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Finding distribution of time values from data
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Finding distribution of time values from data
I have a dataset with time values from a simulation, 10.000+ values. It's a set of REAL values written to a datafile (txt or dat), in the following format, line by line:
[ .]
What I would like to do is the following:
- Find the time distribution
- Present it as a smoothed graph representing the probability density distribution.
I have tried to Google it, read Mathematica references and otherwise try to solve this but couldn't find the solution. I hope someone here can help me. I'm new to Mathematica, as this primarily is a
tool my professor makes us use (ever heard that one before? ;-)). Even a nudge in the right direction is appreciated!
Best regards,
3 Replies
use something like:
data= Select[data,#>0.5&]
data= Select[data,0.3<#<2&]
Thank you very much Darya, your answer is highly appreciated. Much obliged.
But, with this newfound knowledge, a new question showed itself:
How do i remove values from a list that are GreaterEqual than some value? Or LessEqual for that matter.
Lets simulate some data distributed for example normally:
data = RandomVariate[NormalDistribution[3, 2.5], 10^4];
We can use so called SmoothKernelDistribution to build a general smooth distribution curve
SKD = SmoothKernelDistribution[data];
To show visually correspondence with data:
Show[Histogram[data, 20, "PDF"], Plot[PDF[SKD, x], {x, -5, 15}, PlotStyle -> Directive[Red, Thick]]]
But on other hand, if you would have a guess, what kind of known analytic distribution fits your case, you could try to find distribution parameters that fit your data best - and of course they are
very close to the parameters we have chosen for original Gaussian to simulate your data:
params = FindDistributionParameters[data, NormalDistribution[a, b]]
Out[] = {a -> 3.00242, b -> 2.45767}
And you could check whether it is a good choice of distribution - in this case of course it is:
ND = NormalDistribution[a, b] /. params;
tstData = DistributionFitTest[data, ND, "HypothesisTestData"];
{tstData["AutomaticTest"], tstData["TestConclusion"]}
Show[Histogram[data, 20, "PDF"], Plot[PDF[ND, x], {x, -5, 15}, PlotStyle -> Directive[Red, Thick]]]
Now you can easily find various probabilistic and statistical characteristics of this distributions:
Probability[5 < x < 10, x \[Distributed] SKD]
Out[] = 0.207631
Probability[5 < x < 10, x \[Distributed] ND]
Out[] = 0.205963
If you need a practical demonstration of this with real world data take a look at Vitaliy s answer here:
Finding the actual wind power at a location for a given period?
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Decimal to Binary Converter Online Tool
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Online Decimal to Binary Converter
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Explore projects · GitLab
• BSD 3-Clause Open MPI variant
The Algorithms for Lattice Fermions package provides a high-performance, general code for auxiliary-field quantum Monte Carlo simulations.
• Repository for Hamiltonians that are not part of the ALF repository.
• GNU General Public License v3.0 or later
A Python application for simulating electronic band structures of semiconductor devices with k·p theory on a lattice.
• A high-performance Python library for the generation and visualization of hyperbolic tilings
• GNU General Public License v2.0 or later
Python interface for ALF, plus scripts and Jupyter notebooks.
• MARQOV: A modern framework for classical spin models on general topologies
• This is a repository that serves as a central point for storing publicly available docker images.
• Docker images for (py)ALF
• OPEN Tutorials and Presentations on ALF.
• Pseudofermion functional renormalization group solver | {"url":"https://git.physik.uni-wuerzburg.de/explore/projects?sort=latest_activity_desc","timestamp":"2024-11-01T22:54:16Z","content_type":"text/html","content_length":"106974","record_id":"<urn:uuid:e6d17250-3533-4a25-b54d-4ee575b2baba>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00567.warc.gz"} |
Electronic Greens function and overview page for DMFT (#1267) · Issues · nomad-lab / nomad-FAIR · GitLab
Electronic Greens function and overview page for DMFT
This is a continuation of #1156 (closed) . We can prepare the overview page for DMFT entries including:
• Table with input parameters: beta (inverse temperature), interactions (Hubbard U, ratio JH/U), and total filling: resolve from results.method.simulation.dmft. → to be discussed
• Real part of the Green's function in taus: resolve from results.properties.electronic.greens_functions_electronic.real_greens_function_tau.
• Imaginary part of the self-energy in Matsubaras: resolve from results.properties.electronic.greens_functions_electronic.imag_self_energy_iw.
• [DEL:Table with orbital occupations of each correlated orbital and spin, and per atom in the unit cell: resolve from results.properties.electronic.greens_functions_electronic.orbital_occupations.
Table with quasiparticle weights of each correlated orbital and spin, and per atom in the unit cell: resolve from results.properties.electronic.greens_functions_electronic.quasiparticle_weights.
:DEL] → left for the metainfo data tab and the DMFT workflow plots
These will appear in the Electronic Structure part, after the bands, dos, BZ and band gap information. | {"url":"https://gitlab.mpcdf.mpg.de/nomad-lab/nomad-FAIR/-/issues/1267","timestamp":"2024-11-06T01:53:23Z","content_type":"text/html","content_length":"58779","record_id":"<urn:uuid:bc5eaa8b-c6a3-4256-9dd7-b0258db10fea>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00501.warc.gz"} |
How do you calculate the number of moles from volume? | Socratic
How do you calculate the number of moles from volume?
2 Answers
It depends on the substance whose volume you know.
If you have a pure liquid or a solid, you use its density to calculate its mass and then divide the mass by the molar mass.
If you have a solution, you multiply the molarity by the volume in litres.
There are two steps:
1. Multiply the volume by the density to get the mass.
2. Divide the mass by the molar mass to get the number of moles.
How many moles are in 10.00 mL of acetic anhydride? The molar mass of acetic anhydride is 102.1 g/mol and its density is 1.080 g/mL?
The formula for density is
#color(blue)(ul(bar(|color(white)(a/a) ρ = m/V color(white)(a/a)|)))" "#
where #ρ# is the density, $m$ is the mass, and $V$ is the volume of the sample.
We can rearrange the formula to get
#m = V × ρ#
$\text{mass" = 10.00 color(red)(cancel(color(black)("mL"))) × "1.080 g"/(1 color(red)(cancel(color(black)("mL")))) = "10.80 g}$
#n = 10.80 color(red)(cancel(color(black)("g"))) × "1 mol"/(102.1color(red)(cancel(color(black)("g")))) = "0.1058 mol"#
Molarity is the number of moles of a substance in one litre of solution.
$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} M = \frac{n}{V} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$
where $n$ is the number of moles and $V$ is the volume in litres.
We can rearrange this equation to get the number of moles:
#n = M × V#
How many moles of $\text{NaCl}$ are contained in 0.300 L of 0.400 mol/L $\text{NaCl}$ solution?
#n = 0.300 color(red)(cancel(color(black)("L soln"))) × "0.400 mol NaCl"/(1 color(red)(cancel(color(black)("L soln")))) = "0.120 mol NaCl"#
Here are a couple of examples of mole-volume conversions.
The first part shows you how to convert from 0.30mol of CO to liters of CO.
The second part shows the conversion from 18.2L Ne to moles of Ne.
The problems were completed assuming both gases are at STP (standard temperature and pressure conditions).
video from: Noel Pauller
Hope this helps.
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Beyond Maximum Stable Gain of an RF Transistor
Thomas Lancaster
I was plotting constant-gain circles on the Smith chart the other day for a potentially unstable RF transistor.
I recently read on-line at the link below, that for a potentially unstable transistor the maximum gain that can achieved is the MSG or maximum stable gain. MSG is simply the ratio of the magnitude of
S21 over the magnitude of S12.
For my transistor, the MSG in absolute gain 162.5 or 22.1dB as shown on the attached Smith chart.
The on-line article also said that "one should never try to tease more gain from the transistor than the MSG". I have the output stability circle plotted and the unstable region marked as UR on the
Smith chart. Notice that some of the 22.1dB constant-gain circle is located pretty far away and outside of the output stability circle's unstable region.
For the heck of it, I plotted gain circles all the way up to the ludicrous values of 50 and 100dB to see what would happen. The 29dB gain circle is dangerously close to but still outside the output
stability circle's unstable region.
Only when I jumped to 50 and 100dB did the gain circles just align with the output stability circle. Apparently, gain seems to be infinite with respect to a potentially unstable transistor.
I know that for an unconditionally stable transistor the maximum gain that can be had is the Maximum Available Gain or MAG. In my book, 'RF Circuit Design' by Chris Bowick, Bowick discusses Maximum
Available Gain but not Maximum Stable Gain.
When I Plot the constant-gain circles for an unconditionally stable transistor, I found that as the gain is increased, the radius of the circle decreases. When MAG is achieved, the circle becomes a
single point on the Smith chart and the equation for the radius of the constant-gain circle becomes undefined. This isn't the case for a potentially unstable transistor. At MSG, the constant-gain
circle seems to reach its' smallest radius at 22.1dB(MSG) then increases slightly at 26dB, a little more at 29dB, quite a bit more at 50dB and stays at that size through the ridiculous 100dB gain
What is it about MSG that makes it the final word on stable gain when, I'll say, the 23 or even the 26dB gain circle, or at least part of it, seems safely out of the output stability circle's
unstable region and seems like there would still be some safe terminating impedances, although not many, available, right?
So is the maximum gain that can be had from a potentially unstable transistor simply at the point where the constant-gain circle reaches its' smallest radius, at MSG = 22.1dB for my transistor, even
though the 23,26 and 29dB gain circles still have a portion of their locus outside the output stability circle?
In case you guys are wondering, I want to build a small signal RF amplifier to increase the very weak weather band signal on 162.400MHz and 162.550MHz in my area.
001 (175K)
I have read at least one published paper on positive feedback in the second stage of an LNA. Not sure if this is the article, I don't have ieeexoplore access, but the title sounds useful...
I used the technique and convinced myself I could get one dB more gain beyond MSG. I left the company before the chips came out so I never saw the data...
Thanks for the design notes!
UnknownEditor OptionsShare0
When MAG is achieved, the circle becomes a single point on the Smith chart and the equation for the radius of the constant-gain circle becomes undefined.
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Linear Regression Channel is a three-line technical indicator.
Linear Regression Channel (LRC)
Estimated reading time: 2 min
The Linear Regression Channel is a three-line technical indicator used to analyze the upper and lower limits of an existing trend. Linear regression is a statistical tool used to predict the future
from past data. It is used to determine when prices may be overextended.
A Linear Regression Channel gives potential buy and sell signals based on price volatility.
It consists of three parts:
1. Linear Regression Line
2. Upper Channel Line
3. Lower Channel Line
Linear Regression Line
A Linear Regression Line is a straight line that best fits the prices between a starting price point and an ending price point.
A “best fit” means that a line is constructed where there is the least amount of space between the price points and the actual Linear Regression Line.
The Linear Regression Line is used to determine trend direction.
The acts as the midpoint of the trend.
Think of the trend line as the “equilibrium” price, where any move above or below the trendline indicates overzealous buyers or sellers.
When prices deviate above or below the line, you can expect the price to go back towards the Linear Regression Line.
When prices are below the Linear Regression Line, this is considered bullish.
When prices are above the Linear Regression Line, this is considered bearish.
Upper and Lower Channel Lines
The Upper Channel Line is a line that runs parallel to the Linear Regression Line and is usually one to two standard deviations above the Linear Regression Line.
It marks the top of the trend.
The Lower Channel Line is a line that runs parallel to the Linear Regression Line and is usually one to two standard deviations below the Linear Regression Line.
It marks the bottom of the trend.
The Upper and Lower Channel Lines are evenly distanced from the Linear Regression Line
The default standard deviation setting used is “1” which means 68% of all price movements are contained between that the Upper and Lowe Channe Lines,
When the price breaks outside of the channels, buy and sell signals are generated.
There are two types of Linear Regression channels, depending on the direction of the trend:
1. Bullish Linear Regression channel
2. Bearish Linear Regression channel
These two types of regression channels are defined based on their slope.
Bullish Linear Regression Channel
The bullish Linear Regression Channel indicates a bullish trend. The price is increasing and the slope of the Linear Regression is positive.
Bearish Linear Regression Channel
The bearish Linear Regression Channel indicates a bearish trend. The price is decreasing and the slope of the Linear Regression is negative.
How to Draw
To draw the Linear Regression Channel, simply select the beginning of a trend and stretch the indicator to another point of the trend.
The three lines of the Linear Regression Channel will self-adjust depending on the top and bottom of the trend.
The Linear Regression Channel (middle line) will automatically appear between the Upper and Lower Channels.
How to Use
Trading the Linear Regression Channel involves keeping an eye on the price whenever it interacts with one of the three lines.
Each time that the price interacts with the Upper or Lower Channel, you should expect to see a potential turning point on the price chart.
Buy Signal
If you expect a continuation of the trend, and the price falls below the lower channel line, this should be considered a buy signal.
You can wait for confirmation by waiting for the price to move higher and close back inside the Linear Regression Channel.
Sell Signal
If you expect a continuation of the trend, and the price rises above the upper channel line, this should be considered a sell signal.
You can wait for confirmation by waiting for the price to move lower and close back inside the Linear Regression Channel.
Trend Reversals
When price closes outside of the Linear Regression Channel for long periods of time, this is often interpreted as an early signal that the current trend might be ending and a trend reversal might be
The use of standard deviation can give you an idea ono when prices might be overbought or oversold relative to the long-term trend.
If this article seems useful to your then please click the like button below. You can also share your valuable feedback or ask questions in the below comment section. Also, subscribe to our
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What is an example hardy-weinberg equation practice problem? | Socratic
What is an example hardy-weinberg equation practice problem?
1 Answer
A question could look like:
If $m a t h b f 98$ out of $m a t h b f 200$ individuals in a population express the recessive phenotype, what percent of the population would you predict would be heterozygotes?
An explanation, walked through:
The Hardy-Weinberg equilibrium is a mathematical relationship of the alleles and genotypes in a population that meets certain characteristics. The relationships are as follow:
Alleles: $p + q = 1$
$p = \text{frequency of the dominant allele}$
$q = \text{frequency of the recessive allele}$
Genotypes: ${p}^{2} + 2 p q + {p}^{2} = 1$
${p}^{2} = \text{frequency of homozygous dominant genotype}$
$2 p q = \text{frequency of heterozygous genotype}$
${q}^{2} = \text{frequency of homozygous recessive genotype}$
From the question, we know that $98$ of $200$ individuals express the recessive phenotype. This means that these $98$ also have the homozygous recessive genotype, the frequency of which is equal to $
To determine what the actual frequency is, simply divide $\frac{98}{200} = 0.49$. We now know that ${q}^{2} = 0.49$.
However, we wish to find the frequency of the population that is heterozygous, which is equal to $2 p q$. So, we must find both $p$ and $q$.
Finding $m a t h b f q$:
${q}^{2} = 0.49$
Take the square root of both sides.
$q = 0.7$
(This means that #70%# of the alleles in the system are recessive alleles.)
Now that we've found the value of $q$, we can find the value of $p$ using the allele equation.
Finding $m a t h b f p$:
Through the equation $p + q = 1$, substitute in $q = 0.7$.
$p + 0.7 = 1$
Subtract $0.7$ from both sides to see that
$q = 0.3$
Finding the frequency of heterozygotes:
$\text{frequency of heterozygous genotypes} = 2 p q$
Substitute the known values for $p$ and $q$:
$\text{frequency of heterozygous genotypes} = 2 \left(0.3\right) \left(0.7\right) = 0.42$
Converting this into a percent, we see that #42%# of the population is heterozygous.
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What is Optimization ?
Mathematical optimization is the selection of the best element based on a particular criterion from a set of available alternatives.
In simple cases, a specific optimization problem involves minimizing or maximizing or real function systematically by choosing input values within an allotted set and finding the function’s value. It
applies a large area of mathematics for generalizing theory of optimization. Optimization involves determining “best available” values of the particular objective function in a defined domain along
with a variety of different types of objective functions.
Now, let’s have a look at optimization problems. Here, you need to look for the greatest or the smallest value that can be taken by a function. The constraint will be normal that can be described by
an equation.
Also, to remember that the constraint is the quantity that has to be true irrespective of the solution. You will be looking at one quantity that is clear and has a constant value in every problem.
Once you identify the quantity to be optimized, it’s not difficult to proceed.
Optimization Example
Let us see the solved example on optimization concept for better understanding.
Example: A field has to be enclosed with a fence. You have 500 feet of material, and the building is on one side of the field and not required to be fenced. Calculate the dimensions of the field that
will close the largest region.
In these problems, we have two functions. The first is the function that you will optimize, and second is the constraint. You can sketch the situation that can help you to arrive at the equation.
Here, we want to maximize the area of the field, and we have 500 ft material for fencing. Therefore, the area (i.e. area of a rectangle) will be the function that has to be optimized and the
constraint is the amount of fencing. Hence, the two equations are:
Maximize: A = xy
Constraint: 500 = x + 2y
If you solve the constraint for one of the variables, you can substitute it into the area and then get a function of a single variable.
Let’s solve the constraint for x.
x = 500 – 2y
Now, substitute it in the area function that provides a function of y.
A(y) = (500 – 2y)y = 500y – 2y^2
Now, you need to find the largest value on the interval [0, 250]. Note that the interval corresponds to considering y = 250, i.e., no width and two sides and both must be 250 ft to utilize the whole
500 ft. Also, the endpoints of the interval won’t mean anything if we want to enclose some region both would give zero area. However, they give us a set of limits on y and the Extreme Value Theorem,
which says that we will have a maximum value of the area between two endpoints.
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October 2015
Radian in not-a-unit shocker
One of the nice things about scmutils is that it tracks units, so you can’t accidentally add 10 seconds to 5 metres.
(& 10 &second)
(& 5 &meter))
=> Units do not match: + (*with-units* 10 (*unit* SI ... 1)) (*with-units* 5 (*unit* SI ... 1))
When dealing with angles, it initially seems to do the right thing too:
(& pi/2 &radian)
(& 90 °ree))
=> 3.141... (ie. its converting everything to radians)
But this is less cool:
(/ (& pi &radian) (& 1 &second))
=> (& 3.141592653589793 &hertz)
Err, pi radians should be 0.5Hz. The trouble is, scmutils treats radians as a unit-less number.
To check whether this was a reasonable thing to do, I checked my old favourite Frink. In frink’s units.txt files, we have the following:
// Alan’s editorializing:
// Despite what other units programs might have you believe,
// radians ARE dimensionless units and making them their own
// unit leads to all sorts of arbitrary convolutions in
// calculations (at the possible expense of some inclarity if
// you don’t know what you’re doing.)
// If you really want radians to be a fundamental unit,
// replace the above with “angle =!= radian”
// (This will give you a bit of artificiality in calculations.)
// The radian was actually a fundamental base unit in the SI
// up until 1974, when they changed it, making it no longer
// be a special unit, but just a dimensionless number (which
// it is.) See the definition of the “Hz” below for a
// discussion of how this broke the SI’s definitions of
// basic circular / sinusoidal measures, though.
And down a bit, on the section about hertz, we have:
// Alan’s Editorializing: Here is YET ANOTHER place where the SI made a
// really stupid definition. Let’s follow their chain of definitions, shall
// we, and see how it leads to absolutely ridiculous results.
// The Hz is currently defined simply as inverse seconds. (1/s).
// See: http://physics.nist.gov/cuu/Units/units.html
// The base unit of frequency in the SI *used* to be “cycles per second”.
// This was fine and good. However, in 1960, the BIPM made the
// change to make the fundamental unit of frequency to
// be “Hz” which they defined as inverse seconds (without qualification.)
// Then, in 1974, they changed the radian from its own base unit in the SI
// to be a dimensionless number, which it indeed is (it’s a length divided by
// a length.) That change was correct and good in itself.
// However, the definition of the Hz was *not* corrected at the same
// time that the radian was changed. Thus, we have the conflicting SI
// definition of the radian as the dimensionless number 1 (without
// qualification) and Hz as 1/s. (Without qualification.)
// This means that, if you follow the rules of the SI,
// 1 Hz = 1/s = 1 radian/s which is simply inconsistent and violates basic
// ideas of sinusoidal motion, and is simply a stupid definition.
// The entire rest of the world, up until that point, knew that 1 Hz needs to
// be equal to *2 pi* radians/s or be changed to mean *cycles/second* for
// these to be reconcilable. If you use “Hz” to mean cycles/second, say,
// in sinusoidal motion, as the world has done for a century, know that the SI
// made all your calculations wrong. A couple of times, in different ways.
// This gives the wonderful situation that the SI’s Hz-vs-radian/s definitions
// have meant completely different things in the timeperiods:
// * pre-1960
// * 1960 to 1974
// * post-1974
// Thus, anyone trying to mix the SI definitions for Hz and angular
// frequencies (e.g. radians/s) will get utterly wrong answers that don’t
// match basic mathematical reality, nor match any way that Hz was ever used
// for describing, say, sinusoidal motion.
// Beware the SI’s broken definition
// of Hz. You should treat the radian as being correct, as a fundamental
// dimensionless property of the universe that falls out of pure math like
// the Taylor series for sin[x], and you should treat the Hz as being a
// fundamental property of incompetence by committee.
// One could consider the CGPM in 1960 to have made the original mistake,
// re-defining Hz in a way that did not reflect its meaning up to that point,
// or the CGPM in 1974 to have made the absolutely huge mistake that made
// the whole system inconsistent and wrong, and clearly broke the definition
// of Hz-vs-radian/s used everywhere in the world, turning it into a broken,
// self-contradictory mess that it is now.
// Either way, if I ever develop a time machine, I’m going to go back and
// knock both groups’ heads together. At a frequency of about 1 Hz. Or
// better yet, strap them to a wheel and tell them I’m going to spin one group
// at a frequency of 1 Hz, and the other at 1 radian/s and let them try to
// figure out which one of those stupid inconsistent definitions means what.
// Hint: It’ll depend on which time period I do it in, I guess, thanks to
// their useless inconsistent definition changes.
// It’s as if this bunch of geniuses took a well-understood term like “day”
// and redefined it to mean “60 minutes”. It simply breaks every historical
// use, and present use, and just causes confusion and a blatant source of
// error.
// In summary: Frink grudgingly follows the SI’s ridiculous, broken definition
// of “Hz”. You should not use “Hz”. The SI’s definition of Hz should be
// considered harmful and broken. Instead, if you’re talking about circular
// or sinusoidal motion, use terms like “cycles/sec” “revolutions/s”,
// “rpm”, “circle/min”, etc. and Frink will do the right thing because it
// doesn’t involve the stupid SI definition that doesn’t match what any
// human knows about sinusoidal motion. Use of “Hz” will cause communication
// problems, errors, and make one party or another look insane in the eyes
// of the other. | {"url":"https://www.nobugs.org/blog/archives/2015/10/","timestamp":"2024-11-04T11:18:37Z","content_type":"text/html","content_length":"56751","record_id":"<urn:uuid:ffbfa42f-800d-435d-aeeb-f335c3c22f10>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00210.warc.gz"} |
From SA-MP Wiki
This callback is called when a player disconnects from the server.
This callback can also be called by NPC.
(playerid, reason)
playerid The ID of the player that disconnected.
reason The reason for the disconnection. See table below.
Return Values:
• 0 - Will prevent other filterscripts from receiving this callback.
• 1 - Indicates that this callback will be passed to the next filterscript.
• It is always called first in filterscripts.
Some functions might not work correctly when used in this callback because the player is already disconnected when the callback is called. This means that you can't get unambiguous information
from functions like GetPlayerIp and GetPlayerPos.
ID Reason Details
0 Timeout/Crash The player's connection was lost. Either their game crashed or their network had a fault.
1 Quit The player purposefully quit, either using the /quit (/q) command or via the pause menu.
2 Kick/Ban The player was kicked or banned by the server.
Example Usage:
public OnPlayerDisconnect(playerid, reason)
GetPlayerName(playerid, playerName, MAX_PLAYER_NAME);
new szDisconnectReason[3][] =
format(szString, sizeof szString, "%s left the server (%s).", playerName, szDisconnectReason[reason]);
SendClientMessageToAll(0xC4C4C4FF, szString);
return 1;
Related Callbacks
The following callbacks might be useful as well, as they are related to this callback in one way or another. | {"url":"https://sampwiki.blast.hk/wiki/OnPlayerDisconnect","timestamp":"2024-11-08T09:14:59Z","content_type":"application/xhtml+xml","content_length":"14914","record_id":"<urn:uuid:284b8de4-920e-41e2-ac9d-4fd08ecda610>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00102.warc.gz"} |
Yet more on ANOVA
Traditional ANOVA (between-subjects)
aov_car(formula = phit ~ cond + Error(subj), data = prod)
Contrasts set to contr.sum for the following variables: cond
Anova Table (Type 3 tests)
Response: phit
Effect df MSE F ges p.value
1 cond 1, 58 0.04 2.62 .04 .11
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '+' 0.1 ' ' 1
Here’s what each component of the output means:
Contrasts set to contr.sum for the following variables: cond - This refers to a part of the aov_car command that we do not cover in these worksheeets. It can be safely ignored.
Anova Table - An ANOVA table is just a description of the way we report the results of an ANOVA. It is everything from the line which begins Effect until the line ---.
(Type 3 tests) - There’s more than one way of doing an ANOVA. In fact, there are three ways, known as Type 1, Type 2, and Type 3. Statisticians argue about which way is best, and R can calculate them
in any of these three ways. However, psychologists have nearly always used Type 3, mainly because they didn’t realize there were other ways of doing it! So, Type 3 in this context basically just
means “the way psychologists expect it to have been done”. If you are interested in how these types differ, take a look at this blog post.
Response: phit - This just confirms that R is doing what you asked it to, i.e. analyzing the variable phit.
The next two lines are read as a table. So, the first line, Effect... are the labels on the table, and the second line are the relevant numbers. Taking each of these in turn:
Effect: cond - This confirms we are looking at the effect of the variable cond (on phit, see above).
df: 1, 58 - We came across the concept of degrees of freedom (df) before, when looking at traditional t-tests. In that case, there was just one number, and this corresponds to the second number in an
ANOVA - 58 in this case. This second number is a way of talking about the size of the dataset. It isn’t quite the sample size, but it’s related to it. The first number, 1 in this case, is a way of
talking about the number of levels in the factor. There are two levels in the factor we are analyzing (silent vs. read aloud). So, the first df is not quite the number of levels, but it is related to
it. The two degrees of freedom in an ANOVA are also known as the numerator (first number) and denominator (second number) degrees of freedom.
MSE: 0.04 - MSE stands for “mean squared error”. Like a standard deviation, this is a measure of variability around the mean. It’s generally more useful to report ges (covered in the main worksheet)
instead, but some journals insist on MSE, so it’s also provided for completeness.
F: 2.62 - The F value (also known as an F-ratio) is just a number, like a t value. On its own, it does not allow us to draw any conclusions. However, when we also know the degrees of freedom, we can
use it to work out the p value. Decades ago, this was done by looking up the F value in the back of a book. Today, we just let the computer work it out for us.
ges: .04 - Generalized eta-squared, see main worksheet.
p.value: .11 - The p value, see main worksheet.
Signif. codes: ... - This is a ‘key’, i.e. it explains some other part of the output. If the p value had been less than .1, you would have seen one of these symbols next to the F value. These symbols
help the reader quickly spot ‘significant’ results.
This material is distributed under a Creative Commons licence. CC-BY-SA 4.0. | {"url":"https://www.andywills.info/rminr/yet-more-on-anova.html","timestamp":"2024-11-07T09:49:57Z","content_type":"application/xhtml+xml","content_length":"634568","record_id":"<urn:uuid:e48fdf44-9750-4db9-9c71-ebb986de398e>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00679.warc.gz"} |
\overrightarrow b = 0$. What can you conclude about the vectors $\
Hint: Here, we need to draw a conclusion about the vectors $\overrightarrow a $ and $\overrightarrow b $from the statements $\overrightarrow a .\overrightarrow b = 0$ and $\overrightarrow a \times \
overrightarrow b = 0$ by considering $\overrightarrow a .\overrightarrow b = \left| a \right|.\left| b \right|.\cos \theta $and $\overrightarrow a \times \overrightarrow b = \left| a \right|.\left| b
\right|.\sin \theta $.
Complete step-by-step answer:
i. $\overrightarrow a .\overrightarrow b = 0$.
Here, $\overrightarrow a .\overrightarrow b = 0$ is the dot product of the vectors $\overrightarrow a $ and $\overrightarrow b $.As, we know the dot product of two vectors can be written as:
$\overrightarrow a .\overrightarrow b = \left| a \right|.\left| b \right|.\cos \theta \to (1)$
$\left| a \right|$ Is the magnitude of$\overrightarrow a $, $\left| b \right|$is the magnitude of $\overrightarrow b $and $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $.
It is given that $\overrightarrow a .\overrightarrow b = 0$ i.e..,
$\left| a \right|.\left| b \right|.\cos \theta = 0 \to (2)$
So, from equation (2) we can say that the dot product of vectors $\overrightarrow a $ and $\overrightarrow b $is ‘0’ in the following cases.
(i) $\left| a \right| = 0$i.e.., the magnitude of $\overrightarrow a $is zero.
(ii) $\left| b \right| = 0$i.e.., the magnitude of $\overrightarrow b $is zero.
(iii) $\overrightarrow a \bot \overrightarrow b $i.e.., the angle between the vectors is${90^o}$$[\because \cos {90^o} = 0]$.
Hence, we can conclude that $\overrightarrow a .\overrightarrow b = 0$if ‘$\left| a \right| = 0$’or if ‘$\left| b \right| = 0$’or ‘if the vectors are perpendicular to each other.
ii. $\overrightarrow a \times \overrightarrow b = 0$.
Here, $\overrightarrow a \times \overrightarrow b = 0$ is the cross product of the vectors $\overrightarrow a $ and $\overrightarrow b $.As, we know the cross product of two vectors can be written
$\overrightarrow a \times \overrightarrow b = \left| a \right|.\left| b \right|.\sin \theta \to (1)$
$\left| a \right|$ Is the magnitude of$\overrightarrow a $, $\left| b \right|$is the magnitude of $\overrightarrow b $and $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $.
It is given that $\overrightarrow a \times \overrightarrow b = 0$ i.e..,
$\left| a \right|.\left| b \right|.\sin \theta = 0 \to (2)$
So, from equation (2) we can say that the cross product of vectors $\overrightarrow a $ and $\overrightarrow b $is ‘0’ in the following cases
(i) $\left| a \right| = 0$i.e.., the magnitude of $\overrightarrow a $is zero.
(ii) $\left| b \right| = 0$i.e.., the magnitude of $\overrightarrow b $is zero.
(iii)$\overrightarrow a \parallel \overrightarrow b $i.e.., the angle between the vectors is${0^o}$$[\because \sin {0^o} = 0]$.
Hence, we can conclude that $\overrightarrow a \times \overrightarrow b = 0$if ‘$\left| a \right| = 0$’or if ‘$\left| b \right| = 0$’or ‘if the vectors are parallel to each other.
Note: The dot product of two vectors will be $'0'$ if the vectors are perpendicular to each other (in case vectors are non-zero).Similarly, the cross product of two vectors will be $'0'$ if the
vectors are parallel to each other (in case vectors are non-zero). | {"url":"https://www.vedantu.com/question-answer/given-that-overrightarrow-a-overrightarrow-b-0-class-8-maths-cbse-5ee717fd47f3231af27dcfac","timestamp":"2024-11-11T08:04:38Z","content_type":"text/html","content_length":"158800","record_id":"<urn:uuid:5e036e1c-06b5-4db8-9b21-3a62c03b41b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00797.warc.gz"} |
An efficient approach for extracting anharmonic force constants from atomistic simulations
Phonon theory is an important tool when analyzing solid state systems which is of high importance in modern technology. Central to the theory is the so called force constants (FCs) which determines
the thermal behaviour. The super cell method is one way to extract the FCs from the force field of a displaced specimen but they are computationally expensive by requiring a lot of carefully prepared
input data. By constructing an interatomic potential which uses the harmonic and higher order FCss as parameters the FCs can be extracted by a simple fitting procedure. Some small number of quasi
random input configurations and the resulting force fields can contain enough information to extract all FCs with comparably low amount of computation. The underlying symmetries of the lattice must
be used to reduce the number of free parameters in the model. This thesis demonstrates the implementation and application of the Force Constant Model. The model performs well and anharmonic FCs can
be extracted. Given the FCs the model can be used as a potential and molecular dynamics can be performed yielding a direct method for computing thermal properties. Possible applications include
analysis of thermal stabilization and phonon life time determination. | {"url":"https://materialsmodeling.org/theses/2018-An-efficient-approach-for-extracting-anharmonic-force-constants-from-atomistic-simulations/","timestamp":"2024-11-02T17:09:53Z","content_type":"text/html","content_length":"64614","record_id":"<urn:uuid:ce9c1670-4985-4918-9e79-920e280f0961>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00335.warc.gz"} |
One Degree Equals How Many Miles? - Answered
One Degree Equals How Many Miles?
One degree equals how many miles.
How far is a nautical mile in sea of thieves?
A nautical mile is about 3,280 feet.
How do you calculate degrees and minutes on a map?
There is no one definitive answer to this question. Some methods of calculating degrees and minutes on a map include using a compass, using a ruler, or using a timer.
How do you read a compass degree and minutes?
To read a compass degree, look at the top right corner and the number on the left. The degrees are measured in degrees, while minutes are measured in minutes.
Why do sailors use knots?
A knot is a type of knot that is used in sailing. Knots help to keep a sail in the wind and prevent it from getting caught on something.
How many miles is 5 degrees longitude?
There is 5 degrees of longitude.
How do you convert latitude and longitude to distance?
To convert latitude and longitude to distance, divide the latitude by the longitude.
What is 0 lat 0 long called?
The latitude and longitude of a location are both measured in degrees. 0 latitude is the point at the equator, and 0 longitude is the point north of the equator.
Why is a nautical mile different than a mile?
A nautical mile is 1,609.4 feet, while a mile is 1,609.4 inches.
How many minutes is 1 degree on a map?
1 degree is the amount of time it takes for the sun to travel from the equator to the poles.
How does the Earth rotate 1 degree of longitude in 4 minutes?
The Earth rotates 1 degree of longitude in 4 minutes because it takes 4 minutes for the Earth to make a full orbit around the sun.
How do you calculate degrees on a calculator?
There is no one definitive answer to this question as it depends on the calculator and the degree level that the user is working on. Generally speaking, a calculator will allow users to input various
fractions, decimals, and other mathematical operations to calculate degrees. Additionally, many calculators have online calculators that allow users to complete online quizzes and tests to help them
calculate degrees.
What does a degree measure on a world map?
A degree measures a person’s ability to communicate and think abstractly.
How many nautical miles is 1 degree of latitude?
There are a total of 18 degrees of latitude.
How far is a geographic minute?
A geographic minute is about the distance from the Earth’s equator to the sun.
What is the relation between minute and degree?
The relation between minute and degree is 1/60th of a degree.
How do you convert degrees into minutes?
To convert degrees into minutes, divide the degrees by 60. To convert degrees into seconds, divide the degrees by 60.
What are the 5 difference between latitude and longitude?
Latitude and longitude are two of the most important pieces of information you need to know in order to navigate the world. Latitude is the angle north or south of the equator, while longitude is the
angle east or west of the prime meridian. Latitude and longitude can be used to calculate distances and find locations, as well as to find directions in a text or map.
How many minutes is a degree of latitude?
There is no definitive answer to this question since it depends on a variety of factors, such as climate, terrain, and population. However, a good estimate would be around 18 minutes.
What distance is 1 second of latitude?
1 second of latitude is equal to the angle between the equator and the prime meridian.
How do you calculate distance using longitude?
Longitude can be calculated by adding the latitude to the longitude.
How many miles are between latitudes?
There is about 100 miles between latitude lines.
How many degrees are there in longitude?
There are 3 degrees in longitude: 0 degrees north, 0 degrees south, and 1 degree east.
How far is 1 degree on a map?
1 degree is the distance from the equator to the poles.
How many miles is 1 degree longitude at the equator?
The equator is about 93.3 miles longitude at the equator.
How do you read degrees on a map?
There are a few ways to read degrees on a map. One way is to use a map to find the school or college where the person receiving the degree was born. Another way is to use a map to find the city or
town where the person receiving the degree was living.
What are the 2 main lines of longitude?
The two main lines of longitude are Greenwich Mean Time (GMT) and Universal Time (UT). GMT is the standard time in the United Kingdom, Europe, and most of Asia. UT is the standard time in the United
States, Canada, and most of Mexico.
How many miles is 360 degrees?
There is 360 degrees in a circle.
How many miles around the Earth in a straight line?
There is no definitive answer to this question as it depends on the latitude of the location and the length of the straight line. In general, the closer you are to the equator, the more miles you
will travel in a straight line.
How many miles is 60 degrees longitude?
There is no definitive answer to this question as it is dependent on a number of factors, including the latitude of the location and the time of day. However, according to the National
Geospatial-Intelligence Agency, the average length of a straight line drawn from the North Pole to the South Pole is about 60 degrees. Therefore, the latitude of the location would affect the number
of miles.
What length is a nautical mile?
A nautical mile is a unit of distance that is equal to 12,500 feet. | {"url":"https://www.nftartrank.com/one-degree-equals-how-many-miles/","timestamp":"2024-11-09T14:02:38Z","content_type":"text/html","content_length":"125922","record_id":"<urn:uuid:ad4ceafc-8dfd-4f9c-99fe-af1306999137>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00204.warc.gz"} |
#ML concepts - Regularization, a primer
Regularization is a fundamental concept in Machine Learning (#ML) and is generally used with activation functions . It is the key technique that help with overfitting.
Overfitting is when an algorithm or model ‘fits’ the training data too well - it seems to good to be true. Essentially overfitting is when a model being trained, learns the noise in the data instead
of ignoring it. If we allow overfitting, then the network only uses (or is more heavily influenced) by a subset of the input (the larger peaks), and doesn’t factor in all the input.
The worry there being that outside of the training data, it might not work as well for ‘real world’ data. For example the model represented by the green line in the image below (credit: Wikipedia),
follows the sample data too closely and seems too good. On the other hand, the model represented by the black line, which is better.
Regularization helps with overfitting (artificially) penalizing the weights in the neural network. These weights are represented as peaks, and this reduces the peaks in the data. This ensure that the
higher weights (peaks) don’t overshadow the rest of the data, and hence getting it to overfit. This diffusion of the weight vectors is sometimes also called weight decay.
Although there are a few regularization techniques for preventing overfitting (outlined below), these days in Deep Learning, L1 and L2 regression techniques are more favored over the others.
• Cross validation: This is a method for finding the best hyper parameters for a model. E.g. in a gradient descent, this would be to figure out the stopping criteria. There are various ways to do
this such as the holdout method, k-fold cross validation, leave-out cross validation, etc.
• Step-wise regression: This method essentially is a serial step-by-step regression where one reduces the weakest variable. Step-wise regression essentially does multiple regression a number of
times, each time removing the weakest correlated variable. At the end you are left with the variables that explain the distribution best. The only requirements are that the data is normally
distributed, and that there is no correlation between the independent variables.
• L1 regularization: In this method, we modify the cost function by adding the sum of the absolute values of the weights as the penalty (in the cost function). In L1 regularization the weights
shrinks by a constant amount towards zero. L1 regularization is also called Lasso regression.
• L2 regularization: In L2 regularization on the other hand, we re-scale the weight to a subset factor - it shrinks by an amount that is proportional to the weight (as outlined in the image below).
This shrinking makes the weight smaller and is also sometimes called weight decay. To get this shrinking proportional, we take a <strong>squared mean of the weights</strong> , instead of the
sum. At face value it might seem that the weight eventually get to zero, but that is not true; typically other terms cause the weights to increase. L2 regularization is also called Ridge
• Max-norm: This enforces a upper bound on the magnitude of the weight vector. The one area this helps is that a network cannot ’explode’ when the learning rates gets very high, as it is
bounded. This is also called projected gradient descent.
• Dropout : Is very simple, and efficient and is used in conjunction with one of the previous techniques. Essentially it adds a probably on the neuron to keep it active, or ‘dropout’ by setting it
to zero. Dropout doesn’t modify the cost function; it modifies the network itself as shown in the image below.
• Increase training data: Whilst one can artificially expand the training set theoretically possible, in reality won’t work in most cases, especially in more complex networks. And in some cases one
might think also to artificially expand the dataset, typically it is not cost effective to get a representative dataset.
L1 Regularization
L2 Regularization
Between L1 and L2 regularization, many say that L2 is preferred, but I think it depends on the problem statement. Say in a network, if a weight has a large magnitude, L2 regularization shrink the
weight more than L1 and will better. Conversely, if the weight is small then L1 shrinks the weight more than L2 - and is better as it tends to concentrate the weight in fewer but more important
connections in the network.
In closing, the key aspect to appreciate - the small weights (peaks) in a regularized network essentially means that as our input changes randomly (i.e. noise), it doesn’t have a huge impact to the
network and its output. So this makes it difficult for the network to learn the noise and respond to that. Conversely, in an unregularized networks, that has higher weights (peaks), small random
changes to those weights can have a larger impact to the behavior of the network and the information it carries. | {"url":"https://blog.desigeek.com/post/2018/09/ml-concepts-regularization-a-primer/","timestamp":"2024-11-02T21:29:49Z","content_type":"text/html","content_length":"34940","record_id":"<urn:uuid:d978d58b-6e07-4457-a9d1-3fbb04dc4092>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00277.warc.gz"} |
Brain Teaser Of the Day: Only Sharp Eyes Can Find the 3 Hidden Words in this Image in 10 Seconds? Viral Picture IQ Test Quiz - EduViet Corporation
Brain Teaser Of the Day: Only Sharp Eyes Can Find the 3 Hidden Words in this Image in 10 Seconds? Viral Picture IQ Test Quiz
brain teasers
Brain teasers are puzzles that need to be solved. Solving brainteasers requires lateral thinking. You need to look at a puzzle from different angles. Archimedes was a Greek mathematician who became
famous in his early years as a fan of brain teasers because he was very good at solving difficult problems with ease. Honestly, most brainteasers are tests of IQ and require you to use your
intelligence to solve the problem.
The benefits of brain teasers
Brainteasers have many benefits, including:
You are watching: Brain Teaser Of the Day: Only Sharp Eyes Can Find the 3 Hidden Words in this Image in 10 Seconds? Viral Picture IQ Test Quiz
1. Mental stimulation: Brain teasers challenge your cognitive abilities and force you to think critically, which helps keep your mind sharp.
2. Improves memory: Solving brainteasers improves your memory because you have to remember patterns and information in order to solve them.
3. Improves problem-solving skills: Brainteasers require you to think creatively and develop new problem-solving strategies, which can improve your problem-solving skills in other areas of life.
4. Reduce stress: Solving brain teasers is a fun and relaxing activity that can take your mind off the stress in your life.
5. Improves Mood: The feeling of accomplishment that comes from solving a brainteaser can improve your mood and boost your confidence.
6. Increases productivity: Taking short breaks to solve brain teasers can actually increase your productivity because it gives your brain a chance to rest and recharge.
Overall, brainteasers are a great way to keep your mind sharp and engaged, and they can provide a variety of mental and emotional benefits.
Daily Brainteaser: Only the sharp-eyed can find the 3 hidden words in this image in 10 seconds?Viral Picture IQ Test Quiz
Solving these brainteasers requires you to think creatively and allow yourself to think of easy ways to solve the problem. Brainteasers are very simple if you hack the problem-solving process by
applying various strategies that help solve the problem. For example, thinking outside the box allowed you to easily find the solution to this brain teaser.
If you’re excited about solving brainteasers and want to try your hand at solving a problem, we have a solution you can try, pictured below. You only have a few seconds to solve this brain teaser
picture puzzle. Do your best and try to find it within the given seconds. Take a close look at the picture above and try to answer it. You still have a few seconds.
Daily Brainteaser: Only the sharp-eyed can find the 3 hidden words in this image in 10 seconds?Viral Picture IQ Test Quiz – Solutions
Finding the answers to the brain teasers is not a difficult task. In just a few seconds you can get the answer from these pictures. You just need to concentrate and look carefully at the pictures we
have given below to solve this brain teaser. You’ll feel like you’ve won the jackpot after finding the answer to the brainteaser.
The picture below shows the solution to this picture puzzle. The picture below shows the solution to this brain teaser. The highlighted area in this picture is the answer to this brain teaser picture
90% of people can’t answer “5+2×5-2=?” – are you smart enough to solve this math puzzle?
Many people struggle to solve the mathematical puzzle “5 + 2 × 5 – 2 = ?”, resulting in a 90% failure rate. It tests understanding of the sequence of operations.
The key is to follow the BODMAS order of operations, where multiplication precedes addition and subtraction. First, we perform multiplication: 2 × 5 = 10. We then add 5 to the result: 5 + 10 = 15.
Finally, we subtract 2 from 15: 15 – 2 = 13. So, the answer is 13.
Brainteaser Math Quiz: Solve 5+5×5+5=?
See more : Test Visual Acuity: If you have Eagle Eyes Find the B in 12 Secs
Stimulate your thinking with this brainteaser math quiz: Can you solve the answer to 5 + 5 × 5 + 5=? Remember, understanding the order of operations is critical to meeting this challenge.
Following the order of operations of BODMAS, we first multiply 5 by 5 to get 5 + 25 + 5=?. Then, we add 5 and the result is 5 + 30 = ?. Finally, add the last 5, and the total is 35.
Brainteaser: Find the next term in 9, 19, 21, 43, 45,?
Can you spot the pattern and predict the next item in the sequence 9, 19, 21, 43, 45…? Uncovering underlying patterns is key to solving this puzzle.
Each term follows an interesting pattern where alternating (x 2 + 1) and (x 1 + 2) apply this 9, 9(x2+1)=19, 19(x1+2)=21, 21(x2 +1 )=43, 43(x1+2)=45, the next term is obtained from 45(x2+1)=91.
Brain teaser math speed test: 35÷5x(4+9)=?
This brainteaser shows a mathematical expression involving division, multiplication, and addition: 35 ÷ 5 x (4 + 9). It is critical to follow the order of operations (BODMAS) to correctly solve this
To solve this expression, first evaluate the value inside the brackets: 4 + 9 = 13. Then, divide 35 by 5 to get 7. Finally, multiply 7 by 13 to get the answer, 35 ÷ 5 x (4 + 9 ) = 91. Remember that
parentheses dictate the order of calculations to ensure correct calculations.
Brain teaser math test: 1+6=7, 2+7=16, 3+8=27, 6+11=?
This brainteaser shows a pattern where the sum of two numbers is calculated in a unique way. In the equation 1+6=7, 2+7=16, 3+8=27, 6+11=? The challenge is to decipher this pattern and apply it to
the next equation. The pattern involves multiplying the second number by the first number and then adding the second number. Apply the pattern 1+6 = (6×1)+1 = 6+1=7 to the fourth equation 6+11=
(11×6)+6 = 66+6=72. Therefore, the answer to this brainteaser is 72.
Disclaimer: The above information is for general information purposes only. All information on this website is provided in good faith, but we make no representations or warranties, express or
implied, as to the accuracy, adequacy, validity, reliability, availability or completeness of any information on this website.
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Category: Brain Teaser
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Various important events and important contributions were made in the period 500 BC – 400 AD which we call it as Post Vedic period. Religions like Buddhism, Jainism were formed; first Indian scripts-
Brahmi scripts (derived from Kharoshti scripts) were recorded and which led the foundation for Devanagari numbers which ultimately led to the numbers which we follow today, epics like Ramayana and
Mahabharata were written, complete treatise on Sanskrit grammar which was prepared by Panini and is even followed today, etc.
[wptabs style=”wpui-narrow” effect=”slide” mode=”horizontal”][wptabtitle] Jaina & Buddhist Mathematics[/wptabtitle] [wptabcontent] Jaina mathematics is one of the least understood chapters of Indian
mathematics, mainly because of the scarcity of the work. Jain religion was formed by Varadhamana (Mahavira) around in the period 570 BC. Jaina mathematics is termed as mathematics which was followed
by those who followed Jainism and was seen in the period 400 BC – 200 AD. Jaina mathematics could be seen written in various Sutras like Surya Prajnapati, Bhagabati Sutra, Sthananga Sutra, Jambudvipa
Prajnapti , Vaishali Ganit, Uttaradhayyan Sutra, Anuyoga Dwara Sutra, Tiloyapannatti, etc. Many topics on mathematics were discussed in Sthananga Sutra: arithmatic operations like multiplication,
division, subtraction, addition; Number Theory; Geometry;Mensuration of solid objects like sphere, cylinders,etc; fractions; Solving simple, quadratic, cubic, biquadratic equations, laws of indices
and Permutation & Combinations. Like people from Vedic period, Jaina mathematicians were also interested in cosmology and large numbers.
Jain cosmology says universe, as an uncreated entity, existing since infinity,the shape of the universe as similar to a man standing with legs apart and arm resting on his waist. This Universe,
according to Jainism, is narrow at the top, broad at the middle and once again becomes broad at the bottom.
• Infinty:
This brings the topic of infinity. They classified infinity in 5 types: infinity in 1 direction, in 2 directions, in an area, everywhere and perpetually infinite. They were able to calculate that
cosmology contains a time period of 2588 years= a number with 178 digits which is a very large number and that many number of years calculated as the time period of cosmology which can be
considered almost infinity.
• Laws of Indices:
In their work, laws of indices were also seen. Eg: The first square root multiplied by the second square root is the cube of the second square root. = (√a).(√√a) = (√√a)^3.
• Classification of Numbers:
Jaina mathematicians classified numbers in 3 groups: Enumerable (that can be counted), Innumerable (large numbers to count) and Infinite (difficult to count).
• Geometry:
Geometry: Various solid and general geometrical figures and terms were used like circle, arc, chord, ellipse, sphere, cylinder, etc. They calculated value of pi as sqrt of 10.
Pingala was born in the 4th century of BC and was the younger brother of Panini. He used the concepts of mathematics in his work Chandahsastra, a Sanskrit poetry.
Panini born around 6th century BC is well known for his formulation of rules, grammar, semantics of Sanskrit language written in his treatise ‘Ashtadhyayi’. Panini’s scientific theory of Sanskrit is
marked as the beginning of Classical era of Sanskrit and his work was so complete and correct that it is followed still today.
In Pingala’s work Chandahsastra following things can be seen:
• First ever description of Binary number system(which is the language of computers) consisting of patterns which are in terms of long and short syllables.
• Binomial theorem
• Pascal’s triangle
• Fibonacci series
He followed the Patterns of long and short (modern patterns are with 0’s and 1’s) Example: In his pattern of 4 syllables, he used them as
│Syllables │Possibilities/Patterns │No. of Patterns│Combinations│
│4 short │SSSS │1 │^4C[0] │
│3 short & 1 long│SSSL, SSLS, SLSS, LSSS │4 │^4C[1] │
│2 short & 2 long│SSLL, LLSS, SLSL, LSSL, LSLS, SLLS │6 │^4C[2] │
│1 short & 3 long│SLLL, LSLL, LLSL, LLLS │4 │^4C[3] │
│4 long │LLLL │1 │^4C[0] │
• Column 2 in modern binary system is written in terms of 0’s and 1’s.
• Patterns in column 3 are same as that of coefficients of terms with a power of 4[Binomial Distribution: (a+b)^4].
• Column 4 shows the knowledge of Combinations Theory.
Likewise, various combinations of patterns can be seen in his work. Again these patterns were not for pure mathematical purpose but were used in writing Sanskrit poems. The patterns were named
separately and were used as words in poems and these patterns are called as metres/rhythms in terms of poems. Eg: LSS was called as Bha, SSL as Sa, L as Gu, S as La, SLS as Ja, etc.
Fibonacci Series:
Fibonacci series can also be seen in these patterns. Taking 1 unit for S and 2 units for L we have.
│Units│Patterns │No. of Patterns│
│1 │S │1 │
│2 │SS, L │2 │
│3 │SSS, SL, LS │3 │
│4 │SSSS, SSL, SLS, LSS, LL │5 │
│5 │SSSSS, SSSL, SSLS, SLSS, LSSS, LLS,SLL, LSL │8 │
│6 │SSSSSS, SSSSL, LSSSS, SLSSS, SSLSS, SSSLS, LLSS, SSLL, LSSL, LLL, SLSL, LSLS, SLLS │13 │
Last column shows the Fibonacci series pattern.
These patterns were discovered by Gopala and Hemachandra from the poems of Sanskrit around 1000 years later and are named them as Hemachandra-Gopala pattern of numbers. But these numbers today are
known as Fibonacci series as these numbers were introduced to the western world by Fibonacci, born around 50 years after them, through his book “Liber Abaci”.
[/wptabcontent] [wptabtitle] Numerals & Manuscripts[/wptabtitle] [wptabcontent]
Brahmi Numerals:
Brahmi numerals were the ancient Indian numeral system which were followed between 300 BC and 500 AD. This numeral system was based on decimal system (base 10) which we follow today. Brahmi numerals
were modified many times in this period and finally arriving at the numerals which we use today. Brahmi numerals are the ancestors of Hindu/Hindu-Arabic numerals. Brahmi numerals are derived from
Kharoshti Numerals which were present in the period between 400 and 300 BC. In kharoshti numerals, 1/2/3 were indicated by vertical lines like that of Roman while in Brahmi numerals, they were
indicated by horizontal lines.
