text:
Preparing NOJ

# Digital Display

1000ms 65536K

## Description:

There is a suspicion that some officers cheat when reporting exact time of various events. They
antedate important documents and gain enormous profit by that. To eliminate such practices,
ACM decided to install big digital clock on the wall of each office. Thank to this, everyone can
easily see the current time and verify that it is correctly recorded on documents.
Your task is to write a firmware for this digital clock. Digits will be shown using the classical
seven-segment display.

## Input:

On each line of the input, there are exactly five characters: two digits, a colon (“ : ”) and another
two digits. These characters specify a time between 00:00 and 23:59, inclusive.
The last time is followed by a line containing the word “ e n d”.

## Output:

For each input line, print a display consisting of seven rows of exactly 29 characters each.
The only characters allowed are dashes (“ -”) for horizontal segments, pipes (“ | ”) for vertical
segments, plus signs (“ +”) for corners, small latin letter “ o” for the colon between hours and
minutes, and spaces (“ ”).
The area for each digit is exactly 5 characters wide and there are two empty columns (spaces)
between neighboring digit areas or between a digit area and the colon. Exact digit shapes and
placements can be seen in Sample Output.
Always print two empty lines after each time displayed. Also, print the word “ e n d” at the end
of the output.
Notice In this problem, the meaning of evaluation system responses is a little bit different
than usual. If you get a Presentation Error, it most probably means that you did not follow the
output format exactly. Note that it does not necessarily imply that your answer is close to the
correct one. On the other hand, a Wrong Answer will be issued if the format seems basically
all-right but the contents of the display are not correct. If your output contains both errors, the
decision between Wrong Answer and Presentation Error is solely at the discretion of the system.
Therefore, you should never try to derive any important conclusions from the response.

## Sample Input:

1 6 : 4 72 3 : 5 90 0 : 0 8e n d

## Sample Output:

+ + - - - + + + + - - - +| | | | || | o | | |+ + - - - + + - - - + +| | | o | || | | | |+ + - - - + + ++ - - - + + - - - + + - - - + + - - - +| | | | || | o | | |+ - - - + + - - - + + - - - + + - - - +| | o | || | | |+ - - - + + - - - + + - - - + + - - - ++ - - - + + - - - + + - - - + + - - - +| | | | | | | || | | | o | | | |+ + + + + + + - - - +| | | | o | | | || | | | | | | |+ - - - + + - - - + + - - - + + - - - +

## Note:

undefined

Info

Provider NOJ

Code NOJ1561

Tags

Submitted 3

Passed 1

AC Rate 33.33%

Date 04/20/2019 10:03:10

Related

Nothing Yet


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Page 1
Aside

## e - like everywhere

Everybody doing any kind of geometry, calculus or higher algebra will have stumbled over this number:

2.71828182845904523536028747135266249775724709369995...

More commonly it is known as $e$ , Euler's number. You may wonder, where does this number come from and why is it so important? Well... If you are like me that is: waiting for simulations to finish, having 20 minutes of spare time on your hands, and having a tenacious drive to understand things. So come along as I explain two different ways that result explain where this beautiful number comes from.

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Standard

## MATLAB® and Objects

Working with objects once a project becomes sufficiently complex makes it much easier to keep track of things. MATLAB® (a.k.a. MATrix LABoratory) is a nice product by a company called MathWorks that makes matrix manipulation incredibly easy. For us engineers and academics it is the standard goto, regardless if we need to rapid prototype some functionality or develop a complex and convoluted system. Only recently did I discover a whole hidden side of  MATLAB®: classes! And because I am an incredibly big fan of those beasts I decided to write a quick summary of how to use them.

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## My trip to Germany and back

When I last drove to Munich and back, in order to visit my family and girlfriend, I decided to do something especially geeky: track my driving progress. The reason behind me tracking myself was two folded. For one I wanted to see if I could write a full app, front- and back-end to track myself across Europe, but mostly I wanted to keep my folks in the loop of where I am. Usually this was done by repetitively calling or, when having a break, posting or texting many to notify them of my progress. But this is neither live, precise, nor is it a pretty solution. Over a few posts (still to come) I will show you how I:

• designed the iOS app (in Swift with MVC paradigm),
• made a horribly insecure web server to track the (sometimes out of order) data, and
• displayed the entire journey (in correct order) on a Google Map on my webpage.

But for now, here are the tracking results and me waffling about my journey.

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Aside

## Workaround MATLAB's intellisense/auto-complete issues for user defined classes

This will be a short post, but a little wakeup call for myself, and others that may get annoyed by the same issue. The issue at hand is, that MATLAB does not seem to have fully functioning intellisense/auto-complete when dealing with a class definition that spans over multiple files. Let me elaborate...

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Aside

## Fix to MicroPython's Memory Allocation Problem on ESP8266

I've been trying to write code that allows me to display text on my ST7735 display. The font module I found is around 4kB of source code that needs to be loaded by Python, compiled into byte code and executed by the Python Virtual Machine, or VM (for more details refers here). This process works somewhat like Java's VM approach, but on devices with limited memory, i.e. the ESP8266, can become troublesome. Loading, compiling and storing the byte code takes up precious memory, which may not always be allocatable; thus throwing an MemoryError: memory allocation failed, allocating XXX bytes.. In this little post I want to show you how you can pre-compile modules for the ESP8266, so you will avoid wasting space. And if that option is too cumbersome for you, I also present a little hack to increase the ESP's heap memory, but that's more of a bodge-job for lazy people: i.e. me :p

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# How to get sensor covariance matrix from CSI values?

I try to use musicdoa() function in Matlab to estimate the direction of arrival of the paths of signals received by the receiver in a wireless link. The function needs sensor covariance matrix as one of the arguments.

I have a matrix of Channel State Information (CSI) that is obtained from a WiFi card. The CSI values come in a matrix shown below:

$$C=\begin{bmatrix}c_{1,1} & c_{1,2} & ... & c_{1,30}\\ c_{2,1} & c_{2,2} & ... & c_{2,30}\\ c_{3,1} & c_{3,2} & ... & c_{3,30}\end{bmatrix}$$

In which, $$c_{i,j}$$ is the CSI value for $$i^{th}$$ antenna on $$j^{th}$$ subcarrier.

How can I get the sensor covariance matrix using the C matrix?

The Channel State Information (CSI) matrix is calculated over a packet of known signals that is sent as part of the communications protocol.

The CSI matrix contains two things (per carrier). The bearing of the "target" and the "strength" of transmission along the line of that bearing. So basically, this can tell you where is the transmitter with respect to the receiver array.

The CSI matrix is calculated via cross correlation between the signals received by each antenna and correlation is normalised covariance.

So, strictly speaking, it is impossible to derive covariance from the CSI because you need access to the received signals themselves. But, the angle (bearing) component of the CSI could be considered as being proportional to the real covariance for the pilot tones that are used to generate them. You could even obtain some theoretical covariance from the pilot tones by delaying them progressively but that would not include the covariance introduced due to noise.


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## Wednesday, July 7, 2021

### Upcoming features of QSoas and github repository

For the past years, most of the development has happened behind the scene in a private repository, and the code has appeared in the public repository only a couple of months before the release, in the release branch. I have now decided to publish the current code of QSoas in the github repository (in the public branch). This way, you can follow and use all the good things that were developed since the last release, and also verify whether any bug you have is still present in the currently developed version !

#### Upcoming features

This is the occasion to write a bit about the some of the features that have been added since the publication of the 3.0 release. Not all of them are polished nor documented yet, but here are a few teasers. The current version in github has:
• a comprehensive handling of column/row names, which makes it much easier to work with files with named columns (like the output files QSoas produces !);
• better handling of lists of meta-data, when there is one value of the meta for each segment or each Y column;
• handling of complex numbers in apply-formula;
• defining fits using external python code;
• a command for linear least squares (which has the huge advantage of being exact and not needing any initial parameters);
• commands to pause in a script or sort datasets in the stack;
• improvements over previous commands, in particular with eval;
• ... and more...
Check out the github repository if you want to know more about the new features !

As of now, no official date is planned for the 3.1 release, but this could happen during fall.

QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there.

## Sunday, June 13, 2021

### Solution for QSoas quiz #2: averaging several Y values for the same X value

This post describes two similar solutions to the Quiz #2, using the data files found there. The two solutions described here rely on split-on-values. The first solution is the one that came naturally to me, and is by far the most general and extensible, but the second one is shorter, and doesn't require external script files.

#### Solution #1

The key to both solution is to separate the original data into a series of datasets that only contain data at a fixed value of x (which corresponds here to a fixed pH), and then process each dataset one by one to extract the average and standard deviation. This first step is done thus:
QSoas> load kcat-vs-ph.dat
QSoas> split-on-values pH x /flags=data

After these commands, the stacks contains a series of datasets bearing the data flag, that each contain a single column of data, as can be seen from the beginnings of a show-stack command:
QSoas> k
Normal stack:
F  C	Rows	Segs	Name
#0	(*) 1	43	1	'kcat-vs-ph_subset_22.dat'
#1	(*) 1	44	1	'kcat-vs-ph_subset_21.dat'
#2	(*) 1	43	1	'kcat-vs-ph_subset_20.dat'
...

Each of these datasets have a meta-data named pH whose value is the original x value from kcat-vs-ph.dat. Now, the idea is to run a stats command on the resulting datasets, extracting the average value of x and its standard deviation, together with the value of the meta pH. The most natural and general way to do this is to use run-for-datasets, using the following script file (named process-one.cmds):
stats /meta=pH /output=true /stats=x_average,x_stddev

So the command looks like:
QSoas> run-for-datasets process-one.cmds flagged:data

This command produces an output file containing, for each flagged dataset, a line containing x_average, x_stddev, and pH. Then, it is just a matter of loading the output file and shuffling the columns in the right order to get the data in the form asked. Overall, this looks like this:
l kcat-vs-ph.dat
split-on-values pH x /flags=data
output result.dat /overwrite=true
run-for-datasets process-one.cmds flagged:data
l result.dat
apply-formula tmp=y2;y2=y;y=x;x=tmp
dataset-options /yerrors=y2

The slight improvement over what is described above is the use of the output command to write the output to a dedicated file (here result.dat), instead of out.dat and ensuring it is overwritten, so that no data remains from previous runs.

