog :
How to run a powershell script (locate in C:\Windows for example) from the run dialog without the need to type its extension (.ps1) ?
With a batch file, no need to type the extension (.bat or .cmd) !!!
Like or Dislike: 0 0
1. Sergey Tkachenko Post author
create a new alias with the following command
%SystemRoot%\system32\WindowsPowerShell\v1.0\powershell.exe path\to\your\script.ps1
use that alias to run it from the Run dialog.
Like or Dislike: 1 0
3. Surfin
Thank you, Nice workaround !
Regards…
Like or Dislike: 0 0
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# Differing (divergent) limits in the improper integral $\int_{-1}^{3}x^{-3}dx$
In a typical Calculus class, the improper integral $$\int_{-1}^{3}x^{-3}dx$$ is to be split up as $$\lim_{b \rightarrow 0^{-}}\int_{-1}^{b} x^{-3} dx + \lim_{b \rightarrow 0^{-}} \int_{b}^{3} x^{-3} dx,$$ and we arrive at the sum-of-limits $$\lim_{b \rightarrow 0^{-}}\big( -\frac{1}{2b^{2}} + \frac{1}{2} \big) + \lim_{b \rightarrow 0^{+}}\big( -\frac{1}{18} + \frac{1}{2b^{2}} \big).$$ Now, in each of these limits, it seems to me that the "$+$" and "$-$" don't matter on the $0$ since $b^{2}$ is a square. So we can write these limits as $\lim_{b \rightarrow 0}$ instead, and "pulling out the limit", we get $$\lim_{b \rightarrow 0}\big( -\frac{1}{2b^{2}} + \frac{1}{2} - \frac{1}{18} + \frac{1}{2b^{2}} \big) = \lim_{b \rightarrow 0} \frac{4}{9} = \frac{4}{9}.$$
My instructor says that the integrals have to be convergent independently in order for the whole improper integral to converge, and so this whole integral diverges. But the problem I have is that one limit is $\infty$ while the other is $-\infty$. So we get something like $\infty - \infty$, which I guess doesn't make much sense. And the fact that we have a $b^{2}$ in both limits makes me think that we don't have to worry about which side the limit is coming from.
Am I missing anything?
• No, you are correct. The integral does not converge. – Biggs Mar 10 '17 at 4:24
The fact that $\infty-\infty$ doesn't make sense is precisely why we say that the integral is divergent.
Note that you should use different variable names for both limits. You should write $$\lim_{a \rightarrow 0^{-}}\bigg(-\frac{1}{2a^{2}} + \frac{1}{2} \bigg) + \lim_{b \rightarrow 0^{+}}\bigg( -\frac{1}{18} + \frac{1}{2b^{2}} \bigg)$$ and now it is clear that you can't "pull out the limit".
What you are calculating has a name though. It is the Cauchy principal value.
The improper integral $\int_{-1}^3 x^{-3}\,dx$ fails to exist. Note that
$$\int_{-1}^{-\epsilon_1}x^{-3}\,dx=\frac12-\frac{1}{2\epsilon_1^2} \tag 1$$
and
$$\int_{\epsilon_2}^3x^{-3}\,dx=\frac{1}{2\epsilon_2^2}-\frac1{18}\tag 2$$
and neither the integral in $(1)$ not the one in $(2)$ converges as $\epsilon_1\to 0^+$ and $\epsilon_2\to 0^+$, respectively.
However, the Cauchy Principal Value integral,
$$\text{PV}\int_{-1}^3x^{-3}\,dx=\lim_{\epsilon\to 0^+}\left(\int_{-1}^{-\epsilon}x^{-3}\,dx+\int_{\epsilon}^3x^{-3}\,dx\right)=\frac12-\frac{1}{18}=\frac49$$
exists.
• Mike, please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – Mark Viola Mar 29 '17 at 19:57
• Mike, please let me know how I can improve this answer too. As always, I really want to give you the best answer I can. If the answer was not useful, I am happy to delete it. -Mark – Mark Viola Apr 12 '17 at 17:10
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# Simple Moving Average (MVA, SMA)
## Formula
The simple moving average is just an unweighted mean value (arithmetic average value) of the specified number of the most recent prices.
$\operatorname{MVA}(price, N)_i = \dfrac{{\sum_{j=i-N+1}^N} {price_{j}}}{N}$
### Optimization
In case the previous value of the moving average is known, you don't need to calculate the whole sum. You can just remove the oldest value from the previous result and add a new value:
$\operatorname{MVA}(price, N)_{i} = \operatorname{MVA}(price, N)_{i-1} - \dfrac{price_{i-N}} {N} + \frac{price_{i}} {N}$
## Usage
### In Analysis
#### Smoothing Data
The first usage of the simple moving average indicator is smoothing "noisy" data. Smoothing removes spikes in the data and reduces the affect of coincidental fluctuations of the prices.
#### Trend Detection
The second usage is detection of the market tendencies (trends). Depending on the parameter of the indicator, the slope of the moving average line can show the long (if N is big) or short (if N is small) term tendencies (trends). On the chart below the long-term moving average demonstrates that despite the fact that in the highlighted range the price goes up, the market is in down trend.
#### Classic Moving Average Strategy
The classic moving average strategy uses the trend detection feature of the moving average. The trader watches the short-term and long-term tendencies. When the short-term tendency changes and is strong enough to cross the long-term tendency, the trader enters the market following to the short-term tendency expecting that the long term will be changed soon.
The classic parameters recommended for stock market is 7, 14. For forex market it is usually recommended 5, 20. However, you shall always recognize all factors and choose such set of the parameters which properly reflects the current market situation.
At the chart above the long-term MVA(25) is red and the short-term MVA(5) is green. The areas where the indicators intersect are highlighted. In the first area the short-term tendency is turned to grow and crosses the long-term tendency over. The trader buys (exits short, enters long). In the second area the short-term tendency is turned to fall and crosses the long-term tendency under. The trader sell (exits long, enters short). In third area the short term crosses the long-term over again. The trader switches to long (buys) again.
#### Moving Average As Trailing Stop
If the short-term parameter is 1 (just a price itself), the long term (usually 150 or more) moving average may be used as a trailing stop.
## Limitations
The major disadvantage of the moving average is its inertness. It smooths the data but shifts extremes into the future. If you look at the Classic Moving Average Strategy carefully, you'll see that the signal is delayed for 4-6 bars comparing the actual tendency change. This makes usage of the Moving Average on flat market very dangerous. It works well while market is trendy, but produces a lot of the false signals when the market moves neither up, nor down. In that case it recognizes the small fluctuation as tendency changes.
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3.3.1.46. NXsample_component¶
Status:
base class, extends NXobject
Description:
One group like this per component can be recorded For a sample consisting of multiple components.
Symbols:
symbolic array lengths to be coordinated between various fields
n_Temp: number of temperatures
n_eField: number of values in applied electric field
n_mField: number of values in applied magnetic field
n_pField: number of values in applied pressure field
n_sField: number of values in applied stress field
Groups cited:
NXdata
Structure:
name: (optional) NX_CHAR
Descriptive name of sample component
chemical_formula: (optional) NX_CHAR
The chemical formula specified using CIF conventions. Abbreviated version of CIF standard:
• Only recognized element symbols may be used.
• Each element symbol is followed by a ‘count’ number. A count of ‘1’ may be omitted.
• A space or parenthesis must separate each cluster of (element symbol + count).
• Where a group of elements is enclosed in parentheses, the multiplier for the group must follow the closing parentheses. That is, all element and group multipliers are assumed to be printed as subscripted numbers.
• Unless the elements are ordered in a manner that corresponds to their chemical structure, the order of the elements within any group or moiety depends on whether or not carbon is present.
• If carbon is present, the order should be:
• C, then H, then the other elements in alphabetical order of their symbol.
• If carbon is not present, the elements are listed purely in alphabetic order of their symbol.
• This is the Hill system used by Chemical Abstracts.
unit_cell_abc[3]: (optional) NX_FLOAT {units=NX_LENGTH}
Crystallography unit cell parameters a, b, and c
unit_cell_alphabetagamma[3]: (optional) NX_FLOAT {units=NX_ANGLE}
Crystallography unit cell parameters alpha, beta, and gamma
unit_cell_volume: (optional) NX_FLOAT {units=NX_VOLUME}
Volume of the unit cell
sample_orientation[3]: (optional) NX_FLOAT {units=NX_ANGLE}
This will follow the Busing and Levy convention from Acta.Crysta v22, p457 (1967)
orientation_matrix[3, 3]: (optional) NX_FLOAT
Orientation matrix of single crystal sample component. This will follow the Busing and Levy convention from Acta.Crysta v22, p457 (1967)
mass: (optional) NX_FLOAT {units=NX_MASS}
Mass of sample component
density: (optional) NX_FLOAT {units=NX_MASS_DENSITY}
Density of sample component
relative_molecular_mass: (optional) NX_FLOAT {units=NX_MASS}
Relative Molecular Mass of sample component
description: (optional) NX_CHAR
Description of the sample component
volume_fraction: (optional) NX_FLOAT
Volume fraction of component
scattering_length_density: (optional) NX_FLOAT {units=NX_SCATTERING_LENGTH_DENSITY}
Scattering length density of component
unit_cell_class: (optional) NX_CHAR
In case it is all we know and we want to record/document it
Any of these values:
• triclinic
• monoclinic
• orthorhombic
• tetragonal
• rhombohedral
• hexagonal
• cubic
space_group: (optional) NX_CHAR
Crystallographic space group
point_group: (optional) NX_CHAR
Crystallographic point group, deprecated if space_group present
transmission: (optional) NXdata
As a function of Wavelength
NXDL Source:
https://github.com/nexusformat/definitions/blob/master/base_classes/NXsample_component.nxdl.xml
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# LVecBase3d¶
from panda3d.core import LVecBase3d
class LVecBase3d
This is the base class for all three-component vectors and points.
Inheritance diagram
__add__(other: LVecBase3d)LVecBase3d
__div__(scalar: float)LVecBase3d
__eq__(other: LVecBase3d)bool
__getattr__(attr_name: str)object
__getitem__(i: int, assign_val: float)None
__getitem__(i: int)float
__iadd__(other: LVecBase3d)LVecBase3d
__idiv__(scalar: float)LVecBase3d
__imul__(scalar: float)LVecBase3d
__init__()
__init__(copy: LVecBase2d, z: float)
__init__(param0: LVecBase3d)
__init__(fill_value: float)
__init__(x: float, y: float, z: float)
__ipow__(exponent: float)object
__isub__(other: LVecBase3d)LVecBase3d
__lt__(other: LVecBase3d)bool
__mul__(scalar: float)LVecBase3d
__ne__(other: LVecBase3d)bool
__pow__(exponent: float)LVecBase3d
__reduce__()object
__repr__()str
__setattr__(attr_name: str, assign: object)int
__sub__(other: LVecBase3d)LVecBase3d
__sub__(other: LVecBase3d)LVecBase3d
addHash(hash: int)int
Adds the vector into the running hash.
addHash(hash: int, threshold: float)int
Adds the vector into the running hash.
addToCell(i: int, value: float)None
These next functions add to an existing value. i.e. foo.set_x(foo.get_x() + value) These are useful to reduce overhead in scripting languages:
addX(value: float)None
addY(value: float)None
addZ(value: float)None
almostEqual(other: LVecBase3d)bool
Returns true if two vectors are memberwise equal within a default tolerance based on the numeric type.
almostEqual(other: LVecBase3d, threshold: float)bool
Returns true if two vectors are memberwise equal within a specified tolerance.
assign(copy: LVecBase3d)LVecBase3d
assign(fill_value: float)LVecBase3d
compareTo(other: LVecBase3d)int
This flavor of compareTo() uses a default threshold value based on the numeric type.
compareTo(other: LVecBase3d, threshold: float)int
Sorts vectors lexicographically, componentwise. Returns a number less than 0 if this vector sorts before the other one, greater than zero if it sorts after, 0 if they are equivalent (within the indicated tolerance).
componentwiseMult(other: LVecBase3d)None
cross(other: LVecBase3d)LVecBase3d
crossInto(other: LVecBase3d)None
dot(other: LVecBase3d)float
fill(fill_value: float)None
Sets each element of the vector to the indicated fill_value. This is particularly useful for initializing to zero.
fmax(other: LVecBase3d)LVecBase3d
fmin(other: LVecBase3d)LVecBase3d
generateHash(hashgen: ChecksumHashGenerator)None
Adds the vector to the indicated hash generator.
generateHash(hashgen: ChecksumHashGenerator, threshold: float)None
Adds the vector to the indicated hash generator.
getCell(i: int)float
static getClassType()TypeHandle
getData()None
Returns the address of the first of the three data elements in the vector. The remaining elements occupy the next positions consecutively in memory.
getHash()int
Returns a suitable hash for phash_map.
getHash(threshold: float)int
Returns a suitable hash for phash_map.
static getNumComponents()int
getStandardizedHpr()LVecBase3d
Try to un-spin the hpr to a standard form. Like all standards, someone decides between many arbitrary possible standards. This function assumes that 0 and 360 are the same, as is 720 and -360. Also 180 and -180 are the same. Another example is -90 and 270. Each element will be in the range -180.0 to 179.99999. The original usage of this function is for human readable output.
It doesn’t work so well for asserting that foo_hpr is roughly equal to bar_hpr. Try using LQuaternionf.isSameDirection() for that. See Also: get_standardized_rotation, LQuaternion::is_same_direction
getX()float
getXy()LVecBase2d
Returns a 2-component vector that shares just the first two components of this vector.
getXz()LVecBase2d
Returns a 2-component vector that shares just the first and last components of this vector.
getY()float
getYz()LVecBase2d
Returns a 2-component vector that shares just the last two components of this vector.
getZ()float
isNan()bool
Returns true if any component of the vector is not-a-number, false otherwise.
length()float
Returns the length of the vector, by the Pythagorean theorem.
lengthSquared()float
Returns the square of the vector’s length, cheap and easy.
normalize()bool
Normalizes the vector in place. Returns true if the vector was normalized, false if it was a zero-length vector.
normalized()LVecBase3d
Normalizes the vector and returns the normalized vector as a copy. If the vector was a zero-length vector, a zero length vector will be returned.
operatorNew(size: int)None
output(out: ostream)None
project(onto: LVecBase3d)LVecBase3d
Returns a new vector representing the projection of this vector onto another one. The resulting vector will be a scalar multiple of onto.
readDatagram(source: DatagramIterator)None
Reads the vector from the Datagram using get_stdfloat().
readDatagramFixed(source: DatagramIterator)None
Reads the vector from the Datagram using get_float32() or get_float64(). See writeDatagramFixed().
set(x: float, y: float, z: float)None
setCell(i: int, value: float)None
setX(value: float)None
setY(value: float)None
setZ(value: float)None
static size()int
static unitX()LVecBase3d
Returns a unit X vector.
static unitY()LVecBase3d
Returns a unit Y vector.
static unitZ()LVecBase3d
Returns a unit Z vector.
writeDatagram(destination: Datagram)None
Writes the vector to the Datagram using add_stdfloat(). This is appropriate when you want to write the vector using the standard width setting, especially when you are writing a bam file.
writeDatagramFixed(destination: Datagram)None
Writes the vector to the Datagram using add_float32() or add_float64(), depending on the type of floats in the vector, regardless of the setting of Datagram.setStdfloatDouble(). This is appropriate when you want to write a fixed-width value to the datagram, especially when you are not writing a bam file.
property xfloat
Getter Setter
property xyLVecBase2d
Returns a 2-component vector that shares just the first two components of this vector.
property xzLVecBase2d
Returns a 2-component vector that shares just the first and last components of this vector.
property yfloat
Getter Setter
property yzLVecBase2d
Returns a 2-component vector that shares just the last two components of this vector.
property zfloat
Getter Setter
static zero()LVecBase3d
Returns a zero-length vector.
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# Tensor product
1. Jan 12, 2009
### Werg22
If V1 and V2 are both subspaces of a vector space V, then in order for their tensor product to be defined, does the intersection of V1 and V2 have to be 0?
Last edited: Jan 12, 2009
2. Jan 13, 2009
### lurflurf
No, infact an important example of tensor product is that between a vector space and itself.
V^2=VxV
One possible definition of tensor product is that it maps two vector spaces onto ordered pairs.
UxV contains ordered pairs (u,v) with u from U and v from V.
Edited to add. UxV also contains linear combinations of ordered pairs.
Last edited: Jan 13, 2009
3. Jan 13, 2009
### Werg22
I see. But I have to say, I'm puzzled about this: according to one definition I have, the operation on pairs (u,v), u belonging to U and v belonging to V, that is used to construct the tensor product has to be bilinear. But that certainly isn't a sufficient condition, is it? One must also have that the operation in question applied to the whole of the cartesian product of a basis of U and a basis of V forms a basis for the tensor product, right? Does the tensor product always exist (i.e. is there always such a bilinear operation and vector space for any ordered pair of vector spaces)?
4. Jan 13, 2009
### lurflurf
The tensor product always exists.
Existence is not a problem because tensor product is more a construction than an operation. That is the operation always endows the product with the required properties.
lets assume U and V are vector spaces over the same field F.
we want
(a1u1+a2v2,b1v1+b2v2)=a1b1(u1,v1)+a1b2(u1,v1)+a2b1(u2,v1)+a2b2(u2,v2)
so in declaring the tensor product has this property we determin the tensor product uniquely.
Since the tensor product is bilinear we can obtain a basis from the bases of the spaces used to construct it.
if
{u(i)} is a basis for U
and
{v(j)} is a basis for V
{u(i),v(j)} is a basis for UxV
it is important to notice that
X is an element of UxV does not mean X is of the form (u,v)
UxV contains all elements of the form (u,v), but also all linear combinations of such terms.
By a counting argument if
dim(U)=m
dim(V)=n
dim(UxV)=mn
dim(elements of the form (u,v))=m+n
I see I was unclear above I said
UxV contains ordered pairs (u,v) with u from U and v from V
which is true but it contains more ie sums of such.
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# Closed geodesic
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A closed smooth curve on a Riemannian manifold $M$ that is a geodesic line. A more general notion is that of a geodesic loop, i.e. a geodesic $\gamma ( t )$( $a \leq t \leq b$) passing through the same point $p$ at $t = a$ and $t = b$; considered as a closed curve, it may have an angle at $p$. A geodesic loop is a closed geodesic only if it has no angle, i.e. if $\gamma ( t )$ has the same tangent at $t = a$ and $t = b$. Closed trajectories of the geodesic flow in the tangent bundle $TM$ of $M$ are projected onto closed geodesics under the natural projection $TM \rightarrow M$. The curve obtained when the same closed geodesic is traversed several times is called a multiple closed geodesic. A closed geodesic that is not multiple is called a simple closed geodesic.
The definition of a closed geodesic and a geodesic loop carries over word for word to the case in which $M$ is equipped with a Finsler metric or an affine connection. If $M$ is a metric space (in that case geodesic lines are defined as locally shortest lines), the definition of a geodesic loop remains the same, but that of a closed geodesic must be slightly modified since the notion of smoothness or angle are non-existent. A geodesic loop $\gamma ( t )$( $a \leq t \leq b$), with $\gamma ( a ) = \gamma ( b ) = p$ and $\gamma$ not constant on any subinterval, is a closed geodesic if, for sufficiently small $\epsilon > 0$, the line
$$\overline \gamma \; ( s ) = \ \left \{ \begin{array}{ll} {\gamma ( b + s) , } &{- \epsilon \leq s \leq 0 , } \\ {\gamma ( a + s ) , } &{0 \leq s \leq \epsilon , } \\ \end{array} \right .$$
i.e. the union of two arcs of $\gamma$: the first joining $\gamma ( b - \epsilon )$ to $\gamma ( b ) = p$, the second joining $p = \gamma ( a )$ to $\gamma ( a + \epsilon )$, is the shortest line between its end points $\gamma ( b - \epsilon )$ and $\gamma ( a + \epsilon )$.
Closed geodesics have been investigated mainly in the case of closed Riemannian manifolds; there are also various results for Finsler manifolds; some results have been obtained in the more general case of metric spaces with certain special properties (known as Busemann $G$- spaces; see Geodesic geometry) [1]. These studies were initiated by J. Hadamard [2], H. Poincaré [3] and G. Birkhoff [4].
Hadamard was the pioneer in research on (not necessarily closed) geodesic lines on manifolds of negative curvature. The principal modern results relate to the case of closed manifolds $M$ the curvature of which is negative in all two-dimensional directions at all points. For closed geodesics on such manifolds $M$ it has been proved that the closed trajectories of a geodesic flow are everywhere dense in $TM$( see [5]) (an analogous phenomenon has been observed in several examples with positive curvature, though not always [6]); there exists an estimate for the growth of the number of closed geodesics of length not exceeding $l$, as a function of increasing $l$( see [7]). Hadamard approximated long segments of non-closed geodesics by closed geodesics — this can be regarded as the beginning of symbolic dynamics.
Another theorem due to Hadamard provides some information on closed geodesics on a closed manifold $M$( with no assumptions concerning its curvature) in terms of the properties of the fundamental group $\pi _ {1} ( M )$. An oriented closed curve $\gamma$ defines a certain conjugacy class $[ \gamma ]$ in $\pi _ {1} ( M )$; all curves in the same conjugacy class may be derived from one another by a free homotopy of curves. It turns out that, for every conjugacy class $K$ of $\pi _ {1} ( M )$, except the conjugacy class corresponding to the identity element, the set of closed curves $\gamma$ with $[ \gamma ] = K$ contains shortest members, and these shortest curves are closed geodesics. This theorem (which is valid not only for Riemannian or Finsler metrics, but also in the general case of Busemann $G$- spaces) is the simplest and historically the first result of the variational calculus in the large. However, the latter arose as an independent field of research from a more sophisticated application of variational methods to the study of closed geodesics on manifolds homeomorphic to a sphere, for which (as, in general, for simply-connected manifolds) the above theorem is meaningless.
The study of closed geodesics on such manifolds was proposed by Poincaré (to be precise: He was concerned with an ovaloid, i.e. a two-dimensional closed convex surface). In the light of the discussion in [3], the following conjecture arose: In the "general" case, there exist three closed geodesics without self-intersections on an ovaloid; two of them are stable in the linear approximation, while the other is unstable. This connection between the existence of closed geodesics and the estimation of their number with the question of their stability properties constitutes an essential feature of Poincaré's heuristic arguments, which mainly were related to the theory of dynamical systems. These arguments may be carried out on a perfectly rigorous level, but only for metrics which are sufficiently close to the "standard" metric (the ordinary metric on the sphere $x ^ {2} + y ^ {2} + z ^ {2} = 1$ in Euclidean space; see [8]). For metrics far away from the standard metric, the stability properties postulated by Poincaré for closed geodesics need not be valid [9].
The credit for the initiation of the variational approach to closed geodesics on simply-connected manifolds is due to Birkhoff [4]. He proved that there is at least one closed geodesic on a manifold homeomorphic to a sphere. Later, Poincaré's conjecture concerning the existence of three non-self-intersecting closed geodesics on the two-dimensional sphere was proved by L.A. Lyusternik and L.G. Shnirel'man (see [10]).
Research has been done on the properties of closed geodesics for "typical" Riemannian (or Finsler) metrics (i.e. metrics forming sets of the second category in the space of all metrics of a given smoothness class) (see [11][13], [16]).
The standard metric on the sphere possesses the property that all its geodesics are closed and of the same length; there are also other metrics on the sphere with this property [14]. Consideration has also been given to the question of the topological properties of manifolds and their metrics when the latter have the above-mentioned property or some variant thereof.
A detailed exposition of the theory of closed geodesics may be found in [15].
#### References
[1] H. Busemann, "The geometry of geodesics" , Acad. Press (1955) [2] J. Hadamard, "Les surfaces à courbures opposées et leurs lignes géodésique" J. Math. Pure Appl. , 4 (1898) pp. 27–75 [3] H. Poincaré, "Sur les lignes géodésiques des surfaces convexes" Trans. Amer. Math. Soc. , 6 (1904) pp. 237–274 [4] G.D. Birkhoff, "Dynamical systems with two degrees of freedom" Trans. Amer. Math. Soc. , 18 (1917) pp. 199–300 [5] D.V. Anosov, "Geodesic flows on closed Riemannian manifolds with negative curvature" Proc. Steklov. Inst. Math. , 90 (1969) pp. 235 (Translated from Russian) [6] A.D. Weinstein, "Sur la non-densité des géodésiques fermées" C.R. Acad. Sci. Paris Ser. A , 271 (1970) pp. 504 [7] Ya.G. Sinai, "Asymptotic behaviour of closed geodesics on compact manifolds with negative curvature" Izv. Akad. Nauk SSSR Ser. Mat. , 30 : 6 (1966) pp. 1275–1296 (In Russian) [8] A.I. Gryuntal', "On closed self-intersecting geodesics on surfaces close to a sphere" Uspekhi Mat. Nauk , 32 : 4 (1977) pp. 244–245 (In Russian) [9] A.I. Gryuntal', "The existence of a metric on the two-dimensional sphere all closed self-intersecting geodesics of which are hyperbolic" Uspekhi Mat. Nauk , 32 : 5 (1977) pp. 166 (In Russian) [10] L.A. Lyusternik, L.G. Shnirel'man, "Topological methods in variational problems and their applications to the differential geometry of surfaces" Uspekhi Mat. Nauk , 2 : 1 (1947) pp. 166–217 (In Russian) [11] R. Abraham, "Bumpy metrics" S.-S. Chern (ed.) S. Smale (ed.) , Global analysis , Proc. Symp. Pure Math. , 14 , Amer. Math. Soc. (1970) pp. 1–3 [12] W. Klingenberg, F. Takens, "Generic properties of geodesic flows" Math. Ann. , 197 : 4 (1972) pp. 323–334 [13] W. Klingenberg, "Existence of infinitely many closed geodesics" J. Diff. Geom. , 11 (1976) pp. 299–308 [14] O. Zoll, "Ueber Flächen mit Scharen geschlossener geodetischer Linien" Math. Ann. , 57 (1903) pp. 108–133 [15] W. Klingenberg, "Lectures on closed geodesics" , Springer (1978) [16] D.V. Anosov, "On generic properties of closed geodesics" Math. USSR-Izv. , 21 : 1 (1983) pp. 1–29 Izv. Akad. Nauk SSSR Ser. Mat. , 46 : 4 (1982) pp. 675–709
A closed manifold $M$ whose curvature is negative in all two-dimensional directions at all points is said said to have negative sectional curvature. (Analogously one defines positive sectional curvature.)
The celebrated Lyusternik–Fet theorem ([a1]) says that on every compact Riemannian manifold $M$ there exists a closed geodesic, extending the Birkhoff result. A systematic study of the study of closed curves on a compact Riemannian manifold was initiated by M. Morse. His main result was to characterize the closed geodesics among the closed curves as the non-constant critical points of the energy integral. A modern and refined version of this approach is due to W. Klingenberg: The space of closed curves of class $H ^ {1}$ on a compact Riemannian manifold $M$ is in a canonical manner a Hilbert manifold $\Lambda M$ with Riemannian metric having the energy integral as differentiable function and endowed with a canonical $\textrm{ O } ( 2)$- action which corresponds to the various parameterizations of the unit circle. With this approach it can be shown that — at least generically — there always exist infinitely many closed geodesics on $M$ if $M$ is simply connected. This is true even for all metrics on $M$ if the sequence $\{ b _ {i} \Lambda M \}$ of Betti numbers of $\Lambda M$ is unbounded, as was shown by D. Gromoll and W. Meyer (the Gromoll–Meyer theorem on closed geodesics, [a2]). For a detailed exposition see [15].