2nd Image shows the Transition of Brahmi Numerals over the years.Zero was not yet invented.
Bakhashali Manuscript:
This manuscript is the ancient Indian manuscript written on birch bark which was found near the village of Bakhshali which is in Pakistan today. Not all barks are found, some are still missing while
many are destroyed/scrapped. Bakhshali manuscript is considered as the oldest Ancient Indian Mathematical manuscript. These manuscripts are from 200 BC – 400 AD. (End date is not known exactly but it
is surely before Aryabhatta’s time as in the scripts 0 was denoted by Dot).
• Arithmetic operations. But the words were used instead of symbols. Eg ‘bha’ was used to indicate division
• Fractions
• Technique for calculating square roots
• Solving Equations
• Negative numbers
Fractions were not so different to that used today, written with one number below the other. No line appears between the numbers as we would write today. Another unusual feature is, the sign + was
placed after a number to indicate -(subtraction)/negative number.
Squareroot Formula:
They used formula formula to calculate square root and surprisingly it was very close to accuracy. √Q = √(A^2 + b) = A + b/2A – (b/2A)^2/[2(A + b/2A)] Where A = square root of perfect square just
less than that Q, b = Q-A.
Q = 41, then A = 6, b = 5
Bakhshali formula gives 6.403138528.
Correct answer is 6.403124237
Here 8 decimal values are correct.
Bakhshali formula gives 582.2447938796899
Correct answer is 582.2447938796876
Here 11 decimal values are correct.
[/wptabcontent] [/wptabs]
1. Lola says
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Applied Mathematics
2024-2025 Undergraduate Course Catalog
Applied Mathematics, BS
Department Chair: Graham J. Leuschke, 215 Carnegie Building,
, 315-443-1478
Associate Chair for Undergraduate Studies: Leonid Kovalev, 311C Carnegie Building,
, 315-443-1487
S.P. Diaz, L. Kovalev, S. Wehrli
Uday Banerjee, Pinyuen Chen, Dan Coman, Steven Diaz, Shukai Du, Nicole M.L. Fonger, Thomas John, Pierre Yves Gaudreau Lamarre, Jack E. Graver, Duane Graysay, Pawel Grzegrzolka, Lee Kennard,
Hyune-Ju Kim, Leonid Kovalev, Graham J. Leuschke, Wei Li, Jianxuan Liu, Adam Lutoborski, Rachana Maharian, Joanna O. Masingila, Moira McDermott, Jeffrey Meyer, Claudia Miller, Jani Onninen,
Josh Pollitz, Declan Quinn, Lixin Shen, Gregory Verchota, Stephan Wehrli, William Wylie, Yuan Yuan, Yiming Zhao
B.S. in Applied Mathematics is recommended for those students who intend to pursue a career in a field that requires advanced mathematical training, often beyond an undergraduate degree.
Students who plan to pursue graduate study in applied mathematics should obtain the B.S. degree and consider taking at least one first-year graduate (600-level) course.
For all Arts and Sciences|Maxwell students, successful completion of a bachelor’s degree in this major requires a minimum of 120 credits, 96 of which must be Arts and Sciences|Maxwell
credits, completion of the Liberal Arts Core requirements, and the requirements for this major that are listed below.
Dual Enrollments:
Students dually enrolled in Newhouse* and Arts and Sciences|Maxwell will complete a minimum of 122 credits, with at least 90 credits in Arts and Sciences|Maxwell coursework and an Arts and
Sciences|Maxwell major.
*Students dually enrolled in the College of Arts and Sciences|Maxwell as first year students must complete the Liberal Arts Core. Students who transfer to the dual program after their first
year as singly enrolled students in the Newhouse School will satisfy general requirements for the dual degree program by completing the Newhouse Core Requirements.
Student Learning Outcomes
1. Demonstrate facility with the techniques of single and multivariable calculus and linear algebra
2. Effectively communicate mathematical ideas orally and in writing
3. Make symbolic calculations by hand and numerical calculations using MATLAB (or other appropriate software)
4. Reproduce essential assumptions, definitions, examples, and statements of important theorems
5. Solve standard science and engineering problems by selecting and applying and appropriate mathematical model
6. Solve problems using advanced undergraduate methods from each of the following areas: differential and linear equations, analysis, probability and statistics
Preliminary and Extra-disciplinary Requirements
Preliminary and extra-disciplinary requirements for both B.A and B.S. degrees
1. Complete 18 credits in the following classes with no grade below a C: MAT 295 , MAT 296 , MAT 331 , MAT 397 , and MAT 375 or CIS 375 . These courses are prerequisites for most
upper-division courses. The following sequence is recommended: MAT 295 in the first semester; MAT 296 in the second semester; MAT 331 , MAT 397 in the third semester; and MAT 375 /CIS
375 when appropriate. However, a student with knowledge of trigonometry and a year of high school calculus may be able to enter the sequence at MAT 296 or even MAT 397 ; a student with
less preparation may be advised to complete MAT 194 before beginning the calculus sequence. Students considering becoming mathematics majors are strongly encouraged to talk to a
mathematics major advisor as soon as possible to ensure appropriate selection of courses.
2. Complete a course in computing such as CPS 196 , ECS 102 or ECS 104 . A similar course may be substituted with advisor’s approval.
3. Complete two sequences of two approved science courses for a total of four different courses outside of the mathematics department, such as BIO 121 - BIO 123 , CHE 106 (CHE 107 ) - CHE
116 (CHE 117 ), PHY 211 (PHY 221 ) - PHY 212 (PHY 222 ), ECN 101 - ECN 102 , or ECN 101 - ECN 203 , ECS 221 - ECS 222 , ELE 231 , or another, more advanced sequence with the approval
of a mathematics major advisor. This requirement is waived if the student earns either (i) a minor in Applied Statistics, Biology, Chemistry, Computer Science, Economics, Engineering
disciplines, Information Management & Technology, or Physics, or (ii) a major in one of natural sciences, engineering/technology disciplines, economics or finance.
B.S. Degree Requirements
Students interested in pursuing the B.S. degree in Applied Mathematics obtain, in advance, the approval of a mathematics major advisor and the department chair of a petition to the effect
that the upper-division courses to be taken satisfy the requirement for a B.S. degree.
In addition to the preliminary and extra-disciplinary requirements described above, the student must earn credit for the following courses, with a grade average of at least 2.0.
Table 1: sequences for B.S. in Applied Mathematics*
*First course in each of the sequences in Table 1. Second course in one of the sequences in Table 1.
Linear Transforms Sequence:
Second course:
Analysis Sequence:
Probability and Statistics Sequence:
9 credits of elective mathematics courses numbered 490 or higher, except MAT 503. With prior approval of the student’s major advisor, mathematically rich 500+ level courses in other
departments may be substituted for the mathematics electives.
Additional Information
Distinction in Applied Mathematics is awarded by the Mathematics Department upon completion of a B.S. in Applied Mathematics with a minimum cumulative GPA of 3.4, a minimum GPA of 3.6 in
mathematics (MAT) courses at the 300+ level, and either an A or A- in the Senior Seminar or a high-quality Capstone Thesis. See the Mathematics Department undergraduate advisor for additional | {"url":"https://courses.syracuse.edu/preview_program.php?catoid=38&poid=19009&returnto=4779","timestamp":"2024-11-14T23:56:32Z","content_type":"text/html","content_length":"68371","record_id":"<urn:uuid:905abc01-8ee1-4778-8faf-bb650b87904c>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00347.warc.gz"} |
Calories (15°C,CIPM, 1950) to Foot-poundals
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
The International Committee for Weights and Measures have defined one cal15 is the ammount of heat energy needed to raise the temperature of 1g of water from 14.5°C to 15.5°C
A foot poundal is the work done by a force of one poundal acting through a distance of one foot, in the direction of the force. It equates to 0.0421401100938048 J
Calories (15°C,CIPM, 1950) Foot-poundals
0cal15 0.00ft-pd
1cal15 99.32ft-pd
2cal15 198.65ft-pd
3cal15 297.97ft-pd
4cal15 397.29ft-pd
5cal15 496.62ft-pd
6cal15 595.94ft-pd
7cal15 695.26ft-pd
8cal15 794.59ft-pd
9cal15 893.91ft-pd
10cal15 993.23ft-pd
11cal15 1092.56ft-pd
12cal15 1191.88ft-pd
13cal15 1291.20ft-pd
14cal15 1390.53ft-pd
15cal15 1489.85ft-pd
16cal15 1589.17ft-pd
17cal15 1688.50ft-pd
18cal15 1787.82ft-pd
19cal15 1887.15ft-pd
Calories (15°C,CIPM, 1950) Foot-poundals
20cal15 1986.47ft-pd
21cal15 2085.79ft-pd
22cal15 2185.12ft-pd
23cal15 2284.44ft-pd
24cal15 2383.76ft-pd
25cal15 2483.09ft-pd
26cal15 2582.41ft-pd
27cal15 2681.73ft-pd
28cal15 2781.06ft-pd
29cal15 2880.38ft-pd
30cal15 2979.70ft-pd
31cal15 3079.03ft-pd
32cal15 3178.35ft-pd
33cal15 3277.67ft-pd
34cal15 3377.00ft-pd
35cal15 3476.32ft-pd
36cal15 3575.64ft-pd
37cal15 3674.97ft-pd
38cal15 3774.29ft-pd
39cal15 3873.61ft-pd
Calories (15°C,CIPM, 1950) Foot-poundals
40cal15 3972.94ft-pd
41cal15 4072.26ft-pd
42cal15 4171.58ft-pd
43cal15 4270.91ft-pd
44cal15 4370.23ft-pd
45cal15 4469.55ft-pd
46cal15 4568.88ft-pd
47cal15 4668.20ft-pd
48cal15 4767.52ft-pd
49cal15 4866.85ft-pd
50cal15 4966.17ft-pd
51cal15 5065.49ft-pd
52cal15 5164.82ft-pd
53cal15 5264.14ft-pd
54cal15 5363.46ft-pd
55cal15 5462.79ft-pd
56cal15 5562.11ft-pd
57cal15 5661.44ft-pd
58cal15 5760.76ft-pd
59cal15 5860.08ft-pd | {"url":"https://www.metric-conversions.org/energy-and-power/calories-15-degc-cipm-1950-to-foot-poundals.htm","timestamp":"2024-11-02T15:41:44Z","content_type":"text/html","content_length":"56974","record_id":"<urn:uuid:f9b423af-4095-4226-8966-26847fb01630>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00443.warc.gz"} |
The Group of Differentiable Real-Valued Functions
The Group of Differentiable Real-Valued Functions
Recall from the Groups page that a group is a set $G$ paired with a binary operation $\cdot : G \times G \to G$ where:
• 1) For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ (Associativity of $\cdot$).
• 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).
• 3) For all $a \in G$ there exists a $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (The existence of inverses for each element in $G$).
We will now look at the group of differentiable real-valued functions on a fixed interval $[a, b]$.
Let $C^1 [a, b]$ denote the set of differentiable real-valued functions on the interval $[a, b]$ and let $+$ be the operation of function addition defined for $f, g \in C^1 [a, b]$ by:
\quad (f + g)(x) = f(x) + g(x)
Let $f, g, h \in C^1 [a, b]$. We know by Calculus that the sum of two differentiable functions is a differentiable function, so $(f + g) \in C^1 [a, b]$ and hence $C^1 [a, b]$ is closed under $+$.
Furthermore, since $f$, $g$, and $h$ are real-valued functions, then $f(x), g(x), h(x) \in \mathbb{R}$ for all $x \in [a, b]$, and from the associativity of real numbers, we have that:
\quad (f + [g + h])(x) = f(x) + [g(x) + h(x)] = [f(x) + g(x)] + h(x) = ([f + g] + h)(x)
Therefore $+$ is associative.
The identity element is the differentiable function $z(x) = 0$ (which of course is differentiable for any interval $[a, b]$).
For each $f \in C^1 [a, b]$ we know from calculus that for any $k \in \mathbb{R}$ that $kf$ is differentiable on $[a, b]$, that is, $kf \in C^1 [a, b]$. For $k = -1$ we have that $-f \in C^1 [a, b]$
and this function is the inverse of $f$ with respect to $+$ since:
\quad (f + (-f))(x) = f(x) - f(x) = 0 = z(x)
\quad ((-f) + f)(x) = -f(x) + f(x) = 0 = z(x)
Hence $(C^1 [a, b], +)$ is a group. In fact, if we define $C^n [a, b]$ to be the set of $n$-times differentiable real-valued functions on the interval $[a, b]$ then it can be shown that $(C^n[a, b],
+)$ is also a group.
Furthermore, we note that the set of continuous real-valued functions on the interval $[a, b]$ denoted $C[a, b]$ and the sets of $n$-differentiable real-valued functions on $[a, b]$, $C^n [a, b]$,
are such that:
\quad C[a, b] \supset C^1 [a, b] \supset C^2 [a, b] ... \supset C^n[a, b] \supset ...
Hence we can say that $(C^n [a, b], +)$ is a subgroup of $(C[a, b], +), (C^1[a, b], +), ..., (C^{n-1} [a, b], +)$. | {"url":"http://mathonline.wikidot.com/the-group-of-differentiable-real-valued-functions","timestamp":"2024-11-04T22:13:36Z","content_type":"application/xhtml+xml","content_length":"17904","record_id":"<urn:uuid:4dbc32a7-450a-4ae6-86ef-3cc7f766e186>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00128.warc.gz"} |
Probabilistic Data Structures And Algorithms In Big Data | HackerNoon
Shocker: Big Data derives its name not just from the size.
The datasets of Big Data are larger, more complex information bits fetched from new data sources. These massive volumes of data can be used to address business problems more intelligently. However,
traditional data processing algorithms fall flat to handle this magnitude.
Deterministic data structures like HashSet do the trick with smaller amounts of data. But when we have to deal with something like streaming applications, those structures cannot process everything
in one pass and assist in incremental updates.
That is why we need more space-efficient and fast algorithms. Thus, probabilistic data structures are a great fit for modern Big Data applications.
With that said, let’s have a look at probabilistic data structures and algorithms as well as their common use.
Deterministic Vs Probabilistic Data Structure
Deterministic data structures are common for a techie. Thus, we often bump into Array, List, HashTable, HashSet, etc. The latter is suitable for a wide variety of operations including insert, find,
and delete (provided you have specific key values). As a result of such operations, we get deterministic or accurate results.
However, probabilistic data structures work according to their name. Probabilistic data structures cannot give you a definite answer, instead, they give you a reasonable approximation of the answer
and a way to approximate that estimate.
How do they work?
These data structures are a great fit for a large data set. The most prominent examples of operations may include identifying some unique or frequent items. To complete the operation, probabilistic
data structures use hash functions to randomize items.
Because they ignore collisions, they keep the size constant. Yet, this is also the reason why they cannot give you exact values. The higher the number of hash functions is, the more accurate
determination you get.
The main use cases of probabilistic data structures include:
1. Huge datasets.
2. Statistical analysis.
3. Mining tera-bytes of data sets, and others.
Examples of probabilistic data structures are as follows:
1. Membership query (Bloom filter, Bloom count filter, private filter, cuckoo filter).
2. Power (linear counting, probabilistic counting, LogLog, HyperLogLogLog, HyperLogLog++).
3. Frequency (Counting sketch, Counting-minimal sketch).
4. Similarity (LSH, MinHash, SimHash), and others.
Let’s have a look at the most widely used data structures within this realm.
Bloom Filter
The Bloom filter is an implementation of a probability set, invented by Burton Bloom in 1970. This approximate member data query structure allows you to compactly store elements and check if a given
element belongs to the set.
In this case, you can get a false positive (the element is not in the set, but the data structure says it is), but not a false negative. Bloom's filter can use any memory size, predefined by the
user. And the larger it is, the lower the probability of a false positive.
The operation of adding new elements to the set is supported. However, you can’t delete the existing ones.
Bloom filter allows you to perform three kinds of operations:
• add an item to the set
• check whether the element belongs to the set
• check whether the element does not belong to the set
When the structure flags the element as Found/Present, there is a small chance that it’s lying. But if we’re talking about the Not Found/Not Present category, the Bloom filter boasts 100% accuracy
plus space-saving perks.
Its hype can be attributed to the fact that Bloom filters have this powerful combo of simplicity and multi-purpose nature. In layman’s terms, they support operations similar to the hash tables but
use less space.
Apache Cassandra, for example, benefits from these structures to process massive amounts of information. This storage system taps into Bloom filters to find out whether an SSTable has data for a
specific partition.
HyperLogLog is a beautiful, yet simple algorithm for handling cardinality estimation. It excels when dealing with sets of data with a huge number of values.
Let’s say, you have a massive dataset of elements with duplicate entries. The latter is taken from a set of cardinality n and you are required to find n, which is the number of unique components in
the set.
This can be helpful when identifying the amount of Google searches performed by end-users in a day. If you try to squeeze all the data into the memory, you’ll need storage proportionate to the number
of Google searches per day.
Thereby, the HyperLogLog data structure turns the data into a hash of random numbers that represent the data's cardinality, allowing it to solve the problem with as little as 1.5kB of RAM.g
HyperLogLog operates by predicting an approximate count of distinct elements through a function known as APPROX_DISTINCT.
• Insert: add an element to the data structure
• Merge: generate a structure which is a combination of two structures
• Cardinality: calculate the total number of elements inserted
Count-min Sketch
The counting-minimal sketch is another efficient algorithm for counting streaming data. This data structure boils down to keeping track of the count of things. Therefore, by performing this
algorithm, you can find out how many times an element is met in the set. You can also easily test if a given member has been observed before.
Just like with Bloom Filters, Count-min sketch saves a lot of space by using probabilistic techniques. To implement a counting mechanism, you need to use a hash function.
Overall, the count-min structure works great whenever you’re looking for just approximate counts of the most essential elements.
• Insert: add an element to the data structure
• Merge: generate a structure which is a combination of two structures
• Cardinality: estimate the number of times a specific estimate was inserted
As for the prominent applications, AT&T leverages the structure in network switches to analyze traffic in memory-constrained environments. The structure is also implemented as part of Twitter's
Algebird library.
To Sum It Up
Working with Big Data is a challenge itself, let alone digging up answers from it. In this case, the best you can count on is an approximate answer. Probabilistic data structures allow you to conquer
the beast and give you an estimated view of some data characteristics. You lose the accuracy of results, but get an enormous amount of storage space in exchange for that. | {"url":"https://hackernoon.com/probabilistic-data-structures-and-algorithms-in-big-data","timestamp":"2024-11-07T17:29:17Z","content_type":"text/html","content_length":"233186","record_id":"<urn:uuid:7f233140-b20c-425e-9132-a8a8df0281a6>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00734.warc.gz"} |
Exploring scikit-learn
For my current NEAT project, I read the paper A Comparative Study of Supervised Machine Learning Algorithms for Stock Market Trend Prediction to learn how others approached stock market prediction
using supervised machine learning algorithms. This paper compared the accuracy between Support Vector Machine, Random Forest, K-Nearest Neighbor, Naive Bayes, and SoftMax algorithms. After reading
about these different algorithms, I took it upon myself to learn more about the scikit-learn python library. I am learning how to use scikit-learn because it will give me the ability to implement
some the algorithms outlined in the paper. The algorithms I have covered so far include:
• Random Forest
• K-Nearest Neighbors
• Support Vector Machine
Random Forest
The first algorithm I sought to better understand was the Random Forest algorithm. This algorithm is made up of multiple decision trees. A decision tree learns simple decision rules from the data and
produces a output. Each tree is used on a different sub sample of the dataset and averaging is used to improve accuracy. The decision trees all “vote” by producing outputs, and whatever the most
popular output is becomes the algorithms output.
I first imported my data as a Pandas data frame and split it up into a training section and a testing section. I then created a Random Forest algorithm using the scikit-learn random forest
classifier. After running the algorithm a few times, I noticed that the accuracy was unusually high. After some investigation, I discovered that the algorithms were predicting the stock price
movement for the current day instead of one day in the future. After fixing the problem, the accuracy on the testing dataset stayed around 50%. Here is the code that I wrote:
1 import csv
2 import pandas as pd
3 from sklearn.model_selection import train_test_split
4 from sklearn.ensemble import RandomForestClassifier
5 from sklearn import metrics
6 from sklearn.metrics import classification_report
8 #this opens the file and puts the data in a pandas dataframe
9 data = pd.read_csv('RandomForest.csv')
10 data.head()
12 #setting the size of the training dataset
13 training_size = .7
14 split = int(len(data)*.7)
16 #Identifying all of the inputs and the expected output
17 X=data[['Open','High','Low','Close','Volume','Accumulation Distribution Line','MACD','Chaikan Oscillator (CHO)','Highest closing price (5 days)',
18 'Lowest closing price (days)','Stochastic %K (5 days)','%D','Volume Price Trend (VPT)','Williams %R (14 days)','Relative Strength Index','Momentum (10 days)',
19 'Price rate of change (PROC)','Volume rate of change (VROC)','On Balance Volume (OBV)']]
20 y=data['Outputs']
22 #splitting the dataframe into a training and testing dataset
23 X_train = X[:split]
24 X_test = X[split:]
25 y_train = y[:split]
26 y_test = y[split:]
28 #create a the random forest with 500 decision trees
29 clf=RandomForestClassifier(n_estimators=500)
31 #train the model on the training data
32 clf.fit(X_train,y_train)
34 #making predictions on the testing data
35 y_pred=clf.predict(X_test)
37 #printing the testing accuracy
38 print("Accuracy:",metrics.accuracy_score(y_test, y_pred))
K-Nearest Neighbors
The next algorithm I studied was the K-Nearest Neighbors algorithm. The nearest neighbor method looks for the most similar trading days. The predefined number of the most similar trading days become
the nearest neighbors. Just like the decision trees in the Random Forest algorithm, the nearest neighbors “vote”. Instead of using decision trees to produce an output, the K-Nearest Neighbor
algorithm uses the actual expected output for the most similar trading days. The most popular output becomes of the neighbors becomes the algorithms output. If 5 of the nearest neighbors had the
closing stock price go up and 3 neighbors had the stock price go down, the prediction will be that the stock price goes up.
While studying this algorithm, I learned that K-Nearest Neighbor algorithms performs better with a lower number of features/inputs. To reduce the size of my dataset, I first normalized it with the
standard scaler tool from scikit-learn. I then used PCA to shrink the number of inputs in my dataset down to two while preserving the variance in the dataset. Here is my K-Nearest Neighbors code:
1 import csv
2 import pandas as pd
3 from sklearn import metrics
4 from sklearn.metrics import classification_report
5 from sklearn.neighbors import KNeighborsClassifier
6 from sklearn.decomposition import PCA
7 from sklearn.preprocessing import StandardScaler
9 #this opens the CSV File
10 data = pd.read_csv('RandomForest.csv')
11 data.head()
13 #preprocessing and normalizing
14 Features = ['Open','High','Low','Close','Volume','Accumulation Distribution Line','MACD','Chaikan Oscillator (CHO)','Highest closing price (5 days)',
15 'Lowest closing price (days)','Stochastic %K (5 days)','%D','Volume Price Trend (VPT)','Williams %R (14 days)','Relative Strength Index','Momentum (10 days)',
16 'Price rate of change (PROC)','Volume rate of change (VROC)','On Balance Volume (OBV)']
18 X = data.loc[:, Features].values
19 Y = data.loc[:,['Outputs']].values
21 #normalizing the dataset
22 X = StandardScaler().fit_transform(X)
24 #performing PCA
25 PCA = PCA(n_components=2)
26 Components = PCA.fit_transform(X)
27 ComponentDf = pd.DataFrame(data=Components)
29 #setting the size of the training set
30 training_size = .8
31 split = int(len(data)*training_size)
33 #splitting up the dataset
34 X_train = pd.DataFrame(data=ComponentDf[:split])
35 X_test = pd.DataFrame(data=ComponentDf[split:])
36 Y_train = pd.DataFrame(data=Y[:split])
37 Y_test = pd.DataFrame(data=Y[split:])
39 #create the K Nearest Neighbor Algorithm and running the algorithm
40 clf=KNeighborsClassifier(n_neighbors=5)
41 clf.fit(X_train, Y_train.values.ravel())
42 Y_pred=clf.predict(X_test)
44 #printing the testing accuracy
45 print("Accuracy:",metrics.accuracy_score(Y_test, Y_pred))
After performing principal component analysis on my dataset, I graphed the principal components to better understand the data. After creating the graph, it became clear why the accuracy of my
K-Nearest Neighbor algorithm stayed around 50%. Both expected outputs (Stock price going up or down) are clustered together.
Support Vector Machine
A Support Vector Machine is a machine learning tool that uses a hyperplane to define data points. Instead of looking at the nearest neighbors like a KNN, I like to think that Support Vector Machines
split up data points into different neighborhoods. For example, if we use the two components from my PCA analysis, a SVM will create a one dimensional hyperplane splitting up the data into two
“neighborhoods”. The hyperplane will create decision boundaries that maximize the margins from both expected outputs. The graph below shows the data points and the decision boundaries created by my
Any data point that falls into the blue background gets categorized as a 0 (closing stock price goes down). To create this graph, I used a Support Vector Machine with a RBF kernel. Kernel functions
allow SVMs to create hyperplanes in high dimensional data without having to calculate the coordinates of the data in that space. Here is my Support Vector Machine Code:
1 #support vector machine
3 import csv
4 import pandas as pd
5 from sklearn import metrics
6 from sklearn.metrics import classification_report
7 from sklearn import svm
8 from sklearn.decomposition import PCA
9 from sklearn.preprocessing import StandardScaler
10 import matplotlib.pyplot as plt
11 import numpy as np
13 #this opens the CSV File
14 data = pd.read_csv('RandomForest.csv')
15 data.head()
17 #setting the size of the training dataset
18 training_size = .7
19 split = int(len(data)*.7)
21 #Identifying all of the inputs and the expected output
22 X=data[['Open','High','Low','Close','Volume','Accumulation Distribution Line','MACD','Chaikan Oscillator (CHO)','Highest closing price (5 days)',
23 'Lowest closing price (days)','Stochastic %K (5 days)','%D','Volume Price Trend (VPT)','Williams %R (14 days)','Relative Strength Index','Momentum (10 days)',
24 'Price rate of change (PROC)','Volume rate of change (VROC)','On Balance Volume (OBV)']]
25 y=data['Outputs']
27 #splitting the dataframe into a training and testing dataset
28 X_train = X[:split]
29 X_test = X[split:]
30 y_train = y[:split]
31 y_test = y[split:]
33 #create the support vector machine
34 svc = svm.SVC(kernel='rbf', C=1.0)
35 svc.fit(X_train, y_train)
36 y_pred = svc.predict(X_test)
38 #printing the testing accuracy
39 print("Accuracy:",metrics.accuracy_score(y_test, y_pred))
41 ##################################################
42 #The next section transforms the dataset#
43 #And graphs the components with decision boundries
45 #Normalizing the dataset
46 X = StandardScaler().fit_transform(X)
48 #performing PCA
49 PCA = PCA(n_components=2)
50 Components = PCA.fit_transform(X)
51 ComponentDf = pd.DataFrame(data=Components,columns = ['principal component 1', 'principal component 2'])
53 #transforming the dataset
54 X = ComponentDf.to_numpy()
55 y = y.ravel()
57 h = 0.2
59 # create a mesh to plot in
60 x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
61 y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
62 xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
63 np.arange(y_min, y_max, h))
65 #create the support vector machine
66 svc = svm.SVC(kernel='rbf', C=1.0).fit(X, y)
68 #graphing the data
69 Z = svc.predict(np.c_[xx.ravel(), yy.ravel()])
71 # Put the result into a color plot
72 Z = Z.reshape(xx.shape)
73 plt.contourf(xx, yy, Z, cmap=plt.cm.coolwarm, alpha=0.8)
75 # Creating the plot
76 plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.coolwarm)
77 plt.xlabel('Principal Component 1')
78 plt.ylabel('Principal Component 2')
79 plt.xlim(xx.min(), xx.max())
80 plt.ylim(yy.min(), yy.max())
81 plt.xticks(())
82 plt.yticks(())
83 plt.title('Support Vector Machine Graph')
84 plt.legend()
86 # Saving the Plot
87 plt.savefig("matplotlib.png")
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MyOpenMath Assessment
Try Another Version of This Question
A population grows according to an exponential growth model. The initial population is `P_0 = 17`, and the growth rate is `r = 0.35`.
`P_1` =
`P_2` =
Find an explicit formula for `P_n`. Your formula should involve `n`.
`P_n` =
Use your formula to find `P_9`
`P_9` =
Give all answers accurate to at least one decimal place
Box 1: Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172
Enter DNE for Does Not Exist, oo for Infinity
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Enter DNE for Does Not Exist, oo for Infinity
Box 3: Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
Be sure your variables match those in the question
Box 4: Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172
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Delta Math Algebra 2 Answer Key: Your Ultimate GuideDelta Math Algebra 2 Answer Key: Your Ultimate Guide
Delta Math Algebra 2 Answer Key: Your Ultimate Guide
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3.md.7b Worksheets | Free - Distance Learning, worksheets and more: CommonCoreSheets
Previous Standard
Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths.
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Math 2001: Introduction to Discrete Math (and Proofs)
General Course Notes
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Mathematics for Engineers, 3rd edition
Mathematics for Engineers, 3rd edition
A Modern Interactive Approach
By Anthony Croft and Robert Davison
September 2008
Pearson / Prentice Hall
Distributed By Trans-Atlantic Publications
ISBN: 9780132051569
1173 pages, Illustrated
$115.00 Paper Original
Understanding key mathematical concepts and applying them successfully to solve problems are vital skills that all engineering students must acquire. Mathematics for Engineers teaches, develops
and nurtures these skills. Practical, informal, and accessible, it beings with the foundations and gradually builds upon this knowledge as it introduces more complex concepts until you have
learned everything you will need for your first year engineering maths course, together with introductory material for even more advanced topics.
Key Features:
--Hundreds of Interactive examples are provided in the context of various engineering disciplines, so you are able to both engage with the problems and also see the relevance of the maths to your
wider studies
--A wealth of practice and revision exercises with solutions help test your knowledge
--Key points and important results are highlighted throughout
--Computer and calculator examples and exercises are incorporated in relevant sections
Specimen examination papers give further opportunity to practise for high stake assessments
--A foundation section gives you a firm base in arithmetic, the building block of many high level mathematical topics
--An accompanying website is available, which contains supporting resources for both students and teachers.
Guided Tour
Publisher's acknowledgements
Using mathematical software packages
1. Arithmetic
2. Fractions
3. Decimal numbers
4. Percentage and ratio
5. Basic algebra
6. Functions
7. Polynomial equations, inequalitites, partial fractions and proportionality
8. Logartihms and exponentials
9. Trigonometry
10. Further trigonometry
11. Complex numbers
12. Matrices and determinants
13. Using matrices and determinants to solve equations
14. Vectors
15. Differentiation
16. Techniques and applications of differentiation
17. Integration
18. Applications of integration
19. Sequences and series
20. Differential equations
21. Functions of more than one variable and partial differentiation
22. The Laplace transform
23. Statistics and probability
24. An introduction to Fourier series and the Fourier transform
Typical examination papers
Appendix: SI units and prefixes
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4,851 research outputs found
An efficient all-electron G$^0$W$^0$ method and a quasiparticle selfconsistent GW (QSGW) method for molecules are proposed in the molecular orbital space with the full random phase approximation. The
convergence with basis set is examined. As an application, the ionization energy ($I$) and electron affinity ($A$) of a series of conjugated molecules (up to 32 atoms) are calculated and compared to
experiment. The QSGW result improves the G$^0$W$^0$ result and both of them are in significantly better agreement with experimental data than those from Hartree-Fock (HF) and hybrid density
functional calculations, especially for $A$. The nearly correct energy gap and suppressed self-interaction error by the HF exchange make our method a good candidate for investigating electronic and
transport properties of molecular systems.Comment: 4 pages, 2 figures, 1 tabl
The accurate but expensive product of geminals ansatz may be approximated by a geminal power, but this approach sacrifices size consistency. Here we show both analytically and numerically that a size
consistent form very similar to the product of geminals can be recovered using a network of location specific Jastrow factors. Upon variational energy minimization, the network creates particle
number projections that remove the charge fluctuations responsible for size inconsistency. This polynomial cost approach captures strong many-electron correlations, giving a maximum error of just 1.8
kcal/mol during the double-bond dissociation of H2O in an STO-3G atomic orbital basis.Comment: Updated the original arXiv submission to include improvements resulting from journal peer review. 5
pages, 4 figures, 1 tabl
In a paper recently published in Phys. Rev. A [arXiv:1010.4223], Schirmer has criticized an earlier work of mine [arXiv:0803.2727], as well as the foundations of time-dependent density functional
theory. In Ref.[2], I showed that the so-called "causality paradox" - i.e., the failure of the exchange-correlation potential derived from the Runge-Gross time-dependent variational principle to
satisfy causality requirements - can be solved by a careful reformulation of that variational principle. Fortunately, the criticism presented in Ref.[1] is based on elementary misunderstandings of
the nature of functionals, gauge transformations, and the time-dependent variational principle. In this Comment I wish to point out and clear these misunderstandings.Comment: 4 pages. Accepted for
publication in Phys. Rev.
Following a recent work [Gal, Phys. Rev. A 64, 062503 (2001)], a simple derivation of the density-functional correction of the Hartree-Fock equations, the Hartree-Fock-Kohn-Sham equations, is
presented, completing an integrated view of quantum mechanical theories, in which the Kohn-Sham equations, the Hartree-Fock-Kohn-Sham equations and the ground-state Schrodinger equation formally stem
from a common ground: density-functional theory, through its Euler equation for the ground-state density. Along similar lines, the Kohn-Sham formulation of the Hartree-Fock approach is also
considered. Further, it is pointed out that the exchange energy of density-functional theory built from the Kohn-Sham orbitals can be given by degree-two homogeneous N-particle density functionals (N
=1,2,...), forming a sequence of degree-two homogeneous exchange-energy density functionals, the first element of which is minus the classical Coulomb-repulsion energy functional.Comment: 19 pages;
original manuscript from 2001 (v1) revised for publication, with presentation substantially improved, some errors corrected, plus an additional summarizing figure (Appendix B) include
A systematic and detailed investigation of the finite-temperature behavior of small sodium clusters, Na_n, in the size range of n= 8 to 50 are carried out. The simulations are performed using
density-functional molecular-dynamics with ultrasoft pseudopotentials. A number of thermodynamic indicators such as specific heat, caloric curve, root-mean-square bond length fluctuation, deviation
energy, etc. are calculated for each of the clusters. Size dependence of these indicators reveals several interesting features. The smallest clusters with n= 8 and 10, do not show any signature of
melting transition. With the increase in size, broad peak in the specific heat is developed, which alternately for larger clusters evolves into a sharper one, indicating a solidlike to liquidlike
transition. The melting temperatures show irregular pattern similar to experimentally observed one for larger clusters [ M. Schmidt et al., Nature (London) 393, 238 (1998) ]. The present calculations
also reveal a remarkable size-sensitive effect in the size range of n= 40 to 55. While Na_40 and Na_55 show well developed peaks in the specific heat curve, Na_50 cluster exhibits a rather broad
peak, indicating a poorly-defined melting transition. Such a feature has been experimentally observed for gallium and aluminum clusters [ G. A. Breaux et al., J. Am. Chem. Soc. 126, 8628 (2004); G.
A.Breaux et al., Phys. Rev. Lett. 94, 173401 (2005) ].Comment: 8 pages, 11 figure
A simple and completely general representation of the exact exchange-correlation functional of density-functional theory is derived from the universal Lieb-Oxford bound, which holds for any
Coulomb-interacting system. This representation leads to an alternative point of view on popular hybrid functionals, providing a rationale for why they work and how they can be constructed. A similar
representation of the exact correlation functional allows to construct fully non-empirical hyper-generalized-gradient approximations (HGGAs), radically departing from established paradigms of
functional construction. Numerical tests of these HGGAs for atomic and molecular correlation energies and molecular atomization energies show that even simple HGGAs match or outperform
state-of-the-art correlation functionals currently used in solid-state physics and quantum chemistry.Comment: v2: Major revison. Added information on relation to the gradient expansion and to local
hybrids, improved discussion of size consistency and of performance relative to other functional
Electrons in one-dimension display the unusual property of separating their spin and charge into two independent entities: The first, which derive from uncharged spin-1/2 electrons, can travel at
different velocities when compared with the second, built from charged spinless electrons. Predicted theoretically in the early sixties, the spin-charge separation has attracted renewed attention
since the first evidences of experimental observation, with usual mentions as a possible explanation for high-temperature superconductivity. In one-dimensional (1D) model systems, the spin-charge
separation leads the frequencies of Friedel oscillations to suffer a 2k_F -- 4k_F crossover, mainly when dealing with strong correlations, where they are referred to as Wigner crystal oscillations.
In non-magnetized systems, the current density functionals which are applied to the 1D Hubbard model are not seen to reproduce this crossover, referring to a more fundamental question: Are the Wigner
crystal oscillations in 1D systems non-interacting v-representable? Or, is there a spin-independent Kohn-Sham potential which is able to yield spin-charge separation? Finding an appropriate answer to
both questions is our main task here. By means of exact and DMRG solutions, as well as, a new approach of exchange-correlation potential, we show the answer to be positive. Specifically, the
v-representable 4k_F oscillations emerge from attractive interactions mediated by positively charged spinless holes -- the holons -- as an additional contribution to the repulsive on-site Hubbard
We demonstrate the existence of different density-density functionals designed to retain selected properties of the many-body ground state in a non-interacting solution starting from the standard
density functional theory ground state. We focus on diffusion quantum Monte Carlo applications that require trial wave functions with optimal Fermion nodes. The theory is extensible and can be used
to understand current practices in several electronic structure methods within a generalized density functional framework. The theory justifies and stimulates the search of optimal empirical density
functionals and effective potentials for accurate calculations of the properties of real materials, but also cautions on the limits of their applicability. The concepts are tested and validated with
a near-analytic model.Comment: five figure
The random phase approximation (RPA) for the correlation energy functional of density functional theory has recently attracted renewed interest. Formulated in terms of the Kohn-Sham (KS) orbitals and
eigenvalues, it promises to resolve some of the fundamental limitations of the local density and generalized gradient approximations, as for instance their inability to account for dispersion forces.
First results for atoms, however, indicate that the RPA overestimates correlation effects as much as the orbital-dependent functional obtained by a second order perturbation expansion on the basis of
the KS Hamiltonian. In this contribution, three simple extensions of the RPA are examined, (a) its augmentation by an LDA for short-range correlation, (b) its combination with the second order
exchange term, and (c) its combination with a partial resummation of the perturbation series including the second order exchange. It is found that the ground state and correlation energies as well as
the ionization potentials resulting from the extensions (a) and (c) for closed sub-shell atoms are clearly superior to those obtained with the unmodified RPA. Quite some effort is made to ensure
highly converged RPA data, so that the results may serve as benchmark data. The numerical techniques developed in this context, in particular for the inherent frequency integration, should also be
useful for applications of RPA-type functionals to more complex systems.Comment: 11 pages, 7 figure
Density-functional theory requires ever better exchange-correlation (xc) functionals for the ever more precise description of many-body effects on electronic structure. Universal constraints on the
xc energy are important ingredients in the construction of improved functionals. Here we investigate one such universal property of xc functionals: the Lieb-Oxford lower bound on the
exchange-correlation energy, $E_{xc}[n] \ge -C \int d^3r n^{4/3}$, where $C\leq C_{LO}=1.68$. To this end, we perform a survey of available exact or near-exact data on xc energies of atoms, ions,
molecules, solids, and some model Hamiltonians (the electron liquid, Hooke's atom and the Hubbard model). All physically realistic density distributions investigated are consistent with the tighter
limit $C \leq 1$. For large classes of systems one can obtain class-specific (but not fully universal) similar bounds. The Lieb-Oxford bound with $C_{LO}=1.68$ is a key ingredient in the construction
of modern xc functionals, and a substantial change in the prefactor $C$ will have consequences for the performance of these functionals.Comment: 10 pages, 3 figure | {"url":"https://core.ac.uk/search/?q=author%3A(Parr%20R.%20G.)","timestamp":"2024-11-12T16:09:58Z","content_type":"text/html","content_length":"150151","record_id":"<urn:uuid:2b893f0a-55e4-45e6-b5be-e0053a449fc7>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00333.warc.gz"} |
Student's T-Test: Don't Ignore These Secrets
Student’s T-Test: Don’t Ignore These Secrets
The Students t-test is a statistical analysis that compares the means of two groups to determine their equalities, assuming normality and equal variances.
Did you know that the incorrect use of statistical tests, including the Student’s t-test, is more common than one might think and can lead to drawing wrong conclusions? This is problematic because
incorrect conclusions can result in disastrous decisions — to avoid falling into this trap, read this article to the end.
The wide availability, simplicity, and ease of understanding of the Students t-test have made it one of the most commonly used statistical tests worldwide.
However, this popularization has brought some issues, such as the incorrect analysis application.
This concern is so significant that several recent systematic review papers have been published — and in particular studies, up to 80% of published articles contain errors in statistical analysis!
Since solutions to humanity’s problems often depend on these studies, it is troubling to think that such a large proportion of them have been analyzed incorrectly and may have led to erroneous
The secrets of correctly conducting a Student’s t-test are not complicated — we only need to understand some basic analysis features!
The Student’s t-test determines whether the means of two populations are equal. Therefore, carefully following the five steps listed below is essential for a proper and unquestionable application of
a t-test.
1. What type of t-test should I use?
Independent samples t-test is the most commonly used type, assessing whether the means of samples collected from two independent populations are equal. For example, we can check if the muscle mass
between men and women in a given company is similar.
Paired samples t-test determines whether the mean of a sample collected before an intervention is equal to that of the sample collected — from the same elements — after the intervention. For example,
we may want to check if older women’s blood pressure in a city is the same before and after taking a specific medication. Since we will measure each older woman’s blood pressure twice — before and
after the intervention — each sample unit is paired.
One sample t-test, the least used type, checks whether the mean of the sample collected from a population is equal to an already known mean. For example, we may wan
2. Student’s T-test sample size calculation
After defining which type of t-test to use, we need to determine how many elements to collect.
Consistent sample sizes often solve many problems in statistics.
The best way to find the appropriate sample size for our test is to use a specific calculation, with a different one for each statistical test.
For the t-test, we need the two studied populations’ approximate or estimated mean and standard deviation.
In the statistical software, we should enter these values in the sample size calculator for the appropriate t-test type. It then returns the correct number of elements to collect.
But where do we obtain the means and standard deviations of the studied populations? Usually from other pilot studies or similar studies!
Check out this article to learn how to calculate sample size!
Golden tip: G*Power is an excellent free statistical package for calculating sample size.
3. The observations must be independent!
The independence of observations assumption is one of the most important, yet often ignored, assumptions in statistics.
The Students t-test and ALL other statistical tests rely on this assumption, which can bias the analysis and generate misleading results when violated.
To avoid this violation, the observations in your dataset must be independent. Therefore, each element of the sample should not show dependence on others.
The concept of randomness is crucial here. A great way to avoid violating the assumption of independence of observations is to randomly collect your data in a universe suitable for your study.
To gain a better grasp of this idea, let’s take a look at the following example:
Suppose we want to identify the genetic profile of inhabitants in a municipality to answer specific questions. After defining the sample size, we must randomly sample individuals in the city to avoid
violating the assumption of independence of observations! Using the same example, a sampling method that would violate this assumption would be if we sampled only individuals from the same family to
represent the municipality. In this case, the observations would be dependent on each other since the genetic profile is heritable and exhibits a remarkable similarity between families.
4. Normality
The normality of residuals is an assumption applicable to all so-called parametric inferential analyses.
This assumption means that your residuals (errors) must conform to the famous normal distribution — the Gaussian distribution — to use the t-test.
The normal distribution presents a bell-shaped curve, symmetric around its midpoint, unimodal.
But how do we know if our residuals conform to the normal distribution?
The good news is that statistical analyses called normality tests identify this pattern — the main ones are Shapiro-Wilk, Anderson-Darling, and Kolmogorov-Smirnov.
Be cautious when interpreting these tests’ results to avoid misinterpreting their indications.
If the analyses indicate that the assumption of normality has been violated, there are three possibilities:
(a) Use the t-test even with the assumption of normality violated, claiming it is a robust test for these deviations and that we have an adequate sample size.
(b) Try a data transformation and then recheck for normality.
(c) Lastly, use a nonparametric analysis equivalent to the t-test.
We don’t recommend the first possibility. After all, if you plan to use the t-test regardless of the normality of residuals, why perform these analyses?
We usually choose the second option, and if the normality assumption remains violated, we proceed to the third.
Parametric tests and their nonparametric equivalents:
• Independent samples t-test equivalent is the Mann-Whitney U-test.
• Paired samples t-test equivalent is the Wilcoxon signed-rank test.
• One sample t-test equivalent is the one-sample Wilcoxon signed-rank test.
5. Homoscedasticity
We only need to proceed to this final step if the residuals have not violated the normality assumption. Therefore, we will use a Student’s t-test.
Despite its complex name, the meaning of this assumption is simple to understand.
Homoscedasticity means nothing more than similar variances.
To apply the t-test in its standard format, the variances of the two sampled populations’ residuals must be similar.
To check whether this assumption is being violated, inferential tests are widely available in statistical software!
The F-test, Bartlett’s, and Levene’s tests are among the most commonly used tests to verify the equality of variances. However, it only makes sense to apply them if the residuals have not violated
the normality assumption.
This assumption usually poses little problem. If it is being violated — the populations present unequal variances or heteroscedasticity — a minor correction in the Students t-test resolves the issue.
Most statistical software will automatically return the t-test value corrected for heteroscedasticity.
Carefully review the results of the Students t-test and homoscedasticity analyses for correct interpretation.
The Student’s t-test is one of the most widely used inferential statistical analyses worldwide.
Many studies have misapplied it, casting doubt on its results and conclusions.
You can perform an unquestionable analysis by following the five straightforward steps.
To recap, we must:
[1] choose the t-test type, [2] calculate the sample size, [3] conduct proper random sampling, [4] check the normality of the residuals, and [5] verify the equality of variances between populations.
After completing these steps, all that remains is to run the analysis!
If you still need to figure out how to interpret the p-value, check out this excellent article!
Can Standard Deviations Be Negative?
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Illustrative Mathematics
Reflections and Equilateral Triangles II
Alignments to Content Standards: G-CO.B
Suppose $ABC\,$ is an equilateral triangle. Let $D$ be the midpoint of side $\overline{AB}$ and $E$ the midpoint of side $\overline{BC}$ as pictured below:
The line through $C$ and $D$ is denoted $L$ and the line through $E$ and $A$ is denoted $M$. Let $r_L$ denote the reflection of the plane about the line $L$ and $r_M$ reflection of the plane about
the line $M$.
1. Where do the reflections $r_L$ and $r_M$ map the vertices $A,B,$ and $C$ of triangle $ABC$? Explain.
2. Where do the vertices $A,B,$ and $C$ end up if the plane is first reflected about line $M$ and then about $L$?
3. Where do the vertices $A,B,$ and $C$ end up if the plane is first reflected about line $L$ and then about $M$? Is the answer the same as (b) or different?
IM Commentary
This task examines some of the properties of reflections of the plane which preserve an equilateral triangle: these were introduced in ''Reflections and Isosceles Triangles'' and ''Reflection and
Equilateral Triangles I''. The task gives students a chance to see the impact of these reflections on an explicit object and to see that the reflections do not always commute, that is $r_L(r_M(P))$
is not the same as $r_M(r_L(P))$ unless $P$ is the center of the triangle: the first of these rotates 120 counterclockwise while the other rotates 120 degrees clockwise. The solution to this task
uses the fact, established in ''Reflections and triangles I, II,'' that the lines $L$ and $M$ are the perpendicular bisectors of segments $\overline{AB}$ and $\overline{BC}$ respectively.
In order to help visualize how the different reflections of the plane influence the vertices of a triangle, students may wish to use manipulatives: this can either take the form of paper triangle
cutouts with labelled vertices which can either be folded along the line of reflection or actually reflected. Both approaches are important: if using the latter, however, students need to be careful
to record where the vertices of the triangle were prior to the reflection. This will also be important as they continue to reflect because the three lines of reflection of this problem are all
determined by the original triangle rather than its reflected images.
The problem can be naturally extended. Suppose $F$ is the midpoint of side $\overline{AC}$ and $K$ the line joining $B$ and $F$. Let $r_K$ denote reflection about $F$. Then the students can be asked,
for example, to write $r_F$ in terms of $r_L$ and $r_M$. Geometrically, if the triangle is rotated counterclockwise by $120$ degrees about its center then line $K$ gets mapped to line $L$. So
rotating $120$ degrees counterclockwise and then reflecting about $L$ will have the same effect as reflecting about $K$. If $r_K$ denotes reflection about $K$ this means that $$ r_K(x) = r_L(r_M(r_L
(x))) $$ for every $x$ in the triangle $ABC$.
In fact, all symmetries of the triangle $ABC$, that is all rigid transformations of the plane which preserve triangle $ABC$, can be expressed in terms of the reflections $r_L$ and $r_M$ as was done
in the previous paragraph for $r_K$. This task is primarily intended for instructional purposes, providing a context for students to explore how reflections of the plane behave under composition. If
used for assessment, it is vital that students have already mastered the material from ''Reflections and Isosceles Triangles'' and ''Reflection and Equilateral Triangles I.''
1. The line $L$ passes through vertex $C$ of the triangle. Since points on $L$ are all fixed when reflecting about $L$, we have $r_L(C) = C$. Since $L$ is the perpendicular bisector of segment $\
overline{AB}$, reflection about $L$ will map $B$ to $C$ and $C$ to $B$ so $r_L(B) = A$ and $r_L(A) = B$.
2. In order to see what happens when two successive reflections are performed, we first check the impact of reflection about $M$ on the vertices. Since $A$ lies on $M$ we know that $r_M(A) = A$.
Since $M$ is the perpendicular bisector of $\overline{BC}$ we have $r_M(B) = C$ and $r_M(C) = B$. We have calculated in part (a) the impact of $r_L$ on the three vertices so we can calculate: \
begin{eqnarray} r_M(r_L(A)) &=& r_M(B) = C \\ r_M(r_L(B)) &=& r_M(A) = A\\ r_M(r_L(C)) &=& r_M(C) = B. \end{eqnarray} Sending $A$ to $C$, $C$ to $B$, and $B$ to $A$ can also be accomplished by
rotating the triangle by $120$ clockwise about its center (the point where $L$ and $M$ meet). So the net effect of reflecting about $M$ and then reflecting about $L$ is a rotation.
3. Reversing the order of the two reflections from part (b) we find \begin{eqnarray} r_L(r_M(A)) &=& r_L(A) = B \\ r_L(r_M(B)) &=& r_L(C) = C\\ r_L(r_M(C)) &=& r_L(B) = A. \end{eqnarray}
The calculations above show that performing reflection about $M$ and then reflection about $L$ does not have the same impact on the triangle $ABC$ as performing reflection about $L$ and then
reflection about $M$: the order in which the reflections are performed influences the outcome.
Note that performing $r_L$ and then $r_M$ undoes the transformation obtained by performing $r_M$ and then $r_L$. This is because performing a reflection about the same line twice moves all points
back to their original position: \begin{eqnarray} r_L(r_M(r_M(r_L(x)))) &=& r_L(r_L(x)) = x \\ r_M(r_L(r_L(r_M(x)))) &=& r_M(r_M(x)) = x \end{eqnarray} for any $x$ in the plane.
Reflections and Equilateral Triangles II
Suppose $ABC\,$ is an equilateral triangle. Let $D$ be the midpoint of side $\overline{AB}$ and $E$ the midpoint of side $\overline{BC}$ as pictured below:
The line through $C$ and $D$ is denoted $L$ and the line through $E$ and $A$ is denoted $M$. Let $r_L$ denote the reflection of the plane about the line $L$ and $r_M$ reflection of the plane about
the line $M$.
1. Where do the reflections $r_L$ and $r_M$ map the vertices $A,B,$ and $C$ of triangle $ABC$? Explain.
2. Where do the vertices $A,B,$ and $C$ end up if the plane is first reflected about line $M$ and then about $L$?
3. Where do the vertices $A,B,$ and $C$ end up if the plane is first reflected about line $L$ and then about $M$? Is the answer the same as (b) or different? | {"url":"http://tasks.illustrativemathematics.org/content-standards/HSG/CO/B/tasks/984","timestamp":"2024-11-08T04:31:24Z","content_type":"text/html","content_length":"30093","record_id":"<urn:uuid:93623501-22d1-4e7a-938a-984dfab24b9a>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00540.warc.gz"} |
011 - DMA RO Circular no. 011 - Guidelines for demonstrating the buoyancy of open fishing vessels
• Guidelines on open vessels (viking ships) of 20 June 1994, item 5.2: “... that the ves- sel, when filled with water and fully loaded, floats without considerable trim and with a freeboard of a
minimum of 50 mm.”