#### Solution #2

The second solution is almost the same as the first one, with two improvements:
This yields the following, smaller, solution:
l kcat-vs-ph.dat
split-on-values pH x /flags=data
stats /meta=pH /accumulate=* /stats=x_average,x_stddev /buffers=flagged:data
pop
apply-formula tmp=y2;y2=y;y=x;x=tmp
dataset-options /yerrors=y2

QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there.

## Sunday, May 30, 2021

### QSoas quiz #2: averaging several Y values for the same X value

This second quiz may sound like the first one, but in fact, the approach used is completely different. The point is to gather some elementary statistics from a series of experiments performed under different conditions, but with several repeats at the same conditions.

#### Quiz

You are given a file (which you can download there) that contains a series of pH value data: the X column is the pH, the Y column the result of the experiment at the given pH (let's say the measure of the catalytic rate of an enzyme). Your task is to take this data and produce a single dataset which contains, for each pH value, the pH, the average of the results at that pH and the standard deviation. The result should be identical to the following file, and should look like that:
There are several ways to do this, but all ways must rely on stats, and the more natural way in QSoas is to take advantage of split-on-values, which is a very powerful command but somehow hard to master, which is the point of this Quiz.
By the way, the data file is purely synthetic, if you look in the GitHub repository, you'll see how it was generated.

QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there.

## Sunday, May 16, 2021

### Tutorial: analyze redox inactivations/reactivations

Redox-dependent inactivations are actually rather common in the field of metalloenzymes, and electrochemistry can be an extremely powerful tool to study them, providing one can analyze the data quantitatively. The point of this point is to teach the reader how to do so using QSoas. For more general information about redox inactivations and how to study them using electrochemical techniques, the reader is invited to read the review del Barrio and Fourmond, ChemElectroChem 2019.

This post is a tutorial to learn the analysis of data coming from the study of the redox-dependent substrate inhibition of periplasmic nitrate reductase NapAB, which has the advantage of being relatively simple. The whole processed is discussed in Jacques et al, BBA, 2014. What you need to know in order to follow this tutorial is the following:

• the whole inactivation/reactivation process can be modelled by a simple reversible reaction: $$\mathrm{A} \rightleftharpoons \mathrm{I}$$ A is the active form, I the inactive form;
• the forward rate constant is $$k_i(E)$$ (dependent on potential) and the backward rate constant is $$k_a(E)$$, also dependent on potential;
• the experiment is done in a series of 5 steps at 3 different potentials: $$E_0$$ then $$E_1$$ then $$E_2$$ then $$E_1$$ then, finally, $$E_0$$;
• the enzyme is assumed to be fully active at the beginning of the first step;
• a single experiment is used to obtain the values of $$k_i$$ and $$k_a$$ for the three potentials (although not reliably for the value at $$E_0$$
• the current given by the active species depends on potential (and it is negative because the enzyme catalyzes a reduction), and the inactive species gives no current;
• in addition to the reversible reaction above, there is an irreversible, potential-dependent loss.
You can download the data files from the GitHub repository. Before fitting the data to determine the values of the rate constants at the potentials of the experiment, we will first subtract the background current, assuming that the respective contributions of faradaic and non-faradaic currents is additive. Start QSoas, go to the directory where you saved the files, and load both the data file and the blank file thus:
QSoas> cd
QSoas> S 1 0

(after the first command, you have to manually select the directory in which you downloaded the data files). The S 1 0 command just subtracts the dataset 1 (the first loaded) from the dataset 0 (the last loaded), see more there. blanc is the French for blank...

Then, we remove a bit of the beginning and the end of the data, corresponding to one half of the steps at $$E_0$$, which we don't exploit much here (they are essentially only used to make sure that the irreversible loss is taken care of properly). This is done using strip-if:

QSoas> strip-if x<30||x>300

Then, we can fit ! The fit used is called fit-linear-kinetic-system, which is used to fit kinetic models with only linear reactions (like here) and steps which change the values of the rate constants but do not instantly change the concentrations. The specific command to fit the data is:
QSoas> fit-linear-kinetic-system /species=2 /steps=0,1,2,1,0

The /species=2 indicates that there are two species (A and I). The /steps=0,1,2,1,0 indicates that there are 5 steps, with three different conditions (0 to 2) in order 0,1,2,1,0. This fits needs a bit of setup before getting started. The species are numbered, 1 and 2, and the conditions (potentials) are indicated by #0, #1 and #2 suffixes.
• The I_1 and I_2 are the currents for the species 1 and 2, so something for 1 (active form) and 0 for 2 (inactive form). Moreover, the parameters I_2_#0 (and _#1, _#2) should be fixed and not free (since we don't need to adjust a current for the inactive form).
• The k_11 and k_22 correspond to species-specific irreversible loss. It is generally best to leave them fixed to 0.
• k_12 is the formation of 2 (I) from 1 (A), and k_21 is the formation of A from I. Their values will be determined for the three conditions. The default values should work here.
• The k_loss parameters are the rates of irreversible loss that apply indiscriminately on all species (unlike k_11 and k_22). They are adjusted and ther default values should work too.
• alpha_1_0 and alpha_2_0 are the initial concentrations of species 1 and 2, so they should be fixed to 1 and 0.
• Last, the xstart_a and (_b, _c, _d and _e) correspond to the starting times for the steps, here, 0, 60, 120, 210 and 270.
For the sake of simplicity, you can also simply load the starting-parameters.params parameters to have all setup the correct way. Then, just hit Fit, enjoy this moment when QSoas works and you don't have to... The screen should now look like this:
Now, it's done ! The fit is actually pretty good, and you can read the values of the inactivation and reactivation rate constants from the fit parameters.

You can train also on the 21.oxw and 21-blanc.oxw files. Usually, re-loading the best fit parameters from other potentials as starting parameters work really well. Gathering the results of several fits into a real curve of rate constants as a function of potentials is left as an exercise for the reader (or maybe a later post), although you may find these series of posts useful in this context !

QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there.

## Thursday, March 11, 2021

### All tips and tricks about QSoas

I've decided to post regular summaries of all the articles written here about QSoas; this is the first post of this kind. All the articles related to QSoas can be found here also.

The articles written here can be separated into several categories.

#### Tutorials to analyze real data

These are posts about how to reproduce the data analysis of published articles, including links to the original data so you can fully reproduce our results.
These posts all have the label tutorial.

QSoas has a particularly powerful interface for non-linear least square minimisations (fits):
• See here to learn how you can take advantage of the fit interface to explore easily how a parameter influences the shape of a function.
• See here how you can produce smooth curves for a fit on jaggy data.

#### Meta-data

Meta data describe the conditions in which experiments were performed.
• You can see here how to use them in data analysis, to plot for instance peak position as a function of meta-data;
• You can see here how to permanently store meta-data for files.

#### Quiz and their solutions

Quiz are small problems that take some skill to solve; they can teach you a lot about how to work with QSoas.
• Quiz #1: computing the standard deviation of spectra, along with the solution. This quiz can teach you a lot about combining data from different datasets and manipulating data row-by-row.

#### Other tips and tricks

• See here how to find the 0s of experimental data.
• See here how one can save just the selected points in a baseline, and reuse them.
• See here to learn how to generate many datasets in one go from a mathematical formula.
• See here how to define a custom mathematical function using Ruby.
• See here how one can take advantage of Ruby, the underlying programming language to sum columns, extend the values of columns with lots of missing values and rename datasets using a pattern.
• See here how you can use QSoas without starting the graphical interface !

#### Release annoucements

These have generally lot of general information about the possibilities in QSoas:

QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there.

## Tuesday, February 16, 2021

### QSoas tips and tricks: permanently storing meta-data

It is one thing to acquire and process data, but the data themselves are most often useless without the context, the conditions in which the experiments were made. These additional informations can be called meta-data. In a previous post, we have already described how one can set meta-data to data that are already loaded, and how one can make use of them.

QSoas is already able to figure out some meta-data in the case of electrochemical data, most notably in the case of files acquired by GPES, ECLab or CHI potentiostats. However, only a small number of constructors are supported as of now[1], and there are a number of experimental details that the software is never going to be able to figure out for you, such as the pH, the sample, what you were doing...

The new version of QSoas provides a means to permanently store meta-data for experimental data files:

QSoas> record-meta pH 7 file.dat
This command uses record-meta to permanently store the information pH = 7 for the file file.dat. Any time QSoas loads the file again, either today or in one year, the meta-data will contain the value 7 for the field pH. Behind the scenes, QSoas creates a single small file, file.dat.qsm, in which the meta-data are stored (in the form of a JSON dictionnary).

You can set the same meta-data to many files in one go, using wildcards (see load for more information). For instance, to set the pH=7 meta-data to all the .dat files in the current directory, you can use:

QSoas> record-meta pH 7 *.dat
You can only set one meta-data for each call to record-meta, but you can use it as many times as you like.

