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# Listings package latex ing

I recommend you to use apt-file to search for the package that contains a specific file.. If you invoke. apt-file search valdostamac.com you should find the package that contains listings.. On my system it's contained in texlive-latex-recommended that you have already installed.. To play it safe i would execute. I stumbled upon a similar problem today and found another solution to this rather old question. The listings package allows us to define delimiters inside of which the package will still process keywords and other stuff. From the listings documentation. You can even let the package detect keywords, comments, strings, and other de- limiters inside the contents. However, running the example through LATEX shows whether the listings package is installed correctly.!The example doesn’t work. Are the two packages listings and keyval installed on your system? Consult the administration tool of your TEX distribution, your system administrator.

# Listings package latex ing

## Watch Now Listings Package Latex Ing

LaTeX Programming : 031 : Writing Source Codes Part 1 of 4 : Verbatim and Listing, time: 10:01
Tags: Typeahead min js file ,Games of gta 4 for pc , How to pdf from safari on iphone , Juve misca din buric fileshare, Xu so rong battle realms games However, running the example through LATEX shows whether the listings package is installed correctly.!The example doesn’t work. Are the two packages listings and keyval installed on your system? Consult the administration tool of your TEX distribution, your system administrator. Download the contents of this package in one zip archive (M).. listings – Typeset source code listings using L a T e X. The package enables the user to typeset programs (programming code) within L a T e X; the source code is read directly by T e X —no front-end processor is needed. Keywords, comments and strings can be typeset using different. Location: CTAN Packages listings listings – Typeset source code listings using L a T e X. Keywords, comments and strings can be typeset using different styles (default is bold for keywords, italic for comments and no special style for strings).

### 2 comments on “Listings package latex ing”

• Zulkijinn says:

It absolutely agree with the previous message

• Kenos says:

I can recommend to come on a site on which there are many articles on this question.


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## oracle 11g – ORA-27369: failed to work with exit code: no child process to wait

I have an existing .bat file where I call FDSLoader64.exe daily to update the following tables.

FF_BASIC_SAF
FF_BASIC_AF

Once the update is successful, FDSLoader64.exe updates the metadata tables in which one can know if the update was successful or not.

To do this I have created a PROGRAM in Oracle 11g as follows

TO START
DBMS_SCHEDULER.DISABLE (name => FDS_DEMO.FF_V3, force => TRUE);
DBMS_SCHEDULER.ENABLE (name => FDS_DEMO.FF_V3 & # 39;)
END;

Create a schedule for this, it should work every day at 11:30.

TO START
DBMS_SCHEDULER.set_attribute (name => "FDS_DEMO". "DAILY_9AM", attribute => "repeat_interval", value => & # 39; FREQ = DAILY; BYDAY = MON, FUTURE, MARI, GAME , FRI; BY, FRI; BYHOUR = 11; BYMINUTE = 30; BYSECOND = 0;);
END;

Finally I have created a job who runs the program using the calendar.

TO START
DBMS_SCHEDULER.set_attribute (name => "FDS_DEMO". "FF_V3_GENESIS", attribute => job_priority, value => 1);
END;
/

The calendar was successfully run at 11:30 am on June 24, 2019. According to the metadata tables, we can confirm that the data was successfully uploaded to the tables, but when I see the content of user_scheduler_job_run_details, it displays the results as following.

** The complete error code is as follows. **
"ORA-27369: EXECUTABLE job failed with exit code: there is no job
process children to wait.
STANDARD_ERROR = "2019/06/24 11:30:16 INFO>: config.xml is at the latest
version 2.10, in progress.
2019/06/24 11:30:21 INFO>: FDSLoader version 2.13.1.0
2019/06/24 11:30:21 INFO>: Dump directory D:  FDSLoader  d ""

## customs and immigration – entry / exit by train – UK / Schengen

I am not a citizen of the European Union, I will travel from Paris to Manchester next month. I was wondering therefore, since most countries follow the dates of entry and departure, if I use the train to change country and that they do not control it, my Schengen visa will look like an entry YES but no departure. Same for the UK, I will not have a date of entry but I will leave,

Will it be problems later? Do they control that in the trains? If not, is there anything I should do before?

## How to cancel empty exit errors in Google Sheets?

I'm using the following code in many cells of the sheet,

= SPLIT (REGEXREPLACE ("" & S2, "( d)", "\$ 1,"), ",")

not all will contain data at once. How can I prevent empty cell value notification from appearing?

## ios – the command of the linker failed with the exit code

Thank you for contributing an answer to Game Development Stack Exchange!

• Please make sure to respond to the question. Provide details and share your research!

But to avoid

• Make statements based on the opinion; save them with references or personal experience.

Use MathJax to format equations. MathJax reference.

## Help using 2 parameters to exit a specific category

I am looking to be able to enter an "AGE" and a "V02" in a formula that indicates in which "category" the person would be in ie; good, excellent

## locktime – How can I lock the exit time of a transaction?

Yes, you can use OP_CHECKLOCKTIMEVERIFY opcode in the lock script of the output. This opcode takes as parameter to lock either blocks, or a time of Unix (seconds since January 1, 1970). If the parameter is different from zero and less than 500 million, it is interpreted as a block height. If it is greater than or equal to 500 million, it is interpreted as a Unix time stamp. When you want to reference this UTXO when you spend, you must define the transaction nLocktime equal or larger than the OP_CHECKLOCKTIMEVERIFY. the nLocktime parameter while spending should be identical to what was specified in the CLTV; that is, if the CLTV lock setting was in blocks, it should be specified in blocks, otherwise in the Unix Epoch timestamp.

Since you want it to be usable after January 1, 2030, this equates to a time stamp of 1 893 456 000 (or 0x70danga80). So, your lock script will concatenate:

OP_DATA (size of CLTV_expiry): 0x04
CLTV (little endian): 0x80d8db70
OP_CLTV: 0xb1
OP_DROP: 0x75
OP_DUP: 0x76
OP_HASH160: 0xa9
OP_DATA (size of the address): 0x14
Address:
OP_EQUALVERIFY: 0x88
OP_CHECKSIG: 0xac

Two things to note: (1) When you spend this transaction, the entry referencing this UTXO must have nSequence value between 0xF0000000 and 0xFFFFFFFE. (2) Median_time (time of the last 11 blocks) is used to check nLocktime when the script is evaluated, which is usually an hour late compared to the actual time of your wall clock.

## exit – Where is the problem with Length?

j & # 39; uses

Length[Reduce[ans2[[1]]<= y <= ans1[[1]], y, integers]]

count the integers from ans2 to ans1.

This code

F[x_] : = x ^ 3 - 3 * x;
m = -1;
M = 2;
g[x] = -f[x];
ans1 = Maximize[{F[g[{F[g[{F[g[{f[g[x]], m <= x <= M}, x]// Simplify
ans2 = minimize[{F[g[{F[g[{F[g[{f[g[x]], m <= x <= M}, x]// Simplify
Length[Reduce[ans2[[1]]<= y <= ans1[[1]], y, integers]]

I have the right exit 5. When I change

F[x_] : = x ^ 3 - 3 * x;
m = -1;
M = 5;
g[x] = -f[x];
ans1 = Maximize[{F[g[{F[g[{F[g[{f[g[x]], m <= x <= M}, x]// Simplify
ans2 = minimize[{F[g[{F[g[{F[g[{f[g[x]], m <= x <= M}, x]// Simplify
Length[Reduce[ans2[[1]]<= y <= ans1[[1]], y, integers]]

The output is wrong. Then I use it by hand

year1[[1]]- ans2[[1]]+1

Where is the problem in the order?

Length[Reduce[ans2[[1]]<= y <= ans1[[1]], y, integers]]

## How to have different conditions of entry and exit for a context in Tasker?

I'm using Tasker with a set of conditions on a context.

I want the context to be entered only when the screen is off.

However, I do not want the context to be got out by simply turning on the screen; only the other conditions must be relevant to get out of the context.

What is the "right" way to do this in Tasker? Is there an elegant solution?

## ulimit – Using the full Apache2 processor after exit with error (it's impossible to create a worker thread)

I have my outgoing Apache2.4 (mpm_worker) instance with the following log:

[Mon May 27 11:27:33.196177 2019] [core:warn] [pid 567365:tid 139765793668032]    AH00045: The child process 567368 has still not been closed, sending a SIGTERM
[Mon May 27 11:27:35.198179 2019] [core:warn] [pid 567365:tid 139765793668032]    AH00045: The child process 567368 has still not been closed, sending a SIGTERM
[Mon May 27 11:27:37.200177 2019] [core:error] [pid 567365:tid 139765793668032]    AH00046: child process 567368 still has not stopped, sending a SIGKILL
[Mon May 27 11:27:38.297736 2019] [core:warn] [pid 569972:tid 140126180117440]    AH00098: pid file /var/run/apache2/apache2.pid overwritten - Uncleaned shutdown of the last Apache runtime?
[Mon May 27 11:27:38.300264 2019] [mpm_worker:notice] [pid 569972:tid 140126180117440]    AH00292: Apache / 2.4.29 (Ubuntu) OpenSSL / 1.1.0g configured - resume normal operations
[Mon May 27 11:27:38.300296 2019] [core:notice] [pid 569972:tid 140126180117440]    AH00094: Command line: / usr / sbin / apache2 & # 39;
[Mon May 27 11:27:38.303713 2019] [mpm_worker:alert] [pid 569975:tid 140126052628224]    (11) Resource temporarily unavailable: AH03142: apr_thread_create: unable to create a worker thread
[Mon May 27 11:27:38.359830 2019] [mpm_worker:alert] [pid 569973:tid 140126180117440]    (11) Resource temporarily unavailable: AH00282: apr_thread_create: Can not create a worker thread.
[Mon May 27 11:27:38.371350 2019] [mpm_worker:alert] [pid 569974:tid 140126060156672]    (11) Resource temporarily unavailable: AH03142: apr_thread_create: unable to create a worker thread
[Mon May 27 11:27:40.302219 2019] [mpm_worker:alert] [pid 569972:tid 140126180117440]    AH02325: A shortage of resources or some other irremediable failure has been encountered before successful initialization of a child process ... httpd closes!

When reading the last line of the journal, I assumed that apache would be dead now.
Instead of top gives me a 320% CPU usage for Apache. It lasts forever (at least several days).
By reading the front lines of the journal, it is clear that after apache pretends to leave, it is impossible to shut down or kill a child process for any reason.