To document that the vessel is capable of keeping afloat with a freeboard of a minimum of 50 mm when filled with water.
Method for demonstrating buoyancy
Documentation of buoyancy can be made either through a calculation or through a practical test.
Calculations of buoyancy must be carried out using a stability program approved by the Danish Maritime Authority (DMA). Consideration must be paid to both the longitudinal and the transversal trim
caused by asymmetrical filling. When making the calculations, the same densities must be used as in connection with normal stability calculations.
The practical test must be carried out as described below in these guidelines.
Irrespective of the choice of method, the buoyancy of the vessel must be documented. As regards the practical buoyancy test, the result is to be documented through a certificate as described in
section 7. Demonstration of buoyancy through calculations is to be documented in report form.
As regards type-approved vessels, it is evident from the approval how much deadweight, including how many persons, the vessel is approved for. The buoyancy must be demonstrated by means of the
maximum permissible deadweight on board. The deadweight includes all loose objects such as the crew, outboard motor, fuel, fishing gear, catch and the like.
If the buoyancy is documented by means of a weight smaller than the one evident from the vessel’s type approval, this reduced weight will in the future constitute the vessel’s maximum permissible
Preparation for practical buoyancy test
The weight of the fishing gear on board is decided. If there is no fishing gear on board, steel/lead is instead placed in the locations on board the vessel where the gear is normally stored –
typically at the bottom. The weight of steel/lead must amount to at least 60% of the weight of the gear.
The weight of the expected catch is decided. If there is no catch on board, steel/lead is in- stead placed in the locations on board the vessel where the catch is normally stored. The weight of steel
/lead must amount to at least 60% of the weight of the catch.
The weight of an outboard motor, fuel tank, instruments and the like is decided. The outboard motor, instruments and similar equipment can be dismounted before carrying out the practical buoyancy
test. If this equipment is fully or partly taken ashore, steel/lead is instead placed in the locations on board the vessel from where the equipment is taken ashore. The weight of steel/lead must
amount to at least 100% of the weight of the equipment taken ashore.
The weight of a person is normally considered as 75 kg. When the vessel is swamped, a part of the person’s body will be lowered in the water aboard the vessel. During the test, 40 kg weight will be
distributed on the thwarts instead of each person.
See also annex 1 with examples of calculations.
Carrying out a practical buoyancy test
The vessel must be afloat with slack moorings.
The test is carried out in the following way:
1. Deadweights placed in the vessel are noted as regards their location, weight and material.
2. The vessel is filled with water by means of a pump.
3. The filling of water is stopped when the water starts running out of the vessel.
4. When the vessel has found an equilibrium where water no longer runs out of the vessel, the freeboard is measured at the lowest point.
5. The vessel remains lying filled with water for ten minutes to ensure that water does not penetrate into the watertight sections of the hull. The vessel must remain lying for the said ten minutes
with unchanged freeboard and trim.
6. Following the ten minutes’ with water, the filling of water is repeated twice as described in items 2-4 to ensure that the vessel stabilizes immediately with the same freeboard and trim each
The three measured freeboards must all be of more than 50 mm in order for the test to be approved.
After the practical buoyancy test, a certificate is issued documenting that the vessel has been subject to a buoyancy test with a satisfactory result (an example is given in annex 2).
The following must be evident from the certificate:
1. The vessel’s identification (name, port of registry, port registration number, call sign).
2. Type, construction material.
3. Place and date of the test.
4. Maximum deadweight, cf. the vessel manufacturer.
5. List of deadweights and their location during the test.
6. Measured freeboard at water-fillings nos. 1 + 2 +3.
7. Average freeboard of the three water-fillings.
8. Position of the freeboard measured.
9. Other remarks.
The certificate is issued and signed by a person authorised for this by the DMA. The original certificate is handed out to the owner of the vessel and a copy of the certificate is sent to the DMA.
Examples of how the deadweight is decided at the test
Example 1:
The gear and the catch amount to 300 kg at present.
With a density of 2 kg/litres, the deadweights will amount to a volume of 150 litres.
Steel has a density of 8 kg/litre. Consequently, the 300 kg deadweights can be replaced by 180 kg iron (22.5 litre) + 150-22.5 = 127.5 litre = 127.5 kg water.
Thus, the contribution from iron and water will give the necessary volume of 150 litres and the weight will amount to 180 kg + 127.50 = 307.50 kg.
Example 2:
A vessel is approved for seven persons and a maximum in total of 630 kg deadweight.
The seven persons of 75 kg, corresponding to 525 kg, will when seated on the thwarts be partly submerged when the vessel is filled with water. Consequently, these seven persons can instead be
replaced by 7 X 40 kg weight on the thwarts.
If the outboard motor and the tank weigh 50 kg, iron is placed instead of this weight. Then, the residual weight for gear and catch will amount to 630 – 525 – 50 = 55 kg. These 55 kg can be replaced
by 0.6 x 55 = 33 kg iron placed at the bottom.
The 630 kg for which the vessel is approved are thus converted into steel with the following weight:
7 persons x 40 kg, 280 kg placed on thwarts. Outboard motor, 1 x 50 kg, 50 kg located astern. Gear/catch 0.6 x 55 kg, 33 kg located at the bottom. A total of 363 kg iron/steel.
Example 3:
The point of departure is the same vessel as the one used in example 2, but during fishing activities the vessel will, in practice, often carry a maximum of 2 persons corresponding to 150 kg, whereby
the vessel’s proportion for fishing gear and catch can be increased to 630 – 150 – 50 = 430 kg. In order to reflect this condition, the deadweights during the test are com- pounded thus:
2 persons x 40 kg, 80 kg located on the thwarts. Outboard motor, 1 x 50 kg, 50 kg located astern. Gear/catch 0.6 x 430 kg, 258 kg located at the bottom. A total of 388 kg iron/steel
Certificate of buoyancy at practical buoyancy test
For open fishing vessels below 15 metres
│Name of ship│Distinctiv number of letters │Port registration number│Port of registry│
Dimensions and tonnages
│Length overall │Bredth│Depth│GT│NT│Scantings │
Freeboard measurements
│Freeboard measurement no. 1│Freeboard measurement no. 2│Freeboard measurement no. 3│Freeboard average│
│Position for freeboard measurement │ │ │
│Date and place of the bouyancy test │ │ │
│ │Persons │ │ │
├───────────────────────────────────────┼──┬───────┤Other deadweights (kg)│Total deadweight (kg)│
│ │No│Weights│ │ │
│Deadweights according to type approval │ │ │ │ │
│Deadweights at buoyancy test │ │ │ │ │
│Vessel's max permissible deadweight │ │ │ │ │
Certificate issued at .............................. date.................. 20....
It is hereby certified that this buoyancy test is carried out in accordance with the guidelines for documenting the buoyancy of open fishing vessels.
The signatory is authorised by the Danish Maritime Authority to issue this document.
(Underskrift og stempel) (Signature and Stamp)
│ │Number │Position │Substitue weight at test (kg) │Actual weight (kg) │
│Persons │ │ │ │ │
│Motor and fuel tanks, etc. ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│(Substitute weights are equal to 100% of the actual weights)├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│ ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│ ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│Gear, catch and the like ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│(Substitute weights are equal to 60% of the actual weights) ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│ ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│ ├────────────────┼─────────────────────────────────────────────────────────┼────────────────────────────────────┤
│Sum of substitute weights used during the test │ │ │
│Vessel’s max. permissible deadweight (to be transferred to the front page) │ │
│Conversion between the actual weight and the substitute weight: │
│The actual weight of one person i 75 kg. The substitute weight during the test is for one person 40 kg when placed on the thwarts. │
│The substitute weight during the test for motor and fuel tank is the similar weight in steel/lead. │
│The substitute weight during the test for gear and catch is 60% of the actual weight when the sub- stitute weight consists of steel/lead. │
│Other conditions: │
│After water-filling no. 1, the vessel must remain lying for ten minutes filled with water without any change to the trim/freeboard. │
│The freeboard must be measured to the lowest point, for example at the gunwale in the side, at the │
│upper edge of the well hatchway, at the stern, at the upper edge of the motor trunk. │
│Supplementary vessel information: │
│Construction yard │Hull no. │Construction material │Model if relevant │
│Remarks │ │ | {"url":"https://www.dma.dk/publications/2011/november/011-dma-ro-circular-no-011-guidelines-for-demonstrating-the-buoyancy-of-open-fishing-vessels-3","timestamp":"2024-11-11T08:11:08Z","content_type":"text/html","content_length":"24105","record_id":"<urn:uuid:f68d3e28-2665-47de-9c91-2af6abd84922>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00552.warc.gz"} |
How Many Laps In A Pool Is A Mile 2024 Best Swimming Info
How Many Laps in a Pool Is a Mile: 2024 Best Swimming Pool Info
Today’s topic is How Many Laps In A Pool Is A Mile. Swimming is a fantastic form of exercise, providing a full-body workout that improves cardiovascular health and builds endurance. Many people
wonder how many laps they need to swim in a pool to cover a mile. The distance covered in a mile swim is determined by the length of the pool and the swimmer’s ability. In this article, we’ll delve
into the process of calculating the equivalent of a mile in a swimming pool, the number of swimming laps that make one mile, and the factors impacting how many swimming laps a mile adds up to.
How Many Laps In A Pool Is A Mile
How to Measure a Mile in a Swimming Pool – How Many Laps In A Pool Is A Mile?
Using a 25-Yard Pool
If you are using a 25-yard pool, which is a common size for competitive swimming, a mile would equate to 70.4 laps. This computation becomes valid when we consider one mile equals 1,760 yards, and in
a yard pool where one length equals 25 yards, you would need to complete 70.4 lengths (or laps) to total 1,760 yards.
Also Read: How To Make Espresso Without A Machine
Using a 50-Meter Pool
In a 50-meter pool, which is often used in international swimming competitions, a mile is equivalent to 66 laps. This is because a mile consists of 1,609 meters, and in a 50 meter pool, you would
need to swim 66 laps (each lap is made up of two lengths of the pool) to total 1,650 meters, marginally less than the real mile distance.
Considering Different Pool Sizes
It’s important to note that not all swimming pools are standard sizes. If you are swimming in a pool with non-standard measurements, the distance covered for a mile will vary. You can calculate the
number of laps needed based on the specific measurements of the pool.
Determining the Number of Laps for One Mile
Calculating Laps in a 25-Meter Pool
In a 25-meter pool, a mile would correspond to 64 laps or two lengths of the pool, since 1,609 meters make up a mile, and 25 meters (which determines one lap in a meter pool) multiplied by 64 laps
equals 1,600 meters, just short of the accurate mile distance.
.Also Read: How Much Does A Human Head Weigh
Finding Laps in a 25-Yard Pool
If you are swimming laps in a 25-yard pool, you would need to swim approximately 88 of these laps to cover a mile. This is because 1,760 yards make up a mile, and 25 yards multiplied by 88 laps
totals 2,200 yards, slightly over the true mile distance.
Adjusting for Open Water Swimming
When swimming in open water, such as in a lake or the ocean, the distance covered in a mile swim can be affected by factors such as currents and waves. Swimmers should be prepared to adjust their
stroke technique and exert more effort when swimming in open water to cover the mile distance.
Factors Affecting the Number of Laps in a Mile
Effect of Different Stroke Styles
The number of laps you need to swim to cover a mile can vary based on your stroke style and the length of the pool you choose, such as a 25-yard pool or a 50 meter pool. For example, the butterfly
stroke may require more energy and therefore more laps to cover a mile compared to the freestyle stroke.
Also Read: How Much Does A Private Investigator Cost
Impact of Swimmer’s Stride Length
A swimmer’s stride length can also impact the number of laps needed to cover a mile. Swimmers with longer strides may cover more distance with each stroke, requiring fewer laps to complete a mile
compared to swimmers with shorter strides.
Influence of Pool Length
The length of the pool directly affects the number of laps required to swim a mile. A standard 25-yard pool will necessitate more laps compared to a 50-meter pool due to the difference in distance
covered with each lap.
As you can see, there are several factors that contribute to determining the number of laps required to cover a mile in a swimming pool. By understanding these factors, swimmers can better plan their
workouts and training sessions to achieve their mileage and fitness goals effectively.
Q: How many laps in a 25-yard pool is a mile?
A: In a 25 yard pool, a mile measures 1650 yards, and you would have to swim 66 lengths of the pool, or equivalently, 66 swimming laps to cover a mile.
Also Read: What Is A Rehydration Clause In Boxing
Q: How many laps in a 25-meter pool is a mile?
A: In a 25-meter pool, a mile is equivalent to 1609.34 meters, so you would need to swim 64 laps to complete a mile.
Q: How many laps in a 50-meter pool is a mile?
A: In a 50-meter pool, a mile is equivalent to 1610 meters, so you would need to swim 32 laps to complete a mile.
Q: How many lengths in a 25-yard pool is a mile?
A: In a 25-yard pool, which is one length of a pool that’s 25 yards long, a mile is equivalent to 1650 yards, or 66 lengths of the pool.
Q: How many lengths in a 25-meter pool is a mile?
A: In a 25-meter pool, a mile is equal to 1609.34 meters, which would be 65 lengths of the pool.
Q: How many laps makes a mile in swimming?
A: The number of swimming laps required to complete a mile in swimming is dependent on the length of the pool, whether it’s a yard pool or a meter pool. In a 25-yard pool, it would be 66 laps, in a
25-meter pool, it would be 64 laps, and in a 50-meter pool, it would be 32 laps.
Also Read: How Much Does Disneyland Make A Day
Q: How do I know how many laps I need to swim to make a mile?
A: To determine how many laps you need to swim to make a mile, you can calculate the distance of the pool (in yards or meters) and then divide the distance of a mile by the length of the pool.
Q: Do I have to start swimming from the end of the pool to complete a mile?
A: No, you are free to begin swimming from either end of the pool to complete a mile; it does not affect the number of swimming laps required to make a mile. The important factor is the total
distance covered, not the starting point.
Q: What size of pool is typically used for swimming a mile?
A: A 25-yard or 25-meter pool, each one being the standard for a yard pool and meter pool, respectively, is widely used for swimming one mile, as these pools allow swimmers to reach the required
distance without having to turn excessively.
Q: Why is a mile in swimming different from a true mile?
A: A “mile” in swimming is a standardized distance set by the sport’s governing bodies. It may be slightly shorter or longer than a true mile, depending on the type of pool (short course or long
course) being used for the swim.
Hope you enjoy reading our article How Many Laps In A Pool Is A Mile 2024 Best Swimming Info. | {"url":"https://www.snifor.com/how-many-laps-in-a-pool-is-a-mile/","timestamp":"2024-11-01T23:31:47Z","content_type":"text/html","content_length":"77605","record_id":"<urn:uuid:4d9f7efd-281f-438c-af3a-4fe95f97f66f>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00282.warc.gz"} |
Model Evaluation
Carlos Salas
Portfolio Manager and Data Scientist
Evaluating model performance for accuracy is crucial. But models are a simplified version of reality and are prone to model error. Join Carlos Salas in this video as he explores the 3 kinds of model
error and how to find the right balance.
Evaluating model performance for accuracy is crucial. But models are a simplified version of reality and are prone to model error. Join Carlos Salas in this video as he explores the 3 kinds of model
error and how to find the right balance.
Access this and all of the content on our platform by signing up for a 7-day free trial.
Key learning objectives:
• Identify the 3 types of model error
• Understand how to fix model errors
Measuring a model’s performance is one of the most important aspects of the data science workflow. There are 3 types of model error to be aware of: model bias (originating from erroneous assumptions
in the learning algorithm), model variance (how sensitive the model is to small fluctuations in the utilised training dataset) and random error. The right path for a data scientist is to always try
to find a model that has a balance between accuracy and precision, called the bias-variance trade-off. Fixing high bias can be achieved by adding more input features or by increasing the complexity
of the model. Fixing high variance models can be achieved by using more training data and reducing the number of features so that only the most important are considered.
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What are the 3 types of model error?
1. Model bias. An error originating from erroneous assumptions in the learning algorithm. Models suffering from high bias can cause an algorithm to miss the relevant relations between features and
the response variable, commonly referred to as underfitting.
2. Model variance. This is related to the instability of the model - how sensitive the model is to small fluctuations in the utilised training dataset. Models suffering from high variance can cause
an algorithm to miss relevant relations, commonly referred to as overfitting.
3. Random error. A model is a simplification of reality and can never be perfect, so it will therefore always suffer from random errors. A different way of interpreting the random error is by looking
at it as a measure of the amount of data noise present.
How can you fix model error?
As random error is inevitable, the only two error components that can be minimised are bias and variance. The right path for a data scientist is to always try to find a model that has a balance
between accuracy and precision, called the bias-variance trade-off.
Fixing high bias can be achieved by adding more input features or by increasing the complexity of the model. Fixing high variance models can be achieved by using more training data and reducing the
number of features so that only the most important are considered.
Subscribe to watch
Access this and all of the content on our platform by signing up for a 7-day free trial.
Carlos Salas
Carlos Salas is a professional investor passionate about the lifelong development of an investment process that blends man and machine. Over the last 15 years, he has worked in investment roles for
firms such as Santander AM, BNP Paribas, Jefferies, and LCAM. He is currently pursuing three careers simultaneously - as an investment manager, consultant and lecturer.
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Questions and Answers about Julia
Here is a summary of my answers to the questions people asked in chat on September 3 about the Julia programming language.
Does the math text in Jupyter have the same structure as LaTeX?
The mathematical equations in JupyterLab are intentionally formatted using the same commands as in LaTeX. Note, however, that the formatting of the rest of the document follows Markdown conventions.
Although full LaTeX may be important when writing an involved scientific report, the compromise of Markdown with LaTeX equations will be sufficient for this class. Since JupyterLab can serve as a
stepping stone for LaTeX, I think it is reasonable to learn JupyterLab first.
While it is possible to create a project report using LaTeX or even Microsoft Word, since you will be working in groups it is important everyone in the group agrees to this and is fully able to
participate. For simplicity my recommendation is to use JupyterLab. I will be using it in class, able to answer questions and people will be familiar with it.
There is less of a problem using Microsoft Word, LaTeX or any other software on homework as long as the final result can be uploaded as a pdf file. However, for questions involving a pencil-and-paper
calculation, I would prefer that people turn in pdf scans of their pencil-and-paper work, unless a special disability makes this difficult.
Can Julia be a substitute for Maple?
Maple is a computer algebra system while Julia is intended for numeric computation. Though both are related to mathematics, they are different.
Julia was designed as a replacement for MATLAB. There is a symbolic toolbox add-on for MATLAB based off of the MuPAD computer algebra system designed at the University of Paderborn that includes many
features of Maple; however, the most common use of MATLAB is for numeric computations.
Maple can be accessed on campus in the Math Center, in the ECC Lab and rented by students for a semester to use at home for about $25. Maple is not available through the UNR remote desktop. However,
a different computer algebra system called Mathematica is available. When I need a CAS in this course I will use Mathematica.
Mathematica also comes free with a Raspberry Pi, which is a $35 computer designed to teach computer science throughout the primary and secondary schools in the UK. It is also useful for certain
university courses. For example, you may already have a Raspberry Pi if you took or are taking CS219 Computer Architecture. While not suitable for Zoom-based distance learning, in addition to
Mathematica the Raspberry Pi can also be used to run Julia, JupyterLab and do all the homework for this class.
What's the relationship between Julia and Python?
Jupyter is an amalgamation of the names of the Julia, Python and R programming languages.
• Julia is a domain specific language designed to replace MATLAB for performing numeric computation with a convenient notation for dealing with vectors. It features a just-in-time compiler that
creates an easy-to-use interactive environment while at the same time resulting in much faster performance than MATLAB.
• Python is a general purpose language designed to replace BASIC for teaching computer programming that has become popular for many other things. It can be much slower than Julia being an
interpreted language, but Python could work for the programming done in this course. My lectures will focus on Julia, but I am okay grading homework and programming projects that were done in
Are we allowed to use Python for our projects?
Yes. However, since you will be working in groups it is extremely important that everyone in the group agrees to this. The reason I will teach with Julia rather than Python in class is because Julia
is faster and easier to use for numeric computations, especially when working with vectors and matrices.
Can we use MATLAB?
MATLAB is a common choice for an introductory course in numerical methods. I have taught Math 466/666 using MATLAB in the past but chose to use Julia this semester. Julia has a syntax which retains
the convenience of MATLAB while fixing many idiosyncrasies related to the way MATLAB evolved over time.
At this point Julia is faster, easier to understand, feature complete and one of the three main software components behind JupyterLab. It is also free. In my opinion it is likely Julia will be
increasingly used for engineering, science and mathematics in the future. Therefore, even if you already know MATLAB, I think it is worthwhile to learn Julia.
While it is possible to use MATLAB for this course, as with Python, it is important when working in a group that everyone agrees. Since I will be teaching the course using Julia and because anyone
who already knows MATLAB should find Julia easy to learn, my recommendation is to use Julia.
Last Updated: Fri Sep 4 13:25:54 PDT 2020 | {"url":"https://renomath.org/466-20/qanda.html","timestamp":"2024-11-07T19:24:41Z","content_type":"text/html","content_length":"5554","record_id":"<urn:uuid:a9d8b99f-c0a0-496d-96dc-73755a6b0076>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00713.warc.gz"} |
How to measure the size of any plot
The process is simple to measure the size of any plot. Given below, some useful tricks for measurements.
The size of any land is essential to provide manure (or chemical fertiliser) at the right rate.
Initially, pace the length (L) and width (W) of the land, and note down the number of paces in each. Suppose, each pace is approximately 1m. Now, the number of paces for the length should be
multiplied with the number of paces for the width.
Thus, the area in square metres (m^2) will be determined.
Now, the number of paces for the length should be multiplied with the number of paces for the width to obtain the area in square metres (m^2).
It ensures that the plot is in rectangular or square shape. If it contains constant length but dissimilar width, or constant width but dissimilar length, then a marginally different method should be
employed to determine the area.
Under this situation, calculate the width (W) of the land by pacing across it at a right angle (90?). Then, find out the length (L) by pacing along the length of the plot on both sides. It will
provide two plot lengths.
Assume, L1 ? 8 paces and L2 ? 12 paces.
Now, the average length of the plot will be computed by dividing the total of L1 and L2 by 2 i.e. 8m + 12m ? 2 = 10m.
To work out the area, multiply this average length (10m) with the width (5m) i.e. 10m x 5m = 50m2.
The same method can be used where the length is constant but the width fluctuates. Calculate the length, then obtain two measures of the width, W1 + W2, and perform the estimation.
To change the area of any land to hectares (ha), the area in square metres should be divided with 10 000. In this case, it will become 50m2 ? 10 000 = 0,005ha. | {"url":"https://quantity-takeoff.com/measure-the-size-of-any-plot.htm","timestamp":"2024-11-05T10:35:44Z","content_type":"text/html","content_length":"23279","record_id":"<urn:uuid:fec8fa48-a256-4b36-8e0d-90dbbe3226a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00672.warc.gz"} |
Methods of Proof - Discrete Mathematics
Methods of Proof – Discrete Mathematics
Understanding what constitutes a sound mathematical argument—that is, a proof—is essential to understanding written mathematics. This necessitates an understanding of proof-building strategies. The
strategies for creating proofs that we will examine are also applied in other areas of computer science, including the principles that computers use to reason, the methods for confirming the accuracy
of programs, etc. In mathematics, a lot of theorems are implications: p → q. Different methods of evidence arise from different ways of proving implications.
The implication p →q can be proved by showing that if p is true, the q must also be true. This shows that the combination p true and q false never occurs. A proof of this kind is called a direct
1. An integer n is even if, and only if, n = 2k for some integer k.
2. An integer n is odd if, and only if, n = 2k + 1 for some integer k.
3. An integer n is prime if, and only if, n > 1 and for all positive integers r and s, if n = r·s, then r = 1 or s = 1.
4. An integer n > 1 is composite if, and only if, n = r·s for some positive integers r and s with r ≠ 1 and s ≠ 1.
5. A real number r is rational if, and only if, a b for some integers a and b with b≠0.
6. If n and d are integers and d ≠0, then d divides n, written d | n if, and only if, n = d.k for some integers k.
7. An integer n is called a perfect square if, and only if, n = k2 for some integer k.
Prove that the sum of two odd integers is even.
Let m and n be two odd integers. Then by definition of odd numbers
m = 2k + 1 for some k ∈Z
n = 2l + 1 for some l ∈ Z
Now m + n = (2k + 1) + (2l + 1)
= 2k + 2l + 2
= 2 (k + l + 1)
= 2r where r = (k + l + 1) ∈Z
Hence m + n is even. | {"url":"https://universitydistancelearning.com/methods-of-proof/","timestamp":"2024-11-14T22:13:25Z","content_type":"text/html","content_length":"105717","record_id":"<urn:uuid:39865e4d-5f28-4579-9830-3607634beb9f>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00711.warc.gz"} |
Mean Mode Median and Range Worksheets 3 | Free Maths Worksheets
Mean Mode Median and Range Worksheets 3: Find the median, mode, mean and range of the following series of numbers. Click the button below to try all our FREE PDF printable median, mode, mean and
range worksheets.
How to find the Median, Mode, Mean and Range! Try all Median, Mode, Mean and Range Worksheets! Read more
Try this online practice worksheet on finding the median, mode, mean and range of a series of numbers and hit the button ‘Check Answers’ to correct your answers. Click ‘Download/Print Worksheet’
button to download or print this worksheet.
Find the median, mode, mean and range of the following series of numbers. | {"url":"https://k8schoollessons.com/mean-mode-median-and-range-worksheets-3/","timestamp":"2024-11-14T22:10:51Z","content_type":"text/html","content_length":"45858","record_id":"<urn:uuid:89992c40-5781-4386-9dc6-0e939f22e26e>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00795.warc.gz"} |
Tirupati seva in one day tour pack from cennai
Tirumala Tirupati SEVA from Chennai balaji travels
Seva is also called as “sevai” in Tamil.
1. Suprabhatam
2. Thomala Seva
3. Ekantha Seva
4. Kalyanotsavam
5. Arjitha Brahmotsavam
6. Unjal Seva
7. Sahasra Dipalankara Seva
8. Koil Alwar Thirumajanam
9. Vasanthotsavam
10. Pavithrotsavam
11. Pushpayagam
12. Aabhidhyayaka Abhishekam
13. Padmavathi Parinayam
14. Pushpa Palaki
15. Ashtadalapada Padmaradhanasahasra Kalabhishekamthirupavada Seva
16. Purna Abhishekam
chennai balaji travels seva
1. This is the very first seva.Which held’s in between 2:30 – 3:00
2. Believed that lord awake to the sound of Vedic solka
3. The cost of this ticket is rs 200 can be purchased one day before
1. This seva will be held between 3:45 – 4:30
2. Lord is decorated with floral decoration, garland, and Tulasi when the ablution is performed.
3. The cost of the ticket is rs 220 can be collected at the beginning of the Vijaya bank counter.
1. Every day after sarvadarsanam the lord needs to sleep in this seva.
2. The ticket price for this seva is rs 120
3. It can be purchased daily at the beginning of Vijaya bank.
4. But no reservation for this seva.
1. This is very special Seva. The Seva for lord venkateshwara and the goddess Lakshmi with Padmavati are known as kalyanotsavam.
2. The cost of this seva is rs 1000.
3. Only two persons allowed husband and wife.
1. Here the lord is venerated with both spouses in vahanam sesha, garuda vahanam, and hanumantha vahanam.
2. The cost of the ticket for this seva is Rs.1000.
3. 5 members are allowed to attend this seva.
1. This seva is held by 3:30 pm
2. The lord and her husband on unjal (swing) worshipped.
3. The ticket cost for this seva is Rs.1000
4. The number of persons allowed is 6
5. Money has to be sent in advance at least 90 days ahead
1. This seva is conducted in kolavu mandapam
2. Where the lord utchava murthy is beautifully decorated and his wives
3. The cost of the ticket for this seva is Rs.1000
4. The number of people allowed is 5
1. This seva is performed only 4 Tuesdays a year
2. The cost of a ticket for this seva is rs 3000
3. The number of people allowed is 10
4. Money has to be sent in advance at least 90 days ahead
1. This seva is conducted in the month of April
2. Except for Thursdays and Fridays
3. The ticket for this seva is rs 3000
4. The number of people allowed is 10
1. This Seva is conducted in the month of august
2. The ticket cost is Rs.5000
3. The number of people allowed is 10
1. Pushpayagam seva is a yagam dedicated to the lord using different kinds of flowers
2. The cost of this seva is rs 3500
3. The number of people allowed is 5
1. This seva is conducted in the month of June
2. The cost of the ticket is Rs.2000
3. The number person allowed is 5
1. This Seva is conducted in the month of may
2. The ticket charge is rs 5000
3. The numbers of person allowed are 5
1. This Seva is held in the month of July
2. The cost of the ticket is Rs.1000
3. The number of people allowed is 5
Ashtadalapada padmaradhana
1. This Archana is only held on tuesdays
2. The lord is venerated with 108 lotus flowers of gold
3. The tickets cost Rs.2500
4. The number of people is 5
1. This is a specil arjitha seva
2. Performed only on Wednesday
3. The cost of the ticket is rs 5000
4. The number of people allowed is 6
1. This is the only seva on Thursday
2. The ticket cost Rs.5000
3. The number of people allowed is 6
1. This seva is performed only on Friday
2. Ablutions gave to the lord in different boats
3. The cost of the ticket is Rs.750
4. The number of people allowed is 1 | {"url":"https://tirupaticarpackages.com/chennai-balaji-travels-seva-package/","timestamp":"2024-11-14T04:28:04Z","content_type":"text/html","content_length":"204995","record_id":"<urn:uuid:148e8d6e-959d-45bd-901b-ae7bcb345f4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00101.warc.gz"} |
Order of Operations and RPN
02-09-2020, 06:11 PM
Post: #6
toml_12953 Posts: 2,192
Senior Member Joined: Dec 2013
RE: Order of Operations and RPN
(02-09-2020 02:41 PM)rprosperi Wrote:
(02-09-2020 12:32 PM)toml_12953 Wrote: I don't get the expression on pg 4 and 5. How is (9-5)+2 the same as (9+-5)*1/2 ?
The initial equation is actually (9-5) / 2, though using the obelus division symbol (where I show the slash), which unfortunately looks like a plus sign because the font used is fuzzy.
Thanks for the explanation! I thought they were using "New Math"!
Tom L
Cui bono?
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The Stacks project
Lemma 59.31.1. Let $S$ be a scheme. Let $\mathcal{F}$ be a subsheaf of the final object of the étale topos of $S$ (see Sites, Example 7.10.2). Then there exists a unique open $W \subset S$ such that
$\mathcal{F} = h_ W$.
Proof. The condition means that $\mathcal{F}(U)$ is a singleton or empty for all $\varphi : U \to S$ in $\mathop{\mathrm{Ob}}\nolimits (S_{\acute{e}tale})$. In particular local sections always glue.
If $\mathcal{F}(U) \not= \emptyset $, then $\mathcal{F}(\varphi (U)) \not= \emptyset $ because $\{ \varphi : U \to \varphi (U)\} $ is a covering. Hence we can take $W = \bigcup _{\varphi : U \to S, \
mathcal{F}(U) \not= \emptyset } \varphi (U)$. $\square$
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Math Colloquia - Quantum Dynamics in the Mean-Field and Semiclassical Regime
The talk will review a new approach to the limits of the quantum N-body dynamics leading to the Hartree equation (in the large N limit) and to the Liouville equation (in the small Planck constant
limit). This new strategy for studying both limits involves functionals that are analogous to the Wasserstein distance of exponent 2 used in optimal transport. (Work in collaboration with Clément
Mouhot and Thierry Paul). | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=colloquia&page=10&l=en&sort_index=speaker&order_type=desc&document_srl=800698","timestamp":"2024-11-08T22:15:51Z","content_type":"text/html","content_length":"45812","record_id":"<urn:uuid:476f3dbd-8f65-4a79-91d1-87cc6af0a35d>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00707.warc.gz"} |
Coordinate Geometry Formulas
The list of all coordinate geometry formulas for class 9, 10, 11 is provided here to help the students. To recall, coordinate geometry is the study of geometry using the coordinate points. In
coordinate geometry, the position of a point can be easily defined using coordinates.
Download Coordinate Geometry Formulas PDF: Download Now!
Coordinate Geometry Formulas List for Class 9, 10 and 11
Coordinate geometry is an integral topic in class 9, 10 and 11. All the important coordinate geometry formulas for class 9, class 10 and class 11 are given below.
All Formulas of Coordinate Geometry
General Form of a Line Ax + By + C = 0
Slope Intercept Form of a Line y = mx + c
Point-Slope Form y − y[1]= m(x − x[1])
The slope of a Line Using Coordinates m = Δy/Δx = (y[2 ]− y[1])/(x[2 ]− x[1])
The slope of a Line Using General Equation m = −(A/B)
Intercept-Intercept Form x/a + y/b = 1
Distance Formula |P[1]P[2]| = √[(x[2 ]− x[1])^2 + (y[2 ]− y[1])^2]
For Parallel Lines, m[1 ]= m[2]
For Perpendicular Lines, m[1]m[2] = -1
Midpoint Formula M (x, y) = [½(x[1 ]+ x[2]), ½(y[1 ]+ y[2])]
Angle Formula tan θ = [(m[1 ]– m[2])/ 1 + m[1]m[2]]
Area of a Triangle Formula ½ |x[1](y[2]−y[3])+x[2](y[3]–y[1])+x[3](y[1]–y[2])|
Distance from a Point to a Line d = [|Ax[0 ]+ By[0 ]+ C| / √(A^2 + B^2)]
Topics Related to Coordinate Geometry Formulas
More Math Formulas
Practice Questions Using Coordinate Geometry Formulas
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Zastosowanie metody różnic skończonych do zagadnień pełzania kołowo cylindrycznych powłok
Engineering Transactions, 18, 4, pp. 631-674, 1970
Zastosowanie metody różnic skończonych do zagadnień pełzania kołowo cylindrycznych powłok
The numerical approach to creep deformation in pressurized circular cylindrical shells is developed by way of the finite difference method, and some features of the deformation in the shells are
In Part I of the paper, the transient creep analysis of circular cylindrical shell is developed on the basis of the power creep law and the creep theories of Mises-Mises, Tresca-Mises and
Tresca-Tresca type. Use is made of the strain-hardening hypotheses. The creep deformation and the associated state of stress is investigated for various shell geometries and various magnitudes of
internal pressure. The difference between the creep theories and hardening hypotheses as applied to the present problem is also discussed. Calculations are performed for constant as well as variable
internal pressures. Part II is concerned with the analysis of the steady-state crcep of a circular cylindrical shell according to the power creep law nad the creep theory of Mises-Mises type. An
iterative procedure
combined with the finite-difference method is proposed. The effect of shell geometry and the creep exponent on the state of stress and rate of deformation is investigated. The rigorous results
obtained are also compared with the previous solution on the basis of a sandwich shell, and the validity of the assumption of sandwich construction is discussed.
Copyright © Polish Academy of Sciences & Institute of Fundamental Technological Research (IPPT PAN).
W. OLSZAK and A. SAWCZUK, Inelastic Behaviour in Shells, Noordhoff, Groningen 1967.
J.F. BESSELING, Creep in Structures, Proc. IUTAM Colloquium, p. 174, Springer, Berlin 1962.
G. ROSENGREN, Arch. Mech. Stos., 4, 16 (1964), 959.
B. EINARSSON, J. Engng Math., 2, 2 (1968), 123.
S. MURAKAMI and M. TANI, Proc. Romanian Nat. Conf. Apl. Mech., Bucharest 1969, in press.
R. K. PENNY, Int. J. Mech. Sci., 9 (1967), 373.
A. E. JOHNSON and B. KHAN, Int. J. Mech. Sci., 7 (1965), 791.
I. FINNIE and W. R. HELLER, Creep of Engineering Materials, McGraw-Hill, New York 1959.
F. K. G. ODQVIST and J. HULT, Kriechfestigkeit metallischer Werkstoffe, Springer, Berlin 1962.
F. GAROFALO, Fundamentals of Creep and Creep-Rupture in Metals, McMillan, New York 1965.
A. M. WAHL, Trans. ASME, Ser. D, J. Basic Engng, 1, 85 (1963), 66. 12. A. M. WAHL, Proc. 4th U. S. Nat. Cong. Appl. Mech., ASME, New York 1962, 18.
A. E. GEMMA, G. H. ROWE and R. J. SPAHL, Trans, ASME, Ser. D, J. Basic Engng, 83 (1961), 545.
S. P. TIMOSHENKO and S. WOINOWSKY-KRIEGER, Theory of Plates and Shells, 2nd ed., McGraw-Hill, New York 1959.
G. E. FORSYTHE and W.R. WASOW, Finite-Difference Methods for Partial Differential Equations, John Wiley, New York 1966.
A. RALSTON and H.S. WILE, Mathematical Methods for Digital Computers, John Wiley, New York 1960.
YU.N. RABOTNOV, Creep of Constructional Elements, Nauka (1966).
E. T. ONAT and H. YÚKSEL, Proc. 3rd U.S. Nat. Cong. Appi. Mech., ASME, New York 1958, 625.
M. P. BIENIEK and A. M. FREUDENTHAL, J. Aerospace Sci., 1, 27 (1960), 763.
V. I. ROZENBLUM, Prikl. Math. Mekh., 1, 27 (1963), 154.
C. R. CALLADINE, Proc. Symp. Nuclear Reactor Containment Building and Pressure Vessels, Butterworth 1960, 411.
C. R. CALLADINE, Nonclassical Shell Problems, North-Holland, 1964, 384.
C. R. CALLADINE and D. C. Drucker, Quart. Appl. Math., 20 (1962), 79.
I. G. TEREGULOV, Prikl. Math. Mekh., 26 (1962), 492.
I. G. TEREGULOV, Izv. AN SSSR, Mekh. i Mashinostr., No 6, 1963, 169.
YU. N. RABOTNOV, Prikl. Math. Mekh., 1, 27 (1963), 154.
YU. N. RABOTNOV, Applied Mechanics, Proc, 11th Internat. Cong. Appl. Mech., Springer, 1966, 415.
YU. M. VOLCHOV, Izv. AN SSSR, Mekh. i Mashinostr., No. 5, 1966, 118.
A. E. GEMMA, T. Aerospace Sci., 12, 27 (1960), 953.
A. B. GEMMA, J. Aerospace Sci., 3, 29 (1962), 352.
F. A. COZZARELLI, S. A. PATEL and B. VENKATRAMAN, AIAA J., 7, 3 (1965), 1298.
T. P. BRYNE and A.C. MACKenzIe, J. Mech. Eng. Sci., 8 (1966), 216. 33. N. J. HOFF, Quart. Appl. Math., 1, 12 (1954), 49.
H. B. KELLER and E. L. REISS, Comm. Pure and Appl. Math., 3, 11 (1958), 273.
H. B. KELLER and E. L. REISS, J. Aerospace Sci., 10, 26 (1959), 643. | {"url":"https://et.ippt.pan.pl/index.php/et/article/view/2590","timestamp":"2024-11-02T15:52:21Z","content_type":"text/html","content_length":"22874","record_id":"<urn:uuid:5cb49005-a2dc-41e5-b8c6-6916e19a95a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00464.warc.gz"} |
/n Linear regression assignment
The dataset contains information about
/n Linear regression assignment The dataset contains information about several major liquor brands, their advertising revenue information and sales. We are interested to see the impact of different
types of advertising expenses, sales and market shares to current year sales. For each of the question, paste visualization plots when necessary. Paste the R code at the end of the paper. Discuss the
statistics in your own words when asked. 1) Compute a new variable by adding magazine, newspaper, outdoor, broadcast and print advertising expenditure, and name it TotalAds. (10 pts). 2) Check
histogram distributions of the following variables: • TotalSales ● TotalAds ● PriceRerUnit . MarketShare Do the histograms resemble normal distribution? Submit your visualization plots, and discuss.
(10 pts) 3) Conduct a correlation plot with all the variables. Submit your visualization plots, and describe the relationships among the variables as you learn from the plot. (20pts) 4) We are
interested to see impact of different types of advertising expenses on total sales. Run scatterplot charts with the predictor and output variables first to understand the patterns. Then, conduct a
regression analysis with the following variables: Output variable TotalSales Predictor variable Mag News Outdoor Broad Print Which advertising medium are significant predictor of sale? Mention
relevant important statistics from the regression output, including the Beta, the P value and the R square. (30 pts) 5) We are interested to see the impact of different types of ad expenses, market
share and pricing on sales. Run scatterplot charts with the predictor and output variables first to understand the patterns. Then, conduct a regression analysis with the following variables: Output
variable TotalSales Predictor variable TotalAds PriceRerUnit Marketshare Which advertising mediums are significant predictors of sales? Mention relevant important statistics from the regression
output, including the Beta, the P value and the R square. (30 pts) Helping R code #See column names of the file. colnames(data) #Compute new data columns data<-data %>% mutate(TotalAds = Mag +
News+Outdoor+Broad+Print) #Please use the previous codes from week 1 to do data exploration and visualization. #Create and visualize Correlation plot library(corrplot) M = cor(data) corrplot(M,
method = 'number') options(scipen=999) #Create regression equation model <- Im(y ~ x1 + x2 + x3 + x4 + x5, data = data) summary(model) | {"url":"https://tutorbin.com/questions-and-answers/n-linear-regression-assignment-the-dataset-contains-information-about-several-major-liquor-brands-their-advertising","timestamp":"2024-11-07T00:53:21Z","content_type":"text/html","content_length":"65846","record_id":"<urn:uuid:04926d2f-b6a8-4a91-8344-77d91e0f45df>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00436.warc.gz"} |
What is noise?
👈 Home
What is noise?
September 2019
I typically think of noise as “the mysterious, unseen force that causes your A/A tests to look like they’ve found a winner on day 3.”
But that definition is a bit hand-wavey, and a bit circular. I’ll try to do a little better.
Noise is any departure from complete uniformity of data
So your conversion rate is 4%? Cool, I’ll assume that means that 4 out of every 100 visitors make a purchase. Or for that matter, 2 out of every 50 … or 1 out of every 25.
In fact, I’ll assume that every 25th visitor converts, and the other 24 never convert.
Of course I’m completely wrong about that.
So while you may consistently observe a 4% conversion rate month over month, when you zoom in on a subset of that data, things get messier.
Noise is more pronounced in smaller datasets
Here are results from 100 coin tosses:
49 Heads, 51 Tails. That’s pretty uniform.
But look at those long blocks of blue - for example, that very suspicious run of 6 Heads in the third column 🤔.
If you observed this exciting experiment just long enough to see those 6 tosses, you’d probably think the coin was rigged.
Of course you’d be completely wrong about that. If you stuck around for the whole experiment 😴 you’d have enough data to reject that theory.
Noise is data that obscures the “true” nature of the dataset
The true nature of a coin toss is that half the outcomes should be Heads, half Tails. The true nature of an A/A test is that each variation should have the same conversion rate.
But despite your fairly consistent 4% conversion rate, sometimes 2 or 3 visitors convert in a row. And despite the fact that your experimentation tool should assign visitors to different variations
each time, sometimes it chooses the same variation a few times in a row.
If both of these random occurrences converge, you’ve got a variation with 3 conversions where the other ones have 0.
It’s like that run of 6 Heads in a row; in time, the noise will get drowned out by the signal, and all of your A/A test variations will have a 4% conversion rate. But only in time.
Noise is real data; noise is random; noise is measurable
It’s worth noting that the noise we’re referring to here isn’t junk data. It’s real, it’s true; it just doesn’t conform perfectly to statistical models.
And it’s random, which means the noisiness of your data will vary from trial to trial, from week to week.
The good news is that it can be measured, and understood. You can make your peace with it.
In fact, you already know how. If I come to you raving and screaming about how I flipped a coin 3 times in a row and got Heads every single time OMG what is going on the laws of physics have been
overturned … you’d calmly counsel me to give it a few more flips.
How many more flips? Or in the case of a website experiment, how many more days before we conclude that something’s up?
These numbers are calculable, or at least estimable. More to come!
Next time I write something, I'll let you know.
© 2024 Brian David Hall | {"url":"https://briandavidhall.com/what-is-noise/","timestamp":"2024-11-14T23:29:59Z","content_type":"text/html","content_length":"10427","record_id":"<urn:uuid:916e2336-dd8c-48a3-8384-0e143ea7bee2>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00688.warc.gz"} |
Seminars and Colloquia by Series
Friday, November 12, 2021 - 14:30 for 1 hour (actually 50 minutes)
Skiles 169
Luca Dieci – Georgia Institute of Technology – dieci@math.gatech.edu
We review some theoretical and computational results on locating eigenvalues coalescence for matrices smoothly depending on parameters. Focus is on the symmetric 2 parameter case, and Hermitian 3
parameter case. Full and banded matrices are of interest. | {"url":"https://math.gatech.edu/seminars-and-colloquia-by-series?series_tid=55&page=1","timestamp":"2024-11-06T18:33:13Z","content_type":"text/html","content_length":"56474","record_id":"<urn:uuid:c3a6eaf2-0c07-4f7f-a7fb-f5b9540f6f83>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00511.warc.gz"} |
What is Bubble Sort Algorithm?
Bubble Sort Algorithm
Bubble sort is a simple sorting algorithm that works by repeatedly comparing adjacent elements and swapping them if they are in the wrong order. The algorithm starts at the beginning of the array and
compares the first two elements. If the first element is greater than the second element, they are swapped.
The algorithm then moves on to the next two elements and repeats the process. This continues until the end of the array is reached.
Does Bubble Sort Algorithm Efficient
The bubble sort algorithm is a very simple algorithm to understand and implement. However, it is not very efficient for large arrays. The worst-case time complexity of bubble sort is O(n^2), where n
is the number of elements in the array. This means that the time it takes to sort the array increases quadratically as the number of elements increases.
Bubble sort is not a very good algorithm to use for sorting large arrays. However, it can be useful for sorting small arrays or for educational purposes.
Step-by-step Bubble Sort Algorithm Explanation
Here’s a step-by-step explanation of the Bubble Sort algorithm:
1. Start with an unsorted list of elements.
2. Compare the first element with the second element. If the first element is greater than the second element, swap them.
3. Move to the next pair of elements (2nd and 3rd), and continue comparing and swapping until the end of the list is reached.
4. At this point, the largest element is in its correct position at the end of the list.
5. Repeat steps 2-4 for all other elements, except the last one.
6. After each iteration, the next largest element will be in its correct position at the end of the list.
7. Continue the iterations until the entire list is sorted.
Let’s say we have an unsorted list: [5, 3, 8, 2, 1]
1st Iteration
• Comparing 5 and 3: Swap [3, 5, 8, 2, 1]
• Comparing 5 and 8: No swap [3, 5, 8, 2, 1]
• Comparing 8 and 2: Swap [3, 5, 2, 8, 1]
• Comparing 8 and 1: Swap [3, 5, 2, 1, 8]
Result: [3, 5, 2, 1, 8]
2nd iteration
• Comparing 3 and 5: No swap [3, 5, 2, 1, 8]
• Comparing 5 and 2: Swap [3, 2, 5, 1, 8]
• Comparing 5 and 1: Swap [3, 2, 1, 5, 8]
Result: [3, 2, 1, 5, 8]
3rd iteration
• Comparing 3 and 2: Swap [2, 3, 1, 5, 8]
• Comparing 3 and 1: Swap [2, 1, 3, 5, 8]
Result: [2, 1, 3, 5, 8]
4th iteration
• Comparing 2 and 1: Swap [1, 2, 3, 5, 8]
Result: [1, 2, 3, 5, 8]
Bubble Sort Algorithm Usage Areas
Bubble sort is a simple sorting algorithm that repeatedly steps through a list of elements and compares adjacent elements, swapping them if they are in the wrong order. While bubble sort is not
efficient for large data sets, it can still find its usage in certain scenarios where simplicity and ease of implementation are prioritized over efficiency. Here are some areas where bubble sort may
be used:
1. Educational Purposes: Bubble sort is often used in computer science and programming courses as an introductory sorting algorithm. Its straightforward implementation helps beginners understand the
concept of sorting and algorithm analysis.
2. Small Data Sets: Bubble sort can be suitable for sorting small data sets, such as arrays or lists with only a few elements. In such cases, the performance difference between bubble sort and more
complex algorithms is negligible.
3. Partially Sorted Data: If the input data is already partially sorted, bubble sort can be relatively efficient. It has a best-case time complexity of O(n) when the input is already sorted, as it
only requires a single pass to confirm that the list is sorted.
4. Online Sorting: Bubble sort can be used in situations where new elements are continuously added to an already sorted list. Since bubble sort performs comparisons and swaps adjacent elements, it
can easily accommodate new elements while maintaining the sort order.
5. Teaching and Demonstrating Sorting Algorithms: Bubble sort is often used to illustrate the concept of sorting algorithms and their performance characteristics. It provides a simple and visual way
to demonstrate how sorting algorithms work and how their efficiency can vary.
Bubble Sort Algorithm in Programming Languages
The bubble sort algorithm can be implemented in any programming language.
Python Bubble Sort Algorithm
def bubble_sort(arr):
n = len(arr)
# Traverse through all array elements
for i in range(n):
# Last i elements are already in place
for j in range(0, n - i - 1):
# Traverse the array from 0 to n-i-1
# Swap if the element found is greater than the next element
if arr[j] > arr[j + 1]:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
return arr
# Example usage:
array = [5, 3, 8, 2, 1]
Javascript Bubble Sort Algorithm
function bubbleSort(arr) {
var len = arr.length;
var swapped;
do {
swapped = false;
for (var i = 0; i < len - 1; i++) {
if (arr[i] > arr[i + 1]) {
// Swap elements
var temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
swapped = true;
// Optimized: Reduce the number of iterations on each pass
} while (swapped);
return arr;
// Example usage:
var array = [5, 3, 8, 2, 1]; | {"url":"https://juniortoexpert.com/en/bubble-sort-algorithm/","timestamp":"2024-11-07T15:50:00Z","content_type":"text/html","content_length":"45875","record_id":"<urn:uuid:dee02c69-2c23-44c9-9cdb-1f374acb0642>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00662.warc.gz"} |
ACT Math: Pacing & Strategy Overview
The ACT Math Test is always the second test and comes right after the English Test. It is always 60 minutes long and consists of 60 questions. Concepts tested include arithmetic, algebra, coordinate
geometry, plane geometry, and (unfortunately) trigonometry. The math concepts get harder as you progress, so timing is an important part of the test. As you proceed, you must make sure not to spend
more than 1-2 minutes on any single question.
Once you are in the middle of the section and begin to encounter more challenging concepts, make sure you don’t let any one question frustrate you or waste your time. You’ll need a little extra time
to handle those challenging questions at the end, so make sure to move through the first third of the test quickly and confidently but don’t rush yourself.
If a question is taking you more than a minute, skip it and move on. You can always come back! If you run out of time at the end and don’t have time to come back to it, make sure to fill in something
on your answer grid. There is no wrong answer penalty on the ACT, so you definitely want to make sure to answer every single question!
In terms of general strategy, when you read each Math question, don’t just automatically reach for the calculator and start crunching numbers. Sometimes taking a few extra seconds to consider the
best approach to solve a problem will actually save you time in the long run. Many Math problems on the ACT can be solved in a few different ways and often one way will be much faster.
Be thoughtful and deliberate when reading each question stem. Look for patterns and shortcuts. If you are stuck on a problem and can’t find a way to approach it, study the answer choices. Eliminating
four wrong answer choices is just as effective as finding the right answer! Also look for opportunities to use one of the following two strategies:
Backsolving is an excellent strategy to use when there are numbers in the answer choices. Instead of setting up your own equation, assume each answer choice is correct. This is a great strategy to
use when you are stuck because it lets you check your work as you go! Let’s look at an example:
Claire took a fifty-question algebra test and answered every question, scoring a 10. Her teacher calculated the score by subtracting the number correct from three times the number incorrect. How many
questions did Claire answer correctly?
A. 30
B. 34
C. 36
D. 40
E. 42
Since this is a word problem and there are numbers in the answer choices, this question is a great candidate for Backsolving. Let’s start with answer choice (C) 36. Since the answer choices on the
ACT are ranked from smallest to greatest, starting with answer choice (C) will help us eliminate three answers at once.
If Claire answered 36 correctly, then she must have answered 14 incorrectly (because it was a 50 question-test). Three times the number incorrect (14) = 42. We then subtract the number correct (36)
to get a score of 6. However, the problem told us that her score was a 10. A 6 is too small, so we can eliminate choices (A), (B) and (C).
The next logical choice to try is (D). If she answered 40 correctly, then she must have answered 10 incorrectly. Three multiplied by 10 = 30. Then we subtract 30 from the number correct (40). Our
answer is 10, which matches the score mentioned in the question stem, so we know we have our correct answer!