Finally, you can use the /for-which option to load or browse to select only the files which have the meta you need:

QSoas> browse /for-which=$meta.pH<=7 This command browses the files in the current directory, showing only the ones that have a pH meta-data which is 7 or below. [1] I'm always ready to implement the parsing of other file formats that could be useful for you. If you need parsing of special files, please contact me, sending the given files and the meta-data you'd expect to find in those. #### About QSoas QSoas is a powerful open source data analysis program that focuses on flexibility and powerful fitting capacities. It is released under the GNU General Public License. It is described in Fourmond, Anal. Chem., 2016, 88 (10), pp 5050–5052. Current version is 3.0. You can download its source code there (or clone from the GitHub repository) and compile it yourself, or buy precompiled versions for MacOS and Windows there. ## Wednesday, January 13, 2021 ### Taking advantage of Ruby in QSoas First of all, let me all wish you a happy new year, with all my wishes of health and succes. I sincerely hope this year will be simpler for most people as last year ! For the first post of the year, I wanted to show you how to take advantage of Ruby, the programming language embedded in QSoas, to make various things, like: • creating a column with the sum of Y values; • extending values that are present only in a few lines; • renaming datasets using a pattern. #### Summing the values in a column When using commands that take formulas (Ruby code), like apply-formula, the code is run for every single point, for which all the values are updated. In particulier, the state of the previous point is not known. However, it is possible to store values in what is called global variables, whose name start with an $ sign. Using this, we can keep track of the previous values. For instance, to create a new column with the sum of the y values, one can use the following approach:
QSoas> eval $sum=0 QSoas> apply-formula /extra-columns=1$sum+=y;y2=$sum  The first line initializes the variable to 0, before we start summing, and the code in the second line is run for each dataset row, in order. For the first row, for instance, $sum is initially 0 (from the eval line); after the execution of the code, it is now the first value of y. After the second row, the second value of y is added, and so on. The image below shows the resulting y2 when used on:
QSoas> generate-dataset -1 1 x

#### Extending values in a column

Another use of the global variables is to add "missing" data. For instance, let's imagine that a files given the variation of current over time as the potential is changed, but the potential is only changed stepwise and only indicated when it changes:
## time	current	potential
0	0.1	0.5
1	0.2
2	0.3
3	0.2
4	1.2	0.6
5	1.3
...

If you need to have the values everywhere, for instance if you need to split on their values, you could also use a global variable, taking advantage of the fact that missing values are represented by QSoas using "Not A Number" values, which can be detected using the Ruby function nan?:
QSoas> apply-formula "if y2.nan?; then y2=$value; else$value=y2;end"

Note the need of quotes because there are spaces in the ruby code. If the value of y2 is NaN, that is it is missing, then it is taken from the global variable $value else $value is set the current value of y2. Hence, the values are propagated down:
## time	current	potential
0	0.1	0.5
1	0.2	0.5
2	0.3	0.5
3	0.2	0.5
4	1.2	0.6
5	1.3	0.6
...

Of course, this doesn't work if the first value of y2 is missing.

#### Renaming using a pattern

The command save-datasets can be used to save a whole series of datasets to the disk. It can also rename them on the fly, and, using the /mode=rename option, does only the renaming part, without saving. You can make full use of meta-data (see also a first post here)for renaming. The full power is unlocked using the /expression= option. For instance, for renaming the last 5 datasets (so numbers 0 to 4) using a scheme based on the value of their pH meta-data, you can use the following code:
QSoas> save-datasets /mode=rename /expression='"dataset-#{$meta.pH}"' 0..4  The double quotes are cumbersome but necessary, since the outer quotes (') prevent the inner ones (") to be removed and the inner quotes are here to indicate to Ruby that we are dealing with text. The bit inside #{...} is interpreted by Ruby as Ruby code; here it is $meta.pH, the value of the "pH" meta-data. Finally the 0..4 specifies the datasets to work with. So theses datasets will change name to become dataset-7 for pH 7, etc...


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## $$\frac{2}{5}(e-1)$$

Vector Calculus

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so for us to evaluate this line, integral. One thing that we should notice is that here we have a d y as opposed to a D s. So what we would want to do is essentially first find what is D Y by D? T. So, like, this is equal to something. And then we could just kind of multiply the DT over and we get de y busy, but whatever. So that's essentially what we're going to do in this case, eso was find that so if we take the derivative of this with respect to t, we end up with two t multiply that over, we get d Y is equal to t d t. So we're just going to replace D Y with that. And then we can make our other substitution for X and Z as well. Yeah, we get it. Just integrate from 01 So let's do all that here. So our new balance is going to be from 0 to 1. X is going to be t y is ah t squared e is eso e r e e to the y z eso again. Why is t squared? Z is t cubed and then d wise was to be two times t Then we have DT actually, just grow simplify the stone a little So zero toe one looks like to t to the fourth times e raised to the to t to the fifth and we can go ahead and use a substitution of you is opportunity to fifth, which would give us the differential. Both this here eso We need a five here, So let's just multiply this by five that weaken divide by five as well. And now that turns into d you. So if I come over here maybe 1/5 integral of so if I plug zero in that zero plug in 11 um and then we'd have And actually let me just pull that two out as well. So it would be 2/5 and then the e to the you, do you and we'll evaluating that would just give us eat the youth. So you have to fifth c e to the u evaluated from 01 eso we plug in one, which is just gonna be e plug in zero. That's once a minus one. And so this here should be what we get for evaluating that line. Integral

University of North Texas

#### Topics

Vector Calculus

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

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2020-10-19, 12:34   #3367
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

29·167 Posts

Quote:
 Originally Posted by LaurV Oh, you mean the other way around... Never tried that, because I always stop after class, but it should be easy to try, just take a factor line already reported and edit the asterisk. I will try when I reach home in about one hour (still at job).
A) what are you talking about? "edit the asterisk"?
B) shouldn't be necessary to manually intervene. What about people who use Misfit or mfloop etc.?
C) ... especially since completing the bit level is the mfaktc default configuration. (But not mfakto)
mfaktc ini file as distributed in zip file:
Code:
# Allow to split an assignment into multiple bit ranges.
# 0 = disabled
# 1 = enabled
# Enabled Stages make only sense when StopAfterFactor is 1 or 2.
# Do not change this in the middle of a run which spans over multiple
# bitlevels, in this case mfaktc will ignore the checkpoint file and
# restarts from the beginning.
#
# Default: Stages=1

Stages=1

# possible values for StopAfterFactor:
# 0: Do not stop the current assignment after a factor was found.
# 1: When a factor was found for the current assignment stop after the
#    current bitlevel. This makes only sense when Stages is enabled.
# 2: When a factor was found for the current assignment stop after the
#    current class.
#
# Default: StopAfterFactor=1

StopAfterFactor=1
mfakto ini file as distributed in zip file:
Code:
# Allows you to split an assignment into multiple bit ranges.
# Note that many mfakto-kernels can handle only one bit level at a time,
# i.e. in order to use them, Stages needs to be enabled.
# 0 = disabled
# 1 = enabled
# Enabled Stages make only sense when StopAfterFactor is 1 or 2.
#
# Default: Stages=1

Stages=1

# possible values for StopAfterFactor:
# 0: Do not stop the current assignment after a factor was found.
# 1: When a factor was found for the current assignment stop after the
#    current bitlevel. This makes only sense when Stages is enabled.
# 2: When a factor was found for the current assignment stop after the
#    current class.
#
# Default: StopAfterFactor=2

StopAfterFactor=2
D) Finishing the bit level makes more sense to me, because it leaves the database cleaner, with completed bit levels, not partially completed, so the folks that like to overfactor or completely factor Mersenne numbers don't need to either repeat trial factoring work already performed, or gamble by skipping part of a bit level.
E) The assignment is by bit level. Finish the assignment seems like the right thing to do.
F) The first factor found is not necessarily the smallest, since factoring on gpus is done class by class, not in ascending k order.
Attached Thumbnails

Last fiddled with by kriesel on 2020-10-19 at 13:27


text:
# How do you solve 6 + t - 8 >= -7?

Jun 29, 2017

See a solution process below:

#### Explanation:

First, group and combine like terms on the left side of the inequality:

$6 + t - 8 \ge - 7$

$t + 6 - 8 \ge - 7$

$t + \left(6 - 8\right) \ge - 7$

$t + \left(- 2\right) \ge - 7$

$t - 2 \ge - 7$

Now, add $\textcolor{red}{2}$ to each side of the inequality to solve for $t$ while keeping the inequality balanced:

$t - 2 + \textcolor{red}{2} \ge - 7 + \textcolor{red}{2}$

$t - 0 \ge - 5$

$t \ge - 5$


text:
# Consider fully developed laminar flow in a circular pipe of a fixed length:1) The friction factor is inversely proportional to Reynolds number2) The pressure drop in the pipe is proportional to the average velocity of the flow in the pipe3) The friction factor is higher for a rough pipe as compared to a smooth pipe4) The pressure drop in the pipe is proportional to the square of average of flow in the pipe. Which of the above statements are correct?

1. 1 and 4
2. 3 and 4
3. 2 and 3
4. 1 and 2

Option 4 : 1 and 2

## Laminar Flow through Pipe MCQ Question 1 Detailed Solution

Explanation

Friction factor of laminar flow and turbulent flow is given as,

$$for\;Laminar\;flow,\;f = \frac{{64}}{{Re}}$$

∴ The friction factor is inversely proportional to Reynold's Number

$$for\;Turbulent\;flow,\;f = \frac{{0.316}}{{R{e^{1/4}}}}$$

The pressure drop in a fully developed laminar flow through pipe is given as,

$$\Delta P = \frac{{32\mu {V_{mean}}L}}{{{D^2}}}$$

∴ ΔP is proportional to mean velocity or average velocity in pipe.

In laminar flow, the friction factor is independent of roughness because boundary layer covers the roughness.

# For laminar flow, Kinetic energy correction factor is:

1. 1
2. 1.33
3. 2
4. 2.7

Option 3 : 2

## Laminar Flow through Pipe MCQ Question 2 Detailed Solution

Explanation:

Kinetic energy correction factor(α):

• It is defined as the ratio of kinetic energy/second based on actual velocity to the kinetic energy/second based on average velocity.
• $$\alpha = \frac{1}{A}\mathop \smallint \limits_A^{} {\left( {\frac{u}{V}} \right)^3}dA$$
• where A = area, V= average velocity, u= local velocity at distance r.
• In circular pipe laminar flow α = 2

# A fluid of viscosity 7 poise and density 1300 kg/m3 is flowing through a circular pipe of diameter 200 mm. The maximum shear stress at the pipe wall is 200 N/m2. What will be the pressure loss mere of pipe length?