Googlerunable to create a worker thread& # 39; Direct me to the limits of the server resources and adjustment of the correct settings of apache2. Here is the Ulimit for www-data:

### Ulimit for Apache

root @ xxx: ~ # sudo -u www-data bash -c-ulimit -a & # 39;
size of the base file (blocks, -c) 0
data segment size (kilobytes, -d) unlimited
planning priority (-e) 0
unlimited file size (blocks, -f)
pending signals (-i) 1546431
max memory locked (kilobytes, -l) 16384
maximum memory size (kilobytes, -m) unlimited
open files (-n) 105000
pipe size (512 bytes, -p) 8
POSIX message wait queues (bytes, -q) 819200
priority in real time (-r) 0
stack size (kilobytes, -s) 8192
CPU time (seconds, -t) unlimited
Maximum number of user processes (-u) 10000
virtual memory (kilobytes, -v) unlimited
Unlimited file locks (-x)

### mpm_worker.conf

StartServers 2
MaxRequestWorkers 100
MaxConnectionsPerChild 50

### Server information:

VPS: 6core 2Ghz
RAM: 8 GB
Usage: One instance of domain / apache for Nextcloud.

### # 1

How can I debug what's going on with Apache after saying that it was abandoned, but that it was causing high CPU usage? (repair would be even better)

### # 2

How can I adjust the worker settings to the case of use? (I guess that's the root cause)

Any help would be appreciated!
Thank you

## el capitan – I can not exit recovery mode (OS X Utilities) by resetting or reinstalling the OS

I have purchased a used iMac and, when I turn it on, the OS X Utilities screen appears. I would like to install the operating system but I can not do it.

I have tried everything on any other questions on this topic:

1. Back to factory settings

That is to say. formatting the hard drive (as indicated by Apple) and reboot. According to the Apple website, this was supposed to display the country selection screen, but this is not the case. It just takes me back to OS X Utilities.

1. OS X Resintall

There is an option to reinstall OS X El Capitan. After being connected to iCloud (with a new account, I had to create this – I did not have one before), the error message "This item is temporarily unavailable". I've googled this and tried the many fixes to no avail. Apparently, the iCloud account of the original user should be used, but I do not have access to it and I do not want them to be linked to this computer either.

1. Other macOS recovery modes

Apple gives key combinations for other recovery options. I have tried all three as shown, but I am still asked to enter the details of iCloud.

1. Start from CD

I do not have the original CD with the operating system on it, nor access to a computer capable of burning a bootable disc. To be honest, this is not an option I want to consider – resetting an Apple computer does not require it, according to the many guides and videos on this subject.)

Screenshots of the formatting of the hard disk:

Screen capture of the iCloud error:

format: plain


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rate of decay of fourier coefficients vs smoothness

suppose $f \in L1$, 2pi periodic and that the fourier coefficients decay with order $|n|^{-k}, k \gt 2$

show that the derivative of f is continuous

i read that the rate of decay of fourier coefficients relates that the "smoothness" of the function. but im not sure how to formalize my argument for this question

any help would be much appreciated

-
possible duplicate of Differentiability and decay of magnitude of fourier series coefficients. Prof. Blatter's hints in his answer should give enough info to start working. – t.b. Mar 31 '11 at 4:04
@Theo: On the face of it, I don't see anything in that thread (or in the one linked to in its comments) that goes in the direction from Fourier coefficients to continuity/differentiability. If any of the statements made there in the other direction can be turned around, I think this should be made explicit to answer the present question. – joriki Mar 31 '11 at 6:04
@joriki: I'm thinking about writing this stuff up once and for all. First, it's not that hard at all and second it's at least the fifth time I see it coming up in one direction or the other (some of these have been deleted in the meantime). – t.b. Mar 31 '11 at 10:02

[Notation. $a_n \lesssim b_n$ means: there exists some positive constant $C$ s.t. $a_n \le C b_n$.]

let

$$c_n=\frac{1}{2 \pi} \int_{-\pi}^{\pi}f(y)e^{-i n y}\, dy$$

and write

$$f(x)=\sum_{n\in \mathbb{Z}} c_n e^{i n x}\quad (1)$$

The decay condition on $c_n$ implies uniform convergence of this series:

$$\lvert c_n e^{i n x} \rvert \le \lvert n\rvert^{-k}\lvert n^k c_n \rvert \lesssim \lvert n\rvert^{-k}$$

and $\sum_{n \in \mathbb{Z}} \lvert n\rvert^{-k}$ is a convergent numerical series. Now differentiate (1) termwise: you get

$$\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$

which is again a uniformly convergent series:

$$\lvert i n c_n \rvert \lesssim \lvert n\rvert^{1-k}.$$

So (1) is a uniformly convergent series whose term-by-term derivative is uniformly convergent. This implies that $f$ is differentiable and

$$f'(x)=\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$

so that, again by uniform convergence, $f'$ is also continuous.

-
You don't have to exclude $0$ in the summation; the corresponding term is $0$ anyway. – joriki Mar 31 '11 at 8:00
@joriki: Of course. Thank you. – Giuseppe Negro Mar 31 '11 at 8:23


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# American Institute of Mathematical Sciences

November  2013, 12(6): 3047-3071. doi: 10.3934/cpaa.2013.12.3047

## Pullback exponential attractors for evolution processes in Banach spaces: Theoretical results

 1 Departamento de Matemática, Instituto de Ciências Matemáticas e de Computação, Caixa postal 668, 13560-970 São Carlos, São Paulo, Brazil 2 BCAM Basque Center for Applied Mathematics, Mazarredo 14, E-48009 Bilbao, Basque Country, Spain

Received  October 2012 Revised  February 2013 Published  May 2013

We construct exponential pullback attractors for time continuous asymptotically compact evolution processes in Banach spaces and derive estimates on the fractal dimension of the attractors. We also discuss the corresponding results for autonomous processes.
Citation: Alexandre Nolasco de Carvalho, Stefanie Sonner. Pullback exponential attractors for evolution processes in Banach spaces: Theoretical results. Communications on Pure & Applied Analysis, 2013, 12 (6) : 3047-3071. doi: 10.3934/cpaa.2013.12.3047
##### References:
 [1] T. Caraballo, A. N. Carvalho, J. A. Langa and L. F. Rivero, Existence of pullback attractors for pullback asymptotically compact processes, Nonlinear Anal., 72 (2010), 1967-1976. doi: 10.1016/j.na.2009.09.037.  Google Scholar [2] A. N. Carvalho, J. A. Langa and J. C. Robinson, "Attractors for Infinite-Dimensional Non-Autonomous Dynamical Systems," Appl. Math. Sci., 182, Springer, 2012. Google Scholar [3] J. W. Cholewa, R. Czaja and G. Mola, Remarks on the fractal dimension of bi-space global and exponential attractors, Boll. Unione Mat. Ital., 1 (2008), 121-146. Google Scholar [4] D. N. Cheban, P. E. Kloeden and B. Schmalfuss, The relationship between pullback, forward and global attractors of nonautonomous dynamical systems, Nonlinear Dyn. and Syst. Theory, 2 (2002), 9-28. Google Scholar [5] V. Chepyzhov and M. Vishik, "Attractors for Equations of Mathematical Physics," Amer. Math. Soc., Providence, RI, 2002. Google Scholar [6] R. Czaja and M. A. Efendiev, Pullback exponential attractors for nonautonomous equations part I: Semilinear parabolic equations, J. Math. Anal. Appl., 381 (2011), 748-765. doi: 10.1016/j.jmaa.2011.03.053.  Google Scholar [7] A. Eden, C. Foias, B. Nicolaenko and R. Temam, "Exponential Attractors for Dissipative Evolution Equations," Research in Applied Mathematics, Masson, Paris, John Wiley & Sons, Ltd., Chichester, 1994. Google Scholar [8] M. A. Efendiev, A. Miranville and S. Zelik, Exponential attractors for a nonlinear reaction-diffusion system in $\R ^3$, C. R. Acad. Sci. Paris Sr. I Math., 330 (2000), 713-718. doi: 10.1016/S0764-4442(00)00259-7.  Google Scholar [9] M. A. Efendiev, A. Miranville and S. Zelik, Exponential attractors and finite-dimensional reduction for nonautonomous dynamical systems, Proc. R. Soc. Edinburgh Sect.A, 135A (2005), 703-730. doi: 10.1017/S030821050000408X.  Google Scholar [10] M. A. Efendiev, Y. Yamamoto and A. Yagi, Exponential attractors non-autonomous dissipative systems, J. Math. Soc. Japan, 63 (2011), 647-673. doi: 10.2969/jmsj/06320647.  Google Scholar [11] J. A. Langa, A. Miranville and J. Real, Pullback exponential attractors, Discrete Contin. Dyn. Syst., 26 (2010), 1329-1357. doi: 10.3934/dcds.2010.26.1329.  Google Scholar [12] J. A. Langa and B. Schmalfuss, Finite dimensionality of attractors for non-autonomous dynamical systems given by partial differential equations, Stoch. Dyn., 4 (2004), 385-404. doi: 10.1142/S0219493704001127.  Google Scholar [13] P. Marín-Rubio and J. Real, On the relation between two different concepts of pullback attractors for non-autonomous dynamical systems, Nonlinear Anal., 71 (2009), 3956-3963. doi: 10.1016/j.na.2009.02.065.  Google Scholar [14] R. Temam, "Infinite Dimensional Dynamical Systems in Mechanics and Physics," 2nd edition, Appl. Math. Sci. 68, Springer Verlag, New York, 1997. Google Scholar