Anytime you see variables in a question stem or in the answer choices, you can Pick Numbers. Let’s try an example question:
There are m students total in the classroom taking the ACT. n of those students will score above the 90% percentile. Which expression represents the number of students who will NOT score in the 90
1. n (m – n)
2. n + n
3. m / n + n
4. n (m – n) / n
5. m x n
Here we have two variables: m and n. We can pick numbers for those variables as long as they make sense within the context of the word problem. We know that m must be a bigger number than n since m
is the total number of students and n is a portion of that. Let’s say m = 4 and n = 2. I chose these numbers because they are low and easy to work with – remember that Picking Numbers should make the
math simpler. Why pick 4,567 for m when we can pick 4?
If m = 4 and n = 2, we know that 2 students will NOT score in the 90^th percentile. Let’s plug in our picked numbers into the answer choices and see which once yields 2! The answer here is (D).
Start looking for opportunities to Backsolve and Pick Numbers when you practice and remember, the smart ACT Math test-taker excels in both time-management and strategy!
0 0 admin http://wpapp.kaptest.com/wp-content/uploads/2020/09/kaplan_logo_purple_726-4.png admin2016-11-28 06:00:032020-09-11 20:42:32ACT Math: Pacing & Strategy Overview | {"url":"https://wpapp.kaptest.com/study/act/act-math-test-pacing-strategy-overview/","timestamp":"2024-11-05T10:41:10Z","content_type":"text/html","content_length":"189024","record_id":"<urn:uuid:ec6f9390-8c1f-458d-ac22-468e716b3508>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00091.warc.gz"} |
• modular form: (1) modularity, (2) holomorphic, (3) holomorphic at cusps.
How they are generalized or restricted ?
• cusp form: (3) vanishes at cusps.
• non-holomoprhic modular form(?): (2) real analytic, and eigenfunction of Casimir operator.
• weakly holomorphic modular form: (3) allowing finite-order pole at cusps.
• harmonic Maass form: (2) real analytic, and annihilated by weight-$k$-Laplacian. (3) at most linear exponential growth at cusps.
• weak Maass form: (2) real analytic, and eigenfunction of weight-$k$-Laplacian. (3) at most linear exponential growth at cusps.
• mock modular form: ``holomorphic part'' of harmonic Maass form.
• meromorphic modular form:
• weakly holomorphic cusp form: weakly holomorphic, and its Fourier expansion has no $q^0$ terms. (still may allow $q^n$ with negative $n$.)
• quasi-modular form
Generalization to another direction
• modular form with half integral weight
• Siegel modular form
• Jacobi form
• $M_k$ modular form
• $S_k$ cusp form
• $M^!_k$ weakly holomorphic modular form
• $S^!_k$ weakly holomorphic cusp form
• $QM_k$ quasi-modular form | {"url":"https://user.math.kyushu-u.ac.jp/?ochiai/mock","timestamp":"2024-11-12T05:44:09Z","content_type":"application/xhtml+xml","content_length":"6088","record_id":"<urn:uuid:e6a24008-59e8-44a0-b85b-9be75c36d0d1>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00252.warc.gz"} |
Motivating simple electric models (2013)
Working Content > Electric Field
Though much biochemistry and lots of cellular biology are the result of electrical forces, there are very few biological situations where forces between a small number of charges is what's going on.
And since Coulomb's law tells us that only the force between individual point charges has a simple 1/r^2 dependence, for any complex system, we have to add up the results of lots of different
charges. Even if we use electric potential energy instead of electric force (1/r dependence and no vectors) it can still be very complicated. Well, sometimes that's just what you have to do. And
computers can help (see, for example, The water-coat potential). But often, we can use one of a few simple models as an approximation. This can give us a good starting point for understanding a
complicated situation.
The three models that we can analyze in a pretty straightforward way (or, be more complicated and actually carry out the integrals) are (1) an infinite uniform line of charge; (2) an infinite uniform
plane of charge; (3) a spherical charge distribution.
Though of course we never have "infinite" distributions of charge, letting these become infinite makes the calculations easier and suppresses "edge effects" -- changes that occur when we get to the
end of a finite line or sheet of charge. And we will see just what the conditions are that let us treat a finite system as if it went on forever. These model distributions will much improve our
ability to discuss a variety of complicated situations.
Joe Redish 2/14/12
You don't have permission to comment on this page. | {"url":"http://umdberg.pbworks.com/w/page/73084682/Motivating%20simple%20electric%20models%20(2013)","timestamp":"2024-11-06T22:04:43Z","content_type":"application/xhtml+xml","content_length":"24139","record_id":"<urn:uuid:be963d7c-fba7-4731-b8ec-37ce56fbc85c>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00052.warc.gz"} |
How Much Does Transmit Correlation Affect the Sum-Rate of MIMO Broadcast Channels?
In this talk, we consider the effect of spatial correlation between transmit antennas on the sum-rate capacity of the MIMO broadcast channel (i.e., downlink of a cellular system). Specifically, for a
system with a large number of users n, we analyze the scaling laws of the sum-rate for the dirty paper coding (DPC) and for different types of beamforming transmission schemes. When the channel is
i.i.d., it has been shown that for large number of users n, the sum rate is equal to M*loglog(n) + M*log SNR where M is the number of transmit antennas.
When the channel exhibits some spatial correlation with a covariance matrix R, we show that this results in an SNR hit that depends on 1) the multiuser broadcast technique and 2) on the eigenvalues
of the correlation matrix R. In this talk, we quantify the hit resulting from spatial correlation to the sum-rate of DPC, beam-forming with channel whitening, random beam-forming, and deterministic
This is a joint work with Masoud Sharif (Boston University) and Babak Hassibi (California Institute of Technology) | {"url":"https://faculty.kfupm.edu.sa/EE/naffouri/talks/5.html","timestamp":"2024-11-03T09:45:25Z","content_type":"text/html","content_length":"3575","record_id":"<urn:uuid:58211444-ba53-4ed5-ae85-2e3dfa60e084>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00340.warc.gz"} |
Finding Initial velocity given certain variables
• Thread starter Ryang
• Start date
In summary, the problem is to find the magnitude and direction of the initial velocity of a baseball given its final velocity, displacement on the Y axis, and acceleration data. The equations
relevant to this problem are the magnitude and direction equations for initial velocity. The attempted solution involved finding the vertical and horizontal components of the initial velocity using
trigonometry and the kinematic equations. The correct solution involves setting up a system of equations using the displacement and final velocity equations and solving for time, then using time to
find the vertical component of the initial velocity.
Homework Statement
To Find initial velocity, given final velocity, and displacement on Y axis.
Hello. I've spent four hours trying different solutions to this following problem, none of which have worked. I've just started my undenominated science degree, and since I havn't started lectures
yet I've worked ahead in my physics book, to avoid getting stuck later on in the year. Help with workings would be much appreciated.Given:
"an baseball has velocity of 36 m/s, at an angle of 28 from the horizontal (x axis). The positive directions are to the right of the horizontal, and the top of the vertical axis. Ignoring air
resistance, find the magnitude and direction of the initial velocity of the object."
- The objects final position is displaced +7.5 metres on the Y axis from its original position. -- Final velocity is 36 m/s, 28 degrees down from the horizontal.
- Acceleration on the X axis is obviously = 0m/s
- Acceleration on the Y axis is = -9.80 m/s
The Unknown variables are the:
Magnitude of initial velocity.
Direction of initial velocity.
Homework Equations
These are the equations I imagine are relevant.
Initial Velocity
Magnitude = V[o] = √V[ox]^2 + V[oy]^2
Direction = Theta = Tan inverse (V[oy]/V[ox])
The Attempt at a Solution
Sorry, I'm not great with using the mathmatical symbols on this board. I'll do my best.
V = 36m/s
Vx and Vy form a right angle triangle to V.
36/Sin 90 = V[y]/ Sin 28
36 m/s (Sin 28) = V[y ]
Vy = 16.90
36^2 = 16.9^2 + 19.1^2
Vx = √19.1 = 4.37
- - - - - - - - - - - -
To find V[oy], I tried [oy] = + √(-36 sin 28)^2 - 2 (-9.80)(7.5) = 20.8
V[oy] = 20.80 m/s
V[ox] = 4.37 m/s
- - - - -- - - - - - - - - - - -
using pythagorases theorum for a right triangle, I found that the velocity V[0] was = 21.25, but my book says that the magnitude for the velocity should be:
Where am I going wrong?? I would really appreciate help, very sorry if I didn't provide sufficient workings but I think I did. And sorry if my inevitably idiotic error is frustrating for you to see,
I'm not very mathematically inclined, which is the reason I'm practicing so much. Thanks
Last edited:
I agree with your 16.9 for the Vfy, but I get
Vfx = 36*cos(28) = 31.79 instead of your 4.37.
You don't say how you got the initial speed. I used
d = Vit+½at² and Vf = Vi+at simultaneously to get the vertical Vi, which resulted in a combined initial speed of 38.
Thanks, I've redone the calculation and managed to get the Vf
and Vf
values correct. I've tried the d = Vit+½at² equation before, and got:
7.5m = (V[SUB]i[/SUB])(t) - 4.9 (t)[SUP]2[/SUP].
to find time I rejigged
a = 0.5 (V[SUB]ox[/SUB] + V[SUB]t[/SUB])t
to give
t = a/0.5(V[SUB]ox [/SUB]+ V[SUB]x[/SUB])
2t = (2)(-9.80)/(X + 16.9m/s)
But their is both an X and T variable in this equation, so I have no idea how to solve for V
- - - - - -
I also tried:
V[SUB]y[/SUB][SUP]2[/SUP] = V[SUB]oy[/SUB][SUP]2[/SUP] + 2 (a[SUB]y[/SUB])(y)
31.8[SUP]2[/SUP] = V[SUB]oy[/SUB][SUP]2[/SUP] + 2(-9.80)(7.5)
31.8 = √voy[SUP]2[/SUP] - √147
31.8 + 12.1 = V[SUB]oy[/SUB]
43.9 m/s = V[SUB]oy[/SUB]
the initial velocity is only 38m/s, and pythagoras's theorem implies that V
< V
. So this solution is false. Where am I going wrong.
I appreciate the help, thanks.
- - - - - - - - - - - - - -
Last edited:
7.5m = (Vi)(t) - 4.9 (t)²
This looks good. You also have
Vf = Vi + at
16.9 = Vi - 9.8t
If you solve the system of these 2 equations, you'll get Vi (vertical).
To solve
½(-9.80)(t) + (vi)(t) = 7.5m
vi = (16.9-9.8t)
-4.9t + (16.9-9.8t)(t) = 7.5m
quadratic equation
9.8t² + 12t – 7.5 = 0
{-b ± (square root)b² - 4ac} / 2a
- 12 ± (square root) 144 – 4 (9.8)(-7.5) / 2(9.8)
{12 ± 20.9}/19.6
1.6 or – 0.5
Vi = (16.9 – 9.8t)
Vi = (16.9 – 9.8 (1.6)) < = > 16.9 – 9.8 (1.6)) ≤ 32 m/s
Vi = 1.2
Looking at the quadratic equation, I reckon the solution for t lies somewhere around 0.15s, which is definitely too small a time period to be correct.
Gravity acts at -9.8m/s² in this case, and if the equation only ran for 0.5 seconds the ball wouldn’t even begin to move towards the negative y, given the initial velocity is 42m/s. It’d probably
take at least take 3 seconds for the Vy = 0.
The equation 9.8t² + 12t – 7.5 = 0 is obviously, non – linear, since when this thing essentially describes a differential curve, but I don’t now how to solve non linear equations. Are they solved
using matrixes or something??
Sorry for wasting so much time with this stupid question, but it is pretty important that I learn how to derive Vo’s from V’s, and visa versa, it’s the fundamental element of kinematics really. I
would trouble my lecturers, but we havn’t even started Vectors yet in Physics, and in a class of 400 they hardly want to be bothered by a student.
You have 7.5m = (Vi)(t) - 4.9 (t)² [1]
and Vf = Vi + at
-16.9 = Vi - 9.8t [2]
Solve [2] for Vi = 9.8t - 16.9 and sub in [1]:
7.5 = 9.8t² - 16.9 t - 4.9t²
0 = 4.9t² - 16.9t - 7.5
Once you get t, you can use [2] again to find the vertical Vi.
Solved it. Thanks a million for the help, I really appreciate it:D
FAQ: Finding Initial velocity given certain variables
1. What are the variables needed to find the initial velocity?
The variables needed to find the initial velocity are displacement, time, and acceleration. These variables can be represented by the equation v = u + at, where v is the final velocity, u is the
initial velocity, a is the acceleration, and t is the time.
2. Can initial velocity be negative?
Yes, initial velocity can be negative. This indicates that the object is moving in the opposite direction of the positive direction chosen in the equation.
3. How do you calculate initial velocity if acceleration is not constant?
If acceleration is not constant, you can still use the equation v = u + at, but you will need to use the average acceleration over a specific time interval instead of a single value. This can be
calculated by dividing the change in velocity by the change in time.
4. What if the object starts from rest?
If the object starts from rest, the initial velocity (u) will be 0. This means the equation v = u + at simplifies to v = at.
5. How does the angle of launch affect the initial velocity?
The angle of launch affects the initial velocity by changing the direction of the velocity vector. The initial velocity can be calculated using the components of the velocity in the horizontal and
vertical directions, which are determined by the angle of launch. | {"url":"https://www.physicsforums.com/threads/finding-initial-velocity-given-certain-variables.430427/","timestamp":"2024-11-09T13:19:06Z","content_type":"text/html","content_length":"112035","record_id":"<urn:uuid:edabcc05-1d86-4166-a253-8f74b50dc84c>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00054.warc.gz"} |
A car travels 348 kilometers in 7 hours (with a constant speed). How much time will it tak...
A car travels 348 kilometers in 7 hours (with a constant speed). How much time will it take traveling 396 kilometers?
STEP 1
1. The car travels at a constant speed.
2. The distance traveled in 7 hours is 348 kilometers.
3. We want to find the time it takes to travel 396 kilometers at the same speed.
STEP 2
First, we need to find the speed of the car. Speed is defined as the distance traveled divided by the time taken.
$Speed = \frac{Distance}{Time}$
STEP 3
Now, plug in the given values for the distance and time to calculate the speed.
$Speed = \frac{348\, kilometers}{7\, hours}$
STEP 4
Calculate the speed of the car.
$Speed = \frac{348}{7} = 49.714\, \text{kilometers per hour}$
STEP 5
We can round the speed to a more convenient number for calculation purposes. Since the speed is a constant, we can use this rounded speed to find the time for the new distance.
$Speed \approx 50\, \text{kilometers per hour}$
STEP 6
Now we need to find the time it takes to travel 396 kilometers at the speed of 50 kilometers per hour. Time is defined as the distance traveled divided by the speed.
$Time = \frac{Distance}{Speed}$
STEP 7
Plug in the values for the new distance and the rounded speed to calculate the time.
$Time = \frac{396\, kilometers}{50\, \text{kilometers per hour}}$
STEP 8
Calculate the time it takes to travel 396 kilometers.
$Time = \frac{396}{50} = 7.92\, \text{hours}$
STEP 9
Since we rounded the speed up to 50 kilometers per hour, the actual time will be slightly more than 7.92 hours. To find the precise time, we should use the exact speed we calculated earlier.
$Time = \frac{396\, kilometers}{49.714\, \text{kilometers per hour}}$
Calculate the precise time using the exact speed.
$Time = \frac{396}{49.714} \approx 7.96\, \text{hours}$
The car will take approximately 7.96 hours to travel 396 kilometers. | {"url":"https://studdy.ai/shared-solution/d0001451-6b57-47a0-b86a-5b91370b8fe6","timestamp":"2024-11-04T20:52:45Z","content_type":"text/html","content_length":"109030","record_id":"<urn:uuid:e4c7dfb5-b685-4cda-bfc0-b0158108f1a9>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00316.warc.gz"} |
The Stacks project
Lemma 52.16.9. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Let $A \to A'$ be a flat ring map. Set $X' = \mathop{\mathrm{Spec}}(A')$, let $U' \
subset X'$ be the inverse image of $U$, and denote $g : U' \to U$ the induced morphism. Set $(\mathcal{F}'_ n) = (g^*\mathcal{F}_ n)$, see Cohomology of Schemes, Lemma 30.23.9. If $(\mathcal{F}_ n)$
canonically extends to $X$, then $(\mathcal{F}'_ n)$ canonically extends to $X'$. Moreover, the extension found in Lemma 52.16.8 for $(\mathcal{F}_ n)$ pulls back to the extension for $(\mathcal{F}'_
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Equity & Fixed Income Securities - Custom Scholars
Equity & Fixed Income Securities
I was wondering if someone could help me with questions 3 and 4? I keep running formula’s but I the answers I’m getting just can’t be right.. Please review the ‘overview’ tab. I need the fx function
formulas for PV, FV and RATE for sure just regular formulas for the rest. Thanks so much!
Input boxes in tan
Output boxes in yellow
Given data in blue
Calculations in red
Answers in green
NOTE: Some functions used in these spreadsheets may require that
the “Analysis ToolPak” or “Solver Add-In” be installed in Excel.
To install these, click on the Office button
then “Excel Options,” “Add-Ins” and select
“Go.” Check “Analyis ToolPak” and
“Solver Add-In,” then click “OK.”
The concept of time value of
money and bonds go hand in hand.
In these examples, you will be able
to solve time value of money
problems with Excel.
These components are required
for each problem that you solve.
To begin a function, start with the
fx on your toolbar
Once you select fx
on your tool bar,
search for the PV
Once you select
on your tool bar,
search for the
ce you select fx
your tool bar,
arch for the PV
Once you select fx
on your tool bar,
search for the FV
Enter any payment amount.
Usually 0, or the coupon
payment if calculating bond
Once you select fx
on your tool bar,
search for the RATE
Make sure you
calculate the correct
number of
Sample problem
Present Value
Output area:
Input area:
Present value
Interest rate Future value
5% $
4% $
7% $
2% $
Future Value
Output area:
Input area:
Future value
Interest rate Present value
5% $
4% $
7% $
2% $
Interest Rate
Output area:
Input area:
Interest Rate
Years Future Value Present value
2,183 $
33,323 $
49,179 $
2,229 $
The present value will always come
out negative (red or in parenthesis) in
Excel. Think of it as you are taking the
amount out of your pocket right now,
so you would be in the negative!
The Piedmont Flyers have issued a 30-year bond with a face value of $100
The Piedmont Flyers have issued a 30-year bond with a face value of $1000
The Piedmont Flyers have issued a 8-year bond with a face value of $1000.
The Piedmont Flyers have issued a 30-year bond with a selling price of $75
ith a face value of $1000 and a 5% coupon rate. The market is yielding 7%, what is the current selling price
th a face value of $1000 that pays semiannually with a 4% coupon rate. The market is yielding 6%, what is
h a face value of $1000. The price of the bond is $880. The market rate is 10% annually. What is the coupon
th a selling price of $750 and a coupon payment of $100. The face value is $1000. What is the return on th
is yielding 6%, what is the current selling price of the bond? | {"url":"https://customscholars.com/equity-fixed-income-securities/","timestamp":"2024-11-08T18:25:30Z","content_type":"text/html","content_length":"54456","record_id":"<urn:uuid:0c1c3383-6f1c-49e8-acdd-615d8c5c772b>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00800.warc.gz"} |
In abstract algebra, a unital map on a C*-algebra is a map ${\displaystyle \phi }$ which preserves the identity element:
${\displaystyle \phi (I)=I.}$
This condition appears often in the context of completely positive maps, especially when they represent quantum operations.
If ${\displaystyle \phi }$ is completely positive, it can always be represented as
${\displaystyle \phi (\rho )=\sum _{i}E_{i}\rho E_{i}^{\dagger }.}$
(The ${\displaystyle E_{i}}$ are the Kraus operators associated with ${\displaystyle \phi }$). In this case, the unital condition can be expressed as
${\displaystyle \sum _{i}E_{i}E_{i}^{\dagger }=I.}$
• Paulsen, Vern I. (2002). Completely bounded maps and operator algebras. Cambridge: Cambridge University Press. ISBN 0-511-06103-X. OCLC 228110971. | {"url":"https://www.knowpia.com/knowpedia/Unital_map","timestamp":"2024-11-14T20:30:58Z","content_type":"text/html","content_length":"72923","record_id":"<urn:uuid:0bef0d37-5838-4f1f-975a-5afb704672bd>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00685.warc.gz"} |
Intersection of Two Linked Lists
Problem statement:
Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.
For example, the following two linked lists begin to intersect at node c1:
The test cases are generated such that there are no cycles anywhere in the entire linked structure.
Note that the linked lists must retain their original structure after the function returns.
Custom Judge:
The inputs to the judge are given as follows (your program is not given these inputs):
* intersectVal - The value of the node where the intersection occurs. This is 0 if there is no intersected node.
* listA - The first linked list.
* listB - The second linked list.
* skipA - The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
* skipB - The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.
The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be
Example 1:
Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Intersected at '8'
Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
- Note that the intersected node's value is not 1 because the nodes with value 1 in A and B (2nd node in A and 3rd node in B) are different node references. In other words, they point to two
different locations in memory, while the nodes with value 8 in A and B (3rd node in A and 4th node in B) point to the same location in memory.
Example 2:
Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Intersected at '2'
Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
Example 3:
Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: No intersection
Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary
Explanation: The two lists do not intersect, so return null.
* The number of nodes of listA is in the m.
* The number of nodes of listB is in the n.
* 1 <= m, n <= 3 * 104
* 1 <= Node.val <= 105
* 0 <= skipA < m
* 0 <= skipB < n
* intersectVal is 0 if listA and listB do not intersect.
* intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.
Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?
Solution in C++
Solution in Python
Solution in Java
Solution in Javascript
Solution explanation
The algorithm uses two pointers, one for each linked list head (headA and headB). The pointers traverse through the lists such that they traverse the sum of the node count of both lists. If the lists
intersect at a node, the pointers will eventually meet at this intersection node because they will have traversed the same number of total nodes. If the lists do not intersect, the pointers will meet
at the end of the lists, resulting in both pointers being null, and the function will return null. | {"url":"https://www.freecodecompiler.com/tutorials/dsa/intersection-of-two-linked-lists","timestamp":"2024-11-07T12:37:10Z","content_type":"text/html","content_length":"44338","record_id":"<urn:uuid:742eb5e2-0fe8-434b-9fb7-76bdc504d317>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00264.warc.gz"} |
€155,500 a Year After-Tax is How Much a Month, Week, Day, an Hour?
How much is €155,500 a Year After Tax in Ireland?
In the year 2024, in Ireland, €155,500 a year gross salary, after-tax, is €88,564 annual, €7,380 monthly, €1,697 weekly, €339.47 daily, and €42.43 hourly gross based on the information provided in
the calculator above.
Check the table below for a breakdown of €155,500 a year after tax in Ireland.
Yearly Monthly Weekly Daily Hourly
Gross Salary €155,500 €12,958 €2,980 €596.04 €74.51
Tax €51,440 €4,287 €985.87 €197.17 €24.65
USC €9,276 €772.99 €177.78 €35.56 €4.44
PRSI €6,220 €518.33 €119.21 €23.84 €2.98
Tax Credit €3,400 Per Year
Taxable Salary €152,100 Per Year
Take-Home Pay (NET) €88,564 €7,380 €1,697 €339.47 €42.43
If you're interested in discovering how much you would earn a year with an extra €100 added to your annual salary, you can explore the calculations for a €155,600 annual income. This will provide you
with a better understanding of the difference that additional €100 can make to your annual earnings.
€155,500 a Year is How Much a Month?
When evaluating a €155,500 a year after tax income, the corresponding monthly earnings can be determined:
• Take-home (NET) monthly income: €7,380
In order to discover €155,500 a year is how much a month? - simply divide the annual amount by 12, resulting in a monthly income of €7,380.
€155,500 a Year is How Much a Week?
When assessing a €155,500 a year after tax salary, the associated weekly earnings can be calculated:
• Take-home (NET) weekly income: €1,697
To answer €155,500 a year is how much a week? - divide the annual sum by 52, resulting in a weekly income of €1,697.
€155,500 a Year is How Much a Day?
When examining a €155,500 a year after tax income, the corresponding daily earnings can be determined:
• Take-home (NET) daily income: €339.47 (assuming a 5-day work week)
To find out €155,500 a year is how much a day? - divide the annual figure by 260 (52 weeks * 5 days), resulting in a daily income of €339.47.
€155,500 a Year is How Much an Hour?
When analyzing a €155,500 a year after-tax salary, the associated hourly earnings can be calculated:
• Take-home (NET) hourly income: €42.43 (assuming a 40-hour work week)
To answer €155,500 a year is how much an hour? - divide the annual amount by 2,080 (52 weeks * 40 hours), resulting in an hourly income of €42.43.
Is €155,500 a Year a Good Salary?
To answer if €155,500 a year is a good salary. We need to compare it to the national median. After calculation using ongoing year (2024) data, the salary of €155,500 a year is 3.53 times or 71.64%
higher than the national median. So, is €155,500 a year a good salary?
Based on comparison to the national median, yes, in our opinion, it is a very good salary in Ireland.
We think these three links are helpful and related to the €155,500 After Tax IE: Taxing Wages - Ireland, Tax rates, bands and reliefs, and Taxation in the Republic of Ireland. | {"url":"https://incomeaftertax.com/155500-after-tax-ie","timestamp":"2024-11-10T12:16:47Z","content_type":"text/html","content_length":"114737","record_id":"<urn:uuid:f376656a-2413-4452-9ccd-8f7310a58e50>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00464.warc.gz"} |
Free Online Rounding to the Nearest Hundred Thousandths Calculator with Steps
Example 1:
Round 106.456342 to its nearest hundred thousandths.
Given decimal is 106.456342
Here, 4 is the hundred thousandths digit and 2 is the next digit which is considered to round off the decimals to the nearest 100000th.
2 is less than 5 so round down to hundred thousandths by remaining the digit same in the fifth place after the decimal point and removing all the digits next to it.
Hence, the Rounded value of 106.456342 to its nearest hundred thousandths is 106.45634
Example 2:
What is the 127.267158 round-off decimal to the nearest 100000th?
Given decimal is 127.267158
To round off decimal to nearest 100000th, identify the digit in the fifth place after the decimal point. Here 5 is the digit at hundred thousandth place.
Now, look at the next digit which is 8. Here 8 is greater than 5 so round up the hundred thousandth value. Add 1 to the digit and remove all the digits.
Therefore, 127.26716 is the rounded decimal to the nearest 100000th. | {"url":"https://roundingcalculator.guru/rounding-to-the-nearest-hundred-thousandth-calculator/","timestamp":"2024-11-10T18:55:47Z","content_type":"text/html","content_length":"47527","record_id":"<urn:uuid:071ddea5-5b09-46cf-8e24-e94dae33baf8>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00823.warc.gz"} |
VA vs Watts: Difference and ComparisonVA vs Watts: Difference and Comparison
The current is the flow of electrons from a point. The current is measured to determine the power of the electricity being supplied by the wires.
The measure of current can be measured in many unit systems. Two of them included are the VA and Watts.
The current, when being transferred on a large scale, is measured in the unit Kilo Watt Hour.
Key Takeaways
1. Volt-amperes (VA) represent apparent power, while watts (W) indicate real power in electrical systems.
2. VA measures the total power in an AC circuit, including both active and reactive power, whereas watts only account for the active power consumed by a device.
3. Power factor, the ratio between real and apparent power, is essential to convert VA to watts or vice versa, which helps determine the efficiency of electrical systems.
VA vs Watts
The difference between VA and Watt is that VA refers to the apparent power, and Watt refers to actual power. The Watt can be calculated by adding all the current linearly, whereas the VA can not be
measured by just adding the values. Many other factors, like heat loss and current loss during conduction, need to be measured.
The VA stands for the Volts Ampere. It is used as a unit of measure for the apparent current.
The absolute value of current can not be measured because of the other factors. Like during transportation, the current produces a heat effect therefore, there is a loss of power.
Other factors like loss of current during distribution or due to leakage can also be seen. Thus, the absolute current can not be measured.
The Watt is also the measure of the current. It is the value of the real current. The Watt is used to measure the work done by the current.
Therefore, the current can be added linearly as the loss of power is not considered when we are calculating the work done by the power. Thus, the absolute value of work done by the current can be
Comparison Table
Parameters of VA Watts
Full-Form The VA stands for the Volt-Ampere The Watt is itself a full form of the abbreviation “W”.
Definition The VA is used to calculate the apparent power in the circuit. The Watt is used to calculate the work done by the current thus can measure the real power.
Formula The VA is the product of the Voltage and Current in the circuit. ( VA=V×I) The Watt can be measured by the product of the voltage and current in the circuit. (W=V×I)
Alternate Current The VA is applied to the Alternate current. The Watt is applicable on the Alternate current circuit.
Direct Current The VA is not feasible in the Direct Current circuit. The Watt is also applicable on the Direct Current circuit.
Uses The VA can be used to calculate the current of the product that can be conducted The Watt is used to measure the amount of electricity spent in a house or company. It can
through it if we know the voltage. provide an estimation of the bill too.
What is VA?
The VA stands for Volt-Ampere. It is used to measure the apparent current in the circuit.
VA is greater than or equal to the Watt. The VA is significant in knowing the sizes of the wires to be used in the circuit.
The VA is used in the Direct current circuit. The apparent values of the direct current circuit are almost equal to the real power measured in the Watt.
The VA can be calculated if we know the RMS current and voltage of the product. The product of the Voltage of RMS and the Current of RMS is the value of the VA.
The VA in a circuit can not be added linearly and know the whole value as during transportation there is a loss of power or some amount is even lost as heat during conduction.
The VA can be used in a circuit to know the value of the current that the wire can transport. In some circuits, due to overload, the wires or the products are affected.
Therefore, the VA current is used to measure the load a circuit wire can afford and therefore maintain the current to avoid any mishap.
What is Watts?
The Watt is used to measure the work done by the power. It is the real value of the current in the circuit.
The Watt can be used to measure current in both AC and DC circuits. The Watt can be added linearly in the circuit as the heat loss or leakage is not included in the Watt.
The Watt is calculated if we know the circuit’s voltage and current. The product of the circuit’s voltage and current equals the circuit’s power in watts.
The 1W is equal to the 1J/s. If we are calculating the Power and considering it, the average current value needs to be considered, as the Instantaneous current is always changing.
The Watt is used to calculate the power in the AC and DC circuits. Thus, the Watt provides an estimation of the power that is being utilized by the company or in the home.
This includes all the appliances that are spending electricity. This provides an estimation of the bill to be paid.
Main Differences Between VA and Watts
1. The VA is used to calculate the apparent current, and Watt is used to measure the real current value in the circuit.
2. The VA is used in the AC, while the Watt is used in both AC and DC.
3. VA is always greater than or equal to the Watt.
4. The Watt can be measured by adding linearly in a circuit, but this does not apply to VA.
5. The VA is used to measure the power consumed, while Watt is used to measure the work done by the current.
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Law or principle of bivalence - HKT Consultant
Law or principle of bivalence
Theory that every proposition is either true or false.
Possible objections are of two kinds.
First, can we decide what counts as a proposition in the relevant sense?
Second, might not the principle fail for some presumably genuine propositions; for example, ‘Jones was brave’ (where Jones died peacefully after a life entirely devoid of danger)?
The law of bivalence is not necessarily the same as that of excluded middle.
MAE Dummett, ‘Truth’, Proceedings of the Aristotelian Society (1958-59)
The lawsedit
For any proposition P, at a given time, in a given respect, there are three related laws:
For any proposition P, P is either true or false.
• Law of the excluded middle:
For any proposition P, P is true or ‘not-P’ is true.
• Law of non-contradiction:
For any proposition P, it is not the case that both P is true and ‘not-P’ is true.
Bivalence is deepestedit
Through the use of propositional variables, it is possible to formulate analogues of the laws of non-contradiction and the excluded middle in the formal manner of the traditional propositional logic:
• Excluded middle: P ∨ ¬P
• Non-contradiction: ¬(P ∧ ¬P)
In second-order logic, second-order quantifers are available to bind the propositional variables, allowing one to formulate closer analogues:
• Excluded middle: ∀P(P ∨ ¬P)
• Non-contradiction: ∀P¬(P ∧ ¬P)
Note that both the aforementioned logics assume the law of bivalence. The law of bivalence itself has no analogue in either of these logics: on pain of paradox, it can be stated only in
the metalanguage used to study the aforementioned formal logics.
Analogues of excluded middle are not valid in intuitionistic logic; this rejection is founded in intuitionists’ constructivist as opposed to Platonist conception of truth and falsity. On the other
hand, in linear logic, analogues of both excluded middle and non-contradiction are valid,^[1] and yet it is not a two-valued (i.e., bivalent) logic.
Why these distinctions might matteredit
These different principles are closely related, but there are certain cases where we might wish to affirm that they do not all go together. Specifically, the link between bivalence and the law of
excluded middle is sometimes challenged.
Future contingentsedit
Main article: Problem of future contingents
A famous example is the contingent sea battle case found in Aristotle’s work, De Interpretatione, chapter 9:
Imagine P refers to the statement “There will be a sea battle tomorrow.”
The law of the excluded middle clearly holds:
There will be a sea battle tomorrow, or there won’t be.
However, some philosophers wish to claim that P is neither true nor false today, since the matter has not been decided yet. So, they would say that the principle of bivalence does not hold in such a
case: P is neither true nor false. (But that does not necessarily mean that it has some other truth-value, e.g. indeterminate, as it may be truth-valueless). This view is controversial, however.
Multi-valued logics and fuzzy logic have been considered better alternatives to bivalent systems for handling vagueness. Truth (and falsity) in fuzzy logic, for example, comes in varying degrees.
Consider the following statement.
The apple on the desk is red.
Upon observation, the apple is a pale shade of red. We might say it is “50% red”. This could be rephrased: it is 50% true that the apple is red. Therefore, P is 50% true, and 50% false. Now consider:
The apple on the desk is red and it is not red.
In other words, P and not-P. This violates the law of noncontradiction and, by extension, bivalence. However, this is only a partial rejection of these laws because P is only partially true. If P
were 100% true, not-P would be 100% false, and there is no contradiction because P and not-P no longer holds.
However, the law of the excluded middle is retained, because P and not-P implies P or not-P, since “or” is inclusive. The only two cases where P and not-P is false (when P is 100% true or false) are
the same cases considered by two-valued logic, and the same rules apply.
Of course, it may be stated that bivalence must always be true, and that multi-valued logic is simply by definition a vague state of perception. That is, multi-valued logic is a convenient way of
saying, “This instance has not been observed in enough detail to determine the truth value of P.” In other words, if a pale apple is 50% red (where red is noted as P), then P can be said to be 100%
true, noting that bivalence makes little delineation as to the nature of not-P aside from the given, meaning that the apple might very well be 50% white as well (when white is noted as not-P),
meaning that P and not-P can both be true, but in separate instances, as they both exist as separate colours, which combine in a larger instance set in perhaps an unobservable, exceedingly subtle way
to create the shade of pale red. In this case, the apple might be set S, which consisted of P and not-P to greater or lesser or equal respective degrees, as long as it is acknowledged that P and
not-P are separate instances within a set instance. In this way, bivalence simply states that white cannot be red, and makes no claim about the colour of the set instance, to which is applied
multi-value logic, in which case multi-value logic is simply derivative of bivalence as well.
One thought on “Law or principle of bivalence”
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4x ^ 2 – 5x – 12 = 0 | Your Guide To Quadratic Equations
Quadratic equations are vital in both mathematics and real-world applications. The equation \(4x ^ 2 – 5x – 12 = 0\) is a classic example, showcasing how to find the values of \(x\) where the
equation holds.
This blog will walk you through several methods to solve this quadratic equation, including factoring, using the quadratic formula, completing the square, and graphing. By mastering these techniques,
you’ll be well-prepared to tackle a wide range of quadratic equations confidently.
What is 4x ^ 2 – 5x – 12 = 0?
What is 4x ^ 2 – 5x – 12 = 0?
The equation \(4x^2 – 5x – 12 = 0\) is a classic quadratic equation where the variable \(x\) is raised to the power of two. In algebra, quadratic equations are crucial as they model numerous
phenomena and problems. This specific equation falls into the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
Here, \(a = 4\), \(b = -5\), and \(c = -12\). Solving this equation involves finding the values of \(x\) that satisfy the equation, meaning they make the left side equal to zero. These values, known
as the roots of the equation, are critical because they represent key points where the function \(y = 4x^2 – 5x – 12\) crosses the x-axis.
Understanding and solving such equations not only aids in mathematical analysis but also in understanding real-world situations like predicting the optimal points of a parabolic path, calculating
maximum and minimum values in various contexts, and even optimizing designs and strategies in engineering and economics.
Step-by-Step Guide to Solving “4x^2 – 5x – 12 = 0”
Factoring Method:
Factoring is a technique where we express the quadratic equation as a product of two binomials. This method is proper when the equation can be easily broken down into factors.
1. Identify the coefficients \(a\), \(b\), and \(c\): In \(4x^2 – 5x – 12 = 0\), these are \(a = 4\), \(b = -5\), and \(c = -12\).
2. Find two numbers that multiply to \(ac\) and add up to \(b\): Here, we need numbers that multiply to \(4 \times -12 = -48\) and add up to \(-5\).
• List pairs of factors of \(-48\) (like \((1, -48), (-1, 48), (2, -24), (-2, 24)\), etc.) and check which pair sums to \(-5\).
• The correct pair should be \(8\) and \(-6\) because \(8 \times -6 = -48\) and \(8 – 6 = 2\). Since this doesn’t sum to \(-5\), we might reconsider our factoring method or move to another
3. Rewrite the middle term: Given that the correct pair doesn’t easily factor, we can try to split the middle term or opt for another method like the quadratic formula.
4. Factor by grouping: If we had a workable pair, we could rewrite \(bx\) using our pairs and then factor by grouping. For example, rewriting \(bx\) in \(4x^2 + bx + c = 0\) and factoring by grouping
could simplify the equation.
Given the complexity of finding straightforward factors, the quadratic formula or completing the square may be more efficient here.
Quadratic Formula:
The quadratic formula is a versatile method to solve any quadratic equation. It’s handy when factoring is challenging or not possible.
1. Identify the coefficients \(a\), \(b\), and \(c\): For \(4x^2 – 5x – 12 = 0\), these are \(a = 4\), \(b = -5\), and \(c = -12\).
2. Substitute into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]
Substitute \(a = 4\), \(b = -5\), and \(c = -12\):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 – 4 \cdot 4 \cdot (-12)}}{2 \cdot 4} \]
3. Simplify the expression under the square root (the discriminant):
\[ \Delta = (-5)^2 – 4 \cdot 4 \cdot (-12) = 25 + 192 = 217 \]
4. Calculate the roots:
\[ x = \frac{5 \pm \sqrt{217}}{8} \]
• This results in two solutions:
• \( x_1 = \frac{5 + \sqrt{217}}{8} \)
• \( x_2 = \frac{5 – \sqrt{217}}{8} \)
Completing the Square Method:
Completing the square is a method that rewrites the quadratic equation as a perfect square trinomial, making it easier to solve. It’s useful for equations that don’t factor neatly.
1. Rewrite the equation: Start by isolating the quadratic and linear terms on one side:
\[ 4x^2 – 5x = 12 \]
2. Divide by the coefficient of \(x^2\): Simplify by dividing the entire equation by 4 (the coefficient of \(x^2\)):
\[ x^2 – \frac{5}{4}x = 3 \]
3. Complete the square: Add and subtract \(\left(\frac{-5/4}{2}\right)^2\) inside the equation:
• Calculate \(\left(\frac{-5/4}{2}\right)^2 = \left(\frac{-5}{8}\right)^2 = \frac{25}{64}\).
\[ x^2 – \frac{5}{4}x + \frac{25}{64} = 3 + \frac{25}{64} \]
4. Simplify the equation: Combine the terms on the right:
\[ x^2 – \frac{5}{4}x + \frac{25}{64} = \frac{217}{64} \]
5. Express as a perfect square:
\[ \left(x – \frac{5}{8}\right)^2 = \frac{217}{64} \]
6. Solve for \(x\) by taking the square root of both sides:
\[ x – \frac{5}{8} = \frac{\pm \sqrt{217}}{8} \]
• This provides two solutions:
\[ x = \frac{5}{8} + \frac{\sqrt{217}}{8} \quad \text{and} \quad x = \frac{5}{8} – \frac{\sqrt{217}}{8} \]
\[ x_1 = \frac{5 + \sqrt{217}}{8} \quad \text{and} \quad x_2 = \frac{5 – \sqrt{217}}{8} \]
Graphical Method:
Graphing the quadratic function allows us to visually identify the roots, or the x-values where the function intersects the x-axis. This method provides a graphical understanding of the solutions.
1. Rewrite the equation as a function:
\[ y = 4x^2 – 5x – 12 \]
2. Plot the function: Use a graphing calculator or software to plot the function’s curve \(y = 4x^2 – 5x – 12\).
3. Identify the x-intercepts: Look for the points where the curve crosses the x-axis. These points correspond to the solutions of the equation \(4x^2 – 5x – 12 = 0\).
• The x-intercepts should match the roots calculated algebraically: \( x = \frac{5 \pm \sqrt{217}}{8} \).
4. Verify solutions: Use the graph to confirm the accuracy and reasonableness of the algebraic solutions. If the graph intersects the x-axis at the calculated points, it confirms the correctness of
the solutions.
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Common Problems Faced in Solving “4x^2 – 5x – 12 = 0”
Difficulty Factoring:
Factoring can be especially tricky when the quadratic equation involves more significant coefficients or non-integer solutions.
• Check for patterns: Before deciding that an equation can’t be factored, look for common factoring patterns like the difference of squares or perfect square trinomials.
• Use trial and error: Experimenting with different pairs of factors can help find the correct pair. Be patient and systematic.
• Alternative methods: If factoring proves too difficult, switch to the quadratic formula or complete the square, which is more generally applicable.
Complex Roots:
When the discriminant (\(b^2 – 4ac\)) is negative, the quadratic equation has complex roots, meaning the solutions involve imaginary numbers.
• Recognize the nature of the roots: If the discriminant is negative, expect the solutions to be complex.
• Use the quadratic formula: Calculate the roots using the formula, acknowledging that the square root of a negative number introduces an imaginary unit \(i\). For example, if \(\Delta\) were
negative, the solution would involve terms like \(\sqrt{-1} = i\).
• Practice with imaginary numbers: Familiarize yourself with handling \(i\) and complex number arithmetic to comfortably solve these problems.
Mistakes in Algebraic Manipulation:
Errors during algebraic manipulation can lead to incorrect solutions. These mistakes often occur in steps involving factoring, expanding, or applying the quadratic formula.
Mistakes in Algebraic Manipulation
• Double-check your work: Always review each step carefully. Use parentheses to keep track of negative signs and ensure you apply operations correctly.
• Work methodically: Take time with each step, and don’t rush through algebraic manipulations.
• Verify solutions: Substitute your solutions into the original equation to confirm they satisfy them.
Graphical Inaccuracies:
Graphing can sometimes be imprecise, especially when dealing with irrational or close roots.
• Use graphing tools: Advanced graphing calculators or software can provide more accurate plots and allow you to zoom in on the points of interest
• Combine methods: Use graphing to approximate the solutions and confirm with algebraic methods for exact solutions.
• Understand limitations: Recognize that graphical methods are helpful for visualization but might not always provide precise numerical answers.
Frequently Asked Questions (FAQs) About “4x^2 – 5x – 12 = 0”
What is a quadratic equation, and why is it important?
A quadratic equation is an equation where the highest power of the variable \(x\) is 2. It takes the general form \(ax^2 + bx + c = 0\), where \(a \neq 0\). Quadratic equations are crucial because
they model many real-world situations, such as projectile motion, optimization problems, and natural phenomena. Understanding how to solve them helps in predicting outcomes and analyzing various
How do I know if an equation is quadratic?
An equation is quadratic if the highest exponent of the variable \(x\) is 2. It will always fit the form \(ax^2 + bx + c = 0\) with \(a\) not equal to zero. If \(a = 0\), the equation reduces to a
linear equation.
What are the different methods to solve “4x^2 – 5x – 12 = 0”?
You can solve quadratic equations using:
Factoring: Expressing the quadratic as a product of binomials.
Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\) to find the roots.
Completing the Square: Rewrite the equation to make it a perfect square trinomial.
Graphing: Plotting the function and finding the x-intercepts where the function crosses the x-axis.
What if I can’t factor the quadratic equation?
If factoring is difficult or impossible, use the quadratic formula or complete the square method. These methods work for any quadratic equation, whether it factors neatly or not.
What are complex roots, and how do I deal with them?
Complex roots arise when the discriminant (\(b^2 – 4ac\)) is negative, resulting in the square root of a negative number. These roots are expressed in the form \(a + bi\) where \(i\) is the imaginary
unit (\(\sqrt{-1}\)). To deal with complex roots:
1. Use the quadratic formula and simplify the expression under the square root to include \(i\).
2. Understand basic operations with complex numbers to handle calculations effectively.
Can I solve quadratic equations without using algebra?
Yes, you can use graphical or numerical methods like Newton-Raphson. Graphing provides a visual representation and can approximate the solutions, while numerical methods use iterative techniques to
find approximate solutions. However, algebraic methods give precise answers and are fundamental for a deep understanding.
Why is it essential to double-check my work when solving quadratic equations?
Double-checking ensures that you haven’t made any mistakes in the algebraic manipulations or misused the methods. It’s easy to overlook calculation errors, especially with complex steps involving
square roots or multiple terms. Verifying your solutions by substituting them into the original equation helps confirm their accuracy.
How can I improve my skills in solving quadratic equations?
Practice regularly: The more you solve, the more familiar you become with different quadratic equations.
Review introductory algebra: Strengthen your understanding of algebraic principles and operations.
Study different methods: Learn and practice various methods to solve quadratics, including factoring, completing the square, and using the quadratic formula.
Check your work: Always verify your solutions to ensure they are correct.
Use educational resources: Online tutorials, videos, and interactive tools can provide additional practice and explanations.
Quadratic equations, such as \(4x^2 – 5x – 12 = 0\), are integral to algebra and frequently appear in academic and practical applications. Mastering various methods to solve these equations, like
factoring, the quadratic formula, completing the square, and graphing, equips you with valuable mathematical tools.
Whether you encounter straightforward or complex problems, understanding these techniques ensures you can find the solutions accurately and efficiently.
Bonus Tips for Quick Solutions:
1. Memorize the quadratic formula:
Having this formula at your fingertips lets you quickly solve any quadratic equation.
2. Check for simple factoring opportunities:
Before diving into more complex methods, see if the equation can be factored easily.
3. Use a calculator or graphing software:
These tools can verify your solutions and provide insights into the function’s behavior.
4. Practice solving different forms of quadratic equations:
Familiarity with various forms will make recognizing and applying the most efficient solution method easier.
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What is a multiplication chart?
A multiplication chart is a table that helps you learn and remember multiplication facts. It shows the products of numbers when multiplied together. For example, if you want to know what 3 times 4
is, you find the number 3 on one side of the chart and the number 4 on the other side. Where the row and column meet, you will see the answer, which is 12. It's a helpful tool for practicing and
understanding multiplication.
What is the history of the multiplication chart?
The history of the multiplication chart goes back a long time. People have been using multiplication for thousands of years. The ancient Egyptians and Babylonians used methods to multiply numbers. In
China, around 2,200 years ago, they created early multiplication tables. The multiplication chart as we know it today became popular in schools in the 19th century. It has been a useful way to teach
children multiplication and help them with math.
Who should use a multiplication chart?
A multiplication chart is good for anyone who wants to learn or practice multiplication. It is very helpful for kids who are learning how to multiply numbers. Kids can use a multiplication chart to
learn how to multiply numbers. It helps them see patterns and remember the answers. For example, they can quickly find out that 2 times 3 is 6 by looking at the chart.
At what age should kids start using a multiplication chart?
Kids can start using a multiplication chart around the age of 7 or 8, when they begin learning multiplication in school. However, younger kids can also benefit from seeing the patterns and practicing
with the chart.
How do you use a multiplication chart?
To use a multiplication chart, click on the product you want to find. The related factors (multiplicands and multipliers) will change color along with the product, making it easy to see the
relationship between the numbers. Additionally, the multiplication formula will be displayed below.
Can a multiplication chart help with division?
Yes, a multiplication chart can help with division. By knowing the multiplication facts, you can use the chart to see the relationships between numbers and solve division problems more easily.
How to generate a dynamic multiplication chart?
To generate a dynamic multiplication chart, you can append "/2-10" to your domain URL. This will create a multiplication chart for the range of numbers from 2 to 10. Similarly, adding "/3-13" to the
domain will generate a multiplication chart for numbers 3 to 13. Users can modify the URL to dynamically generate charts for any range between 1 and 100. For example, "/5-20" will create a
multiplication chart for numbers 5 to 20. This feature allows users to explore multiplication tables for any desired range within the specified limits. | {"url":"https://multiplicationschart.com/","timestamp":"2024-11-03T07:32:41Z","content_type":"text/html","content_length":"232159","record_id":"<urn:uuid:6124f3d1-ba42-4f6e-a12b-6d91e8863824>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00483.warc.gz"} |
Perpendicular Relation Between a Line and a Plane
012. Perpendicular Relation Between a Line and a Plane
Given a sphere $S$ with a radius of $2$ and two lines $l$ and $m$ which are not parallel, let $A$ and $B$ be the intersection points between $S$ and $l$, and let $P$ and $Q$ be the intersection
points between $S$ and $m$ as shown below. When $APQ$ is an equilateral triangle with side $2\sqrt{3}$, $\overline{AB} = 2\sqrt{2}$, and $\angle ABQ = \pi / 2$, find $\cos \theta$ where $\theta$ is
the angle between the plane $APB$ and the plane $APQ$.
With the given conditions, Pythagorean theorem, and the fact that $A$, $B$, $P$, and $Q$ are on $S$, the lengths of most lines are revealed. Let $O$ be the origin of $S$. Also, let $M$ be the
midpoint of $\overline{AQ}$. Then, it can be found that the equilateral triangle $APQ$ includes $O$ since $O$ is the centroid of this triangle. Besides, $BMO$ is a right triangle by Pythagorean
Here is the most important observation. The line $PM$ is perpendicular to the line $AQ$ and $BM$. When a line $l$ is perpendicular some two different lines which means they are not parallel, a plane
containing these two lines is also perpendicular to $l$.
This fact implies that the line $PM$ is perpendicular to $ABQ$. That is, $APB$ is projected into $APH$ where $H$ is the perpendicular from $B$ to the line $AC$. Therefore, $\cos \theta$ is the ratio
of these two triangles’ areas. For $APB$, $\overline{BP} = 2\sqrt{3}$ by Pythagorean theorem of $BMP$, so $APB$ is an isosceles triangle such that $\overline{PA} = \overline{PB}$. \begin{aligned} \
text{Area of } APB = \frac{1}{2} \cdot 2\sqrt{2} \cdot \sqrt{(2\sqrt{3})^2 - (\sqrt{2})^2} = 2\sqrt{5} \end{aligned}
For $APH$, find $\overline{BH}$ first. From the area of $ABQ$, \begin{aligned} \text{Area of } ABQ = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \frac{1}{2} \cdot 2\sqrt{3} \cdot \overline{BH} \implies \
overline{BH} = \frac{2\sqrt{6}}{3} \end{aligned}
As such, the area of $APH$ is \begin{aligned} \text{Area of } APH = \frac{1}{2} \cdot 3 \cdot \sqrt{(2\sqrt{2})^2 - \left(\frac{2\sqrt{6}}{3}\right)^2} = 2\sqrt{3} \end{aligned}
Therefore, $\cos \theta = 2\sqrt{3} / (2\sqrt{5}) = \sqrt{15}/5$. | {"url":"https://jeesunkim.com/maths/problems/012/","timestamp":"2024-11-08T17:11:01Z","content_type":"text/html","content_length":"101357","record_id":"<urn:uuid:48a6cbd7-42c0-4ede-ae90-c2d05c3f83a0>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00630.warc.gz"} |
OpenStax College Physics for AP® Courses, Chapter 5, Problem 8 (Problems & Exercises)
Show that the acceleration of any object down a frictionless incline that makes an angle $\theta$ with the horizontal is $a = g \sin{\theta}$ . (Note that this acceleration is independent of mass.)
Question by
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Final Answer
Please see the solution video.