1. 400 N/m2
2. 4000 N/m2
3. 6000 N/m2
4. 8000 N/m2

Option 2 : 4000 N/m2

## Laminar Flow through Pipe MCQ Question 3 Detailed Solution

Concept:-

Maximum shear stress in circular pipe flow is given by:

$$τ = - \left( {\frac{{dp}}{{dx}}} \right)\times\frac{R}{2}$$

where

τ = Shear stress, $$\frac{{dp}}{{dx}}$$= pressure gradient, R = Radius of pipe

Calculation:

Given:

τ = 200 N/m2, D = 200 mm ⇒ R = 100 mm = 100 × 10-3 m

μ = 7 poise ⇒ 0.7 Ns/m2 (∵ 1 poise =0.1 Ns/m2)

$$τ = - \left( {\frac{{dp}}{{dx}}} \right)\frac{R}{2}$$

$$\;200 = - \left( {\frac{{dp}}{{dx}}} \right)\frac{{100 × {{10}^{ - 3}}}}{2}$$

$$\frac{{dp}}{{dx}} = \; - 4000\frac{N}{{{m^2}}}$$

$$τ = - \left( {\frac{{dp}}{{dx}}} \right)\frac{R}{2}$$

Shear stress Distribution across a section is linear in laminar flow.

The maximum velocity in a pipe is given by:

$${U_{max}} = - \frac{1}{{4\mu }}\left( {\frac{{dp}}{{dx}}} \right){R^2}$$

Velocity distribution across a section is parabolic in laminar flow.

# For laminar flow through a long pipe, the pressure drop per unit length is

1. In direct proportion to the cross-sectional area
2. In proportion to the diameter of the pipe
3. In inverse proportion to the square of the cross-sectional area
4. In inverse proportion to cross-sectional area

Option 4 : In inverse proportion to cross-sectional area

## Laminar Flow through Pipe MCQ Question 4 Detailed Solution

Explanation:

The pressure drop per unit length in a fully developed laminar flow through a pipe is given as,

$$\Delta P = \frac{{32\mu {V_{mean}}}}{{{D^2}}}$$

We know Q = A × V

so V = $$\frac{Q}{A}$$

∴ $$\Delta P = \frac{{32\mu {Q}}}{{{AD^2}}}$$

$$\Delta P\propto\frac{1}{A}$$

For laminar flow through a long pipe, the pressure drop per unit length is in inverse proportion to the cross-sectional area.

# Laminar developed flow at an average velocity of 5 m/s occurs in a pipe of 10 cm radius. The velocity at 5 cm radius is

1. 7.5 m/s
2. 10 m/s
3. 2.5 m/s
4. 5 m/s

Option 1 : 7.5 m/s

## Laminar Flow through Pipe MCQ Question 5 Detailed Solution

Concept:

Distribution of velocity profile:

In a fully developed laminar pipe flow, the average velocity is one half of the maximum velocity i.e. Umax = 2Uavg

$$\begin{array}{l} U (r) = U_{max}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}$$

$$\begin{array}{l} U (r) = 2{U_{av}}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^2}} \right] \end{array}$$

Calculation:

Given:

Uavg = 5 m/s, R = 10 cm, r = 5 cm

$$U = 2 \times 5\left[ {1 - {{\left( {\frac{5}{{10}}} \right)}^2}} \right] = 7.5\;m/s$$

# If the laminar flow of the fluid having the viscosity μ is flowing through the pipe of the diameter D. Pressure gradient in the flow is $$\left( { - \frac{{\partial p}}{{\partial x}}} \right)$$. What is the average velocity of flow?

1. $$V = \frac{1}{{32\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$
2. $$V = \frac{1}{{8\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$
3. $$V = \frac{1}{{16\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$
4. $$V = \frac{1}{{4\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$

Option 1 : $$V = \frac{1}{{32\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$

## Laminar Flow through Pipe MCQ Question 6 Detailed Solution

Explanation:

The pressure drop in a laminar flow through a circular pipe is given by:

$${{{P_1} - {P_2}}}{{}} = \frac{{32μ VL}}{{{D^2}}}$$

where μ = viscosity of fluid, L = length of pipe, D = diameter of pipe, V = average velocity of flow

After rearranging

$$\frac{{{P_1} - {P_2}}}{{L}} = \frac{{32\mu V}}{{{D^2}}}$$-------(1)

for a small length ∂x, the pressure gradient is $$-\frac{{{\partial P}}}{{\partial x}}$$

∴ Equation 1 becomes,

$$-\frac{{{\partial P}}}{{\partial x}} = \frac{{32\mu V}}{{{D^2}}}$$

Average velocity is:

$$V = \frac{1}{{32\mu }}\left( { - \frac{{\partial p}}{{\partial x}}} \right){D^2}$$

# In case of laminar flow, the loss of pressure head is proportional to

1. Velocity
2. Square of velocity
3. Cube of velocity
4. Half of velocity

Option 1 : Velocity

## Laminar Flow through Pipe MCQ Question 7 Detailed Solution

Explanation:

For pipe:

The loss of pressure head in case of laminar flow is given by:

$${h_{L}} = \frac{{32{μ \bar UL}}}{{ \rho gd^2}}$$

Parallel plate:

$${h_L} = \frac{{12\mu ̅ UL}}{{\rho g{t^2}}}$$

where μ = viscosity, L = length of pipe/plate, U̅ = average velocity, d = diameter of the pipe.

Thus, the loss of the pressure head is proportional to the velocity of flow in laminar flow.

Reynolds Number:

• It is a dimensionless number.
• It is the ratio of inertial forces to viscous forces occurring in a fluid flow.
• It determines whether the flow is laminar or turbulent.
• It is generally used in case of flow in closed conduit i.e. flow through pipes

In pipe flow

$$Re = \frac{{Inertia\;Force}}{{Viscous\;Force}} \Rightarrow \frac{{ρ VD}}{μ }$$

where V = average flow velocity, D = Diameter of Pipe, μ = dynamic viscosity of the fluid, and ρ = density of the fluid.

If Re is less than 2000 then the flow is laminar.

If Re is greater than 4000 then flow is turbulent

# Hydraulic gradient line (HGL) represents the sum of

## Laminar Flow through Pipe MCQ Question 8 Detailed Solution

Concept:

Total energy line (T.E.L) – Line representing the sum of pressure head, datum head, and velocity head.

# A fully developed laminar viscous flow through a circular tube has the ratio of maximum velocity to average velocity as

1. 3.0
2. 2.0
3. 2.5
4. 1.5

Option 2 : 2.0

## Laminar Flow through Pipe MCQ Question 9 Detailed Solution

Explanation:

For a fully developed laminar viscous flow through a circular pipe, the maximum velocity is equal to twice the average velocity.

i.e. $${U_{max}} = 2 \times {U_{avg}}$$

$${U_{max}} = \frac{1}{{4\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}$$

$$U_{avg}=\bar U = \frac{1}{{8\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}$$

Hence $$\frac{{{U_{max}}}}{{\bar U}} = 2$$

For the fully developed laminar flow through the parallel plates, the maximum velocity is equal to the (3/2) times of the average velocity.

i.e. $${U_{max}} = \frac{3}{2} \times {U_{avg}}$$

# In laminar flow, maximum velocity at the centre of the pipe is how many times the average velocity?

1. Two
2. Three
3. Four
4. Five

Option 1 : Two

## Laminar Flow through Pipe MCQ Question 10 Detailed Solution

Velocity distribution of a flow of viscous fluid through circular pipe:

$$u\left( r \right) = - \frac{1}{{4\mu }}\frac{{\partial p}}{{\partial x}}\left[ {{R^2} - {r^2}} \right]$$

The velocity is maximum when r = 0 i.e. at the centre of pipe.

$${u_{max}} = - \frac{1}{{4\mu }}\frac{{\partial p}}{{\partial x}}{R^2}$$

Average velocity:

$${u_{avg}} = \left( {\frac{1}{{8\mu }}} \right)\left( { - \frac{{\partial p}}{{\partial x}}} \right){R^2}$$

$$\frac{{{u_{max}}}}{{{u_{avg}}}} = 2.0$$

# In a pipe network,

1. the algebraic sum of discharges around each elementary circuit must be zero.
2. the head at each node be the same.
3. the piezometric head loss in each line of the circuit is the same
4. the algebraic sum of the piezometric head drops around each elementary circuit is zero

Option 4 : the algebraic sum of the piezometric head drops around each elementary circuit is zero

## Laminar Flow through Pipe MCQ Question 11 Detailed Solution

Concept:

In any pipe network following two rules must be satisfied:

1. The algebraic sum of the piezometric head (pressure head + datum head) drops around each elementary circuit (or each closed loop) must be zero, i.e. there cannot be any discontinuity in piezometric heads.

2. The flow entering a junction (or node) must be equal to the flow leaving the same junction, i.e. the law of continuity must be satisfied.

Note:

• The algebraic sum of discharges around each elementary circuit cannot be zero because if it would be zero then no net discharge would be available to pass into next elementary circuit i.e. flow in the network would not be possible.
• The head at each node cannot be same because if there is no head difference, the flow would not occur in the pipe network.
• The piezometric head loss in each line of the circuit cannot be the same as it depends upon the datum height, length and diameter of the line.

In a pipe network the algebraic sum of the piezometric head drops around each elementary circuit must be zero.

# A sphere of diameter 30 cm is moving with a uniform velocity of 4 m/s. The dynamic viscosity and specific gravity of the liquid are given as 0.8 poises and 0.9 respectively. What is the value of the Reynolds number?

1. 135
2. 10000
3. 13500
4. 15000

Option 3 : 13500

## Laminar Flow through Pipe MCQ Question 12 Detailed Solution

Concept:

Reynolds Number Re is defined as the ratio of the Inertia force to viscous force.

Inertia force (Fi) = mass × acceleration

$${{\rm{F}}_{\rm{i}}} = {\rm{ρ }} \times {\rm{volume}} \times \frac{{{\rm{velocity}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{velocity}} \times \frac{{{\rm{volume}}}}{{{\rm{time}}}} = {\rm{ρ }} \times {\rm{V}} \times {\rm{AV}}$$

Viscous force (Fv) = shear stress × area $$= {\rm{\;}}\tau \times {\rm{A}} = \mu \times \frac{{\rm{V}}}{{\rm{L}}} \times {\rm{A}}$$

Reynolds Number (Re) = $$\frac{{ρ \times {\bf{V}} \times {\rm{L}}}}{\mu }$$

Calculation:

$${{\rm{R}}_{\rm{e}}} = \frac{{{\rm{ρ }} \times {\rm{V}} \times {\rm{d}}}}{{\rm{\mu }}}$$

ρ = 0.9 g/cm3 = 900 kg/m3, V = 4 m/s, d = 30 cm = 0.3 m, μ = 0.8 poise = 0.08 Ns/m2

$${{\rm{R}}_{\rm{e}}} = \frac{{900 \times 4 \times 0.3}}{{0.08}} = 13500$$

# An oil of kinematic viscosity of 0.5 stokes flows through a pipe of 20 cm diameter. What is the velocity if the flow is critical?