show all references

##### References:
 [1] T. Caraballo, A. N. Carvalho, J. A. Langa and L. F. Rivero, Existence of pullback attractors for pullback asymptotically compact processes, Nonlinear Anal., 72 (2010), 1967-1976. doi: 10.1016/j.na.2009.09.037.  Google Scholar [2] A. N. Carvalho, J. A. Langa and J. C. Robinson, "Attractors for Infinite-Dimensional Non-Autonomous Dynamical Systems," Appl. Math. Sci., 182, Springer, 2012. Google Scholar [3] J. W. Cholewa, R. Czaja and G. Mola, Remarks on the fractal dimension of bi-space global and exponential attractors, Boll. Unione Mat. Ital., 1 (2008), 121-146. Google Scholar [4] D. N. Cheban, P. E. Kloeden and B. Schmalfuss, The relationship between pullback, forward and global attractors of nonautonomous dynamical systems, Nonlinear Dyn. and Syst. Theory, 2 (2002), 9-28. Google Scholar [5] V. Chepyzhov and M. Vishik, "Attractors for Equations of Mathematical Physics," Amer. Math. Soc., Providence, RI, 2002. Google Scholar [6] R. Czaja and M. A. Efendiev, Pullback exponential attractors for nonautonomous equations part I: Semilinear parabolic equations, J. Math. Anal. Appl., 381 (2011), 748-765. doi: 10.1016/j.jmaa.2011.03.053.  Google Scholar [7] A. Eden, C. Foias, B. Nicolaenko and R. Temam, "Exponential Attractors for Dissipative Evolution Equations," Research in Applied Mathematics, Masson, Paris, John Wiley & Sons, Ltd., Chichester, 1994. Google Scholar [8] M. A. Efendiev, A. Miranville and S. Zelik, Exponential attractors for a nonlinear reaction-diffusion system in $\R ^3$, C. R. Acad. Sci. Paris Sr. I Math., 330 (2000), 713-718. doi: 10.1016/S0764-4442(00)00259-7.  Google Scholar [9] M. A. Efendiev, A. Miranville and S. Zelik, Exponential attractors and finite-dimensional reduction for nonautonomous dynamical systems, Proc. R. Soc. Edinburgh Sect.A, 135A (2005), 703-730. doi: 10.1017/S030821050000408X.  Google Scholar [10] M. A. Efendiev, Y. Yamamoto and A. Yagi, Exponential attractors non-autonomous dissipative systems, J. Math. Soc. Japan, 63 (2011), 647-673. doi: 10.2969/jmsj/06320647.  Google Scholar [11] J. A. Langa, A. Miranville and J. Real, Pullback exponential attractors, Discrete Contin. Dyn. Syst., 26 (2010), 1329-1357. doi: 10.3934/dcds.2010.26.1329.  Google Scholar [12] J. A. Langa and B. Schmalfuss, Finite dimensionality of attractors for non-autonomous dynamical systems given by partial differential equations, Stoch. Dyn., 4 (2004), 385-404. doi: 10.1142/S0219493704001127.  Google Scholar [13] P. Marín-Rubio and J. Real, On the relation between two different concepts of pullback attractors for non-autonomous dynamical systems, Nonlinear Anal., 71 (2009), 3956-3963. doi: 10.1016/j.na.2009.02.065.  Google Scholar [14] R. Temam, "Infinite Dimensional Dynamical Systems in Mechanics and Physics," 2nd edition, Appl. Math. Sci. 68, Springer Verlag, New York, 1997. Google Scholar
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# Which number can I erase?

All positive integers greater than $$2$$ are written on a board. First we erase number $$3$$ and $$5$$.

With 4 positive integers $$a,b,c,d$$ satisfying $$a+b=c+d$$, if $$ab$$ is erased, then $$cd$$ can be erased, otherwise $$cd$$ cannot be erased.

For example, $$3=3 \times 1$$, $$3+1=4=2+2$$ , then $$2 \times 2 = 4$$ is erased.

a. What are the conditions of a number that can be erased ?

b. If not only $$3$$ and $$5$$, but every prime number is erased at the beginning, can all other numbers be erased as well? If not, what are the conditions of a number to be erased ?

(Sorry for my last question, English is my second language)

• What makes you say you can't erase $7$? Surely you could erase $7=7\times1$, then say $7+1=6+2$, so you can erase $12$. I don't understand. I also think the question is poorly worded and is missing other relevant information. You haven't specified that $a,b\neq c,d$. Can we choose any number to erase if it has the specified property? Or are we given a starting number, $ab$, to erase and have to find all other numbers, $cd$, that we are then allowed to erase? – Jam Oct 16 '18 at 12:28
• @Jam : thanks for your advice. I will edit my question. – apple Oct 16 '18 at 12:31
• Thanks for the edit, it makes a lot more sense :) – Jam Oct 16 '18 at 14:49

A prime $$p>5$$ can be erased if $$m=2(p-1)$$ has been erased, as $$2+(p-1)=p+1$$. Note that $$m=2\times(p-1)=4\times\frac{p-1}{2},$$ where $$p-1$$ is even because $$p>5$$ is prime. This factorization of $$m$$ shows that it can be erased if $$\frac{p+5}{2}$$ has been erased, because $$\frac{p-1}{2}+4=\frac{p+5}{2}+1,$$ where clearly $$\frac{p+5}{2} becase $$p>5$$.

A composite number $$n=uv$$ with $$u,v>1$$ can be erased if $$m=u+v-1$$ has been erased, as $$1+(u+v-1)=u+v.$$ Of course $$m because $$u,v>1$$.

In particular, an integer $$n>5$$ can be erased if all integers less than $$n$$ have been erased, so you can use induction.

Somewhat a long comment (not a complete answer).

The first thing that I would suggest is that you should work through some examples and see what happens. I'll help with some of the initial steps.

To think about what you're erasing, let $$n$$ be a number that has been erased. Then, for each pair of factors $$ab$$ so that $$ab=n$$, compute $$a+b=m$$. Next, you find all other ways to write $$m$$ as a sum $$m=c+d$$ and then delete each $$cd$$ found in this way. Then, you continue until there isn't anything more to delete (or you see the pattern).

In your set-up, you start with $$3$$ and $$5$$ deleted.

Deleted: $$3$$ and $$5$$.

• Using the deleted $$3$$, you know that $$3$$ factors as $$3=1\cdot 3$$. Therefore, $$a+b=1+3=4$$. The only sums equal to $$4$$ are $$1+3$$ and $$2+2$$. Taking $$c=2$$ and $$d=2$$ gives $$cd=4$$, so $$4$$ is deleted. Since we've considered all pairs of factors of $$3$$, there is nothing more that $$3$$ can tell us.

Deleted: $$3$$, $$4$$, and $$5$$.

• Using the deleted $$4$$, you know that $$4$$ factors as $$4=1\cdot 4$$ or $$2\cdot 2$$. In the first case, $$1+4=5$$, and $$5$$ can be written as $$5=1+4$$ or $$5=2+3$$. The first case leads to $$1\cdot 4=4$$, which is already deleted, but $$2\cdot 3=6$$, which can now be deleted. On the other hand, using $$2\cdot 2=4$$, we have two ways to write $$4$$ as a sum, $$4=1+3$$ or $$4=2+2$$. In either case, the product is $$1\cdot 3=3$$ or $$2\cdot 2=4$$, both of which have already been deleted. Therefore, there is nothing more that $$4$$ can tell us.

Deleted: $$3$$, $$4$$, $$5$$, and $$6$$.

• Using the deleted $$5$$, you know that $$5$$ factors as $$1\cdot 5=5$$. Therefore, the sum of these two is $$1+5=6$$. There are three ways to write $$6$$ as a sum, $$1+5=6$$, $$2+4=6$$, and $$3+3=6$$. The first pair has product $$1\cdot 5=5$$, which has already been deleted. The second pair has product $$2\cdot 4=8$$, which can now be deleted. The third pair has product $$3\cdot 3=9$$, which can also be deleted.

Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, and $$9$$.

• Using the deleted $$6$$, you know that $$6$$ factors as $$1\cdot 6=6$$ and $$2\cdot 3=6$$. Therefore, the corresponding sums are $$1+6=7$$ or $$2+3=5$$. There are three sums to get $$7$$, $$1+6=7$$, $$2+5=7$$, and $$3+4=7$$. There are two sums to get $$5$$, $$1+4=5$$ and $$2+3=5$$. These, combined allow one to delete $$6$$, $$10$$, $$12$$, $$4$$, and 6\$.

Deleted: $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$9$$, $$10$$, and $$12$$.

Continue this until you find a pattern.


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# Difference between revisions of "2012 AIME II Problems/Problem 14"

## Problem 14

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$.

## Solution

Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'. This will split it into four cases: Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)

Case 1: To create our groups of three, there are $\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}$. In general, the number of ways we can arrange people within the rings to count properly is $\dfrac{(n-1)!}{2}$, since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities. Thus, each of the three cases has $\dfrac{(3-1)!}{2}=1$ arrangements. Therefore, for this case, there are $\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280$

Case 2: For three and six, there are $\dbinom{9}{6}=84$ sets for the rings. For organization within the ring, as before, there is only one way to arrange the ring of three. For six, there is $\dfrac{(6-1)!}{2}=60$. This means there are $(84)(1)(60)=5040$ arrangements.

Case 3: For four and five, there are $\dbinom{9}{5}=126$ sets for the rings. Within the five, there are $\dfrac{4!}{2}=12$, and within the four there are $\dfrac{3!}{2}=3$ arrangements. This means the total is $(126)(12)(3)=4536$.

Case 4: For the nine case, there is $\dbinom{9}{9}=1$ arrangement for the ring. Within it, there are $\dfrac{8!}{2}=20160$ arrangements.

Summing the cases, we have $280+5040+4536+20160=30016 \to \boxed{016}$.


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chapter  1
45 Pages

## The topography of the Periplus

Whoever really began the work, there see1ns little doubt that, the original canal not having progressed very far from its starting point a little above Bubastis, it was continued by the Persians under Darius Hystaspes in the 6th century B.C., thus making an attribution to N ekho II out of the question. In the 6th century the canal reached the Bitter Lakes, where the work

M.fOITfRRANEANSEA

was suspended on account of political troubles, and it bccatne choked with sand. About 274 B.C. Ptoletny II Philadelphos reconstructed it and took it to Arsinoc, a village at the south end of the Bitter Lakes, but still not to the Red Sea. At length, about A.D. 106, the etnperor Trajan had a new canal cut, this time frorn a different starting point, near the town of Babylon opposite Metnphis, taking it n1ore or less parallel with the Nile through Hcliopolis and Vicus Iudaeorum (Tell el Yahudieh) to Thou, where it joined the earlier canal, and turned cast through Hcroopolis and Serapiutn, going thence through the Bitter Lakes to Arsinoe, with a cut joining the latter with Klusma (Suez) on the Red Sea. This canal was accompanied by a road, on which the Antonine Itinerary gives the following stations: 1

Babylonia . . . . . . . . . . . . . . . . . . . . . . . . mpm XII Heliu (Heliopolis} . . . . . . . . . . . . . . . . . . mpn1 XII Scenas V eteranorum . . . . . . . . . . . . . . . . n1pm XXII Vico Iudaeorun1 . . . . . . . . . . . . . . . . . . . . mptn XII Thou . . . . . . . . . . . . . . . . . . . . . . . . . . . . mpm XII Hero (Heroopolis} . . . . . . . . . . . . . . . . . . mpm XXIIII Serapiun1 . . . . . . . . . . . . . . . . . . . . . . . . . . rnpm XVIII Clysma . . . . . . . . . . . . . . . . . . . . . . . . . . mpm L

Trajan began his canal much higher up-stream than the earlier one, so that the fresh water which filled it could be used to irrigate the land through which it passed; and here there were in fact three lines of communication: the Nile, the canal, and the road just described. This canal was still working seven hundred years afterTrajan's day. 2

A Copto Bereniccn usque Poeniconon ....................... . Didin1c ......................... . Afrodito ......................... . Compasi ......................... . louis ............................. . Aristonis ......................... . Falacro ........................... . Apollonos ....................... . Cabalsi ........................... . Cenon H ydremna ................. . Berenicen ....................... .