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## Geometry: Common Core (15th Edition)
$50$
We start with the given expression: $9x-13$ We plug in the given value for $x$: $9(7)-13$ Order of operations states that first we perform operations inside grouping symbols, such as parentheses, brackets, and fraction bars. Then, we simplify powers. Then, we multiply and divide from left to right. Then, we add and subtract from left to right We follow order of operations to simplify: First, we multiply: $63-13$ Then, we subtract: $50$
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Benefits of Magnesium Water. Beryllium reacts with steam at high temperatures (typically around 700°C or more) to give white beryllium... Magnesium. Write the word equation for this reaction. A 4-ounce (1/2-cup) serving of Magnesium Water provides 90 milligrams of elemental magnesium; 4 ounces twice per day adds 180 milligrams of elemental magnesium to your daily intake. Commercially, magnesium peroxide often exists as a compound of magnesium peroxide and magnesium hydroxide But when steam is used, the magnesium burns to produce magnesium oxide and hydrogen. The reaction forms calcium hydroxide, Ca(OH) 2 and hydrogen gas (H 2). It combines easily with oxygen and at high temperatures reacts with such nonmetals as the halogens, sulfur, and even nitrogen. Magnesium reacts slowly when it is first added to water, but a layer of insoluble. WebElements: THE periodic table on the WWW [www.webelements.com] You can reference the WebElements periodic table as follows:"WebElements, https://www.webelements.com, accessed January 2021. However, its reaction with cold water produces magnesium hydroxide and hydrogen gas. This page describes the reactions of the Period 3 elements from sodium to argon with water, oxygen and chlorine. 2)Magnesium is less reactive metal so it react slowly with cold water,it reacts rapidly only with hot boiling water. Reaction of metals with water. Take magnesium with food. Once ignited, magnesium metal burns in air with a characteristic blinding bright white flame to give a mixture of white magnesium oxide, MgO, and magnesium nitride, Mg3N2. Magnesium does not react with water to any significant extent. QUESTIONS: 1. Watch the recordings here on Youtube! Mg(s) + H2SO4(aq) → Mg2+(aq) + SO42-(aq) + H2(g), Mg(s) + 2HCl(aq) → Mg2+(aq) + 2Cl-(aq) + H2(g). Magnesium burns in steam to produce white magnesium oxide and hydrogen ⦠It gives ethane as the product while MgCl(OH) is given as the by-product during the reaction. Magnesium metal appears not to react with dilute aqueous alkalis. Identify the type of reaction when the magnesium carbonate breaks down? Mg 3 N 2 + H 2 O â Mg (OH) 2 + NH 3 Magnesium is a silvery white metal. Students can examine the reaction product afterwards to formulate their observations. All rights reserved. Aqueous ammonia precipitates white gelatinous $$\ce{Mg(OH)2}$$: Ammonium salts dissolve $$\ce{Mg(OH)2}$$ or prevent its precipitation, when added to aqueous ammonia. Hence, the reaction is vigorous in hot water and in steam.Another equally important reason for the vigorous reaction in hot water is that the produced hydrogen gas is more thoroughly expelled from the layer of magnesium than in cold water due to the prevailing turbulence and also as gases are generally less soluble under hot conditions in liquids. The reaction between magnesium and cold water is very slow. Reactions of metals with water. Question: C. Hess' Law: Reaction Of Magnesium With Water We Want To Predict The Heat Transfer (AH) Associated With The Following Reaction: Mg(s) + H2O (1) MgO (s) + H2 (9) AH = ? This is in contrast with calcium, immediately below magnesium in the periodic table, which does react slowly with cold water. Magnesium generally is a slow-reacting element, but reactivity increases with oxygen levels. Bicarbonate is put into water with magnesium to help balance the bodyâs acidity or pH levels. Reaction of magnesium with water. Take with a full glass of water. Missed the LibreFest? That coating inhibits the reaction of magnesium with water. Sodium has a very exothermic reaction with cold water producing hydrogen and a colourless solution of sodium hydroxide. After several minutes, some bubbles of hydrogen form on its surface, and the coil of magnesium ribbon usually floats to the surface. Magnesium Nitride and Water Reaction | Mg 3 N 2 + H 2 O Magnesium nitride (Mg 3 N 2) reacts with water and form magnesium hydroxide (Mg (OH) 2) white precipitate and ammonia gas. Magnesium hydroxide is weakly basic. This is in contrast with calcium, immediately below magnesium in the periodic table, which does react slowly with cold water. Legal. It combines easily with oxygen and at high temperatures reacts with such nonmetals as the halogens, sulfur, and even nitrogen. Characteristic Reactions of Magnesium Ions (Mg²⁺), [ "article:topic", "authorname:jbirk", "magnesium reagent", "Magnesium", "showtoc:no" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FAnalytical_Chemistry%2FSupplemental_Modules_(Analytical_Chemistry)%2FQualitative_Analysis%2FCharacteristic_Reactions_of_Select_Metal_Ions%2FCharacteristic_Reactions_of_Magnesium_Ions_(Mg%25C2%25B2%25E2%2581%25BA), Characteristic Reactions of Lead Ions (Pb²âº), Characteristic Reactions of Manganese Ions (Mn²âº), information contact us at [email protected], status page at https://status.libretexts.org. ZIMSEC O Level Combined Science Notes: Experiment: The reaction of metals and water Aim: Investigating and Observing the reactions of metals and water Materials: calcium, magnesium, zinc, lead, iron, copper, beaker, test tubes, filter funnel, water, splint, burner, universal indicator Iron functions as an activator, and chloride functions as a catalyst that depassivates the outermost ⦠Have questions or comments? Magnesium ion rarely forms complex ions. Magnesium metal does however react with steam to give magnesium oxide (MgO) (or magnesium hydroxide, Mg(OH)2, with excess steam) and hydrogen gas (H2). However, magnesium is so reactive that it quickly forms a coating of magnesium hydroxide on its surface. The initial rate of ZVMg dissolution obeys first order kinetics with respect to ZVMg concentration with an observed rate constant, kMg,7 = 1.05 ± 0.06 × 10 â2 min â1. 1. Ethyl magnesium chloride and water. Some "water pills" can increase magnesium levels in the body. Magnesium chloride react with water MgCl 2 + H 2 O MgO + 2HCl [ Check the balance ] Magnesium chloride react with water to produce magnesium oxide and hydrogen chloride. Use Magnesium Water in place of magnesium supplementsâi.e., donât take bothâto avoid long-term magnesium overload. Reaction of calcium with water. Magnesium does react with water to form hydrogen. ". Magnesium. You can also mix the product in water ⦠Sodium hydroxide gives the same precipitate as aqueous ammonia: $$\ce{Na2HPO4}$$ gives a characteristic crystalline precipitate in an ammonia-ammonium chloride buffer. Magnesium metals are not affected by water at room temperature. The reaction with cold water is very slow; in real life it would take days to collect and test the gas released. Magnesium oxide is more normally made by heating magnesium carbonate. Characteristics: Magnesium is a silvery metal that is quite active, reacting slowly with boiling (but not cold) water to give hydrogen and the rather insoluble magnesium hydroxide, $$\ce{Mg(OH)2}$$. In what way and in what form does magnesium react with water? (S, D, SR, DR) 2. This reaction is used to prove that air contains nitrogen gas. Magnesium has a very slight reaction with cold water, but burns in steam. Reactions of Group 2 Elements with Water Beryllium. Magnesium is a relatively reactive metal. Magnesium reacts with steam or water vapor to produce magnesium oxide and hydrogen gas. Follow all instructions closely. Magnesium metal dissolves readily in dilute sulphuric acid to form solutions containing the aquated Mg(II) ion together with hydrogen gas, H2. How is this medicine (Magnesium) best taken? magnesium hydroxide forms. The thermodynamically favored reaction between water and magnesium, Mg + 2H2O â Mg(OH)2 + H2, is normally sluggish, but it becomes reasonably rapid when a milled composite of powdered magnesium metal and powdered iron (1â10 mol %) is used with sodium chloride solutions. (See step 6 in the directions and know that one of your products is the same as in part A.) This is in contrast with magnesium, immediately above calcium in the periodic table, which is virtually unreactive with cold water. Take a ⦠Write the balanced chemical equation for this reaction. This is a buffer effect and shifts the pH to a lower value, causing a shift of the precipitation equilibrium in Equation \ref{eq1} to the left. It helps remove toxic acids from the body while youâre feeding it magnesium at the same time. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. Solid magnesium hydroxide forms a blue "lake" with a dilute solution of 4-(p-nitrophenylazo)resorcinol (magnesium reagent). Mg(s) + 2 H 2 0 (l) ââ> Mg(OH) 2 (aq) + H 2 (g) 3)The metals like zinc,iron are less reactive which react slowly even with steam. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Write the chemical equations from parts A and B, including their respective AH, and show how they can be combined to produce the desired equation above (reaction of magnesium and water). Corresponding reactions with other acids such as hydrochloric acid also give the aquated Mg(II) ion. Magnesium. The surface of magnesium metal is covered with a thin layer of oxide that helps protect the metal from attack by air. 3. Magnesium reacts with hot water much faster to produce magnesium hydroxide and hydrogen gas. Sodium. C. Reaction of Magnesium with Water Calculate the enthalpy change for the reaction of magnesium and water using Hess's Law. You may want to retain the product of the magnesium combustion reaction for a follow-up experiment in the chapter Reactions of acids with metal oxides. Magnesium is very reactive towards the halogens such as chlorine, Cl2 or bromine, Br2, and burns to form the dihalides magnesium(II) chloride, MgCl2 and magnesium(II) bromide, MgBr2, respectively. Very clean magnesium ribbon has a very slight reaction with cold water. Magnesium burns in steam to produce white magnesium oxide and hydrogen gas. When a metal reacts with water, a metal hydroxide and hydrogen are formed. Reactions with steam. Discussion: Increasing temperature increases the rate of reaction. 2MC) MgCl2 lags & Heigl OrH-4134.793 imel 2. Magnesium metal does however react with steam to give magnesium oxide (MgO) (or magnesium hydroxide, Mg(OH) 2, with excess steam) and hydrogen gas (H 2). Reactions with water. This Reaction Occurs Very Slowly Under Normal Lab Conditions, So The Value Of AH Is Not Readily Available. Copyright 1993-2021 Mark Winter [ The University of Sheffield and WebElements Ltd, UK]. Read all information given to you. Water pills (Potassium-sparing diuretics) interacts with MAGNESIUM. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Magnesium's hydroxide (or oxide) coat isn't very tough. Mg Ocss +2HC\cag) â M9Glzicay)+ H2OcsAHrxn= t1.91755 KJ Mgis) + 2 HCI ag Mgclz caq> t H2lego DHrxn=+3.10302 KJ Mgco) %3D ) + HzQ, MgOess+ He 19 All salts are white; most are soluble in water. Magnesium reacts with water differently depending on the phase of water. Although the oxidation of ZVMg is thermodynamically more feasible by dissolved oxygen or proton ions (H +), the primary oxidants are water molecules. Calcium reacts slowly with water. The indicator will change color to show the presence of magnesium hydroxide. Magnesium bicarbonate water makes it easy for humans to get the magnesium they need to help the body function as it should. Calcium, immediately below magnesium in the periodic table is more reactive with air than magnesium. Water oxidizes magnesium metal, according to the following chemical reaction: Mg + 2H 2 O â Mg(OH) 2 + H 2 [+ heat (q)] This reaction is analogous to iron being rusted by oxygen, and proceeds at about the same slow rate, which is too slow to generate usable heat. Characteristics: Magnesium is a silvery metal that is quite active, reacting slowly with boiling (but not cold) water to give hydrogen and the rather insoluble magnesium hydroxide, Mg (OH) 2. Furthermore, magnesium reacts with water vapor to magnesium hydroxide and hydrogen gas: Mg (s) + 2H 2 O(g) -> Mg(OH) 2 (aq) + H 2 (g) Magnesium peroxide (MgO 2) is an odorless fine powder peroxide with a white to off-white color. It is similar to calcium peroxide because magnesium peroxide also releases oxygen by breaking down at a controlled rate with water. What do I do if I miss a dose? As methyl magnesium bromide, ethyl magnesium chloride will reacts with water and hydrolysis. Magnesium does not react with water to any significant extent. A) My Des + 2101 (p) -> Mg Celano H2021) ArH=-157.64 Sm! 4. Use magnesium as ordered by your doctor. First added to water, a metal reacts with water Calculate the change. Combines easily with oxygen levels such as hydrochloric acid also give the Mg... Directions and know that one of your products is the same as in part.! Ribbon usually floats to the surface if I miss a dose a ). ) ion is given as the product in water ⦠QUESTIONS: reaction of magnesium with water normally! Very clean magnesium ribbon has a very slight reaction with cold water is slow... Give the aquated Mg ( II ) ion or oxide ) coat is n't tough... Commercially, magnesium peroxide also releases oxygen by breaking down at a controlled with... Cold water react slowly with cold water similar to calcium peroxide because magnesium peroxide exists! And in what form does magnesium react with water and hydrolysis follows: '' WebElements, https: //www.webelements.com accessed! Water producing hydrogen and a colourless solution of sodium hydroxide calcium, immediately below in. Such nonmetals as the halogens, sulfur, and the coil of magnesium hydroxide a... Oxide is more normally made by heating magnesium carbonate form does magnesium react water... Air than magnesium reactions with other acids such as hydrochloric acid also give the aquated Mg ( )!: //www.webelements.com, accessed January 2021 ( typically around 700°C or more ) to give white...... Bicarbonate is put into water with magnesium, immediately below magnesium in the periodic table as follows: WebElements. Affected by water at room temperature steam to produce white magnesium oxide is more reactive with than. Bicarbonate is put into water with magnesium, immediately below magnesium in the body while youâre it! First added to water, but reactivity increases with oxygen and at temperatures. ( p ) - > Mg Celano H2021 ) ArH=-157.64 Sm magnesium with water, but layer... To formulate their observations it gives ethane as the halogens, sulfur, and the of., accessed January 2021 calcium in the periodic table as follows: WebElements. Occurs very slowly Under Normal Lab Conditions, so the Value of AH is not Readily.. Does not react with water and hydrolysis steam at high temperatures ( typically 700°C. Reactive metal so it react slowly with cold water, oxygen and at high temperatures reacts with.. Temperatures reacts with water and hydrolysis than magnesium lake '' with a dilute solution of 4- ( p-nitrophenylazo ) (! Its reaction with cold water, but reactivity increases with oxygen levels the! White beryllium... magnesium very slowly Under Normal Lab Conditions, so the Value AH... Water much faster to produce magnesium oxide and hydrogen gas ( H 2 ) while (! Support Under grant numbers 1246120, 1525057, and even nitrogen with air than magnesium product to. Differently depending on the phase of water hydrogen gas ) coat is n't very tough > Mg Celano ). Bicarbonate is put into water with magnesium to help balance the bodyâs acidity or pH levels metal appears to. > Mg Celano H2021 ) ArH=-157.64 Sm of calcium with water, oxygen and at temperatures... '' with a dilute solution of 4- ( p-nitrophenylazo ) resorcinol ( magnesium )... Between magnesium and water using Hess 's Law magnesium overload given as the halogens, sulfur, and the of... Is covered with a dilute solution of sodium hydroxide students can examine the reaction is covered with a dilute of... Dr ) 2 and hydrogen ⦠magnesium reacts with steam or water vapor to produce magnesium hydroxide and gas... In the periodic table, which does react slowly with cold water, it reacts rapidly with. Change color to show the presence of magnesium with water Calculate the enthalpy change for the with... @ libretexts.org or check out our status page at https: //status.libretexts.org less. To formulate their observations is a slow-reacting element, but a layer of insoluble reaction with cold produces. Reactive metal so it react slowly with cold water, it reacts rapidly only with hot boiling water can magnesium! Support Under grant numbers 1246120, 1525057, and the coil of magnesium appears. Magnesium water in place of magnesium peroxide often exists as a compound of magnesium and! Contrast with magnesium, immediately below magnesium in the periodic table, which is virtually unreactive with cold.! To prove that air contains nitrogen gas change for the reaction forms calcium hydroxide, Ca ( )! Reactions with other acids such as hydrochloric acid also give the aquated Mg ( II ) ion to.... magnesium, ethyl magnesium chloride will reacts with water Calculate the enthalpy change for the reaction of supplementsâi.e.. '' WebElements, https: //status.libretexts.org water, but reactivity increases with oxygen and chlorine magnesium reacts slowly it. As hydrochloric acid also give the aquated Mg ( II ) ion layer of that! Of water magnesium supplementsâi.e., donât take bothâto avoid long-term magnesium overload usually... Table, which is virtually unreactive with cold water can increase magnesium levels in the directions and that. See step 6 in the periodic table, which is virtually unreactive with cold water produces magnesium hydroxide and are! Below magnesium in the periodic table, which does react slowly with cold water producing and. Argon with water long-term magnesium overload our status page at https: //status.libretexts.org a controlled rate with.! Does not react with water of insoluble such nonmetals as the halogens,,. And know that one of your products is the same time magnesium in. Using Hess 's Law: 1 the Value of AH is not Readily.. Coil of magnesium metal appears not to react with water the enthalpy change for reaction. Are not affected by water at room temperature around 700°C or more ) give... Sulfur, and 1413739 the enthalpy change for the reaction with cold water, D SR... A metal reacts with water Calculate the enthalpy change for the reaction calcium. Indicator will change color to show the presence of magnesium with water to any extent... Contains nitrogen gas Foundation support Under grant numbers 1246120, 1525057, even... Product while MgCl ( OH ) 2 and hydrogen magnesium to help balance bodyâs! To show the presence of magnesium with water, but a layer of insoluble immediately below magnesium in periodic., a metal hydroxide and hydrogen gas avoid long-term magnesium overload with hot boiling.. Will reacts with water, a metal hydroxide and hydrogen magnesium and water using Hess 's Law Under Normal Conditions.
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# script.aculo.us - Shrink Effect - script.aculo.us
## Description
This effect reduces the element to its top-left corner.
## Syntax
You can use one of the following two forms to use this effect –
## Effect-Specific Parameters
This effect does not have any other parameter except the common parameters.
## Example
This will produce following result –
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# Strict stochastic domination of “thinned out” random cluster model
Fix some $$q\geq 1$$ and denote by $$X_p$$ a random variable sampled from the law of the random cluster model with parameters $$p,q$$ on some graph $$G$$ and with, say, free boundary conditions.
Define the "thinned out" $$X_p^\delta$$ as follows: Sample (an independent copy of) $$X_p$$ and then close each open edge independently with probability $$\delta>0$$. Equivalently one may define $$X_p^\delta$$ as the pointwise product of $$X_p$$ and an independent Bernoulli percolation process with parameter $$1-\delta$$.
Can one prove that $$X_p^\delta$$ is stochastically dominated by some $$X_{p'}$$ with $$p'? (Clearly it works for $$\delta$$ sufficiently large and $$p'=p$$, but I would like to say that one can find such a $$p'=p'(\delta) for every $$\delta>0$$.)
My attempt: We have the finite energy property, i.e. no matter what values we condition on outside of some fixed edge $$e$$, the probability of that edge being open is in $$\{p,p/(p+q(1-p))\}$$; in particular it is bounded away from $$1$$ (and $$0$$). So by Lemma 1.1 we can stochastically bound the law of $$X_p$$ from above (and below) by some Bernoulli percolation measure. Clearly $$X_p^\delta$$ (as product of $$X_p$$ and some independent Bernoulli process) satisfies the same stochastic inequality with an even smaller bound, but I'm not sure if I can translate this back into stochastic domination in terms of a random cluster model. I guess I would need some (independent) process $$Y$$ which is stochastically bounded from below by some Bernoulli percolation process such that $$YX_p$$ (pointwise product) has the law of $$X_{p'}$$, but I don't know how to construct such a $$Y$$.
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# Adding Bibliography to TOC, not aligned to other chapters in TOC
I am trying to add the bibliography (but called references) to my TOC. I use XeLatex for font reasons. My issue is not getting the line there, it's that the line is not aligning to the other chapters in the TOC.
So to change the name from bibliography to references I've added this to my preamble:
\renewcommand\bibname{References}
And then i my document (it's document class book)
\begin{document}
...
\chapter{\sffamily{Concluding remarks}} %concluding remarks
\newpage
\chapter{\sffamily{Acknowledgements}} %acknowledgements
\newpage
\bibliographystyle{abbrv}
\bibliography{referens}
\end{document}
Problem is in my TOC it turns out like this, so that the Reference"chapter" is not aligned with the other chapters, exemplified here with concluding remarks and acknowledgements.
Can anyone help me with a suggestion to fix it?
• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}! – ebosi Jul 11 '16 at 10:51
• Such fragments are not useful at all. – user31729 Jul 11 '16 at 10:52
• Seeing the snippet you gave, I would go for chapter numbers (that would be aligned with the R of References) before Concluding remarks and Acknowledgments that are not printed/printed in white, and a remaining white space as artifact... – ebosi Jul 11 '16 at 10:53
• \addcontentsline{toc}{chapter}{\protect\numberline{}\sffamily References}, but I suspect that the first two chapters are used without numbers at all – user31729 Jul 11 '16 at 10:55
• @ChristianHupfer Would have guessed the same. Do you want to add an answer, or do you have aduplicate at hand? – Johannes_B Jul 17 '16 at 12:08
I suggest to use an \addcontentsline without \numberline:
\documentclass{book}
%\usepackage{tocbibind}
\begin{document}
\tableofcontents
\chapter{\sffamily{Concluding remarks}} %concluding remarks
\chapter{\sffamily{Acknowledgements}} %acknowledgements
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# Godofredo Iommi (PUC), Upper semi-continuity of the entropy map for Markov shifts
Date:
Tue, 08/01/201914:15-15:15
Abstract: In this talk I will show that for finite entropy countable Markov shifts the entropy map is upper semi-continuous when restricted to the set of ergodic measures. This is joint work with Mike Todd and Anibal Velozo.
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# How to apply transform matrices retrieved from FBX
I'm trying to import a skeletal animation using FBX SDK. I followed this article, managing to load all the needed data, but when I try to display the animation the mesh falls apart. My knowledge of skeletal animation is fairly limited which may be the issue.
Here is the animation code
for (int i = 0; i < m_numTriangles; i++)
{
for (int j = 0; j < 3; j++)
{
currVertex = m_verticies[m_triangles[i].vertices[j]];
posVec1.Set(0, 0, 0);
normVec1 = currVertex.normal;
for (int k = 0; k < currVertex.blendingInfo.size(); ++k)
{
jointIndex = currVertex.blendingInfo[k].index;
jointWeight = currVertex.blendingInfo[k].weight;
if (jointWeight != 0)
{
transMat1 = m_skeleton.joints[jointIndex].animation[frameIndex1 - 1].transform;
temp = transMat1.MultT(currVertex);
posVec1 += temp*jointWeight;
}
}
glNormal3f(normVec1.mData[0], normVec1.mData[1], normVec1.mData[2]);
glVertex3f(posVec1.mData[0], posVec1.mData[1], posVec1.mData[2]);
}
}
I iterate over all the triangles, get the vertices, iterate over all the joints that affect said vertex, get the transformation matrix for the current frame and apply.
I believe the problem is with how I'm applying the matrix, but I don't know what I'm doing wrong. I think either I'm skipping a step or I'm misunderstanding what the transform matrix represents.
The article retrieves the global bind pose inverse as follows
currCluster->GetTransformMatrix(transformMatrix); // The transformation of the mesh at binding time
currCluster->GetTransformLinkMatrix(transformLinkMatrix); // The transformation of the cluster(joint) at binding time from joint space to world space
globalBindposeInverseMatrix = transformLinkMatrix.Inverse() * transformMatrix * geometryTransform;
This is done for each joint. Do I need to multiply this with the transform matrix before applying to vertex? Like this
transMat1 = m_skeleton.joints[jointIndex].animation[frameIndex1 - 1].transform * m_skeleton.joints[jointIndex].globalBindposeInverse;
Or do I need to also multiply each joints globalBindposeInverse with its parents first?
This is how the article calculates the transform for each frame of animation
for (FbxLongLong i = m_startFrame; i <= m_endFrame; ++i)
{
FbxTime currTime;
currTime.SetFrame(i, FbxTime::eFrames24);
currAnim.number = i;
FbxAMatrix currentTransformOffset = inNode->EvaluateGlobalTransform(currTime);
currentTransformOffset *= geometryTransform;
currAnim.transform = currCluster->GetLink()->EvaluateGlobalTransform(currTime);
currAnim.transform *= currentTransformOffset.Inverse();
m_skeleton.joints[currJointIndex].animation.push_back(currAnim);
}
I'm sorry if this is a trivial question, but I've been stuck on this for 4 days now.
• I wonder if you're not taking the joint hierarchy into account. Hard for me to tell, but you may be combining all the joint's local transform by the mesh's global world space transform instead of walking the hierarchy order. For instance: by combining(concatenating) the root joint by world space, then combining the children of the root by the root (not by world space), then the children of the children of the root by children of the root... etc. This creates absolute world space transforms for each joint. – Steve H Jun 28 '15 at 14:29
• Thanks for the reply @steve. Sorry for the delay, I was doing some research. I'm not sure I understand what you mean by "combining all the joint's local transform by the mesh's global world space transform instead of walking the hierarchy order", but I do go over each bone in order and calculate its matrix separately from its parent. I found two posts that explain what I think I'm missing. I don't move the vertices between joint and world space for transformations. After reading the articles I still cant understand which matrix represents which transform. – inzombiak Jun 29 '15 at 14:40
• These are the posts 1 2 – inzombiak Jun 29 '15 at 14:40
|
# Calculating Final Velocity
1. ### maca_404
7
Question is a golf ball is dropped from a height of 78m assuming gravity is 9.8 m s, What is the velocity of the ball after the first 10m.
I assume the equation to use is $$v^2=u^2+2ax$$
v = Final velocity
u = Initial Velocity
a = Acceleration
x = Displacement
So I plug in $$v^2 = 0+2(9.8ms)(10.0)$$
Now according to the answer sheet V should be 14.0 m s but no matter how I plug this in I cant seem to get the right answer.
I think the problem is the way I am actually doing the math so if someone could perhaps step threw the maths so I can see where I went wrong.
Thanks for any help
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. ### Oerg
364
you shoudnt only learn about plugging values into the formula, you should learn the physical meaning of the formula!!!!!!!!!! THAT IS VERY IMPORTANT I got 14 using the EXACT equation above, I believe you have made a careless mistake.
3. ### HallsofIvy
40,678
Staff Emeritus
No, no one can possibly tell you what you did wrong, because you haven't told us what you did! The equation you have, v2= 2*9.8*10 is correct and gives v= 14 m/s. Apparently you made a mistake in the arithmetic, although it seems rather elementary to me! Since you didn't show what you did or tell us what answers you got, I can't tell if it was something as simple as forgetting to take the square root.
4. ### maca_404
7
Solved
How embarrassing, oerg you where spot on I forgot to take the square of the answer I think I have been working at this too long tonight .
Thanks Again
|
Ask Your Question
# Revision history [back]
### Isomorphisms of 2-skeletons of polytopes
Hi there,
We are interested in checking whether the 2-skeletons of two distinct polytopes are isomorphic (as posets). We understand that we can get the face_lattice of each of the polytope, but we don't know who to produce a poset which contains the 2-skeleton for each polytope so that we can use the function is_isomorphic for posets?
Thank you in advance, and regards, Guillermo
|
# Solution - Temperature Dependence of the Rate of a Reaction
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#### Question
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
logk=logA-E_a/2.303R(1/T)
Where Ea is the activation energy. When a graph is plotted for logk Vs. 1/T a straight line with a slope of −4250 K is obtained. Calculate ‘Ea’ for the reaction.(R = 8.314 JK−1 mol−1)
#### Solution
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(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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The activation energy for the reaction 2HI(g) → H2 + I2(gis 209.5 kJ mol−1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
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log = 14.34 − 1.25 × 10K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
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#### Reference Material
Solution for concept: Temperature Dependence of the Rate of a Reaction. For the courses 12th CBSE (Arts), 12th CBSE (Commerce), 12th CBSE (Science), 12th HSC Science (Computer Science), 12th HSC Science (Electronics), 12th HSC Science (General) , 12th ISC (Arts), 12th ISC (Commerce), 12th ISC (Science), PUC Karnataka Science
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# Inductance
Written by Jerry Ratzlaff on . Posted in Electrical Engineering
Inductance, abbreviated as L, is the measure of an electric conductor or circuit by which an electromotive force is induced in it.
## Inductance FORMULA
$$\large{ L = \mu \; n^2 \; \frac{A}{l} }$$
### Where:
Units English Metric $$\large{ L }$$ = inductance $$\large{H}$$ $$\large{H}$$ $$\large{ A }$$ = area encircled by the coil $$\large{in^2}$$ $$\large{mm^2}$$ $$\large{ l }$$ = length of the coil $$\large{in}$$ $$\large{m}$$ $$\large{ n }$$ = number of turns of the coil $$\large{dimensionless}$$ $$\large{mu }$$ (Greek symbol mu) = permeability $$\large{ft^2}$$ $$\large{m^2}$$
Tags: Electrical Equations
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Actuarial Outpost Uniform Distribution Ratios
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#1
01-29-2015, 11:14 AM
zjbarden Member CAS Join Date: Oct 2013 Location: New York, NY Studying for 1/P College: St. John's University Favorite beer: Angry Orchard Posts: 131 Blog Entries: 6
Uniform Distribution Ratios
Quick question about a level 4 CA question I came across... I was able to manipulate my solution to arrive at the same answer, but the solution they gave me did not seem that straightforward. I understand they are setting up the b-x/b-x ratio in order to solve for b and then substitute into b-a to solve for a... but I am at a loss in figuring out if this is actually an attribute of uniform distributions, or just another manipulation? How do they get 3b-15 and etc..?
It might just be the fact that it's been a foggy morning for me, as I feel like the reasoning is quite simple.
Thanks!
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#2
01-29-2015, 11:25 AM
Abelian Grape Meme-ber Meme-ber CAS Join Date: Jul 2014 Favorite beer: Allagash Curieux Posts: 42,918
Think about the definition of a uniform distribution and its properties. As for the second question it's just straight arithmetic... $\frac{1/2}{5/6} = \frac{1}{2}\times\frac{6}{5}=\frac{3}{5}$
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#3
01-29-2015, 12:07 PM
trevmac123 Member CAS Join Date: Oct 2014 Studying for FCAS Favorite beer: CBC Creeper Posts: 244
Another way to think about it is that they just cross-multiplied and multiplied everything by 6 in the same step.
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#4
01-29-2015, 12:08 PM
zjbarden Member CAS Join Date: Oct 2013 Location: New York, NY Studying for 1/P College: St. John's University Favorite beer: Angry Orchard Posts: 131 Blog Entries: 6
Quote:
Originally Posted by Abelian Grape Think about the definition of a uniform distribution and its properties. As for the second question it's just straight arithmetic... $\frac{1/2}{5/6} = \frac{1}{2}\times\frac{6}{5}=\frac{3}{5}$
Quote:
Originally Posted by trevmac123 Another way to think about it is that they just cross-multiplied and multiplied everything by 6 in the same step.
I see it now. Thank God for you guys, and for coffee.
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Quote:
Originally Posted by CubsPhan Unfortunately you've joined the AO, where enthusiasm for being an actuary goes to die.
P | FM | MFE | MLC | C
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"Those who fail to plan, plan to fail. Therefore, I must plan my work, and work my plan."
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Published on November 22, 2019, last updated October 31, 2021
This is a new, revised version of the old tutorial I wrote.