1. 0.5 m/s
2. 0.9 m/s
3. 1.5 m/s
4. 2 m/s

Option 1 : 0.5 m/s

## Laminar Flow through Pipe MCQ Question 13 Detailed Solution

Concept:

Reynold's number is given by:

$$Re=\frac{\rho VD}{\mu}=\frac{VD}{\nu}$$

For the flow through a circular pipe,

Laminar flow – Reynold’s Number is less than 2000

Turbulent Flow – Reynold’s Number is greater than 4000

Transition Flow – Reynold’s Number is between 2000 to 4000

For critical condition, Reynold’s Number is 2000

Calculation:

Given:

ν  = 0.5 × 10-4 m2/s, D = 20 cm = 0.2 m, Re = 2000 (critical condition)

Therefore,

Reynold's Number is:

$$Re= \frac{{VD}}{ν } = 2000 \Rightarrow V =\frac{{2000 \times {\nu}}}{{D{{}{ }}}}= \frac{{2000 \times 0.5 \times {{10}^{ - 4}}}}{{20 \times {{10}^{ - 2}}}} = \frac{{50}}{{100}} = 0.5\;m/s$$

# The velocity distribution in a laminar flow trough a pipe follows the law:

1. Exponential law
2. Parabolic law
3. logarithmic law
4. linear law

Option 2 : Parabolic law

## Laminar Flow through Pipe MCQ Question 14 Detailed Solution

Explanation:

In the case of a pipe,

Velocity distribution for laminar flow is given by:

$$u = - \frac{1}{{4\mu }}\frac{{\partial p}}{{\partial x}}\left[ {{R^2} - {r^2}} \right]$$

∴ Velocity follows a parabolic law in case of laminar flow.

Important Points

Velocity distribution for turbulent flow in smooth pipe:

$$u = \frac{{{u_\infty }}}{k}\ln y + c$$

In the case of turbulent flow, the velocity distribution is logarithmic in nature.

# The hydraulic diameter, Dh of a rectangular duct with side ‘a’ and ‘b’ is

1. $${D_h} = \frac{{4ab}}{{2a + b}}$$
2. $${D_h} = \frac{{2ab}}{{a + b}}$$
3. $${D_h} = \frac{{2ab}}{{2a + b}}$$
4. $${D_h} = \frac{{4ab}}{{a + b}}$$

Option 2 : $${D_h} = \frac{{2ab}}{{a + b}}$$

## Laminar Flow through Pipe MCQ Question 15 Detailed Solution

Concept:

Hydraulic diameter (Dh): It is defined as $${{\rm{D}}_{\rm{h}}} = \frac{{4{{\rm{A}}_0}}}{{\rm{P}}}$$ where, A0 is minimum flow area on one fluid side of conduit and P is the wetted perimeter of flow passages of that side.

Therefore, $$D_h = \frac{{4A}}{P} = \frac{{4 \times ab}}{{2\left( {a + b} \right)}} = \frac{{2ab}}{{a + b}}$$

 Shape Hydraulic diameter Diagram Circular Pipe $${{\rm{D}}_{\rm{h}}} = \frac{{4\left( {\frac{{\rm{\pi }}}{4}{{\rm{D}}^2}} \right)}}{{{\rm{\pi D}}}} = {\rm{D}}$$ Square Duct $${{\rm{D}}_{\rm{h}}} = \frac{{4{{\rm{a}}^2}}}{{4{\rm{a}}}} = {\rm{a}}$$ Rectangular Duct $${{\rm{D}}_{\rm{h}}} = \frac{{4{\rm{ab}}}}{{2\left( {{\rm{a}} + {\rm{b}}} \right)}} = \frac{{2{\rm{ab}}}}{{{\rm{a}} + {\rm{b}}}}$$

# Glycerine (μ = 1.50 Pa.s: ρ = 1260 kg/m3) flows at a velocity of 6.0 m/s in 10 cm diameter pipe. Head loss in a length of 7 m pipe will be (g = 10 m/s2).

1. 14 m
2. 16 m
3. 7 m
4. 8 m

Option 2 : 16 m

## Laminar Flow through Pipe MCQ Question 16 Detailed Solution

Concept:

Hagen Poiseuille equation

Loss of pressure head  $$= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32μ VL}}{{ρ g{D^2}}} = {h_f}$$

Calculation:

Given:

Viscosity μ = 1.50 Pa.s, diameter d = 0.1 m, V = 6 m/s, L = 7 m, ρ = 1260 kg/m3

Loss of pressure head  $$= \dfrac{{{p_1} - {p_2}}}{{ρ g}} = \dfrac{{32 \times 6\times1.50\times 7 }}{{1260\times 0.01{}}} = 16 ~m$$

# For the laminar flow through a circular pipe the ratio of maximum velocity to average velocity is

1. 1.5
2. 2.5
3. 2.0
4. 1.0

Option 3 : 2.0

## Laminar Flow through Pipe MCQ Question 17 Detailed Solution

Average velocity for laminar flow through circular pipe,

$$\bar U = \frac{1}{{8\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}$$

Maximum velocity for laminar flow through circular pipe,

$${U_{max}} = \frac{1}{{4\mu }}\left( { - \frac{{\partial P}}{{\partial x}}} \right).{R^2}$$

∴ $$\frac{{{U_{max}}}}{{\bar U}} = 2$$

# The coefficient of skin friction for laminar flow in a pipe is given as-(where Re = Reynolds number)

1. Re/32
2. 32/Re
3. Re/16
4. 16/Re

Option 4 : 16/Re

## Laminar Flow through Pipe MCQ Question 18 Detailed Solution

Explanation:

Skin friction coefficient

• Skin friction drag is a component of profile drag that occurs differently depending on the type of flow over the lifting body
• Within smooth pipes the value of the skin friction coefficient varies with the Reynolds number and depends on whether the flow is laminar or turbulent.
• It is possible mathematically to model a laminar flow situation in a pipe in order to establish an analytical value for the skin friction coefficient.
• The coefficient of skin friction for laminar flow in a pipe is given as, $${C_{fx}} = \frac{{16}}{{{R{e_x}} }}$$
• There is no equivalent analytical approach to establish a skin friction coefficient for turbulent flow.
• All turbulent flow data are based on experimental studies.
• For turbulent flow in smooth pipes with values of the Reynolds number up to 105, the value of the skin friction coefficient, Cf, is given by, Cf = 0.079 (Re)-1/4
• This relationship is known as the Blasius equation.

Important Points

• The local Skin friction coefficient for the boundary layer over a flat plate is given as
• $${C_{fx}} = \frac{{0.664}}{{\sqrt {R{e_x}} }}$$

# For the laminar flow through a circular pipe

1. the maximum velocity = 1.5 times the average velocity
2. the maximum velocity = 2.0 times the average velocity
3. the maximum velocity = 2.5 times the average velocity
4. none of the above

Option 2 : the maximum velocity = 2.0 times the average velocity

## Laminar Flow through Pipe MCQ Question 19 Detailed Solution

For the laminar flow through a circular pipe, velocity distribution is parabolic.

Velocity profile is given by, $$V = - \frac{1}{{4\mu }} \times \frac{{\partial P}}{{\partial x}} \times \left( {{R^2} - {r^2}} \right)$$

Maximum velocity is at the center line of the pipe i.e. r = 0

Maximum velocity, $${U_{maximum}} = - \frac{1}{{4\mu }} \times \frac{{\partial P}}{{\partial x}} \times \left( {{R^2}} \right)$$

Average velocity, $${U_{average}} = - \frac{1}{{8\mu }} \times \frac{{\partial P}}{{\partial x}} \times \left( {{R^2}} \right)$$

Therefore $${U_{maximum}} = 2 \times {U_{average}}$$

# The relationship governing the laminar flow of water through the capillary tube is known as?

1. Newton's Law
2. Archimedes' Law
3. Poiseuille’s Law
4. Mohr's Law

Option 3 : Poiseuille’s Law

## Laminar Flow through Pipe MCQ Question 20 Detailed Solution

Explanation:

Poiseuille's Law:

• Real fluids have viscosity, So a pressure difference is required between the ends of the tube or pipe for the flow to happen.
• The relation between pressure difference and discharge in the capillary tube is given by Poiseuille's law.

$$Q = \frac{\pi R^{4}\left ( P_{1}-P_{2} \right )}{8η L}$$

Where, Q = Discharge in the pipe/tube, R = Radius of pipe, P1, P2 are the pressure at point 1 and 2 respectively, η = Dynamic viscosity of the fluid, L = Length of pipe between point 1 and 2

Assumptions for the use of Poiseuille's law

1. The fluid must be incompressible.
2. It should be laminar between the given boundaries


text:
Compose two lenses to produce a new lens which represents focussing first with the first lens, then with the second. A view using the resulting composite lens will first view using the first, then the second, while an set will view via the first lens, set into the resulting piece with the second, and then replace the updated structure in the first with set. Lens composition is analogous to the . syntax of object-oriented programming or to a flipped version of function composition.

l %.% m

## Arguments

l the first lens (or an oscope) the second lens

## Examples

  lst <- list(b = c(3,4,5))
lns <- index_l("b") %.% index_l(2)
lst %>% view(lns)                 # returns 4#> [1] 4  lst %>% set(lns, 1)              # returns list(b = c(3,2,5))#> $b #> [1] 3 1 5 #> lst # returns list(b = c(3,4,5))#>$b
#> [1] 3 4 5
#>


text:
# Can I create a standalone section after having chapters?

I have a document which contains 4 'chapters'. In the end, I am trying to add a Bibliography section but it is getting filed under the last chapter.

I want it as a standalone section.