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# Finance question 1 answer below »

Consider two firms, With and Without, that have identical assets that generate identical cash flows. Without is an all-equity firm, with 1 million shares outstanding that trade for a price of $24 per share. With has 2 million shares outstanding and$12 million dollars in debt at an interest rate of 5%. According to MM Proposition 1, what is the stock price for With?

Wahid A

As per MM Proposition 1

VL = VU

VL = Value of levered firm

VU = Value of unlevered firm

VL = 2...

Looking for Something Else? Ask a Similar Question


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# Do three linearly independent vectors in R4 with their fourth entries as 0 span R3?

So, I understand in general that if you have three linearly independent vectors in $\mathbb{R}^4$, they span a space in $\mathbb{R}^4$, not all of $\mathbb{R}^3$, because vectors with four entries live in $\mathbb{R}^4$ not $\mathbb{R}^3$. However, what if you have three linearly independent vectors like:

$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix}$

Do these span $\mathbb{R}^3$? It seems like they would occupy the same space as $\mathbb{R}^3$ no matter how many zeroes you add.

• Only a subset of a space can span that space. So these don't span $\mathbb R^3$, instead the span the vector space $\{ (x,y,z,0)\}$, where $x,y,z \in \mathbb R$. As it turns out, this subspace of $\mathbb R^4$ is isomorphic to $\mathbb R^3$, but it is incorrect to call it $\mathbb R^3$ itself, although this is often done in textbooks, with the meaning to be implicitly understood. – астон вілла олоф мэллбэрг May 2 '17 at 0:50
• They span a three dimensional subspace of $\Bbb R^4$ which is isomorphic to $\Bbb R^3$, however the space that they span contains no vectors from $\Bbb R^3$, those are all vectors which live in $\Bbb R^4$. Note that $\Bbb R^4\cap \Bbb R^3=\emptyset$ – JMoravitz May 2 '17 at 0:50


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## Want to keep learning?

This content is taken from the Raspberry Pi Foundation & National Centre for Computing Education's online course, How Computers Work: Demystifying Computation. Join the course to learn more.
1.8

## Raspberry Pi Foundation

Skip to 0 minutes and 1 second In this step, we’re going to look at working with binary. As you saw in a previous step, a Turing machine is capable of operating using only two symbols, 1 and 0. And the rather bold statement was made that using just these symbols, any solvable problem could be solved. How can this be? How can you even add two numbers, say 12 and 6, if you only have the ability to use ones and zeros? To answer this question, we should take a brief look at the nature of numbers. When you were learning to do basic arithmetic when you were a child, you probably learned about counting in ones, tens, hundreds, and thousands. This is called base 10, or denary.

Skip to 0 minutes and 45 seconds When writing down numbers, you begin at 0, count all the way up to 9, then go back to 0 again, but have an additional digit in the next column. You could ask yourself, why is it that we have 10 separate symbols to represent all our numbers? Why not five? Then counting would look like this. The answer is most probably because we have 10 fingers and thumbs, although it’s not really known. Some cultures around the world count in different bases, and most people are fairly happy using base 12 and base 60 when telling the time. The Turing machine uses base 2, or what is often called binary. Here’s an example of counting in binary.

Skip to 1 minute and 31 seconds You might recall that this is exactly what your Turing machine did in the last step. So to answer the original question, how could a Turing machine add the numbers 12 and 6, the answer is that the machine would use binary representations of these numbers. Here’s an example of simultaneously counting in both binary and denary. So how can we convert from one to the other? We said earlier that when you were learning about numbers as a child, you probably used the terms ones, tens, hundreds, and thousands. These are all powers of 10. You can draw a table using all these values as headings.

Skip to 2 minutes and 12 seconds You can write in a number, in this case 09032, and we can see that it is made up of zero 10,000s, nine 1,000s zero 100s, three 10s, and two 1s. In binary, we can do the same thing, but using powers of 2.

Skip to 2 minutes and 35 seconds Again, a table can be drawn up using these numbers as headings.

Skip to 2 minutes and 40 seconds As we did with denary, we can write in a binary number, such as 10111. Now we can see it is made up of one 16, zero 8s, one 4, one 2, and one 1. So if this was in denary, it would be the number 23. You’re now going to have a go at doing some binary to denary conversions yourself. So you might like to replay this video and have a look over it again. Pause the video here and have a practise on the following binary numbers, and see if you can work out what they would be in denary. There’ll be a quiz on this in the next section. Converting numbers back the other way isn’t much more difficult.

Skip to 3 minutes and 23 seconds If we wanted to convert the number 21 into binary, we could start out with an empty table like this. Then look in each of the columns to see what needs to be placed there using only ones and zeros. In this case, we would need one 16, zero 8s, one 4, zero 2s, and one 1.

Skip to 3 minutes and 54 seconds Reading off from the table, we can see that 21 in binary is 10101. Have a go with converting the following denary numbers into binary. Don’t forget to share your answers and help each other out in the comments section.

# Working with binary

In this step we’re going to look at working with binary.

As you saw in a previous step, a Turing machine is capable of operating using only two symbols – 1 and 0 – and the rather bold statement was made that, using just these symbols, it can do any computation!

How can this be? How can you even add up two numbers, say 12 and 6, if you only have the ability to use 1s and 0s?

To answer this question, we should first take a brief look at the nature of numbers. When you were learning to do basic arithmetic, when you were a child, you probably learned about counting in ones, tens, hundreds and thousands.

This is called base 10. When writing down numbers, you begin at 0, count up to 9, then go back to 0 again, but add an additional digit in the next column.

You could ask yourself, why is it we have 10 separate symbols to represent all our numbers? Why not 5? Then counting would look like this:

The answer is most probably because we have 10 fingers and thumbs, although it’s not really known. Some cultures around the world count in different bases, and most people in the world are fairly happy using base 12 and base 60 when telling the time.

The Turing machine uses base 2, or what is often called binary. Here’s an example of counting in binary

You might recall that this is exactly what your Turing machine did in the last step.

So to answer the original question – how could a Turing machine add the numbers 12 and 6? – the answer is that the machine would use the binary representations of these numbers.

## Converting binary to denary

Here’s an example of simultaneously counting in both binary and denary.

So how can we convert from one to another?

We said earlier that when you were learning about numbers as a child, you probably used the terms ones, tens, hundreds, thousands. These are all powers of 10.

We can then arrange these values horizontally to be used as headings.

Now we can write in a number – in this case 09032 – and we can see that it is made up of 0 x 10000, 9 x 1000, 0 x 100, 3 x 10 and 2 x 1:

In binary we can do the same thing, but using powers of 2.

Again, a the values are arranged as headings.

As we did with denary, we can write in a binary number such as 10111.

Now we can see it is made up of 1 x 16, 0 x 8, 1 x 4, 1 x 2 and 1 x 1. So if this was in denary it would be the number 23.

Have a practice on the following binary numbers and see if you can work out what they would be in denary. There’ll be a quiz on this in the next section.

1010
1001
10001

Converting numbers back the other way isn’t much more difficult. If we wanted to convert the number 21 into binary we would do it like this:

We keep dividing the number by a power of two, and then using the remainder for the next division.

Reading off from the table, we can see that 21 in binary is 10101.

Have a go at converting the following denary numbers into binary.

7
15
40

You can compare your answers in the comments section, or even challenge each other to do more binary to denary and denary to binary calculations.


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Location of Repository

## Differential Spaces, Vector Fields, and Orbit-Type Stratifications

### Abstract

Let $G$ be a Lie group, and let $(M,\omega)$ be a symplectic manifold. If $G$ admits a Hamiltonian action on $(M,\omega)$ with momentum map $\mu$, then $M$, the zero-level set of $\mu$, the orbit space, and the corresponding symplectic quotient all have induced stratifications. We push this setting into the language of differential spaces, and as a consequence we find that the stratifications are intrinsic to the ring of smooth functions on each space.Comment: 44 pages. Earlier versions of this paper were supposed to prove a result regarding a de Rham complex of differential forms on the symplectic quotient. A crucial lemma in the proof was incorrect. The theory used remains useful for studying the stratifications mentioned in the abstract from the point of view of differential spaces. Erroneous parts remove

Topics: Mathematics - Symplectic Geometry, Mathematics - Differential Geometry
Year: 2013
OAI identifier: oai:arXiv.org:1104.4084


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# Camera rotation deforms objects

I'm working on a ray-tracer in C++ with an adjustable camera for a university assignment. I'm entirely new with graphics and I'm struggling with one thing.

I have a matrix as a mat4 object (4x4 matrix with each cell saved in a 1D array) that I save in a Camera object that handles transformations of every point in the scene (which is currently just two spheres). Translations seem to be working fine, but when I attempt rotations, things break. The spheres deform, and it only gets worse the more I attempt to rotate the camera.

This is where I set the camera matrix:

mat4 Renderer::setCameraMatrix(const Camera &camera)
{
mat4 cameraToWorld = mat4(); // Initialise 4x4 matrix.

// Translation.
cameraToWorld.cell[3] = camera.pos.x;
cameraToWorld.cell[7] = camera.pos.y;
cameraToWorld.cell[11] = camera.pos.z;

// Rotation.
cameraToWorld.cell[0] = camera.orientmat.cell[0], cameraToWorld.cell[1] = camera.orientmat.cell[1], cameraToWorld.cell[2] = camera.orientmat.cell[2];
cameraToWorld.cell[4] = camera.orientmat.cell[4], cameraToWorld.cell[5] = camera.orientmat.cell[5], cameraToWorld.cell[6] = camera.orientmat.cell[6];
cameraToWorld.cell[8] = camera.orientmat.cell[8], cameraToWorld.cell[9] = camera.orientmat.cell[9], cameraToWorld.cell[10] = camera.orientmat.cell[10];

return cameraToWorld;
}

This matrix is then used to set the origin and direction of the camera in the world space later on in the code:

vec3 orig = transformpoint(vec3(0), cameraToWorld);
/*...*/
vec3 dir = transformvector(vec3(x, y, -1), cameraToWorld) - orig;
dir.normalize();

The transformpoint and transformvector functions basically multiply a vector v by a matrix M (M.v), except the former takes into consideration translations, and the latter doesn't. Their code can be found here.