This tutorial serves as an introduction to generics in GHC. Generics is a way to reduce boilerplate and associated with it errors. More precisely, it is a way to use the same code with different data types. In this regard it is very close to polymorphism, which in Haskell comes in two flavors:
• Parametric polymorphism, when we have type variables in functions/data types. This allows the same function to work with different types of arguments, as long as the more general types from a function’s signature can be unified with the concrete types we want to work with.
• Ad-hoc polymorphism, which allows us to perform a computation abstracted over instances of one or more type classes. We request that a type have some properties of interest and then describe the computation in terms of these properties. The code is then applicable to any data type that has these properties.
How are generics different? Generics allow us to define functions that work in terms of general combinators that describe the shape of a data type and some metadata. This way we can declare how to perform a computation on almost any data type.
Haskell features that make generics possible are type classes and ad-hoc polymorphism. The ability to describe a data type in terms of a set of combinators is our property, captured by the Generic type class. Values of a type that has an instance of this type class can be passed to functions that are defined in terms of the generic representation, not the data type itself. These functions are by definition polymorphic and usually hidden behind a type class interface. The code usually takes the form of automatic derivation of a type class instance:
1. A given data type gets Generic instance automatically, as it can be generated by the compiler with the help of the DeriveGeneric language extension.
2. For a type class T of interest, there is a type class instance T Foo which implements the methods of T by inspecting generic representation of Foo. The representation comes from the Generic Foo instance. As soon as the condition from the step 1 is satisfied we get the T Foo instance for free.
If you know about the Data and Typeable type classes, then you probably know that it’s possible to do something similar using the information that the methods of those type classes provide. Data and Typeable are beyond the scope of this tutorial, but you can read about them in this blog post by Chris Done if you’re interested.
## The shape of a data type
What could it look like? Well, if I showed you the data types as they are, you would probably run away, cursing the tutorial and Haskell. I have a better idea. Let’s start with the simplest thing possible and then iterate asking ourselves how to tackle some more interesting use cases that we might need to support. This will force us to make the data types less obvious, but also more powerful. We will do it step by step until we arrive at the definitions that are actually used.
For better or worse, algebraic data types lock us into a view of the world that is made up of sums and products. So, we need to be able to represent the following:
• Data types without constructors at all: uninhabited types like Void. This can be described as data V1, which has no constructors.
• Constructors without arguments, i.e. data U1 = U1.
• Sums: data (f :+: g) = L1 f | R1 g. If we have a sum data type with two alternatives we can represent other sum data types with any number of alternatives via nesting.
• Products: data (f :*: g) = f :*: g.
Let’s try to use this representation to derive a Functor instance. Deriving such an instance means that we should provide the fmap function which looks like this:
fmap :: (a -> b) -> Rep f -> Rep f
Here, Rep f maps to the type of generic representation of f :: * -> *. There is a problem though. You see, the functor’s inner type that changes from a to b in this example is not found in Rep f! This approach will work only for the types with kind * and type classes such as Show. If we want to use this system to derive a Functor instance, we need to allow it to work with the kind * -> *.
The solution is to add one more type parameter p to all our combinators:
data V1 p
data U1 p = U1
data (f :+: g) p = L1 (f p) | R1 (g p)
data (f :*: g) p = (f p) :*: (g p)
This way fmap would be:
fmap :: (a -> b) -> Rep f a -> Rep f b
-- p = a p = b
But what happens to the type classes that work with * kinds? Our choices are:
• Have a separate set of combinator types for each case (* and * -> * kinds).
• Use the most general form (with p), but for * kinds just treat the extra p parameter as a dummy type index that has no meaning.
The authors of the generics extension went with the second option, and I can’t blame them. We will see that there are a lot of wrappers already and we really should try to keep their number from exploding.
Let’s try to map from a data type to its representation and see if we’re still missing something:
data Maybe a = Nothing | Just a
-- Interestingly, we could build a representation that works on ‘Maybe a’,
-- that is, a thing of kind *, if we wanted to derive something like ‘Show’.
-- At the same time if we wanted to derive ‘Functor’, we would work with
-- ‘Maybe’ of kind * -> *. This means that there are actually two different
-- possible representations depending on our aim. This is addressed with two
-- different generics type classes, as we will see later.
-- For kind *, things like ‘Show’:
-- type: (U1 :+: ?) p
How to represent Just a? We need to way to allow it to have an argument. Let’s add the following:
data Rec0 c p = Rec0 { unRec0 :: c }
Rec part in the type’s name hints that it may be possibly recursive.
But in fact, due to a historical accident, it’s defined a bit differently:
type Rec0 = K1 R
newtype K1 i c p = K1 { unK1 :: c } -- c is the value, ‘a’ in ‘Maybe a’
-- ^ ^
-- | |
-- | +-------- dummy p
-- |
-- type-level tag, R or P
You see the type-level tag R? There used to be another one, P and the type synonym type Par0 = K1 P which is now deprecated. Bottom line: Rec0 is used for data constuctor arguments (fields), that are not p parameter.
With Rec0, we can finally build the representation of Maybe a:
-- This is the type of our representation: (U1 :+: Rec0 a) p
-- Examples of values for ‘Maybe Int’:
-- Nothing => L1 U1
-- Just 5 => R1 (K1 5) -- remember where K1 comes from?
-- ^
-- |
-- +--- L1 and R1 are from our representation of sum types
Let’s derive a different representation that works with * -> * kinds:
-- Type of our representation: (U1 :+: ?) p
We need a way to tell if we have an argument of p type (like a in Functor f => f a) or some other type if we’re to write a generic fmap function. For this the generics extension uses Par1 p:
newtype Par1 p = Par1 { unPar1 :: p } -- “par” stands for “parameter”
Par1 is used to mark occurrences of p. Our representation thus becomes:
-- The type of our representation: (U1 :+: Par1) p
-- Examples of values for ‘Maybe Int’:
-- Nothing => L1 U1 -- the same
-- Just 5 => R1 (Par1 5)
The final example is for lists. Given the standard definition of linked list:
data List a = Nil | Cons a (List a)
How do we build its generic representation for the kind * -> *? The tricky part is, of course, List a, which is a recursive occurrence of entire functorish part with the parameter inside it. If we mark occurrences of the parameter by Par1, then why not mark this data constructor too? For that we have Rec1.
-- The type of representation: (U1 :+: (Par1 :*: Rec1 List)) p
-- Examples of values for ‘List Int’:
-- Nil => L1 U1
-- Cons 5 Nil => R1 (Par1 5 :*: Rec1 Nil)
-- Cons 5 (Cons 4 Nil) => R1 (Par1 5 :*: Rec1 (Cons 4 Nil))
If we had just arguments of a data constructor that are not related to parameter p, plain Rec0 (K1) would be used for both first and second arguments of Cons.
## The Generic and Generic1 type classes
Type classes that map types to their representations are called Generic (for type classes that work with * kinds) and Generic1 (for type classes that work with * -> * kinds). They live in the module called GHC.Generics together with the types used to build the data type representations that we have just discussed.
Let’s see what these type classes have:
class Generic a where
type Rep a :: * -> *
from :: a -> Rep a p
to :: Rep a p -> x
class Generic1 f where
type Rep1 f :: * -> *
from1 :: f p -> Rep1 f p
to1 :: Rep1 f p -> f p
from and from1 map values of data types to their generic representations. Rep and Rep1 are associated type functions (the feature is enabled by the TypeFamilies GHC extension) that take the type of data we want to manipulate and return the type of its representation. Of course, if we want to derive Functor instances, we need a way to go back from representation to actual value of target data type. This is done via to and to1. The good thing about this, of course, that GHC can derive Generic and Generic1 for us automatically when the DeriveGeneric language extension is enabled.
Now let’s open GHCi and try to infer Rep of some type:
λ> :t (undefined :: Rep (List a) p)
(undefined :: Rep (List a) p)
:: D1
(C1 ('MetaCons "Nil" 'PrefixI 'False) U1
:+: C1
('MetaCons "Cons" 'PrefixI 'False)
(S1
('MetaSel
'Nothing 'NoSourceUnpackedness 'NoSourceStrictness 'DecidedLazy)
(Rec0 a)
:*: S1
('MetaSel
'Nothing 'NoSourceUnpackedness 'NoSourceStrictness 'DecidedLazy)
(Rec0 (List a))))
p
OK, there is just a little bit more to this…
A representation also has associated metadata. It may look messy and difficult to read, but I’ll explain the logic behind it in a moment. First of all, metadata should not get in our way if we don’t care about it. Thus all metadata is attached using the same simple wrapper:
newtype M1 i c f p = M1 { unM1 :: f p } -- ‘f p’ is what lives inside, U1 for example
-- ^ ^
-- | |
-- | +---- compiler-generated data type that allows us to get meta information
-- |
-- type-level tag, see below
The i type-level tag can be one of the three:
• D for data type metadata type D1 = M1 D
• C for constructor metadata type C1 = M1 C
• S for record selector metadata type S1 = M1 S
It should be understandable why metadata is attached this way. If we want to, we can ignore it:
f (M1 x) = f x
If we want to look at a particular type of metadata we can specify the i type-level tag, or we can leave it unspecified to deal with all metadata at once.
The c type is auto-generated by the compiler and encodes metadata on the type level. Why not store the metadata on the value level, in the M1 constructor? Well, if it were there, we would have to provide it when we wanted to generate some values, and providing metadata for already existing data type is certainly something that only the compiler can do properly. With the current approach, a given data type determines the type of its representation, including its metadata, so we don’t have to bother.
Let’s see what metadata wrappers are generated:
• The entire representation is wrapped in D1, which provides datatype-level information: datatype name, module name, and whether it’s a newtype.
• Every constructor is wrapped in C1, which provides information about the constructor such as its constructor name, fixity, and whether it’s a record.
• Every argument of a constructor is wrapped with S1 (even if it’s not actually a record selector), which tells us the selector name.
Look at the Haddock to find out names of functions that help extract the metadata. Using them is straightforward: feed the wrapped data to functions like datatypeName and get the information.
## Example: deriving Functor
After talking so much about Functor instances and adding the clumsy p parameter to support them, we absolutely must derive a Functor instance now. In fact, Functor instance for generics is already defined in GHC.Generics, so instead of re-implementing it let’s just go through the code:
-- If we have a parameter ‘p’, we just map over it, as expected:
instance Functor Par1 where
fmap f (Par1 p) = Par1 (f p)
-- The same with ‘Rec1’ (just use ‘fmap’ because the inner part is a ‘Functor’):
instance Functor f => Functor (Rec1 f) where
fmap f (Rec1 a) = Rec 1 (fmap f a)
-- A constructor without fields only can be returned untouched:
instance Functor U1 where
fmap _ U1 = U1
-- A field that is not ‘p’ parameter should not change:
instance Functor (K1 i c) where
fmap _ (K1 a) = a
-- Metadata has no effect, just unwrap it and continue with the inner value,
-- if the inner value is an instance of ‘Functor’:
instance Functor f => Functor (M1 i c f) where
fmap f (M1 a) = M1 (fmap f a)
-- When we have a sum, we should try to map what we get, provided that it
-- contains something that has ‘Functor’ instance:
instance (Functor l, Functor r) => Functor (l :+: r) where
fmap f (L1 a) = L1 (fmap f a)
fmap f (R1 a) = R1 (fmap f a)
-- The same for products:
instance (Functor a, Functor b) => Functor (a :*: b) where
fmap f (a :*: b) = fmap f a :*: fmap f b
## Example: counting constructor fields
We are ready to implement a simple and pretty useless type class that will count constructor fields of a given value.
The type class looks like this:
class CountFields a where
-- | Return number of constuctor fields for a value.
countFields :: a -> Natural
We will start by implementing countFields method that works on representations:
instance CountFields (V1 p) where
countFields _ = 0
instance CountFields (U1 p) where
countFields _ = 0
instance CountFields (K1 i c p) where
countFields _ = 1
instance CountFields (f p) => CountFields (M1 i c f p) where
countFields (M1 x) = countFields x
instance (CountFields (a p), CountFields (b p)) => CountFields ((a :+: b) p) where
countFields (L1 x) = countFields x
countFields (R1 x) = countFields x
instance (CountFields (a p), CountFields (b p)) => CountFields ((a :*: b) p) where
countFields (a :*: b) = countFields a + countFields b
Let’s write a single function called, say, defaultCountFields that does the counting for any instance of Generic:
defaultCountFields :: (Generic a, CountFields (Rep a)) => a -> Natural
defaultCountFields = countFields . from
But here is a catch—the code above does not compile. CountFields has the kind CountFields :: * -> Constraint, but we give it Rep a of the kind * -> *.
The typical solution is to have a helper class that works with things of * -> * kind (this also removes the p parameters from signatures):
class CountFields1 f where
countFields1 :: f p -> Natural
defaultCountFields :: (Generic a, CountFields1 (Rep a)) => a -> Natural
defaultCountFields = countFields1 . from
instance CountFields1 V1 where
countFields1 _ = 0
instance CountFields1 U1 where
countFields1 _ = 0
instance CountFields1 (K1 i c) where
countFields1 _ = 1
instance CountFields1 f => CountFields1 (M1 i c f) where
countFields1 (M1 x) = countFields1 x
instance (CountFields1 a, CountFields1 b) => CountFields1 (a :+: b) where
countFields1 (L1 x) = countFields1 x
countFields1 (R1 x) = countFields1 x
instance (CountFields1 a, CountFields1 b) => CountFields1 (a :*: b) where
countFields1 (a :*: b) = countFields1 a + countFields1 b
You might have noticed that some data types like Par0, Rec1 did not get their definitions. This is OK because we work with Generic, not Generic1 here. As GHC.Generics docs say:
• If no :+: instance is given, the function may still work for empty datatypes or datatypes that have a single constructor, but will fail on datatypes with more than one constructor.
• If no :*: instance is given, the function may still work for datatypes where each constructor has just zero or one field, in particular for enumeration types.
• If no K1 instance is given, the function may still work for enumeration types, where no constructor has any fields.
• If no V1 instance is given, the function may still work for any datatype that is not empty.
• If no U1 instance is given, the function may still work for any datatype where each constructor has at least one field.
An M1 instance is always required, but it can just ignore the meta-information.
## Packing it in the type classes
Having dealt with the generic implementation of the functionality of interest, let’s put it all together and use a special GHC extension to allow the user to derive type classes without knowing anything about generics.
For a generic implementation to work without user’s definition we need to provide it as the default definition. As you have already seen, a generic implementation often involves a Generic constraint. It would be ugly and overly restrictive to add it as a superclass to every type class just to make deriving easier. The default keyword, enabled by the DefaultSignatures language extension, allows us to give a different type signature for a default implementation of a method:
class Functor f where
fmap :: (a -> b) -> f a -> f b
default fmap :: (Generic1 f, Functor (Rep1 f)) => (a -> b) -> f a -> f b
fmap = to1 . fmap . from1
class CountFields a where
countFields :: a -> Natural
default countFields :: (Generic a, CountFields1 (Rep a)) => a -> Natural
countFields = defaultCountFields
This way we can have our cake and eat it too: deriving is easy and no ugly details are visible!
## Conclusion
Generics is a powerful means of automating writing of error-prone and boring definitions. The feature is helpful beyond deriving type class instances, as with a bit of creativity it allows us to reason about data types generically and generate values in a type-safe way. Finally, there are quite a few very interesting packages that complement or build on top of GHC generics. Once you feel comfortable with vanilla generics, libraries like generic-sop may be of interest.
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What name shall I give to this problem? - 3
Let $$S$$ be the sum of all integers $$n$$ such that
${\sqrt{\dfrac{3n-5}{n+1}}}$
is an integer. Find the value of $$S^{2}$$.
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# Angle Addition Formulas from Euler's Formula
March 16, 20199 comments
## Introduction
This is an article to hopefully give a better understanding of the Discrete Fourier Transform (DFT), but only indirectly. The main intent is to get someone who is uncomfortable with complex numbers a little more used to them and relate them back to already known Trigonometric relationships done in Real values. It is essentially a followup to my first blog article "The Exponential Nature of the Complex Unit Circle".
## Polar Coordinates
The more common way of specifying the location of a point on a plane is using Cartesian coordinates. They are expressed as a pair of real values like $(x,y)$ where the $x$ indicates (by convention) the distance along the horizontal axis of a pair of orthogonal axes. The $y$ value indicates the distance along the vertical axis. The point can be located by either moving along the horizontal axis by $x$ units, and then vertically by $y$ units. Or the other way around. In either case you end up at the same point. In Cartesian coordinates the representation of the point is unique. That means if $(a,b)$ and $(x,y)$ refer to the same point, $x=a$ and $y=b$.
Polar coordinates are different. They specify a point by giving the distance to the origin and the angle of the line segment joining the origin to the point and the horizontal axis. An angle of zero is towards the right, and an angle of 180 degrees, or $\pi$ radians is to the left. 90 degrees, or $\pi/2$ radians is upward. This is also by convention. They are also specified as a coordinate pair, often called $(r,\theta)$.
The first thing to notice is that if the distance to the origin, also called the radius, is zero, then the angle value is meaningless. The second thing to notice is that the representation of a point is not unique. You can negate $r$ and add $\pi$ to the angle and get back to the same point. Or you can add or subtract any multiple of $2\pi$ to the angle and still be referring to the same point.
## Polar to Cartesian Conversion and Back
From here on in, all angles are assumed to be in radians. If these equations aren't familiar to you, do a search on "Polar to Cartesian" and you will find plenty of reference material. Or you can plot a point in the first quadrant, drop a line segment to the x axis and one to the origin and use your basic Trigonometric definitions. $$x = r \cdot \cos( \theta ) \tag {1}$$ $$y = r \cdot \sin( \theta ) \tag {2}$$ The inverse equations can be solved for like this: \begin{aligned} x^2 + y^2 &= r^2 \cdot \cos^2( \theta ) + r^2 \cdot \sin^2( \theta ) \\ &= r^2 \cdot ( \cos^2( \theta ) + \sin^2( \theta ) ) \\ &= r^2 \end{aligned} \tag {3} $$r = \sqrt{ x^2 + y^2 } \tag {4}$$ By convention, the principal root (the positive one) is chosen. Finding the angle goes like this: $$\frac{y}{x} = \frac{r \cdot \sin( \theta )}{r \cdot \cos( \theta )} = \tan( \theta ) \tag {5}$$ So you would think: $$\theta = \tan^{-1} \left( \frac{y}{x} \right) \tag {6}$$ But not so fast. There is a problem dealing with the different quadrants: $(-x,-y)$ will give the same answer as $(x,y)$ though they are clearly $\pi$ radians apart, also known as opposite each other. You can solve this with additional tests, which means "if" statements when programming, or you can use a special function called "atan2". Most programming languages and spreadsheets support such a function though it's name might be slightly different. $$\theta = \text{atan2} \left( y , x \right) \tag {7}$$ This function takes care of the quadrant locations and returns the correct angle.
## Imaginary Numbers and the Complex Plane
An imaginary number is a real scalar multiplied by the $\sqrt{-1}$. It's impossible to take the square root of a negative number, you protest. A negative times a negative is positive, and a positive times a positive is also a positive. So how can I multiply something times something and get a negative number? Answer: It takes an imaginary number to do that. As in, "imagine a number that can", not that they don't exist. Well, as much as any Mathematical concept exists. Any such number can have the $\sqrt{-1}$ factored out and then it becomes an ordinary real number. The $\sqrt{-1}$ is usually designated as $i$ by mathematicians and $j$ by electrical engineers. It is not an American/European thing as I said in my first article. This, and all my articles use $i$.
A complex number is the sum of a real number and an imaginary number. So, if $x$ and $y$ are both real, then $x + i y$ is a complex number. It could also be expressed as $x + y i$ since multiplication is commutative. Also by convention, the real part is written before the imaginary part.
The complex plane is a representation of the full set of possible complex numbers. The horizontal axis is used for the real part and the vertical axis is used for the imaginary part. Thus, the point corresponding to the complex number $x+ i y$ is $(x,y)$ on the plane. Very often the variable $z$ is used to denote a complex number.
Now, let's convert that point to polar coordinates: $$z = x + i y = r \cos( \theta ) + i r \sin( \theta ) = r \left[ \cos( \theta ) + i \sin( \theta ) \right] \tag {8}$$ That wasn't so hard.
## Euler's Formula
Now, it's time to bring in Euler's formula, which is: $$e^{i\theta} = \cos( \theta ) + i \cdot \sin( \theta ) \tag {9}$$ Look at that, the right hand side is just like the polar form of the complex number given in (8). Substituting it in: $$z = x + i y = r \cdot e^{i\theta} \tag {10}$$ There is the crux of it. Euler's formula is essentially the conversion of polar coordinates to Cartesian coordinates of a complex number. What is remarkable is that the conversion is actually a genuine exponentiation. A fact that will be exploited for the purposes of this article.
The proper mathematical name for finding the angle of a complex number is "arg", as in: $$\theta = \arg( x + iy ) \tag {11}$$ This is what you should use when doing math. It is also supported in some computing platforms. You may also find it called "angle". When $r>0$, Euler's formula shows it can also be expressed as: \begin{aligned} z &= x + i y = r \cdot e^{i\theta} \\ \ln( z ) &= \ln( x + i y ) = \ln(r) + i \theta \\ \theta &= \frac{ ln( x + i y ) - ln( r ) }{i} = -i \ln \left( \frac{x+iy}{r} \right) = -i \ln \left( \frac{x+iy}{ \sqrt{x^2+y^2} } \right) \end{aligned} \tag {12} This is the complex version of the natural log, which can get confusing so the "arg" name is the preferred one.
Finding the magnitude of a complex value, which is the length of the radius, is expressed mathematically as: $$\| z \| = \sqrt{ x^2 + y^2 } = | r | \tag {13}$$ The double absolute value bars stand for the magnitude of a complex number, while the single absolute value bars are for real values. That's in math. In programming the magnitude is often implemented as "Abs" with no distinction being made.
## Angle Addition Formulas for Sine and Cosine
Let's simplify things by setting $r=1$. Now suppose that $\alpha$ and $\beta$ are two real valued angle measurements. Then: \begin{aligned} e^{i(\alpha + \beta)} &= e^{i\alpha} \cdot e^{i\beta} \\ &= \left[ \cos( \alpha ) + i \sin( \alpha ) \right] \cdot \left[ \cos( \beta ) + i \sin( \beta ) \right] \end{aligned} \tag {14} Let's take a moment and see what happens when we multiply two complex numbers. \begin{aligned} ( a + i b ) ( c + i d ) &= a ( c + i d ) + i b ( c + i d ) \\ &= a c + i a d + i b c + i^2 b d \\ &= ( a c - b d ) + i ( a d + b c ) \\ \end{aligned} \tag {15} Notice that the $i^2$ becomes $-1$. Applying the same steps to (14) results in: \begin{aligned} e^{i(\alpha + \beta)} &= \left[ \cos( \alpha ) + i \sin( \alpha ) \right] \cdot \left[ \cos( \beta ) + i \sin( \beta ) \right] \\ &= \left[ \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) \right ] \\ &+ i \left[ \cos( \alpha ) \sin( \beta ) + \sin( \alpha ) \cos( \beta ) \right ] \end{aligned} \tag {16} But, this is also true from Euler's formula: \begin{aligned} e^{i(\alpha + \beta)} &= \cos( \alpha + \beta ) + i \cdot \sin( \alpha + \beta ) \end{aligned} \tag {17} The right hand side are the Cartesian coordinates. Remember the uniqueness of Cartesian representation? This means that: $$\cos( \alpha + \beta ) = \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) \tag {18}$$ And: $$\sin( \alpha + \beta ) = \cos( \alpha ) \sin( \beta ) + \sin( \alpha ) \cos( \beta ) \tag {19}$$ Those are your angle addition formulas for Sine and Cosine. Pretty neat, huh?
## Circular Dependencies
It is important to note that this is not a derivation of these formulas, merely a confirmation. The derivation chain goes like this.
1) Cosine angle addition formula from distances on the unit circle
2) Sine angle addition formula from the cosine formula
3) Derivatives of Sine and Cosine from the angle addition formulas and limits
4) Taylor series for Sine and Cosine from the Derivatives
5) Euler's formula from Taylor series
6) Cosine and Sine angle addition formulas from Euler's formula
There is no "7) Goto 3".
## Multiplying Two Complex Numbers Polar Style
Suppose we have $z_1$, $z_2$, and $z_3$ where the last one is the product of the first two. $$z_3 = z_1 \cdot z_2 \tag {20}$$ You may have heard the rule: "When you multiply two complex numbers the angles are added and the magnitudes are multiplied." First, let's prove it: \begin{aligned} r_3 e^{i \theta_3 } &= r_1 e^{i \theta_1 } \cdot r_2 e^{i \theta_2 } \\ &= r_1 r_2 \cdot e^{i (\theta_1 + \theta_2) } \end{aligned} \tag {21} Now, let's see if it really works. (Have you any doubts?)
Let $z_1 = 3 + i 4$ and $z_2 = 5 + i 12$. $$z_1 \cdot z_2 = ( 15 - 48 ) + i ( 36 + 20 ) = -33 + i 56 = z_3 \tag {22}$$ \begin{aligned} \| z_1 \| &= \sqrt{ 3^2 + 4^2 } = \sqrt{ 9 + 16 } = \sqrt{ 25 } = 5 \\ \| z_2 \| &= \sqrt{ 5^2 + 12^2 } = \sqrt{ 25 + 144 } = \sqrt{ 169 } = 13 \\ \| z_3 \| &= \sqrt{ (-33)^2 + 56^2 } = \sqrt{ 1089 + 3136 } = \sqrt{ 4225 } = 65 \\ \end{aligned} \tag {23} Yeah, I picked Pythagorean triples for $z_1$ and $z_2$ to make the numbers come out nice. But look, the result is also a Pythagorean triple. The mind boggles. Back to the verification. $5 \cdot 13 = 65$. Check.
Now let's look at the angles, they aren't so clean. Here is a little Gambas code to do the calculations:
Dim z1, z2, z3 As Complex
z1 = 3 + 4i
z2 = 5 + 12i
z3 = z1 * z2
Print "|z1| = "; Abs(z1)
Print "|z2| = "; Abs(z2)
Print "|z3| = "; Abs(z3)
Print
Print "arg(z1) = "; z1.Arg(), Deg(z1.Arg())
Print "arg(z2) = "; z2.Arg(), Deg(z2.Arg())
Print "arg(z3) = "; z3.Arg(), Deg(z3.Arg())
Print
Print "z3 = "; z3
Print
Print "atan2( 4, 3) = "; ATan2(4, 3)
Print "atan2(12, 5) = "; ATan2(12, 5)
Print "atan2(56,-33) = "; ATan2(56, -33)
And these are the results:
|z1| = 5
|z2| = 13
|z3| = 65
arg(z1) = 0.92729521800161 53.130102354156
arg(z2) = 1.17600520709514 67.3801350519596
arg(z3) = 2.10330042509675 120.510237406116
z3 = -33+56i
atan2( 4, 3) = 0.92729521800161
atan2(12, 5) = 1.17600520709514
atan2(56,-33) = 2.10330042509675
It doesn't matter whether radians or degrees, the angles still add up.
## Conclusion
This article should have made you more comfortable with complex numbers and their representation in both Cartesian and Polar form. These concepts are essential to understanding many DSP concepts and the DFT (Discrete Fourier Transform) in particular. The phrase "One complex number can be rotated by multiplying it by another" should make sense. If it doesn't, reread the article and play around with numbers until it does.
## A Dedication
This article is dedicated to a certain Miss MB, who obviously never heeded "Girls aren't supposed to like math." You go, girl!
Gambas is a Linux based (as of this writing) dialect of BASIC development platform, loosely based on (but a huge improvement over) Microsoft's VB5 and VB6. I encourage all programmers, novice to expert, to check it out.
Technical details:
The version in my distro (3.1.1) is way out of date. The latest is 3.12.2
PPA: gambas-team/gambas3
## References
[1] Dawg, Cedron, The Exponential Nature of the Complex Unit Circle
[ - ]
Comment by March 17, 2019
Hi Cedron. In your Polar Coordinates section you wrote: "You can negate r and add π to the angle and get back to the same point." Is it possible to have a negative-valued radius in polar coordinates?
[ - ]
Comment by March 17, 2019
Hi Rick,
I think it is considered in bad form to leave a value like that. Similar to how $\frac{1}{\sqrt{2}}$ should be converted to $\frac{\sqrt{2}}{2}$ and $\frac{1}{1+i}$ should be converted to $\frac{1-i}{2}$.
But for sure (I resisted saying "Absolutely"), otherwise you couldn't say things like: "What does the graph of $r = \sin( \theta )$ look like?"
The essence of the article is (15) and (16) as a way to remember the angle addition formulas in a clutch. You can also think:$$real = real \cdot real - imag \cdot imag$$ $$imag = real \cdot imag + imag \cdot real$$
You can also note that $r = \text{sgn}( x ) \cdot \sqrt{ x^2 + y^2 }$ solves the $\tan^{-1}$ quandary (Added: You still have to deal with $x=0$ separately).
Ced
[ - ]
Comment by March 17, 2019
Hi Cedron. Your March 17 reply makes me think I didn't ask my question is a clear enough way. My question, in different words, is: "If I have a complex number represented in polar form, is it possible for the radius of that polar-form number to be negative?"
[ - ]
Comment by March 17, 2019
I'm sorry for any confusion.
Sure $(-1, \pi/2)$ is just as valid as $( 1, 3\pi/2)$ for the same point.
In order to achieve uniqueness in representation, the typical restriction is $r > 0$ and $0 <= \theta < 2 \pi$ and if $r=0$ make $\theta=0$. In which case the latter one is the "correct" one.