\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fancyhdr}
\usepackage{imakeidx}
\usepackage[nottoc]{tocbibind}
\usepackage{nomencl}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage{parskip}
\usepackage{makeidx}
\usepackage{tikz}

These are the packages I am using.

This is my Index Page

\tableofcontents

\pagestyle{fancy}
\cfoot{Page \thepage\, of $29$}

Can anybody help ? I have two standalone sections titles - Acknowledgements and Certificate but these are there before any of the chapters start.

How can I do the same for the Bibliography part?

• You can try with \chapter*{Bibliography}. In any case, we don't know which class are you using and don't know how you include bibliography in your document. How can we answer? – Ignasi Jul 9 '15 at 11:06
• it's not clear, can you explain more the problem? and what do you mean by filed under the last chapter? can you edit your code to be complete? – touhami Jul 9 '15 at 11:17
• Why is this tagged as latex-project? – user31729 Jul 9 '15 at 11:58
• ordinarily, in a document with chapters, the bibliography would also be a chapter. if you insist that it be a section, at the end of the last chapter, add \cleardoublepage and then \section*{Bibliography}. you will also have to adjust the running heads if it is more than one page long. – barbara beeton Jul 9 '15 at 14:06
• Try \setcounter{chapter}{0} right before the bibliography – John Kormylo Jul 9 '15 at 16:43

Using a combination of \front..., \main... and \backmatter as well as using pagestyle and \counterwithin and \counterwithout (package chngcntr this is no problem at all.

\documentclass{book}

\usepackage{fancyhdr}
\usepackage[nottoc]{tocbibind}
\usepackage[english]{babel}
\usepackage{chngcntr}
\usepackage{lastpage}

\pagestyle{fancy}
\cfoot{Page \thepage{} of \pageref{LastPage}}

\usepackage{blindtext}

\begin{document}
\frontmatter
\counterwithout{section}{chapter}
\pagestyle{plain}

\section{Acknowledgements}
\section{Certificate}

\mainmatter
\counterwithin{section}{chapter}
\pagestyle{fancy}
\cleardoublepage
\chapter{First}

\cite{Lam94}
\chapter{Second}
\chapter{Third}
\chapter{Fourth}
\section{First}
\blindtext[10]

\backmatter
\bibliographystyle{alpha}
\bibliography{biblio}

\end{document}


text:
# In a railway track at a curve, the outer track/edge is comparatively raised in height than the inner edge in order to _______.

This question was previously asked in
Karnataka PSC Prelims 2017 Paper 2 Official Paper
View all KPSC KAS Papers >
1. Avoid frictional force between the track and wheels
2. Obtain the necessary centrifugal force from the horizontal component of the weight of the train
3. Obtain the necessary centripetal force from the horizontal component of the weight of the train
4. Decrease the speed of the train

## Answer (Detailed Solution Below)

Option 3 : Obtain the necessary centripetal force from the horizontal component of the weight of the train

## Detailed Solution

The correct answer is Obtain the necessary centripetal force from the horizontal component of the weight of the train.

Key Points

Cant or superelevation:

• Cant or superelevation is the amount by which one rail is raised above the other rail.
• It is positive when the outer rail on a curved track is raised above the inner rail and is negative when the inner rail on a curved track is raised above the outer rail.
• It is done to obtain the necessary centripetal force from the horizontal component of the weight of the train. Hence statement 3 is correct.
• It also helps to compensate for the centrifugal force which causes irregular stressing of rails and other track components.

Useful terms related to Cantor superelevation:

• Equilibrium speed is the speed at which the centrifugal force developed during the movement of the vehicle on a curved track is exactly balanced by the cant provided.
• Cant deficiency- Cant deficiency occurs when a train travels around a curve at a speed higher than the equilibrium speed. It is the difference between the theoretical cant required for such higher speed and actual cant provided.
• Cant excess - Cant excess occurs when a train travels around a curve at a speed slower than the equilibrium speed. It is the difference between the actual cant and the theoretical cant required for such a lower speed.


text:
# Properties

 Label 9025.2.a.c Level $9025$ Weight $2$ Character orbit 9025.a Self dual yes Analytic conductor $72.065$ Analytic rank $1$ Dimension $1$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$9025 = 5^{2} \cdot 19^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 9025.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$72.0649878242$$ Analytic rank: $$1$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 95) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

 $$f(q)$$ $$=$$ $$q - q^{2} - q^{4} - 2 q^{7} + 3 q^{8} - 3 q^{9} + O(q^{10})$$ $$q - q^{2} - q^{4} - 2 q^{7} + 3 q^{8} - 3 q^{9} - 4 q^{11} - 2 q^{13} + 2 q^{14} - q^{16} - 4 q^{17} + 3 q^{18} + 4 q^{22} + 6 q^{23} + 2 q^{26} + 2 q^{28} + 6 q^{29} + 4 q^{31} - 5 q^{32} + 4 q^{34} + 3 q^{36} - 10 q^{37} + 10 q^{41} - 2 q^{43} + 4 q^{44} - 6 q^{46} + 6 q^{47} - 3 q^{49} + 2 q^{52} + 10 q^{53} - 6 q^{56} - 6 q^{58} + 2 q^{61} - 4 q^{62} + 6 q^{63} + 7 q^{64} + 8 q^{67} + 4 q^{68} - 4 q^{71} - 9 q^{72} - 4 q^{73} + 10 q^{74} + 8 q^{77} - 4 q^{79} + 9 q^{81} - 10 q^{82} + 18 q^{83} + 2 q^{86} - 12 q^{88} + 2 q^{89} + 4 q^{91} - 6 q^{92} - 6 q^{94} + 6 q^{97} + 3 q^{98} + 12 q^{99} + O(q^{100})$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 0
−1.00000 0 −1.00000 0 0 −2.00000 3.00000 −3.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$5$$ $$-1$$
$$19$$ $$-1$$

## Inner twists

This newform does not admit any (nontrivial) inner twists.

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 9025.2.a.c 1
5.b even 2 1 9025.2.a.h 1
5.c odd 4 2 1805.2.b.c 2
19.b odd 2 1 475.2.a.c 1
57.d even 2 1 4275.2.a.e 1
76.d even 2 1 7600.2.a.l 1
95.d odd 2 1 475.2.a.a 1
95.g even 4 2 95.2.b.a 2
285.b even 2 1 4275.2.a.p 1
285.j odd 4 2 855.2.c.b 2
380.d even 2 1 7600.2.a.i 1
380.j odd 4 2 1520.2.d.b 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
95.2.b.a 2 95.g even 4 2
475.2.a.a 1 95.d odd 2 1
475.2.a.c 1 19.b odd 2 1
855.2.c.b 2 285.j odd 4 2
1520.2.d.b 2 380.j odd 4 2
1805.2.b.c 2 5.c odd 4 2
4275.2.a.e 1 57.d even 2 1
4275.2.a.p 1 285.b even 2 1
7600.2.a.i 1 380.d even 2 1
7600.2.a.l 1 76.d even 2 1
9025.2.a.c 1 1.a even 1 1 trivial
9025.2.a.h 1 5.b even 2 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(9025))$$:

 $$T_{2} + 1$$ $$T_{3}$$ $$T_{7} + 2$$ $$T_{11} + 4$$ $$T_{29} - 6$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$1 + T$$
$3$ $$T$$
$5$ $$T$$
$7$ $$2 + T$$
$11$ $$4 + T$$
$13$ $$2 + T$$
$17$ $$4 + T$$
$19$ $$T$$
$23$ $$-6 + T$$
$29$ $$-6 + T$$
$31$ $$-4 + T$$
$37$ $$10 + T$$
$41$ $$-10 + T$$
$43$ $$2 + T$$
$47$ $$-6 + T$$
$53$ $$-10 + T$$
$59$ $$T$$
$61$ $$-2 + T$$
$67$ $$-8 + T$$
$71$ $$4 + T$$
$73$ $$4 + T$$
$79$ $$4 + T$$
$83$ $$-18 + T$$
$89$ $$-2 + T$$
$97$ $$-6 + T$$