The orientmat matrix is calculated from the three separate orientation angles (orient.x, orient.y, orient.z) in this line of code:

cam->orientmat = mat4().rotatexyz(cam->orient.x, cam->orient.y, cam->orient.z);

Using this function:

mat4 mat4::rotatexyz(const float a, const float b, const float c)
{
mat4 M;
const float ca = cosf(a), sa = sinf(a);
const float cb = cosf(b), sb = sinf(b);
const float cc = cosf(c), sc = sinf(c);

M.cell[0] = cb * cc, M.cell[1] = -1 * cb * sc, M.cell[2] = sb;
M.cell[4] = sa * sb * cc + ca * sc, M.cell[5] = -1 * sa * sb * sc + ca * cc, M.cell[6] = -1 * sa * cb;
M.cell[8] = -1 * ca * sb * cc + sa * sc, M.cell[9] = ca * sb * sc + sa * cc, M.cell[10] = ca * cb;

return M;
}

(This function basically creates this transformation matrix from the three camera orientation angles.)

I've tried to keep this question as compact as possible but I can provide more code if needed. I'm completely baffled about why I'm getting this kind of deformation on my spheres, and some insight would be invaluable! Thanks.

• Transpose your matrices. Most APIs use column major ordering – Bálint Dec 2 '17 at 19:03
• @Bálint I'm not using an API. Coding everything from scratch. I will try the transposition though. – Step Dec 2 '17 at 23:30

Well, first off, unlike transformation and scale, you can't just combine orientation matrices in order to get a real orientation matrix out. You have to apply the orientation matrices with respect to a single axis, apply them separately, or use a single complex valued quaternion matrix. You may be getting into something called a gimbal lock or some other strange positional artifact in combination with other bugs in your code.

Since your Y rotation doesn't change after going the other direction if you look at your gif, I suspect there is a bug in the code you aren't showing us, probably not updating rotation when going the opposite direction.

The easiest way to avoid gimbal locking is to just rotate with respect to each axis individually, something like this:

// because I don't want to use silly linear indexing
class mat3 : ...{
...
// incase of inheritance
int width(){
return 3;
}
int height(){
return 3;
}
// can put this in parent class if inherits width and height functions
float& operator () (int row, int col){
// if m_array is row ordered
int linear_index = row * width() + col;
// if m_array is column ordered
int linear_index = col * height() + row;

assert(linear_index < (width()*height()));

return m_array[linear_index];
}

// duplicated for const usage
const float& operator () (int row, int col) const{
// if m_array is row ordered
int linear_index = row * width() + col;
// if m_array is column ordered
int linear_index = col * height() + row;

assert(linear_index < (width()*height()));

return m_array[linear_index];
}
...
};

...

//from https://stackoverflow.com/questions/14607640/rotating-a-vector-in-3d-space
mat3 m;

m(0,0) = 1;
m(0,1) = 0;
m(0,2) = 0;

m(1,0) = 0;

m(2,0) = 0;
}

//from https://stackoverflow.com/questions/14607640/rotating-a-vector-in-3d-space
mat3 m;

m(0,1) = 0;

m(1,0) = 0;
m(1,1) = 1;
m(1,2) = 0;

m(2,1) = 0;
}

//from https://stackoverflow.com/questions/14607640/rotating-a-vector-in-3d-space
mat3 m;

m(0,2) = 0;

I should also point out that translating the vector [0,0,0] by another vector [x,y,z] just returns [x,y,z], because all a translation will do is add x, y, and z to the original xyz elements.


text:
# Tikzpictures in forest nodes without saveboxes

I am able to generate the following tree using the MWE below. It creates tikzpictures using a \drawtape command and puts them into saveboxes, which in turn can be straightforwardly used as the content of nodes in a forest environment.

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{positioning}

\newcommand{\drawtape}[3]{
\begin{tikzpicture}[cell/.style={minimum size=0.5cm,outer sep=0pt,draw,rectangle,node distance=0pt,text height=1.5ex}]
\foreach \x [count=\xi] in {#1} {
\ifnum \xi=1
\node[cell] (\xi) {$\x$};
\else
\pgfmathtruncatemacro{\xil}{\xi - 1}
\node[cell] (\xi) [right=of \xil] {$\x$};
\fi
}
\node[font=\footnotesize] [below=0cm of #2, xshift=0.03em] {$#3$};
\end{tikzpicture}
}

\newsavebox{\treeroot}
\newsavebox{\treeone}
\newsavebox{\treetwo}

\usepackage{forest}

\begin{document}

\savebox{\treeroot}{\drawtape{1, 1, 1}{1}{1}}
\savebox{\treeone}{\drawtape{0, 1, 1, 1}{1}{2}}
\savebox{\treetwo}{\drawtape{1, 1, 1}{2}{1}}

\begin{forest}
[\usebox{\treeroot}
[\usebox{\treeone}]
[\usebox{\treetwo}]
]
\end{forest}

\end{document}

This works okay for small trees like the one pictured, but if I want to create a larger tree then the number of saveboxes I have to manually create quickly becomes incredibly cumbersome. Ideally, I would like to be able to write a forest environment like the one below instead. Is there any way to achieve this?

\begin{forest}
[\drawtape{1, 1, 1}{1}{1}
[\drawtape{0, 1, 1, 1}{1}{2}]
[\drawtape{1, 1, 1}{2}{1}]
]
\end{forest}

• See tex.stackexchange.com/questions/184497, in comparison to this solutions you need to define two commands. – Zarko Feb 20 at 14:32
• If you just want to avoid the \newsavebox command, the xsavebox package could be an option. – AlexG Feb 20 at 16:08
• @Zarko Your diagram at tex.stackexchange.com/questions/184497 looks very good, but I can't figure out how to adapt it to this question. Can you please elaborate? – solisoc Feb 20 at 16:13
• @Zarko I figured out how to do it, so don't worry. – solisoc Feb 21 at 5:33

If you are willing to consider to use a slightly different syntax, you can use simple tabulars along with append after command.

\documentclass{article}
\usepackage{array}
\newcolumntype{C}[1]{>{\centering\arraybackslash}p{#1}}
\usepackage{forest}
\tikzset{drawtape/.style n args={3}{node contents={
\begin{tabular}{|*{10}{@{}C{5mm}@{}|}}
\hline
#1
\hline
\end{tabular}},append after command={
([xshift=#2*5mm-1mm]\tikzlastnode.south west)
node[anchor=north,inner sep=0pt,font=\footnotesize]{#3}}}}

\begin{document}

\begin{forest}
for tree={l sep+=2ex,edge={shorten <=2.5ex}}
[,drawtape={1& 1& 1\\}{1}{1}
[,drawtape={0 & 1 & 1& 1\\}{1}{2}]
[,drawtape={1& 1& 1\\}{3}{1}]
]
\end{forest}
\end{document}

Thanks to Zarko pointing me towards the use of multipart nodes, I arrived at the following solution:

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{shapes.multipart, positioning}

\usepackage{xinttools}

\usepackage{forest}

\def\numname#1{%
\ifcase#1 \or one\or two\or three\or four\or five\or six\or seven\or eight\or nine\or ten\or eleven \or twelve\or thirteen\or fourteen\or fifteen\or sixteen\or seventeen\or eighteen\or nineteen\or twenty\fi%
}

\newcounter{cellnum}
\newcommand{\tape}[1]{%
\setcounter{cellnum}{1}
\xintFor* ##1 in {#1} \do {%
\nodepart[text width=1.5ex, align=center]{\numname{\value{cellnum}}} $##1$%
\stepcounter{cellnum}%
}
}

\begin{document}

\begin{forest}
for tree={
delay={
content={\tape{#1}}
},
rectangle split,
rectangle split horizontal,
rectangle split parts=\value{cellnum},
rectangle split part align=base,
draw,
l sep=6ex,
s sep=2em,
anchor=parent,
edge path={
\noexpand\path[draw]
([yshift=-2.5ex]!u.children) -- ([yshift=1ex].parent);
},
tikz={\node[below=0cm of .\numname{#1} south, font=\scriptsize]{$#2$};}

Unfortunately there is an issue with edge paths being of unequal lengths. I will inquire after a solution and update this answer when I find one. EDIT: Using anchor=parent in for tree solves this.


text:
# SEQUENCES

What is the next term in the sequence: 93, 99, 105, 111…?
What is the missing term in the sequence: 11, 2, -7, __,-25…?
What is the 1st term of the sequence: __, 13, 17, 21…?
What is the common difference of the sequence: -8, -5, -2…?
What is the common difference of the sequence: 1¼, 3½, 5¾, 8...?
What is the nth term of the sequence: 1, -8, -17, -26…?
What are the first four terms of the sequence 8n+6?
What is the next term in the sequence: 1, 5, 15, 45…?
What is the missing term in the sequence: -2, 4, -8, _ ,-32…?
What is the 1st term of the sequence: _ , 1, 4, 16…?
What is the common ratio of the sequence: 8, 2, ½ …?
What is the common ratio of the sequence: 12, 24, 48…?
What is the nth term of the sequence: 1, -3, 9, -27…?
What are the first four terms of the sequence 5x2n?
What are the first four terms of the sequence 8 x (½)n ?


text:
# Error with \footnote{\bibentry{foo}} when using beamer

I am making a beamer presentation and having the following problem:

When I use \footnote{\bibentry{foo}} to cite an article, I get the following error message:

Undefined Control Sequence}

And the footnote does not appear right. I have used this in other beamer presentation before, though I'm not sure what is going wrong in my case. I typed

\usepackage{bibentry}
\nobibliography*

before my title slide (which is before \begin{document}) and I also typed

\begin{frame}
\titlepage
\bibliographystyle{amsplain}
\nobibliography{diss_def}
\end{frame}

Immediately after \begin{document}

Any suggestions on what could be wrong?? Or other suggestions for citing in beamer?

-
Welcome to TeX.sx! A tip: If you indent lines by 4 spaces or enclose words in backticks , they'll be marked as code, as can be seen in my edit. You can also highlight the code and click the "code" button (with "{}" on it). – Marco Daniel Apr 13 '12 at 20:41
Re: other suggestions: Use biblatex. See my answer to tex.stackexchange.com/questions/10547/…; for footnote citations, add the autocite=footnote package option. – lockstep Apr 13 '12 at 20:47
Please, try providing a minimal example, as what I've tried along your lines works. What is the "undefined control sequence"? – egreg Apr 13 '12 at 20:51
@lockstep, thanks! I saw your post on biblatex. What if I also want to include the name of the journal? – dblazevski Apr 14 '12 at 15:59
You could a) switch to an authortitle style b) use \footfullcite` or c) a customized citation command. – lockstep Apr 14 '12 at 16:06


text:
## a triangle maths formula

When i was 6th class in primary school, we had just finished the primary maths curriculum and had a few days to spare, so the teacher decided he’d try teach us a bit of algebra.