Solution video
OpenStax College Physics for AP® Courses, Chapter 5, Problem 8 (Problems & Exercises)
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Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to show that an object sliding down a frictionless incline at an angle of Θ is going to have an acceleration equal to acceleration due
to gravity multiplied by sin Θ. So we have to draw a free-body diagram and there's going to be only... well, I guess there's the normal force here as well but it's only the gravity that's important
because this normal force does not get translated into any friction here as would normally be the case because we assume that the incline is frictionless. Okay! So we have gravity straight down— I
defined the coordinate system so that down the ramp is positive— and we have a y-component of gravity and we have a x-component of gravity or a perpendicular component and a parallel component and
it's this x-component that we are concerned with. And this angle in here I have labeled it as Θ but let's pretend that we don't really know what the angle is: let's call it β and I'll show you why it
equals this incline angle of Θ. So we know this is a right triangle here when we consider this triangle like here and with this being a right triangle, we'll call this angle here α and we'll figure
out what α is when we know that these three angles have to add up to 180 so that means Θ plus α plus 90 equals 180 in which case α is 180 minus this 90 which is 90 and then minus this Θ as well so
that's α. And then we know that α plus β has to work out to 90 because this dotted line is perpendicular to the incline so let's write that down: α plus this unknown angle β equals 90 and then we'll
solve for β by subtracting α from both sides and we have β then is 90 minus α. And now let's substitute in what α is from up here: it is 90 minus Θ. So we have 90 minus 90 minus Θ, the Θ becomes
positive, the 90's cancel and we are left with Θ. So this angle here is Θ so we don't have to call it β anymore, we can call it Θ now. Okay! So the x-component of this gravity is going to equal the
mass times acceleration because there are no other forces acting horizontally and this x-component of gravity then is the gravity F g multiplied by the sin of this angle Θ because we are trying to
find the opposite leg of this right triangle and we use sin of the angle multiplied by the hypotenuse to get it and this gravity is mass times acceleration due to gravity, mg. So we can replace F g x
with mgsin Θ equals to ma and then divide both sides by m and switch the sides around and we get the expression we wanted that acceleration equals g times sin Θ. | {"url":"https://collegephysicsanswers.com/openstax-solutions/show-acceleration-any-object-down-frictionless-incline-makes-angle-theta-0","timestamp":"2024-11-08T17:46:35Z","content_type":"text/html","content_length":"133519","record_id":"<urn:uuid:b5b4ae40-6071-43b8-97c5-e4902470f990>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00786.warc.gz"} |
[Solved] 1. CNA Industrial Group Corp. carries two | SolutionInn
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Verified Expert Solution
1. CNA Industrial Group Corp. carries two products. Total sales are estimated to be 10,000 regular + 5,000 large units. Fixed costs are $20,000.
1. CNA Industrial Group Corp. carries two products. Total sales are estimated to be 10,000 regular + 5,000 large units. Fixed costs are $20,000. Sales price of regular is $70 with variable expenses
of $50. Sales price of large is $100 with variable expenses of $55. a. What is the break-even amount? (10 points) b. How many units must be sold if the company wanted to have an operating income of
$30,000? (5 points)
There are 3 Steps involved in it
Step: 1
Weighted average CM per unit Total contribution marginTotal units ...
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March 3, 2014 - We're back! - The Daily Papert
March 3, 2014 – We’re back!
“Certain notions of mathematics are not sufficiently embedded in the culture for children to learn in their natural way, so they come to school to learn them. Once children are in school, we try
to impose mathematics on them in much the same way it was imposed on us: we begin by making them work at unimportant and uninteresting problems on little squares of paper. If before we ever
allowed children to dance we insisted that they spend hundreds of hours drawing dance steps on square papers, and only when they could pass a test on the ability to draw dance steps on paper
would we let them actually get up and dance, many children would find dancing impossibly difficult. Those gifted in dance would give up. And I think this is exactly what we do with mathematics.
We teach it to the children in a way quite analogous to drawing dance steps on paper, and only those who can survive twelve years of that ever get to use it, to dance with it.”
Papert, S. (1984). Computer as Mudpie. Intelligent Schoolhouse: Readings on Computers and Learning. In D. Peterson. Reston, VA, Reston Publishing Company.
The Daily Papert is a service of Constructing Modern Knowledge, the world’s premiere educational event for educators to learn-by-doing. Learn more about this year’s institute at
5 thoughts on “March 3, 2014 – We’re back!”
1. Fred
With dance you can get up and “do it” but you might be doing it wrong (even if you yourself as a child are having fun) and learning to dance well once you realize what you are doing is just
randomly jumping around takes a lot of time and repetition. Is the equivalent with math to just play with numbers and symbols without worrying if it is wrong or even meaningless until you get to
a point where you want to do it the right way? What would be considered important and interesting problems that could be solved correctly without learning and building up the basic foundations?
Is anyone doing that now 30 years later?
My experience with young students is not that they are either right or wrong, but rather they come with misconceptions. My first misconception was thinking that since 2+2 was 4 and that 2X2
was also 4, that multiplication and addition had the same rules. I soon discovered that was not the case so that misconception went away to be followed by other misconceptions or as Papert
would say I encountered “bugs” in my thinking. Debugging is the most important skill that a young person should have.
2. Joey
My two boys, ages 8 and 12, are learning how to code in Scratch and having a marvelous experience. Computational thinking skills and strategies can be developed by clicking & dragging blocks and
seeing the results of their choices. Making mistakes & learning how to “dance” is safe in this environment and they both want to dig deeper to learn more so they can complete a game or storyline
that they are developing for others to view.
3. German
Welcome back!!. We missed the daily food for thougth.
Seymour predicted educational achievements, that only years after, the technologies are making possible….
Fred’s question is a good one, basically asking Papert to elaborate on the math side of the dancing analogy.
One example is to have a class of students pick any three whole numbers they want and a) add them together and b) multiply them together. We write up everybody’s numbers on two different sides of
the boards and start looking for patterns. Someone notices that all the sums are divisible by three. Someone notices that all the products are divisible by 2 and by 3. Someone ELSE notices that
they’re divisible by 6. Students pick more sets of numbers and add and multiply them to see if they can break these patterns.
Once we’re satisfied, we prove. We’ve taken the number play, the investigation, the pattern-seeking, and locked it down tighter. It’s not enough to have fun noticing, the mathematician says, we
have to know WHY.
That’s one example of how the play can turn into practice. | {"url":"https://dailypapert.com/march-3-2014-were-back/","timestamp":"2024-11-04T20:23:22Z","content_type":"text/html","content_length":"120787","record_id":"<urn:uuid:49cb47b9-d54b-4f65-bc05-aefbdb56a075>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00441.warc.gz"} |
Development of Laryngeal Stroboscopic Effect With Continuous Light Source
The visualization of the vocal folds is important for understanding the physiology of voice production and the pathophysiology of vocal fold disorders [
]. The high-quality static images are able to evaluate the mucosal status and the vasculature of the vocal folds, providing valuable information on vocal fold pathologies [
]. In addition, dynamic images of vocal fold vibrations are important for understanding the physiology of voice production and the pathophysiology of vocal fold disorders [
]. Since Schonharl described the stroboscopic visualization of vocal fold vibrations in 1960 [
], videostroboscopy has been the gold standard for laryngeal imaging, and has been the primary method used to evaluate vocal fold vibrations [
Currently, in addition to videostroboscopy, various other tools with different mechanisms have been introduced for visualizing vocal folds vibration: videolaryngoscopy, videokymography (VKG), high
speed videolaryngoscopy, digital VKG, and two dimensional (2D) VKG [
]. As each tool provides unique information to the laryngologists, two or three methods are often used to evaluate the vocal folds in order to differentiate between voice disorders. Each method has
its own advantages and disadvantages and is able to complement videostroboscopy. However, the application of multiple modalities could cause discomfort to patients and may require considerable time
to complete all examinations. If two or three modalities could be technically joined as one unit, these problems would be improved upon.
Most modalities used a continuous light source, but modern videostroboscopy has adopted a stroboscopic flashing light that is triggered externally by fundamental voice frequencies. This difference of
light source would be an obstable for the integration among the imaging modalities. Also, the flickering light can hinder the examination of tiny lesions, vascular lesions, or suspected malignancies
owing to biases in color and brightness. If laryngeal stroboscopy could be implemented with continuous light sources, it would be better for the image quality and the compatibility with other
Thus, in this study, we established the mathematical algorithm for stroboscopic effect with continuous light and tried to determine the feasibility of laryngeal stroboscopic effect with continuous
illumination in the mechanical model.
Fan motor system
A frequency-adjustable fan motor (Green Electronics, Busan, Korea) system was designed for the purpose of this study, and was connected to the digital multi-meter in order to measure the rotational
frequency. Following this, the letter A and two lines were written on the fan motor. The fan motor is capable of 180 rotation per second (Hz).
Equipments for laryngeal stroboscopic effect with continuous light
A full high definition (HD) charge-coupled device (CCD) camera system (USC-700 HD, U-medical, Busan, Korea) was used to capture images of the rotational fan motor and the human vocal folds. A full HD
CCD image sensor (1920×1080 pixels) and a global shutter camera were connected. A rigid endoscope (7.2 mm, 70 degrees, 8706CA E, Storz, Tuttlingen, Germany) and a 300 W xenon light source (NOVA 300,
Storz, Tuttlingen) were assembled, and the system was used to visualize the vocal folds in their entirety. The camera was used to capture the rotational fan and the entire human vocal folds and the
video was recorded at 25 frames per second. The shutter speed of the camera could be adjusted from 1/25 to 1/50000.
Establishment of the mathematical model for laryngeal stroboscopic effect with continuous light source
In this study, we conceptualized the laryngeal stroboscopic effect using continuous illumination to integrate laryngeal videostroboscopy with other modalities. To theoretically explain this
phenomenon, the main author, who is a laryngologist (W. Cha), suggested the mathematical model including sampling rate and fundamental frequency for stroboscopic effects.
1) The movement of vocal fold mucosal waves is a cyclic movement and was hypothesized to be a rotational cycling model to establish a new algorithm for laryngeal stroboscopic effect using a
continuous light (
Fig. 1
2) The sampling rates are defined as the rate of image acquisition in a camera, and the frequency of rotation (F) is defined as the number of rotations per seconds in a rotational object:
S (frames/sec)=Sampling rateF (Hz, cycle/sec)=Frequency of rotation
3) The sampling interval is the number of frames divided by sampling rate. A single sampling interval is calculated as the inverse of sampling rate:
Sampling interval (sec)=(the number of frames)/ST1 (sec)=one sampling interval=1/S
4) Phase 0 is defined as the initial phase of a rotating object, and its value is set as 0 degree. Phase T[1] is defined as the phase of the first sample of image, which is 360xF multiplied by a
single sampling interval, and could be calculated as 360 multiplied by F divided by S:
Phase 0 (°)=0Phase T1 (°)=360 F T1= 360 F/S
5) Finally, the phasic difference (ΔT[1]) between Phase 0 and T[1] is defined as Phase T[1] modulated by 360, or the remainder of Phase T[1] divided by 360 Hz. It can be calculated as 360 Hz over
sampling rate, modulated by 360:
ΔT1 (°) =the phasic difference between Phase 0 and Phase T1=Phase T1 mod 360 =(360 F/S) mod 360 (mod: modulo operation, the remainder after division)
6) If a minimum of 10 sequential phases were mandatory for realizing the stroboscopic effect, the value of ΔT[1] should range from -36 degrees to +36 degrees. Theoretically, the phasic difference the
optimal frequency range is (multiples of S)±(S/10) and the optimal range is S/5 (Hz):
The optimal frequency ranges for stroboscopic effect=xS±(S/10) (x: integers)
Validation of laryngeal stroboscopic effect using continuous light in a fan motor model
On the basis of the mathematical model, the optimal frequency was expected as 25x±2.5 in our setting. The images were acquired with the camera system at various rotational speeds (frequencies) of the
fan motor model (
Fig. 2
). The serial images of the fan, acquired at approximately 99 Hz using continuous light (shutter speed=1/2000; sampling rate=25 FPS), are presented in
Fig. 3A
. The images of the fan rotated at slow and consistent speeds (approximately 17.8 degrees per frame) are presented in
Fig. 3B
It is important to examine the vibrations of vocal fold mucosa in order to understand the mechanisms of voice production, and to diagnose various vocal fold disorders [
]. Historically, laryngeal videostroboscopy has been the standard tool for vocal fold imaging in clinical practice. However, it cannot be applied to the movement of vocal folds with irregular
phonatory cycles. Various modalities, such as high speed imaging systems, VKG, and 2D VKG have been introduced to complement laryngeal videostroboscopy [
]. Imaging modalities for observing vocal fold mucosal movements have their own advantages and disadvantages. Laryngologists sometimes require several imaging tools to visualize the vocal folds in
order to diagnose and differentiate various vocal fold pathologies. This also means that patients might have repeated nuisance and painful laryngological examinations. Patient-oriented and
user-friendly approaches have recently been trending in medical technology. Several researchers reported the simulataneous uses of two imaging modalities for the clinical convenience [
]. In clinical practice, the integration of multi-modalities for vocal fold evaluation would be an innovation that could make laryngoscopic examinations more time-efficient for physicians and more
comfortable for patients. However, laryngeal videostroboscopy uses a special light source which flashes and flickers [
]. It is likely that the light source of stroboscopy is a big technological obstacle when being integrated with other modalities. Thus, the standardization of light sources is assumed to be an
essential prerequisite for integrating multiple imaging modalities.
In past, it was difficult to acquire the clear image of vibrating vocal folds due to fast movement. Classically, videostroboscopic systems have adopted the periodic light emission method,
synchronizing with the fundamental frequency of vocal fold mucosal vibrations. Since laryngeal videostroboscopy was developed, the technologies of light sources, optics, and camera systems have
remarkably advanced [
]. High power light source and camera technology enables to obtain the clear image within very short period. Thus, we postulated that laryngeal stroboscopic effect using continuos light source would
be feasible with the coupling of the fundamental frequency of the vocal folds and the sampling rate of camera. The mathematical model was established for laryngeal stroboscopic effect. Theoretically,
if the phonated frequency would not range within the optimal range (xS±[S/10]), the images be captured at the unexpected and irregular phases rather than at the slow and continuous serial phases. Our
model could explain the phenomenon of irregular stroboscopic effects observed in videolaryngoscopy with continuous lights
To validate the mathematical model, a rotating fan motor was used. The shutter speed should be shorter than the rotating cycle in order to capture clear images of rotating fans. However, the captured
images could be too dark to be distinguished if the light source is weak or the shutter speed is short. Thus, the images were captured on the conditions of various shutter speeds using high power
light source (300 W). The optimal shutter speed was 1/2000 s and the camera setting was applied in human experiments. The characters and lines on the fans were fixed in the rotating frequencies
(e.g., 75 Hz, 100 Hz, and 125 Hz) of multiples of the sampling rate (25 FPS), and the two aliasing fans were seen in the frequencies (e.g., 87.5 Hz, 112.5 Hz, and 137.5 Hz) between multiples of the
sampling rate. Stroboscopic effects could be examined in the rotating frequencies between 97.5 Hz and 102.5 Hz on the sampling rate of 25 FPS. Our mathematical model could explain this phenomenon. On
the frequency of 99 Hz, the stroboscopic effects could be seen and the phasic difference of 17.8 degree was measured in the image analysis. This result is approximately identical with the calculated
phasic difference (14.4 degrees) by the model.
In this study, we could establish the mathematical model to explain the phenomenon and verify the feasibility of laryngeal stroboscopic effect using a continuous light. This result provides the
possibility for the integration of laryngeal stroboscopy with the other modalities using a continuous light. However, in this study, there was the limitation that the effect of stroboscopy could only
be observed at specific frequencies (25x±2.5Hz, as calculated with the mathematical model), and auditory feedback was necessary to phonate the targeted frequencies. To overcome this problem, we are
currently developing a new camera system that can modify the sampling rate with collaborating engineers. There are two possible methods for better visualizing vocal fold vibrations using continuous
light. The first method is to automatically set the optimal sampling rate according to the real-time analysis of the images. We coined this algorithm as ‘internal triggered mode’. The second method,
which is the classical method, is to automatically set the optimal sampling rate according to the fundamental frequency obtained with external devices such as a microphone, electro-glottography, and
a vibration sensor. This algorithm would be termed as ‘external triggered mode’. Further studies for the system applied with these algorithms will be followed. Moreover, utilizing these mathematical
model, we aim to undertake a comparative scrutiny of vocal fold vibration imagery obtained from male and female cohorts using traditional stroboscopic apparatus, thereby elucidating their respective
merits and demerits.
In some local ENT clinics, the absence of stroboscopic equipment has prompted exploration into alternative methods for observing vocal fold vibrations, such as adjusting shutter speed and voice
frequency. This adaptation may aid in early diagnosis and intervention in various cases. There’s potential for flexible endoscopy to replace traditional laryngoscopy and stroboscopy, pending
sufficient light sources. This transition could reduce hospital setup costs and offer patients time, cost savings, and greater convenience by eliminating the need for repeat invasive examinations.
Laryngeal stroboscopic effect with continuous light source might be feasible. The stroboscopic effect with continuous light would be expected to provide greater compatibility to integrate with the
other imaging modalities for the vocal folds. | {"url":"https://jkslp.org/journal/view.php?number=2228","timestamp":"2024-11-11T00:33:10Z","content_type":"application/xhtml+xml","content_length":"88990","record_id":"<urn:uuid:0c0e7e4a-9bc7-405d-9548-c43ea8d4b3e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00325.warc.gz"} |
Area of a Circle Calculator
Area calculation is one of the most basic measurements done by Mathematics students. Being one of the most basic figures of Geometry, it is important for every student to learn how to find the Area
of a Circle. The process is pretty simple if you know about what this figure is, its parts, and the formula to find the area.
But if you don’t know about it deeply, you can also use the Area of a Circle calculator. To find the area with the help of this Math Calculator, you only need to copy the given input values and
insert them. The tool will automatically perform the calculations and give you the final answer. Here we will guide you on how to use this tool as well as explore the details of this specific
Geometrical figure.
Area of a Circle Definition
In Mathematics, the term Area is used to show the region covered by a specific figure or object. The Area of a Circle means the region that is covered by a circle when drawn on paper or in a
real-life field. In simple words, the region that comes under the boundaries of the circle will be termed its area.
This calculation doesn’t only important for subjective purposes but also has great importance in professional study. It is common to find the area of a specific field designed in a circular shape.
For this purpose, many students look for a handy tool like the area of a circle calculator that can perform this calculation quickly. If you want to know about it deeply for manual calculation, the
upcoming sections will help you a lot.
Parts of a Circle
Like other Geometrical figures, a Circle also has specific parts that we are going to discuss here.
• Center: It is a fixed point that is equidistant from all points of the boundary of the circle.
• Radius: The distance of any point on the boundary of the circle from its center is called Radius.
• Circumference: The boundary of the circle is called Circumference in terms of Mathematics.
• Diameter: It is the distance between two points on the circumference when the joining line is passing from the center of the circle. A Diameter is the largest line (chord) of any circle.
• Arc: Any specific part of the complete circumference is called an Arc.
• Chord: A line connecting two points on the circumference of the circle without passing through the center is called a Chord.
These are some important parts of the circle that you should know to perform different calculations like Area calculation. Keep in mind you need to have basic knowledge about them even when you are
using an Area of a circle calculator.
Formula of Circle Area
Depending on the above-mentioned parts, the Area of a Circle has multiple formulas. Here we have enlisted some important and most-used formulas for this calculation.
Area of a Circle (For Radius):
Area = 𝞹 x r^2
Area of a Circle (For Diameter):
Area = 𝞹 x r^2
Area of a Circle (For Circumference):
Area = C^2/4𝞹
You can use any of these formulas according to the given data to find the area of the circle.
How to Calculate the Area of a Circle?
To find the area of a circle, you only have to put the values in the above-given formulas. Let us share a few examples with you for better understanding.
Example 1:
Find the Area of a circle if its radius is 4 cm.
As we know,
Area = 𝞹 x r^2
= 3.14 x (4)^2
= 50.27 cm^2
Example 2:
Find the Area of a Circle if its Diameter is 16 cm.
As we know,
Area = 𝞹 x (d/2)^2
Area = 3.14 x (16/2)^2
= 3.14 (64)
= 201.06 cm^2
How to use the Area of a Circle Calculator?
Calculator’s Bag has designed one of the finest tools to find the Area of a Circle. Follow these steps to find the Area of a circle with specific values.
• Insert the given measurement (Radius or Diameter)
• The tool will automatically calculate the Area and give the answer
FAQ | Area of a Circle
What are the 2 formulas for the area of a circle?
To find the Area of a circle, the 2 most used formulas are,
Area = 𝞹 x r^2 (When the radius is given)
Area = 𝞹 x (d/2)^2 (When the diameter is given)
How do you find an area with a diameter?
To find the Area of a circle if the diameter is given, we need to use the following formula. Area = 𝞹 x (d/2)^2
How do you find an area with a radius, diameter, and circumference?
To find an Area using radius, diameter, and circumference, you need to use the following formulas respectively.
Area = 𝞹 x r^2
Area = 𝞹 x (d/2)^2
Area = C^2/4𝞹
Is the formula for circumference and area the same?
No, circumference is the boundary of a circle while the area is the whole region that comes under a circle. Both quantities have different formulas and methods to calculate them.
How to find the part of the area of a circle?
It is not possible to find a specific part of the area of a circle. But you can find the measurements for different parts of the circle like radius, diameter, and chord using different approaches. | {"url":"https://calculatorsbag.com/calculators/math/area-of-a-circle","timestamp":"2024-11-12T19:28:20Z","content_type":"text/html","content_length":"51506","record_id":"<urn:uuid:914b04c0-9b1c-4a36-8284-7dfb46cc585a>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00542.warc.gz"} |
Diffusion on fractal lattices and the fractal Einstein relation for Journal of Physics A: General Physics
Journal of Physics A: General Physics
Diffusion on fractal lattices and the fractal Einstein relation
View publication
Finitely ramified fractal lattices show anomalous diffusion with (r 2) varies as t2H. There is a hierarchy of transit times which are shown by Monte Carlo simulation to satisfy ( tau n) varies as
alpha n for large n, where alpha =b1H/ and b is the base of the lattice. The lattice resistivity scales with size as rho varies as Rn where R is characteristic of the lattice. Exact renormalisation
group methods yield alpha and R analytically and verify the Einstein relation for fractals alpha =RN. | {"url":"https://research.ibm.com/publications/diffusion-on-fractal-lattices-and-the-fractal-einstein-relation","timestamp":"2024-11-08T08:32:07Z","content_type":"text/html","content_length":"69449","record_id":"<urn:uuid:248ada7b-f832-4e13-b949-d596f3215e61>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00029.warc.gz"} |
Round to the nearest whole dollar. \(\$ 908.23\)
Short Answer
Expert verified
The dollar amount rounded to the nearest dollar is \( \$ 908 \).
Step by step solution
Locate the Dollar and Cent Separation
First, identify the decimal point which separates the dollars from the cents in the amount \( \$ 908.23 \). The dollar part is 908 and the cent part is 23.
Apply Rounding Rules
Next, apply the standard rule for rounding: If the first digit after the decimal point (representing cents in this context) is 5 or more, round the dollar amount up. If it's less than 5, keep the
dollar amount as is. Here, the digit after the decimal point is 2, which is less than 5.
Final Rounded Amount
Since the cent amount was less than 5, the dollar amount remains as 908. Therefore, the rounded amount to the nearest dollar is \( \$ 908 \).
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decimal Point Rounding
When you come across the need to round a number with a decimal, understanding the process of decimal point rounding is crucial. This is often encountered in situations where precision is not as
important as having a neat, whole number. For instance, take the number \(\$ 908.23\). Here, it is essential to pinpoint the decimal's location - it's the dot that splits the dollar amount from the
cents. In this case, \(908\) is the whole dollar amount, and \(23\) represents the cents.
Rounding to the nearest whole number means assessing the digit immediately after the decimal point, which is \(2\) in our example. According to rounding rules, since \(2\) is less than \(5\), the
amount remains \(\$ 908\) when rounded to the nearest whole dollar. This method is consistent for any numbers where you are rounding to the nearest whole unit, be it currency or units of measure. The
key takeaway is pinpointing the digit after the decimal and determining how it affects the whole number to its left. An understanding of this concept ensures accuracy in everyday situations like
making a budget or adjusting recipe measurements.
Rounding Rules
The mechanism for rounding numbers is governed by a simple set of rules that determine whether to round up or to keep the number the same. These rules hinge on the digit located immediately after the
digit you are rounding to. If this digit, also known as the rounding digit, is \(5\) or greater, you round up. If it's \(4\) or less, you maintain the current digit.
In the case of rounding \(\$ 908.23\) to the nearest whole dollar, the key digit to look at is the cent value \(23\). The first digit of this value is \(2\), which is less than \(5\). Therefore, the
rules state that you keep the dollar amount the same, resulting in a rounded figure of \(\$ 908\). These principles are universal, applying to all manner of rounding whether you're dealing with
money, measurements, or academic grading. For practical understanding, envision a number line where each whole number has a midpoint; numbers that cross this midpoint go up to the next whole number,
while those that do not reach it stay put.
Whole Dollar Rounding
Translating an amount with cents into a rounded whole dollar figure simplifies financial documentation and mental calculations, which is where the concept of whole dollar rounding comes into play.
This is particularly useful when large sums of money are involved or when a rough estimate is more valuable than an exact figure. It's also standard practice in accounting to present figures in whole
dollars for cleaner reporting.
Let's illustrate this with our example of \(\$ 908.23\). After acknowledging the digit after the decimal (\(2\)), and applying the rounding rules, we know that the amount does not have to be adjusted
upward. Hence, the whole dollar rounding results in \(\$ 908\). The process disregards cents and focuses on presenting a rounded number that is easier to work with for budgeting or when providing
estimates. By mastering whole dollar rounding, students and professionals can streamline their calculation processes and ensure clarity in financial communications. | {"url":"https://www.vaia.com/en-us/textbooks/math/algebra-1-concepts-and-skills-1-edition/chapter-5/problem-73-round-to-the-nearest-whole-dollar-90823/","timestamp":"2024-11-03T15:28:57Z","content_type":"text/html","content_length":"247258","record_id":"<urn:uuid:32423717-ad1f-4ef7-ac88-5b4d7e84be4b>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00432.warc.gz"} |
Aya is a stack based programming language originally intended for code golf and programming puzzles. Unlike other stack-based programming languages, it supports user-defined types, macro-like
function definitions, key-value pair dictionaries, and natural variable scoping rules allowing for more complex programs and data structures.
Aya comes with a rapidly-growing standard library written entirely in Aya code. The standard library features types such as matrices, stacks, dates and more. It also features hundreds of functions
for working working on numerical computations, strings, plotting and file I/O. It even features a basic turtle library for creating drawings in the plot window.
Aya also features a minimal GUI that interfaces with Aya's stdin and stdout. The GUI features, plotting, tab-completion for special characters, and an interactive way to search QuickSearch help data.
Basic language features
Aya is a stack based language. The code
1 1 +
will return "2".
All lowercase letters are used for variables. The colon (:) operator is used for assignment.
.# This is a line comment
"Hello" :first
"World!" :snd
Almost everything else is an operator. The :P operator will print the item on the top of the stack to stdout.
first " " + snd + :P
Blocks are used to define functions.
4 double .# will return 8
Blocks may have arguments and local variables. In the example below, a, b, and c are arguments and x, y, and z are local variables.
{a b c : x y z,
[a b c] .# a list with a, b and c inside
the following will call "myfun" and assign 1 to a, 2 to b, and 3 to c within the scope of the function. It will return the list [1 2 3].
1 2 3 myfun
Aya also supports dictionaries. {,} creates an empty dictionary. The operator .: is used for dictionary assignment.
{,} :dict
3 dict.:x
dict.x .# returns 3
Variables may also be assigned within the dictionary literal
{, "hi":a 4:b }:dict
dict.a .# returns "hi"
dict.b .# returns 4
Additional features are outlined in the wiki
Standard library
The Aya standard library consists of type definitions, mathematical functions, string and list operations, plotting tools and even a small turtle graphics library. It also defines functions and
objects for working with colors, dates, files, GUI elements, and basic data structures such as queues, stacks, and sets. The standard library also contains a file which defines extended ASCII
operators for use when code golfing.
The complete standard library can be explored here.
Project Euler problem 1
Find the sum of all the multiples of 3 or 5 below 1000.
A golfed version using very few characters:
A version using procedural programming features:
for 'x (999R) {
x5%0= x3%0= | then {x sum +:sum}
Recursive factorial
This first function is written with a C-style syntax. The backtick (`) operator is used to convert postfix operators into infix ones.
`:factorial {n,
if (n `.< 1) {
} {
n `* ( `factorial(n`-1) )
Another version:
User defined types
Below is a definition of a simple 2D vector type
.# Constructor
{x y, {, x:x y:y} vec MO}:new;
.# Print Override
{self, "<$(self.x),$(self.y)>"}:repr;
.# Member Function
{self, self.x2^ self.y2^ + Mq}:len;
.# Operator Overload
{a b, a.x b.x+ a.y b.y+ vec!}:add
Call constructor using ! operator and print using .repr definition:
aya> 1 2 vec!
Perform operations on the type:
aya> 3 4 vec! :v
aya> v.len
Operator overloading:
aya> 10 10 vec! v +
Basic Ploting
Aya uses JFreeChart to plot expressions and arrays. The standard library file "plot" defines a plot object which is used to easily draw plots.
.# The domain
0 2pi* 600 linspace p.domain
.# Add each function
"sin" 2 colors.red.rgb {sin} p.addexpr
"cos" 2 colors.blue.rgb {cos} p.addexpr
"ln" 2 colors.orange.rgb {ln} p.addexpr
.# Plot title
"A Sample Expression Plot" p.:title
.# Other plot parameters
[(-2) 2] p.:yaxis;
[0 2pi*] p.:xaxis;
1 p.:legend;
"f(x)" p.:ylabel;
400 p.:width;
300 p.:height;
.# Open the plot window
.# Uncomment to save the plot
"sample_plot.png" p.save
Output: A Basic Plot
Plot a Lorenz Attractor
.# Starting parameters
0.01 :x;
0.1 :y;
0.3 :z;
10 :a;
28 :b;
8`/3 :c;
0.01 :dt;
.# List to keep track of state
.# Update the point
y x - a * dt* x+:x
b z - x * y - dt* y+:y
x y * c z * - dt* z+:z;
.# Append to state list
} 5000 %
.# Transpose to separate list of x and y values
transpose ~ plot.line:p
"Lorenz Attractor" p.:title;
Output: Lorenz Attractor
Additional features
Interactive documentation
Add a ? to a line comment operator .# to add the comment to the interactive help. The interactive help can be searched from within the REPL or IDE and can be used to document source files.
aya> .#? A help comment!\n This comment will be added to the interactive help
aya> \? help comment
A help comment!
This comment will be added to the interactive help
Sample documentation from math.aya
{Mp}:primes; .#? N primes\n returns a list containing the primes up to and including N
{{*}U}:product; .#? L product\n product of a list
{.!}:signnum; .#? N signnum \n returns the sign of a number (1,0,-1)
{Mq}:sqrt; .#? N sqrt\n square root
External resources | {"url":"http://esolangs.org/wiki/Aya","timestamp":"2024-11-11T17:12:04Z","content_type":"text/html","content_length":"26277","record_id":"<urn:uuid:18113819-df39-41fe-9532-839f74ac16d6>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00138.warc.gz"} |
Regression fundamentals
3 Regression fundamentals
In this chapter, we provide a short introduction to the fundamentals of regression analysis with a focus on ordinary least-squares (OLS) regression. Our emphasis here is on helping the reader to
build intuition for the mechanics of regression. We demonstrate different ways of achieving various regression results to strengthen the reader’s intuition for what regression is doing. In this
spirit, we close the chapter with a brief discussion of the Frisch-Waugh-Lovell theorem, which provides a way of representing multivariate regression coefficients as the result of a single-variable
While we motivate some of our regressions with a data set that prompts a number of causal questions, we largely sidestep the issue of when OLS regression does or does not produce valid estimates of
causal effects. Thus, while we hint at possible causal interpretations of results in this chapter, the reader should be cautious about these interpretations. We begin our formal analysis of causal
inference with Chapter 4.
Additionally, while we note that OLS will provide noisy estimates of estimands, we do not address issues regarding how precise these estimates are or how to assess the statistical significance of
results.^1 A few p-values and t-statistics will crop up in regression analysis shown in this chapter, but we ignore those details for now. We begin our study of statistical inference in Chapter 5.
The code in this chapter uses the packages listed below. For instructions on how to set up your computer to use the code found in this book, see Section 1.2 (note that Step 4 is not required as we do
not use WRDS data in this chapter). Quarto templates for the exercises below are available on GitHub.
3.1 Introduction
Suppose we have data on variables \(y\), \(x_1\) and \(x_2\) for \(n\) units and we conjecture that there is a linear relationship between these variables of the following form:
\[ y_i = \beta_0 + \beta_1 \times x_{i1} + \beta_2 \times x_{i2} + \epsilon_i \] where \(i \in {1, \dots, n}\) denotes the data for a particular unit. We can write that in matrix form as follows:
\[ \begin{bmatrix} y_1 \\ y_2 \\ \dots \\ y_{n-1} \\ y_{n} \\ \end{bmatrix} = \begin{bmatrix} 1 & x_{11} & x_{12} \\ 1 & x_{21} & x_{22} \\ \dots & \dots & \dots \\ 1 & x_{n-1,1} & x_{n-1, 2} \\ 1 &
x_{n, 1} & x_{n, 2} \end{bmatrix} \times \begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \end{bmatrix} + \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \dots \\ \epsilon_{n-1} \\ \epsilon_{n} \\ \end
{bmatrix} \]
And this can be written even more compactly as:
\[ y = X \beta + \epsilon \] where \(X\) is an \(n \times 3\) matrix and \(y\) is an \(n\)-element vector. It is conventional to denote the number of columns in the \(X\) matrix using \(k\), where \
(k = 3\) in this case.^2
In a regression context, we call \(X\) the regressors, \(y\) the regressand, and \(\epsilon\) the error term. We assume that we observe \(X\) and \(y\), but not \(\epsilon\). If we did observe \(\
epsilon\), then we could probably solve for the exact value of the coefficients \(\beta\) with just a few observations. Lacking such information, we can produce an estimate of our estimand, \(\beta\)
. Our estimate (\(\hat{\beta}\)) is likely to differ from \(\beta\) due to noise arising from the randomness of the unobserved \(\epsilon\) and also possibly bias. There will usually be a number of
estimators that we might consider for a particular problem. We will focus on the ordinary least-squares regression (or OLS) estimator as the source for our estimates in this chapter.
OLS is a mainstay of empirical research in the social sciences in general and in financial accounting in particular. In matrix notation, the OLS estimator is given by
\[ \hat{\beta} = (X^{\mathsf{T}}X)^{-1}X^{\mathsf{T}}y \]
Let’s break this down. First, \(X^{\mathsf{T}}\) is the transpose of \(X\), meaning the \(k \times n\) matrix formed by making the rows of \(X\) into columns. Second, \(X^{\mathsf{T}} X\) is the
product of the \(k \times n\) matrix \(X^{\mathsf{T}}\) and the \(n \times k\) matrix \(X\), which results in a \(k \times k\) matrix. Third, the \(-1\) exponent indicates the inverse matrix. For a
real number \(x\), \(x^{-1}\) denotes the number that when multiplied by \(x\) gives \(1\) (i.e., \(x \times x^{-1} = 1\)). For a square matrix \(Z\) (here “square” means the number of rows equals
the number of columns), \(Z^{-1}\) denotes the square matrix that when multiplied by \(Z\) gives the identity matrix, \(\mathbf{I}\) (i.e., \(Z \times Z^{-1} = \mathbf{I}\)).
The \(3 \times 3\) identity matrix looks like this
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Note that, just as there is no meaningful way to calculate the inverse of \(0\), it’s not always possible to take the inverse of a matrix. But, so long as no column of \(X\) is a linear combination
of other columns and \(n > k\), then we can calculate \((X^{\mathsf{T}}X)^{-1}\) (and there are standard algorithms for doing so). Now, \(X^{\mathsf{T}}y\) is the product of a \(k \times n\) matrix (
\(X^{\mathsf{T}}\)) and a vector with \(n\) elements (this can be thought of as an \(n \times 1\) matrix), so the result will be a \(k \times 1\) matrix. Thus, the product of \((X^{\mathsf{T}}X)^{-1}
\) and \(X^{\mathsf{T}}y\) will be a vector with \(k\) elements, which we can denote as follows:
\[ \begin{bmatrix} \hat{\beta}_1 \\ \hat{\beta}_2 \\ \hat{\beta}_3 \\ \end{bmatrix} \]
So \(X^{\mathsf{T}}X\) is a \(k \times k\) matrix (as is its inverse), and \(X^{\mathsf{T}}y\) is the product of a \(k \times N\) matrix times an \(N \times 1\) matrix (i.e., a vector). So \(\hat{\
beta}\) is a \(k\)-element vector. If we have a single regressor \(x\), then \(X\) will typically include the constant term, so \(k = 2\).
For this chapter, we will assume that the model \(y = X \beta + \epsilon\) is a structural (causal) model. What this means is that if we could somehow increase the value of \(x_{i1}\) by 1 unit
without changing any other part of the system, we would see an increase in the value of \(y_{i}\) equal to \(\beta_1\). This model is causal in the sense that a unit change in \(x_{i1}\) can be said
to cause a \(\beta_1\) change in \(y_i\).
To make this more concrete, let’s consider some actual (though not “real”) data. The following code uses R functions to generate random data. Specifically, we generate 1000 observations with \(\beta
= 1\) and \(\sigma = 0.2\). This is the first of at least a dozen simulation analyses that we will consider in the book.
To make our analysis easier to replicate, we include set.seed(2021) to set the random-number generator used by rnorm() to the same point, so that we can reproduce the analysis ourselves later and so
that others can reproduce it too. For more on random-number generation, type ? set.seed in the R console. Note that the value 2021 is arbitrary and represents nothing more than the year in which this
material was first written. Any value would work here, e.g., 2024, 42, or 20240215 could be used.
We next construct \(X\) as a matrix comprising a column of ones (to estimate the constant term) and a column containing \(x\).
[,1] [,2]
[1,] 1 -0.1224600
[2,] 1 0.5524566
[3,] 1 0.3486495
[4,] 1 0.3596322
[5,] 1 0.8980537
[6,] 1 -1.9225695
Naturally, R has built-in matrix operations. To get \(X^{\mathsf{T}}\), the transpose of the matrix \(X\), we use t(X). To multiply two matrices, we use the matrix multiplication operator %*%. And to
invert \((X^{\mathsf{T}}X)\) to get \((X^{\mathsf{T}}X)^{-1}\), we use the solve() function. Thus, the following calculates the OLS estimator \(\hat{\beta} = (X^{\mathsf{T}}X)^{-1} X^{\mathsf{T}}y\).
[1,] 0.007539896
[2,] 0.997646586
3.2 Running regressions in R
According to the documentation for lm():
The basic function for fitting ordinary multiple models is lm(), and a streamlined version of the call is as follows:
fitted.model <- lm(formula, data = data.frame)
For example,
fm1 <- lm(y ~ x1 + x2, data = production)
would fit a multiple regression model of y on x1 and x2 using data from the data frame production. We use the lm() function—part of base R—to estimate regressions in this chapter.
Note that R was developed by statisticians and thus works in a way consistent with that history. For example, if we run the following code, we actually estimate coefficients on the constant term (the
intercept), x1, x2, and the product of x1 and x2.
fm2 <- lm(y ~ x1 * x2, data = production)
In contrast to other statistical packages, with lm() there’s no need to calculate the product of x1 and x2 and store it as a separate variable and there’s no need to explicitly specify the “main
effect” terms (i.e., x1 and x2) in the regression equation. R (and the lm() function) takes care of these details for us. The third argument to lm() is subset, which allows us to specify as condition
that each observation needs to satisfy to be included in the regression.
Here the first argument to lm() (formula) is “an object of class formula … a symbolic description of the model to be fitted.” As we are regressing \(y\) on \(x\), we use the formula y ~ x here; we
will soon see more complicated formula expressions. The value to the data argument is normally a data frame, so below we put x and y into a tibble. We then call the lm() function, store the returned
value in the variable fm, then show some of the contents of fm:
df <- tibble(y, x)
fm <- lm(y ~ x, data = df)
lm(formula = y ~ x, data = df)
(Intercept) x
0.00754 0.99765
From this output, we see that we get the same results using lm() as we do using matrix algebra.
The data sets we will focus on next are test_scores and camp_attendance, both of which are part of the farr package. The test_scores data frame contains data on test scores for 1000 students over
four years (grades 5 through 8).
# A tibble: 4,000 × 3
id grade score
<int> <int> <dbl>
1 1 5 498.
2 1 6 513.
3 1 7 521.
4 1 8 552.
5 2 5 480.
6 2 6 515.
# ℹ 3,994 more rows
The camp_attendance data set contains data on whether a student attended a science camp during the summer after sixth grade.
# A tibble: 1,000 × 2
id camp
<int> <lgl>
1 1 TRUE
2 2 TRUE
3 3 FALSE
4 4 FALSE
5 5 FALSE
6 6 TRUE
# ℹ 994 more rows
We can also see that exactly half the students in the sample attended the science camp.
camp_scores <-
test_scores |>
inner_join(camp_attendance, by = "id") |>
rename(treat = camp) |>
mutate(post = grade >= 7)
The question we might be interested in is whether attending the science camp improves test performance. The natural first thing to do would be plot the data, which we do in Figure 3.1.
camp_scores |>
group_by(grade, treat) |>
summarize(score = mean(score),
.groups = "drop") |>
ggplot(aes(x = grade, y = score,
linetype = treat, colour = treat)) +
Figure 3.1: Student test scores by grade and camp participation
Figure 3.1 shows that the students who went to the camp had lower scores in grades 5 and 6 (i.e., before the camp), but stronger performance in grades 7 and 8. This provides prima facie evidence of a
positive effect of the science camp on scores.
While the ideal approach might be to randomly assign students to the camp and then compare scores after the camp, in this case, it appears that students with lower scores went to camp. Given our lack
of contextual information, there is no obvious story for what we see in Figure 3.1 that is more plausible than (or at least as simple as) one that attributes a positive effect on test scores of going
to the science camp.
A question that we might now have is: What is the best way to test for an effect of the science camp on test scores? One reasonable hypothesis is that the summer camp has a biggest effect on
seventh-grade scores and that we might compare seventh-grade scores with sixth-grade scores to get the best estimate of the effect of camp on scores.
In the code below, we use filter() to limit our analysis to data related to sixth and seventh grades. We then calculate mean scores for each value of (post, treat). We then use pivot_wider() to put
score values for different levels of post in the same row so that we can calculate change.
test_summ <-
camp_scores |>
filter(grade %in% 6:7L) |>
group_by(post, treat) |>
summarize(score = mean(score), .groups = "drop") |>
pivot_wider(names_from = post, values_from = score) |>
rename(post = `TRUE`, pre = `FALSE`) |>
mutate(change = post - pre)
You may find it helpful in understanding code examples such as this one to look at the intermediate output along the line of pipes. For example, if you highlight the text from test_scores |> through
to the end of the summarize() line just before the pipe at the end of the last line, and click “Run Selected Lines” in the “Code” menu of RStudio (or hit CTRL + Enter on Windows or Linux or ⌘ + Enter
on MacOS) you will see what is coming out at that point of the pipe (it should look like the following):
# A tibble: 4 × 3
post treat score
<lgl> <lgl> <dbl>
1 FALSE FALSE 516.
2 FALSE TRUE 511.
3 TRUE FALSE 520.
4 TRUE TRUE 532.
We strongly recommend that you use this approach liberally when working through this book (and when debugging your own pipelines).
We stored the results of our analysis in test_summ, which we show in Table 3.1.
Table 3.1: Initial analysis of student scores
treat pre post change
FALSE 515.539 519.657 4.118
TRUE 510.810 531.938 21.128
We see in Table 3.1 that the scores of the treated students (i.e., those who went to the summer camp) increased by 21.128, while the scores of the control students increased by 4.118. One approach is
to view the outcome of the control students as the “but-for-treatment” outcome that the treated students would have seen had they not gone to summer camp. With this view, the effect of going to camp
is the difference in the difference in scores, or 17.010. This is the difference-in-differences estimator of the causal effect.
We can also recover this estimator using the lm() function:
fm_dd <- lm(score ~ treat * post, data = camp_scores,
subset = grade %in% 6:7L)
(Intercept) treatTRUE postTRUE treatTRUE:postTRUE
515.539023 -4.728960 4.118259 17.009708
Note that we did not need to specify the inclusion of the main effects of treat and post; R automatically added those when we requested their interaction (treat * post). Also note that we did not
need to convert the logical variables treat and post so that TRUE is 1 and FALSE is 0; in effect, R also did this for us.
A natural question might be whether we could do better using all four years of data. There is no simple answer to this question unless we have a stronger view of the underlying causal model. One
causal model might have it that students have variation in underlying talent, but that there is also variation in industriousness that affects how students improve over time. From the perspective of
evaluating the effect of the camp, variation in industriousness is going to add noise to estimation that increases if we are comparing performance in fifth grade with that in eighth grade.
Another issue is that the effects of the summer camp might fade over time. As such, we might get a larger estimated effect if we focus on seventh-grade scores than if we focus on (or also include)
eighth-grade scores. But from a policy perspective, we might care more about sustained performance improvement and actually prefer eighth-grade scores.
However, if we were willing to assume that, in fact, scores are a combination of underlying, time-invariant individual talent, the persistent effects (if any) of summer camp, and random noise, then
we’d actually do better to include all observations.^3
fm_dd_all <- lm(score ~ treat * post, data = camp_scores)
(Intercept) treatTRUE postTRUE treatTRUE:postTRUE
505.887759 -3.536433 21.764629 15.735916
Another possibility is that scores are a combination of underlying, time-invariant individual talent, the persistent effects (if any) of summer camp, and random noise and the grade in which the test
is taken. For example, perhaps the test taken in seventh grade is similar to that taken in sixth grade but, with an extra year of schooling, students might be expected to do better in the higher
grade assuming the scores are not scaled in any way within grades. (Recall that we just have this data set without details that would allow us to rule out such ideas, so the safest thing to do is to
examine the data.) The easiest way to include grade is as a linear trend, which means that grade is viewed as a number (e.g., 5 or 8):
fm_dd_trend <- lm(score ~ treat * post + grade, data = camp_scores)
(Intercept) treatTRUE postTRUE grade
412.335651 -3.536433 -12.254320 17.009474
Note that we have exactly the same coefficients on treat and treat * post as we had before (\(-3.5364\) and \(15.7359\), respectively). The easiest way to understand the estimated coefficients is to
plug in some candidate \(X\) values to get the fitted values.
Suppose we have a student who went to summer camp. In grade 6, this student’s predicted score would be
\[ 412.3357 + -3.5364 + 6 \times 17.0095 = 510.8561 \]
In grade 7, this student’s predicted score would be
\[ 412.3357 + -3.5364 + 7 \times 17.0095 + 15.7359 = 543.6015 \]
An alternative approach that allows for grade to affect scores would be to estimate a separate intercept for each level that grade takes on. That is, we’d have a different intercept for grade==6, a
different intercept for grade==7, and so on. While we could achieve this outcome by creating variables using an approach such as mutate(grade7 = grade == 7), it is easier to use the R’s support for
As discussed in Chapter 2, factors are a type that is useful for representing categorical variables, which often have no meaningful numerical representation (e.g., “red” or “blue”, or “Australia” or
“New Zealand”) or where we want to move away from a simple numerical representation (e.g., grade 7 may not be simply 7/6 times grade 6).^4 Rather than adding a factor version of grade to the model
above, let’s run a simpler regression.
(Intercept) factor(grade)6 factor(grade)7 factor(grade)8
495.06454 18.11000 30.73311 46.64206
This model estimates fixed effects for each grade without other covariates. Table 3.2 provides the mean scores by grade for comparison.
Table 3.2: Mean scores by grade
grade score
5 495.065
6 513.175
7 525.798
8 541.707
The idea of fixed effects is that there are time-invariant factors that have a constant effect on the outcome (hence fixed effects). In some settings, we would posit fixed effects at the level of the
individual. Here we are positing fixed effects at the grade level. Working through the exercises should provide additional insights into what we are doing here. For more on fixed effects, see
Cunningham (2021, pp. 391–392) and also Chapter 21.
Now, let’s estimate fixed effects for both grade and student (id). This will yield more fixed effects than we have students (we have 1000 students), so we suppress the coefficients for the fixed
effect in the regression output, which is shown in Table 3.3.
fm_id <- lm(score ~ treat * post + factor(grade) + factor(id),
data = camp_scores, x = TRUE)
estimate = "{estimate}{stars}",
coef_omit = "^factor",
gof_map = "nobs",
stars = c('*' = .1, '**' = 0.05, '***' = .01))
Table 3.3: Test-score regressions with grade and id fixed effects
(Intercept) 495.370***
treatTRUE -2.235
postTRUE 38.774***
treatTRUE × postTRUE 15.736***
Num.Obs. 4000
Note that we specified x = TRUE so that the \(X\) matrix used in estimation was returned by lm(). The size of this matrix is given by the dim() function:
We have 1006 columns because we have added so many fixed effects. This means that \((X^{\mathsf{T}}X)^{-1}\) is a 1006 \(\times\) 1006 matrix. As we add more years and students, this matrix could
quickly become quite large and inverting it would be computationally expensive (even more so for some other operations that would need even larger matrices). To get a hint as to a less
computationally taxing approach, let’s see what happens when we “demean” the variables in a particular way.
demean <- function(x) x - mean(x)
camp_scores_demean <-
camp_scores |>
group_by(id) |>
mutate(score = demean(score)) |>
group_by(grade) |>
mutate(score = demean(score))
fm_demean <- lm(score ~ treat * post,
data = camp_scores_demean, x = TRUE)
Results from this analysis are shown in Table 3.4.
estimate = "{estimate}{stars}",
coef_omit = "^factor",
gof_map = "nobs",
stars = c('*' = .1, '**' = 0.05, '***' = .01))
Table 3.4: Regression with demeaned test scores
(Intercept) 3.934***
treatTRUE -7.868***
postTRUE -7.868***
treatTRUE × postTRUE 15.736***
Num.Obs. 4000
The size of the \(X\) matrix is now 4000 \(\times\) 4. This means that \((X^{\mathsf{T}}X)^{-1}\) is now a much more manageable 4 \(\times\) 4 matrix. While we had to demean the data, this is a
relatively fast operation.
3.2.1 Exercises
1. In using pivot_wider() in Chapter 2, we supplied a value to the id_cols argument, but we omitted that in creating test_summ. If we wanted to be explicit, what value would we need to provide for
that argument in the code creating test_summ?
2. What is the relation between the means in Table 3.2 and the regression coefficients in fm_grade?
3. Why is there no estimated coefficient for factor(grade)5 in fm_grade?
4. Now let’s return to our earlier regression specification, except this time we include fixed effects for grade (see code below and output in Table 3.5). We now have two fixed effects omitted:
factor(grade)5 and factor(grade)8. Why are we now losing two fixed effects, while above we lost just one? (Hint: Which variables can be expressed as linear combinations of the grade indicators?)
fm_dd_fe <- lm(score ~ treat * post + factor(grade),
data = camp_scores)
estimate = "{estimate}{stars}",
gof_map = "nobs",
stars = c('*' = .1, '**' = 0.05, '***' = .01))
Table 3.5: Regression results with grade fixed effects
(Intercept) 496.833***
treatTRUE -3.536***
postTRUE 38.774***
factor(grade)6 18.110***
factor(grade)7 -15.909***
treatTRUE × postTRUE 15.736***
Num.Obs. 4000
5. In words, what are we doing to create camp_scores_demean? Intuitively, why might this affect the need to use fixed effects?
6. Can you relate the coefficients from the regression stored in fm_demean to the numbers in Table 3.6? Which of these estimated coefficients is meaningful? All of them? Some of them? None of them?
Table 3.6: Demeaned test scores by grade and treat
grade treat score
5.0000 False 3.3377
5.0000 True -3.3377
6.0000 False 4.5302
6.0000 True -4.5302
7.0000 False -3.9746
7.0000 True 3.9746
8.0000 False -3.8933
8.0000 True 3.8933
7. The feols() function from the fixest package offers a succinct syntax for adding fixed effects and uses computationally efficient algorithms (much like our demeaning approach above) in estimating
these. What is the same in the results below and the two specifications we estimated above? What is different? Why might these differences exist? What is the I() function doing here? What happens
if we omit it (i.e., just include post * treat)?
fefm <- feols(score ~ I(post * treat) | grade + id, data = camp_scores)
3.3 Frisch-Waugh-Lovell theorem
The Frisch-Waugh-Lovell theorem states that the following two regressions yield identical regression results in terms of both the estimate \(\hat{\beta_2}\) and residuals.
\[ y = X_1 \beta_1 + X_2 \beta_2 + \epsilon \] and
\[ M_{X_1} y = M_{X_1} X_2 \beta_2 + \eta \]
where \(M_{X_1}\) is the “residual maker” for \(X_1\) or \(I - P_{X_1} = I - X_1({X_1^{\mathsf{T}}X_1})^{-1}X_1^{\mathsf{T}}\), \(y\) is a \(n \times 1\) vector, and \(X_1\) and \(X_2\) are \((n \
times k_1)\) and \((n \times k_2)\) matrices.