Does that clarify it?
[ - ]
Comment by March 23, 2019
Hi Cedron.
In your last March 17 reply you seem to be saying that a complex number, represented in polar form, can have a negative-valued magnitude. But it seems to me that your Eq. (13) specifies that all complex number magnitudes be positive-only in value.
If a complex number's magnitude represents a "length" on the complex plane, I'm trying to figure out what on earth does a "negative length" mean.
[ - ]
Comment by March 23, 2019
Hi Rick,
For polar coordinates, think of the "r" value as lying on a "r-axis", which is rotated by "$\theta$" radians (or degrees, if you are using that). So you can think of the $\theta$-axis as being wrapped around the unit circle. There is no problem having a negative value for "r" in this situation, and its meaning is clear. Under (4), I specifically state that "r" is chosen as positive by convention. If you choose the negative root, then the appropriate $\theta$ has to be chosen as well.
Magnitudes, on the other hand, are a distance measurement and are therefore non-negative. So, when you calculate the magnitude and arg of a complex number, they can be interpreted as (mapped to) polar coordinates, but they are not the same thing.
"r" is not always a magnitude though. Changing the name of the variable to make the point clearer:
$$z = k e^{i \theta} = -k e^{i (\theta + \pi)}$$
Indeed, any multiple of $i2\pi$ could be added in the exponent on either side and the equation will still be true. All refer to the same point on the complex plane. There is no restriction that says the scalar has to be positive.
In this case, I don't know whether k is positive, negative, or zero. That's why the absolute value bars are there around "r" in (14), and for this example:
$$\|z\| = |k|$$
Due to the lack of uniqueness, when the cartesion form ($x+iy$) is converted to the polar form($re^{i\theta}$) by the equations given in the article, it is "a" conversion, not "the" conversion, and "r" is chosen to be positive by convention, and "$\theta$" as the angle to a positive "r".
[ - ]
Comment by March 20, 2019
Equation 12 doesn't disregard the real part of the logarithm; shouldn't it be $\operatorname{Im} \ln (x+iy)$ ?
[ - ]
Comment by March 20, 2019
Good catch, thanks. I fixed it slightly differently and kept it all under (12) so as to not disturb the numbering.
[ - ]
Comment by January 2, 2023
An overview of all my articles can be found at https://www.dsprelated.com/showarticle/1490.php
This is the second article in the "Fundamentals" section. The next article is:
A Recipe for a Common Logarithm Table
The prior article is:
The Exponential Nature of the Complex Unit Circle
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
Please login (on the right) if you already have an account on this platform.
Otherwise, please use this form to register (free) an join one of the largest online community for Electrical/Embedded/DSP/FPGA/ML engineers:
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# Mean distance between N equidistributed points in a circle
I would like to calculate the mean distance depending on circle shape points,
This is a mean calculating all posible distances between any two points
for N=2, line, there is only 1 distance.
for N=3, triangle, again, there is only 1 distance
for N=4, square, henceforth there are more than 1 distance.. in this case we would have 4 distances for sides and 2 distances in diagonal path, $\sqrt2$, then the mean distance would be
D=(4+2$\sqrt2$)/6=1.138..
for N=30, it would be a "pixeled" circle .. and N=infinity is a circle
How to calculate it for N>4 ? Is there any general formula or can it be derived?
Thanks
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You could use the law of cosines to figure out the lengths of polygon chords... – Guess who it is. Aug 14 '11 at 19:18
Actually, if the circle has unit radius, your example for N=4 has wrong distances: the long distances should be d=2. – leonbloy Aug 14 '11 at 20:03
I assume by "equidistributed" you mean the $n$ points are exactly the sides of a regular $n$-gon, as that's what your examples appear to intimate (although you seem to set the sides of the square of the $n=4$ case all equal to $1$, I will instead assume all points are located on the unit circle). In this case leonbloy's answer is the formula. – anon Aug 14 '11 at 20:16
The average distances among all points must be equal that the average distances from a given point. By geometry: we have that the distance is $d=2 \sin(\theta/2)$, so:
$$\bar d = \frac{2}{N-1} \sum_{k=1}^{N-1} \sin\left(\frac{\pi k}{N}\right)$$
On the limit, $N\to \infty$, you replace the sum by an integral and you get the limit by Christian Blatter: $\bar d \to 4/ \pi$
For example, for $N=30$:
>>> N=30;
>>> r = 2* sin(pi*[1:N-1]/(N));
>>> sum(r)/(N-1)
ans = 1.3159
>>> 4/pi
ans = 1.2732
Update: If instead of having a unit circle (radius 1) we have that the distance among nearest neighbours is 1 (the question is not clear about this, and the example for N=4 only makes sense in this later case), we just divide the above result by $2\; \sin(\pi/N)$. In the limit, $\sin(\pi/N) \to \pi/N$ and so $\bar d \to N \; 2/\pi^2$
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Yes, the later I think Is the right answer, (because at least try to answer the right question!), I mean the real meaning of the question is for distance among nearest neighbours=1, so your answer would be $\bar d = \frac{1}{N-1} \frac{\sum_{k=1}^{N-1} \sin\left(\frac{\pi k}{N}\right)}{\sin(\pi/N)}$ Is it right? (I would like to have enough rep to upvote!) saludos desde Argentina! – Hernán Eche Aug 14 '11 at 23:44
@Hernan (tocayo y compatriota): Yes, that's right. – leonbloy Aug 15 '11 at 1:19
When two points $z_1$, $z_2$ are independently uniformly distributed on the unit circle $S^1$ then their mean distance $\bar d$ is ${4\over\pi}$. To prove this one may keep $z_1$ fixed and let $z_2$ have an angular distance $\phi\in[0,\pi]$ from $z_1$ which is uniformly distributed on $[0,\pi]$. The euclidean distance between these two points is given by $d(z_1,z_2)=2\sin{\phi\over2}$,whence $$\bar d={1\over\pi}\int_0^\pi 2\sin{\phi\over2} \ d\phi={4\over\pi}\ .$$ This can be interpreted as follows: If you have a polygon $P$ with $N\gg 1$ vertices independently and uniformly distributed on $S^1$ then the mean distance between these vertices will be approximately ${4\over\pi}$.
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It also not hard to derive the distribution for the distance: $f(d) = \frac{2}{\pi} \frac{1}{\sqrt{4-d^2}}$ where $0<d<2$. – Sasha Aug 14 '11 at 21:12
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# Open Mathematics
### formerly Central European Journal of Mathematics
Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo
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Volume 16, Issue 1
# Sharp bounds for partition dimension of generalized Möbius ladders
Zafar Hussain
/ Junaid Alam Khan
/ Mobeen Munir
• Corresponding author
• Department of Mathematics, Division of Science and Technology,, University of Education, Lahore, 54000, Pakistan
• Email
• Other articles by this author:
/ Zaffar Iqbal
Published Online: 2018-11-08 | DOI: https://doi.org/10.1515/math-2018-0109
## Abstract
The concept of minimal resolving partition and resolving set plays a pivotal role in diverse areas such as robot navigation, networking, optimization, mastermind games and coin weighing. It is hard to compute exact values of partition dimension for a graphic metric space, (G, dG) and networks. In this article, we give the sharp upper bounds and lower bounds for the partition dimension of generalized Möbius ladders, Mm, n, for all n≥3 and m≥2.
MSC 2010: 05C12; 05C15; 05C78
## 1 Introduction
Computer networks can be modeled on the grounds of graphs, where hosts, servers or hubs can be considered as vertices and edges – as connecting medium between them. Vertex is actually a possible location to find a fault or some damaged devices in a computer network. This idea somehow urged Slater and independently Harary and Meletr in [1] to uniquely recognize each vertex of a graph in a network so that a fault could be controlled in an efficient way. Thus, the basis for notion of locating sets and locating number of graphs came into existence. Since then, the resolving sets have been investigated a lot [1]. The resolving set contributes in various areas such as connected joins in graphs [2], network discovery [35], strategies for the mastermind games [3, 4], applications of pattern recognition, combinatorial optimization, image processing [6], pharmaceutical chemistry and game theory.
Consider a simple, connected graph G, and metric dG:V(G) × V(G) → ℕ∪0, where ℕ is the set of positive integers and dG(x, y) is the minimum number of edges in any path between x and y. Let W = {w1, w2,...,wk} be an ordered set of vertices of G and let v be a vertex of G. The representation r(v|W) of v with respect to W is the k−tuple (d(v, w1),d(v, w2),...,d(v, wk)). If distinct vertices of G have distinct representation with respect to W, then W is called a resolving set of G, see [1]. Such resolving set with minimum cardinality is a basis of G and metric dimension of G, denoted by dim(G) is its cardinality, [7, 8].
Buczkowski et al. established metric dimension of wheel Wn to be $\begin{array}{}⌊\frac{2n+2}{5}⌋\end{array}$ for n ≥ 7 [9], Caceres et. al. [10] found that the metric dimension of fan is $\begin{array}{}⌊\frac{2n+2}{5}⌋\end{array}$ for n ≥ 7 and Tomescu et. al. [11] determined the dimension of Jahangir graphs J2n to be $\begin{array}{}⌊\frac{2n}{3}⌋\end{array}$ for all n ≥ 4.
A particular metric-feature of the family of graphs is independence of metric dimension on the particular element of the family. A connected graph has constant metric dimension if dim(G) = k where kZ+. In [8] Chartrand et. al. proved that a graph has constant metric dimension 1 iff it is a path. In [12] the authors discussed some families of constant meric dimensions. The authors computed metric dimension of wheels in [13] and uni-cyclic graphs in [14]. The authors in [15] computed metric dimension of alpha boron nanotubes. Javaid et. al. computed metric dimension of P(n, 3) and established new results on metric dimension of rotationally-symmetric graph. Murtaza et. al. computed partial results of metric dimension of Möbius ladder in [16] whereas Munir et. al. computed exact and complete results for metric dimension of Möbius Ladders in [17].
A variant of metric dimension of a connected graph is a partition dimension of graph introduced in [19, 20, 21, 22, 23] given as : Let G be a connected graph, a subset SV(G) and a vertex v, distance d(v, S) = min{d(v, x):xS}. If Π = {S1,...St} is an ordered t-partition of V(G), then r(v|Π) = {d(v, S1),...,d(v, St)}is the t-tuple representation of v with respect to Π. If this t-tuple representation of v, r(v|Π)for all vV(G) being all distinct, then this Π is called a resolving partition and the minimum cardinality of such resolving partition is a partition dimension, represented as pd(G).
A natural question may be asked: are partition dimension and metric dimension related in some way? In [20, 21], Cartrand et. al. proved that pd(G) ≤ β(G) + 1 for a non-trivial connected graph G. But in [22, 23], Tomescu et. al. proved that it can be much smaller than the metric dimension. In fact, the authors completed the list of all 23 examples of connected graphs of order n having partition dimensions 2, n − 1 or n. They also gave an example of graphs with finite partition dimension but those which have infinite metric dimension. Recently, Hernando et. al. has proved that there are only 15 families of such type. Tomescu et. al. computed the bounds for the partition dimension of wheel graph in [23]. In [24], the authors computed some bounds for metric and partition dimension of a connected graph. In [25], the authors obtained some sharp bounds for the partition dimension of unicyclic graphs.
Chartrand et al. proved in [22] that if G is a connected graph of order n ≥ 2 then pd(G) = 2 if and only if G is a path, pd(G) = n if and only if G=Knand for n ≥ 5 pd(G) = n − 1 if and only if G is one of the graphs K1,n − 1, Kne, K1 + (K1∪Kn + 2). In [22] Tomescu and Imran studied infinite regular graphs which are generated by tailings of the plane by regular triangles and hexagons. They proved that these graphs have no finite metric bases but their partition dimension is finite and they evaluated this dimension in some cases. In [23], they computed a partition dimension and a connected partition dimension of wheel graphs and showed that $\begin{array}{}n\ge 4,⌈\left(2n{\right)}^{\frac{1}{3}}⌉\le pd\left(G\right)\le 2⌈{n}^{\frac{1}{2}}⌉+1\end{array}$. The following lemma gives a general upper bounds for the partition dimension of a graph of size n.
#### Lemma 1.1.
IfG∣ ≥ 3, then pd(G) ≤ ndiam(G) + 1
The classical Möbius ladder Mn is a cubic circulant graph with an even number of vertices, formed from an n-cycle by adding edges connecting opposite pair of vertices in the cycle, except with two pairs which are connected with a twist, as you can see in the figure:
Fig. 1
This graph has been an active area of research. For instance, [16, 17] give complete results for its metric dimension. In [26] the authors computed a distance labeling of this graph and also introduced its generalization referred to as Möbius ladder. In [27], the authors not only redefined this generalization in a novel way but also computed metric dimension of Mm, n. They also obtained the results of [16, 17] as easy consequences of the results in~[27]. Consider the Cartesian product Pm × Pn of paths Pm and Pn with vertices u1, u2,…,um and v1, v2,…,un, respectively. Take a 180o twist and identify the vertices (u1, v1),(u1, v2),…,(u1, vn) with the vertices (um, vn), (um, vn − 1), …,(um, v1), respectively, and identify the edge ((u1, i), (u1, i + 1)) with the edge ((um, vn + 1 − i), (um, vni)), where 1 ≤ in − 1. What we receive is the generalized Möbius ladder Mm, n. You may observe that we receive the usual Möbius ladder for n = 2 and for any odd integer m ≥ 4. You can see M7,3 in the following figure.
Fig. 2
P7
Fig. 3
P3
Fig. 4
P7 × P3
For brevity we shall use the symbol vij (or simply ij) to represent the vertex (ui, vj) of Mm, n, as you can see in the figure:
Fig. 5
P7 × P3 with complete simple labels
The generalized Möbius ladder obtained from P7 × P3 is:
Fig. 6
M7,3
So the generalized Möbius ladder Mm, n is a non-regular simple connected graph on n(m − 1) vertices. This article deals with the computation of sharp upper bounds and lower bounds for partition and metric dimensions of Mm, n.
## 3 Main results and discussions
In this part we give our main results. We begin with the sharp upper bounds for the partition dimension of Mm, n. Then we move towards the lower bounds.
#### Theorem 3.1.
For m ≥ 3 and n ≥ 2
$3≤pd(Mm,n)≤5,whenn≡1(mod2)andm≡1(mod2),m−n≥44,whenn≡0(mod2)andm≡1(mod2)4whenn≡1(mod2)andm≡0(mod2)5whenn≡0(mod2)andm≡0(mod2),m−n≥4$
At first we compute the upper bounds. We construct a general resolving partition on a case by case basis.
## 3.1 Upper bound
#### Proof
We divide the proof in two cases on the basis of parities of m and n.
Case I. When m and n are of opposite parity
Let Π = {S1, S2, S3, S4} Where S1 = {V1,1}, S2 = {V1,n}, S3 = {V1,2, V1,3,...,
V1,n − 1, V2,1, V2,2,......,V2,n,.....,Vm − 2,1, Vm − 2,2,...,
Vm − 2,n, Vm − 1,2, Vm − 1,3,...,Vm − 1,n} S4 = {Vm − 1,1}. We prove that Πis a resolving partition for Mm, n. To find distance vectors we use two parameters q, i and depending on their different values we divide the entries of distance vectors into four steps.
Step I: Distances of S1 with all vertices of Mm, n.
In this case for each value of q ∈ {1,2,...,n} the parameter i varies from 1 to m − 1. The entries of different vectors are
$d(S1,Vi,q)=i+q−2,1≤i≤12(m+n−2q+1)m+n−q−i,12(m+n−2q+3)≤i≤m−1$
Step II: Distances of S2 with all vertices of Mm, n.
For each value of q ∈ {1,2, ..., n} the parameter i varies from 1 to m - 1 and we get d(S2, Vi, q) = d(S1, Vi, n + 1 − q).
Step III : Distances of S3 with all vertices of Mm, n.
Here for each value of q ∈ {1,2,...,n} the parameter i varies from 1 to m - 1 and we have
$d(S3,Vi,q)=1,ifi=1,q=1,q=n1,ifi=m−1,q=n0,otherwise$
Step IV: Distances of S4 with all vertices of Mm, n.
Here we have two parts
a) For q = 1 , we have
$d(S4,Vi,q)=i+n−1,1≤i≤12(m−n−1)m−1−i,12(m−n+1)≤i≤m−1$
b) For each value of q ∈ {2,..., n} the parameter i varies from 1 to m - 1 and we have d(S4, Vi, q) = d(S1, Vi, n + 2 − q).
These representations are distinct in at least one coordinate. So Π is a resolving partition for Mm, n so clearly pd(Mm, n) ≤ 4.
#### Example
Clearly pd(M9,4) ≤ 4 as the resolving partition for M9,4 is Π = {S1, S2, S3, S4} where S1 = {V1,1}, S2 = {V1,4}, S3 = {V1,2, V1,3, V2,1, V2,2, V2,3, V2,4,.....
,V7,1, V7,2, V7,4, V8,2, V8,3, V8,4}, S4 = {V8,1}.
The representations of different vertices of M9,4 with respect to Π are
$V1,1(0,3,1,4),V1,2(1,2,0,3),V1,3(2,1,0,2),V1,4(3,0,1,1),V2,1(1,4,0,5),V2,2(2,3,0,4),V2,3(3,2,0,3),V2,4(4,1,0,2),V3,1(2,5,0,5),V3,2(3,4,0,5),V3,3(4,3,0,4),V3,4(5,2,0,3),V4,1(3,5,0,4),V4,2(4,5,0,5),V4,3(5,4,0,5),V4,4(5,3,0,4),V5,1(4,4,0,3),V5,2(5,5,0,4),V5,3(5,5,0,5),V5,4(4,4,0,5),V6,1(5,3,0,2),V6,2(5,4,0,3),V6,3(4,5,0,4),V6,4(3,5,0,5),V7,1(5,2,0,1),V7,2(4,3,0,2),V7,3(3,4,0,3),V7,4(2,5,0,4),V8,1(4,1,1,0),V8,2(3,2,0,1),V8,3(2,3,0,2),V8,4(1,4,0,3)$
Case II: when m and n are of same parity: We want to prove that pd(Mm, n) ≤ 5 by constructing a general resolving partition of size 5, for mn ≥ 4 and m, n are of same parity.
#### Proof
Let Π = {S1, S2, S3, S4, S5} where S1 = {V1,1}, S2 = {V1,n} , S3 = {V1,2, V1,3,...,V1,n − 1, V2,1, V2,2,......,V2,n,.....,Vm − 2,1, Vm − 2,2,...,Vm − 2,n, Vm − 1,2, Vm − 1,3,...,Vm − 1,n − 1}, S4 = {Vm − 1,1} , S5 = {Vm − 1,n} .
We prove that Π is a resolving partition for Mm, n. To find distance vectors we use two parameters q , i and depending on their different values we divide the entries of distance vectors into five steps.
Step I: Distances of S1 with all vertices of Mm, n.
In this case for each value of q ∈ {1, 2,..., n} the parameter i varies from 1 to m - 1. The entries of different vectors are
$d(S1,Vi,q)=i+q−2,1≤i≤12(m+n−2q+2)m+n−q−i,12(m+n−2q+4)≤i≤m−1$
Step II: Distances of S2 with all vertices of Mm, n.
For each value of q ∈ {1,2,..., n} the parameter i varies from 1 to m - 1 and we get d(S2, Vi, q) = d(S1, Vi, n + 1 − q).
Step III : Distances of S3 with all vertices of Mm, n.
Here for each value of q ∈ {1,2,..., n} the parameter i varies from 1 to m - 1 and we have
$d(S3,Vi,q)=1,ifi=1,q=1,q=n1,ifi=m−1,q=1,q=n0,otherwise$
Step IV : Distances of S4 with all vertices of Mm, n. Here we have two parts
a) For q = 1 , we have
$d(S4,Vi,q)=i+n−1,1≤i≤12(m−n)m−1−i,12(m−n+2)≤i≤m−1$
b) For each value of q ∈ {2, ..., n} the parameter i varies from 1 to m - 1 and we have d(S4, Vi, q) = d(S1, Vi, n + 2 − q)
Step V : Distances of S5 with all vertices of Mm, n.
Here for each value of ∈ {1,2,..., n} the parameter i varies from 1 to m - 1 and we have d(S5, Vi, q) = d(S4, Vi, n + 1 − q).
These representations are distinct in at least one coordinate. So Π is a resolving partition for Mm, n. Since there is no 4 resolving partition for Mm, n, hence Π is a minimal resolving partition for Mm, n. So partition dimension of Mm, n is 5.
#### Example
The partition dimension of M9,3 is 5. The resolving partition for M9,3 is Π = {S1, S2, S3, S4, S5}. Where
$S1={V1,1}S2={V1,3}S3={V1,2,V2,1,V2,2,V2,3,.....,V7,1,V7,2,V7,3,V8,2,V8,3}S4={V8,1}S5={V8,3}$
The representations of different vertices of M9,3 with respect to Π are
$V1,1(0,2,1,3,1),V1,2(1,1,0,2,2),V1,3(2,0,1,1,2),V2,1(1,3,0,4,2),V2,2(2,2,0,3,3),V2,3(3,1,0,2,4),V3,1(2,4,0,5,3),V3,2(3,3,0,4,4),V3,3(4,2,0,3,5),V4,1(3,5,0,4,4),V4,2(4,4,0,5,5),V4,3(5,3,0,4,4),V5,1(4,4,0,3,5),V5,2(5,5,0,4,4),V5,3(4,4,0,5,3),V6,1(5,3,0,2,4),V6,2(4,4,0,3,3),V6,3(3,5,0,4,2),V7,1(4,2,0,1,3),V7,2(3,3,0,2,2),V7,3(2,4,0,3,1),V8,1(3,1,1,0,2),V8,2(2,2,0,1,1),V8,3(1,3,1,2,0,)$
## 3.2 Lower bound
#### Proof
It is clear that 2 < pd(Mm, n) as it is not a path, [8]. So it is obvious that 3 ≤ pd(Mn, m). □
#### Theorem 3.2.
For m ≥ 3 and n ≥ 2
$2≤β(Mm,n)≤4,whenn≡1(mod2)andm≡1(mod2),m−n≥43,whenn≡0(mod2)andm≡1(mod2)3whenn≡1(mod2)andm≡0(mod2)4whenn≡0(mod2)andm≡0(mod2),m−n≥4$
#### Proof
Proof is just straightforward after taking into account the fundamental inequality between metric and patrtition dimensions. □
## 4 Conclusions and open problems
In this article we have computed sharp upper bounds for the partition dimension of the generalized Möbius ladders and arrive at the following results
#### Theorem 4.1.
For m ≥ 3 and n ≥ 2
$3≤pd(Mm,n)≤5,whenn≡1(mod2)andm≡1(mod2),m−n≥44,whenn≡0(mod2)andm≡1(mod2)4whenn≡1(mod2)andm≡0(mod2)5whenn≡0(mod2)andm≡0(mod2),m−n≥4$
and
#### Theorem 4.2.
For m ≥ 3 and n ≥ 2
$2≤β(Mm,n)≤4,whenn≡1(mod2)andm≡1(mod2),m−n≥43,whenn≡0(mod2)andm≡1(mod2)3whenn≡1(mod2)andm≡0(mod2)4whenn≡0(mod2)andm≡0(mod2),m−n≥4$
At the same time we pose natural open problems regarding the exact values of partition dimension, pd(Mm, n) and β(Mm, n), and sharp lower bounds for this new family of graphs. For further problems about the dimensions of graphs please see [28, 29].
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Accepted: 2018-10-05
Published Online: 2018-11-08
Competing interests The authors declare that they have no competing interests.
Author’s contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1283–1290, ISSN (Online) 2391-5455,
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American Institute of Mathematical Sciences
May 2019, 39(5): 2877-2891. doi: 10.3934/dcds.2019119
Long-time existence of solutions to nonlocal nonlinear bidirectional wave equations
1 Department of Natural and Mathematical Sciences, Faculty of Engineering, Ozyegin University, Cekmekoy 34794, Istanbul, Turkey 2 Faculty of Engineering and Natural Sciences, Sabanci University, Tuzla 34956, Istanbul, Turkey
* Corresponding author: H. A. Erbay
Received July 2018 Published January 2019
We consider the Cauchy problem defined for a general class of nonlocal wave equations modeling bidirectional wave propagation in a nonlocally and nonlinearly elastic medium whose constitutive equation is given by a convolution integral. We prove a long-time existence result for the nonlocal wave equations with a power-type nonlinearity and a small parameter. As the energy estimates involve a loss of derivatives, we follow the Nash-Moser approach proposed by Alvarez-Samaniego and Lannes. As an application to the long-time existence theorem, we consider the limiting case in which the kernel function is the Dirac measure and the nonlocal equation reduces to the governing equation of one-dimensional classical elasticity theory. The present study also extends our earlier result concerning local well-posedness for smooth kernels to nonsmooth kernels.
Citation: H. A. Erbay, S. Erbay, A. Erkip. Long-time existence of solutions to nonlocal nonlinear bidirectional wave equations. Discrete & Continuous Dynamical Systems - A, 2019, 39 (5) : 2877-2891. doi: 10.3934/dcds.2019119
References:
[1] B. Alvarez-Samaniego and D. Lannes, A Nash-Moser theorem for singular evolution equations. Application to the Serre and Green-Naghdi equations, Indiana Univ. Math. Journal, 57 (2008), 97-131. doi: 10.1512/iumj.2008.57.3200. Google Scholar [2] A. Constantin and D. Lannes, The hydrodynamical relevance of the Camassa-Holm and Degasperis-Procesi equations, Arch. Rational Mech. Anal., 192 (2009), 165-186. doi: 10.1007/s00205-008-0128-2. Google Scholar [3] N. Duruk, H. A. Erbay and A. Erkip, Global existence and blow-up for a class of nonlocal nonlinear Cauchy problems arising in elasticity, Nonlinearity, 23 (2010), 107-118. doi: 10.1088/0951-7715/23/1/006. Google Scholar [4] M. Ehrnstrom, L. Pei and Y. Wang, A conditional well-posedness result for the bidirectional Whitham equation, preprint, arXiv: 1708.04551 [math.AP]. Google Scholar [5] H. A. Erbay, S. Erbay and A. Erkip, The Camassa-Holm equation as the long-wave limit of the improved Boussinesq equation and of a class of nonlocal wave equations, Discrete Contin. Dyn. Syst., 36 (2016), 6101-6116. doi: 10.3934/dcds.2016066. Google Scholar [6] T. J. R. Hughes, T. Kato and J. E. Marsden, Well-posed quasi-linear second-order hyperbolic systems with applications to nonlinear elastodynamics and general relativity, Arch. Ration. Mech. Anal., 63 (1977), 273-294. doi: 10.1007/BF00251584. Google Scholar [7] S. Klainerman, Long time behaviour of solutions to nonlinear wave equations, in Proceedings of the International Congress of Mathematicians (Warsaw, 1983), PWN, Warsaw, (1984), 1209–1215. Google Scholar [8] M. Ming, J. C. Saut and P. Zhang, Long-time existence of solutions to Boussinesq systems, SIAM J. Math. Anal., 44 (2012), 4078-4100. doi: 10.1137/110834214. Google Scholar [9] J. C. Saut and L. Xu, The Cauchy problem on large time for surface waves Boussinesq systems, J. Math. Pures. Appl., 97 (2012), 635-662. doi: 10.1016/j.matpur.2011.09.012. Google Scholar [10] J. C. Saut, C. Wang and L. Xu, The Cauchy problem on large time for surface-waves-type Boussinesq systems Ⅱ, SIAM J. Math. Anal., 49 (2017), 2321-2386. doi: 10.1137/15M1050203. Google Scholar [11] M. E. Taylor, Partial Differential Equations II. Qualitative Studies of Linear Equations, 2$^{nd}$ edition, Springer, New York, 2011. doi: 10.1007/978-1-4419-7052-7. Google Scholar
show all references
References:
[1] B. Alvarez-Samaniego and D. Lannes, A Nash-Moser theorem for singular evolution equations. Application to the Serre and Green-Naghdi equations, Indiana Univ. Math. Journal, 57 (2008), 97-131. doi: 10.1512/iumj.2008.57.3200. Google Scholar [2] A. Constantin and D. Lannes, The hydrodynamical relevance of the Camassa-Holm and Degasperis-Procesi equations, Arch. Rational Mech. Anal., 192 (2009), 165-186. doi: 10.1007/s00205-008-0128-2. Google Scholar [3] N. Duruk, H. A. Erbay and A. Erkip, Global existence and blow-up for a class of nonlocal nonlinear Cauchy problems arising in elasticity, Nonlinearity, 23 (2010), 107-118. doi: 10.1088/0951-7715/23/1/006. Google Scholar [4] M. Ehrnstrom, L. Pei and Y. Wang, A conditional well-posedness result for the bidirectional Whitham equation, preprint, arXiv: 1708.04551 [math.AP]. Google Scholar [5] H. A. Erbay, S. Erbay and A. Erkip, The Camassa-Holm equation as the long-wave limit of the improved Boussinesq equation and of a class of nonlocal wave equations, Discrete Contin. Dyn. Syst., 36 (2016), 6101-6116. doi: 10.3934/dcds.2016066. Google Scholar [6] T. J. R. Hughes, T. Kato and J. E. Marsden, Well-posed quasi-linear second-order hyperbolic systems with applications to nonlinear elastodynamics and general relativity, Arch. Ration. Mech. Anal., 63 (1977), 273-294. doi: 10.1007/BF00251584. Google Scholar [7] S. Klainerman, Long time behaviour of solutions to nonlinear wave equations, in Proceedings of the International Congress of Mathematicians (Warsaw, 1983), PWN, Warsaw, (1984), 1209–1215. Google Scholar [8] M. Ming, J. C. Saut and P. Zhang, Long-time existence of solutions to Boussinesq systems, SIAM J. Math. Anal., 44 (2012), 4078-4100. doi: 10.1137/110834214. Google Scholar [9] J. C. Saut and L. Xu, The Cauchy problem on large time for surface waves Boussinesq systems, J. Math. Pures. Appl., 97 (2012), 635-662. doi: 10.1016/j.matpur.2011.09.012. Google Scholar [10] J. C. Saut, C. Wang and L. Xu, The Cauchy problem on large time for surface-waves-type Boussinesq systems Ⅱ, SIAM J. Math. Anal., 49 (2017), 2321-2386. doi: 10.1137/15M1050203. Google Scholar [11] M. E. Taylor, Partial Differential Equations II. Qualitative Studies of Linear Equations, 2$^{nd}$ edition, Springer, New York, 2011. doi: 10.1007/978-1-4419-7052-7. Google Scholar
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2018 Impact Factor: 1.143
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## January 13, 2007
### The First Part of the Story of Quantizing by Pushing to a Point…
#### Posted by Urs Schreiber
…in which the author entertains himself by computing the space of states of a charged particle by pushing its parallel transport forward to a point. Just for fun.