As a test of our out-of-the-box thinking, he posed this little puzzle:

Given a triangle of balls, where there are x balls on the bottom, x-1 above that, x-2 above that, and so on until there is only one on top, how can you figure out x if you are given the overall total y

```    O
O O
O O O
O O O O
O O O O O
x=5, y=15
```

In other words, if x is 5, then there are 5+4+3+2+1=15 balls in total. Given the number 15, how would you figure out that the original x was 5?

I was a bit of a whiz back then and waded in to the problem. About 5 minutes later, I had an answer – but it was for the other way around – given x, I could figure out y with a simple formula:

```y=x((x+1)/2)
```

So, given x=5, y=5((5+1)/2), y=15.

The teacher was a bit surprised at this. After all, I was about 12 years old at the time.

It didn’t solve the problem as it was asked, though.

All through secondary school, I tried reversing the formula to put it in terms of y instead of x. I didn’t succeed until only recently:

```y=x((x+1)/2)
2y=x2+x
2y=(x+1/2)(x+1/2)-1/4 <-- this step only occurred to me a few weeks ago
2y+1/4=(x+1/2)2
(2y+1/4)1/2-1/2=x
```

How a primary school student was supposed to guess that formula, I have no idea!

To test, we plug in the 15 Y example we had:

```(2*15+1.4)1/2-1/2=x
30.251/2-.5=x
5.5-.5=x
x=5
```

Success, but not pretty!

There /is/ a way of doing it which doesn’t involve quite as much maths.

First, you imagine how the triangle looks:

```    O
O O
O O O
O O O O
O O O O O
```

Nothing obvious yet… But, shift all the levels over to the left:

```O
O O
O O O
O O O O
O O O O O
```

Now, it is obvious that the triangle is half of a square.

It’s not /exactly/ half, but it’s close enough that you can easily figure out x by getting the square root of 2y, then rounding down.

```x = Floor[2y1/2]
```

I am almost certain that that’s the method that my teacher expected us to come up with. Took me 15 minutes to figure out the original formula, and 18 years to notice the much more obvious way to do it.


text:
# Proving that under certain conditions, the product of a sequence of holomorphic functions converges uniformly

The following theorem appears in "Princeton Lectures in Analysis, Complex Analysis," by Stein & Shakarchi:

Suppose $\{F_n\}$ is a sequence of holomorphic functions on the open set $\Omega$.
If there exists constants $c_n \gt 0$ s.t. $\sum c_n < \infty$ and $|F_n(z) - 1| \le c_n$, for all $z \in \Omega$, then
The product $\prod_{n=1}^\infty F_n(z)$ converges uniformly in $\Omega$ to a holomorphic function $F(z)$

The idea is that we can write $\prod_{n=1}^N F_n(z)$ as $\prod_{n=1}^Ne^{\text{log}F_n(z)} = e^{\sum_{n=1}^N\text{log}F_n(z)}$
Using the fact that $|\text{log} f| \le 2|f-1|$ for $f$ close enough to $1$, together with $|F_n(z) - 1| \le c_n$ and $\sum c_n < \infty$, it follows that $e^{\sum_{n=1}^N\text{log}F_n(z)}$ converges to some function $F(z)$,
but how do we know that it converges uniformly?

Hint: bound $\left|\sum_{n=M}^N \log F_n(z)\right|$
• I may have some clue at what you're getting at. So I want to write $e^{\sum_{n=1}^{M-1}\text{log}F_n}e^{\sum_{n=M}^{N}\text{log}F_n}$ and then I can show that the second exponential term is close to $1$. But that only allows me to get within a multiplicative constant of my limit function, not an additive constant, right? – Mark Sep 20 '12 at 1:52


text:
# American Institute of Mathematical Sciences

ISSN:
1930-5311

eISSN:
1930-532X

All Issues

## Journal of Modern Dynamics

2022 , Volume 18

Select all articles

Export/Reference:

2022, 18: 1-11 doi: 10.3934/jmd.2022001 +[Abstract](198) +[HTML](71) +[PDF](162.97KB)
Abstract:

Let \begin{document}$a < b$\end{document} be multiplicatively independent integers, both at least \begin{document}$2$\end{document}. Let \begin{document}$A,B$\end{document} be closed subsets of \begin{document}$[0,1]$\end{document} that are forward invariant under multiplication by \begin{document}$a$\end{document}, \begin{document}$b$\end{document} respectively, and let \begin{document}$C : = A\times B$\end{document}. An old conjecture of Furstenberg asserted that any planar line \begin{document}$L$\end{document} not parallel to either axis must intersect \begin{document}$C$\end{document} in Hausdorff dimension at most \begin{document}$\max\{\dim C,1\} - 1$\end{document}. Two recent works by Shmerkin and Wu have given two different proofs of this conjecture. This note provides a third proof. Like Wu's, it stays close to the ergodic theoretic machinery that Furstenberg introduced to study such questions, but it uses less substantial background from ergodic theory. The same method is also used to re-prove a recent result of Yu about certain sequences of sums.

2020 Impact Factor: 0.848
5 Year Impact Factor: 0.815
2020 CiteScore: 0.9


text:
Tricki

## Induction front page

### Example 1: A simple arithmetic identity

Consider the identity

Since the left-hand side of (1) is neither an arithmetic nor a geometric progression, we are unable to use the standard formulas for calculating such sums; and, indeed, it is not obvious how to transform the expression into one that can be manipulated into a formula for the sum. However, this can be proved using induction.

### Example 2: The base case is important

Sometimes one can have the temptation of looking at the base case of induction as a mere formal step that is always trivially true, thinking that the "big deal" is just on the induction hypothesis. But actually it IS important: here we have an example of a general statement which induction hypothesis is true but that is always false... because we can never find a base case!

Suppose true that, for a fixed , . Then, . But we obviously know that this statement is never true, so there can't be a base case.

### Induction on a general ordered set

Strengthen your inductive hypothesis

Transfinite induction is discussed in a separate article.

Using generators and closure properties

### Moved the discussion to the comments

When this article is written, it should have examples of several different styles of inductive proof: the usual kind where you deduce from , the slightly more sophisticated kind where you deduce it from the fact that is true for every , induction over more sophisticated well-ordered sets, use of the well-ordering principle, etc.

### Induction without an indexing set

"The generalized induction is harder because we don't have a nice indexing set like the natural numbers. Is there a way of doing induction without the crutch of an indexing set? Yes there is: you look for a minimal counterexample."

The above is from this article by Prof Gowers: http://www.dpmms.cam.ac.uk/~wtg10/bounded.html.

I feel that this could be put somewhere, but I am not sure where. The connection between the well-ordering principle and induction could also be mentioned (albeit briefly, since it doesn't seem like a trick).


text:
### The isolation and characterization of a senescence associated nodulin cDNA.

##### Description
 Title: The isolation and characterization of a senescence associated nodulin cDNA. Authors: Chan, Christina Chi-Yat. Date: 1995 Abstract: The goal of this thesis is to study cellular senescence with the use of the determinate nodule of soybean as the model system. The determinate nodule is an appropriate model system for the study of senescence due to the homogeneity of infected cells of various developmental stages within the organ. In this model system consisting of Glycine max (L.) Merrill, Maple Arrow, infected with Bradyrhizobium japonicum strain 61A76, we showed that nodule senescence occurs late in nodulation, at the time of flowering and pod filling. We also demonstrated that nodule senescence can be induced with the addition of 20mM ammonium nitrate. The strategy to study the molecular mechanisms of nodule senescence involves the isolation of senescence-associated nodulin (SAN) cDNAs by the differential screening of a cDNA library representing mRNA from senescent nodules. This library constructed in the $\lambda$gt11 vector is screened with two bulk probes made from the PCR amplification of the cDNA inserts of libraries in the $\lambda$gt10 vector: one representing mRNA of healthy nodules that are actively fixing nitrogen and the other of senescent nodules. In this thesis, one SAN cDNA, clone 8-4 was characterized through a Northern analysis to demonstrate the up-regulation of the corresponding gene with senescence. In the natural senescent system, SAN8-4 expression was demonstrated to increase with nodule senescence. A search through nucleotide and protein sequence banks demonstrates that SAN 8-4 may be a novel F3H related protein. (Abstract shortened by UMI.) URL: http://hdl.handle.net/10393/10335http://dx.doi.org/10.20381/ruor-8237 Collection Thèses, 1910 - 2010 // Theses, 1910 - 2010


text:
# Use of Quaternions to replace transformation matrices

This topic is 4072 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

I'm experimenting with shaders, and thought I could use quaternions instead of transformation matrices. This would reduce data sent from system ram to the video card and multiplications done for each transformation. The only thing I'm not sure how to do is translations. Can quaternions represent translations? Thanks. -Nick

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Quote:
 Original post by nickwintersCan quaternions represent translations?
No, not on their own. But that's not a problem: just use a quaternion to represent the rotation, and a separate vector to represent the translation.

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Quaternion cannot also represent scaling. So you would need two vectors and one quaternion to represent a full transformation which you could do with a single matrix. One vector for translation,one for scaling and the quaternion for rotation. But in the end you will still have to convert the quaternion data to a 3x3 matrix.

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Not that it has any practical value, but you can use a (additional) quaternion to represent translation:

Q_old_pos := [0, old_pos.x, old_pos.y, old_pos.z]Q_new_pos := Q_rot * Q_old_pos * ~Q_rot[_, new_pos.x, new_pos.y, new_pos.z] := Q_new_posnew_pos := new_pos + trans

use
Q_old_pos := [0, old_pos.x, old_pos.y, old_pos.z]Q_trans := [0, trans.x, trans.y, trans.z]Q_new_pos := Q_rot * Q_old_pos * ~Q_rot + Q_trans[_, new_pos.x, new_pos.y, new_pos.z] := Q_new_pos

But this is just a "notational trick" and makes no real difference in implementation.

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Quote:
 Original post by BornToCodeQuaternion cannot also represent scaling. So you would need two vectors and one quaternion to represent a full transformation which you could do with a single matrix. One vector for translation,one for scaling and the quaternion for rotation. But in the end you will still have to convert the quaternion data to a 3x3 matrix.