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text:
The usual centre, called just centroid (centre of area) comes from considering the surface of the quadrilateral as having constant density. How many diagonals does a rectangle have? {\displaystyle K={\tfrac {1}{2}}|\tan \theta |\cdot \left|a^{2}-b^{2}\right|. How many diagonals does a quadrilateral have. {\displaystyle m^{2}+n^{2}={\tfrac {1}{2}}(p^{2}+q^{2}). 13. Heptagon Angles. {\displaystyle A} For triangles AOB and COD, angle 1 is equal to angle 2, as they are . [45], For any quadrilateral ABCD with points P and Q the intersections of AD and BC and AB and CD, respectively, the circles (PAB), (PCD), (QAD), and (QBC) pass through a common point M, called a Miquel point. {\displaystyle D} The German mathematician Carl Anton Bretschneider derived in 1842 the following generalization of Ptolemy's theorem, regarding the product of the diagonals in a convex quadrilateral[26]. The most general kite has unequal diagonals, but there is an infinite number of (non-similar) kites in which the diagonals are equal in length (and the kites are not any other named quadrilateral). Answer. In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenter O are collinear in this order, and HG = 2GO. [22]:p.125, In a convex quadrilateral with sides a, b, c and d, the length of the bimedian that connects the midpoints of the sides a and c is, where p and q are the length of the diagonals. d The last trigonometric area formula including the sides a, b, c, d and the angle α (between a and b) is:[citation needed]. A square has 2 diagonals: An octagon has 20 diagonals: A polygon's diagonals are line segments from one corner to another (but not the edges). [47] 2 A _____ is both ‘equiangular’ and ‘equilateral’. Equality holds if and only if θ = 90°. Leversha, Gerry, "A property of the diagonals of a cyclic quadrilateral". where the lengths of the bimedians are m and n and the angle between them is φ. Bretschneider's formula[16][13] expresses the area in terms of the sides and two opposite angles: where the sides in sequence are a, b, c, d, where s is the semiperimeter, and A and C are two (in fact, any two) opposite angles. That's the maximum and also the minimum. [24] The list applies to the most general cases, and excludes named subsets. Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas. A Quadrilateral has four-sides, it is 2-dimensional (a flat shape), closed (the lines join up), and has straight sides. NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals:. − Quadrilaterals appear in four different classifications: Convex -- Each interior angle is less than 180° and the two diagonals are inside the closed space of the quadrilateral Concave -- One interior angle is greater than 180° and one diagonal lies outside the shape A ( Parallelogram. Then[36], Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K, then the following inequality holds:[37], A corollary to Euler's quadrilateral theorem is the inequality. 5. A In a convex quadrilateral, there is the following dual connection between the bimedians and the diagonals:[27], The four angles of a simple quadrilateral ABCD satisfy the following identities:[32]. b EXERCISE 3.2. {\displaystyle B} A | Hence, it has two diagonals. Quadrilaterals are either simple (not self-intersecting), or complex (self-intersecting, or crossed). The shape and size of a convex quadrilateral are fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices. [48] [49]. View Answer. How many diagonals does a quadrilateral have ? where equality holds if and only if the quadrilateral is a parallelogram. Recall that an inscribed (or 'cyclic') quadrilateral is one where the four vertices all lie on a circle. Complete Quadrilateral The figure determined by four lines, no three of which are concurrent, and their six points of intersection. From the figure, we infer that the quadrilateral ABCD is formed by two triangles, i.e. Kite: A quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” … Each vertex has two diagonals, so if you counted each diagonal from every vertex twice, you might think there were 10 diagonals. [44] These points can be used to define an Euler line of a quadrilateral. . The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[13] by the Cayley-Menger determinant, as follows: The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral[22]:p.127 (that is, the four intersection points of adjacent angle bisectors are concyclic) or they are concurrent. The "vertex centroid" is the intersection of the two bimedians. A polygon is such that when all diagonals from one vertex are drawn, eleven triangles are formed. 1 2 Solution for If the diagonals of a quadrilateral are perpendicular bisectors of each other (but not congruent), what can you conclude regarding the… {\displaystyle K={\tfrac {1}{2}}pq} Other names for quadrilateral include quadrangle (in analogy to triangle), tetragon (in analogy to pentagon, 5-sided polygon, and hexagon, 6-sided polygon), and 4-gon (in analogy to k-gons for arbitrary values of k). Every quadrilateral has its VIPs, or Very Important Polygons. The sum of angle measures of all inner angles is 2 pi in radians (or … = with equality only in the case of a square. 2 In a convex quadrilateral, all interior angles are less than 180°, and the two diagonals both lie inside the quadrilateral. Why is legal separation is being discouraged? Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then. D [22]:p. 126 The corresponding expressions are:[23], if the lengths of two bimedians and one diagonal are given, and[23]. Figures will overlap, use the hide button to hide an obstructing figure. b is sometimes denoted as . 4. The list applies to the most general cases, and excludes named subsets. It turns out that there is a relationship between the sides of the quadrilateral and its diagonals. It … for diagonal lengths p and q, with equality if and only if the diagonals are perpendicular. [35]:p.119 This is a direct consequence of the fact that the area of a convex quadrilateral satisfies. 2 Check the below NCERT MCQ Questions for Class 8 Maths Chapter 4 Practical Geometry with Answers Pdf free download. ( It states that. How many sides does the polygon have? 2 These three points are in general not all the same point.[42]. p In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length. Who was prime minister after Winston Churchill? Rashid, M. A. In a regular heptagon, each interior angle is roughly 128.57 °.. Below is the formula to find the measure of any interior angle of a regular polygon (n = number of sides): B C View Answer. If P is an interior point in a convex quadrilateral ABCD, then, From this inequality it follows that the point inside a quadrilateral that minimizes the sum of distances to the vertices is the intersection of the diagonals. a In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as: In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length. No matter how you change the angle they make, their tips form a … Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral. where there is equality if and only if the quadrilateral is cyclic. ◻ Square. [44], Another remarkable line in a convex non-parallelogram quadrilateral is the Newton line, which connects the midpoints of the diagonals, the segment connecting these points being bisected by the vertex centroid. Answer: (b) Question 6. A quadrilateral is said to contain perpendicular diagonals if four 90-degree angles are formed at the intersection of these diagonal lines. This page was last edited on 5 December 2020, at 06:21. The opposite sides … The area of any quadrilateral also satisfies the inequality[35], Denoting the perimeter as L, we have[35]:p.114. In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. If the diagonals of a quadrilateral bisect each other, then it’s a parallelogram (converse of a property). This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of. Then the intersection of the lines OaOc and ObOd is called the quasicircumcenter, and the intersection of the lines HaHc and HbHd is called the quasiorthocenter of the convex quadrilateral. n-2. How far is the title justified the story an astrologers day? by Jennifer Kahle. A square has no diagonals.Improved Answer:-A square has 2 diagonals in common with all other quadrilateralsThe formila is: 0.5*(n2-3n) = diagonals whereas n … 1. The sum of the measures of all the three angles of a triangle is where the lengths of the diagonals are p and q and the angle between them is θ. Which of the following quadrilaterals has two pairs of adjacent sides equal and diagonals intersecting at right angles? In Euclidean plane geometry, a quadrilateral is a polygon with four edges (sides) and four vertices (corners). Archibald, R. C., "The Area of a Quadrilateral". which can also be used for the area of a concave quadrilateral (having the concave part opposite to angle α), by just changing the first sign + to - . Three angles of a quadrilateral are in the ratio 2: 6: 4. 2 Who is the divisor of centigrade thermomoter? | Let exterior squares be drawn on all sides of a quadrilateral. D If all of the diagonals are drawn from a vertex of a quadrilateral, how many triangles are formed? Inclusive definitions are used throughout. . In geometry, the term "diagonal" refers to a segment connecting two vertices that does not form a polygon's side. a Why don't libraries smell like bookstores? + What is the meaning of fruit gatherer in exegesis? [51] Historically the term gauche quadrilateral was also used to mean a skew quadrilateral. In a crossed quadrilateral, the four "interior" angles on either side of the crossing (two acute and two reflex, all on the left or all on the right as the figure is traced out) add up to 720°.[9]. How many sides does the polygon have? Shape of a Diagonal Given here are some figures. [38]:Prop.1 This follows directly from the quadrilateral identity In quadrilateral ABCD, if the angle bisectors of A and C meet on diagonal BD, then the angle bisectors of B and D meet on diagonal AC. }, In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to ⋅ where x is the distance between the midpoints of the diagonals, and φ is the angle between the bimedians. 2. How does wealth maximisation goal takes care of conflict between managers and shareholders? In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 2 : 4 : 5 : 7.  The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral. This is incorrect because you … A quadrilateral has two pairs of non-adjacent vertices. θ K K What is plot of the story Sinigang by Marby Villaceran? In a concave quadrilateral, one interior angle is bigger than 180°, and one of the two diagonals lies outside the quadrilateral. Example 2: A hexagon has $$9$$ diagonals. [22]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law. The area of a quadrilateral ABCD can be calculated using vectors. Note that the two opposite sides in these formulas are not the two that the bimedian connects. The two diagonals of a convex quadrilateral are the line segments that connect opposite vertices. which is half the magnitude of the cross product of vectors AC and BD. [44], In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle. The square is one such quadrilateral, but there are infinitely many others. Josefsson, Martin (2016) ‘100.31 Heron-like formulas for quadrilaterals’. ΔADC and ΔABC. ) Who is the longest reigning WWE Champion of all time? It has rotational symmetry of order two. The VIPs of the trapezoid family are the isosceles trapezoids. .[1][2]. Since cos (A + C) ≥ −1, it also gives a proof of Ptolemy's inequality. H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, MAA, 1967, pp. Diagonals of Quadrilaterals -- Perpendicular, Bisecting or Both. n In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E, where e = AE, f = BE, g = CE, and h = DE.[28]. 2 K Peter, Thomas, "Maximizing the Area of a Quadrilateral", § Remarkable points and lines in a convex quadrilateral, Perpendicular bisector construction of a quadrilateral, "List of Geometry and Trigonometry Symbols", "Quadrilaterals - Square, Rectangle, Rhombus, Trapezoid, Parallelogram", http://www.cleavebooks.co.uk/scol/calrect.htm, "Five Proofs of an Area Characterization of Rectangles", "Characterizations of Orthodiagonal Quadrilaterals", "A New Formula Concerning the Diagonals and Sides of a Quadrilateral", "Properties of equidiagonal quadrilaterals", "On Two Remarkable Lines Related to a Quadrilateral", https://www.austms.org.au/Publ/Gazette/2010/May10/TechPaperMiller.pdf, "A Set of Rectangles Inscribed in an Orthodiagonal Quadrilateral and Defined by Pascal-Points Circles", "Properties of a Pascal points circle in a quadrilateral with perpendicular diagonals", "On Some Results Obtained by the Quaternion Analysis Respecting the Inscription of "Gauche" Polygons in Surfaces of the Second Order", Quadrilaterals Formed by Perpendicular Bisectors, Definitions and examples of quadrilaterals, A (dynamic) Hierarchical Quadrilateral Tree, An extended classification of quadrilaterals, The role and function of a hierarchical classification of quadrilaterals, https://en.wikipedia.org/w/index.php?title=Quadrilateral&oldid=992429983, Short description is different from Wikidata, Articles with unsourced statements from April 2015, Creative Commons Attribution-ShareAlike License. Dao Thanh Oai, Leonard Giugiuc, Problem 12033, American Mathematical Monthly, March 2018, p. 277. In the last two formulas, no angle is allowed to be a right angle, since tan 90° is not defined. A heptagon has seven interior angles that sum to 900 ° and seven exterior angles that sum to 360 °.This is true for both regular and irregular heptagons. A hierarchical taxonomy of quadrilaterals is illustrated by the figure to the right. + = . tan m If they aren't glorified for their congruent base angles and legs, then their diagonals do the talking. C Hence that point is the Fermat point of a convex quadrilateral. The following two formulas express the area in terms of the sides a, b, c, d, the semiperimeter s, and the diagonals p, q: The first reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then pq = ac + bd. 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# Simulation for Inference II — Matching Simulations to Data

27 October 2020 (Lecture 17)

# In our last episodes

• Monte Carlo: figuring out what a model predicts, in detail, by simulating it
• Bootstrap: getting at sampling distributions by simulating from a good approximation to the data-generating distribution
• But how do we estimate a model in the first place…?