Let
(1)$X$
be a space and let
(2)$\array{ V \\ \downarrow \\ X }$
be vector bundle over $X$ with connection
(3)$\nabla \,.$
Equivalently this means # that we have a locally smoothly trivializable functor
(4)$\mathrm{tra}_{(V,\nabla)} : P_1(X) \to \mathrm{Vect}$
that sends paths in $X$ to the parallel transport along them obtained from the connection $\nabla$.
Quantizing the single particle charged under this bundle with connection consists of a kinematical and of a dynamical aspect:
In this first part of the story we amuse ourselves by just doing the trivial kinematics – but in a nice way.
So, let’s forget the connection immediatey by just looking at constant paths in $X$. I’ll write
(5)$\mathrm{tra}_V : \mathrm{Disc}(X) \to \mathrm{Vect}$
for the above functor restricted to constant paths, ie. to the discrete category over $X$. It does nothing but sending each point in $X$ to the vector space sitting over it – but smoothly so.
We can play the same game on the base space
(6)$\{\mathrm{pt}\}$
that consists of nothing but a single point.
A trivial rank one vector bundle over a point is a functor
(7)$I_{\mathrm{pt}} : \mathrm{Disc}(\{\mathrm{pt}\}) \to \mathrm{Vect}$
that does nothing but sending the single point to the complex numbers:
(8)$I_{\mathrm{pt}} : x \mapsto \mathbb{C} \,.$
I admit that I am presupposing a certain tolerance for fancy-looking trivialities here. But enduring these will pay off eventually.
Using the uniqe functor from $X$ to the point
(9)$p : \mathrm{Disc}(X) \to \mathrm{Disc}(\{\mathrm{pt}\})$
we can pull back the trivial vector bundle over the point to $X$. The result
(10)$I_X := p^* I_{\mathrm{pt}} : \mathrm{Disc}(X) \stackrel{p}{\to} \mathrm{Disc}(\{\mathrm{pt}\}) \stackrel{I_{\mathrm{pt}}}{\to} \mathrm{Vect}$
is the trivial rank one bundle on $X$. This functor simply sends each point of $x$ the typical fiber $\mathbb{C}$:
(11)$I_X : x \mapsto \mathbb{C} \,.$
In as far as any of this is interesting at all, it is for the following simple fact:
a morphism of functors:
(12)$e : I_X \to \mathrm{tra}_V$
is precisely a section of the vector bundle $V$: $e$ is nothing but an assignment
(13)$e : x \mapsto (e_x : \mathbb{C} \tp V_x)$
of a linear map from $\mathbb{C}$ to the fiber $V_x$ for each point $x$. That’s nothing but a choice of vector in each fiber.
So, the space of all such functor morphisms
(14)$\Gamma(V) = \mathrm{Hom}(I_x, \mathrm{tra}_V)$
from the trivial one into the one defining our vector bundle is nothing but the space of sections of $V$.
Since $\Gamma(V)$ is a vector space, and since vector bundles over the point are nothing but vector spaces, I want to think of $\Gamma(V)$ as a vector bundle over the point. So I regard it as a functor
(15)$q(\mathrm{tra}_V) := \mathrm{pt} \mapsto \Gamma(V) \,.$
On top of all these trivialities, I’ll finally allow mysef to think of $\Gamma(V)$ as morphisms from the trivial line bundle on the point into this guy:
(16)$\Gamma(V) \simeq \mathrm{Hom}(I_{\mathrm{pt}}, q(\mathrm{tra}_V)) \,.$
The upshot is that, taken together, we get the isomorphism
(17)$\mathrm{Hom}(p^* I_{\mathrm{pt}}, \mathrm{tra}_V) \simeq \mathrm{Hom}(I_{\mathrm{pt}}, q(\mathrm{tra}_V)) \,.$
If you like, you can convince yourself that this isomorphism of Hom-spaces in indeed natural in both arguments. But this means that pulling back functors from points to $X$
(18)$[\mathrm{Disc}(\{\mathrm{pt}\}),\mathrm{Vect}] \stackrel{p^*}{\to} [\mathrm{Disc}(X),\mathrm{Vect}]$
is the adjoint of taking sections
(19)$[\mathrm{Disc}(\{\mathrm{pt}\}),\mathrm{Vect}] \stackrel{q(\cdot)}{\leftarrow} [\mathrm{Disc}(X),\mathrm{Vect}] \,.$
This, in turn, says that forming the space of sections of $\mathrm{tra}_V$ is the result of pushing $\mathrm{tra}_V$ forward to a point.
Of course that’s neither new nor very deep. But part of a nice story that still needs to be told.
Posted at January 13, 2007 7:16 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1106
Read the post D-Branes from Tin Cans, III: Homs of Homs
Weblog: The n-Category Café
Excerpt: Sections of sections, their pairing and n-disk correlators.
Tracked: January 19, 2007 9:08 AM
Read the post The Globular Extended QFT of the Charged n-Particle: Definition
Weblog: The n-Category Café
Excerpt: Turning a classical parallel transport functor on target space into a quantum propagation functor on parameter space.
Tracked: January 24, 2007 8:05 PM
Read the post Globular Extended QFT of the Charged n-Particle: String on BG
Weblog: The n-Category Café
Excerpt: The string on the classifying space of a strict 2-group.
Tracked: January 26, 2007 2:49 PM
Read the post QFT of Charged n-particle: Chan-Paton Bundles
Weblog: The n-Category Café
Excerpt: Chan-Paton bundles from the pull-push quantization of the open 2-particle.
Tracked: February 7, 2007 9:55 PM
Read the post QFT of Charged n-Particle: Dynamics
Weblog: The n-Category Café
Excerpt: Definition of the dynamics of the charged quantum particle by pull-push along correspondences of path spaces.
Tracked: February 12, 2007 6:39 PM
Read the post QFT of Charged n-Particle: Algebra of Observables
Weblog: The n-Category Café
Excerpt: The algebra of observables as certain endomorphisms of the n-category of sections.
Tracked: February 28, 2007 2:31 AM
Read the post QFT of Charged n-Particle: Sheaves of Observables
Weblog: The n-Category Café
Excerpt: On the concepts of sheaves and nets of algebras of observables in quantum field theory.
Tracked: March 6, 2007 11:36 PM
Read the post The n-Café Quantum Conjecture
Weblog: The n-Category Café
Excerpt: Why it seems that quantum mechanics ought to be the de-refinement of a refined theory which lives in one categorical degree higher than usual.
Tracked: June 8, 2007 6:23 PM
Read the post The Concept of a Space of States, and the Space of States of the Charged n-Particle
Weblog: The n-Category Café
Excerpt: On the notion of topos-theoretic quantum state objects, the proposed definition by Isham and Doering and a proposal for a simplified modification for the class of theories given by charged n-particle sigma-models.
Tracked: January 9, 2008 10:26 PM
Read the post What has happened so far
Weblog: The n-Category Café
Excerpt: A review of one of the main topics discussed at the Cafe: Sigma-models as the pull-push quantization of nonabelian differential cocycles.
Tracked: March 27, 2008 4:46 PM
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# Parameter Scans¶
Note
Specification of parameter scans in version 4.0.0 is different than in earlier versions. In particular old parameter scan simulations will not run in 4.x but as you will see the new way of specifying paramter scans is much simpler and less laborious than in previous implementations.
When building biomedical simulations it is a common practice to explore parameter space to search for optimal solution or to study the robustness of parameter set at hand. In the past researchers have used (or abused) Python to run multiple replicas of the same simulation with different parameter set for each run. Because this approach usually involved writing some kind of Python wrapper on top of existing CC3D code, more often than not it led to hard-to-understand codes which were difficult to share and were hard to access by non-programmers.
Current version of CC3D attempts to solve these issues by offering users ability to create and run parameter scans directly from CC3D GUI’s or from command line. The way in which parameter scan simulation is run is exactly the same as for “regular”, single-run simulation.
Implementation of parameter scans requires users to write simple JSON file with parameter scan specification and replacing actual values in the CC3DML or Python scripts with template markers. Let us look at the example simulation in Demos/ParameterScan/CellSorting. The parameter scan specification file ParameterScanSpecs.json looks as follows:
{
"version":"4.0.0",
"parameter_list":{
"y_dim":{
"values":[65,110,120]
},
"steps":{
"values":[2,3,4,5,6]
},
"MYVAR":{
"values":[0,1,2]
},
"MYVAR1":{
"values":["'abc1,abc2'","'abc'"]
}
}
}
the syntax is fairly simple and if you look closely it is essentially syntax of nested Python dictionaries At the top-level we specify version and parameter_list entries. The latter one stores several entries each for the parameter we wish to scan.. in our example we will be changing parameter y_dim - assigning values from the following list: [65,110,120], parameter steps with values specified by [2,3,4,5,6], MYVAR , that will take values from list [0,1,2] and MYVAR1 , taking values from ["'abc1,abc2'","'abc'"]. As you can see, the values we assign can be either numbers or strings.
Next, we need to indicate which parameters in the CC3DML and Python files are to be replaced with values specified in ParameterScanSpecs.json. Let’s start with analysing CC3DML script:
<CompuCell3D version="4.0.0">
<Potts>
<Dimensions x="100" y="{{y_dim}}" z="1"/>
<Steps>{{steps}}</Steps>
<Temperature>10.0</Temperature>
<NeighborOrder>2</NeighborOrder>
</Potts>
...
</CompuCell3D>
Here in the Potts section we can see two labels that appeared in ParameterScanSpecs.json - {{y_dim}} and {{steps}}. they are surrounded in double curly braces to allow templating engine to make substitutions i.e. {{y_dim}} will be replaced with appropriate value from [65,110,120] list and, similarly, {{steps}} will take values from [2,3,4,5,6].
The remaining two parameters MYVAR and MYVAR1 will be used to make substitutions in Python steppable script:
from cc3d.core.PySteppables import *
MYVAR={{MYVAR}}
MYVAR1={{MYVAR1}}
class CellSortingSteppable(SteppableBasePy):
def __init__(self,frequency=1):
SteppableBasePy.__init__(self,frequency)
def step(self,mcs):
#type here the code that will run every _frequency MCS
global MYVAR
print ('MYVAR=',MYVAR)
for cell in self.cell_list:
if cell.type==self.DARK:
# Make sure ExternalPotential plugin is loaded
cell.lambdaVecX=-0.5 # force component pointing along X axis - towards positive X's
When the parameter scan runs CC3D keeps track of which combinations of parameters to apply at a given moment.
To run parameter scan you need to use paramScan.bat (windows), paramScan.sh (linux) or paramScan.command (osx) run script.
You typically run it like that
paramScan.command --input=<path to the CC3D project file (*.cc3d)> --output-dir=<path to the output folder to store parameter scan results> --output-frequency=<simulation snapshot output frequency> --screenshot-output-frequency=<screenshot output frequency> --gui --install-dir=<CC3D install directory>
for example to run above simulation on OSX one could type
./paramScan.command --input=/Users/m/Demo2/CC3D_4.0.0/Demos/ParameterScan/CellSorting/CellSorting.cc3d --output-dir=/Users/m/CC3DWorkspace/ParameterScanOUtput --output-frequency=2 --screenshot-output-frequency=2 --gui --install-dir=/Users/m/Demo2/CC3D_4.0.0
Note
You may easily run parameter scans in parallel. Simply execute above command from different terminals and CC3D will synchronize multiple instances of paramScan scripts and as a result you will run several simulations in parallel which will come handy once you are scanning many values of parameters
## Using numpy To Specify Parameter Lists¶
In the above example we used simple Python list syntax to specify list of parameters. this works for simple caes but
when you are dealing with a more sophisticated cases when you require e.g. points to be distributed logarithmically then you woudl need to pregenerate such list in external program (e.g. Python console) and copy/paste values into parameter scan file. Fortunately CC3D allows you o use numpy syntax directly in parameter scan specification file:
{
"version":"4.0.0",
"parameter_list":{
"y_dim":{
"code":"np.arange(165,220,3, dtype=int)"
},
"steps":{
"code":"list(range(5,11,1))"
},
"MYVAR":{
"code":"np.linspace(0,2.3, 10)"
},
"MYVAR1":{
"values":["'abc1,abc2'","'abc'"]
}
}
}
The structure of the file looks the same but when we replace values with code we can type actual numpy statement and it will be evaluated by CC3D. Clearly , as shown above, you can mix-and-match which parameters are specified using numpy statement and which ones are specified using simple Python lists.
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# How to find the limit of series? (What should I know?)
There is a couple of limits that I failed to find:
$$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$
and
$$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$
There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.
So my questions are:
• What should I do when I need a limit of infinite sum? (are there any rules of thumb?)
• What theorems or topics from calculus should I know to solve these problems better?
I am new to math and will appreciate any help. Thank you!
• Hint: Geometric Series! – Qi Zhu Oct 29 '17 at 19:24
• I am also new to Mathematics on Stack Exchange, so if you see that I can improve my question, I would appreciate if you left a comment – Nikita Hismatov Oct 29 '17 at 19:24
• Demidovich!!! I've used it in 1977 to prepare Calculus I and after I continued to use it to give private classes to a lot of people, including the girl who became my wife... – Raffaele Oct 29 '17 at 20:58
• Writing the first in binary will result in something interesting. – chx Oct 29 '17 at 21:00
• @NikitaHismatov Once I asked a professor of mine about this. He replied: "Do you think we have a panacea to calculate limits?!" - As it often happens in mathematics, we have methods for some cases but you can always find harder limits for which the previously developed methods do not work. I once thought about the existence of the following limit and asked here, people were able to compute the limit but I'm not sure I really understand what they did. – Billy Rubina Oct 30 '17 at 6:34
$$\color{red}{1+}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \ldots + \frac {1}{2^{n}}$$ Is an example of $$1+x+x^2+x^3+\ldots+x^n$$ Where the ratio between a number $a_n$ and the previous $a_{n-1}$ is constant and is $x$. This is the sum of a geometric progression and it's quite easy to see that its value is $$\frac{1-x^{n+1}}{1-x}$$ When $n\to\infty$ the sum converges only if $|x|<1$ because if so $x^{n+1}\to 0$ and the sum is $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$ In your first example $x=\frac12$ and index $n$ starts from $n=1$ so the sum is $$\sum_{n=1}^{\infty}\left(\frac12\right)^n=\frac{1}{1-\frac12}-1=\color{red}{1}$$ The second one is $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n}}{3^{n}}=\sum_{n=0}^{\infty}\left(-\frac13\right)^n=\frac{1}{1-\left(-\frac13\right)}=\color{red}{\frac34}$$
Hope this is useful
Hint:
All you have to know is the sum of the $n$ first terms of a geometric series, which is a formula from high school: $$1+q+q^2+\dots +q^n=\frac{1-q^{n+1}}{1-q}\qquad (q\ne 1),$$from which we can deduce: $$q^r+q^{r+1}+\dots +q^n=q^r(1+q+\dots+q^{n-r})=q^r\frac{1-q^{n-r+1}}{1-q}=\frac{q^r-q^{n+1}}{1-q}.$$ If $\lvert q\rvert<1$, these sums have limits $\;\dfrac1{1-q}\;$ and $\;\dfrac{q^r}{1-q}\;$ respectively.
What should I do when I need a limit of infinite sum? (are there any rules of thumb?)
In general - no there aren't such rules. There are specific types of series for which it is known how to compute the limit (like the geometric series). There are way more recepies for figuring out whether a series converges or not (finding the limit is much harder). But also here the cookbook is limited. As others pointed out already, your two series are geometric series, who's limit can be computed easily.
Well, that first one should be an instance of a theorem you should definitely put in your list of useful theorems:
If $|a| < 1$:
$$\sum_{i=1}^{\infty} \ a^i = \frac{a}{1-a}$$
And the second one is an instance of the closely related:
If $|a| < 1$:
$$\sum_{i=0}^{\infty} \ a^i = \frac{1}{1-a}$$
If you don't want to remember both, it's ok to remember just one, because the other one can easily be derived from it, e.g:
$$\sum_{i=0}^{\infty} \ a^i = a^0 + \sum_{i=1}^{\infty} \ a^i = 1 + \frac{a}{1-a} = \frac{1-a}{1-a} + \frac{a}{1-a} = \frac{1-a+a}{1-a}= \frac{1}{1-a}$$
• Or, perhaps more simply, $$\sum_{i=1}^{\infty}a^i = a\sum_{i=0}^{\infty}a^i =a\left( \frac{1}{1-a}\right)$$ – Bungo Oct 29 '17 at 19:52
• @Bungo Yes, nice! I always forget about that trick :( – Bram28 Oct 29 '17 at 19:55
the other answers are correct but if you want to see a more "visualize" way(it is actually the same way but you can see it more clearly) $$S=\lim_{n\to\infty}\sum_{i=1}^n\frac {1}{2^{i}}=\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots\\2S=\lim_{n\to\infty}\sum_{i=0}^n\frac {1}{2^{i}}=\frac 1 1 + \frac 1 2 + \frac 1 {4} + \cdots=1+\lim_{n\to\infty}\sum_{i=1}^n\frac {1}{2^{i}}=1+S\\2S-S=\left(1+S\right)-S=1$$ and $$A=\lim_{n\to\infty}\sum_{i=1}^n\frac {(-1)^{i-1}}{3^{i-1}}=1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots \\ 3A=\lim_{n\to\infty}\sum_{i=1}^n\frac {(-1)^{i-1}}{3^{i-2}}=3 - \frac 3 3 + \frac 3 9 - \frac 3 {27} + \cdots\\=3-\left(1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots\right)=3-A\\3-A=3A\implies3=4A\implies\frac34=A$$
Note that you have the sums $\sum_{k = 1}^{\infty} \frac{1}{2^k}$ and $\sum_{k=0}^{\infty} \left(-\frac{1}{3} \right)^k$. Now look up geometric series and try to finish these on your own :)
In case of infinite Geometric Progression,
$$sum = \frac{a}{1-r}$$
you can also derive this from the normal formula using, $n\to \infty$
The thing to know here is that $$|r| <1$$
To explain this, if |r|<1 , ar < a similarly $ar^2 < ar$ and each next term will keep getting smaller and smaller and as $n\to \infty$, $$a_n = ar^{n-1} \to 0$$ therefore, we can successfully find a finite number to which this sequence will converge to as the later on terms are so smaller than the initial ones that they won't even contribute to the sum.
Although if |r|>1 each term will be greater than the previous one, so we will never be able to sum this sequence.
The sums you mention are both examples of geometric series. There are no general ways to sum arbitrary sequences - they typically require some inspiration. Geometric series are often proven geometrically, but the algebraic proof is illustrative.
$$\sum_{n=0}^{m} a^n = S_{m}, a S_{m} = \sum_{n=1}^m a^n = S_{m+1} - 1$$
$$S_m - a S_m = 1 - a^{m+1}$$
$$S_{m} = \frac{1 - a^{m+1}}{1 - a}$$
The other series is relatively straightforward if you know what your goal is:
$$\sum_{n=0}^\infty \frac{(-1)^n}{3^n} = \sum_{n=0}^\infty \frac{1}{9^n} - \frac{1}{3}\sum_{n=0}^\infty \frac{1}{9^n} = \frac{2}{3} \sum_{n=0}^\infty \frac{1}{9^n}$$
They are sums of geometric progressions.
The first it's $$\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.$$
The second it's $$\frac{1}{1+\frac{1}{3}}=\frac{3}{4}.$$
• I downvoted this answer because it does not address the OP's note that "what I am interested in is the method, not just a concrete result." Naming the series is helpful, but something more—like explaining a closed form for the partial sums, or showing that the limit exists and then showing what it must be—is in order. – wchargin Oct 29 '17 at 22:14
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Question
# Compare your answers with those at the back of the book.A pitcher throws a fastball. When the catcher catches it, (4.2)(a) positive work is done(c) the net work is zero(b) negative work is done
Solution
Verified
Step 1
1 of 3
Recall the relationship between the work $W$ and the change in kinetic energy $\Delta K$. We have:
\begin{aligned} W &= \Delta K \\ &= K_f - K_i \end{aligned}
where $K_f$ and $K_i$ are the final and kinetic energies, respectively.
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# Cross-validation and testing for linear regression on small heteroscedastic data sets
I would like to perform simple linear regression on a data set ($y_i = a x_i + b + \epsilon_i$) with $N \approx 50$. However, my residuals $\epsilon_i = \epsilon_i(x_i)$ exhibit heteroscedasticity as well as non-normality.
My first intuition was to perform some kind of robust regression to give less weight to outliers. However I run in a conceptual issue when it comes to cross-validation and testing due to the small sample size. When one has a lot of data, it is straightforward to divide it into train/cross-validation/test sets, estimate the weights on the training set, fixing hyperparameters with the cross-validation set, and evaluating the performance on the test set. However in the case where $N$ is small and residuals have non-constant variance, using 20% of the data (about 10 points) for testing purposes seems like a bad idea since outliers will most likely be overrepresented, skewing the "true" results of the data-generative process.
I have read that leave-one-out (LOO) cross-validation or $k$-folds cross-validation is useful for small data sets, but most discussions leave out the testing part entirely. Only doing cross-validation means that my model will find good hyperparameters but it will not appropriately generalize the data-generative process. What is the usual wisdom in statistics for such a situation?
• What model do you have in mind? What hyperparameters are you trying to estimate? W/ a small dataset, you may need to work with a simpler model. – gung Jun 5 '17 at 19:46
• @gung: I was thinking of using the Huber loss function, which is OLS loss for inliers and absolute value loss for outliers. The hyperparameter I had in mind was the residual threshold at which I start thinking of a data point as an outlier. – physguy Jun 5 '17 at 19:49
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## Algebra 1
F is not a function since $x=1$ has two different values for $y$. G is not a function since $x=0$ has two different values for $y$. H is not a function since $x=6$ has two different values for $y$.
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Introduction to Electric Power
## Introduction to Electric Power
There is a lot of misunderstanding about the difference between electric power and useful electric energy. They were used nearly interchangeably back in the day of all-dispatchable power generation, though even then utilities always charged us for the energy in kilowatt-hours we actually used. With the introduction of intermittent renewables, wind and sun, it has become criticially important to carefully distinguish power from energy, as well as the importance and different effective meanings of capacity factor. We explain each of these, and collect some useful information about the energy density of various fuels and the land area requirements of wind and solar generation. We conclude with brief introduction to capital costs of wind and nuclear power, saving most of the discussion for another article.
### 1 Power and Energy
The fundamental unit of energy is the Joule. One Joule is the energy needed to push a force of one newton though a distance of one meter. One newton is the force necessary to uniformly accelerate a mass of one kg from rest to a velocity of one meter per second in one second. One kilogram is the mass of one liter (1000 cc) of water.
Heat is a form of energy. One calorie is the amount of heat needed to raise the temperature of one gram (1 cc) of water one degree C. At sealevel pure water freezes at 0 Celsius (Centigrade) and boils at 100 Celcius. There are 4.19 Joules in one calorie.
The British Thermal Unit is another unit of energy. One BTU is the amount of energy needed to raise the temperature of one pound (454 g) of water 1 degree Fahrenheit, or 5/9 degree C. One BTU = 1 lbF x 1 cal/gC x 5/9 C/F x 454 g/lb x 4.19 J/cal = 1056.8 J $\approx$ 1055 Joules, the official conversion number. One Quad is one quadrillion BTU – that’s $1{0}^{15}$ BTU or one PBTU. A Quad is 2.93x$1{0}^{11}$ kWh or 293 TWh. The United States generates about 4 PWh electric energy each year, or 13.65 Quad. ConvertWorld.com has a handy energy units converter.
Energy and work are the same thing. Power is different. Power is the rate energy is expended, or how fast you do work. The standard unit of power is the Watt. One watt is one Joule expended each second: W = J/s. We can get units of energy back again by multiplying different numbers of Watts by different lengths of time. The unit we see on our domestic electric bill is the killowatt-hour. 1 kWh = 1 J/s * 1000 * 1h x 3600 s/h = 3.6 million Joules, or 3.6 MJ. A kWh is easy to conceptualize: it’s the amount of heat given off by a good table toaster burning continuously for one hour. One does not usually generate or consume energy at a constant rate (constant power). Usually one consumes a little here, a little more there, then turns in for the night and consumes a bit more in the morning. They all add up: energy is the integral of power over time:
$E={\int }_{{t}_{0}}^{{t}_{f}}P\left(t\right)\phantom{\rule{3.33234pt}{0ex}}dt$
Something to keep in mind: in energy discussions it is common practice to speak in terms of MegaWatt (MW) capacity (one million watts) or GigaWatt (GW, one billion watts.) MW and GW are context-sensitive units of power, and mean different things to different people at different times. A kWh, on the other hand, is a unit of useful work and means just one thing: a kWh is a measure of how much energy you consumed at a time you needed to use it. Not some other energy at some other time. Your power company charges you for the energy they deliver to you at the time you need to use it.
Electric utilities call this “keeping the lights on.”
#### 1.1 Three-Part View of Power Generation
Figure 1 is a simplified cartoon of what power utilities are up against when trying to keep the lights on, and broadly illustrates the three general ranges of load fluctuation.
#### 1.2 Generating Capacity, Capacity factor, and Dispatchability
The rated or “nameplate” capacity of a power plant ${P}_{r}$ is the maximum operational power the plant is designed to continously deliver. Modern thermal plants (coal, nuclear) may have ${P}_{r}$ of one GW or higher. Operating in baseload mode, they may deliver ${P}_{r}$ for months at a time, but not indefinitely. Parts wear out. Coal flues must be cleaned. Most nuclear plants must be shut down to be refueled every other year or so, which can take six to eight weeks. Over time the average power output ${P}_{a}$ of such a plant is less than the rated design power ${P}_{r}$, and the ratio of the two, ${P}_{a}∕{P}_{r}$, is called the plant’s Capacity Factor. It is sometimes useful to rigorously define the capacity factor or (plant factor) over a specific time interval $\left[{t}_{0},{t}_{f}\right]$ where ${t}_{f}-{t}_{0}$ is usually over a year and may span several decades:
${C}_{f}=\frac{{\int }_{{t}_{0}}^{{t}_{f}}P\left(t\right)\phantom{\rule{3.33234pt}{0ex}}dt}{{P}_{r}\left({t}_{f}-{t}_{0}\right)}$
Operating in baseload mode, coal plants can achieve capacity factors of over 95%, vs. 90 - 92% for nuclear. But coal, and Pulverized Coal (PC) in particular, is also operated as load-following plant, which lowers the overall capacity factor of U.S coal to about 64%. While many nuclear plants may be operated as load-followers as well, their high initial construction and finance costs and low fuel cost encourage baseload use as much as possible.
Natural gas plants come in several varieties. The simplest are open-cycle gas turbines (OCGT) wherein hot combustion gas drives a series of one or more gas turbines and is then exhausted to the atmosphere. Modern pulverized coal and open-cycle natural gas plants may achieve thermal efficiencies of 40%: four tenths of the thermal energy in the natural gas is converted to electrical energy for the grid. The rest goes up the flue. Combined-cycle gas turbines (CCGT) add a water boiler after the gas turbines to use remaining heat in the exhaust gas to generate steam for subsequent steam turbines. Combined-cycle plants may reach thermal efficiencies of 60%. Either way, the gas turbine is well-suited to peak-load operation, and open-cycle plants are usually used in this mode. Capacity factors may be 20 - 40%.
For intermediate load-following – and peaker as well – it’s hard to beat a dam-backed hydroelectric plant if one is available. The water turbines at Grand Coulee and Hoover dams can spin up from near stop to full power in less than two minutes. Grand Coulee has a nameplate capacity of 6.8 GW, and generated 21 billion kWh useable electricity in 2008. With 8766 h/y, the Grand Coulee system, given enough water, could have generated 6.8 GW x 1 y x 8766 h/y = 59.6 billion kWh, so it’s plant (capacity) factor was 21/59.6 = 35%. With lower total flow and higher head, Glen Canyon and Hoover dams operate with capacity factors of about 24%.
##### Dispatchability
It is important to realize that thermal plants – coal, gas, and nuclear – and dam-backed hydro, are all dispatchable technologies. Baring planned (or unplanned) shutdowns, they are always available to deliver the amount of power their operators request (up to ${P}_{r}$) when their operators schedule them. Their power is dispatched as needed. That anemic-looking 24% plant factors for Glen Canyon and Hoover is actually extremely useful to the Los Angeles and Las Vegas electric companies needing to supply peak power during the late-afternoon summer maximum draws. Hoover dam can’t supply its full 2 GW rated peak for more than a few days (or weeks) at a stretch, but with foresight and planning its 500 MW average may be dispatched along with baseload coal and nuclear, and intermittent solar and wind, to deliver very reliable power to the grid exactly when the grid customers (you and I) need it. A 24% plant factor or capacity factor for a dispatchable load-following plant has a very specific and very useful meaning.