Quaternions =can= be used to represent uniform scalings:

whereas x -> Q.x.Q* with a unit quaternion Q represents a rotation with centre O, the same transformation with arbitrary Q represents a rotation with centre O combined with a uniform magnification (i.e. an isotropic blow-up or shrink-down) with a magnification factor |Q|^2.

Whenever Q <> 0 (or equivalently: |Q| > 0), the rotation part of this transformation is represented by the unit quaternion Q/|Q|.

With Q = 0, all of 3D space collapses into the single point O.

The practical meaning of all this remains to be seen: who has ever met the 3D space transformation group consisting of 3D displacements combined with uniform magnifications with factors > 0? (all orientation-preserving 3D similarity transformations)
Or with factors <> 0, positive and negative? (all 3D similarity transformations, preserving or reversing orientation)

Johan E. Mebius

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Anyone have a code/psuedocode example of how this would be done -- just the rotation. Say I have a cube composed of 4 vertices. How would I rotate it on a given axis using quaternions?

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Quote:
 Original post by drvannostrandAnyone have a code/psuedocode example of how this would be done -- just the rotation. Say I have a cube composed of 4 vertices. How would I rotate it on a given axis using quaternions?
Assuming you already know how to construct a quaternion representing a given axis-angle rotation, the equation to rotate a vector using a quaternion is then:
v' = q*v*q~
Or:
v' = q~*v*q
Here, q is the quaternion in question, ~ is the conjugate, and v is the vector in quaternion form, that is, stored in the imaginary part of the quat with the real part set to an arbitrary value (typically 0). The first form is the 'mathematically correct' form, but the second is also sometimes used (e.g. in the DirectX math library) to mirror the use of row-vector transforms.

In practice, quaternion rotations are rarely applied this way. Instead, the quaternion is converted to a matrix representing the rotation, and the matrix is used to actually apply the transformation.

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text:
# Cross Product Vector Proof

1. Oct 2, 2011

### planauts

Not sure if this is the correct section. I apologize if it's not.

1. The problem statement, all variables and given/known data
For any vectors $\vec{a}, \vec{b}, \vec{c}$ show that:
$(\vec{a} \times \vec{b} ) \times \vec{c}$
lies in the plane of $\vec{a}$ and $\vec{b}$

2. Relevant equations

3. The attempt at a solution

I assigned $\vec{a} = (e,f,g), \vec{b} = (x,y,z), \vec{c} = (s,t,u)$

then I used the cross product formula and got:

(u(gx-ez)-t(ey-fx), s(ey-fx)-u(fz-gy), t(fz-gy)-s(gx-ez))

which expanded comes to:

(gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

I'm not sure if that helps...
Thanks.

2. Oct 2, 2011

### vela

Staff Emeritus
Hint: If a vector lies in the plane of A and B, you can express it as a linear combination of the two vectors. So see if you can write your result in the form

(something)x(e, f, g) + (something)x(x, y, z)

3. Oct 2, 2011

### Dick

Try using what you should know about the geometry of the cross product. axb is perpendicular to a and b, right? So it's a normal to the plane of a and b. Anything that is perpendicular to that normal is in the plane of a and b. Is (axb)xc perpendicular to axb?

4. Oct 2, 2011

### planauts

Yes, that makes sense. I get it :D But is it possible to prove it algebraically?

I don't understand :S Could you explain it? Do you mean:
(something)*(e, f, g) + (something)*(x, y, z)

:S

5. Oct 2, 2011

### Dick

Sure, that's what vela meant. And you can find those two somethings with a little work. Write (axb)xc=i*a+j*b and try to find i and j.

6. Oct 2, 2011

### planauts

Ah... I see...
So it would be like:

(ae+bx, af+by, ag+bz) where a and b are the somethings... right?

The goal is to get something like this:
(gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

right.

the constants s, t, u could be formed by multiplying the a and b "something" right?

Am I on the right track?

-----

EDIT:
Is that the long way of doing it? Is there an easier way? I feel that there is an easier way of solving it algebraically...

7. Oct 3, 2011

### Dick

You already used 'a' and 'b' for the vectors, so I like (axb)xc=i*a+j*b better. But, yes, that's the general idea. That's three linear equations in the two unknown constants i and j. You can solve it systematically. There's a lot of letters flying around but everything is a constant except for i and j. I like the geometrical approach better, but you can pull it off algebraically.


text:
# Chapter 7 - Linear Momentum - Questions: 4

Consider the filled balloon as our system. The air inside the balloon shoots out of the balloon’s opening. The system’s momentum is conserved. The escaping gas gains momentum one way, and the balloon and its remaining gas gain momentum the other way, i.e., they move in the opposite direction.

#### Work Step by Step

This situation may also be analyzed in terms of Newton’s third law.

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.


text:
# On orthogonal decomposition of a Sobolev space

Document Type: Original Article

Author

Bryant & Stratton College

Abstract

The theme of this short article is to investigate an orthogonal decomposition of the Sobolev space $W^{1,2}\left( \Omega \right)$ as $W^{1,2}\left( \Omega \right) =A^{2,2}\left( \Omega \right) \oplus D^{2}\left( W_{0}^{3,2}\left( \Omega \right) \right)$ and look at some of the properties of the inner product therein and the distance defined from the inner product. We also determine the dimension of\ the orthogonal difference space $W^{1,2}\left( \Omega \right) \ominus \left(W_{0}^{1,2}\left( \Omega \right) \right) ^{\perp }$ and show the expansion of Sobolev spaces as their regularity increases.

Keywords

### Histroty

• Receive Date: 11 March 2017
• Revise Date: 07 June 2017
• Accept Date: 07 June 2017


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# gampdf

Gamma probability density function

## Syntax

``y = gampdf(x,a)``
``y = gampdf(x,a,b)``

## Description

example

````y = gampdf(x,a)` returns the probability density function (pdf) of the standard gamma distribution with the shape parameter `a`, evaluated at the values in `x`.```

example

````y = gampdf(x,a,b)` returns the pdf of the gamma distribution with the shape parameter `a` and the scale parameter `b`, evaluated at the values in `x`.```

## Examples

collapse all

Compute the density of the observed value `5` in the standard gamma distribution with shape parameter `2`.

`y1 = gampdf(5,2)`
```y1 = 0.0337 ```

Compute the density of the observed value `5` in the gamma distributions with shape parameter `2` and scale parameters `1` through `5`.

`y2 = gampdf(5,2,1:5)`
```y2 = 1×5 0.0337 0.1026 0.1049 0.0895 0.0736 ```

## Input Arguments

collapse all

Values at which to evaluate the pdf, specified as a nonnegative scalar value or an array of nonnegative scalar values.

• To evaluate the pdf at multiple values, specify `x` using an array.

• To evaluate the pdfs of multiple distributions, specify `a` and `b` using arrays.

If one or more of the input arguments `x`, `a`, and `b` are arrays, then the array sizes must be the same. In this case, `gampdf` expands each scalar input into a constant array of the same size as the array inputs. Each element in `y` is the pdf value of the distribution specified by the corresponding elements in `a` and `b`, evaluated at the corresponding element in `x`.

Example: `[3 4 7 9]`

Data Types: `single` | `double`

Shape parameter of the gamma distribution, specified as a positive scalar value or an array of positive scalar values.

• To evaluate the pdf at multiple values, specify `x` using an array.

• To evaluate the pdfs of multiple distributions, specify `a` and `b` using arrays.

If one or more of the input arguments `x`, `a`, and `b` are arrays, then the array sizes must be the same. In this case, `gampdf` expands each scalar input into a constant array of the same size as the array inputs. Each element in `y` is the pdf value of the distribution specified by the corresponding elements in `a` and `b`, evaluated at the corresponding element in `x`.

Example: `[1 2 3 5]`

Data Types: `single` | `double`

Scale parameter of the gamma distribution, specified as a positive scalar value or an array of positive scalar values.

• To evaluate the pdf at multiple values, specify `x` using an array.

• To evaluate the pdfs of multiple distributions, specify `a` and `b` using arrays.

If one or more of the input arguments `x`, `a`, and `b` are arrays, then the array sizes must be the same. In this case, `gampdf` expands each scalar input into a constant array of the same size as the array inputs. Each element in `y` is the pdf value of the distribution specified by the corresponding elements in `a` and `b`, evaluated at the corresponding element in `x`.

Example: `[1 1 2 2]`

Data Types: `single` | `double`

## Output Arguments

collapse all

pdf values evaluated at the values in `x`, returned as a scalar value or an array of scalar values. `y` is the same size as `x`, `a`, and `b` after any necessary scalar expansion. Each element in `y` is the pdf value of the distribution specified by the corresponding elements in `a` and `b`, evaluated at the corresponding element in `x`.

collapse all

### Gamma pdf

The gamma distribution is a two-parameter family of curves. The parameters a and b are shape and scale, respectively.

The gamma pdf is

`$y=f\left(x|a,b\right)=\frac{1}{{b}^{a}\Gamma \left(a\right)}{x}^{a-1}{e}^{\frac{-x}{b}},$`

where Γ( · ) is the Gamma function.

The standard gamma distribution occurs when b = 1.

## Alternative Functionality

• `gampdf` is a function specific to the gamma distribution. Statistics and Machine Learning Toolbox™ also offers the generic function `pdf`, which supports various probability distributions. To use `pdf`, create a `GammaDistribution` probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. Note that the distribution-specific function `gampdf` is faster than the generic function `pdf`.

• Use the Probability Distribution Function app to create an interactive plot of the cumulative distribution function (cdf) or probability density function (pdf) for a probability distribution.

## Version History

Introduced before R2006a


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#### 2.8 Summary of Research Achievements

In summary, within the scope of SiC oxidation, the following research goals have been achieved:

• 1.  A full set of dry SiC oxidation growth rate coefficients for the $$(0001)$$ Si-, $$(11\bar{2}0)$$ a-, $$(1\bar{1}00)$$ m-, and $$(000\bar{1})$$ C-face has been obtained, which enables high-accuracy simulations of SiC oxidation [72].

• 2.  A novel 3D interpolation method to compute oxidation growth rate coefficients for arbitrary crystal direction has been developed and calibrated. The direction-dependent interpolation method captures oxidation anisotropy of SiC and thus enables the simulation of SiC oxidation for arbitrary crystal orientations and device structures [73].

• 3.  The initial SiC oxidation stages (in the regime of ns) have been investigated in order to extend the understanding of the orientation-dependent oxidation mechanisms and to better estimate initial (i.e., native) oxide thicknesses from first principles simulations [77].


text:
# 4-derivative kinetic term Lagrangian

## Homework Statement

Show that $$L=\phi\Box^2\phi$$ generates negative energy density.