In other words, we have two procedures that we can use to estimate \(\hat{\beta}_2\) in the regression equation above. First, we could simply regress \(y\) on \(X_1\) and \(X_2\) to obtain estimate \
(\hat{\beta}_2\). Second, we could take the following more elaborate approach:
1. Regress \(X_2\) on \(X_1\) (and a constant term) and store the residuals (\(\epsilon_{X_1}\)).
2. Regress \(y\) on \(X_1\) (and a constant term) and store the residuals (\(\epsilon_{y}\)).
3. Regress \(\epsilon_{y}\) on \(\epsilon_{X_1}\) (and a constant term) to obtain estimate \(\hat{\beta}_2\).
The Frisch-Waugh-Lovell theorem tells us that only the portion of \(X_2\) that is orthogonal to \(X_1\) affects the estimate \(\hat{\beta}_2\). Note that the partition of \(X\) into \([X_1 X_2]\) is
quite arbitrary, which means that we also get the same estimate \(\hat{\beta_1}\) from the first regression equation above and from estimating
\[ M_{X_2} y = M_{X_2} X_1 \beta_1 + \upsilon \]
To verify the Frisch-Waugh-Lovell theorem using some actual data, we draw on the data set comp from the farr package and a regression specification we will see in Chapter 24.
As our baseline, we run the following linear regression and store it in fm.
fm <- lm(ta ~ big_n + cfo + size + lev + mtb +
factor(fyear) * (inv_at + I(d_sale - d_ar) + ppe),
data = comp, na.action = na.exclude)
Here the dependent variable is ta (total accruals), big_n is an indicator variable for having a Big \(N\) auditor (see Section 25.1) and the other variables are various controls (use help(comp) or ?
comp for descriptions of these variables). We again use the I() function we saw above and interact factor(fyear) with three different variables.
We then run two auxiliary regressions: one of ta on all regressors except cfo (we store this in fm_aux_ta) and one of cfo on all regressors except cfo (we store this in fm_aux_cfo). We then take the
residuals from each of these regressions and put them in a data frame under the names of the original variables (ta and size respectively). Finally, using the data in aux_data, we regress ta on size.
fm_aux_ta <- lm(ta ~ big_n + size + lev + mtb +
factor(fyear) * (inv_at + I(d_sale - d_ar) + ppe),
data = comp, na.action = na.exclude)
fm_aux_cfo <- lm(cfo ~ big_n + size + lev + mtb +
factor(fyear) * (inv_at + I(d_sale - d_ar) + ppe),
data = comp, na.action = na.exclude)
aux_data <- tibble(ta = resid(fm_aux_ta),
cfo = resid(fm_aux_cfo))
fm_aux <- lm(ta ~ cfo, data = aux_data)
The Frisch-Waugh-Lovell theorem tells us that the regression in fm_aux will produce exactly the same coefficient on cfo and the same residuals (and very similar standard errors) as the regression in
fm, as can be seen in Table 3.7. Here we use modelsummary() from the modelsummary package to produce attractive regression output. We use coef_omit = "(fyear|ppe|inv_at|d_sale)" to focus on
coefficients of greater interest.
modelsummary(list(fm, fm_aux),
estimate = "{estimate}{stars}",
coef_omit = "(fyear|ppe|inv_at|d_sale)",
gof_map = "nobs",
stars = c('*' = .1, '**' = 0.05, '***' = .01))
Table 3.7: Demonstration of FWL theorem
(1) (2)
(Intercept) -0.017 0.000
(0.028) (0.004)
big_nTRUE 0.022*
cfo 0.141*** 0.141***
(0.008) (0.008)
size 0.000
lev -0.066***
mtb 0.000***
Num.Obs. 8850 8850
The Frisch-Waugh-Lovell theorem is an important result for applied researchers to understand, as it provides insights into how multivariate regression works. A side-benefit of the result is that it
allows us to reduce the relation between two variables in a multivariate regression to a bivariate regression without altering that relation. For example, to understand the relation between cfo and
ta embedded in the estimated model in fm, we can plot the data.
We produce two plots. The first—Figure 3.2—includes all data, along with a line of best fit and a smoothed curve of best fit. However, Figure 3.2 reveals extreme observations of the kind that we will
study more closely in Chapter 24 (abnormal accruals more than 5 or less than 5 times lagged total assets!).
So we trim the values of ta at \(-1\) and \(+1\) and produce a second plot.^5 In Figure 3.3, there is no visually discernible relation between size and ta and the line of best fit is radically
different from the curve. If nothing else, hopefully these plots raise questions about the merits of blindly accepting regression results with the messy data that we often encounter in practice.
aux_data |>
filter(!is.na(cfo), !is.na(ta)) |>
ggplot(aes(x = cfo, y = ta)) +
geom_point() +
geom_smooth(method = "lm", formula = y ~ x, se = FALSE, linetype = 2,
colour = "red") +
geom_smooth(method = "gam", formula = y ~ s(x, bs = "cs"), se = FALSE)
Figure 3.2: Illustration of FWL: Total accruals and CFO
aux_data |>
filter(!is.na(cfo), !is.na(ta), abs(ta) < 1) |>
ggplot(aes(x = cfo, y = ta)) +
geom_point() +
geom_smooth(method = "lm", formula = y ~ x, se = FALSE, linetype = 2,
colour = "red") +
geom_smooth(method = "gam", formula = y ~ s(x, bs = "cs"), se = FALSE)
Figure 3.3: Illustration of FWL: Total accruals and CFO excluding outliers
3.3.1 Exercises
1. Verify the Frisch-Waugh-Lovell theorem using big_n and lev in place of cfo and produce plots like Figure 3.2 for each variable. Does the plot with big_n as the independent variable seem less
2. Above we said that the standard errors of the main regression and the auxiliary regression using the Frisch-Waugh-Lovell theorem should be “very similar”. Confirm that the standard errors are
similar across the variants of fm and fm_aux that you calculated for the previous question. (Hint: summary(fm_aux)$coefficients and summary(fm)$coefficients["big_nTRUE", ] should provide access
to the data you want to compare.) Can you guess what might explain any differences? (Hint: Compare fm$df.residual and fm_aux$df.residual and perhaps use sqrt().)
3. In words, what effect does converting fyear into a factor and interacting it with inv_at, I(d_sale - d_ar) and ppe have? (Hint: It may be helpful to visually inspect the more complete regression
output produced without coef_omit = "(fyear|ppe|inv_at|d_sale)".)
3.4 Further reading
This chapter provides a bare minimum introduction to running regressions in R plus some concepts that help develop intuition about what’s going on in OLS regression. Treatments that provide a similar
emphasis on intuition, but go deeper into the details include Angrist and Pischke (2008) and Cunningham (2021). Any econometrics textbook will offer a more rigorous treatment of OLS and its | {"url":"https://iangow.github.io/far_book/reg-basics.html","timestamp":"2024-11-02T22:20:09Z","content_type":"application/xhtml+xml","content_length":"173313","record_id":"<urn:uuid:b08bd962-d467-41a5-b53c-15f3786c22ab>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00032.warc.gz"} |
Multiplication of Single Variable Polynomial : Algorithm and Program - Quescol
Multiplication of Single Variable Polynomial : Algorithm and Program
Multiplication of single-variable polynomials is a fundamental operation in polynomial algebra. The process involves multiplying each term of one polynomial by each term of the other and then
combining like terms. Here’s a detailed explanation of the algorithm followed by a C program to demonstrate it.
Algorithm for Polynomial Multiplication
1. Initialize Resultant Polynomial:
□ Start with an array (or another data structure) to store the result. Initialize all elements to zero. The size of this array should be large enough to hold the highest possible degree, which
is the sum of the degrees of the two input polynomials.
2. Iterate Through Both Polynomials:
□ Loop through each term of the first polynomial.
□ For each term in the first polynomial, loop through each term of the second polynomial.
3. Multiply and Add Terms:
□ In each iteration of the nested loop, multiply the coefficients of the current terms from both polynomials.
□ Calculate the new degree by adding the exponents of these terms.
□ Add the result to the corresponding position (based on the calculated degree) in the resultant polynomial array.
4. Combine Like Terms:
□ As the multiplication proceeds, some terms might have the same degree. Ensure that these like terms are combined by adding their coefficients.
5. Result:
□ The final array represents the resultant polynomial after multiplication.
Certainly! Let’s go through the steps of multiplying two single-variable polynomials step by step with an example. We’ll use the polynomials P(x)=5x^2+3x+7 and Q(x)=6x+2 for our demonstration.
Polynomial Multiplication Example:
• P(x)=5x^2+3x+7 (Degree 2)
• Q(x)=6x+2 (Degree 1)
Steps for Multiplication:
1. Initialize the Resultant Polynomial:
□ The degree of the resultant polynomial will be the sum of the degrees of P(x) and Q(x), which is 2 + 1 = 3.
□ So, the resultant polynomial R(x) will have a degree of 3 and needs 4 terms (including the constant term).
2. Multiply Each Term of P(x) by Each Term of Q(x):
□ Multiply 5x^2 (from P(x)) by each term in Q(x): 5x^2*6x=30x^3 and 5x^2*2=10x^2.
□ Multiply 3x (from P(x)) by each term in Q(x): 3x*6x=18x^2 and 3x*2=6x.
□ Multiply 7(from P(x)) by each term in Q(x): 7*6x=42x and 7*2=14.
3. Combine Like Terms:
□ Collect terms with the same degree: 30x^3 (no other x3 terms), 10x^2+18x^2 (combine x2 terms), 6x+42x (combine x terms), and 14 (constant term).
4. Write the Resultant Polynomial:
□ R(x)=30x^3+10x^2+18x^2+6x+42x+14
□ Simplify: R(x)=30x^3+28x^2+48x+14
Polynomial Multiplication Program
#include <stdio.h>
#define MAX_DEGREE 100
// Multiply two Polynomials
void polynomialMultiplication(int polynomial1[], int polynomial2[], int degree1, int degree2, int result[]) {
// Initialize the result array
for (int i = 0; i <= degree1 + degree2; i++) {
result[i] = 0;
// Multiplication of first polynomial with each term of second polynomial
for (int i = 0; i <= degree1; i++) {
for (int j = 0; j <= degree2; j++) {
result[i + j] += polynomial1[i] * polynomial2[j];
// Function to print the polynomial
void printPolynomial(int polynomial[], int degree) {
for (int i = degree; i >= 0; i--) {
if (polynomial[i] != 0) {
printf("%dx^%d ", polynomial[i], i);
if (i != 0) printf("+ ");
int main() {
// Polynomial P(x) = 5x^2 + 3x + 7
int polynomial1[] = {7, 3, 5}; // Degree 2
int degree1 = 2;
// Polynomial Q(x) = 6x + 2
int polynomial2[] = {2, 6}; // Degree 1
int degree2 = 1;
int result[MAX_DEGREE];
polynomialMultiplication(polynomial1, polynomial2, degree1, degree2, result);
printf("Resultant polynomial after multiplication:\n");
printPolynomial(result, degree1 + degree2);
return 0;
Resultant polynomial after multiplication:
30x^3 + 28x^2 + 48x^1 + 14x^0
Explanation of Polynomial Multiplication Program
This C program is designed to perform the multiplication of two single-variable polynomials and print the resultant polynomial. Let’s go through the program step by step:
Preprocessor Directive and Constants:
#include <stdio.h>
#define MAX_DEGREE 100
• Includes the standard input-output library for functions like printf.
• Defines a constant MAX_DEGREE to set the maximum degree of the polynomial that the program can handle.
Multiplication Function:
void polynomialMultiplication(int polynomial1[], int polynomial2[], int degree1, int degree2, int result[]) {
// Initialize the result array
for (int i = 0; i <= degree1 + degree2; i++) {
result[i] = 0;
// Multiplication of first polynomial with each term of second polynomial
for (int i = 0; i <= degree1; i++) {
for (int j = 0; j <= degree2; j++) {
result[i + j] += polynomial1[i] * polynomial2[j];
• The function polynomialMultiplication multiplies two polynomials.
• It initializes an array result to store the coefficients of the multiplied polynomial.
• It then iteratively multiplies each term of the first polynomial (polynomial1) with each term of the second polynomial (polynomial2) and accumulates the products in the appropriate position in
the result array.
Printing Function:
void printPolynomial(int polynomial[], int degree) {
for (int i = degree; i >= 0; i--) {
if (polynomial[i] != 0) {
printf("%dx^%d ", polynomial[i], i);
if (i != 0) printf("+ ");
• This function printPolynomial prints the polynomial in a readable format.
• It iterates through the array containing the polynomial coefficients, printing each non-zero term along with its degree.
Main Function:
int main() {
int polynomial1[] = {7, 3, 5}; // Degree 2
int polynomial2[] = {2, 6}; // Degree 1
int result[MAX_DEGREE];
polynomialMultiplication(polynomial1, polynomial2, degree1, degree2, result);
printf("Resultant polynomial after multiplication:\n");
printPolynomial(result, degree1 + degree2);
return 0;
• The main function initializes two polynomial arrays (polynomial1 and polynomial2) representing P(x)=5x2+3x+7 and Q(x)=6x+2, respectively.
• It then calls polynomialMultiplication to multiply these polynomials and stores the result in result.
• Finally, it calls printPolynomial to print the resultant polynomial.
Hope this tutorial helped you to understand the multiplication of two polynomial with the help of algorithm and program. | {"url":"https://quescol.com/data-structure/multiplication-single-variable-polynomial-algorithm-program","timestamp":"2024-11-09T06:38:36Z","content_type":"text/html","content_length":"90932","record_id":"<urn:uuid:e49a55e3-016c-4b73-a4f7-b05fbee7ccb2>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00495.warc.gz"} |
Primary 5 math | Syllabus | Practicle
Aligned with the latest 2025 Singapore MOE Math syllabus and endorsed by Education Alliance of Finland.
Hundreds of hours of learning content for all topics covered in P5 to get your child build a strong Math foundation!
P5 is a difficult year for many students.
Having just made it through a trying year of transitioning into upper primary math in P4, they now face the onslaught of PSLE preparations.
Not to worry, because we have plenty of fun and effective learning material to help them build their skills and confidence, and put your mind at ease.
Find out what’s in the syllabus and try some of our sample questions!
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Overview of Primary 5 Math Syllabus
Here are the P5 topics and skills that are covered our system. The 2025 Primary 5 Math Syllabus can be divided into 3 main branches – Numbers, Measurement and Geometry and finally, Statistics.
The Primary 5 (P5) topics that are covered under Number are Whole Number, Fraction, Decimal, Percentage, Ratio and Rate. The following shows the breakdown of the skills in topic:
• Count to 1 000 000
• Place values up to hundred thousands
• Use order of operations
• Divide by 1o, 100, 1000 & their multiples
• Multiply by 10, 100, 1000 & their multiples
• Solve word problems involving 4 operations
• Add & subtract mixed numbers
• Multiply a fraction by a whole number
• Multiply a proper fraction by a proper fraction
• Understand how fractions and division are related
• Divide a whole number by a whole number
• Convert a fraction to a decimals and vice versa
• Multiply 2 improper fractions
• Solve fraction word problems
• Express rate as the amount of quantity per unit of another quantity
• Find rate, total amount or number of units, given any other 2 quantities
• Solve problem sums with proportional reasoning
• Write ratios
• Write equivalent ratios
• Divide a quantity in a given ratio
• Simplify ratios
• Find the ratio of 2 or 3 given quantities
• Solve ratio word problems
• Understand a percentage of the whole
• Convert a fraction to a percentage and vice versa
• Express a decimal as a percentage and vice versa
• Convert a percentage to a decimal and vice versa
• Find discount, GST and annual interest
• Find a percentage part of a whole
• Solve percentage word problems
• Add decimals
• Subtract decimals
• Rounding off decimals
• Multiply decimals
• Divide decimals
• Multiply decimals by 10, 100 1000 and their multiples
• Divide decimals by 10, 100 1000 and their multiples
• Convert measurements of length, mass & volume
• Solve decimal word problems
P5 Math: Measurement and Geometry
The Primary 5 (P5) topics that are covered under Measurement and Geometry are Area of Triangle, Volume and Angles. Refer to the following for the breakdown of the skills in each topic.
• Identify the base or height of a triangle
• Know the properties of an isosceles triangle
• Know the properties of a right-angled triangle
• Know the properties of an equilateral triangle
• Relate the area of a triangle to its rectangle
• Find the area of a triangle using the formula
• Solve area of triangle word problems
• Draw the top/side/front view of a solid
• Find the volume of a cuboid
• Find the volume of a cube
• Find the volume of a liquid in a rectangular tank
• Convert 1 litres to 1 000 cubic mililitres
• Solve volume word problems
• Use the angle rule of angles on a straight line
• Use the angle rule of angles at a point
• Use the angle rule of vertically opposite angles
• Know what is the sum of angles of a triangle
• Find an unknown angle in a triangle
• Know the properties of a parallelogram
• Know the properties of a rhombus
• Know the properties of a trapezium
• Find an unknown angle in a special quadrilateral
• Identify basic shapes in a composite figure
The Primary 5 (P5) topics that are covered under Statistics include Average. Refer below for the breakdown of the skills in that topic.
• Find the average, given total value & the number of data
• Find the total value, given the average & the number of data
• Solve average word problems
Sample P5 questions and video explanations
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.2 – Factoring
Learning Objectives
In this section students will:
5.2.1 – Factor the greatest common factor of a polynomial.
Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in (Figure).
The area of the entire region can be found using the formula for the area of a rectangle.
[latex]$$\begin{array}{ccc}\hfill A& =& lw\hfill \\ & =& 10x\cdot 6x\hfill \\ & =& 60{x}^{2}{\text{ units}}^{2}\hfill \end{array}$$[/latex]
The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of [latex]\,A={s}^{2}={4}^{2}=16\,[/latex]
units^2. The other rectangular region has one side of length [latex]\,10x-8\,[/latex] and one side of length [latex]\,4,[/latex] giving an area of [latex]\,A=lw=4\left(10x-8\right)=40x-32\,[/latex]
units^2. So the region that must be subtracted has an area of [latex]\,2\left(16\right)+40x-32=40x\,[/latex] units^2.
The area of the region that requires grass seed is found by subtracting [latex]\,60{x}^{2}-40x\,[/latex] units^2. This area can also be expressed in factored form as [latex]\,20x\left(3x-2\right)\,[/
latex] units^2. We can confirm that this is an equivalent expression by multiplying.
Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.
5.2.1 – Factoring the Greatest Common Factor of a Polynomial
When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, [latex]\,4\,[/latex] is the GCF of
[latex]\,16\,[/latex] and [latex]\,20\,[/latex] because it is the largest number that divides evenly into both [latex]\,16\,[/latex] and [latex]\,20\,[/latex] The GCF of polynomials works the same
way: [latex]\,4x\,[/latex] is the GCF of [latex]\,16x\,[/latex] and [latex]\,20{x}^{2}\,[/latex] because it is the largest polynomial that divides evenly into both [latex]\,16x\,[/latex] and [latex]
When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.
Greatest Common Factor
The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.
How To
Given a polynomial expression, factor out the greatest common factor.
1. Identify the GCF of the coefficients.
2. Identify the GCF of the variables.
3. Combine to find the GCF of the expression.
4. Determine what the GCF needs to be multiplied by to obtain each term in the expression.
5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.
Example 1 – Factoring the Greatest Common Factor
Factor [latex]\,6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy.[/latex]
First, find the GCF of the expression. The GCF of [latex]\,6,45,[/latex] and [latex]\,21\,[/latex] is [latex]\,3.\,[/latex] The GCF of [latex]\,{x}^{3},{x}^{2},[/latex] and [latex]\,x\,[/latex] is
[latex]\,x.\,[/latex] (Note that the GCF of a set of expressions in the form [latex]\,{x}^{n}\,[/latex] will always be the exponent of lowest degree.) And the GCF of [latex]\,{y}^{3},{y}^{2},[/latex]
and [latex]\,y\,[/latex] is [latex]\,y.\,[/latex] Combine these to find the GCF of the polynomial, [latex]\,3xy.[/latex]
Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]$$\,3xy\left(2{x}^{2}{y}^{2}\right)=6{x}^{3}{y}^{3},$$[/latex] [latex]$$3xy\left(15xy
\right)=45{x}^{2}{y}^{2},$$[/latex] and [latex]$$\,3xy\left(7\right)=21xy.$$[/latex]
Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.
After factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\,\left(3xy\right)\left(2{x}^{2}{y}^{2}+15xy+7\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}
Try It
Factor [latex]\,x\left({b}^{2}-a\right)+6\left({b}^{2}-a\right)\,[/latex] by pulling out the GCF.
Show answer
5.2.2 – Factoring a Trinomial with Leading Coefficient 1
Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]\,{x}^{2}+5x+6\,[/latex] has a GCF of
1, but it can be written as the product of the factors [latex]\,\left(x+2\right)\,[/latex] and [latex]\,\left(x+3\right).[/latex]
Trinomials of the form [latex]\,{x}^{2}+bx+c\,[/latex] can be factored by finding two numbers with a product of [latex]c\,[/latex] and a sum of [latex]\,b.\,[/latex] The trinomial [latex]\,{x}^{2}
+10x+16,[/latex] for example, can be factored using the numbers [latex]\,2\,[/latex] and [latex]\,8\,[/latex] because the product of those numbers is [latex]\,16\,[/latex] and their sum is [latex]
\,10.\,[/latex] The trinomial can be rewritten as the product of [latex]\,\left(x+2\right)\,[/latex] and [latex]\,\left(x+8\right).[/latex]
Factoring a Trinomial with Leading Coefficient 1
A trinomial of the form [latex]\,{x}^{2}+bx+c\,[/latex] can be written in factored form as [latex]\,\left(x+p\right)\left(x+q\right)\,[/latex] where [latex]\,pq=c\,[/latex] and [latex]\,p+q=b.[/
Can every trinomial be factored as a product of binomials?
No. Some polynomials cannot be factored. These polynomials are said to be prime.
How To
Given a trinomial in the form [latex]\,{x}^{2}+bx+c,[/latex] factor it.
1. List factors of [latex]\,c.[/latex]
2. Find [latex]\,p\,[/latex] and [latex]\,q,[/latex] a pair of factors of [latex]\,c\,[/latex] with a sum of [latex]\,b.[/latex]
3. Write the factored expression [latex]\,\left(x+p\right)\left(x+q\right).[/latex]
We have a trinomial with leading coefficient [latex]\,1,b=2,[/latex] and [latex]\,c=-15.\,[/latex] We need to find two numbers with a product of [latex]\,-15\,[/latex] and a sum of [latex]\,2.\,[/
latex] In (Figure), we list factors until we find a pair with the desired sum.
Factors of [latex]\,-15[/latex] Sum of Factors
[latex]1,-15[/latex] [latex]-14[/latex]
[latex]-1,15[/latex] 14
[latex]3,-5[/latex] [latex]-2[/latex]
[latex]-3,5[/latex] 2
Now that we have identified [latex]\,p\,[/latex] and [latex]\,q\,[/latex] as [latex]\,-3\,[/latex] and [latex]\,5,[/latex] write the factored form as [latex]\,\left(x-3\right)\left(x+5\right).[/
We can check our work by multiplying. Use FOIL to confirm that [latex]\,\left(x-3\right)\left(x+5\right)={x}^{2}+2x-15.[/latex]
Does the order of the factors matter?
No. Multiplication is commutative, so the order of the factors does not matter.
Try It
Factor [latex]\,{x}^{2}-7x+6.[/latex]
Show answer
Access these online resources for additional instruction and practice with factoring polynomials.
Key Equations
difference of squares [latex]{a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)[/latex]
perfect square trinomial [latex]{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}[/latex]
sum of cubes [latex]{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex]
difference of cubes [latex]{a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex]
• The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem. See (Example 1).
• Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term. See (Example 2).
factor by grouping
a method for factoring a trinomial in the form [latex]\,a{x}^{2}+bx+c\,[/latex] by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then
factoring out the GCF of the entire expression
greatest common factor
the largest polynomial that divides evenly into each polynomial | {"url":"https://pressbooks.pub/guttmanalgebratrig/chapter/5-2-factoring/","timestamp":"2024-11-01T19:03:38Z","content_type":"text/html","content_length":"77843","record_id":"<urn:uuid:25fa9a31-c6c5-400a-8d55-7a2d92925950>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00875.warc.gz"} |
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The Ring Tree Facility Location Problem
Electronic Notes in Discrete Mathematics (50): 331-336 (2015-12)
In this work we discuss a facility location variant of the capacitated ring tree problem. The new model generalizes vehicle routing problems and Steiner tree problems, yielding applicability in
telecommunication and transportation networks. In this ring tree facility location problem (RTFLP) two layers of networks have to be designed to connect two different types of customers to a central
depot. The first, or inner layer, consists of cycles that intersect in the depot and collect all type 2 customers, and some of the type 1 customers. An outer layer is represented by a forest that
contains the remaining type 1 customers such that each tree shares exactly one vertex with the inner layer. Capacity bounds apply to the number of connected substructures emanating from the depot,
the number of customers in each of these so-called ring trees, and in each tree of the forest. Additional optional Steiner vertices can be used to reduce the overall costs, which are layer-dependent
edge costs and facility location costs at the vertices in which the two layers coincide. Our contribution is the introduction of the RTFLP, the development of two mathematical formulations, and
preliminary computational results for the first RTFLP test set derived from instances from the literature.
Capacitated ring tree problem
Combinatorial optimization
Facility location
Integer programming
Network design
Steiner trees | {"url":"https://tore.tuhh.de/entities/publication/89206ee5-8d62-4c5a-a749-4f5dee88a5a3","timestamp":"2024-11-07T22:29:41Z","content_type":"text/html","content_length":"891907","record_id":"<urn:uuid:72c18c48-3cc3-4c58-ac54-233eb920492b>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00605.warc.gz"} |
mit18 05 s22 rquiz-instructions
R Quiz Instructions
18.05, Spring 2022
The quiz format is identical to that of the R studios and practice quiz.
• READ THE INSTRUCTIONS PDF!
• Read the problems carefully
• Save your work frequently
• You may use any resources except another person
• (This includes, paper, books, code on your computer, code on the internet ...)
• If code is given with a question it is meant to be used. Don’t just ignore it.
• Remember to use print or cat statements to print the values asked for
• Before uploading the code: clear your environment and source the entire file (choose
source from the code menu)
• Make sure that it runs without error and outputs just the answers asked for in the
• Due Friday 5/6 by 5 PM.
Download the zip file
• You should have downloaded the file mit18_05_s22_RQuiz.zip from our MITx site.
• Unzip it in your 18.05 studio folder.
• You should see the following R files
and the following other files
General instructions
• Using the Session menu, set the working directory to source file location.
• Answer the questions in the detailed instructions just below. Your answers should be
put in mit18_05_s22_RQuiz.r
R quiz instructions, Spring 2022
• For each question, you will complete the code for the corresponding function.
• As usual, use the function’s arguments and any other code given in the function.
• Do not print out things that are not asked for.
Note: The file mit18_05_s22_RQuiz-samplecode.r contains only a few lines of code, but
they may be useful to you.
Detailed instructions for this quiz
0. Clean your space
Problem 1 (20 points)
This problem will ask you to do several different short tasks. The parts are not related.
Problem 1a (5 points) Graphing
Here you will finish the code for the function:
rquiz_problem_1a(mu, sigma, w_shape, w_scale, a, b)
mu = mean of the normal pdf to plot
sigma = standard deviation of the normal pdf to plot
w_shape = shape parameter for Weibull pdf
w_scale = scale parameter for Weibull pdf
a, b = endpoints of the range of x for the plot
On the same plot for x between a and b put graphs of the pdfs of:
(i) Norm(mu, sigma)
(ii) Weibull distribution with shape and scale parameters w_shape and w_scale.
The graphs should be in different colors
Every plot should be a line graph (type=’l’)
Problem 1b (5 points) Combination and factorials.
Here you will finish the code for the function: rquiz_problem_1b(n, k, m)
n = see instructions below
k = see instructions below
m = see instructions below
This function should compute and print the following values:
(i) n choose k
(ii) m factorial
(iii) log of n choose k (for this use the function lchoose to avoid overflow)
Problem 1c (10 points) Bayesian success.
Here you will finish the code for the function:
rquiz_problem_1c(theta_values, num_patients, num_cured)
R quiz instructions, Spring 2022
theta_values = List of possible values of 𝜃.
num_patients = The number of patients in the trial.
num_cured = The number of successes in the trial.
A treatment with unknown probability 𝜃 of success is tried on num_patients patients yielding num_cured successes. The possible values of the unknown 𝜃 are given in theta_values.
That is, we only entertain a finite number of hypotheses for the value of 𝜃.
Your code should use the data and do each of the following:
(i) Compute the maximum likelihood estimate (MLE) for 𝜃. (HINT: the function which.max
might be useful.) Print out the MLE
(ii) Suppose there is a flat prior, i.e. each of the possible values of 𝜃 is equally likely. Find
the prior predictive probability that a single patient will be cured. Print out the predictive
probability. (This part does not use the experimental data.)
(iii) Use the data and a flat prior to do a Bayesian update to find the posterior probability
for 𝜃. Print out the posterior.
Problem 2 (20 points)
This problem is on making histograms.
Problem 2a (10 points) Here you will finish the code for the function:
rquiz_problem_2a = function(n_draws, k, bin_width)
n_draws = Number of sample points in the histogram
k = Number of degrees of freedom for the chi-square distribution
bin_width = Bin width for histogram
Your code should simulate n_draws draws from a chi-square distribution with k degrees
of freedom. Use the results to plot a density histogram. Use a bin width of bin_width.
Finally, plot the 𝜒2 (𝑘) probability density function on top of it.
Problem 2b (10 points) Here you will finish the code for the function:
rquiz_problem_2b = function(n_trials, n_draws, k, bin_width)
n_trials = Number of trials
n_draws = Number of sample points in each trial
k = Number of degrees of freedom for the chi-square distribution
bin_width = Bin width for histogram
This problem will illustrate the central limit theorem as follows.
• One trial will consist of drawing a sample of size n_draws from a 𝜒2 (𝑘) distribution.
• Simulate n_trials trials.
• For each trial, compute the standardized mean. It should help to know that the 𝜒2 (𝑘)
distribution has mean 𝑘 and variance 2𝑘.
R quiz instructions, Spring 2022
• Plot a density histogram of the n_trials standardized means. Use bin width bin_width.
• Add a graph of the standard normal pdf to the histogram.
Problem 3 (10 points)
Here you will finish the code for the function
rquiz_problem_3 = function(our_data, alpha)
our_data = data from some experiment.
alpha = Significance level for the Shapiro-Wilk test
You have collected data and before running a t-test, you want to check if the data comes
from a normal distribution. To do this, you run a Shapiro-Wilk test for normality.
Run the Shapiro-Wilk test. Then, print out the null hypothesis, p-value and whether or
not to reject the null hypothesis.
You need to use code to find and print the p-value and decide whether to reject or not.
You shouldn’t just read the p-value off the screen and enter that value. That is, your code
should work correctly even if we change the data or significance level for the test.
Problem 4 (Extra credit: 5 points)
This problem is for extra credit if you have time.
Make sure you have set the WORKING DIRECTORY to the source file location, so R will
find the file. The lists are data from independent random normal trials.
Here you will finish the code for the function
rquiz_problem_4 = function(data_file_name, alpha)
data_file_name = data file
alpha = Significance level for t-test
The given code extracts two lists, x and y from the data file.
Run a two sample t-test with unequal variances to test if x and y are drawn from distributions with the same mean
Print out the p-value and whether or not to reject the null hypothesis at significance level
You need to use code to find and print the p-value and decide whether to reject or not.
You shouldn’t just read the p-value off the screen and enter that value. That is, your code
should work correctly even if we change the data file or significance level for the test.
Testing your code
For each problem, we ran the problem function with certain parameters. You can see
the function call and the output in mit18_05_s22_RQuiz-test-answers.html. If you
call the same function with the same parameters, you should get the same results as in
R quiz instructions, Spring 2022
mit18_05_s22_RQuiz-test-answers.html – if there is randomness involved the answers
should be close but not identical.
For your convenience, the file mit18_05_s22_RQuiz-test.r contains all the function calls
used to make mit18_05_s22_RQuiz-test-answers.html.
Before uploading your code
1. Make sure all your code is in mit18_05_s22_RQuiz.r. Also make sure it is all inside
the functions for the problems.
2. Clean the environment and plots window.
3. Source the file.
4. Call each of the problem functions with the same parameters as the test file
5. Make sure it runs without error and outputs just the answers asked for in the questions.
6. Compare the output to the answers given in mit18_05_s22_RQuiz-test-answers.html.
Upload your code
Use the upload link on our MITx site to upload your code for grading.
Leave the file name as mit18_05_s22_RQuiz.r. (The upload script will automatically add
your name and a timestamp to the file.)
You can upload more than once. We will grade the last file you upload.
Due date: Friday 5/6 at 5 pm
MIT OpenCourseWare
18.05 Introduction to Probability and Statistics
Spring 2022
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | {"url":"https://studylib.net/doc/27307329/mit18-05-s22-rquiz-instructions","timestamp":"2024-11-13T02:01:03Z","content_type":"text/html","content_length":"58287","record_id":"<urn:uuid:1d75706e-dc88-43b6-a183-805dff35154c>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00000.warc.gz"} |
Fractional exponent equation
Algebra Tutorials! Saturday 2nd of November
fractional exponent equation
Related topics:
Home free online usable graphing calculator | exponential form calculator | history of multiplying and dividing integers | softmath.com | 7th grade slope activity |
Rotating a Parabola algebra help (convert from binary to decimal) | dividing multiplying adding and subtracting polynomials worksheets | simplify radical 1800 | multiplying and
Multiplying Fractions dividing powers
Finding Factors
Miscellaneous Equations
Mixed Numbers and Author Message
Improper Fractions
Systems of Equations in Johon-Daym Posted: Thursday 28th of Dec 12:18
Two Variables I have a difficulty with my math that calls for urgent solution. The difficulty is with fractional exponent equation. I have been looking
Literal Numbers out for somebody who can prepare me immediately as my exam is fast approaching . But it's tough to find someone quick enough besides it
Adding and Subtracting being costly. Can anyone direct me? It will be a huge help.
Subtracting Integers Registered:
Simplifying Complex 17.10.2002
Fractions From: the last place
Decimals and Fractions you saw me...
Multiplying Integers
Logarithmic Functions
Multiplying Monomials
Mixed nxu Posted: Friday 29th of Dec 07:46
The Square of a Binomial There are numerous within the whole subject of fractional exponent equation, for example, adding matrices, rational equations or graphing
Factoring Trinomials lines. I am acquainted with various folks who rejected those pricey alternatives for aid as well . Yet , do not give up because I
The Pythagorean Theorem discovered a different resolution that is low-cost , uncomplicated to use and to a greater extent more applicative than I could have ever
Solving Radical imagined . Subsequent to experimentation with demonstrative math computer software products and closely giving up , I encountered
Equations in One Registered: Algebrator. This package has accurately supplied answers to each and every mathematics problem I have furnished to the software program .
Variable 25.10.2006 Just as powerful , Algebrator as well offers all of the interim measures needed to derive the ultimate resolution . In spite of the fact
Multiplying Binomials From: Siberia, that someone could apply the software merely to complete exercises, I am uncertain about anyone should be allowed to employ the program for
Using the FOIL Method Russian Federation tests .
Imaginary Numbers
Solving Quadratic
Equations Using the
Quadratic Formula MichMoxon Posted: Friday 29th of Dec 15:33
Solving Quadratic Thanks for the advice . Algebrator is actually a pretty good math software. I was able to get answers to questions I had about
Equations perpendicular lines, adding functions and binomials. You just have to type in a problem, click on Solve and you get the all the solutions
Algebra you need. You can use it for all types of , like Pre Algebra, Algebra 2 and Remedial Algebra. I would highly recommend Algebrator.
Order of Operations
Dividing Complex Numbers Registered:
Polynomials 21.08.2001
The Appearance of a From:
Polynomial Equation
Standard Form of a Line
Positive Integral
Divisors mopy8gen Posted: Saturday 30th of Dec 11:57
Dividing Fractions Wow, sounds wonderful! I wish to know more about this fabulous product. Please let me know.
Solving Linear Systems
of Equations by
Factoring Registered:
Multiplying and Dividing 12.05.2002
Square Roots From: Port Dickson,
Functions and Graphs Malaysia
Dividing Polynomials
Solving Rational
Numbers erx Posted: Sunday 31st of Dec 07:14
Use of Parentheses or You can download this program from https://gre-test-prep.com/the-cartesian-coordinate-system.html. There are some demos available to see if
Brackets (The it is what want and if you find it good , you can get a licensed version for a nominal amount.
Distributive Law)
Multiplying and Dividing
by Monomials Registered:
Solving Quadratic 26.10.2001
Equations by Graphing From: PL/DE/ES/GB/HU
Multiplying Decimals
Use of Parentheses or
Brackets (The
Distributive Law) Mibxrus Posted: Monday 01st of Jan 08:13
Simplifying Complex gcf, difference of cubes and proportions were a nightmare for me until I found Algebrator, which is truly the best algebra program that I
Fractions 1 have come across. I have used it through several math classes – Basic Math, Basic Math and Basic Math. Just typing in the math problem and
Adding Fractions clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I highly recommend
Simplifying Complex the program.
Fractions Registered:
Solutions to Linear 19.10.2002
Equations in Two From: Vancouver,
Variables Canada
Quadratic Expressions
Completing Squares
Dividing Radical
Rise and Run
Graphing Exponential
Multiplying by a
The Cartesian Coordinate
Writing the Terms of a
Polynomial in Descending
Quadratic Expressions
Solving Inequalities
Solving Rational
Inequalities with a Sign
Solving Linear Equations
Solving an Equation with
Two Radical Terms
Simplifying Rational
Intercepts of a Line
Completing the Square
Order of Operations
Factoring Trinomials
Solving Linear Equations
Solving Multi-Step
Solving Quadratic
Equations Graphically
and Algebraically
Collecting Like Terms
Solving Equations with
Radicals and Exponents
Percent of Change
Powers of ten
(Scientific Notation)
Comparing Integers on a
Number Line
Solving Systems of
Equations Using
Factoring Out the
Greatest Common Factor
Families of Functions
Monomial Factors
Multiplying and Dividing
Complex Numbers
Properties of Exponents
Multiplying Square Roots
Adding or Subtracting
Rational Expressions
with Different
Expressions with
Variables as Exponents
The Quadratic Formula
Writing a Quadratic with
Given Solutions
Simplifying Square Roots
Adding and Subtracting
Square Roots
Adding and Subtracting
Rational Expressions
Combining Like Radical
Solving Systems of
Equations Using
Dividing Polynomials
Graphing Functions
Product of a Sum and a
Solving First Degree
Solving Equations with
Radicals and Exponents
Roots and Powers
Multiplying Numbers
fractional exponent equation
Related topics:
Home free online usable graphing calculator | exponential form calculator | history of multiplying and dividing integers | softmath.com | 7th grade slope activity |
Rotating a Parabola algebra help (convert from binary to decimal) | dividing multiplying adding and subtracting polynomials worksheets | simplify radical 1800 | multiplying and
Multiplying Fractions dividing powers
Finding Factors
Miscellaneous Equations
Mixed Numbers and Author Message
Improper Fractions
Systems of Equations in Johon-Daym Posted: Thursday 28th of Dec 12:18
Two Variables I have a difficulty with my math that calls for urgent solution. The difficulty is with fractional exponent equation. I have been looking
Literal Numbers out for somebody who can prepare me immediately as my exam is fast approaching . But it's tough to find someone quick enough besides it
Adding and Subtracting being costly. Can anyone direct me? It will be a huge help.
Subtracting Integers Registered:
Simplifying Complex 17.10.2002
Fractions From: the last place
Decimals and Fractions you saw me...
Multiplying Integers
Logarithmic Functions
Multiplying Monomials
Mixed nxu Posted: Friday 29th of Dec 07:46
The Square of a Binomial There are numerous within the whole subject of fractional exponent equation, for example, adding matrices, rational equations or graphing
Factoring Trinomials lines. I am acquainted with various folks who rejected those pricey alternatives for aid as well . Yet , do not give up because I
The Pythagorean Theorem discovered a different resolution that is low-cost , uncomplicated to use and to a greater extent more applicative than I could have ever
Solving Radical imagined . Subsequent to experimentation with demonstrative math computer software products and closely giving up , I encountered
Equations in One Registered: Algebrator. This package has accurately supplied answers to each and every mathematics problem I have furnished to the software program .
Variable 25.10.2006 Just as powerful , Algebrator as well offers all of the interim measures needed to derive the ultimate resolution . In spite of the fact
Multiplying Binomials From: Siberia, that someone could apply the software merely to complete exercises, I am uncertain about anyone should be allowed to employ the program for
Using the FOIL Method Russian Federation tests .
Imaginary Numbers
Solving Quadratic
Equations Using the
Quadratic Formula MichMoxon Posted: Friday 29th of Dec 15:33
Solving Quadratic Thanks for the advice . Algebrator is actually a pretty good math software. I was able to get answers to questions I had about
Equations perpendicular lines, adding functions and binomials. You just have to type in a problem, click on Solve and you get the all the solutions
Algebra you need. You can use it for all types of , like Pre Algebra, Algebra 2 and Remedial Algebra. I would highly recommend Algebrator.
Order of Operations
Dividing Complex Numbers Registered:
Polynomials 21.08.2001
The Appearance of a From:
Polynomial Equation
Standard Form of a Line
Positive Integral
Divisors mopy8gen Posted: Saturday 30th of Dec 11:57
Dividing Fractions Wow, sounds wonderful! I wish to know more about this fabulous product. Please let me know.
Solving Linear Systems
of Equations by
Factoring Registered:
Multiplying and Dividing 12.05.2002
Square Roots From: Port Dickson,
Functions and Graphs Malaysia
Dividing Polynomials
Solving Rational
Numbers erx Posted: Sunday 31st of Dec 07:14
Use of Parentheses or You can download this program from https://gre-test-prep.com/the-cartesian-coordinate-system.html. There are some demos available to see if
Brackets (The it is what want and if you find it good , you can get a licensed version for a nominal amount.
Distributive Law)
Multiplying and Dividing
by Monomials Registered:
Solving Quadratic 26.10.2001
Equations by Graphing From: PL/DE/ES/GB/HU
Multiplying Decimals
Use of Parentheses or
Brackets (The
Distributive Law) Mibxrus Posted: Monday 01st of Jan 08:13
Simplifying Complex gcf, difference of cubes and proportions were a nightmare for me until I found Algebrator, which is truly the best algebra program that I
Fractions 1 have come across. I have used it through several math classes – Basic Math, Basic Math and Basic Math. Just typing in the math problem and
Adding Fractions clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I highly recommend
Simplifying Complex the program.
Fractions Registered:
Solutions to Linear 19.10.2002
Equations in Two From: Vancouver,
Variables Canada
Quadratic Expressions
Completing Squares
Dividing Radical
Rise and Run
Graphing Exponential
Multiplying by a
The Cartesian Coordinate
Writing the Terms of a
Polynomial in Descending
Quadratic Expressions
Solving Inequalities
Solving Rational
Inequalities with a Sign
Solving Linear Equations
Solving an Equation with
Two Radical Terms
Simplifying Rational
Intercepts of a Line
Completing the Square
Order of Operations
Factoring Trinomials
Solving Linear Equations
Solving Multi-Step
Solving Quadratic
Equations Graphically
and Algebraically
Collecting Like Terms
Solving Equations with
Radicals and Exponents
Percent of Change
Powers of ten
(Scientific Notation)
Comparing Integers on a
Number Line
Solving Systems of
Equations Using
Factoring Out the
Greatest Common Factor
Families of Functions
Monomial Factors
Multiplying and Dividing
Complex Numbers
Properties of Exponents
Multiplying Square Roots
Adding or Subtracting
Rational Expressions
with Different
Expressions with
Variables as Exponents
The Quadratic Formula
Writing a Quadratic with
Given Solutions
Simplifying Square Roots
Adding and Subtracting
Square Roots
Adding and Subtracting
Rational Expressions
Combining Like Radical
Solving Systems of
Equations Using
Dividing Polynomials
Graphing Functions
Product of a Sum and a
Solving First Degree
Solving Equations with
Radicals and Exponents
Roots and Powers
Multiplying Numbers
Rotating a Parabola
Multiplying Fractions
Finding Factors
Miscellaneous Equations
Mixed Numbers and
Improper Fractions
Systems of Equations in
Two Variables
Literal Numbers
Adding and Subtracting
Subtracting Integers
Simplifying Complex
Decimals and Fractions
Multiplying Integers
Logarithmic Functions
Multiplying Monomials
The Square of a Binomial
Factoring Trinomials
The Pythagorean Theorem
Solving Radical
Equations in One
Multiplying Binomials
Using the FOIL Method
Imaginary Numbers
Solving Quadratic
Equations Using the
Quadratic Formula
Solving Quadratic
Order of Operations
Dividing Complex Numbers
The Appearance of a
Polynomial Equation
Standard Form of a Line
Positive Integral
Dividing Fractions
Solving Linear Systems
of Equations by
Multiplying and Dividing
Square Roots
Functions and Graphs
Dividing Polynomials
Solving Rational
Use of Parentheses or
Brackets (The
Distributive Law)
Multiplying and Dividing
by Monomials
Solving Quadratic
Equations by Graphing
Multiplying Decimals
Use of Parentheses or
Brackets (The
Distributive Law)
Simplifying Complex
Fractions 1
Adding Fractions
Simplifying Complex
Solutions to Linear
Equations in Two
Quadratic Expressions
Completing Squares
Dividing Radical
Rise and Run
Graphing Exponential
Multiplying by a
The Cartesian Coordinate
Writing the Terms of a
Polynomial in Descending
Quadratic Expressions
Solving Inequalities
Solving Rational
Inequalities with a Sign
Solving Linear Equations
Solving an Equation with
Two Radical Terms
Simplifying Rational
Intercepts of a Line
Completing the Square
Order of Operations
Factoring Trinomials
Solving Linear Equations
Solving Multi-Step
Solving Quadratic
Equations Graphically
and Algebraically
Collecting Like Terms
Solving Equations with
Radicals and Exponents
Percent of Change
Powers of ten
(Scientific Notation)
Comparing Integers on a
Number Line
Solving Systems of
Equations Using
Factoring Out the
Greatest Common Factor
Families of Functions
Monomial Factors
Multiplying and Dividing
Complex Numbers
Properties of Exponents
Multiplying Square Roots
Adding or Subtracting
Rational Expressions
with Different
Expressions with
Variables as Exponents
The Quadratic Formula
Writing a Quadratic with
Given Solutions
Simplifying Square Roots
Adding and Subtracting
Square Roots
Adding and Subtracting
Rational Expressions
Combining Like Radical
Solving Systems of
Equations Using
Dividing Polynomials
Graphing Functions
Product of a Sum and a
Solving First Degree
Solving Equations with
Radicals and Exponents
Roots and Powers
Multiplying Numbers
fractional exponent equation
Related topics:
free online usable graphing calculator | exponential form calculator | history of multiplying and dividing integers | softmath.com | 7th grade slope activity | algebra help (convert from
binary to decimal) | dividing multiplying adding and subtracting polynomials worksheets | simplify radical 1800 | multiplying and dividing powers
Author Message
Johon-Daym Posted: Thursday 28th of Dec 12:18
I have a difficulty with my math that calls for urgent solution. The difficulty is with fractional exponent equation. I have been looking out for somebody who can
prepare me immediately as my exam is fast approaching . But it's tough to find someone quick enough besides it being costly. Can anyone direct me? It will be a huge
From: the last place
you saw me...
nxu Posted: Friday 29th of Dec 07:46
There are numerous within the whole subject of fractional exponent equation, for example, adding matrices, rational equations or graphing lines. I am acquainted with
various folks who rejected those pricey alternatives for aid as well . Yet , do not give up because I discovered a different resolution that is low-cost ,
uncomplicated to use and to a greater extent more applicative than I could have ever imagined . Subsequent to experimentation with demonstrative math computer software
products and closely giving up , I encountered Algebrator. This package has accurately supplied answers to each and every mathematics problem I have furnished to the
Registered: software program . Just as powerful , Algebrator as well offers all of the interim measures needed to derive the ultimate resolution . In spite of the fact that
25.10.2006 someone could apply the software merely to complete exercises, I am uncertain about anyone should be allowed to employ the program for tests .
From: Siberia,
Russian Federation
MichMoxon Posted: Friday 29th of Dec 15:33
Thanks for the advice . Algebrator is actually a pretty good math software. I was able to get answers to questions I had about perpendicular lines, adding functions
and binomials. You just have to type in a problem, click on Solve and you get the all the solutions you need. You can use it for all types of , like Pre Algebra,
Algebra 2 and Remedial Algebra. I would highly recommend Algebrator.
mopy8gen Posted: Saturday 30th of Dec 11:57
Wow, sounds wonderful! I wish to know more about this fabulous product. Please let me know.
From: Port Dickson,
erx Posted: Sunday 31st of Dec 07:14
You can download this program from https://gre-test-prep.com/the-cartesian-coordinate-system.html. There are some demos available to see if it is what want and if you
find it good , you can get a licensed version for a nominal amount.
From: PL/DE/ES/GB/HU
Mibxrus Posted: Monday 01st of Jan 08:13
gcf, difference of cubes and proportions were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have used
it through several math classes – Basic Math, Basic Math and Basic Math. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step
solution to the problem, and my algebra homework would be ready. I highly recommend the program.
From: Vancouver,
Author Message
Johon-Daym Posted: Thursday 28th of Dec 12:18
I have a difficulty with my math that calls for urgent solution. The difficulty is with fractional exponent equation. I have been looking out for somebody who can prepare me
immediately as my exam is fast approaching . But it's tough to find someone quick enough besides it being costly. Can anyone direct me? It will be a huge help.
From: the last place
you saw me...
nxu Posted: Friday 29th of Dec 07:46
There are numerous within the whole subject of fractional exponent equation, for example, adding matrices, rational equations or graphing lines. I am acquainted with various
folks who rejected those pricey alternatives for aid as well . Yet , do not give up because I discovered a different resolution that is low-cost , uncomplicated to use and to a
greater extent more applicative than I could have ever imagined . Subsequent to experimentation with demonstrative math computer software products and closely giving up , I
encountered Algebrator. This package has accurately supplied answers to each and every mathematics problem I have furnished to the software program . Just as powerful ,
Registered: Algebrator as well offers all of the interim measures needed to derive the ultimate resolution . In spite of the fact that someone could apply the software merely to complete
25.10.2006 exercises, I am uncertain about anyone should be allowed to employ the program for tests .
From: Siberia,
Russian Federation
MichMoxon Posted: Friday 29th of Dec 15:33
Thanks for the advice . Algebrator is actually a pretty good math software. I was able to get answers to questions I had about perpendicular lines, adding functions and
binomials. You just have to type in a problem, click on Solve and you get the all the solutions you need. You can use it for all types of , like Pre Algebra, Algebra 2 and
Remedial Algebra. I would highly recommend Algebrator.
mopy8gen Posted: Saturday 30th of Dec 11:57
Wow, sounds wonderful! I wish to know more about this fabulous product. Please let me know.
From: Port Dickson,
erx Posted: Sunday 31st of Dec 07:14
You can download this program from https://gre-test-prep.com/the-cartesian-coordinate-system.html. There are some demos available to see if it is what want and if you find it
good , you can get a licensed version for a nominal amount.
From: PL/DE/ES/GB/HU
Mibxrus Posted: Monday 01st of Jan 08:13
gcf, difference of cubes and proportions were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have used it
through several math classes – Basic Math, Basic Math and Basic Math. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step solution to the
problem, and my algebra homework would be ready. I highly recommend the program.
From: Vancouver,
Posted: Thursday 28th of Dec 12:18
I have a difficulty with my math that calls for urgent solution. The difficulty is with fractional exponent equation. I have been looking out for somebody who can prepare me immediately as my exam is
fast approaching . But it's tough to find someone quick enough besides it being costly. Can anyone direct me? It will be a huge help.
Posted: Friday 29th of Dec 07:46
There are numerous within the whole subject of fractional exponent equation, for example, adding matrices, rational equations or graphing lines. I am acquainted with various folks who rejected those
pricey alternatives for aid as well . Yet , do not give up because I discovered a different resolution that is low-cost , uncomplicated to use and to a greater extent more applicative than I could
have ever imagined . Subsequent to experimentation with demonstrative math computer software products and closely giving up , I encountered Algebrator. This package has accurately supplied answers to
each and every mathematics problem I have furnished to the software program . Just as powerful , Algebrator as well offers all of the interim measures needed to derive the ultimate resolution . In
spite of the fact that someone could apply the software merely to complete exercises, I am uncertain about anyone should be allowed to employ the program for tests .
Posted: Friday 29th of Dec 15:33
Thanks for the advice . Algebrator is actually a pretty good math software. I was able to get answers to questions I had about perpendicular lines, adding functions and binomials. You just have to
type in a problem, click on Solve and you get the all the solutions you need. You can use it for all types of , like Pre Algebra, Algebra 2 and Remedial Algebra. I would highly recommend Algebrator.
Posted: Saturday 30th of Dec 11:57
Wow, sounds wonderful! I wish to know more about this fabulous product. Please let me know.