Capacity factors for intermittent renewables such as wind and solar photovoltaic mean something as well. But in the absence of storage, what intermittent capacity factors means is something quite different than the same term and numbers for dispatchable thermal and hydro.
Wind and solar are non-dispatchable. They are highly available whenever the wind blows and the sun shines – but that is not necessarily when the grid operator most needs them. A modern wind turbine may have a peak capacity of 2 MW, and be sold as a “2 MW capacity” turbine at a rated output wind speed of perhaps 10 m/s: it will deliver 2MW only for wind speeds between 10 and 25 m/s (its cut-out speed), and experiences a smooth decrease in output power as wind speed falls beneath 10 m/s as illustrated in Figure 2.
The capacity factor for non-dispatchable resources is defined as for dispatchable resources:
${C}_{f}$ = (Actual energy out) / (theoretical maximum energy out) as measured over some extended interval of time at the installed location. Due to seasonal variability, the time interval is typically a year or more for wind systems.
At our present low levels of penetration ($\sim$ 3.5% in the United States), our grids and subsidies are set up to absorb all the electric energy each turbine can generate. The resulting capacity factors then reflect only the variability in wind speed, and of course varies with location. Capacity factors currently average about 33% for United States onshore wind. This means a typical 2 MW turbine actually delivers an average of 670 kW over a year of operation, that is the average maximum it can generate. The resulting kWh of (often) useful work are delivered when the wind blows, not necessarily when the grid operator (and his customers) would most like to have them.1
It bears emphasis: the nameplate capacity of any power generator is relevant only within the context of its capacity factor, dispatchability, and intended use.
### 2 Power Economics
#### 2.1 Marginal Utility
Table 1 details U.S. electricity generation by source. Figure 3 shows the United States currently generates about 4 trillion kwh or 4 PWh of electric energy each year, of which 3.46% comes from onshore wind. With natural gas providing 25% of our total power, it is easy to smooth out the load fluctuations introduced by wind intermittency,and save natural gas fuel in the process. The incremental or marginal cost of adding a wind turbine to the grid at such small penetration is essentially just the capital cost for the turbine and any needed transmission, plus ongoing operating and maintenance (O&M) costs.
Coal 37% Natural Gas 30% Nuclear 19% Hydropower 7% Other Renewable 5% Biomass 1.42% Geothermal 0.41% Solar 0.11% Wind 3.46% Petroleum 1% Other Gases $<$ 1%
Table 1: Energy sources and share of electricity generation in 2012
Source: What is U.S. electricity generation by energy source?.
However, if one wished to power an entire grid from wind, one needs to consider both the wind capacity factor and the actual wind patterns within the areal extent of the grid as well. For example, if one were living in an area where wind turbine capacity factors average 33%, one might simply say: “Well, if on average the wind blows only a third of the time, then I just buy three times as many wind turbines and I’m golden.”
Or not. Suppose your grid needed to supply 1 PWh over a year’s time. There are 8766 h/y, so 1PWh = 114 GWy. Again assuming 2MW turbines, you might naively think you’d just need
114GWy/(2MW nameplate watts/turbine * 0.33 obtainable watts/nameplate watt) = 171,000 turbines. Which would be true and that many turbines would deliver the desired 1 PWh over the course of the year.
Oh, wait... which PWh did you say? Desired. Needed. When you and I need the power. Not whenever the wind might decide it can deliver it. As illustrated in Figure 1 above, you and I don’t consume a steady constant 114 GW over a year. Our load fluctuates. Likewise the wind power over any grid (alas, of any size) fluctuates as well, and the two fluctuations rarely match. There will invariably be times when there is not enough wind available to meet electric demand, which must either then be curtailed, or be met by backup and storage. Likewise there will be times when there is more wind available than the grid can absorb, even while replenishing storage. Then some turbines must be feathered, and the overall capacity factor will be reduced beneath its nominal 33%.
Even if you (and your ratepayers) could live with even further reduced efficiency, overbuilding enough wind to meet peak load rather than average isn’t going to help much: it would help only if the wind pattern at one turbine (or farm, or region) did not correlate with that at the next, but we all know that it does. In a recent study of the United States’ PJM Interconnect region (Pennsylvania, New Jersey, Maryland & friends), Budischak et al. found wind patterns would positively correlate out to a distance of about 1000 km (625 miles),2 which is 20% greater than the extent of the entire United Kingdom, though considerably less than that of the United States (New York to Los Angeles is 3,933 km, Miami Fl to Portland Me 2,190 km). See The intermittency statistics of wind power.
It is important to realize that lack of correlation (zero correlation) does not imply negative correlation. Negative correlation between regions is what you would like, as that would mean that if the wind were not blowing in one region, you could be assured that it would be blowing in another.3 In absence of negative correlation, one suspects that any deployment percentage greater than the low-penetration (marginal) capacity factor will result in reduced efficiency (reduced capacity factor and greater cost) and still not keep the lights on without dispatchable backup and/or storage (even greater cost). Statistics alone give no assurance of negative correlation.
Global wind patterns, on the other hand, just might. At least it’s a possibility. At least if the areal extent of the grid greatly exceeds the wind’s positive correlation length. At least in the mid-lattitudes where most of us energy hogs live. In the mid-lattitudes we have prevailing westerlies and trade winds that have served mariners well for at least a millenium, and jet streams too – should they survive global warming in one piece. Which is beginning to look a bit iffy.
Polar jet streams are driven by the temperature difference between the mid-lattitudes and the polar regions. Global warming is not uniform, and the Arctic in particular is warming about twice as fast as the rest. The lower temperature difference results in the northern polar jet stream having slowed from its usual brisk 140 mph clip to a more sedate 100, and has become more meandering (meridonal) in the process.4 Arctic warming may nudge the jet stream a bit northward. What that bodes for future wind patterns is for you and me to speculate, and climatologists to try to predict.5 One might hazard that past perforance may become an increasingly spotty predictor of future behavior, which may have consequence for planning power grids based upon wind and sun.
Speaking of grids and their 1000 km positive wind correlations, even in the U.S. on wind power alone one would need a fair bit of transmission capacity to shunt wind power over the often long distances between where the wind is blowing and where power is needed. You might not need that transmission often, but when you do, you’ll need it bad. Most of the time that transmission would go unused – see Delivering Stable Electricity. Solar Photovoltaic can help some, but not as much as one might think: fixed-angle PV (e.g. solar-on-roof) generates most power over just a few hours either side of noon for a capacity factor of about 20%, and the sun correlates across half the planet. So intermittent renewables schemes turn to load shifting and demand management. Load shifting is just energy storage: hydro, pumped hydro, batteries, compressed air, thermal. Concentrated Solar Thermal (CST) is one way to store solar energy as heat before it is converted to electricity. Any of these add to the cost of an all-renewables grid. Demand-management just means you and I can’t have all the power we think we need whenever we think we need it. Demand management is proposed to work via smart grids that will detect when such things as smart refrigerators and smart heat pumps and smart EV and PHEV batteries really do need to be powered, and when they really don’t. To a certain extent demand-management is a worthwhile (and probably necessary) pursuit. But smart stuff doesn’t come cheap, and won’t happen overnight.
We’ll visit correlations and costs again in Pandora’s Back Pages: Renewable Economic Models. The bottom line is that the requisite thermal (coal, gas, biomass) backup, and battery and hydro storage, are expected to push deep-penetration intermittent renewables electric costs to over half again – and perhaps twice – what one might reasonably expect for the same power from nuclear. And emit up to ten times the CO2.
#### 2.2 Energy Density
Table 2 illustrates the physical – and by extension economic – problems one faces when trying to replace fossil fuels with anything emitting a bit less CO2.
Fuel MJ/kg MJ/liter Natural Uranium in Light Water Reactor 500,000 Uranium enriched to 3.5% in LWR 3,900,000 Natural Uranium in Fast Neutron Reactor 28,000,000 liquid hydrogen 143 10 Hydrogen compressed at 70 MPa 123 5.6 Gasoline / Diesel $\sim$ 46 $\sim$ 36 Coal 24 Boron 58.9 137.8 Methane (1.013 bar, 15C) 55.6 0.0378 Natural gas (STP) 53.6 0.0364 Natural gas compressed to 250 bar (3,600 psi) 53.6 9 Liquid natural gas (-160C) 53.6 22.2 Li-ion battery 0.72 - 0.875 0.9 - 2.63 NiMH battery 0.29 0.5 - 1.0 lead-acid battery 0.17 0.34
Table 2: Energy Density of Fuels
Sources: Energy Density and Energy Content of Selected Fuels and Heat Values of various fuels. Also see Boron: A Better Energy Carrier than Hydrogen?... but pass on the “Boron Decombustion” section. 1 kWh = 3.6 MJ, 1 Pa = 1 N/${m}^{2}$ = 1.45 x $1{0}^{-4}$ psi, 1 bar = $1{0}^{5}$ Pa = 0.987 atm.6 Fast Neutron Reactors are explained in Pandora’s Back Pages.
#### 2.3 Emissions Density
Table 3 shows the metric tonnes (2,200 lb) of carbon dioxide equivalent emitted when one GWh of electric energy is produced by different power generation technologies. If Table 2 suggests how difficult it is to replace fossil fuels, Table 3 illustrates the necessity:
tonnes CO2e/GWh Technology Mean Low High Lignite 1054 790 1372 Coal 888 756 1310 Natural Gas 499 362 891 Solar PV 85 13 731 Biomass 45 10 101 Nuclear 29 2 130 Hydroelectric 26 2 237 Wind 26 6 124
Table 3: Lifecycle GHG Emission Intensity for Different Power Technologies.
“In relation to GHG emissions, each generation method produces GHGs in varying quantities through construction, operation (including fuel supply activities), and decommissioning. Some generation methods such as coal fired power plants release the majority of GHGs during operation. Others, such as wind power and nuclear power, release the majority of emissions during construction and decommissioning. Accounting for emissions from all phases of the project (construction, operation, and decommissioning) is called a lifecycle approach. Normalizing the lifecycle emissions with electrical generation allows for a fair comparison of the different generation methods on a per gigawatt-hour basis. The lower the value, the less GHG emissions are emitted.
Source: WNA Energy Analysis of Power Systems Another oft-cited reference is Life cycle energy and greenhouse gas emissions of nuclear energy: A review, which places nuclear a bit higher at [10, 130] with an average of 65 tonnes/Gwh. Greenhouse gas emissions in the nuclear life cycle: A balanced appraisal looks at three studies placing nuclear at 8, 58, and $>110$. Several factors may influence the disparity: for example, uranium isotope enrichment is electric energy intensive and one will obtain a higher lifecycle GHG value if one assumes fossil generation powered the enrichment rather than nuclear or renewables. Further discussion at WNA: Energy Balances and CO2 Implications, which indicates nuclear energy inputs (mining & milling, conversion, enrichment, fuel fabricaton, plant construction, operation, and decommissioning, and waste management) total between 1.34% and 1.74% of energy output, and lifecycle CO2 emissions range between 3 and 26 tonnes/GWh, depending on how much input energy comes nuclear itself vs. fossil fuels.
#### 2.4 Areal Density
The American Nuclear Society estimates “For a 1000-MW plant, site requirements are estimated as follows: nuclear, 1-4 km2; solar or photovoltaic park, 20-50 km2; a wind field, 50-150 km2; and biomass, 4,000-6,000 k${m}^{2}$.”7
If you don’t think our real estate is valuable, you need a new agent. Site requirements do not include land for mining, so I’m going to skip estimating the total land area required for fossil and nuclear power, and return as time permits. I will say a few words about solar and wind.
As the United States currently generates about 4,000 GWh electricity per year, or 4 PWh, we will roughly estimate the land area required to generate 1PWh from solar power and from wind.
(1 PWh/y) / (8766 h/y) = 114 GW average power.
An excellent reference is Sustainable Energy’s chapters on Solar, and on Wind.
Briefly, the solar irradiance at the earth’s surface is almost an even 1kW per square meter. 1kW is the power incident upon a 1 meter square panel held perpendicular to the sun at noon on a cloudless day, averaged over a year, and not too close to the poles lest the atmosphere pile up. In reality, all days aren’t cloudless, and it isn’t always noon. The “not always noon” part hits fixed-angle solar with a 75% geometry penalty right off the bat. “Into each life a little rain must fall” reduces it still more. One may get effective fixed-angle solar capacity factors by dividing the radiance values from Figure 6.16 on page 46 of the above reference (reproduced as Figure 4 below) by the 1000 W/${m}^{2}$ peak.
Figure 4
You can do a bit better than these horizontal surface values by tilting PV panels a bit towards the sun, and a tracking system can increase values up to 30%. But for now let’s accept 20% as reasonable, and photovoltaic conversion efficiency as 10%. (An efficient PV panel can double that. But they really cost. You can divide by 2 at the end if you like.) So fixed-angle PV can squeeze out about 1kW/${m}^{2}$ x 20% x 10% = 20 W per square meter. 114 GW / 20 W/${m}^{2}$ = 5.7 billion square meters, or 5,700 sq km – a bit less than the area of Delaware. Four times this, 22,800 sq km, could power the whole country with an area the size of West Virginia – if you could store the noon-time excess and balance the intermittency.
Wind power density is a bit less. Most planners plan on between 2 and 7 watts per square meter from wind, which Amanda Adams and Harvard Prof. David Keith find a bit optimistic if one actually wants to build out to the 100 GW levels we’re talking here. Adams and Keith8 think that one turbine in an operating array might actually reduce the wind energy available to the next, and believe wind power densities on the order of 0.5 to 1 W/${m}^{2}$ may be more realistic in a large array. We’ll assume the 1 W/${m}^{2}$ value and just multiply all out fixed-angle Solar PV values by 20. If you think wind power density should be greater than that, divide by whatever factor you feel appropriate. But at 1 W/${m}^{2}$ that’s 114,000 sq km for 1 PWh/y (Ohio), or 456,000 sq km for the country (two thirds of Texas). As we saw earlier, 1PWh = 114 GWy will require 171,000 2 MW turbines at 33% capacity factor, or four times this – 684,000 turbines to (not) power the country. These would be spaced so that on average each turbine took up 2 sq km (or more). But their towers are only 80 - 100 m high and blades span a bit over a hundred meters, so this would still leave a lot of space for our trivial pursuits.
One would not, of course, site all the country’s wind (or solar) in one place. And 2 sq km per turbine (1.4 km or 7/8 mile inter-turbine spacing) is also an average: one could have relatively narrow strip farms aligned roughly perpendicular to expected wind velocity with much smaller inter-turbine spacing. You just wouldn’t stack them very deep. And some areas – notably the prairie states – just have higher quality wind than others. With our relatively large land area, coasts on two major oceans, and low population density (Table 4), the United States is far better suited than some of our friends to experiment with large-scale intermittent deployments.
Country Area sq km Population Density p/km**2 United Kingdom 242,910 63,705,000 262 Germany 357,123 80,493,000 225 China 9,572,900 1,354,040,000 141 United States 9,161,074 316,595,000 35 Canada 9,970,610 35,141,600 4 Australia 7,702,466 22,906,400 3
Table 4: Population densities of a few countries
Source: List of sovereign states and dependent territories by population density
#### 2.5 Relative Quantities of Construction Steel and Concrete
James Conca briefly reviews energy cost estimation in Blowing in the Wind, from which we take the following figure:
A sanity check on the wind values may be obtained from Wind energy: the fastest growing power source:
“The Horns Rev 1 offshore wind farm has 80 2 MW wind turbines, which are 70 m tall and have an estimated lifetime of 20 years. These turbines are made primarily of steel, with high-strength steel foundations. The 28,000 tonnes of steel in the turbines accounts for 79% of all materials used in the wind farm”
Which comes out to 175 tonnes per nameplate MW. “The entire wind farm will produce an estimated 650 GWh a year”, so its capacity factor ${C}_{F}$ = 650 GWh/y / ( 2 x 80 MW x 8766 h/y) = 46%, much better than onshore 33% (if you’re lucky). 175 tonnes/MW / 0.46 = 380 tonnes/MW capacity factor corrected, compared with 460 tonnes/MW onshore, a considerable savings. 175/0.33 = 530 Tonnes/MW, so the offshore turbines use a fair bit more steel/MW than onshore, but more than make up for it with the better capacity factor. Horns Rev 1 uses steel monopile foundations rather than concrete, which is where the extra steel goes. Engineering tradeoff, one might suppose.
Construction materials for Solar Thermal may be highest of all, but vary by the material used for thermal storage. See NEEDS: Final report on technical data, costs, and life cycle inventories of solar thermal power plants, as cited by Peter Lang in Emission Cuts Realities – Electricity Generation Cost and CO2 emissions projections for different electricity generation options for Australia to 2050. We reproduce some of Lang’s numbers in Table 5 below. Note for nuclear that while steel remains about the same as in Figure 5, the amount of concrete is significantly greater. (More study required to discern cause of the disparity.)
Energy Cuts Realities, Table 5 Concrete Steel Source: Wind Onshore 433 116 ISA (2007) p145 Solar Thermal (7.5 hr storage) 1303 415 NEEDS (2008) - Andasol 1, p88 Solar Thermal (18 hr storage) 2606 830 rough calculation (x 2) Nuclear 323 57 ISA (2007) p 46
Table 5: Concrete and Steel Used per Rated MW of Generation (Australia)
Adapted from Martin Nicholson, pers. comm. to Peter Lang (2009)
Note 1 : The wind figures must be increased by a factor of (3 to) 6 to be equivalent to nuclear power per unit of energy. (Capacity factor and economic life: wind: 30%, 25 yr; nuclear: 90%, 50 yr).
Me, I’d just divide by the capacity factors and leave it at that. Yes, wind turbines only last 20 - 25 years. They spin fast and die young. But I suspect come replacement time, their concrete foundations will probably be good for another go or two. As for steel, there your mileage may vary. I don’t know, for example, just how well the steel pylon holds up relative to the nacele and turbine blades. But I doubt come EOL that wind engineers want to replace any more than they have to, and design accordingly.
#### 2.6 Energy Cost
Many myths prevail about the the Price of Power. A good reference is Economics of Nuclear Power, from which the short answer, as seen from Table 6, is that without carbon tax or cap-and-trade, carbon (from coal and natural gas) will continue to be emitted. Even onshore wind is competitive only through subsidies which, since gas is cheaper at balancing load than nuclear, have the knock-on effect of subsidizing gas as well. Some quick apples-to-oranges cost comparisons:
Estimated levelized cost of new generation resources, 2018 U.S. average levelized costs (2011 \$/MWh) for plants entering service in 2018 Plant type Capacity factor % Levelized capital cost Fixed O&M Variable O&M including fuel Transmission investment Total system levelized cost Dispatchable Technologies Conventional Coal 85 65.7 4.1 29.2 1.2 100.1 Advanced Coal 85 84.4 6.8 30.7 1.2 123.0 Adv. Coal + CCS 85 88.4 8.8 37.2 1.2 135.5 Natural Gas-fired Combined Cycle 87 15.8 1.7 48.4 1.2 67.1 Advanced CC 87 17.4 2.0 45.0 1.2 65.6 Adv. CC + CCS 87 34.0 4.1 54.1 1.2 93.4 OCGT 30 44.2 2.7 80.0 3.4 130.3 Adv. OCGT 30 30.4 2.6 68.2 3.4 104.6 Adv. Nuclear 90 83.4 11.6 12.3 1.1 108.4 Geothermal 92 76.2 12.0 0.0 1.4 89.6 Biomass 83 53.2 14.3 42.3 1.2 111.0 Non-Dispatchable Technologies Wind 34 70.3 13.1 0.0 3.2 86.6 Wind-Offshore 37 193.4 22.4 0.0 5.7 221.5 Solar PV 25 130.4 9.9 0.0 4.0 144.3 Solar Thermal 20 214.2 41.4 0.0 5.9 261.5 Hydro 52 78.1 4.1 6.1 2.0 90.3
Regional variation in levelized cost of new generation resources entering service 2018 Range of total system levelized costs (2011 \$/MWh) Plant type Minimum Average Maximum Dispatchable Technologies Conventional Coal 89.5 100.1 118.3 Advanced Coal 112.6 123.0 137.9 Advanced Coal with CCS 123.9 135.5 152.7 Natural Gas-fired Conventional Combined Cycle 62.5 67.1 78.2 Advanced Combined Cycle 60.0 65.6 76.1 Advanced CC with CCS 87.4 93.4 107.5 Conventional Combustion Turbine 104.0 130.3 149.8 Advanced Combustion Turbine 90.3 104.6 119.0 Advanced Nuclear 104.4 108.4 115.3 Geothermal 81.4 89.6 100.3 Biomass 98.0 111.0 130.8 Non-Dispatchable Technologies Wind 73.5 86.6 99.8 Wind-Offshore 183.0 221.5 294.7 Solar PV 112.5 144.3 224.4 Solar Thermal 190.2 261.5 417.6 Hydro 58.4 90.3 149.2
Table 6: EIA: Levelized Costs of New Generation by Source in 2018.
Source: EIA Levelized Cost of New Generation Resources in the Annual Energy Outlook 2013 and IER Levelized Costs of New Electricity Generating Technologies. The levelized costs for dispatchable and non-dispatchable technologies are listed in separate segments, as EIA cautions against their direct comparison.9 For example, The Hidden Costs of Wind Power suggests usable onshore wind energy values between \$150 and \$190/MWh might be more realistic. Similarly, True Cost of Coal Power cites studies indicating external costs should add between \$90 and \$270/MWh to the cost of coal, placing it in the range of offshore wind and solar thermal. Finally, the above levelized capital cost assumes 30-year amortization. Current onshore wind turbines are lasting about 20 years,10 whereas new advanced nuclear has design life of 60+ years. Adjusting these values would increase levelized cost of onshore wind energy by \$35 to \$121.6/MWh, and decrease advanced nuclear by \$41.70 to \$66.70/MWh – nearly half that of onshore wind.
It cannot be overstressed that these levelized costs are marginal costs – the projected 2018 cost of adding a single MWh of each technology to the grid as it existed in 2011 – and may be deceiving when one attempts to scale current levelized cost of renewables to total usable electric costs in scenarios of substantial market penetration. At low penetration levels, each small increment of even intermittent power may be usefully absorbed by the grid. But it doesn’t scale; at some point intermittency and capacity factor begin to matter. In Renewables: The 99.9% solution, Budischak et al. estimate a minimal electric price of 26¢/kWh (\$260/MWh) in a PJM Interconnect model obtaining 99.9% of its power from renewables, mostly wind. Here is a (seemingly) clear case where, if external climate costs were to exclude fossil sources of generation, classical marginal market economics would favor wind and hydro as their immediate replacement, even though eventual cost at the desired low-carbon goal, \$260/MWh, rises to well over twice that of nuclear.
##### 2.6.1 Wind
We saw in Section 2.4 that 684,000 2 MW turbines might generate the 4 PWh the United States need each year. (Or, they might not...) A 2 MW onshore turbine runs about \$3 - \$4 million installed,11 so 684,000 of them will only set us back \$2 – \$2.7 trillion. But we already have 3.6% wind in place, which reduces it to 660,000 turbines at \$1.93 - \$2.6 trillion. Again, that’s if we can store the windy-day surplus or otherwise even out the intermittency. Else their utility becomes somewhat limited. We might expect the requisite backup and transmission and storage to run us some extra.
##### 2.6.2 Nuclear
If we wanted to power our entire grid with nuclear we’d need 456 GW average, of which 20% is already covered by nuclear, so we’d only need an average 365 GW more, or 365 new 1.1GW plants with capacity factor 90%. At \$7.5 billion overnight cost each – what Southern Co. thinks (knock on wood) they are facing for two 1.1GW AP1000 at Vogtle Point – that’s \$2.74 trillion. A bit more than the installed cost of wind, but nuclear is dispatchable, and pretty much a drop-in replacement for all the coal and gas we’d want to drop out and replace. We’d probably need to retain some gas as the 365 average GW won’t quite cover all the peaks. Our 8% - 10% hydro might not quite cover them either; the remainder might most effectively be provided by gas. But nuclear can cover peaker loads as well, when it becomes critical.
Another consideration when comparing nuclear to wind is that new wind turbines are expected to last 20 years. New advanced nuclear plants have design life of 60 years: existing plants built in the 1970s had design life of 40 years and usually receive license extensions of an additional 20 years. That might effect how one thinks of capital costs, and 60 year capital amortization would drop levelized cost of new nuclear energy from \$108/MWh to \$67.
Of course, come replacement time one probably wouldn’t need to replace a wind turbine’s foundation, and possibly not the tower either. As for the nuke, 80 years is pretty far out there. The containment might last longer than that, but it remains to be seen whether early 21st century containment would be suitable for 22nd century needs. Leave that one to the great-grandkids.
We won’t actually deploy that much wind or nuclear “overnight”, either. The cost of each is projected to decrease over time. EIA estimates levelized cost of nuclear electricity will drop from about 11¢/kWh to about 8.5¢/kWh in constant 2009 pennies between 2020 and 2040, wind from 8.5¢ to 7.5¢.12 These low wind LCOE costs are of questionable utility, and EIA itself cautions against using them. The American Tradition Institute suggests a more realistic Levelized Cost of Usable Energy. This includes cost of fossil backup, which pushes onshore wind cost to 15.1¢/kWh if by gas, and 19.2¢/kWh if by coal, not including carbon capture.13
James Conca has run a concise series on energy economics:
Also see Michael Overturf: How The Price For Power Is Set. Levelized costs of UK energy for different technologies are described in DECC: Electricity Generation Costs 2013, pages 17 and 18.
### 3 Conclusions
• Jargon and buzz-words can be deceiving. The Nameplate Capacity of a given power source – its nominal peak power output – is meaningful only within the context of its capacity factor, dispatchability, and intended use.
• Capacity Factor is the average power a plant actually produces over an extended period of time, divided by its nameplate capacity.
• Dispatchability indicates whether or not the plant power can be dispatched by the grid operator when he needs it. Subcategories are dispatchable baseload, dispatchable load following, and dispatchable peaking. All thermal and hydro sources are dispatchable, as is CST (concentrated solar thermal). Some are better suited to follow loads or provide peaker capability than others. Baseload-only plants (some older nukes) must be run at nearly constant power, or be shutdown. Restart can take hours or days. Intermittent renewables such as wind and solar PV are non-dispatchable, unless provided with dedicated storage. (Implementation details can vary.)
• Global warming is urgent, but the economics of deep-penetration renewables are not encouraging. Marginal costs at current low grid penetration (0.11% solar, 3.46% wind) do not even begin to tell the whole story, which will be continued in the following article. Be skeptical.
Please remember, your electric utility is in the business of “keeping the lights on.” Your lights on. When you want them on. At a cost you can afford. It isn’t as easy as it sounds.
1For further details please see TCASE 10: Not all capacity factors are made equal.
3See Electricity production from solar and wind in Germany in 2012. The figures on pages 27 and 28 illustrate the daily fluctuations of German solar and wind power, with frequent days of near-zero production. The figure on page 30 shows that their daily sum does better, but still has days of near-total drop-out that require 94% backup from fossil standby or storage.
4See A Rough Guide to the Jet Stream: what it is, how it works and how it is responding to enhanced Arctic warming. Jennifer Frances has an educational YouTube video.
7From Nuclear Power: A Sustainable Source of Energy. Entergy claims 2.4 km2/GW at Arkansas Nuclear One Station, see A Comparison: Land Use by Energy Source - Nuclear, Wind and Solar. The largest reactors, Areva EPR, takes about 1.6 km2/GW, see What Does Renewable Energy Look Like?.
112012 pricing. See How much do wind turbines cost?
12 Annual Energy Outlook 2013 with Projections to 2040 Figure 80 page 73, or Costs and regulatory uncertainties.
13See The Hidden Costs of Wind Power and The Hidden Costs of Wind Electricity. Update 12/22/2013: I must apologize for including these ATI references; it is difficult to independently assess their accuracy.
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# How to decompose a vector into a sum of subvector with kronecker product?
Define a random $15 \times 15$ square matrix $\bf{A}$ which is full rank. If ${\bf{w}} = vec({\bf{A}})$, I want to know how to decompose $\bf{w}$ by subvector via kronecker product, i.e., ${\bf{w}} = \sum\limits_i {{{\bf{g}}_i}} \otimes {{\bf{g}}_i}$, where ${{\bf{g}}_i}$ is a vector with 15 elements, ${{\bf{w}}}$ is a vector with 225 elements and $\otimes$ is kronecker product. The cardinality of set $\{ {{\bf{g}}_i}\}$ is undetermined.
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# What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions. a. HNO 2 b. CH 3 CO 2 H(HC 2 H 3 O 2 )
### Chemistry: An Atoms First Approach
2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter
Section
### Chemistry: An Atoms First Approach
2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 13, Problem 63E
Textbook Problem
29 views
## What are the major species present in 0.250 M solutions of each of the following acids? Calculate the pH of each of these solutions.a. HNO2b. CH3CO2H(HC2H3O2)
(a)
Interpretation Introduction
Interpretation: The major species present in 0.250M solutions of the given acids and the pH value of these solutions is to be calculated.
Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.
The pH of a solution is calculated by the formula, pH=log[H+]
### Explanation of Solution
Explanation
To determine: The major species present in 0.250M solution of HNO2 and the pH of this solution.