## The Attempt at a Solution

The energy density is $$E=\frac{\partial L}{\partial \dot{\phi}}\dot{\phi}-L$$ Also the Lagrangian can be rewritten (using divergence theorem) as $$L=-\partial_\mu\phi\partial_\mu(\Box\phi)$$ So I would get $$E=-\partial_0\phi\partial_0(\Box\phi)+\partial_\mu\phi\partial_\mu(\Box\phi)$$ $$E=-\partial_x\phi\partial_x(\Box\phi)-\partial_y\phi\partial_y(\Box\phi)-\partial_z\phi\partial_z(\Box\phi)$$. Why is this negative necessarily? Thank you!

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Also the Lagrangian can be rewritten (using divergence theorem) as
L=−∂μϕ∂μ(□ϕ)
No, it cannot. You need to remove the d’Alembertian from this expression.

No, it cannot. You need to remove the d’Alembertian from this expression.
I am not sure what you mean. I used $$\int(\phi\Box^2\phi)=\int\partial_\mu(\phi\partial_\mu \Box \phi)-\int(\partial_\mu\phi\partial_\mu \Box \phi)$$ What is wrong? What do you mean by getting rid of the d'Alembertian?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
I am not sure what you mean. I used $$\int(\phi\Box^2\phi)=\int\partial_\mu(\phi\partial_\mu \Box \phi)-\int(\partial_\mu\phi\partial_\mu \Box \phi)$$ What is wrong? What do you mean by getting rid of the d'Alembertian?
This is simply not true. What you are looking for is
$$\phi \Box \phi = \partial_\mu (\phi \partial^\mu \phi) - (\partial_\mu \phi)(\partial^\mu\phi)$$

Orodruin
Staff Emeritus


text:
#### Sum-rate maximization of OFDMA femtocell networks that incorporates the QoS of macro mobile stations

##### Chunguo Li

This paper proposes a power allocation scheme with co-channel allocation for a femto base station (BS) that maximizes the sum-rate of its own femto mobile stations (MSs) with a constraint that limits the degradation of quality of service (QoS) of macro MSs. We have found a closed-form solution for the upper limit on the transmission power of each sub-channel that satisfies the constraint in a probabilistic sense. The proposed scheme is practical since it uses only the information easily obtained by the femto BS. Moreover, our scheme meets the constraint with minimal degradation compared to the optimal sum-rate of the femto MSs achieved without the constraint.

arrow_drop_up


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# Why is carrier wave affected by percentage modulation?

In amplitude modulation, Power of a carrier wave is represented by : $$P_c = \frac{A_c ^2}{2R}$$

Which is independent of modulation index!

Question: At what factor is the power of carrier wave dependent on!?

A conventional AM signal can be written as

$$x_{AM}(t)=A_c\big(1+\alpha\; m(t)\big)\cos(\omega_ct)\tag{1}$$

where $$\alpha$$ is the modulation index and $$m(t)$$ is the message signal. In Eq. $$(1)$$ it is assumed that the message signal $$m(t)$$ is normalized, i.e., $$\max_t|m(t)|= 1$$. The average power of $$x_{AM}(t)$$ is

$$\overline{x^2_{AM}(t)}=\frac{A_c^2}{2}\left(1+\alpha^2\overline{m^2(t)}\right)\tag{2}$$

If you assume now that the power $$(2)$$ is fixed to some value $$P$$, then the total available power must divided between the carrier term and the message-bearing term in $$(2)$$. Equating $$(2)$$ with the available power $$P$$, and expressing the carrier term $$A_c^2/2$$ in terms of the modulation index and the average power of the message signal gives

$$\frac{A_c^2}{2}=\frac{P}{1+\alpha^2\overline{m^2(t)}}\tag{3}$$

In the figure in your question it looks like they assumed $$\overline{m^2(t)}=\frac12$$, which for a maximum modulation index $$\alpha=1$$ results in

$$\frac{A_c^2}{2}=\frac23 P\tag{4}$$

This value corresponds to the right-most point of the curve labeled "carrier".


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# Points in sites (etale, fppf, … )

I asked a part of this in an earlier question, but that part of my question didn't receive precedence.

Etale site is useful - examples of using the small fppf site?

Let $X$ be a scheme (assume it is as nice as you like). There is a description of "points" in the (small) etale site $X_{et}$, and these are the geometric points of $X$. More generally, I've heard that the notion of "points" makes sense in any site (maybe "any" is a little too strong?).

1.) Can you give me a reference defining "points" in other sites. Specifically, I am interested in the small fppf site over a scheme and the big etale site. Is the notion of "points" a useless notion in sites other Zariski and small etale?

2.) What are "points" in other sites "supposed" to do? Is there an analogy that we keep in mind (as to why they are called points)? In the case of the Zariski site, the "points" have a natural structure of a locally ringed space - (the local rings being the stalks in the Zariski site) and this gives a canonically associated locally ringed space to a given site. An analogy similar to this doesn't seem to hold in the small etale site over a scheme.

3.) To whatever a "point" is, I expect one would have a naturally associated local ring. Is this the case in the small fppf site over a (nice?) scheme? This is of course the case in the etale and Zariski site. The small fppf site over a scheme seems a little strange, since limits tend not to be directed.

-
There's an example of Deligne of a site with no points. – arsmath Dec 30 '12 at 8:35
Regarding 1) : in SGA4 you can find the more general definition of a point of a topos, it is simply a morphism from the topos Sets to the given topos. – Niels Dec 30 '12 at 9:22

See SGA 4, Exposé VIII, 7.8, which defines an abstract point of a site as a functor from the topos of that site (i.e. the category of sheaves of sets on that site) to the category of sets that commutes with arbitrary inductive (direct) limits and finite projective (inverse) limits.*

One should think of a functor as being the functor that takes a sheaf to its stalk at that point (the "fiber functor" of the point). Since every fiber functor preserves such limits, these are reasonable axioms to postulate.

Grothendieck then proves in this exposé that in the étale site, every functor satisfying the axiom above comes from a (unique) geometric point (up to isomorphism in a suitably-defined sense).

Furthermore, and I can only find this referenced in Brian Conrad's unavailable draft book on the Ramanujan conjecture, in a (classical) topological space where every irreducible subset has a unique generic point (also known as a sober space), such as a normal Hausdorff space or the underlying space of a scheme, every abstract point of the category of sheaves on the topological space corresponds to a unique classical point. This is Proposition 2, Section 3, Chapter IX of Sheaves in Geometry and Logic.

Why characterize points in terms of their functors? This fits in with the general Tannakian formalism, which emphasizes fiber functors, and in particular, defines the fundamental group to be the group of automorphisms of the fiber functor. I.e., we should think of a point as the functor that assigns to a sheaf its stalk at that point.

This also fits into a more general philosophy that we should ignore the site and focus on the topos entirely. One can in fact put a Grothendieck topology on the topos so that it is equal to the category of sheaves on itself.

*Edit: One can also find this in Sheaves in Geometry and Logic, by Maclane and Moerdijk, Chapter VII, Section 5.

-
Spaces of the form you mention in your fourth paragraph are called "sober", and the correspondence between points of the topos and points of the space is a straightforward corollary of the fact that sober spaces can be recovered up to homeomorphism from their frame of open subsets, which is manifestly frame of subterminal objects of the corresponding sheaf topos. This can be found in the usual topos theory references such as Sheaves in geometry and logic or Sketches of an elephant. – Zhen Lin Dec 30 '12 at 11:03


text:
Composition of Functions

The “composition of functions” is the process of combining two or more functions in a way where the output of one function is the input for the next function.

Step by step guide to solve Composition of Functions

• The term “composition of functions” is simply the combination of two or more functions where the output from one function becomes the input for the next function.
• The notation used for composition is: $$\color{blue}{(f \ o \ g)(x)=f(g(x))}$$

Example 1:

Using $$f(x)=x-2$$ and $$g(x)=x$$, find: $$f(g(2))$$

Solution:

$$(f \ o \ g)(x)=f(g(x))$$
Then: $$(f \ o \ g)(x)=f(g(x))=f(x)=x-2$$
Substitute $$x$$ with $$2: (f \ o \ g)(2)=2-2=0$$

Example 2:

Using $$f(x)=x+8$$ and $$g(x)=x-2$$, find: $$g(f(4))$$

Solution:

$$(f \ o \ g)(x)=f(g(x))$$
Then: $$(g \ o \ f)(x)=g(f(x))=g(x+8)$$, now substitute $$x$$ in $$f(x)$$ by $$x+8$$. Then: $$g(x+8)=(x+8)-2=x+8-2=x+6$$
Substitute $$x$$ with $$4: (g \ o \ f)(4)=g(f(x))=4+6=10$$

Example 3:

Using $$f(x)=x+2$$ and $$g(x)=4x$$, find: $$f(g(1))$$

Solution:

$$(f \ o \ g)(x)=f(g(x))$$
Then: $$(f \ o \ g)(x)=f(g(x))=f(4x)=4x+2$$
Substitute $$x$$ with $$1: (f \ o \ g)(1)=4+2=6$$

Example 4:

Using $$f(x)=5x+4$$ and $$g(x)=x-3$$, find: $$g(f(3))$$

Solution:

$$(f \ o \ g)(x)=f(g(x))$$
Then: $$(f \ o \ g)(x)=g(f(x))=g(5x+4)$$, now substitute $$x$$ in $$g(x)$$ by $$5x+4$$. Then: $$g(5x+4)=(5x+4)-3=5x+4-3=5x+1$$
Substitute $$x$$ with $$3: (g o f)(x)=g(f(x))=5x+1=5(3)+1=15=1=16$$

Exercises

Using $$f(x) = 5x + 4$$ and $$g(x) = x – 3$$, find:

• $$\color{blue}{f(g(6))}$$
• $$\color{blue}{f(f(8))}$$
• $$\color{blue}{g(f(– 7))}$$
• $$\color{blue}{g(f(x))}$$

• $$\color{blue}{19}$$
• $$\color{blue}{224}$$
• $$\color{blue}{-34}$$
• $$\color{blue}{5x+1}$$

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text:
# Math Help - arcten2+arctan3

1. ## arcten2+arctan3

Dear all!
Please, how do I get that for x>1 all $arctan\frac{x+1}{x-1}+arctan(x)=arctan2 + arctan3$

Many thanks!

2. ## Re: arcten2+arctan3

Originally Posted by Boo
Dear all!
Please, how do I get that for x>1 all $arctan\{x+1}{x-1}=arctan2 + arctan3$

Many thanks!
Again, more clarification. x + 1x -1 = 2x - 1. Possibly x^2 + x - 1?

Please edit these before you post!

-Dan