Posted: Sunday 31st of Dec 07:14
You can download this program from https://gre-test-prep.com/the-cartesian-coordinate-system.html. There are some demos available to see if it is what want and if you find it good , you can get a
licensed version for a nominal amount.
Posted: Monday 01st of Jan 08:13
gcf, difference of cubes and proportions were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have used it through several math classes
– Basic Math, Basic Math and Basic Math. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I
highly recommend the program. | {"url":"https://gre-test-prep.com/algebra-1-practice-test/exponent-rules/fractional-exponent-equation.html","timestamp":"2024-11-02T20:48:11Z","content_type":"text/html","content_length":"118157","record_id":"<urn:uuid:3a5b3bc2-a8e5-4237-a6ac-c6f9f232e8e3>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00132.warc.gz"} |
relative pseudo-complement — definition, examples, related words and more at Wordnik
relative pseudo-complement
from Wiktionary, Creative Commons Attribution/Share-Alike License.
• noun mathematics The residual operation of a Heyting algebra when considered as a residuated lattice whose monoid operation is the meet operation. Equivalently, the relative pseudo-complement of
a with respect to b is the supremum of the set of all z such that , where denotes the meet operation of the given Heyting algebra.
Sorry, no etymologies found.
Sorry, no example sentences found.
Log in or sign up to get involved in the conversation. It's quick and easy. | {"url":"https://wordnik.com/words/relative%20pseudo-complement","timestamp":"2024-11-10T01:31:55Z","content_type":"text/html","content_length":"71535","record_id":"<urn:uuid:7421a12f-ff75-4fbf-93f8-f278a1d70819>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00706.warc.gz"} |
Error "time step too small in opamp"
Not open for further replies.
i have used op27A opamp part in converter circuit it shows time step too small error, so kindly help me to solve the issue.
I don't have this opamp in my library so I am unable to run it. Most probably your opamp is going into saturation. What you can do is test this opamp in a new file with the required input voltages
and power supply. The issue is likely with the power supply you are giving. Also, see from the datasheet its maximum operational frequency and whether you are within the specified range.
Well-Known Member
Most Helpful Member
The problem may not lie with the opamp model. Sometimes giving voltage sources a finite impedance, or slightly altering capacitor or inductor properties, is enough to get round the problem.
Well-Known Member
Most Helpful Member
i have used op27A opamp part in converter circuit it shows time step too small error, so kindly help me to solve the issue.
There is a model used in your simulation that is not in the library. You will have to supply the model if you want other people to be able ot run your simulation.
You will get time step errors when the derivatives in the numerical solution become too large or unstable. What this means is that even with the smallest time step being used the derivatives can not
be contained properly or change too rapidly.
A solution that works sometimes is to make the maximum step size smaller. This sometimes works because it allows the simulator algorithm to be able to control the derivatives leading up to the
problem derivative(s) and thus when it gets there again it finds a naturally smoother solution and so is able to solve it with a reasonable time step.
This doesnt always work because sometimes the solution changes too fast (could be a 'shock') no matter what we do with the algorithm. The alternative is to find the problem time constant and try to
slow it down a little without changing the overall simulation by too much. The odds are that slowing it down slightly wont change the overall results too much because the time frame over which the
solution exists is typically very small compared to the larger time constants.
Last edited:
There is a model used in your simulation that is not in the library. You will have to supply the model if you want other people to be able ot run your simulation.
You will get time step errors when the derivatives in the numerical solution become too large or unstable. What this means is that even with the smallest time step being used the derivatives can
not be contained properly or change too rapidly.
A solution that works sometimes is to make the maximum step size smaller. This sometimes works because it allows the simulator algorithm to be able to control the derivatives leading up to the
problem derivative(s) and thus when it gets there again it finds a naturally smoother solution and so is able to solve it with a reasonable time step.
This doesnt always work because sometimes the solution changes too fast (could be a 'shock') no matter what we do with the algorithm. The alternative is to find the problem time constant and try
to slow it down a little without changing the overall simulation by too much. The odds are that slowing it down slightly wont change the overall results too much because the time frame over which
the solution exists is typically very small compared to the larger time constants.
One specific case of this means that there is something that wants to change instantly in your simulation (vertical line on a graph) when the physical component does not allow instantaneous changes
(like current through an inductor) and the simulation can't deal with it. Sometimes the way to fix this is to approximate the vertical step by a steep slope that the simulator can deal with.
Well-Known Member
Most Helpful Member
One specific case of this means that there is something that wants to change instantly in your simulation (vertical line on a graph) when the physical component does not allow instantaneous
changes (like current through an inductor) and the simulation can't deal with it. Sometimes the way to fix this is to approximate the vertical step by a steep slope that the simulator can deal
But how do you tell the simulator to approximate the vertical step with a steep slope. You cant, all you can do is slow down the time constant.
But how do you tell the simulator to approximate the vertical step with a steep slope. You cant, all you can do is slow down the time constant.
You do it by modifying the circuit, not by literally telling the simulator to make treat a particular vertical slope as ramp. Same thing as slowing the time constant I guess, but slightly different
mental approach which is what I did the last time I had to do this.
I have also received this error simulating certain types of feedback loops that resulted in race conditions because it was an ideal simulation environment with no noise or disturbances to unsettle
the circuit and knock it into equilibrium.
Last edited:
Well-Known Member
Most Helpful Member
time step too small error,
This is a common problem in LT SPICE. Now I wish I could remember......
I think it is from running a very long sim. (in your case 5 seconds) and having a very short Max Time Step.
Example; If you want a time step of 0.1nS and you want 5 seconds of those you have too many data points and run out of memory.
Google "time step too small error LTSPICE" That is what I did when I first saw this.
AND What Mr Al said is usually right.
i have used op27A opamp part in converter circuit it shows time step too small error, so kindly help me to solve the issue.
A few of the symbol and model files are missing.
Some of the symbols will be in the autogenerated folder.
Share your symbols and model files and we'll take a look at it.
Well-Known Member
Most Helpful Member
i'm having trouble running it because of missing symbols. however, the most often cause for "time step too small", is if you get positive feedback where you should have negative feedback. this makes
it impossible for LTSpice to get a stable operating point. somewhere you have an inverting and noninverting input swapped.
Not open for further replies. | {"url":"https://www.electro-tech-online.com/threads/error-time-step-too-small-in-opamp.153093/","timestamp":"2024-11-13T05:31:49Z","content_type":"text/html","content_length":"137643","record_id":"<urn:uuid:46bae71e-8937-4c0b-a001-e5bd856a9340>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00518.warc.gz"} |
Using Writing in Math to Deepen Understanding
When we evaluate student work artifacts — the actual work students are asked to do in the classroom — one thing we count as an important strength is evidence of writing in math class.
Until relatively recently, the use of writing in math was almost non-existent. The Common Core helped usher in some use of writing in math, and we are seeing more of it, though its use is by no
means universal or even consistent within or between schools in the same district. In general, we see it more in the elementary grades, and less and less in the secondary grades — which is
unfortunate, because the increased complexity of secondary mathematics makes the use of writing to organize ideas and justify solutions even more valuable.
Why don’t we do more of this? When the first studies about math and writing began to be published in the 1990s and earlier, there was tremendous pushback from math teachers who felt that writing had
no place in math: math was exclusively a world of numbers and formulas. Also, there was some predictable backlash on the grading front: would math teachers have to grade writing as though they were
— gasp! — English teachers? Would English teachers be asked to grade math’s papers if they required writing? Sounds silly, I know, but I sat through these exact debates in the mid 90s when there was
a big push for reading and writing across content areas. To say it has taken a while to catch on is a massive understatement.
Writing to Learn. Writing uses both hemispheres of the brain: one part generates ideas and the other organizes them and places them in a structure. Both halves work together to make meaning of new
content by describing, ordering, evaluating, and justifying. We see this all the time in the marriage of writing and reading; it works exactly the same way in math. Asking kids to write about their
mathematical understanding and justify their solutions requires them to use different parts of their brains than are usually employed in math processes — it helps them get to the conceptual level of
the math, rather than staying at the level of execution. If math were only about execution, we’d all be math geniuses, memorizing formulas and firing them off with abandon. But problem-solution
isn’t always a simple one-to-one correspondence. As math becomes more complex, there may be many ways to arrive at the same answer. If this is so, then being able to justify your solution is as
important as arriving at it in the first place. There’s also a direct correlation between a student’s ability to explain math concepts and his/her ability to understand and apply them.
Accessing Higher Order Thinking. Writing in math, especially writing to justify a solution, pushes the student’s thinking into higher levels of cognition. A lot of mathematical thinking hovers
around Understanding and Applying. Asking a child to write about math and justify how s/he arrived at a solution or how s/he knows it’s correct pushes the thinking into Analyzing and Evaluating,
while the act of writing about it and synthesizing the understanding into a paragraph qualifies as Creating. Using multiple means to represent understanding also helps consolidate learning from
several directions at once, boosting retention. In two of the three examples here, the student has combined writing and pictures to clarify thinking and demonstrate how the solution works. Kids
should begin doing this as early as possible. For pre-writers, dictation combined with drawing is a good way to help them begin to capture their understanding of math concepts on paper.
One thing that’s important here is that the writing in math is not graded as it would be in language arts. The goal is not grammatical correctness and precision punctuation, it’s consolidating
learning by expressing thought processes and mathematical ideas. We’re looking for whether the student can justify or explain thinking and solutions, not whether he or she capitalizes
appropriately. To a large extent, the child needs to be free of most of the constraints that might hinder expression and we absolutely don’t want to deduct points because they spelled something wrong
(like remainder in the first example). They need opportunity to use mathematics vocabulary (like congruent, mean, quadrilateral, algorithm) in low-stakes environments.
Kids can get better at math….by writing. | {"url":"https://curriculumsolutions.net/blog/2021/04/15/using-writing-in-math-to-deepen-understanding/","timestamp":"2024-11-05T05:35:06Z","content_type":"text/html","content_length":"30910","record_id":"<urn:uuid:20e1b793-3ff8-43f4-801f-4c79f38d213c>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00278.warc.gz"} |
Re: Select from Tuplet using logical expression
• To: mathgroup at smc.vnet.net
• Subject: [mg116856] Re: Select from Tuplet using logical expression
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Wed, 2 Mar 2011 04:36:30 -0500 (EST)
On 3/1/11 at 5:20 AM, lt648 at hszk.bme.hu (Lengyel Tamas) wrote:
>Skip if needed:
>///I am working on a part combinatorical problem with sets of 3
>differently indexed values (e.g. F_i, F_j, F_k, F denoting frequency
>channels) which are subsets of many values (e.g 16 different
>frequency channels, denoted F_0, F_1 ... F_15).
>Now, I need to select triplets from these channels, I used Tuplets.
>So far so good. From these I need those combinations where indexes
>i!=k and/or j!=k, and i=j is allowed (e.g {i,j,k} = {12, 12, 4} is=
>valid channel combination, but {3, 12, 3} is not).///
>So basically I need to generate triplets from a range of integer
>numbers, where the first and second elements of these triplets do
>not match the third. I thought Select would help, but I don't know
>if there exists an option to control elements' values in a
>From then on I must use these triplets' elements in a function.
>But first I am asking your help in generating thos triplets of
Perhaps something like
DeleteCases[Tuples[Range[3], 3], {a_, a_, a_} | {a_, _, a_}]
will do well enough for you. But note, doing Tuples[Range[3]]
generates all possible triplets with the integers from 1 to 3,
including those you don't want. So, if your range of integers is
large, this method may not be satisfactory as it will consume
quite a bit of memory. It is a brute force rather than efficient method. | {"url":"http://forums.wolfram.com/mathgroup/archive/2011/Mar/msg00050.html","timestamp":"2024-11-10T09:26:26Z","content_type":"text/html","content_length":"31386","record_id":"<urn:uuid:696aaaa2-77fc-432d-ba0a-14ead3daa8b4>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00625.warc.gz"} |
Shormann Interactive Math: Prealgebra 1 through Calculus
David Shormann has taught Saxon Math courses for many years. When he decided to create his own math courses, he drew upon some of John Saxon's best ideas and combined those with his own ideas to
create high-school courses for Algebra 1, Algebra 2, Precalculus, Calculus I, and Calculus II. There's also a Prealgebra course available.
As in the original Saxon Math series, Shormann Math integrates geometry with algebra within courses titled Algebra 1 and Algebra 2. There is no separate geometry course.
In addition, courses teach in increments with plenty of practice and continual review, as did John Saxon's courses. For example, when a student learns a new concept, they will learn some new rules
and definitions, and practice a few examples. The new concept is repeated in multiple practice sets, then on a timed quiz, and finally on a quarterly exam. Later, another lesson will teach a new
concept that builds on the previous concept.
Shormann’s courses are presented through streaming videos. Practice problems and problems on quizzes and exams are worked on paper, but answers are entered into the program where students get
immediate feedback. Solutions are on videos and in PDF files, and an automated grading and tracking system records student progress. Many students and parents will also appreciate that Shormann Math
courses have fewer problems per lesson than Saxon Math courses.
Shormann incorporates technology, both in the presentation of the course material and in the teaching itself. While not absolutely required, a graphing calculator and geometry software like Geogebra
(geogebra.org) can be used throughout the courses. All students will need at least a scientific calculator. Precalculus and Calculus students have the opportunity to learn Onshape^®, a cloud-based,
computer-aided design (CAD) program.
Distinctive Features
There are four distinctive features of Shormann Math that help students make connections to other areas of life.
First, the courses teach mathematical history as students learn about the development of mathematical concepts and the people who came up with them.
A second distinctive feature is that a Christian worldview is incorporated throughout each course. Dr. Shormann's main goal in writing these courses was to present math in a way that connects
students to their world and their Creator. For instance, each course begins by defining mathematics as "the language of science and a God-given tool for measuring and classifying pattern and shape."
A third distinctive feature is art. For example, Algebra 1 and 2 students learn how to create one-point perspective drawings and how to identify perspective in famous paintings such as Da Vinci's
"The Last Supper."
The fourth distinctive feature is test preparation. Every problem #15 in Algebra 1 and 2 is a sample problem from an exam—the PSAT™, CLEP^®, College Algebra, CLEP College Math, the SAT^®, or the ACT^
®. (The source exam is shown at the beginning of each problem.) In addition, Shormann Algebra 2 includes a short, optional CLEP prep course called CLEP Professor College Algebra. Based on math taught
up through Shormann Algebra 2 and this CLEP prep course, students should be prepared to take either the CLEP College Math or CLEP College Algebra exam to earn college credits. Students should also be
well prepared for the latest PSAT, SAT, and ACT exams. To earn even more college credits, students completing Shormann Precalculus or Calculus can also use the included CLEP Professor Precalculus
prep lessons to prepare for the CLEP exams for both topics or the two AP^® exams for calculus.
How Lessons Work
For each lesson, students begin by reading through rules and definitions in the online textbook. The textbook lessons are linked for each lesson, so they are simple to access. While students could
also read through the day’s lesson in the online textbook, they should not do so in most cases since the content is thoroughly taught in the video presentations. Shormann believes that the student's
time is better spent trying to solve problems rather than reading.
You are given permission to print out the reading assignment in the textbook for each enrolled student). You might have students copy definitions and rules into a two-inch-thick, three-ring binder,
which they will need for each course as they accumulate their notes and coursework. The physical act of writing out definitions and rules is a strong reinforcement, so I recommend this step. These
notes should be used for reference and review as well. If you choose not to do this, you should probably print out the textbook material for reference so students have a written form of the
information to study. All courses also have digital flashcards for reviewing the rules and definitions, offering yet another way to reinforce key material.
After reading the rules and definitions and writing them out (if you have instructed them to do so), students will watch the video presentation on a computer or a tablet. Videos vary in length, with
some running more than 30 minutes and others running fewer than 10 minutes. The first 25 lessons in each course cover fundamentals and review material, and their videos typically run longer than
videos for the rest of the lessons. Students should take notes as they watch each video, pausing it as needed. During the lectures, after each example problem is presented, the student is directed to
pause the lecture and attempt to solve it on their lecture notes so they can immediately see whether they are understanding the lesson.
A set of practice problems is next. Most regular practice sets (not review lessons) have at least one problem with a real-life application of math. Practice problems are worked out on paper. Then
answers are submitted through the program. There are links above sets of practice problems that go to one or more examples of similar problems already worked out, as well as to the video lecture
where the concept was originally taught. If they are struggling, students should use these links to try to work out their own solution before giving up on a problem.
When students miss a problem, Shormann outlines a step-by-step procedure for them to follow. They can use the video or PDF solutions that become available to students on a “results” page after they
have submitted their answers. They should then rework missed problems on a “corrections” page in their notebook. If they are still stuck at this point, they can submit an email to Dr. Shormann for
assistance. All of this means that students should never get stuck on a concept or skill if they follow the instructions.
After every fourth lesson, there is a four-question, “open-book” quiz. The quizzes are timed, but the time allowed should be more than sufficient if students don’t take a break in the middle. After
every 25 lessons, there is an exam. Two practice exams precede the actual exam, so students will spend a good amount of time reviewing and preparing for each exam.
The program tracks student scores on practice sets, quizzes, and exams, weighting them appropriately. Parents can factor in a grade for student notes if they wish; a complete explanation of how to do
that is included.
For Parents to Consider...
Shormann Math works well in traditional homeschools as well as for co-op classes. In a co-op setting, a "flipped classroom" approach is used. Students watch lectures and complete homework at home,
attending class once or twice weekly for discussion.
Courses should take 30 or more weeks to complete. Because geometry is taught within the Shormann Algebra 1 and 2 courses, it is possible to stretch those two courses out over three years. With each
course subscription, a student has access for 24 months. That should give them plenty of time if they want to stretch out courses for whatever reason.
Interestingly, Shormann recommends that students devote a particular amount of time to math each day rather than committing to the completion of one lesson per day. The courses allow students to stop
in the middle of a practice set and pick up there the next day by simply pressing “save without submitting.” Shormann recommends that students work on math four or five days per week for one to one
and a half hours per day.
Parents really should read through the PDF teacher guide information before a student begins a course. A Getting Started video at the beginning of each course covers much of this information for
students. Critical information such as the “Practice Set Instructions Sheet” is repeated a few places so it won’t be missed.
Courses are challenging, but you can reduce the intensity of a course by following Shormann's recommendation that students work for only a set amount of time each day. Because concepts are taught
incrementally, students gradually master even the more complex concepts. All Shormann Math courses can be used as either standard or honors courses. The teacher guide for each course explains how to
make adjustments for either option.
The first 25 lessons of each Shormann Math course review or introduce math concepts and skills that will be needed throughout the rest of the course. If your student is deficient in prerequisite
skills, it should show up in those first 25 lessons. Because brief reteaching and review are included, strong math students can often pick up what they previously missed and continue on through the
course. Since most students who struggle in any Algebra 1 course are usually struggling with fractions, decimals, and percents, Shormann Algebra 1 includes a link to five-minute drills to develop
fluency in these essential skills. A student who has already taken Algebra 1 and Geometry using resources from another publisher should be able to transition into Shormann Algebra 2, although quite a
bit of geometry will be repeated. Students who come from another publisher’s Algebra 1 course will probably do best starting with Shormann Algebra 1. This should strengthen their algebra skills while
teaching the geometry they haven’t covered.
Some parents prefer to choose courses with textbooks so that they can reuse the course with younger siblings. The publisher of Shormann Math recognizes this, so they offer a very significant sibling
discount that helps to offset the cost of purchasing an entirely new course for subsequent students. There is no time limit on purchasing the sibling course at a discount. It just has to be a course
already purchased at the regular price for a previous student in the family.
When you consider that Shormann Math condenses three courses into two by integrating geometry into the algebra courses and then provides a generous 24 months for students to complete each course, you
can see that the cost per course credit might be significantly less than it first appears. If your student is able to prepare for and pass CLEP or AP exams as well, even better in terms of
Shormann Math courses demonstrate a keen understanding of independent study and what it takes to help students be successful in such courses. These courses are challenging and require students to be
responsible to work through courses properly to get the most out of them. Even while providing content comparable to other challenging courses, Shormann Math courses offer structured avenues to
ensure student success without parental assistance. | {"url":"https://cathyduffyreviews.com/homeschool-reviews-core-curricula/math/grades-9-12/shormann-interactive-math-prealgebra-1-through-calculus","timestamp":"2024-11-06T11:39:51Z","content_type":"text/html","content_length":"57779","record_id":"<urn:uuid:749ee276-cf17-4ffd-b73c-ec7a25d91975>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00122.warc.gz"} |
Can you help me understand this formula?
Apr 7, 2014
=IF(COUNT.IF($GR$3:$GR$18;"<="&GR3)=0;" ";(COUNT.IF($GR$3:$GR$18;"<="&GR3)))
Hi everybody. I'm having trouble understanding the above formula. If there are different letters in the range $GR$3:$GR$18, this formula will give a number of how the letter in question (in cell GR3)
relates to the others, what the letter's number would be if we were to sort the letters from smallest to largest.
However, if I enter numbers in that range, the formula returns nothing. I'd like for it to work with both letters and numbers. Any ideas?
Last edited:
Excel Facts
How to total the visible cells?
From the first blank cell below a filtered data set, press Alt+=. Instead of SUM, you will get SUBTOTAL(9,)
not an anser to your question but change in the formula.
COUNT.IF to countif (without a dot).
Dec 30, 2008
Office Version
1. 365
1. Windows
not an anser to your question but change in the formula.
COUNT.IF to countif (without a dot).
That would cause an error.
Ben, could you give some examples of cell content and expected results?
As far as I can see, the only time it should return nothing would be if GR3 was empty.
Last edited:
Dutch (AANTAL.ALS) English (COUNTIF)
Please explain why that cause an error.
I don't understand, i realy appreciate it if you could explain it to me.
Dec 30, 2008
Office Version
1. 365
1. Windows
There are differences between various English versions of excel, one uses COUNT.IF instead of COUNTIF, using the wrong one would return a #NAME? error, the same as if you tried using English
functions in your Dutch settings, etc.
There are differences between various English versions of excel
I never seen / heared about this.
Thanks for the reply, and point me on this one.
Realy appreciated.
Dec 30, 2008
Office Version
1. 365
1. Windows
I've been trying to find exactly which version with no success, maybe the OP can answer that for us.
Jan 15, 2007
I note that COUNT.IF($GR$3:$GR$18;"<="&GR3)=0 will always be FALSE , unless this formula is changed by dragging.
The OP could get the same result by using the formula =COUNT.IF($GR$3:$GR$18;"<="&GR3)
and formatting the cell with the custom format General;General;""
Apr 7, 2014
Oeldere, I'm using Excel 2013, the language is not English, hence the COUNT.IF, which I translated so people would know what sort of a function I'm talking about. I didn't alter anything else in the
But my point was that I wanted it to work with numbers as well as letters.
jasonb75, the formula is in cells GQ3-GQ18 (dragged down from the top one).
If I put:
A in GR3
B in GR4
C in GR5
Then the results of the formula will be:
14 in GQ3
15 in GQ4
16 in GQ5
However, if I enter numbers in GR-cells, there are no results in the GQ-cells. I'd like to sort numbers in the same fashion as well.
mikerickson, thanks for your input.
May 2, 2008
Office Version
1. 365
1. Windows
2. MacOS
I suspect GR3:GR18 are formatted as Text. Are the numbers aligned to the left of the cells? | {"url":"https://www.mrexcel.com/board/threads/can-you-help-me-understand-this-formula.802147/","timestamp":"2024-11-03T02:51:16Z","content_type":"text/html","content_length":"149228","record_id":"<urn:uuid:e38deeaa-c503-408f-b587-e4b91f4c46d1>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00141.warc.gz"} |
The Stacks project
Remark 42.29.6. Let $X \to S$, $\mathcal{L}$, $s$, $i : D \to X$ be as in Definition 42.29.1 and assume that $\mathcal{L}|_ D \cong \mathcal{O}_ D$. In this case we can define a canonical map $i^* :
Z_{k + 1}(X) \to Z_ k(D)$ on cycles, by requiring that $i^*[W] = 0$ whenever $W \subset D$ is an integral closed subscheme. The possibility to do this will be useful later on.
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MA3351 Transforms and Partial Differential Equations Notes, Questions
Important question Notes question bank Question Paper R-2021 Syllabus
MA3351 Transforms and Partial Differential Equations
MA3351 Transforms and Partial Differential Equations
Anna University – MA3351 Transforms and Partial Differential Equations Regulation 2021 Syllabus , Notes , Important Questions, Question Paper with Answers Previous Year Question Paper.
UNIT I PARTIAL DIFFERENTIAL EQUATIONS MA3351 Transforms and Partial Differential Equations Syllabus
Formation of partial differential equations –Solutions of standard types of first order partial differential equations – First order partial differential equations reducible to standard types-
Lagrange’s linear equation – Linear partial differential equations of second and higher order with constant coefficients of both homogeneous and non-homogeneous types.
UNIT II FOURIER SERIES MA3351 Transforms and Partial Differential Equations Notes
Dirichlet’s conditions – General Fourier series – Odd and even functions – Half range sine series and cosine series – Root mean square value – Parseval’s identity – Harmonic analysis.
UNIT III APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS MA3351 Transforms and Partial Differential Equations Important Questions
Classification of PDE – Method of separation of variables – Fourier series solutions of one dimensional wave equation – One dimensional equation of heat conduction – Steady state solution of two
dimensional equation of heat conduction (Cartesian coordinates only).
UNIT IV FOURIER TRANSFORMS MA3351 Transforms and Partial Differential Equations Question Paper
Statement of Fourier integral theorem– Fourier transform pair – Fourier sine and cosine transforms – Properties – Transforms of simple functions – Convolution theorem – Parseval’s identity.
UNIT V Z – TRANSFORMS AND DIFFERENCE EQUATIONS MA3351 Transforms and Partial Differential Equations Question Bank
Z-transforms – Elementary properties – Convergence of Z-transforms – – Initial and final value theorems – Inverse Z-transform using partial fraction and convolution theorem – Formation of difference
equations – Solution of difference equations using Z – transforms.
Syllabus Click Here
Notes Click Here
Important Questions Click Here
Previous Year Question Paper Click Here
Question Bank Click Here
TEXT BOOKS For MA3351 Transforms and Partial Differential Equations PDF
1. Grewal B.S.,“Higher Engineering Mathematics” 44thEdition, Khanna Publishers,New Delhi, 2018.
2. Kreyszig E, “Advanced Engineering Mathematics “, 10th Edition, John Wiley, New Delhi, India,
1. Andrews. L.C and Shivamoggi. B, “Integral Transforms for Engineers” SPIE Press, 1999.
2. Bali. N.P and Manish Goyal, “A Textbook of Engineering Mathematics”, 10th Edition, Laxmi
Publications Pvt. Ltd, 2015.
3. James. G., “Advanced Modern Engineering Mathematics”, 4thEdition, Pearson Education, New
Delhi, 2016.
4. Narayanan. S., Manicavachagom Pillay.T.K and Ramanaiah.G “Advanced Mathematics for
Engineering Students”, Vol. II & III, S.Viswanathan Publishers Pvt. Ltd, Chennai, 1998.
5. Ramana. B.V., “Higher Engineering Mathematics”, McGraw Hill Education Pvt. Ltd,
New Delhi, 2018.
6. Wylie. R.C. and Barrett . L.C., “Advanced Engineering Mathematics “Tata McGraw Hill Education
Pvt. Ltd, 6th Edition, New Delhi, 2012.
Related Links
Anna University Syllabus Regulation 2021
Anna University Regulation 2021 Study Materials
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Re: st: xtnbreg fe
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
Re: st: xtnbreg fe
From SamL <[email protected]>
To Stata Listserve <[email protected]>
Subject Re: st: xtnbreg fe
Date Thu, 16 Jun 2005 10:57:00 -0700 (PDT)
I sent a similar question about a week ago. There was one response. Both
are copied below; my question is interspersed with Scott Merryman's
Parenthetically, I am still puzzled why stata would change the
widely-understood meaning of fixed effect; is there some statistical
reason why fixed effects as normally understood would be a problem for
nbreg with a varying dispersion parameter? If anyone can answer that, I'd
appreciate it. However, anyway, hope the info below helps.
Date: Tue, 7 Jun 2005 18:16:34 -0500
From: Scott Merryman <[email protected]>
Reply-To: [email protected]
To: [email protected]
Subject: st: RE: "Regular" fixed effects and nbreg, in stata8.2?
> -----Original Message-----
> From: [email protected] [mailto:owner-
> [email protected]] On Behalf Of SamL
> Sent: Tuesday, June 07, 2005 9:24 AM
> To: Stata Listserve
> Cc: [email protected]
> Subject: st: "Regular" fixed effects and nbreg, in stata8.2?
> The manual states that for the command -xtnbreg- with the -fe- option, the
> term "fixed effects" applies "to the distribution of the dispersion
> parameter, and not to the xB term in the model." I am understanding this
> to mean that there is no fixed effect--in the usual sense of a beta
> coefficient for the unit (i) to which the observation belongs. I have two
> questions.
> 1)Is my understanding correct?
> 2)If my understanding is correct, may one obtain a fixed effects nbreg
> model that *does* have a fixed effect, in the usual xB sense, for the unit
> to which each observation belongs and, if so, how? (I don't need to see
> the fixed effect--if it drops out of the conditional model, that's fine--I
> just need for the usual fixed effect to drop out, not the dispersion
> parameter. For my analysis an equal dispersion parameter across units
> would be fine.).
You could use -poisson- with dummy variables. Unlike other nonlinear
estimators, it does not suffer from an incidental parameters problem. See
Cameron and Trivedi 1998, "Regression Analysis of Count Data" p. 280-282.
If you a large number of observations per cross section (enough that you
would be willing to estimate each cross section separately) you may be able
to get away with the use dummy variables (the inconsistency disappears as
T-> infinity). On a similar topic, you may find the post by Bill Gould,
Vince Wiggins, and David Drukker helpful:
On Thu, 16 Jun 2005, Khan, Nasreen wrote:
> Hi everyone,
> I need some expertise.
> I have an unbalanced panel data with around 68,000 observation and 23,000
> individuals (or groups). The dependent variable is in counts. And i am
> controlling for time invarient individual heterogenity by introducing
> fixed effect.
> Ideally, i would like to use poisson with robust standard errors (there is
> overdispersion in the DV) as i find use of dispersion parameter in neg
> binomial somewhat arbitary. However, stata 8 SE does not allow the robust
> option with xtpoisson. So i am restricted to negative binomial. However
> in the nbreg model variables (like race and sex) do not get absorbed in
> the fixed effect (they do so in xtpoisson). My questions are
> 1. Why doesnt the time invarient variables gets absorb in fixed effect in
> xtnbreg?
> 2. If the fe is for dispersion parameter, is there a way to introduce the
> individual level fe in nbreg ?
> 3. Is there any program available to adjust the SE in fixed effect xtpoisson?
> Thank you
> Nancy, PhD. Candidate
> University of Illinois at Chicago
> *
> * For searches and help try:
> * http://www.stata.com/support/faqs/res/findit.html
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2005-06/msg00482.html","timestamp":"2024-11-02T14:00:54Z","content_type":"text/html","content_length":"12180","record_id":"<urn:uuid:19ac5b69-64db-41c9-b645-97e731245bc2>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00318.warc.gz"} |
This number is a prime.
The only prime that can be represented as the smallest side in a Super-Heronian Triangle with 5-digit sides, i.e., (37633, 37634, 37635). [Loungrides]
The number of partitions of {1,...,7} into any number of lists, where a list means an ordered subset. [Honaker]
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Comparing Rainfall Erosivity Estimation Methods Using Weather Radar Data for the State of Hesse (Germany)
Institute of Physical Geography and Landscape Ecology, Leibniz Universität Hannover, Schneiderberg 50, 30167 Hannover, Germany
Hessian Agency for Nature Conservation, Environment and Geology, 65203 Wiesbaden, Germany
Institute of Bio- and Geosciences IBG-3, Forschungszentrum Jülich GmbH, 52425 Jülich, Germany
Author to whom correspondence should be addressed.
Submission received: 9 April 2020 / Revised: 12 May 2020 / Accepted: 14 May 2020 / Published: 16 May 2020
Rainfall erosivity exhibits a high spatiotemporal variability. Rain gauges are not capable of detecting small-scale erosive rainfall events comprehensively. Nonetheless, many operational instruments
for assessing soil erosion risk, such as the erosion atlas used in the state of Hesse in Germany, are still based on spatially interpolated rain gauge data and regression equations derived in the
1980s to estimate rainfall erosivity. Radar-based quantitative precipitation estimates with high spatiotemporal resolution are capable of mapping erosive rainfall comprehensively. In this study,
radar climatology data with a spatiotemporal resolution of 1 km^2 and 5 min are used alongside rain gauge data to compare erosivity estimation methods used in erosion control practice. The aim is to
assess the impacts of methodology, climate change and input data resolution, quality and spatial extent on the R-factor of the Universal Soil Loss Equation (USLE). Our results clearly show that
R-factors have increased significantly due to climate change and that current R-factor maps need to be updated by using more recent and spatially distributed rainfall data. Radar climatology data
show a high potential to improve rainfall erosivity estimations, but uncertainties regarding data quality and a need for further research on data correction approaches are becoming evident.
1. Introduction
The R-factor is a measure of rainfall erosivity and an important input variable for estimating soil losses by water using the Universal Soil Loss Equation (USLE) and its many variations [
]. Based on the documented relationship between the amount of soil erosion and the kinetic energy of precipitation, the rainfall erosivity can be derived directly from temporally highly resolved
precipitation time series [
]. The R-factor of one event is defined as the product of the kinetic energy and the maximum 30-min intensity of an erosive rainfall event. The R-factors of all events throughout a year are added to
obtain the annual R-factor, which is usually averaged over a period of at least ten years as an input to the USLE.
In the past, measurement data from rain gauges or, more recently, from automated rain gauges were used for estimating rainfall erosivity. Still today, the R-factors calculated from these point-scale
data for every station are spatially interpolated to derive maps of rainfall erosivity. This approach has also been recently applied to generate a European erosivity map [
]. However, due to the small spatial extent of convective precipitation cells and a high variability of precipitation intensity within these cells, which contributes significantly to rainfall
erosivity, the spatial recording of the rainfall erosivity is incomplete and patchy [
]. Rain gauges are not capable of detecting the spatial distribution of local heavy rainfall hot spots or individual heavy rainfall events, which are highly relevant for erosion modelling.
Interpolating R-factors calculated from point measurements therefore results in a smoothing and an underestimation of erosivity [
]. In order to capture the highly variable spatiotemporal distribution of rainfall intensity during erosive rainfall events, highly resolved precipitation data, both spatial and temporal, are needed.
Weather radars are capable of providing such data, but the number of studies deriving erosivity directly from such highly resolved datasets is still rather low [
In practice, R-factor maps are frequently derived by regression equations from spatially interpolated summer precipitation sums or annual precipitation sums in order to obtain comprehensive erosivity
information. This methodology is much easier to apply than the direct event-based derivation of the R-factor from gauge data, but it suffers from representativity issues. Again, data smoothing by
spatial interpolation and regression equations lead to smoothed R-factors. High R-factors often remain limited to mountain tops, while the actual occurrence of heavy rainfall as a consequence of
convective events in the lowlands is not taken into account [
In Germany, for instance, the R-factor is derived by regional authorities for each federal state according to the technical standard DIN 19708 [
], whereby most federal states use regional adjusted regression equations. The derived erosivity maps serve inter alia as an input for soil erosion modelling in order to evaluate the fulfilment of EU
Cross-Compliance soil protection regulations. Based on these evaluation outcomes, income support for farmers is calculated and requirements for erosion control are imposed. However, the applied
regression equations were usually derived based on data from a few rain gauge recorders (usually < 20) integrating rainfall data from the 1960s to the 1980s [
]. The regression equations are only rarely updated (e.g., in North-Rhine Westfalia [
]) or, in many federal states, not at all. However, several studies indicate spatial and temporal changes in precipitation distribution and quantities as well as an increase and intensification of
heavy rainfall and thus an increase in precipitation erosivity due to climate change [
]. Consequently, the validity of the currently applied regression equations, which were determined based on precipitation data of the last climate period or even older data, must also be questioned,
especially in regard to the current atmospheric conditions.
In the German federal state of Hesse, a lot of information on soil quality and degradation, including the R-factor, is collected in the technical information system “Erosion Atlas Hesse” [
]. The erosion atlas is an important instrument for precautionary soil protection in Hesse since it shows areas with a high risk of erosion and helps farmers to plan erosion control measures.
Furthermore, it supports urban land-use planning through the identification of sites that require additional protection measures. The estimation of the R-factor for the erosion atlas is currently
based on a regression equation derived in 1981 from data of 18 rain gauges in Bavaria, which comprise time series of up to 14 years throughout the period of 1958–1977 [
]. The precipitation data used for calculating the R-factor are spatially interpolated mean summer precipitation sums (May to October) for the period of 1971–2000 on a 1 km
grid [
]. There is evidence that rainfall distribution and intensity has changed since this time period [
], emphasising the need for updated precipitation datasets and methods that estimate rainfall erosivity.
The radar climatology dataset RADKLIM (“RADarKLIMatologie”) [
] addresses the need for updating precipitation data. RADKLIM is a radar-based quantitative precipitation estimation dataset provided by the German Weather Service (Deutscher Wetterdienst, DWD). It
is available for the whole of Germany starting from 2001 with a high spatial (1 km
) and temporal (up to 5 min) resolution [
]. The largely comprehensive nationwide detection of all precipitation events indicates a high potential for the derivation of spatial information to calculate the R-factor. The high temporal
resolution of the data as well as recent advances in computer hardware enable the direct event-based calculation of the R-factor. However, the differences in measurement method and scale between
radar and rain gauges, especially in detecting heavy rainfall, must be taken into account when interpreting the results. The precipitation totals in radar climatology tend to be slightly lower than
the precipitation amounts measured by rain gauges and this underestimation by radar climatology is particularly pronounced for high precipitation intensities [
]. This is due to the averaging of precipitation over the area of the radar pixels and path-integrated rainfall-induced attenuation of the radar beam [
For the direct event-based calculation of the R-factor based on radar data, Fischer et al. [
] found similar effects and derived correction factors to compensate for the underestimation of the R-factor calculated with radar climatology data. The proposed factors include a spatial scaling
factor, which reflects the attenuation of intensity peaks by averaging the precipitation over the radar pixel area, and a method factor, which should compensate for the systematic underestimation of
erosion by the radar data compared to rain gauge measurements.
In addition, several studies have recently investigated the influence of the temporal resolution of precipitation data on the calculation of the R-factor [
]. In principle, the authors agree that the level of the R-factor decreases with decreasing temporal resolution. The intensity peaks, which are decisive for determining the kinetic energy of the
precipitation, are detected less accurately with decreasing temporal resolution and are thus attenuated. However, authors disagree about the correction of this effect, since the level of any
correction factor depends on the temporal resolution of the rainfall data that is used as a reference. Based on rain gauge and RADKLIM data for Germany, Fischer et al. [
] use one minute as the highest possible resolution for a factor value of 1. Panagos et al. [
], on the other hand, use a reference of 30 min as factor value of 1 in their European-wide study based on rain gauge data. For the RADKLIM product with a 5-min resolution, this results in a temporal
correction factor of 1.05 [
] or 0.7984 [
], and for the RADKLIM product with hourly resolution, the temporal correction factors are 1.9 and 1.5597, respectively.
The goal of this study was to compare the performance of different calculation methods for the R-factor using rain gauge and radar rainfall data. The impacts, advantages, disadvantages and correction
approaches for several input datasets were analysed; additionally, updated regression equations were derived. Taking the improvement in monitoring systems through a higher coverage by measurements
and discrepancies concerning methodology, input data quality and resolution, observation period and correction approaches into account, the paper proposed these hypotheses for the derivation of
R-factors from radar climatology and rain gauge data for the period 2001–2016:
• The newly calculated R-factors from both datasets are higher than the R-factors from earlier calculations due to changes in climate, interannual rainfall distribution and rainfall intensity.
• Since radar data include small-scale convective cells without gaps, the R-factors derived from the radar climatology should be higher on average than those calculated from rain gauge
measurements. At the same time, the radar measurements underestimate the maximum precipitation intensities. The latter can be compensated by the correction factors according to Fischer et al. [
• The spatial distribution of the R-factors derived from the radar climatology deviates from the patterns of the R-factors calculated and interpolated by means of the regression equation due to the
comprehensive coverage of all heavy rainfall events.
2. Materials and Methods
2.1. Study Area
For this study, the federal state of Hesse was selected as the investigation area due to its central location within Germany and its complex terrain, which allow for a good transferability of the
outcomes. The federal state of Hesse has a total area of approximately 21,115 km
. The area is characterised by a diverse topography with several low mountain ranges and highlands crossed by depressions and river valleys (see
Figure 1
). The highest elevation is 950 m.a.s.l., whereas the lowest elevation is about 73 m.a.s.l. A large portion of the intensively used agricultural areas in the lowlands are oriented in Rhenish
direction (SSW-NNO) [
]. The study area is located in the humid midlatitudes in a transition zone between a maritime climate in north-western Germany and a more continental climate in the south and east of Germany.
Westerly winds influence the distribution of precipitation and, thus, many of the intensively used agricultural areas are located in the rain shadow on the lee side east of the mountain ranges.
2.2. Data Basis
2.2.1. Radar Climatology Data
The DWD currently operates 17 ground-based C-band weather radars. The nationwide coverage was established in 2001. In 2018, the DWD published the radar climatology dataset RADKLIM, which consists of
gridded nationwide quantitative precipitation estimate composites with a spatial resolution of 1 km
and a temporal resolution of up to 5 min starting from 2001. For this study, we used the YW product in 5-min resolution [
] and the RW product [
] in hourly resolution for the period 2001–2016. Their derivation procedure comprises various correction algorithms to compensate for typical radar-related errors and artefacts such as clutter,
spokes, signal attenuation and bright band effects. Ground clutter can be caused by non-meteorological objects such as mountains, buildings, wind energy plants or trees that disturb the radar signal
and cause non-precipitation echoes. If the radar beam is blocked in whole or in part by such objects, the sector behind these obstacles is shielded, which causes a linear artefact, the so-called
negative spoke. Signal attenuation may cause significant underestimation of rainfall rates. It can be caused by a wet radome, by heavy precipitation events that shield the sector behind or by range
degradation at far range from the radar. Bright Band effects occur in the melting layer where the comparatively large surface of melting snowflakes is covered by a film of water, which may cause very
strong radar signals.
For the derivation of the radar climatology, the reflectivity is converted to rain rates, and the local radar station data are merged and transformed to a cartesian grid. Aggregated hourly rain rates
are adjusted to ground-truth automated rain gauge measurements, which yields the RW product. Finally, the hourly rain rates are disaggregated to the original 5-min intervals in order to obtain the
quasi-adjusted YW product [
]. For disaggregation, the hourly precipitation sum of the adjusted RW product is distributed to the twelve 5-min intervals based on the temporal rainfall distribution throughout the respective hour.
The data processing was conducted by DWD. In the state of Hesse, only the stations operated by DWD are used for radar data adjustments.
2.2.2. Rain Gauge Data
For this study, we combined two different rain gauge datasets in 1-min resolution. We used data from 76 automated rain gauges throughout Hesse operated by DWD, which are freely available in the DWD
Open Data Portal [
], as well as from 52 rain gauges of the Hessian monitoring network operated by HLNUG, which are not publicly available. Both datasets were carefully checked for plausibility and a cleaning procedure
was implemented to remove erroneous values. For a detailed description of the data processing and cleaning procedure please refer to [
In general, the DWD rain gauge data are available for the period 2001–2016, whereas those of the HLNUG stations only cover the period 2001–2015. However, the time series of the combined rain gauge
dataset varies strongly between stations. In this study, 21 stations with time series shorter than nine years were excluded. The final dataset used for analysis consisted of 110 rain gauge stations.
Finally, the 1-min rain gauge data were aggregated to a temporal resolution of 5 min in order to match the temporal resolution of the radar climatology data.
2.3. Methodology
2.3.1. R-factor Calculation According to DIN 19708
The R-factors were calculated according to the specifications of DIN 19708 [
] for the RADKLIM YW product and the rain gauge data, both in 5-min resolution. According to DIN 19708 [
], which is based on the results of Schwertmann et al. [
], erosive precipitation events have a precipitation sum of at least 10 mm or a precipitation intensity exceeding 10 mm/h within a time window of 30 min (i.e., an actual precipitation quantity of 5
mm in 30 min). The maximum precipitation sum occurring within a 30-min window of a rainfall event is identified by applying a moving window of six 5-min intervals and is related to one hour by
doubling it. This value is referred to as maximum 30-min intensity
. As defined by DIN 19708 [
], the total amount of precipitation is doubled and assigned to
if an event lasts less than 30 min. Rainfall events are separated by a precipitation pause of at least 6 h.
The R-factor of a specific precipitation event results from the product of the maximum 30-min intensity
[mm/h] and the kinetic energy
] of the total rainfall during the event.
The kinetic energy
of an erosive rainfall event was calculated with the following equation from DIN 19708:
$E = ∑ i = 1 , n i = n E i$
$E i = ( 11.89 + 8.73 · l o g 10 ( I i ) ) · N i · 10 − 3 f o r 0.05 ≤ I i ≤ 76.2$
$E i = 0 f o r I i < 0.05$
$E i = 28.33 · N i · 10 − 3 f o r I i > 76.2$
Thereby is
$i$ 5-min interval of the rainfall event
$E i$ kinetic energy of the rainfall in period i [kJ/m^2]
$N i$ rainfall depth in period i, [mm]
$I i$ rainfall intensity in period i, [mm/h], that is $I i = N i · 60 Min 5 Min$
Finally, the R-factor per year for a given location is the sum of the
products [kJ/m
mm/h = N/(ha a)] of all erosive rainfall events in a year. Due to the great interannual variability of erosivity, it is recommended to average the annual R-factors over a period of at least ten years
]. For the calculations based on the radar climatology this criterion was fulfilled everywhere, whereas the time series of five rain gauges was limited to nine years.
For the calculation of the R-factor from both data sets, the development of new routines was necessary. One difficulty is the large data volume of the YW product for the whole of Hesse, which
required a balancing of memory requirements and computing efficiency. The developed Python routines are based on the HDF5 file format [
] with monthly pandas [
] DataFrames introduced by Kreklow [
]. This enables a continuous calculation of the R-factor over all days of a month. However, for reasons of efficiency, no smooth transitions between months were implemented. The routine assumes an
end of the precipitation event at the end of each month and carries out the calculation for the amount of precipitation that has fallen up to that point. Thus, long lasting nightly precipitation
events may be divided into two events or one event can be classified as non-erosive due to the interruption. However, since erosivity shows a clearly pronounced maximum in the late afternoon [
], when convection is usually strongest, the inaccuracy in the calculation due to the interruption at the turn of the month was regarded as negligible.
2.3.2. R-factor Calculation Using Regression
For the erosion atlas Hesse [
], the R-factor was derived using the following regression equation from the mean long-term precipitation of the summer months May–October
$R E A = 0.141 · N S u − 1.48$
For comparison of methodologies and effects of precipitation changes, additional R-factors were calculated using this regression equation based on the hourly RW product of the radar climatology and
the condensed rain gauge dataset. In conjunction with the R-factors calculated according to DIN 19708 (see
Section 2.3.1
) and the erosion atlas Hesse, these additional R-factor estimates based on regression allow to compare different combinations of input data and derivation methods.
All calculated R-factor derivatives are summarised in
Table 1
Since the R-factor is only important for estimating soil loss from agricultural land and not in forests or urban areas, we conducted an additional analysis of all of the abovementioned R-factor
derivatives that only considered cropland areas. For this, all data pairs for which the respective RADKLIM pixel contained less than ten hectares of cropland were removed. The resulting datasets are
marked by the
” in the R-factor index, e.g.,
Consequently, the analyses of this study cover three different spatial extents for which data pairs of all available datasets were created in order to enable meaningful comparisons for similar
spatial scales:
all 1 km^2 pixels within Hesse (n = 23,320)
all pixels containing at least ten hectares of cropland (n = 11,555)
all rain gauge stations (n = 110)
In addition, the summer precipitation sums of RADKLIM and the rain gauges and their respective R-factor derivatives R[YW,DIN] and R[G,DIN] were used to determine two new regression equations. These
serve to assess the following: the changes in the correlation between rainfall erosivity and precipitation sums, changes in comparison to the existing regression equation used for the erosion atlas,
and the impact of sample size.
2.3.3. Application of Scaling Factors
Recent studies propose various scaling/correction factors to compensate for the temporal resolution of the input data and the differences between rain gauge and radar data. In order to be able to
estimate the influence of the correction factors and to compensate for the presumed underestimation of the R-factor by the radar climatology, these factors were applied to the R-factors that were
calculated according to DIN 19708.
The scaling according to [
] for the R-factors calculated from radar climatology results from
$R Y W , F = R Y W , D I N · ( ( s p a t i a l s c a l i n g + m e t h o d f a c t o r ) · t e m p o r a l s c a l i n g )$
$s p a t i a l s c a l i n g f a c t o r = 1.13 ; for a spatial resolution of 1 km ²$
$m e t h o d f a c t o r = 0.35$
$t e m p o r a l s c a l i n g f a c t o r = 1.05 ; for a temporal resolution of 5 min$
For the rain gauge data, the scaling reduces to
$R G , F = R G , D I N · t e m p o r a l s c a l i n g f a c t o r$
$t e m p o r a l s c a l i n g f a c t o r = 1.05 ; for a temporal resolution of 5 min$
In order to include the strongly deviating temporal correction factor proposed by Panagos et al. [
], a further calculation was performed for
$R G , P = R G , D I N · t e m p o r a l s c a l i n g f a c t o r$
$t e m p o r a l s c a l i n g f a c t o r = 0.7984 ; for a temporal resolution of 5 min$
3. Results
3.1. Statistical Comparison of the Calculated R-factors
The R-factor
calculated from the original unscaled RADKLIM YW product according to DIN 19708 ranges between 28.8 and 173.2 kJ/m
mm/h with an average value of 58.0 kJ/m
mm/h (see
Table 2
Figure 2
). It is thus 6.4% higher on average than the values of the erosion atlas
whereas its range is 263.7% higher and its standard deviation is 122.7% higher.
shows thus a much higher variability than the strongly smoothed
which was derived from spatially interpolated rainfall data using a regression equation (Equation (3)).
The R-factor calculated from the gauge dataset R[G,DIN] has an average of 80.6 kJ/m^2 mm/h, which is 47.8% higher than the average value of R[EA] and 39% higher than the average of R[YW,DIN]. At 107
of 110 stations the rain gauges show higher R-factors than the corresponding pixels of the radar climatology. The average R-factor difference for all point-pixel pairs amounts to 20.5 kJ/m^2 mm/h
between R[YWG,DIN] and R[G,DIN]. For the 72 stations operated by DWD, which were used for radar data adjustments, the average difference between R[YWG,DIN] and R[G,DIN] amounts to 19.1 kJ/m^2 mm/h,
whereas the average difference at the 38 stations operated by HLNUG is slightly higher with 23.1 kJ/m^2 mm/h. Compared to the erosion atlas, all 110 rain gauge stations show higher R values with an
average difference of 24.7 kJ/m^2 mm/h.
Using the regression equation from the erosion atlas (Equation (3)) and the RADKLIM RW product to derive
yielded comparable values as
with a slightly lower mean and maximum, significantly lower minimum, but a slightly higher median and standard deviation. For
, all statistical values were slightly higher than for
. Consequently, before scaling, the rain gauge dataset consistently produces the highest R-factors, but the magnitude of the differences is governed by the derivation method. The input dataset has
little influence on the statistical characteristics of the outcome when using a regression equation and the major differences between these regression-based derivatives are the spatial resolutions
and spatial distributions (see
Section 3.2
). When grouping all R-factor derivatives by the calculation method—irrespective of input data and spatial extent—the mean of those R-factors derived according to DIN 19708 (without scaling) is 9.1
mm/h higher than the mean of all R-factors derived using the regression equation. Furthermore, with 15.8 kJ/m
mm/h, the DIN method group showed on average a 122.2% higher standard deviation than the regression method group (7.1 kJ/m
mm/h), which underlines the smoothing effect that can be obtained by using a regression equation instead of the event-based method according to DIN 19708. The difference between both methods is
particularly well illustrated by the very steep empirical cumulative distribution functions (ECDF) of all regression-based derivatives (see
Figure 3
Selecting pixels with cropland leads to an average decrease of
by 3.8 kJ/m
mm/h (−6.6%). The minimum did not change, while the maximum decreased by 27.1 to 146.1 kJ/m
mm/h (see
Figure 2
Figure 3
). Taking into account only the pixels with cropland and rain gauges, the count was reduced to 54 (a total of 54 rain gauges are located in radar pixels with cropland), the average R-factor (
) decreased also by 3.8 to 56.3 kJ/m
mm/h and the maximum decreased by 12.5 to 92.2 kJ/m
mm/h. For
, the impact of the data selection on the statistical distribution is considerably higher due to the smaller sample size. Its average decreased by 6.1 to 74.5 kJ/m
mm/h, whereby the maximum decreased by 42.5 to 114 kJ/m
mm/h when selecting only pixels with cropland. Consequently, the removal of many high erosivity values in the mountainous regions (see
Figure A1
), for which the uncertainty and underestimation of the radar data is particularly high, leads to a slightly better agreement of the R-factors calculated according to DIN 19708 from RADKLIM and rain
gauge data. Grouping the nine R-factor derivatives based on
by spatial extent resulted in a mean of 55.2 kJ/m
mm/h for all pixels of the study area, 52.9 kJ/m
mm/h for pixels with cropland and 56.3 kJ/m
mm/h for pixels with a rain gauge.