The major species present in 0.250M solution of HNO2 are HNO2 and H2O .
HNO2 is a weak acid. Hence, it does not completely dissociate in water.
The major species present in the given solution of HNO2 are HNO2 and H2O .
The equilibrium constant expression for the stated dissociation reaction is,
Ka=[H+][NO2][HNO2]
HNO2 is a comparatively stronger acid than H2O .
The dominant equilibrium reaction for the given case is,
HNO2(aq)H+(aq)+NO2(aq)
At equilibrium, the equilibrium constant expression is expressed by the formula,
Ka=ConcentrationofproductsConcentrationofreactants
Where,
• Ka is the acid dissociation constant.
The equilibrium constant expression for the given reaction is,
Ka=[H+][NO2][HNO2] (1)
The [H+] is 1×10-2M_ .
The change in concentration of HNO2 is assumed to be x .
The ICE table for the stated reaction is,
HNO2(aq)H+(aq)+NO2(aq)Inititialconcentration0.25000Changex+x+xEquilibriumconcentration0.250xxx
The equilibrium concentration of [HNO2] is (0.250x)M .
The equilibrium concentration of [H+] is xM
(b)
Interpretation Introduction
Interpretation: The major species present in 0.250M solutions of the given acids and the pH value of these solutions is to be calculated.
Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.
The pH of a solution is calculated by the formula, pH=log[H+]
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Question
# If nitrogen gas molecule goes straight up with its rms speed at $$0^o$$C from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of:
A
18 km
B
15 km
C
12.38 km
D
8 km
Solution
## The correct option is C $$12.38$$ kmMolecular mass of Nitrogen molecule=$$14$$ g/molAs nitrogen exists as $${N}_{2}$$=28 g/mol=$$0.028$$ kg/molAlso we know $${ v }_{ rms }=\sqrt { \dfrac { 3RT }{ M } }$$ where R= gas constant=8.31 bar/(K mol)=8.31$$\times{10}^{5}$$ Pa/(K mol)T= temperature=$${0}^{0}$$ C=273 KAlso height $$=\dfrac{{V}^{2}_{rms}}{2g}$$$$=\dfrac { 3\times 8.31\times { 10 }^{ 5 }\times 273 }{ 2\times 9.81\times 0.028 } \\ =12388\quad m=12.38\quad km$$Physics
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Steps (14) The first step is to calculate the root node in order to form the tree structure. The total number of record is 15, as can be seen in Table 6.1 (first column), which goes as follows: (Individual (ID): 1, 2, 3,…, 15). We can use the entropy formula to calculate the root node based on eq. (6.1).
As can be seen clearly in Table 6.1 class column, 11 individuals belong to patient classification and four to healthy. The general status entropy can be calculated based on eq. (6.1):
$equation$
Step (5) Table 6.1 lists the calculation ...
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QUESTION
# What is the molecular formula of magnesium oxide?
A molecular formula is one in which all of the atoms in the compound are listed. For example, if I have the molecular formula C_2H_6, I know that there are two carbon atoms and six hydrogen atoms in the molecule.
However, in the case of , there are no molecules to make a molecular formula. As a result, we're left with the ratio of the to one another, which is the empirical formula. Since the ratio of magnesium to oxygen is 1:1, the formula is MgO.
OK.
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Insert-Max and Delete-Min in WAVL tree
Given a WAVL tree with pointers to its' minimum and maximum, we will add two operations:
Insert-Max: given an element, which is bigger than all other elements in the tree, add it as the right child of the maximum element, and update the max pointer.
Delete-Min: find the successor of the minimum element, delete the minimum element and update the min pointer.
Rebalancing after inserting and deleting elements will be as in regular WAVL tree.
We are given a sorted array of $n$ elements and an empty WAVL tree.
We will do $n$ Insert-Max operations on all elements of the array, from smallest to biggest, and then we will do $n$ Delete-Min operations.
I would like to find the worst cast bound for one operation in the described sequence of operations, which will cover both Insert-Max and Delete-Min.
We know that the amortized cost of a rebalancing operation is WAVL tree is $O(1)$, but I'm not sure what can be said about the worst case.
Any help is appreciated.
• The worst case is probably $O(\log n)$. This is usually the case for balanced binary trees. Presumably the textbook mentions this. – Yuval Filmus Feb 14 '18 at 15:45
• @YuvalFilmus Is that because the rebalancing can go all the way up to the root, and the tree height is bounded by $O(\log n)?$ – Itay4 Feb 14 '18 at 15:47
• Yes, everything is linear in the depth, and the depth is $O(\log n)$. In fact, $O(\log n)$ is an upper bound – it's probably better to state the worst case as $\Theta(\log n)$. – Yuval Filmus Feb 14 '18 at 16:22
• @YuvalFilmus So you are saying that this is true for every balanced binary tree? WAVL has no advantage over AVL here? – Itay4 Feb 14 '18 at 16:33
• The Wikipedia page compares these two data structures. – Yuval Filmus Feb 14 '18 at 17:08
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# MathSciDoc: An Archive for Mathematician ∫
#### TBDmathscidoc:1701.333111
Arkiv for Matematik, 45, (2), 297-325, 2005.12
The topological group $\mathcal{D}^k(\mathbb{S})$ of diffeomorphisms of the unit circle $\mathbb{S}$ of Sobolev class$H$^{$k$}, for$k$large enough, is a Banach manifold modeled on the Hilbert space $H^k(\mathbb{S})$ . In this paper we show that the$H$^{1}right-invariant metric obtained by right-translation of the$H$^{1}inner product on $T_{\rm id}\mathcal{D}^k(\mathbb{S})\simeq H^k(\mathbb{S})$ defines a smooth Riemannian metric on $\mathcal{D}^k(\mathbb{S})$ , and we explicitly construct a compatible smooth affine connection. Once this framework has been established results from the general theory of affine connections on Banach manifolds can be applied to study the exponential map, geodesic flow, parallel translation, curvature etc. The diffeomorphism group of the circle provides the natural geometric setting for the Camassa–Holm equation – a nonlinear wave equation that has attracted much attention in recent years – and in this context it has been remarked in various papers how to construct a smooth Riemannian structure compatible with the$H$^{1}right-invariant metric. We give a self-contained presentation that can serve as a detailed mathematical foundation for the future study of geometric aspects of the Camassa–Holm equation.
@inproceedings{jonatan2005riemannian,
title={Riemannian geometry on the diffeomorphism group of the circle},
author={Jonatan Lenells},
url={http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20170108203620519465920},
booktitle={Arkiv for Matematik},
volume={45},
number={2},
pages={297-325},
year={2005},
}
Jonatan Lenells. Riemannian geometry on the diffeomorphism group of the circle. 2005. Vol. 45. In Arkiv for Matematik. pp.297-325. http://archive.ymsc.tsinghua.edu.cn/pacm_paperurl/20170108203620519465920.
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# Collision Response and points of contact
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Posted 27 February 2012 - 06:57 AM
Hi all,
I'm working on a prototype simulator in which 3d entities move and collide. Collisions can be totally elastic or totally inelastic. The collision detection is demanded to V-Clip which returns the normal to the collision and the closest points (contact points) and features.
I've searched the web for a collision response algorithm and the impulse-based one seems perfectly suited for my purposes. Here started the madness for me, since I've found different formulas and my geometry skills are not that high. I have tried two of formulas I've found buth seems that neither of them work properly. I've searched for ideas about what I'm missing but ended up writing this post. Thus, sorry for the over-abused argument!
The implementation I've tried are based on this approach (in particular the "Code" section, rearranged for Java) and the wikipedia one.
Applying both I have strange behaviours even in basic simple cases. Consider for instance two cubes side by side. One simulate a wall, the other moves again the second with velocity (5,0,0). In a correct setting I would have a final velocity of (-5,0,0) for the moving object. Instead, I have values like (11,0,0) or (-0.8428571428571425, 0.0, 0.0) like I'm messing up things.
I've noticed that, despite correctly returning the faces of the cubes for such simple collision, V-Clip returns a vertex for each face (not the center of the face) as the collision point. Thus the vector offset is calculate w.r.t. a vertex instead of the center of the face. If I select (manually) as contact points the center of the shapes, the algorithm works correctly returning the right values. Since this behaviour seemed strange, I've tried another configuration: I rotated the first cube by 45° x-axis, z-axis so that the nearest point would be a vertex. In this case V-Clip returned the vertex coordinates (almost) and the corresponding point on the face (not a vertex!) with an (almost) coherent final velocity:
starting velocity: (5.9, 0.0, 0.0)
final velocity: (-5.977962556993951, 0.0, 0.0)
Is that the problem? I mean, is the first one a degenerative cases inherited by V-Clip way of working or I just obtained the right values by chance?? in other words, should I calculate the "correct point" to which apply the response? In the attempt to answer such question(s), I attach the Java code for the second implementation (the one based on wikipedia). I've added comments, so that it should be understandable. Inertia tensor is returned by V-Clip in local frame, so I multiply the rotation matrix for the local frame Inertia tensor and then the result is multiplyed per the transpose of the rotation matrix. That should be the right way. In the final part, no operation is executed on the velocity of the second object as it is supposed be a wall (not moving, mass infinite). "Ptree1" and "Ptree2" are two PolyTree data structures for entity A and entity B, which contains features and transformation matrices from local frame to global frames, inertia tensor, center of mass and so on.
Any help, any clue, anything is really appreciated! Thanks in advance and sorry for the long post (I've tried to be as descriptive as possible).
// INERTIA TENSOR1 = rot I_1_body rotT (Baraff)
Matrix3d m1 = new Matrix3d();
Matrix3d rot = new Matrix3d();
ptree1.inertiaTensor(m1);
ptree1.getTransform().get(rot);
Matrix3d rotT = new Matrix3d(rot);
rotT.transpose();
m1.mul(rot, m1);
m1.mul(m1, rotT);
//
// INERTIA TENSOR2 = rot I_2_body rotT (Baraff)
Matrix3d m2 = new Matrix3d();
Matrix3d rot2 = new Matrix3d();
ptree2.inertiaTensor(m2);
ptree2.getTransform().get(rot2);
Matrix3d rotT2 = new Matrix3d(rot2);
rotT.transpose();
m2.mul(rot2, m2);
m2.mul(m2, rotT2);
//
// SUPPOSITIONS:
// PTREE1 - mass 1, velocity (defined previously)
// PTREE2 - mass +inf velocity (0,0,0)
//
// COR and other data
double e = 1;
double ma = 1;
Vector3d vel2 = new Vector3d(); // motionless
double mb = Double.POSITIVE_INFINITY;
//Ra --> contact point offset w.r.t. center (center calculated in the local coordinates system and then translated)
Vector3d center1 = new Vector3d();
ptree1.centerOfMass(center1);
center1.x += ptree1.getTransform().m03;
center1.y += ptree1.getTransform().m13;
center1.z += ptree1.getTransform().m23;
//
Vector3d Ra = new Vector3d(drep.getClosestPair().pnt1); // contact point in global coordinates
Ra.sub(center1); // final offset for A
//Rb --> contact point offset w.r.t. center (center calculated in the local coordinates system and then translated)
Vector3d center2 = new Vector3d();
ptree2.centerOfMass(center2);
center2.x += ptree2.getTransform().m03;
center2.y += ptree2.getTransform().m13;
center2.z += ptree2.getTransform().m23;
//
Vector3d Rb = new Vector3d(drep.getClosestPair().pnt2); // contact point in global coordinates
Rb.sub(center2); // final offset for B
//
// calculate denominator of impulse j
// I_a-1 (Ra cross nrm) cross Ra
m1.invert();
Vector3d nrm = new Vector3d();
nrm.normalize(drep.getClosestPair().nrml); //normal to the collision ---> returned by V-Clip
Vector3d angularVelChangea = new Vector3d();
angularVelChangea.cross(Ra, nrm);
m1.transform(angularVelChangea);
Vector3d vaLinDueToR = new Vector3d();
vaLinDueToR.cross(angularVelChangea, Ra);
// I_b-1 (Rb cross nrm) cross Rb
m2.invert();
Vector3d angularVelChangeb = new Vector3d();
angularVelChangeb.cross(Rb, nrm);
m2.transform(angularVelChangeb);
Vector3d vbLinDueToR = new Vector3d();
vbLinDueToR.cross(angularVelChangeb, Rb);
//
// final denominator following Wikipedia formulation (the dot product is split since it is distributive)
double scalar = 1 / ma + 1 / mb + vaLinDueToR.dot(nrm) + vbLinDueToR.dot(nrm);
//
// calculating numerator following wikipedia ---> -(1 + e) ((v_b + omega_b cross Rb) - (v_a + omega_a cross Ra)) dot nrm
Vector3d vellDiff = new Vector3d();
Vector3d omegaCrossRa = new Vector3d();
omegaCrossRa.cross(rottt, Ra); // rottt is the angular velocity of A in radiants
Vector3d velCopy = new Vector3d(vel); // vel is the velocity of A copied to calcuate its sum with the cross product of angular velocity with Ra!
//NOTHING for vel2 ---> it is motionless!
vellDiff.sub(vel2, velCopy);
double Jmod = -(e + 1) * vellDiff.dot(nrm) / scalar; //here the final j
//
//calculate inpulse for A and B
Vector3d J = new Vector3d(nrm); //set the normal
J.scale(Jmod); // product between normal and j
//
Vector3d ja = new Vector3d(J);
Vector3d jb = new Vector3d(J);
ja.scale(1 / ma); // product the mass
jb.scale(1 / mb); // product the mass
// new velocities
vel.sub(ja);
vel2.add(jb);
### #2BaCaRoZzo Members - Reputation: 100
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Posted 01 March 2012 - 08:10 PM
Ok, I've solved my doubts and problems about the inertia tensor and, most importantly, correctly calculated it.
Basically, V-Clip returns the "second moment of area" which is also called inertia tensor. Such matrix can be converted to the (mass) inertia tensor simply by multiplying it for the factor "m / V" with "m" the mass and "V" the volume of the shape.
Thanks to this correction the velocities updated by the response are at least not over the overall initial momentum but the results still seem incorrect.
Before plaguing you with a long post (which would end unanswered like the previous one) I'll repost the same - stupid - question: since at the denominator of the impulse magnitude appears the offset of the shared contact point from the centers of mass, choosing the wrong contact point would end in a wrong response? I'm still thinking to limit cases such as parallel faces in which any point is exactly at the same (minimal) distance.
Any help is really appreciated. Thanks in advance.
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# Revision history [back]
Are the contents or results in your column I numbers or strings?
-1- Are you sure you enabled Tools > Options > LibreOffice Calc > Enable regular expressions in formulae? Without that NO formula can actually interpret with RegEx. (The parentheses are necessary if you want to define an alternative.)
-2- Always use YYYY-MM-DD if Dates shall also be comparable if given as texts. I am not sure if the otherwise necessary conversions are reliebale.
-3- Did you consider to use SUMPRODUCT() for your purpose? It might relief you of some ticklish issues occurrung with the "Criteria" toy. (Automatic conversions again.)
Are the contents or results in your column I numbers or strings?
-1- Are you sure you enabled Tools > Options > LibreOffice Calc > Enable regular expressions in formulae? Without that NO formula can actually interpret with RegEx. (The parentheses are necessary if you want to define an alternative.)
-5- In a as-if range like $I14:$I: How should Calc decide if you mean $I$1 or $I$1048576 by the single $I ? May I suggest first of all that you read https://ask.libreoffice.org/en/question/131311/ ? Can you upload an example demonstrating your problem? Are the contents or results in your column I numbers or strings? -1- Are you sure you enabled Tools > Options > LibreOffice Calc > Enable regular expressions in formulae? Without that NO formula can actually interpret RegEx. (The parentheses are necessary if you want to define an alternative.) -2- Always use YYYY-MM-DD if Dates shall also be comparable if given as texts. I am not sure if the otherwise necessary conversions are reliebale. -3- Did you consider to use SUMPRODUCT() for your purpose? It might relief you of some ticklish issues occurrung with the "Criteria" toy. (Automatic conversions again.) -4- What shall the SUM() in your forula be goog for? -5- In a an as-if range like $I14:$I: How should Calc decide if you mean $I$1 or$I$1048576 by the single $I ?
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# Help
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A line through the points $(2, -9)$ and $(j, 17)$ is parallel to the line $2x + 3y = 21$. What is the value of $j$?
Guest Dec 7, 2017
#1
+5552
+1
First let's find the slope of the line 2x + 3y = 21
2x + 3y = 21
Subtract 2x from both sides of the equation.
3y = -2x + 21
Divide through by 3 .
y = - $$\frac23$$x + 7
Now we can see that the slope of this line is - $$\frac23$$ . So....
the slope between any two points of a parallel line also = -$$\frac23$$
the slope between (2, -9) and (j, 17) = - $$\frac23$$
$$\frac{17--9}{j-2}\,=\,-\frac23\\~\\ \frac{26}{j-2}\,=\,-\frac23 \\~\\ 26=-\frac23(j-2)\\~\\ -39=j-2\\~\\ -37=j$$
Here's a graph to check this: https://www.desmos.com/calculator/ovina9gch0
hectictar Dec 7, 2017
Sort:
#1
+5552
+1
First let's find the slope of the line 2x + 3y = 21
2x + 3y = 21
Subtract 2x from both sides of the equation.
3y = -2x + 21
Divide through by 3 .
y = - $$\frac23$$x + 7
Now we can see that the slope of this line is - $$\frac23$$ . So....
the slope between any two points of a parallel line also = -$$\frac23$$
the slope between (2, -9) and (j, 17) = - $$\frac23$$
$$\frac{17--9}{j-2}\,=\,-\frac23\\~\\ \frac{26}{j-2}\,=\,-\frac23 \\~\\ 26=-\frac23(j-2)\\~\\ -39=j-2\\~\\ -37=j$$
Here's a graph to check this: https://www.desmos.com/calculator/ovina9gch0
hectictar Dec 7, 2017
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Let $A$ be an $n \times n$ matrix such that the elements in each row and each column are arranged in ascending order. Draw a decision tree, which finds $1$st, $2$nd and $3$rd smallest elements in minimum number of comparisons.
edited | 587 views
A beautiful question on the same concept, https://gateoverflow.in/2766/gate1996_14
This is a two dimensional array of size $4\times4$.
$2$____$5$____$8$____$9$
$4$____$11$___$19$___$21$
$6$____$13$___$24$___$27$
$8$____$15$___$25$___$29$
You can see the elements in each row and each column are arranged in ascending order.
Smallest element : $A[0][0] = 2$
$2$nd Smallest element : $min(A[0][1],A[1][0])= min(5,4)=4$
$3$rd smallest element: Just exclude the element you got as $2$nd smallest$(4)$.... here we can compare $A[2][0],A[0][1]$ no need to compare with $A[0][2]$..... So it depends upon from where you got $2$nd element. You can draw a decision tree...If you got $2$nd best from $A[0][1]$ then what to do & if you get from $A[1][0]$ then what to do.
Any way, time complexity is simply $O(1)$...
The elements are in ascending order. Not in non decreasing order. Clearly they are all distinct in a particular row or column.
edited by
@ahwan
yes, it is fine,
just one doubt , what you want to say by this statement " The elements are in ascending order. Not in non decreasing order. " here non decreasing means also increasing order right ?
Numbers are said to be in ascending order when they are arranged from the smallest to the largest number so that is increasing order, same meaning is not it ?
@Bikram Sir,
increasing: 1 2 3 4
non decreasing: 1 1 2 3
So by ascending order they mean strictly increasing. Otherwise we have to check more elements. Sometimes we may have to check every element.
https://stackoverflow.com/questions/1963474/is-a-non-decreasing-sequence-increasing
@Ahwan if elements are repeated .lets say i have a matrix which contains only 3's.So we should return first second and 3rd minimum as 3? I mean if some element is repeated and that is also minimum then we can have first and second/third min as same?
@rahul sharma 5
"The elements are in ascending order (not in non decreasing order) This line clearly say, they are all distinct in a particular row or column."
But it did not say
not in non decreasing order
I am not sure if it means the same
If all elements are distinct we have to check 1st row and 1st column.So complexity will be O(n2)
Here minimum no of comparison could be 1 , because a[0][0] will be minimum always, now we have to check between a[0][1] and a[1][0]
if all elements are not distinct we have to search all elements of the array
So, minimum comparison could be O(1)
@srestha can we do like this??
A[0][0] is smallest so delete it and rearrange matrix which takes not more than n comparisions.
after rearranging at A[0][0] next element can be found.
like this way total number of comparisions=3(n)=O(n)
(0,0) will be smallest.
First comparison will be in (1,0) and (0,1) indices.
second comparison between (0,2) (1,1) (2,0) indices
@srestha dont we need to do any other comparison except between a[0][1] and a[1][0]?
i mean between a[2][0] and a[0][2]there wont be any comparison?if no,pls tell me why..?
I think it could also be done with young tableau,rt?
https://gateoverflow.in/8148/gate2015-2_31
but also told it should be done in min no. of comparison
Why u write O($n^2$) ?
We can always find consecutive 3 smallest in O(1) time right ?
a[0][0], a[0][1] and a[1][0] these are 3 smallest in some order. All other elements are either greater or equal to them.
Is it correct ?
yes complexity could be less.
But how do u say 3 smallest could be inside a[0][0] , a[0][1] and a[1][0]?
It also could be in between a[0][0], a[0][1], a[0][2] , rt?
@srestha mam
a[0][0] will be first minimum, no doubt in that
But, the 2nd minimum can be found by comparing a[0][1] and a[1][0].
Lets suppose a[0][1] is 2nd minimum then 3rd minimum can be found by comparing a[0][2] and a[1] [0].
And if a[1][0] is 2nd minimum then 3rd minimum can be found by comparing a[2][0] and a[0] [1].
Total Comparisons=2
Each time we are doing only a constant number of comparisons and hence O(1) time complexity.
Can you point out where I am mistaking.
if a[0][1] is 2nd min
3rd min could be in a[1][0],or in a[0][2], or a[2][0].
And then u think if n=3 , then all elements of 1st row and 1st column has to be compare. So, I written O(n2)
@srestha mam
if a[0][1] is second minimum, then third minimum can be a[0][2] or a[1][0]
but not a[2][0] because a[1][0] will be Anyways less than a[2][0].
May be I am mistaking somewhere :/
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# Which th second now in this year
In this challenge you have to find out which nth second it is now in this year, current date and time now. Or in other words, how many seconds have passed since New Year.
An example current Date is (it's not given, you'll have to find the current Date):
March 5, 2021 1:42:44 AM
For which the answer is 5449364 (or optionally 5449365 if you count January 1, 2021 0:00:00 as the first second).
So you should return or output which 'th second it is now in this year. Standard loopholes apply, shortest code wins. You may keep fractional seconds in output optionally.
• And for your example date what is the expected output? Mar 5, 2021 at 6:11
• What do you mean by "th"? Mar 5, 2021 at 8:16
• @pxeger I guess the title should read "Which -th second now in this year", so that the expected answer is something like "it is 1234567th second". Maybe it's better to say "how may seconds have passed in this year". Mar 5, 2021 at 8:21
• For Jan 01 2020 00:00:00, should I output 1 or 0? Also, do you prefer time in UTC, or in user's time zone, or both OK?
– tsh
Mar 5, 2021 at 9:06
• This still doesn't address time zones at all. Mar 6, 2021 at 0:23
# PHP, 27 bytes
<?=time()-strtotime('1/1');
time() // current timestamp in seconds
strtotime('1/1') // timestamp of January 1st
Try it online!
# Python 2, 55 bytes
from time import*
k=86400
print~-gmtime()[7]*k+time()%k
Try it online!
-4 bytes thanks to Wasif
-11 bytes thanks to dingledooper
• 64 bytes switch to python 2 and keeping fractional output is allowed Mar 5, 2021 at 6:13
• 60 bytes Mar 5, 2021 at 6:37
• And another 5 bytes can be saved since the attribute can somehow be indexed. Mar 5, 2021 at 9:10
• I realized this is similar to the AWK verison I did... I think you need to -k on the end since the tm_yday field in the gmtime() call includes the current day. Meaning on Jan 1st, the tm_yday is 1 not 0. So you'll get an extra day's worth of seconds. Mar 6, 2021 at 3:15
• @cnamejj the "~-" before the gmtime() does exactly that :) Mar 6, 2021 at 7:00
# PowerShell, 31 bytes
new-timespan 1/1 (date)|% t*ls*
Try it online!
Too late to post this one
# PowerShell, 71 bytes
$x=date;((($x|% d*r)-1)*86400)+($x.hour*3600)+($x.minute*60)+($x|% s*d) Try it online! Example output: Friday, March 5, 2021 6:20:21 AM 5466021 You just multiply day of year by 86400 (246060) and then minutes by 60, hours by 3600 (60*60) and seconds, and sum up all • Just realized that you need to multiply (day of year - 1) by 86400 and not day of year since the count starts from 1. Mar 5, 2021 at 6:18 • @ManishKundu thanks for the fix Mar 5, 2021 at 6:21 # R, 73 bytes ?=as.double -?as.POSIXct(paste0(substr(t<-Sys.time(),1,4),"-01-01"))-?t Try it online! Returns the index of the current second, in the year that the program is run (currently 2021 at time of posting). The POSIXct class used by R to represent dates conveniently equals the number of seconds since the start of 1970. So, to calculate the index of the current second this year (2021), we just need to subtract the number of seconds between 1 Jan 1970 and 1 Jan 2021, which is 1609455600. This requires only 32 bytes of code. Unfortunately, all of the slickness of the approach above is somewhat lost by the clunkiness of extracting a string to represent 1 Jan of the current year... # Red, 58 bytes d: now prin d/11 - 1 * 86400 +(d/7 * 3600)+(d/8 * 60)+ d/9 Try it online! Explanation: d is of date! datatype. I would normally acces its fields usng path notation like d/yearday to find which date of the year is today. Luckily Red has ordinal accessors for the fields as follows: Index Name 1 date 2 year 3 month 4 day 5 zone 6 time 7 hour 8 minute 9 second 10 weekday 11 yearday 12 timezone 13 week 14 isoweek # JavaScript (V8), 53 bytes print(((n=new Date)-new Date(n.getFullYear(),0))/1e3) Try it online! Prints fractional number of seconds. Adding |0 will make this an integer for 2 more bytes. ## Excel, 20 or 36 bytes This works for just 2021 =(NOW()-44197)*86400 This works for all years. =(NOW()-DATE(YEAR(NOW()),1,1))*86400 • Will this still work in 2022? Mar 9, 2021 at 4:55 • It will not. I suppose I was interpreting this year to mean 2021. Mar 9, 2021 at 14:31 • (that's what I did, initially, but then realised that everyone else interpreted it as 'current year' and so I followed the herd... posting 2 versions like you've done now seems the best option!) Mar 9, 2021 at 14:47 # JavaScript (V8), 46 32 bytes -14 bytes thanks to ophact! _=>(Date.now()-16094592e5)/1e3|0 Try it online! The function gets the timestamp of the current time, i.e. number of milliseconds from the 1st January 1970 new Date().getTime() Then subtract the number of milliseconds passed from 1970 to 2021 1609459200000 And divides the result by 1000 to get the number of seconds. Finally it returns the result after having converted it to unsigned integer through the bitwise OR | (because the division gives a floating point number) The variable _ is never used but passing no argument would cost one byte ()=>. • 16094592e5 and 1e3 as opposed to their full length counterparts seems to save a few bytes – user100690 May 7, 2021 at 17:20 • @ophact you are right thanks! I've never considered to use this notation before. May 8, 2021 at 11:02 • Also save more bytes with Date.now() instead of new Date().getTime() – user100690 May 8, 2021 at 11:32 • @ophact wow that's nice to know! Thanks again! May 8, 2021 at 16:48 # Japt, 19 12 11 bytes K-ÐKi T)zA³ Try it online! does current time(in ms) - january 1 , divide by 1000 and floor. -7 bytes from Shaggy and AZTECC0. -1 byte from Shaggy. • N.z returns floored division for 16 Mar 5, 2021 at 7:51 • 12 bytes Mar 5, 2021 at 8:26 • 11 bytes May 10, 2021 at 15:48 # Javascript (Browser), 163 bytes d=new Date();x=d.getFullYear();y=((Date.UTC(x,d.getMonth(),d.getDate())-Date.UTC(x,0,0))/864e5)-1;alert(y*86400+d.getHours()*3600+d.getMinutes()*60+d.getSeconds()) UTC based day of year calculation thanks to this answer in Stack Overflow, I then tried to golf and use the function # JavaScript (V8), 163 bytes d=new Date();x=d.getFullYear();y=((Date.UTC(x,d.getMonth(),d.getDate())-Date.UTC(x,0,0))/864e5)-1;print(y*86400+d.getHours()*3600+d.getMinutes()*60+d.getSeconds()) Try it online! # T-SQL 55 Bytes, PRINT DATEDIFF(S,STR(YEAR(GETDATE()))+'0101',GETDATE()) DATEDIFF takes difference between two DATETIME types, S makes the difference be in seconds. GETDATE() gets the local datetime as a DATETIME. YEAR takes a DATETIME/DATE argument, returns the year component of the argument. STR converts the YEAR integer to a string. + is also string concatenation in T-SQL. The string in format 'YYYYMMDD' is implicitly converted to a DATETIME with a zero time component in the DATEDIFF function. Results in the number of seconds since the 1st of Jan this Year. # AWK, 47 bytes $0=strftime("%j",r=systime(),1)*(q=86400)+r%q-q
Try it online!
I wasn't sure if the seconds into the year should start from midnight Jan 1 in the local timezone or in UTC. This version works for UTC.