# Agenda

• Reminder about estimation in general
• The method of simulated moments
• Auxiliary models and indirect inference
• Model checking

# How we estimate, in general

$\hat{\theta}_n = \argmin_{\theta}{M_n(\theta)}$

If $$M_n(\theta) \rightarrow m(\theta)$$ as $$n\rightarrow\infty$$

and $$m(\theta)$$ has a unique minimum at the true $$\theta^*$$

then (generally) $$\hat{\theta}_n \rightarrow \theta^*$$ (consistency)

If $$m$$ has a well-behaved interior minimum, $\hat{\theta}_n \approx \theta^* - (\nabla \nabla M_n(\theta^*))^{-1} \nabla M_n(\theta^*)$ and so $\Var{\hat{\theta}_n} \approx (\nabla\nabla m(\theta^*))^{-1} \Var{\nabla M_n(\theta^*)} (\nabla\nabla m(\theta^*))^{-1}$

# The Progress of Statistical Methods

1. Calculate likelihood, solve explicitly for MLE
2. Can’t solve for MLE but can still write down likelihood, calculate it, and maximize numerically
3. Even calculating the likelihood is intractable
• Outstanding example: hidden or latent variables $$Y_1, Y_2, \ldots$$ plus observed $$X_1, X_2, \ldots$$

# Why Finding the Likelihood Becomes Hard

• Likelihood become an integral/sum over all possible combinations of latent variables compatible with observations: $\begin{eqnarray*} \Probwrt{\theta}{X_1^n = x_1^n} & = & \int{\Probwrt{\theta}{X_1^n = x_1^n, Y_1^n=y_1^n} d y_1^n }\\ &= & \int{\Probwrt{\theta}{Y_1^n=y_1^n}\left(\prod_{i=1}^{n}{\Probwrt{\theta}{X_i=x_i|Y_1^n=y_1^n,X_1^{i-1}=x_1^{i-1}}}\right) d y_1^n} \end{eqnarray*}$
• Evaluating this sum-over-histories is, itself, a hard problem
• Exponentially hard in $$n$$ (literally)
• One approach: Expectation-Maximization algorithm, try to simultaneously estimate latent variables and parameters (Neal and Hinton 1998)
• Standard, clever, often messy
• Another: simulate!

# Method of simulated moments

Assume: the data really came from our simulation model, at some true value $$\theta^*$$ of the adjustable parameters $$\theta$$

1. Pick your favorite $$q$$-dimensional vector of statistics $$B$$ (“generalized moments”)
2. Calculate from data, $$B(obs) = B(X)$$
3. Pick a staring parameter value $$\theta$$
1. simulate multiple times, say $$s$$, getting $$\tilde{X}_1, \ldots \tilde{X}_s$$
2. calculate average of $$B$$, $$\overline{B}(\theta, n, s) \equiv \frac{1}{s}\sum_{i=1}^{s}{B(\tilde{X}_i)}$$
3. For large $$n$$, $$\overline{B}(\theta, n, s) \approx \Expectwrt{\theta}{B}$$
4. Adjust $$\theta$$ so expectations are close to $$B(obs)$$

# Method of simulated moments

• How this fits into the general theory: $\begin{eqnarray} M_n(\theta) & = & \| \overline{B}(\theta,n,s) - B(obs)\|^2\\ \hat{\theta}_{MSM} & = & \argmin_{\theta}{M_n(\theta)} \end{eqnarray}$
• Assume $$\overline{B}(\theta, n, S) \rightarrow b(\theta)$$ as $$n\rightarrow\infty$$
• Assume $$B(obs) \rightarrow b(\theta^*)$$ as $$n\rightarrow\infty$$
• Then $$M_n(\theta) \rightarrow m(\theta) \equiv \| b(\theta) - b(\theta^*) \|^2$$

# Method of simulated moments

• Works if the expectations of $$B$$ are enough to characterize the parameter
• i.e., if the function $$\theta \mapsto b(\theta)$$ has an inverse, so $$b^{-1}$$ is well-defined
• though for optimization we’d really prefer it to have a smooth inverse
• and if the sample versions converge on the expectations
• so we’re relying on ergodicity again, at least for these moments
• Why expectations rather than medians, modes, … ?
• Basically: easier to prove convergence
• The mean is not always the most probable value!

# A challenging example

• The logistic map: $X(t+1) = 4 r X(t) (1-X(t))$
• Given $$X(1)$$, this is a deterministic dynamical system
• We’ll typically chose $$X(1)$$ as uniform on $$[0,1]$$
• For small $$r$$, $$X(t) \rightarrow 0$$
• For large $$r$$, near 1, $$X(t)$$ is chaotic
• Deterministic behavior
• Sensitive dependence on initial conditions = small perturbations get amplified exponentially (at least to start with)
• Long-run behavior is ergodic (not easy to prove!)

# The logistic map in R

# Do one step of the logistic map, $x(t+1) = 4rx(t)(1-x(t))$
# Inputs: initial condition (x)
# logistic parameter (r)
# Output: next value of the logistic map
# Presumes: x is a single numerical value in [0,1]
# r is a single numerical value in [0,1]
logistic.map <- function(x,r) {
return(4*r*x*(1-x))
}

# Generate a time series from the logistic map, $x(t+1)=4rx(t)(1-x(t))$
# Inputs: length of desired time series (timelength)
# logistic parameter (r)
# value of x(1) (initial.cond, uniformly randomly generated if omitted)
# Output: vector of length timelength
# Presumes: r is a single numerical value in [0,1]
# timelength is a positive integer
# initialcond is a single numerical value in [0,1], or omitted
# logistic.map() is defined as above
logistic.map.ts <- function(timelength, r, initial.cond=NULL) {
x <-vector(mode="numeric", length=timelength)
if(is.null(initial.cond)) {
x[1] <- runif(1)
} else {
x[1] <- initial.cond
}
for (t in 2:timelength) {
x[t] = logistic.map(x[t-1],r)
}
return(x)
}

# Sensitive dependence on initial conditions

traj1 <- logistic.map.ts(1000, r=0.9, initial.cond=0.5001)
traj2 <- logistic.map.ts(1000, r=0.9, initial.cond=0.4998)
plot(1:100, traj1[1:100], ylim=c(0,1), xlab="t", ylab="X(t)", type="b")
points(1:100, traj2[1:100], pch=2, col="blue", type="b")

# Sensitive dependence on initial conditions

plot(traj1, traj2)
rug(x=traj1, side=1, col="black", ticksize=-0.01) # negative sizes for pointing outward
rug(x=traj2, side=2, col="blue", ticksize=-0.01)

# Long-run stability

par(mfrow=c(1,2))
hist(traj1)
hist(traj2)`


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## A Fenwick Tree Problem

Points: 15 (partial)
Time limit: 1.0s
Memory limit: 256M

Problem type

Bob is playing with an array!

Having studied the Fenwick Tree recently, Bob is curious whether he'd be able to solve the following types of operations to the array:

• Q1 i j: Calculate and output the sum of the subarray between the indices $$i$$ and $$j$$ $$(1 \le i \le j \le N)$$ inclusive.
• Q2 L R: Count and output the number of elements in the array with values in between $$L$$ and $$R$$ $$(0 \le L \le R \le 10^5)$$ inclusive.
• U i v: Change the value of the $$i$$-th element $$(1 \le i \le N)$$ in the array to $$v$$ $$(0 \le v \le 10^5)$$.

Unfortunately, Bob needs a little bit of help. Can you do it for him instead?

#### Input Specification

The first line contains two integers $$N$$ and $$Q$$ $$(1 \le N, Q \le 10^5)$$, the size of the array and the number of operations to the array.
The second line contains $$N$$ nonnegative integers, each separated by a space. The value of these elements are nonnegative and will never exceed $$10^5$$.
The next $$Q$$ lines each contain an operation to be performed on the array.

The values inside the array will always be between $$0$$ and $$10^5$$ inclusive.

#### Output Specification

On a new line, output an integer for every Q1 and Q2 query representing the answer to the respective queries.

#### Sample Input

5 2
3 1 0 5 2
Q1 1 3
Q2 2 4

#### Sample Output

4
2

#### Explanation of Output

The sum of the subarray can be described as $$3 + 1 + 0 = 4$$ (These are the first three elements in the array).

Overall, there are $$2$$ elements that fall in the range $$[2, 4]$$, which are the first and last elements in the array.


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location:  Publications → journals → CJM
Abstract view

# Transfer of Fourier Multipliers into Schur Multipliers and Sumsets in a Discrete Group

Published:2011-08-03
Printed: Oct 2011
• Stefan Neuwirth,
Laboratoire de Mathématiques, Université de Franche-Comté, 25000 Besançon, France
• Éric Ricard,
Laboratoire de Mathématiques, Université de Franche-Comté, 25000 Besançon, France
 Format: LaTeX MathJax PDF

## Abstract

We inspect the relationship between relative Fourier multipliers on noncommutative Lebesgue-Orlicz spaces of a discrete group $\varGamma$ and relative Toeplitz-Schur multipliers on Schatten-von-Neumann-Orlicz classes. Four applications are given: lacunary sets, unconditional Schauder bases for the subspace of a Lebesgue space determined by a given spectrum $\varLambda\subseteq\varGamma$, the norm of the Hilbert transform and the Riesz projection on Schatten-von-Neumann classes with exponent a power of 2, and the norm of Toeplitz Schur multipliers on Schatten-von-Neumann classes with exponent less than 1.
 Keywords: Fourier multiplier, Toeplitz Schur multiplier, lacunary set, unconditional approximation property, Hilbert transform, Riesz projection
 MSC Classifications: 47B49 - Transformers, preservers (operators on spaces of operators) 43A22 - Homomorphisms and multipliers of function spaces on groups, semigroups, etc. 43A46 - Special sets (thin sets, Kronecker sets, Helson sets, Ditkin sets, Sidon sets, etc.) 46B28 - Spaces of operators; tensor products; approximation properties [See also 46A32, 46M05, 47L05, 47L20]

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