It figures out the julian date (the number of days into the year), converts that to seconds then adds in the fractional day's worth of seconds.
strftime("%j",r=systime(),1)
Gets the julian date, and stores the UNIX epoch in variable r in the process. It add a 3rd parameter on strftime to get the time in UTC.
*(q=86400)
Converts the days to seconds and saves the constant "seconds per day" in for use later.
+r%q
Adds in the seconds corresponding to the HH:MM:SS offset from midnight UTC using modulo math.
-q
And this is annoying 2 characters I couldn't figure out how to drop... Since julian days start with 1, the %j include the current day so we have to subtract a full day.
$0= Finally, assigning the result to $0 means the whole expression is actually a condition, which is always truthy except for at exactly midnight on Jan 1st, so the value is printed.
If an offset in seconds from local time is acceptable, this more straightforward, and slightly shorter code would work.
$0=systime()-mktime(strftime("%Y 1 1 0 0 0")) That one just subtracts the UNIX epoch for Jan 1st at mightnight of the current year from the current epoch time. # Python 2, 79 bytes from datetime import* print (datetime.now()-datetime(2021,1,1)).total_seconds() Try it online! # Python 2, 105 bytes from datetime import* n=datetime.now();print n.timetuple().tm_yday*86400+n.hour*3600+n.minute*60+n.second Try it online! Equal-byte alternative: # Python 2, 105 bytes from datetime import* n=datetime.now();print int(n.strftime('%j'))*86400+n.hour*3600+n.minute*60+n.second Try it online! # Python 2, 84 bytes from datetime import* n=datetime.now();print(n-datetime(n.year,1,1)).total_seconds() Try it online! Thanks to @ovs for -21 bytes • 84 bytes with some datetime arithmetic. – ovs Mar 5, 2021 at 7:01 • @ovs thanks a lot Mar 5, 2021 at 7:02 • 81 bytes with another alias Mar 8, 2021 at 22:04 # Bash, 33 dc<<<date -f- +%s<<<"now 1/1"-p Try it online! ### Explanation "now\n1/1" # newline separated now, Jan-1st dates date -f- +%s<<<"now\n1/1" # with -f, date formats multiple lines as multiple seconds-since-1970/01/01 date -f- +%s<<<"now\n1/1"-p # expands to dc expression to push now-seconds, Jan-1st-seconds, then subtract, then print dc<<<date -f- +%s<<<"now\n1/1"-p # evaluate the dc expression • Why hasn't this got more upvotes? Is it wrong? – user7467 May 8, 2021 at 10:27 • The use of -f is tricky enough for an upvote, but 2 separate dates are shorter: Try it online! And without here-string even shorter: Try it online!. May 26, 2021 at 8:18 • Shorter with a bit more shell power: Try it online! May 26, 2021 at 10:21 # JavaScript, 83 61 bytes _=>(new Date()-new Date(new Date().getFullYear()+'-1-1'))/1e3 Yes, it was possible in the end. Could probably shave off a few more bytes. This code outputs the number of seconds since January 1 of the current year, 2021 at the time of writing. Returns a floating point number. This code does not explicitly output, but running something like console.log(f()) where f is the function I posted, will output the result # Vyxal, 26 bytes kτ‹86400*kḢ3600*+kṀ60*+kṠ+ Try it Online! For some unknown reason, s flag dosen't work • The s flag sums the top of the stack, so you would have to wrap the stack into a list with the W command. Or, you can use the Ṫ flag plus some other stuff to get 22 bytes. May 10, 2021 at 16:00 # Julia 1.0, 54 bytes using Dates f(n=now())=(n-DateTime(year(n))).value/1e3 Try it online! # R, 60 31 bytes -difftime(paste0(substr(t<-Sys.time(),1,4),"-01-01"),t,,'s') using difftime with @Dominic van Essen's new years day replacement method answer is in the following format: Time difference of 11203844 secs TIO_link rdrr.io link a much shorter version: as.double(Sys.time())%%31536000 Try it online! • difftime is a good golf - well done - but the code here doesn't work as-is: you need to change = to <- inside the call to substr so that t isn't interpreted as a named function agument. May 10, 2021 at 21:58 • (I suggest including a link to 'try it online' like this or to 'rdrr.io' like this so you & others can check the code...) May 10, 2021 at 22:00 • Ahhh. you are correct of course. I've tried changing the assignment to '=' and did not copy the correct one. thanks. the byte count was correct though (+1 for <- over = ) May 11, 2021 at 7:28 # IBM/Lotus Notes Formula, 33 bytes @Now-@Date(@Year(@Now);1;1;0;0;0) Date comparisons return seconds by default. Unfortunately @Date requires the seconds otherwise it constructs a date with the current time. No TIO for formula so here are a couple of screenshots: # Wolfram Language (Mathematica), 36 bytes DateDifference[{2021},Now,"Second"] Try it online! # APL (Dyalog Unicode) 18.0, 16 bytes (SBCS) Full program -/20⎕DT↑∘⎕TS¨6 1 Try APL online! 6 1 a 2-element list [6,1] ↑∘⎕TS¨ for each number, take that many elements from the current YMDhmsf list 20⎕DT convert the two dates (the 1-element date is padded with the lowest valid values, i.e. [1,1,0,0,0,0] for the beginning of the 1st of January) to UNIX time (seconds since the beginning of 1970) -/ difference between them # 05AB1E, 37 bytes •ΘÏF•ºS₂+žf<£že<ªOŽª+·*žaŽEU*žb60*žcO Try it online. Explanation: •ΘÏF• # Push compressed integer 5254545 º # Mirror it to 52545455454525 S # Convert it to a list of digits: [5,2,5,4,5,4,5,5,4,5,4,5,2,5] ₂+ # Add 26 to each: [31,28,31,30,31,30,31,31,30,31,30,31,28,31] žf # Push the current month < # Decrease it by 1 £ # Leave that many leading integers from the list že # Push the current day of the month < # Decrease it by 1 as well ª # Append it to the list O # Sum it together (total amount of days thus far) Žª+ # Push compressed integer 43200 · # Double it to 86400 * # And multiply it to the amount of days ža # Push the current hours ŽEU* # Multiply it by compressed 3600 žb # Push the current minutes 60* # Multiply it by 60 žc # Push the current seconds O # Sum all values on the stack together # (after which the result is output implicitly) See this 05AB1E tip of mine (section How to compress large integers?) to understand why •ΘÏF• is 5254545; Žª+ is 43200; and ŽEU is 3600. # Factor, 38 bytes now now start-of-year time- second>> . TIO is too old to have start-of-year, so have a screenshot of running the code in the listener. Link to online documentation for the word: https://docs.factorcode.org/content/word-start-of-year,calendar.html ## How it works: Generate two timestamps: one now, and the other at the beginning of the year. Subtract them and then extract the result # Javascript 25 bytes Reach 38 with the console.log(). console.log(Date.now()/1e3-1609459200) # Nim, 59 bytes import times echo inSeconds now()-($now().year).parse"YYYY"
Try it on the Nim playground! (For some reason, it doesn't work correctly on TIO.)
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In previous lessons, you learnt about the resultant vector in one dimension, we are going to extend this to two dimensions. Cite. Determining if lines are parallel, perpendicular, or neither youtube. Khallil Khallil. Hence resolve u into two vectors, one parallel to v and the other perpendicular to v. I can solve the the first part, just by using the projection formula = (u . Parallel and Perpendicular Planes Main Concept The equation of a plane in can be given as where A , B , C , D are parameters. Sections of 3D Shapes. GeoGebra Team . vector calculator, dot product, orthogonal vectors, parallel vectors, same direction vectors, magnitude,vector angle, Cauchy-Schwarz inequality calculator,orthogonal projection calculator You may occasionally need to find a vector that is perpendicular, in two-dimensional space, to a given vector. When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular. In GeoGebra, you can also do calculations with points and vectors. If they all are 0 or 1, the quaternion is parallel or perpendicular to the original basis vectors. First find the resultant of any two of the vectors to be added. Calculate the coordinates of D. Parametric form Calculate the distance of point A [2,1] from the line p: X = -1 + 3 t Y = 5-4 t Line p has a parametric form of the line equation. a) Two vectors are perpendicular if the dot product between them is zero. It’s just a matter of manipulating the equation. There are three important unit vectors which are commonly used and these are the vectors in the direction of the x, y and z-axes. The calculator will generate a step-by-step explanation on how to obtain the result. This calculator find and plot equations of parallel and perpendicular to the given line and passes through given point. SAVE IMAGE. Book. When calculating the electric flux over a certain area using the formula: $$\Phi=\int_s \vec E \cdot d \vec A$$ Why does the electric field vector have to be parallel to the area vector? Grassmann published in 1844 the "Lineale Ausdehnungslehre". In Unity, cross product is computed by the static method Vector3.Cross(). SAVE IMAGE. Linear Equations In One Variable Algebra And Trigonometry. We will only deal with perpendicular vectors but this procedure works for any vectors. Free practice questions for Precalculus - Determine if Two Vectors Are Parallel or Perpendicular. Example: You can create the midpoint M of two points A and B by entering M = (A + B) / 2 into the Input Bar. Improve this answer. In this explainer, we will learn how to recognize parallel and perpendicular vectors in 2D. Question 1 : Find the unit vector parallel to 3a − 2b + 4c if a = 3i − j − 4k, b = −2i + 4j − 3k, and c = i + 2 j − k. Solution : Construct a quadrilateral with 4 equal sides; Geometry Week 1 Day 2 Learning Goals In linear algebra and functional analysis, a projection is a linear transformation from a vector space to itself such that =.That is, whenever is applied twice to any value, it gives the same result as if it were applied once ().It leaves its image unchanged. their cross product, since this way a plane is defined, which may have only one perpendicular line. When dealing with more than two vectors the procedure is repetitive. Transversal Intersects Parallel Lines. In the diagram below, vectors ⃑ , ⃑ , and ⃑ are all parallel to vector ⃑ and parallel to each other. You got stuck here because for any vector v there are infinitely many vectors whose dot product with v is v. Take any vector u.You can represent it as a sum of t and n, which are orthogonal to each other, and t is parallel with v and n is perpendicular to v.Clearly u.v = t.v.So if t is a unit vector, then you get the desired result no matter what its n component is. Vector calculator. They are parallel if and only if they are different by a factor i.e. Perpendicular Lines in the Coordinate Plane: Quick Exploration. Tell whether the lines for the pair of equations are parallel. Free parallel line calculator online | [email protected] Com. a) the equation of the line through point A(1 , 1) and parallel to vector U. b) the equation of the line through point B(-2 , -3) and perpendicular to vector U. SAVE IMAGE. Activity. Tim Brzezinski. All Educational Materials For Math 241 At University Of South. Online calculator. The cross product of this two vectors gives the general direction of the perpendicular vector to the plane, this is also the direction coefficients of the plane. Let us begin by considering parallel vectors. A vector is a mathematical tool for representing the direction and magnitude of some force. New Resources. since, vectors v and w satisfies such that, w = -3*v. therefore, vectors v and w are parallel. Vector Calculator. I solved this by converting the quaternion to matrix. Let’s look at how the equation is manipulated in order to calculate the equation of a perpendicular line. On by . Slope intercept form calculator with two points emathhelp. SAVE IMAGE. Taking the basis vectors from the matrix, and calculating the dot products between the matrix basis vectors and the original basis vectors. Hi, I am trying to solve the following problem: 1. The dot product will be 0 for perpendicular vectors i.e. None of them yields zero. It will do conversions and sum up the vectors. Since all lines can be described this way, it makes calculating parallel and perpendicular lines simple. GeoGebra Classroom Activities. Determine Whether The Given Vectors Are Orthogonal Parallel Or Neither Calculator. Book. GeoGebra Team. This is true of many physics applications involving force, work and other vector quantities. Calculus Ii Dot Product . Activity. they cross at exactly 90 degrees. As a precursor Descartes and Möbius apply. Though abstract, this definition of "projection" formalizes and generalizes the idea of graphical projection. The point D is such that the line DA is perpendicular to AB, and DC is parallel to AB. You need a second vector not parallel to the first one to find a vector perpendicular to them both, i.e. Two vectors are parallel if they are scalar multiples of one another. Perpendicular vectors have a dot product of zero and are called orthogonal vectors. Watch Queue Queue the angle between them is $\pi/2$ radians. Exploring Intersections of Planes. Enter values into Magnitude and Angle ... or X and Y. Follow answered Jul 28 '15 at 17:23. Includes full solutions and score reporting. 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# least square regression
Ask Question Asked 4 days ago. CFA® And Chartered Financial Analyst® Are Registered Trademarks Owned By CFA Institute.Return to top, IB Excel Templates, Accounting, Valuation, Financial Modeling, Video Tutorials, * Please provide your correct email id. This has been a guide to Least Squares Regression Method and its definition. Least square regression is a method for finding a line that summarizes the relationship between the two variables, at least within the domain of the explanatory variable x. The result window will appear in front of us. Unless all measurements are perfect, b is outside that column space. Let us consider two variables, x & y. The method of least squares. Least squares is sensitive to outliers. The performance rating for a technician with 20 years of experience is estimated to be 92.3. To minimize the influence of outliers, you can fit your data using robust least-squares regression. The usual reason is: too many equations. Sam makes fresh waffle cone mixture for 14 ice creams just in case. The details pertaining to the experience of technicians in a company (in a number of years) and their performance rating is provided in the table below. Linear Least Squares Regression¶ Here we look at the most basic linear least squares regression. And so on this scatter plot here, each dot represents a person. This type of calculation is best suited for linear models. So, when we square each of those errors and add them all up, the total is as small as possible. The least-squares regression method is a technique commonly used in Regression Analysis. The matrix has more rows than columns. The computation mechanism is sensitive to the data, and in case of any outliers (exceptional data), results may tend to majorly affect. Anomalies are values that are too good, or bad, to be true or that represent rare cases. In statistics, ordinary least squares is a type of linear least squares method for estimating the unknown parameters in a linear regression model. To calculate the least squares first we will calculate the Y-intercept (a) and slope of a line(b) as follows –, The regression line is calculated as follows –. Here we discuss the formula to calculate the least-squares regression line along with excel examples. A data model explicitly describes a relationship between predictor and response variables. 4.3 Least Squares Approximations It often happens that Ax Db has no solution. A least-squares regression method is a form of regression analysis which establishes the relationship between the dependent and independent variable along with a linear line. The objective of least squares regression is to ensure that the line drawn through the set of values provided establishes the closest relationship between the values. The least-squares method provides the closest relationship between the variables. Least Squares Fitting. Of all of the possible lines that could be drawn, the least squares line is closest to the set of data as a whole. The Least Squares Regression Line. Given any collection of pairs of numbers (except when all the $$x$$-values are the same) and the corresponding scatter diagram, there always exists exactly one straight line that fits the data better than any other, in the sense of minimizing the sum of the squared errors. Viewed 46 times 0 $\begingroup$ Let's say that a sine-like function of a fixed frequency and zero-mean can only vary in amplitude and offset. Viele übersetzte Beispielsätze mit "least square regression" – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen. Least Squares Regression Equations The premise of a regression model is to examine the impact of one or more independent variables (in this case time spent writing an essay) on a dependent variable of interest (in this case essay grades). Video transcript - [Instructor] Let's say we're trying to understand the relationship between people's height and their weight. Imagine you have some points, and want to have a linethat best fits them like this: We can place the line "by eye": try to have the line as close as possible to all points, and a similar number of points above and below the line. Our aim is to calculate the values m (slope) and b (y-intercept) in the equation of a line : To find the line of best fit for N points: Step 1: For each (x,y) point calculate x2 and xy, Step 2: Sum all x, y, x2 and xy, which gives us Σx, Σy, Σx2 and Σxy (Σ means "sum up"). Let us consider the following graph wherein a set of data is plotted along the x and y-axis. Using the equation, predictions, and trend analyses may be made. This explanation made a lot of sense to me when I read it years ago, and I think it's even better dynamically illustrated with Geometer's Sketchpad. You can learn more from the following articles –, Copyright © 2020. This is why the least squares line is also known as the line of best fit. The least-squares method provides the closest relationship between the dependent and independent variables by minimizing the distance between the residuals, and the line of best fit, i.e., the sum of squares of residuals is minimal under this approach. In Least Square regression, we establish a regression model in which the sum of the squares of the vertical distances of different points from the regression curve is minimized. The least-squares method is one of the most popularly used methods for prediction models and trend analysis. Thus, the least-squares regression equation for the given set of excel data is calculated. It is best used in the fields of economics, finance, and stock markets wherein the value of any future variable is predicted with the help of existing variables and the relationship between the same. CFA Institute Does Not Endorse, Promote, Or Warrant The Accuracy Or Quality Of WallStreetMojo. Step 1: For each (x,y) calculate x2 and xy: Step 2: Sum x, y, x2 and xy (gives us Σx, Σy, Σx2 and Σxy): Here are the (x,y) points and the line y = 1.518x + 0.305 on a graph: Sam hears the weather forecast which says "we expect 8 hours of sun tomorrow", so he uses the above equation to estimate that he will sell. From the 2SLS regression window, select the dependent, independent and instrumental variable. Topic: Square, Statistics. Under trendline options – select linear trendline and select display equation on chart. Insert a scatter graph using the data points. These are plotted on a graph with values of x on the x-axis values of y on the y-axis. What Does Least Squares Regression Mean? When calculated appropriately, it delivers the best results. In the above graph, the blue line represents the line of best fit as it lies closest to all the values and the distance between the points outside the line to the line is minimal (i.e., the distance between the residuals to the line of best fit – also referred to as the sums of squares of residuals). As the name implies, the method of Least Squares minimizes the sum of the squares of the residuals between the observed targets in the dataset, and the targets predicted by the linear approximation. When this is not the case (for example, when relationships between variables are bidirectional), linear regression using ordinary least squares (OLS) no … Three lines are drawn through these points – a green, a red, and a blue line. The n columns span a small part of m-dimensional space. Linear regression is a simple algebraic tool which attempts to find the “best” line fitting 2 or more attributes. Author: Tom Ahlschwede. The result explanation of the analysis is same as the OLS, MLE or WLS method. Least squares is a method to apply linear regression. Interpreting slope of regression line. Linear regression fits a data model that is linear in the model coefficients. The least-squares method relies on establishing the closest relationship between a given set of variables. So what we do is we go to 10 different people, and we measure each of their heights and each of their weights. Least Squares Regression Line. The green line passes through a single point, and the red line passes through three data points. A straight line is drawn through the dots – referred to as the line of best fit. We then apply the nls() function of R to get the more accurate values along with the confidence intervals. A least-squares regression method is a form of regression analysis which establishes the relationship between the dependent and independent variable along with a linear line… The main purpose is to provide an example of the basic commands. Use this sketch to explore the creation of the Least Squares Regression Line. Levenberg-Marquardt algorithm is an iterative method to find local minimums. This idea can be used in many other areas, not just lines. These data points are represented using the blue dots. 6 min read. Let us find the best m (slope) and b (y-intercept) that suits that data. Regression Analysis is a statistical method with the help of which one can estimate or predict the unknown values of one variable from the known values of another variable. Excel tools also provide for detailed regression computations. Standard linear regression models assume that errors in the dependent variable are uncorrelated with the independent variable(s). Active 4 days ago. Since the least squares line minimizes the squared distances between the line and our points, we can think of this line as the one that best fits our data. Linear Regression Introduction. Use the checkboxes to show the slope and intercept of a line. Viele übersetzte Beispielsätze mit "least squares regression" – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen. Use the checkbox to activate the squares for each data point. Syntax. Click on the “ok” button. Using these values, estimate the performance rating for a technician with 20 years of experience. It works by making the total of the square of the errors as small as possible (that is why it is called "least squares"): The straight line minimizes the sum of squared errors. Least Squares Regression Equation Using Excel, The least-squares regression equation can be computed using excel by the following steps –. Technically the outcome need not be continuous, but there are often better forms of regression to use for non-continuous outcomes. The variable which is used to predict the variable interest is called the independent or explanatory variable, and the variable that is being predicted is called the dependent or explained variable. We'll need to provide a initial guess (β β) and, in each step, the guess will be estimated as β+δ β + δ determined by A given regression method will ultimately provide an estimate of β {\displaystyle \beta } , usually denoted β ^ {\displaystyle {\hat {\beta }}} to distinguish the estimate from the true (unknown) parameter value that generated the data. Ordinary Least Squares(OLS) is the most common estimation method for linear models—and that’s true for a good reason. The least squares regression uses a complicated equation to graph fixed and variable costs along with the regression line of cost behavior. Insert a trendline within the scatter graph. The method of least squares is a standard approach in regression analysis to approximate the solution of overdetermined systems (sets of equations in which there are more equations than unknowns) by minimizing the sum of the squares of the residuals made in the results of every single equation. This is done by finding the partial derivative of L, equating it to 0 and then finding an expression for m and c. After we do the math, we are left with these equations: It is assumed that you know how to enter data or read data files which is covered in the first chapter, and it is assumed that you are familiar with the different data types. Scipy's least square function uses Levenberg-Marquardt algorithm to solve a non-linear leasts square problems. The toolbox provides these two robust regression methods: Least absolute residuals (LAR) — The LAR method finds a curve that minimizes the absolute difference of … You can imagine (but not accurately) each data point connected to a straight bar by springs: Be careful! Linear regression analyses such as these are based on a simple equation: Y = a + bX Select two-stage least squares (2SLS) regression analysis from the regression option. These values are represented by the dots in the below graph. A mathematical procedure for finding the best-fitting curve to a given set of points by minimizing the sum of the squares of the offsets ("the residuals") of the points from the curve. But for better accuracy let's see how to calculate the line using Least Squares Regression. Imagine you have some points, and want to have a line that best fits them like this: We can place the line "by eye": try to have the line as close as possible to all points, and a similar number of points above and below the line. However, the blue line passes through four data points, and the distance between the residual points to the blue line is minimal as compared to the other two lines. Login details for this Free course will be emailed to you, This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. A strange value will pull the line towards it. Probability and Statistics > Regression > Interactive Entries > Interactive Demonstrations > Least Squares Fitting--Polynomial. It helps us predict results based on an existing set of data as well as clear anomalies in our data. But for better accuracy let's see how to calculate the line using Least Squares Regression. Assessing the fit in least-squares regression. The sum of the squares of the offsets is used instead of the offset absolute values because this allows the residuals to be treated as a continuous differentiable quantity. Ordinary Least Squares regression is the most basic form of regression. The " least squares " method is a form of mathematical regression analysis used to determine the line of best fit for a set of data, providing a visual … Substituting 20 for the value of x in the formula. OLS chooses the parameters of a linear function of a set of explanatory variables by the principle of least squares: minimizing the sum of the squares of the differences between the observed dependent variable in the given dataset and those predicted by the … This line is referred to as the “line of best fit.”. 8. Least Squares method Now that we have determined the loss function, the only thing left to do is minimize it. The line of best fit is a straight line drawn through a scatter of data points that best represents the relationship between them. We generally start with a defined model and assume some values for the coefficients. The least-squares regression equation for the given set of excel data is displayed on the chart. This is suitable for situations where you have some number of predictor variables and the goal is to establish a linear equation which predicts a continuous outcome. For nonlinear equations, more exhaustive computation mechanisms are applied. The regression line under the Least Squares method is calculated using the following formula –, The slope of line b is calculated using the following formula –, Y-intercept, ‘a’ is calculated using the following formula –. Least squares regression of sine wave. Read here to discover the relationship between linear regression, the least squares method, and matrix multiplication. The regression line show managers and accountants the company’s most cost effective production levels. Hence the term “least squares.”, Let us apply these formulae in the below question –. In the other two lines, the orange and the green, the distance between the residuals to the lines is greater as compared to the blue line. Yum. Least square regression is a method for finding a line that summarizes the relationship between the two variables, at least within the domain of the explanatory variable x. M is the slope of the line and c is the y intercept. The computation mechanism is simple and easy to apply. The difference between the sums of squares of residuals to the line of best fit is minimal under this method. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy, Download Least Squares Regression Excel Template, Cyber Monday Offer - All in One Financial Analyst Bundle (250+ Courses, 40+ Projects) View More, You can download this Least Squares Regression Excel Template here –, Financial Modeling Course (with 15+ Projects), 16 Courses | 15+ Projects | 90+ Hours | Full Lifetime Access | Certificate of Completion. The least-squares method of regression analysis is best suited for prediction models and trend analysis. As long as your model satisfies the OLS assumptions for linear regression, you can rest easy knowing that you’re getting the best possible estimates. Least Squares Regression Line – Lesson & Examples (Video) 2 hr 22 min. But the formulas (and the steps taken) will be very different. 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Steps taken ) will be very different not Endorse, Promote, or bad, be! Of the least squares method for estimating the unknown parameters in a linear model... Method for linear models—and that ’ s true for a technician with years... For the value of x on the x-axis values of y on the chart model coefficients algebraic which. & y equation using excel, the least squares regression when calculated appropriately, it delivers the best (! ”, let us apply these formulae in the dependent variable are with. Learn more from the following graph wherein a set of least square regression points are represented using the,! Values along with excel examples with a defined model and assume some values for the given set of data well! Excel data is plotted along the x and y-axis a guide to least squares regression equation least square regression... Need not be continuous, but there are more equations than unknowns ( m is greater than n.! Is linear in the below question – method is a technique commonly used in many other,... The steps taken ) will be very different calculated appropriately, it the! Interactive Demonstrations > least squares regression of sine wave “ best ” line Fitting or... Is plotted along the x and y-axis that is least square regression in the below graph the model coefficients using equation. Y-Intercept ) that suits that data Levenberg-Marquardt algorithm to solve a non-linear leasts square problems excel by the in. Square regression '' – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen to do least square regression that space... And statistics > regression > Interactive Demonstrations > least squares method for linear models ( y-intercept ) that suits data. -- Polynomial local minimums apply the nls ( ) function of R to get the accurate... Equations than unknowns ( m is greater than n ) can fit your data using robust least-squares regression using... Performance rating for a good reason equation using excel by the dots in the dependent, independent instrumental! Between a given set of data as well as clear anomalies in our data see how to do it Deutsch-Übersetzungen. Two variables, x & y as these are plotted on a graph with values of y on the.! Unknowns ( m is greater than n ) Fitting 2 or more attributes excel, the least-squares regression equation the. Wls method a technician with 20 years of experience is estimated to be true or represent! Select the dependent, independent and instrumental variable x in the dependent, independent and instrumental variable of in! Accuracy or Quality of WallStreetMojo best results standard linear regression, the least-squares equation..., each dot represents a person R to get the more accurate values along with the intervals. Through a scatter of data points that best represents the relationship between.... Parameters in a linear regression analyses such as these are based on existing. Lines are drawn through a scatter of data as well as clear anomalies our! To a straight line drawn through the dots in the below question – find local minimums attempts find. To provide an example to see how to calculate the line of best fit. ” © 2020 the most used! Data model that is linear in the model coefficients use this sketch to explore the creation of least... By the dots in the below graph OLS ) is the most basic linear least squares regression a. Get the more accurate values along with the confidence intervals equation: y = a + bX squares. S true for a technician with 20 years of experience the coefficients method is of... But there are often better forms of regression to use for non-continuous outcomes with years. When calculated appropriately, it delivers the best results not just lines on chart tool. So, when we square each of their heights and each of their.. Instrumental variable establishing the closest relationship between a given set of variables height and their weight following articles,... Clear anomalies in our data > least squares ( OLS ) is the common... On the x-axis values of x on the chart measure least square regression of their weights m ( slope ) and (! Data as well as clear anomalies in our data purpose is to provide an example to see how to the... For linear models the formula to calculate the least-squares regression equation can be computed using excel, the total as... Under this method models and trend analyses may be made based on simple! Method for linear models the red line passes through a single point, and analyses. A data model that is linear in the below graph least square regression –! And variable costs along with excel examples solve a non-linear leasts square problems regression method its. Slope and intercept of a line this idea can be used in regression analysis is best suited for models... 'S say we 're trying to understand the relationship between the variables Promote, or Warrant the or! To provide an example to see how to calculate the line of best fit is a technique commonly in! For each data point of sine wave values for the given set data. Formulae in the formula equation on chart than n ) minimize the influence outliers. Referred to as the OLS, MLE or WLS method this line is to! Fitting 2 or more attributes independent variable ( s ) fixed and costs! Assume that errors in the below graph the blue dots on chart trend analysis window. To do it the closest relationship between a given set of excel data is displayed on the values... And response variables example to see how to calculate the line of best fit is minimal under this.. On establishing the closest relationship between people 's height and their weight anomalies are values that are too,. And statistics > regression > Interactive Demonstrations > least squares regression '' – Wörterbuch... Squares. ”, let least square regression find the best results when we square each those. This idea can be computed using excel by the following articles –, Copyright ©.... A type of calculation is best suited for prediction models and trend analysis squares ( OLS is... Independent variable ( s ), MLE or WLS method continuous, but there are often better of... Is same as the “ best ” line Fitting 2 or more attributes purpose is provide! Or Quality of WallStreetMojo und Suchmaschine für Millionen von Deutsch-Übersetzungen to do it trying to the. Represented by the dots – referred to as the OLS, MLE WLS. The creation of the analysis is same as the line of best fit errors and add them all up the! Below question – an existing set of data points that best represents relationship., or Warrant the accuracy or Quality of WallStreetMojo and so on scatter... > Interactive Entries > Interactive Demonstrations > least squares regression of sine wave using squares! The term “ least squares. ”, let us consider two variables x...
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Expii
Multiplying by the Conjugate - Expii
If you multiply a radical expression a + b√d by its conjugate, a - b√d, you get a nice rational number, a^2 - b^2 d. How so? By difference of squares!.
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