In regard to the data source, the results showing an underestimation of rainfall erosivity by the radar climatology compared to rain gauge data are in line with the outcomes of Fischer et al. (2018),
thus the application of the proposed correction factors was considered to be useful and necessary. After scaling, the R-factors of the radar climatology and rain gauges correspond much better (see
Figure 2
Figure 3
). The difference between the two datasets shifts in favour of the radar climatology, since on average
is 8.8 kJ/m
mm/h higher than R
Table 2
). In comparison to
, both R-factors were significantly higher after scaling. On average,
was 65.3% higher and
was 58.5% higher than the R-factor
of the erosion atlas. Although the correction factor proposed by Panagos et al. [
] reduces the R-factor to a level close to
still showed an 18.2% higher mean than
. Irrespective of the dataset used for derivation and the application of correction procedures, an increase of the R-factor compared to
can thus be determined without doubt.
3.2. Spatial Distribution
For erosion control applications at a federal state scale which aim to identify regions with a particularly high risk of erosion, the spatial distribution of rainfall erosivity is actually more
relevant than the absolute erosivity values. The lowest values of
occur in the north of Hesse, around the West Hesse Depression, in an area for which no radar measurements were available during some months of the years 2007 and 2014 due to radar hardware upgrades.
The average value of the annual R-factor without these two years shows that the minimum is nevertheless located in this area. This is therefore in accordance with the R-factor
(calculation based on regression), which also shows a minimum in this area (see
Figure 4
). The areas of relatively low R-factors northwest of Fulda and in the Upper Rhine Plain correspond well in both datasets, too. In the north-east of Hesse, however, the newly calculated R-factor
showed slightly lower erosivity over a large area with a similar spatial distribution. Both datasets showed an increase of the R-factor with increasing terrain height, whereby
showed significantly higher values over a large area, especially in the Odenwald, Taunus, Westerwald and at Vogelsberg. However, at Vogelsberg, a weakness of the radar climatology to correctly
quantify precipitation at higher altitudes was evident as the increase of the R-factor in the lower slope areas was considerably higher than in the summit area. In the area of Wetterau, a negative
spoke of the Frankfurt Radar was clearly visible in
and all other R-factors derived from RADKLIM. Still, in this area an increase of the R-factor compared to the erosion atlas can be seen in most of the grid cells, in some places even up to 45% (see
Figure 5
The scaling is able to compensate for the underestimation of the R-factor by the radar climatology, which becomes particularly obvious in the northern parts of Hesse where the difference to the
erosion atlas shows mostly positive values except for a few single pixels (see
Figure 5
c). Moreover,
Figure 5
b shows much better conformity with
in the entire study area, which has already been indicated in the previous section.
3.3. Derivation of Updated Regression Equations
The statistical comparisons in
Section 3.1
show consistently lower values for all R-factors derived by means of regression. Besides the method itself and the input data source, the observation time period of the data used for the derivation
of the regression equation might play a role due to climate change, which is why we derived an updated regression equation for comparison.
The new regression equations derived from the rain gauge data and the radar climatology both show a strong correlation between summer precipitation and R-factor. The fitted regression line has a
considerably higher slope than the original one used for the erosion atlas (Equation (3)) (see
Figure 6
). Some data points of
, which are mainly located in the area of the radar gap in northern Hesse, are still below the regression line from the erosion atlas. For
, however, all data points are above. Consequently, for the period 2001–2016, the regression equation used in the erosion atlas provides a value deviating from the R-factor according to DIN 19708 for
all of the rain gauges.
When considering the newly derived regression equations from the radar climatology, it is striking that the equations for the entire data set and the pixels at the rain gauge locations are almost
identical (see
Figure 6
). Consequently, the spatial distribution of the rain gauge locations can be regarded as very representative for mapping the overall distribution of rainfall erosivity in Hesse.
Another striking difference with regard to the sample size, however, is a series of several very high values of R[YW,DIN] in the range between 400 and 500 mm summer precipitation. These are only
included in the R-factor of the entire radar climatology dataset, but are not significantly reflected in the regression due to their relatively small number. Therefore, it can be assumed that
extraordinarily intensive individual events have a strong impact due to the comparatively short time series. These events could only be detected by the high spatial resolution of the radar
climatology and are not included in the rain gauge dataset.
Using the new regression equation derived from the rain gauges (R = −43.22 + 0.3 N[Su]) with the summer precipitation sums of the RADKLIM RW product for the federal state of Hesse leads to a R-factor
value range between 29.7 and 123.9 kJ/m^2 mm/h with an average of 73.2 kJ/m^2 mm/h. It has thus a significantly lower maximum than all event-based R-factor derivatives. Its mean value is slightly
lower than that of R[G,DIN] (80.6 kJ/m^2 mm/h) due to the slight overall underestimation of precipitation by the radar climatology, and lies approximately in the centre between the averages of R
[YW,DIN] (58 kJ/m^2 mm/h) and the corrected R[YW,F] (90.1 kJ/m^2 mm/h).
4. Discussion
An evaluation of the radar climatology dataset revealed that it slightly underestimates precipitation quantities. This underestimation is particularly pronounced at higher altitudes and at high
rainfall intensities [
]. In particular, the latter plays a decisive role for rainfall erosivity since rainfall intensity is directly linked to the kinetic energy of rainfall and, thus, its ability to detach soil
particles. The assumption that the R-factor calculated from the radar climatology according to DIN 19708 without input data correction is too low could be confirmed by comparing it with the R-factor
derived from the rain gauge dataset. However, irrespective of the dataset used for derivation, the spatial scale and the application of correction procedures, an increase of the R-factor compared to
which is currently used in the technical information system erosion atlas Hesse [
] can be determined without doubt. This result highlights the need of updated R-factor methods for consultation and planning in Hesse.
The R-factors calculated by the regression equation from the erosion atlas, the summer precipitation sums of radar climatology, and rain gauges showed only slightly higher average values than the
erosion atlas. Considering the significant differences to the R-factor derivations according to DIN 19708, this indicates that the regression equation used for the erosion atlas, which was derived
from precipitation data of the 1960s, 70s and 80s, is no longer representative of the current climate conditions. Apparently, although there has only been a small increase in summer precipitation,
there is a change in the heavy rainfall characteristics and/or in the relationship between erosive rainfall and total precipitation amount. This observation is in line with the projected changes in
precipitation characteristics with regard to climate change. For most of Europe, it is expected that precipitation will increase during winter and decrease during summer [
]. Furthermore, the number of wet days is expected to decrease, whereas the intensity and the return levels of daily precipitation events will increase [
]. The combination of increasingly intense heavy rainfall and the reduced water infiltration capacity of dry soils is expected to amplify the risk of floods [
] and is also very likely to increase soil erosion. These observations indicate that the validity of regression equations for R-factor calculation might decrease, particularly if mean summer
precipitation sums are used instead of mean annual sums. An additional influencing factor for higher R-factors calculated from rain gauge data could be the better recording of intensity peaks by more
accurate modern rain gauges as opposed to the less accurate rain gauges used to collect the data for the 1971–2000 dataset [
Despite the discussed limitations, the regression-based approach has the advantage that it is much easier to apply in practice than the method according to DIN 19708, which is computationally much
more expensive, especially when using it on spatially highly resolved data such as the radar climatology. Moreover, the use of a regression equation with precipitation sums always leads to a certain
smoothing and is thus more robust against outliers than the event-based method when only comparatively short precipitation time series are available. However, as our results have clearly shown, the
regression approach also requires frequent updates of the equations and hence a certain maintenance of the methodology. Obviously, updates to the equations rely on the availability of rain gauge
data. For Germany, this is not a major issue anymore since temporally highly resolved rain gauge data are freely available at the DWD Open Data Portal. In other countries, however, this may be a
greater obstacle.
With regard to the scaling of the R-factors which was proposed in recent studies [
], it should be noted that a correction that increases the RADKLIM R-factor is undoubtedly necessary to compensate for the systematic underestimation of precipitation data obtained from radar
climatology. However, the degree of correction is difficult to estimate due to a lack of reference. If the scaled R-factor of the rain gauge dataset
is regarded as a correct reference for validation, the correction applied for
appears somewhat too high, especially when looking at
Figure 2
. When considering the identical sample size and the largely consistent location of the point-pixel data pairs of
, the advantage of the radar and the fact that more events tend to be recorded hardly matters. However, the median of
almost corresponds to the third quartile of
. Here, a direct transferability of the correction factors, which were derived from a four-year series of measurements of 12 rain gauges within one square kilometre in Bavaria [
], may be limited. Further research efforts and measurements to extend these time series and derive correction factors of higher spatial representativity from more than one single raster cell would
have the potential to significantly reduce the uncertainty when using radar climatology data—not only for rainfall erosivity estimation but for applications related to heavy rainfall in general.
In contrast, the scaling according to Panagos et al. [
] to compensate for the temporal resolution of the input data provides very questionable results. Taking into account the conducted plausibility check of the radar climatology and the comparisons
with the rain gauge data by Kreklow et al. [
], an underestimation of the R-factor by the radar data is clearly demonstrated. Since the correction factor proposed by Panagos et al. [
] reduces the R-factor of the rain gauges to a level almost identical to that of the radar climatology, a correction factor that is too low must be assumed. The correction factor does not appear to
be representative for Hesse, due to the fact that its derivation is based on a rain gauge dataset for the whole of Europe and equally includes data from maritime, continental, temperate, subpolar and
Mediterranean climates. Already for the two neighbouring countries Austria and Italy, Fiener et al. [
] found significant differences in the magnitude and monthly distribution of the R-factor, which indicates a lack of spatial representativity of the temporal scaling factor proposed by Panagos et al.
]. Such representativity issues have been subject to discussion between the authors [
]. In addition, the original methodology for the calculation of the R-factor is based on continuous precipitation recordings, which were aggregated to intervals of constant intensity [
]. Consequently, a temporal resolution of 1 min as a lowest reference chosen by Fischer et al. [
] is much closer to the original method than the reference resolution of 30 min used by Panagos et al. [
]. The much lower reference resolution used by Panagos et al. [
] thus explains the significantly lower temporal correction factor compared to the factor proposed by Fischer et al. [
With regard to practical application, it is recommended that the R-factor map currently used in the erosion atlas should be updated. Our results show that the first and most important step is to use
more recent precipitation data for derivation, which are more representative under current climate conditions. Obviously, using the event-based method according to DIN 19708 with radar climatology,
which was proposed by Auerswald et al. [
], provides the R-factor with the highest spatial detail, but it may be locally biased by some extreme rainfall events or radar artefacts which are not balanced out in the comparatively short radar
time series. Moreover, a correction of R-factors derived from radar climatology according to DIN 19708 is necessary to compensate for underestimation, but the level of correction required is still
subject to discussion. However, the radar climatology time series is still considerably longer than the time series used for deriving the original regression equations by Sauerborn [
], of which one was also used in the erosion atlas Hesse. Consequently, during a transition period, the most robust and easy-to-use approach to obtain updated R-factors is by using an updated
regression equation derived from recent rain gauge data with summer precipitation sums calculated from radar climatology data. On the one hand, this approach accounts for climate change by increasing
the R-factors according to reliable rain gauge observations. On the other hand, it makes use of the high spatial resolution of radar data and comprises a certain smoothing, since precipitation sums
are less biased by local extreme events and by the underestimation of high rainfall intensities by weather radar in comparison to the event-based R-factors derived according to DIN 19708. Moreover,
due to less snowfall and thus fewer uncertainties in the radar climatology data during the summer half-year, the use of radar-based summer precipitation sums increases the robustness of the
recommended method compared to the use of radar-based annual precipitation sums.
Due to the central location of Hesse within Germany, the recommended updated regression equation based on rain gauge data for Hesse (R = −43.22 + 0.3 N[Su]) has a high transferability for most of
Germany. However, for federal states in northern and eastern Germany which have a more maritime or continental climate, regional regression equations should be calculated from recent local rain gauge
5. Conclusions
In this study, we compared several derivation approaches for the R-factor of the USLE and evaluated the performance of radar climatology and rain gauge data for different methods and three spatial
extents. Moreover, two correction factors proposed in other studies were tested and updated regression equations were derived for the German federal state of Hesse.
Regarding the three hypotheses put forward at the beginning of this study, our results can be summarised as follows:
1. The newly derived R-factors from rain gauge and radar climatology data are indeed higher than the R-factors from existing calculations due to climate and weather changes. For the study period of
2001–2016, the regression equation used in the erosion atlas provides a lower R-factor than DIN 19708 for all of the rain gauges.
2. The contradiction between the theoretically higher R-factor of the radar climatology due to the more complete recording of all erosive rainfall events on the one hand and the underestimation of
the R-factor due to the attenuation of intensity peaks, on the other hand, could be established. In the spatial average as well as when looking at the point-pixel data pairs, which largely eliminates
the influence of the higher spatial resolution of the radar climatology data, the R-factors of the rain gauges are significantly higher. However, when looking at the entire radar data set, some
strikingly high R-factor values, which were not captured by the rain gauges, become apparent. Due to their comparatively small number, however, they have no significant influence on the spatial mean
value. In addition, these extraordinary high R-factors can also be a result of very intensive rainfall events in the comparatively short observation period that might be smoothed by prolonging the
radar climatology dataset. The correction of the R-factors according to Fischer et al. [
] provides an improvement of the results for the radar climatology, although a possible overcorrection cannot be excluded.
3. The spatial distribution of the newly calculated R-factor according to DIN 19708 and that from the erosion atlas show a relatively good conformity with minima and maxima in similar regions as well
as a consistent mapping of a relief dependency. In the northeast of Hesse, the R-factor calculated from the uncorrected radar climatology according to DIN 19708 shows comparatively lower values than
the erosion atlas. In contrast, it also shows large areas of higher R-factor values than the erosion atlas, especially in the ridges of the low mountain ranges and in the central lowland areas of
Hesse, for example, the Wetterau. The updated regression equations, which are almost identical for all radar pixels and the point-pixel data pairs, indicate that the rain gauge locations are very
representative for mapping the overall spatial distribution of rainfall erosivity in the study area.
The results of this study clearly indicate that the R-factor map currently used in the erosion atlas should be updated. For a transition period until the radar climatology time series is long enough
to compensate for bias from extraordinarily intensive rainfall events, it is recommended to apply a new regression equation derived from recent rain gauge measurements with summer precipitation sums
calculated from radar climatology data.
With the progressive improvement of the data basis (time series, quality and correction), however, radar climatology data will be further incorporated into operational applications such as risk
management and erosion consulting.
Author Contributions
Conceptualisation, J.K. and B.T.; methodology, J.K.; software, J.K.; validation, J.K., B.T. and B.S.-K.; formal analysis, J.K.; data curation, J.K.; writing—original draft preparation, J.K.;
writing—review and editing, B.S.-K., B.T., J.K. and K.F.; visualisation, J.K.; funding acquisition, B.T. and K.F. All authors have read and agreed to the published version of the manuscript.
This research was funded by the Hessian Agency for Nature Conservation, Environment and Geology (HLNUG) within the project “KLIMPRAX–Starkregen,” working package 1.4.
The authors are grateful to DWD for providing open access radar and rain gauge data. Thank you to Erik Jähnke and Angie Faust for proofreading. Finally, the authors would like to thank the four
anonymous reviewers for their proficient and constructive comments which helped improve the manuscript.
Conflicts of Interest
The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the
decision to publish the results.
Appendix A
Figure A1. R-factor percentage change of R[YW,DIN,Agri] against R[EA,Agri] (a), and percentage change of R[YW,F,Agri] against R[EA,Agri] (b).
1. Wischmeier, W.H.; Smith, D.D. Predicting Rainfall Erosion Losses. A Guide to Conservation Planning; Agriculture handbook number 537; U.S. Department of Agriculture: Washington, DC, USA, 1978.
2. Schwertmann, U.; Vogl, W.; Kainz, M. Bodenerosion durch Wasser. Vorhersage des Abtrags und Bewertung von Gegenmaßnahmen, 2. Aufl.; Ulmer: Stuttgart, Germany, 1990; ISBN 3800130882. [Google
3. Deutsches Institut für Normung. DIN 19708. Bodenbeschaffenheit—Ermittlung der Erosionsgefährdung von Böden durch Wasser mithilfe der ABAG; Normenausschuss Wasserwesen im DIN: Berlin, Germany,
2005; (19708). [Google Scholar]
4. Panagos, P.; Ballabio, C.; Borrelli, P.; Meusburger, K.; Klik, A.; Rousseva, S.; Tadic, M.P.; Michaelides, S.; Hrabalikova, M.; Olsen, P.; et al. Rainfall erosivity in Europe. Sci. Total Environ.
2015, 511, 801–814. [Google Scholar] [CrossRef] [PubMed] [Green Version]
5. Fischer, F.; Hauck, J.; Brandhuber, R.; Weigl, E.; Maier, H.; Auerswald, K. Spatio-temporal variability of erosivity estimated from highly resolved and adjusted radar rain data (RADOLAN). Agric.
For. Meteorol. 2016, 223, 72–80. [Google Scholar] [CrossRef]
6. Auerswald, K.; Fischer, F.K.; Winterrath, T.; Brandhuber, R. Rain erosivity map for Germany derived from contiguous radar rain data. Hydrol. Earth Syst. Sci. 2019, 23, 1819–1832. [Google Scholar]
[CrossRef] [Green Version]
7. Tetzlaff, B.; Friedrich, K.; Vorderbrügge, T.; Vereecken, H.; Wendland, F. Distributed modelling of mean annual soil erosion and sediment delivery rates to surface waters. CATENA 2013, 13–20. [
Google Scholar] [CrossRef]
8. Deutsches Institut für Normung. DIN 19708. Bodenbeschaffenheit—Ermittlung der Erosionsgefährdung von Böden durch Wasser mit Hilfe der ABAG, 2017–08; Beuth Verlag GmbH: Berlin, Germany, 2017. [
Google Scholar]
9. Sauerborn, P. Die Erosivität der Niederschläge in Deutschland. Ein Beitrag zur quantitativen Prognose der Bodenerosion durch Wasser in Mitteleuropa. Ph.D. Thesis, Inst. für Bodenkunde, Bonn,
Germany, 1994. [Google Scholar]
10. Elhaus, D. Erosionsgefährdung. Informationen zu den Auswertungen der Erosionsgefährdung durch Wasser, Germany. 2015. Available online: https://www.gd.nrw.de/zip/erosionsgefaehrdung.pdf (accessed
on 11 March 2020).
11. Burt, T.; Boardman, J.; Foster, I.; Howden, N. More rain, less soil: Long-term changes in rainfall intensity with climate change. Earth Surface Process. Landf. 2016, 41, 563–566. [Google Scholar]
[CrossRef] [Green Version]
12. Fiener, P.; Neuhaus, P.; Botschek, J. Long-term trends in rainfall erosivity–analysis of high resolution precipitation time series (1937–2007) from Western Germany. Agric. For. Meteorol. 2013,
171-172, 115–123. [Google Scholar] [CrossRef]
13. Friedrich, K.; Schmanke, M.; Tetzlaff, B.; Vorderbrügge, T. Erosionsatlas Hessen. In Proceedings of the Tagungsband d. Jahrestagung der DBG/BGS, Kommission VI, “Erd-Reich und Boden-Landschaft“,
Bern, Switzerland, 24–27 August 2019. DBG/BGS, Ed.. [Google Scholar]
14. Hessisches Landesamt für Naturschutz, Umwelt und Geologie (HLNUG). BodenViewer Hessen. Available online: http://bodenviewer.hessen.de/mapapps/resources/apps/bodenviewer/index.html?lang=de
(accessed on 21 January 2020).
15. Rogler, H.; Schwertmann, U. Erosivität der Niederschläge und Isoerodentkarte Bayerns. Z. für Kult. und Flurberein. 1981, 22, 99–112. [Google Scholar]
16. Hessisches Landesamt für Naturschutz, Umwelt und Geologie (HLNUG). R-Faktor. Available online: https://www.hlnug.de/themen/boden/auswertung/bodenerosionsbewertung/bodenerosionsatlas/r-faktor
(accessed on 16 January 2020).
17. Donat, M.G.; Alexander, L.V.; Yang, H.; Durre, I.; Vose, R.; Dunn, R.J.H.; Willett, K.M.; Aguilar, E.; Brunet, M.; Caesar, J.; et al. Updated analyses of temperature and precipitation extreme
indices since the beginning of the twentieth century: The HadEX2 dataset. J. Geophys. Res. Atmos. 2013, 118, 2098–2118. [Google Scholar] [CrossRef]
18. Winterrath, T.; Brendel, C.; Hafer, M.; Junghänel, T.; Klameth, A.; Lengfeld, K.; Walawender, E.; Weigl, E.; Becker, A. Radar climatology (RADKLIM) version 2017.002 (YW); gridded precipitation
data for Germany. Available online: https://search.datacite.org/works/10.5676/dwd/radklim_yw_v2017.002 (accessed on 25 June 2019).
19. Winterrath, T.; Brendel, C.; Hafer, M.; Junghänel, T.; Klameth, A.; Walawender, E.; Weigl, E.; Becker, A. Erstellung einer radargestützten Niederschlagsklimatologie; Berichte des Deutschen
Wetterdienstes No. 251; 2017; Available online: https://www.dwd.de/DE/leistungen/pbfb_verlag_berichte/pdf_einzelbaende/251_pdf.pdf?__blob=publicationFile&v=2 (accessed on 29 March 2019).
20. Kreklow, J.; Tetzlaff, B.; Burkhard, B.; Kuhnt, G. Radar-Based Precipitation Climatology in Germany—Developments, Uncertainties and Potentials. Atmosphere 2020, 11. [Google Scholar] [CrossRef] [
Green Version]
21. Bronstert, A.; Agarwal, A.; Boessenkool, B.; Crisologo, I.; Fischer, M.; Heistermann, M.; Köhn-Reich, L.; López-Tarazón, J.A.; Moran, T.; Ozturk, U.; et al. Forensic hydro-meteorological analysis
of an extreme flash flood: The 2016-05-29 event in Braunsbach, SW Germany. Sci. Total Environ. 2018, 977–991. [Google Scholar] [CrossRef]
22. Fischer, F.K.; Winterrath, T.; Auerswald, K. Temporal- and spatial-scale and positional effects on rain erosivity derived from point-scale and contiguous rain data. Hydrol. Earth Syst. Sci. 2018,
22, 6505–6518. [Google Scholar] [CrossRef] [Green Version]
23. Panagos, P.; Borrelli, P.; Spinoni, J.; Ballabio, C.; Meusburger, K.; Beguería, S.; Klik, A.; Michaelides, S.; Petan, S.; Hrabalíková, M.; et al. Monthly Rainfall Erosivity: Conversion Factors
for Different Time Resolutions and Regional Assessments. Water 2016, 8, 119. [Google Scholar] [CrossRef] [Green Version]
24. Semmel, A. Hessisches Bergland. In Physische Geographie Deutschlands, 2nd ed.; Liedtke, H., Marcinek, J., Eds.; Justus Perthes Verlag: Gotha, Germany, 1995; pp. 340–352. ISBN 3-623-00840-0. [
Google Scholar]
25. Winterrath, T.; Brendel, C.; Hafer, M.; Junghänel, T.; Klameth, A.; Lengfeld, K.; Walawender, E.; Weigl, E.; Becker, A. Radar climatology (RADKLIM) version 2017.002 (RW); gridded precipitation
data for Germany. Available online: https://search.datacite.org/works/10.5676/dwd/radklim_rw_v2017.002 (accessed on 25 June 2019).
26. Deutscher Wetterdienst Open Data Portal. Rain Gauge Precipitation Observations in 1-Minute Resolution. Available online: https://opendata.dwd.de/climate_environment/CDC/observations_germany/
climate/1_minute/precipitation/ (accessed on 20 February 2020).
27. Kreklow, J.; Tetzlaff, B.; Kuhnt, G.; Burkhard, B. A Rainfall Data Intercomparison Dataset of RADKLIM, RADOLAN, and Rain Gauge Data for Germany. Data 2019, 4. [Google Scholar] [CrossRef] [Green
28. The HDF Group. Hierarchical Data Format. Available online: https://portal.hdfgroup.org (accessed on 18 December 2018).
29. Mckinney, W. pandas: A Foundational Python Library for Data Analysis and Statistics. Python High-Perform. Sci. Comput. 2011, 14. [Google Scholar]
30. Kreklow, J. Facilitating radar precipitation data processing, assessment and analysis: A GIS-compatible python approach. J. Hydroinformatics 2019, 21, 652–670. [Google Scholar] [CrossRef] [Green
31. Kreklow, J.; Tetzlaff, B.; Burkhard, B.; Kuhnt, G. Radar-Based Precipitation Climatology in Germany - Developments, Uncertainties and Potentials. Available online: https://www.preprints.org/
manuscript/202002.0044/v1 (accessed on 10 February 2020).
32. Field, C.B.; Barros, V.R.; Dokken, D.J.; Mach, K.J.; Mastrandrea, M.D. Climate Change 2014 – Impacts, Adaptation and Vulnerability: Part A: Global and Sectoral Aspects; Working Group II
Contribution to the IPCC Fifth Assessment Report; IPCC: Geneva, Switzerland, 2015; Available online: https://www.cambridge.org/core/books/
climate-change-2014-impacts-adaptation-and-vulnerability-part-a-global-and-sectoral-aspects/1BE4ED76F97CF3A75C64487E6274783A (accessed on 7 February 2018).
33. Giorgi, F.; Bi, X.; Pal, J. Mean, interannual variability and trends in a regional climate change experiment over Europe. II: climate change scenarios (2071–2100). Clim. Dyn. 2004, 23, 839–858. [
Google Scholar] [CrossRef]
34. Semmler, T.; Jacob, D. Modeling extreme precipitation events—A climate change simulation for Europe. Glob. Planet. Chang. 2004, 44, 119–127. [Google Scholar] [CrossRef]
35. Frei, C.; Schöll, R.; Fukutome, S.; Schmidli, J.; Vidale, P.L. Future change of precipitation extremes in Europe: Intercomparison of scenarios from regional climate models. J. Geophys. Res. 2006,
111, 224. [Google Scholar] [CrossRef] [Green Version]
36. Kyselý, J.; Beranová, R. Climate-change effects on extreme precipitation in central Europe: Uncertainties of scenarios based on regional climate models. Theor. Appl. Clim. 2009, 95, 361–374. [
Google Scholar] [CrossRef]
37. Quirmbach, M.; Einfalt, T.; Langstädtler, G.; Janßen, C.; Reinhardt, C.; Mehlig, B. Extremwertstatistische Untersuchung von Starkniederschlägen in NRW (ExUS). Korresp. Abwasser und Abfall 2013,
60, 591–599. [Google Scholar]
38. Fiener, P.; Auerswald, K. Comment on “The new assessment of soil loss by water erosion in Europe” by Panagos et al. (Environmental Science & Policy 54 (2015) 438–447). Environ. Sci. Policy 2016,
57, 140–142. [Google Scholar] [CrossRef]
39. Panagos, P.; Meusburger, K.; Ballabio, C.; Borrelli, P.; Beguería, S.; Klik, A.; Rymszewicz, A.; Michaelides, S.; Olsen, P.; Tadić, M.P.; et al. Reply to the comment on "Rainfall erosivity in
Europe" by Auerswald et al. Sci. Total Environ. 2015, 532, 853–857. [Google Scholar] [CrossRef] [PubMed] [Green Version]
Figure 1. (a) Location, height above sea level [m] and selected landscape units of the federal state of Hesse, (b) spatial distribution of cropland areas in the study area.
Figure 2.
Boxplots of all R-factor derivatives grouped by spatial extent. In the lower subplots, the average of the rain gauges (
) and the rain gauges in pixels with cropland (
) have been added as a ground-truth reference. See
Table 1
for explanation of the used abbreviations.
Figure 3. Empirical cumulative distribution functions (ECDF) for all spatially highly resolved R-factor derivatives. The ECDFs for the rain gauges (R[G,DIN]) and the rain gauges in pixels with
cropland (R[G,DIN,Agri]) have been added as a ground-truth reference.
Figure 5. R-factor percentage change of R[YW,DIN] against R[EA] (a), scaled R-factors R[YW,F] and R[G,F] (b) and percentage change of R[YW,F] against R[EA] (c).
Figure 6. Comparison of regression models between different R-factors and the respective mean summer precipitation sums.
Name Derivation Method Input Dataset Spatial Extent n
R[YW,DIN] DIN 19708 RADKLIM YW (5 min) All radar pixels in Hesse (1 × 1 km) 23,320
R[YW,DIN, Agri] DIN 19708 RADKLIM YW (5 min) Radar pixels containing ≥ 10 ha of cropland 11,555
R[G,DIN] DIN 19708 Rain gauge data (5 min) All rain gauges 110
R[YWG,DIN] DIN 19708 RADKLIM YW (5 min) Pixels containing a rain gauge 110
R[EA] $0.141 · N S u − 1.48$ Interpolated rain gauge data (1971–2000) 1 × 1 km grid for Hesse 23,320
R[EA,Agri] $0.141 · N S u − 1.48$ Interpolated rain gauge data (1971–2000) Grid cells containing ≥ 10 ha of cropland 11,555
R[RW,Reg] $0.141 · N S u − 1.48$ RADKLIM RW (1 h) All radar pixels in Hesse 23,320
R[RW,Reg,Agri] $0.141 · N S u − 1.48$ RADKLIM RW (1 h) Radar pixels containing ≥ 10 ha of cropland 11,555
R[G,Reg] $0.141 · N S u − 1.48$ Rain gauge data All rain gauges in Hesse 110
R[RWG,Reg] $0.141 · N S u − 1.48$ RADKLIM RW (1 h) Pixels containing a rain gauge 110
R[EAG] $0.141 · N S u − 1.48$ Interpolated rain gauge data (1971–2000) Grid cells containing a rain gauge 110
R[YW,F] $R D I N · ( ( 1.13 + 0.35 ) · 1.05 )$ RADKLIM YW (5 min) All radar pixels in Hesse 23,320
R[YW,F,Agri] $R D I N · ( ( 1.13 + 0.35 ) · 1.05 )$ RADKLIM YW (5 min) Radar pixels containing ≥ 10 ha of cropland 11,555
R[G,F] $R D I N · 1.05$ Rain gauge data All rain gauges 110
R[YWG,F] $R D I N · ( ( 1.13 + 0.35 ) · 1.05 )$ RADKLIM YW (5 min) Pixels containing a rain gauge 110
R[G,P] $R D I N · 0.7984$ Rain gauge data All rain gauges 110
R-factor n Method Data Source Mean Standard Deviation Min Median Max
R[YW,DIN] 23,320 DIN 19708 RADKLIM 58.0 14.7 28.8 54.6 173.2
R[YW,DIN,Agri] 11,555 DIN 19708 RADKLIM 54.2 12.0 28.8 52.3 146.1
R[G,DIN] 110 DIN 19708 Gauges 80.6 20.6 53.4 75.3 157.2
R[YWG,DIN] 110 DIN 19708 RADKLIM 60.1 15.8 31.0 57.8 104.7
R[EA] 23,320 Regression Erosion atlas 54.5 6.6 42.1 52.8 81.8
R[EA,Agri] 11,555 Regression Erosion atlas 52.8 5.3 42.1 51.7 81.0
R[RW,Reg] 23,320 Regression RADKLIM 53.2 6.8 32.8 53.0 77.0
R[RW,Reg,Agri] 11,555 Regression RADKLIM 51.9 6.4 32.8 52.1 71.4
R[G,Reg] 110 Regression Gauges 57.0 8.8 44.7 55.0 84.7
R[RWG,Reg] 110 Regression RADKLIM 53.1 7.8 35.9 52.4 73.0
R[EAG] 110 Regression Erosion atlas 55.9 8.1 45.2 53.7 81.8
R[YW,F] 23,320 DIN scaled RADKLIM 90.1 22.8 44.5 84.8 269.1
R[YW,F,Agri] 11,555 DIN scaled RADKLIM 84.2 18.6 44.5 81.3 227.0
R[G,F] 110 DIN scaled Gauges 84.6 21.6 56.1 79.1 165.1
R[YWG,F] 110 DIN scaled RADKLIM 93.4 24.6 48.0 89.8 162.7
R[G,P] 110 DIN scaled Gauges 64.4 16.4 42.6 60.1 125.5
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zhbevd.f - Linux Manuals (3)
zhbevd.f (3) - Linux Manuals
zhbevd.f -
subroutine zhbevd (JOBZ, UPLO, N, KD, AB, LDAB, W, Z, LDZ, WORK, LWORK, RWORK, LRWORK, IWORK, LIWORK, INFO)
ZHBEVD computes the eigenvalues and, optionally, the left and/or right eigenvectors for OTHER matrices
Function/Subroutine Documentation
subroutine zhbevd (characterJOBZ, characterUPLO, integerN, integerKD, complex*16, dimension( ldab, * )AB, integerLDAB, double precision, dimension( * )W, complex*16, dimension( ldz, * )Z, integerLDZ,
complex*16, dimension( * )WORK, integerLWORK, double precision, dimension( * )RWORK, integerLRWORK, integer, dimension( * )IWORK, integerLIWORK, integerINFO)
ZHBEVD computes the eigenvalues and, optionally, the left and/or right eigenvectors for OTHER matrices
ZHBEVD computes all the eigenvalues and, optionally, eigenvectors of
a complex Hermitian band matrix A. If eigenvectors are desired, it
uses a divide and conquer algorithm.
The divide and conquer algorithm makes very mild assumptions about
floating point arithmetic. It will work on machines with a guard
digit in add/subtract, or on those binary machines without guard
digits which subtract like the Cray X-MP, Cray Y-MP, Cray C-90, or
Cray-2. It could conceivably fail on hexadecimal or decimal machines
without guard digits, but we know of none.
JOBZ is CHARACTER*1
= 'N': Compute eigenvalues only;
= 'V': Compute eigenvalues and eigenvectors.
UPLO is CHARACTER*1
= 'U': Upper triangle of A is stored;
= 'L': Lower triangle of A is stored.
N is INTEGER
The order of the matrix A. N >= 0.
KD is INTEGER
The number of superdiagonals of the matrix A if UPLO = 'U',
or the number of subdiagonals if UPLO = 'L'. KD >= 0.
AB is COMPLEX*16 array, dimension (LDAB, N)
On entry, the upper or lower triangle of the Hermitian band
matrix A, stored in the first KD+1 rows of the array. The
j-th column of A is stored in the j-th column of the array AB
as follows:
if UPLO = 'U', AB(kd+1+i-j,j) = A(i,j) for max(1,j-kd)<=i<=j;
if UPLO = 'L', AB(1+i-j,j) = A(i,j) for j<=i<=min(n,j+kd).
On exit, AB is overwritten by values generated during the
reduction to tridiagonal form. If UPLO = 'U', the first
superdiagonal and the diagonal of the tridiagonal matrix T
are returned in rows KD and KD+1 of AB, and if UPLO = 'L',
the diagonal and first subdiagonal of T are returned in the
first two rows of AB.
LDAB is INTEGER
The leading dimension of the array AB. LDAB >= KD + 1.
W is DOUBLE PRECISION array, dimension (N)
If INFO = 0, the eigenvalues in ascending order.
Z is COMPLEX*16 array, dimension (LDZ, N)
If JOBZ = 'V', then if INFO = 0, Z contains the orthonormal
eigenvectors of the matrix A, with the i-th column of Z
holding the eigenvector associated with W(i).
If JOBZ = 'N', then Z is not referenced.
LDZ is INTEGER
The leading dimension of the array Z. LDZ >= 1, and if
JOBZ = 'V', LDZ >= max(1,N).
WORK is COMPLEX*16 array, dimension (MAX(1,LWORK))
On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
LWORK is INTEGER
The dimension of the array WORK.
If N <= 1, LWORK must be at least 1.
If JOBZ = 'N' and N > 1, LWORK must be at least N.
If JOBZ = 'V' and N > 1, LWORK must be at least 2*N**2.
If LWORK = -1, then a workspace query is assumed; the routine
only calculates the optimal sizes of the WORK, RWORK and
IWORK arrays, returns these values as the first entries of
the WORK, RWORK and IWORK arrays, and no error message
related to LWORK or LRWORK or LIWORK is issued by XERBLA.
RWORK is DOUBLE PRECISION array,
dimension (LRWORK)
On exit, if INFO = 0, RWORK(1) returns the optimal LRWORK.
LRWORK is INTEGER
The dimension of array RWORK.
If N <= 1, LRWORK must be at least 1.
If JOBZ = 'N' and N > 1, LRWORK must be at least N.
If JOBZ = 'V' and N > 1, LRWORK must be at least
1 + 5*N + 2*N**2.
If LRWORK = -1, then a workspace query is assumed; the
routine only calculates the optimal sizes of the WORK, RWORK
and IWORK arrays, returns these values as the first entries
of the WORK, RWORK and IWORK arrays, and no error message
related to LWORK or LRWORK or LIWORK is issued by XERBLA.
IWORK is INTEGER array, dimension (MAX(1,LIWORK))
On exit, if INFO = 0, IWORK(1) returns the optimal LIWORK.
LIWORK is INTEGER
The dimension of array IWORK.
If JOBZ = 'N' or N <= 1, LIWORK must be at least 1.
If JOBZ = 'V' and N > 1, LIWORK must be at least 3 + 5*N .
If LIWORK = -1, then a workspace query is assumed; the
routine only calculates the optimal sizes of the WORK, RWORK
and IWORK arrays, returns these values as the first entries
of the WORK, RWORK and IWORK arrays, and no error message
related to LWORK or LRWORK or LIWORK is issued by XERBLA.
INFO is INTEGER
= 0: successful exit.
< 0: if INFO = -i, the i-th argument had an illegal value.
> 0: if INFO = i, the algorithm failed to converge; i
off-diagonal elements of an intermediate tridiagonal
form did not converge to zero.
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
November 2011
Definition at line 215 of file zhbevd.f.
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Interaction graph connectivity
One of the most powerful features of entropy quantum computing, and optical devices in general, is the fact that since light particles move, it is relatively easy to couple them to each other
arbitrarily. This is well documented and understood in the literature, since light is constantly moving and light pulses can easily be divided into parts, realizing full connectivity is conceptually
straightforward. An effective scheme can be to simply pull off a small part of one pulse and add it to another (see figure 1 of this paper from 2016 for example). The interactions can also be
simulated using classical feedback, by measuring and using a laser to return what the feedback pulse should be. Although such an approach disrupts entanglement, it provides an easy way to experiment
with large-scale highly-connected devices. Most, if not all, real industrial optimization problems have highly connected interactivity graphs, and the overhead of mapping to less-connected hardware
can be huge, a quick-back-of-the envelope calculation (explained later in this document) shows that under realistic constraints mapping to a non-fully-connected chip based architecture, means that a
highly connected problem with a few hundred variables (still small in terms of optimization problems), will take on the order of 10,000 qubits to map. This ratio gets worse as problems get bigger.
More typical 10,000 variable problems would require the order of 100 million physical qubits to map. Our hardware does not have this issue, one qubit is one variable in an optimization problem.
In our paradigm we are allowed to interact any qubit or qudit with any other one. This means that the mapping of problem variables to the physical qubits/dits is a trivial one-to-one mapping. A
single binary variable maps to a single qubit, and a single discrete variable maps to a single qudit. If we were restricted in which ones could interact and wanted to solve a problem which is not
naturally compatible with those constraints, then the only way to do it is through a one-to-many mapping, where each variable is mapped to many qubits or qudit. There are a few known ways to do this,
which we will discuss in passing, but won’t be the focus because they aren’t needed on our hardware. What we will do however is to show that lower connectivity can introduce fundamental limitations
in how problems can be mapped. The headline result here is that for realistic restrictions from placement on a two-dimensional chip, the size of a highly-connected problem that can be solved goes as
the square root of the size of the device, whereas on our hardware it scales directly as the size. As an example, we examine the TRIAD method to map a complete interaction graph to quasi-planar
hardware proposed in this paper. The qubit estimate based on this proposal, takes another parameter which is the number of qubits each can interact with, denoted as d, in this example we consider as
examples d=6 and d=20 (note that this estimate is not necessarily accurate when $n\approx d$):
Interaction graphs
A graph (in the formal computer science sense) consists of a set of nodes connected by edges, the nodes can be visualized as circles, and the edges can be visualized as lines connecting the circles.
For analog hardware like ours, we can think of an interactivity graph, which is the graph of what can directly interact with what, as a graph with edges joining all of the nodes, in other words any
qubit or qudit can be made to interact with any other one. A kind of graph known as a complete graph. This seems trivial, but not all devices have full connectivity, and in many cases the way a
device is physically built can limit this graph to have specific structures.
As a concrete example, in superconducting circuits, each variable often corresponds to a physical circuit element, the physics constrains the size of these and they have to be placed somewhere.
Likewise the coupler circuits can only be so long. This means that the interaction graph for such devices cannot be fully connected, it must instead be a type of graph we call quasi-planar. In a
quasi-planar graph nodes (representing the qubits), can only have edges (be allowed to interact) with others which are nearby in a 2D plane.
Interaction graphs of optimization problems
We can now think from the other direction, how much interactivity is needed to solve a given problem. This isn’t something which people often consider in conventional optimization, but it is
important when mapping to devices with limited connectivity. Fortunately it is straightforward to figure out from a QUBO (or QUDO, our discrete extension of a QUBO) representation of a problem. We
can assign a node to each variable, and then assign edges between them whenever there is a non-zero off-diagonal element in the QUBO matrix. A simple example here is the interaction graph of a simple
maximum independent set problem expressed as a QUBO, the QUBO appears on the left, and the maximum independent set problem on the right, with the elements corresponding to different edges color
coded. The solution is also color coded in both, with the nodes not included coloured in light blue, and corresponding light blue columns in the matrix. In both case we can see that the remaining
nodes/variables are independent, on the graph we can see that no grey nodes share an edge, and in the matrix we can see that the light blue rectangle covers all off-diagonal elements.
As an example of how problem connectivity can scale, we can consider a traveling salesperson problem. In the conventional formulation each variable represents a particular city being visited at a
particular point in the tour. Constraints need to be added that each city is visited once, (thus interacting all variables which represent the same city), furthermore only one city can be visited at
a time (thus interacting all variables which represent the same position in the tour). Finally penalties need to be added for the distances between cities, these add edges between a variable
representing one position with those representing the next and previous position in the tour. While not quite fully connected, it is clear that the interaction graph representing a traveling
salesperson problem is highly connected.
Mapping to a less connected graph
I have so far claimed there are methods to map a more connected graph to a less connected graph but haven’t discussed what they are. Since our hardware has full connectivity and therefore these
techniques are never needed on our systems we will not spend much time on it, but it is conceptually useful to have a picture of how this is done. The most used method is one called minor embedding,
where connected groups of qubits (“graph minors” in technical terms) are strongly coupled together so they act as a single variable. A visualization of this process is shown below. The right graphic
shows the embedding thick lines represent strong coupling between the qubits, while the thin lines show couplings which could be used to create interactions between the variables.
Technically minor embedding isn’t the only way to map a more connected graph to a less connected one (this paper for example), but it is the most used and the conceptually easiest to grasp. However,
as I will show later, there are fundamental mathematical limitations on how efficiently a problem with a highly connected interactivity graph can be mapped to any hardware with a quasi-planar
interactivity graph.
Graph theory: treewidth and tree decompositions
Fortunately, rather than try to argue about every possible method which could be used to map to hardware, we can approach the problem at a higher level. Graph theory, which (unsurprisingly) is the
theory which allows us to understand graphs (recall that we are using a formal computer science definition of a “graph”, edges connected by nodes, not a plot of a function which is often called a
“graph” in high-school mathematics) at a more abstract level, gives us tools to understand the limits of how well any mapping technique could perform. In particular, graph theory gives us tools to
measure properties related to graph connectivity which allow us to reason about what is or isn’t possible. The important property here is one called treewidth, which relates to a mathematical
technique called tree decomposition.
A tree decomposition is a way of mapping any graph (which may in general have loops) to a kind of graph known as a “tree”, which has no loops. This decomposition consists of combining nodes of the
original graphs into a new graph which must follow the following rules:
1. Any nodes connected by edges in the original graph must appear in the same node at least once in the new graph
2. All nodes which contain the same node of the original graph in the new graph must be connected by edges (although these connections can be indirect, does not need to be fully connected)
3. The new graph must be a tree, it cannot have any loops
4. The goal is to minimize the number of nodes of the original graph which appear in the “largest” node of the new graph
From this decomposition, we can define a property known as the treewidth, this property is defined as the number of variables in the largest node of the tree decomposition minus one. The convention
of defining it as one less is to guarantee that a graph without loops (a tree) will have a treewidth of one.
The key property of treewidth is that it is not possible to map a graph with a given treewidth into a graph with a smaller treewidth, this is hopefully somewhat intuitive. Tree decompositions are
also important in algorithms, and in fact certain dynamical programming algorithms can solve NP-hard problems on graphs with low treewidth efficiently, even if the problem involves a large number of
variables. This implies that if a mapping were possible from high to low treewidth graphs, it would show P=NP. In general finding the most optimal tree decomposition is a computationally hard
problem, but in some cases it is possible to make simple arguments about the treewidth.
Treewidth of different graphs
From the definition of treewidth, the most optimal tree decomposition for a fully connected graph is to place every node of the original graph into the same node of the new graph, there is no other
way to do it, this makes the treewidth of a fully connected graph of size n equal to n-1. I won’t prove it here, but it should also be intuitively clear that a highly connected graph should also have
a treewidth close to the number of nodes.
For quasi-planar graphs, the situation is different, a node can only share an edge with nodes within a certain distance of them. We could therefore imagine a strategy of tree decomposition where we
divide the graph into overlapping rows which each have a size of twice the interaction range, and spaced by the interaction range. Each of these rows will overlap only with the row above it and the
row below it so therefore the graph resulting from this decomposition will have no loops. For an n-by-n quasi-planar graph (where the interaction range is much smaller than the graph size), the
number of nodes in each row is proportional to n but the total number of nodes is proportional to n^2, therefore the size of highly connected graphs which can be mapped scales as the square root of
the size of the hardware interactivity graph if it is quasi-planar. Below we show a visualization of a decomposition process on a quasi-planar graph.
In this example, the graph has n=25 nodes, but we have shown the treewidth cannot be more than 9. This treewidth implies that it is impossible to map a fully connected problem with more than 10
nodes to it by any method. Note that a more optimal (but more complicated to draw) decomposition is possible, but this simple one demonstrates the necessary scaling. The treewidth of this graph is
actually only 6. Can you figure out the corresponding decomposition with the largest grouping containing only 7? | {"url":"https://quantumcomputinginc.com/learn/module/the-analog-quantum-advantage/interaction-graph-connectivity","timestamp":"2024-11-13T06:18:36Z","content_type":"text/html","content_length":"273933","record_id":"<urn:uuid:8cd17b65-5b77-4642-b363-a282924447e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00083.warc.gz"} |
Revision history
Dear Guillermo,
The answer to your question is yes. Here is an illustration with the octahedron:
sage: P = polytopes.octahedron()
sage: Facets = {f.ambient_V_indices():f for f in P.faces(P.dimension() - 1)} # Create the vertices-indices-to-facets dictionary
sage: P.is_simplicial() # Check that P is simplicial, so we can get the boundary complex
sage: C = P.boundary_complex() # Create the boundary complex
sage: Order = C.is_shellable(True); Order # Get a shelling order
((0, 1, 2),
(1, 2, 5),
(1, 3, 5),
(0, 2, 4),
(0, 1, 3),
(0, 3, 4),
(2, 4, 5),
(3, 4, 5))
Finally, we can get the actual sequence of facet using our dictionary.
sage: Facet_shelling = [Facets[indices.tuple()] for indices in Order]; Facet_shelling
[A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices,
A 2-dimensional face of a Polyhedron in ZZ^3 defined as the convex hull of 3 vertices] | {"url":"https://ask.sagemath.org/answers/52641/revisions/","timestamp":"2024-11-07T03:13:18Z","content_type":"application/xhtml+xml","content_length":"15780","record_id":"<urn:uuid:03cdefed-d28a-455b-8f73-e8cb3f9cca4f>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00354.warc.gz"} |
How is Baseball Slugging Percentage Calculated? (Breakdown and Insights) | Sport Tasty
How is Baseball Slugging Percentage Calculated? (Breakdown and Insights)
To calculate a baseball player’s slugging percentage, you divide the total number of bases they have earned by their total number of at-bats. Total bases include singles, doubles, triples, and home
runs. The formula is slugging percentage = (1B + (2 * 2B) + (3 * 3B) + (4 * HR)) / at-bats. This statistic gives insight into a player’s power and ability to hit for extra bases.
Calling all baseball fans!
Ever wondered how those impressive slugging percentages are calculated?
Today, we delve into the world of baseball analytics to demystify Slugging Percentage.
Follow along as we dissect its significance, decode the formula, and reveal why it’s crucial for player and team evaluation.
Let’s step up to the plate and uncover the secrets of this game-changing stat!
Understanding the Significance of Slugging Percentage in Baseball
Hey there, fellow baseball enthusiasts!
Today, I want to dive deep into the world of baseball statistics and unravel the mystery behind a crucial metric: slugging percentage.
Let’s break it down together, shall we?
What Exactly is Slugging Percentage?
Imagine this: you’re at a baseball game, watching your favorite player step up to the plate.
The pitcher winds up, releases the ball, and crack!
Your favorite player smacks a powerful hit into the outfield.
That hit, my friends, contributes to a stat known as slugging percentage.
In simple terms, slugging percentage measures a player’s ability to hit for extra bases.
It takes into account not just their ability to get on base but also their knack for hitting doubles, triples, and home runs.
In essence, it’s a measure of a player’s raw power at the plate.
How is Slugging Percentage Calculated?
Now, here comes the fun part – the math behind the magic!
Calculating slugging percentage is straightforward yet enlightening.
It’s determined by total bases divided by at-bats.
Yup, it’s that simple: total bases earned from hits divided by the number of at-bats.
Let’s break it down with an example:
– If a player gets 1 single, 2 doubles, and 1 home run in 10 at-bats, their total bases would be 1 (single) + 4 (2 doubles) + 4 (home run) = 9.
– So, their slugging percentage would be 9 total bases divided by 10 at-bats, resulting in a slugging percentage of .900.
The Importance of Slugging Percentage in Baseball
Now, you might be wondering, “Why does slugging percentage matter?” Well, dear reader, slugging percentage offers a unique perspective on a player’s offensive capabilities.
While batting average tells you how often a player gets on base, slugging percentage goes a step further by revealing how impactful those hits are.
Consider this: Player A has a batting average of .350 but a slugging percentage of .400.
Player B, on the other hand, has a batting average of .300 but a slugging percentage of .500.
Intriguing, right?
In this scenario, Player B may have fewer hits but makes them count by frequently hitting for extra bases, showcasing their power and ability to drive in runs.
This highlights the value of slugging percentage in evaluating a player’s overall offensive performance.
Final Thoughts
As we wrap up our discussion on slugging percentage, I hope you’ve gained a newfound appreciation for this fundamental baseball metric.
Remember, while batting average paints a picture of consistency, slugging percentage unveils the raw power and impact a player brings to the plate.
Next time you’re watching a game, keep an eye out for those powerful hits that contribute to a player’s slugging percentage.
It’s these electrifying moments that truly make baseball a game of skill, strategy, and sheer excitement.
Stay tuned for more insightful baseball discussions coming your way soon.
Until next time, keep swinging for the fences, my fellow baseball aficionados!
Demystifying the Calculation of Baseball Slugging Percentage
Welcome to the breakdown of the formula for calculating slugging percentage in baseball.
This fundamental statistic provides valuable insights into a player’s performance at the plate, showcasing their ability to deliver powerful hits and drive in runs efficiently.
What Is Slugging Percentage?
Before diving into the formula, let’s quickly recap what slugging percentage represents.
In essence, slugging percentage is a metric that measures a player’s total bases achieved per at-bat.
It gives more weight to extra-base hits like doubles, triples, and home runs, providing a comprehensive view of a player’s offensive impact.
The Formula: Unveiling the Magic
To calculate slugging percentage, the formula is refreshingly simple: divide the total bases a player has amassed by the number of at-bats they have taken.
Mathematically, it looks like this:
Slugging Percentage = (1B + 2Bx2 + 3Bx3 + HRx4) / AB
Let’s break it down even further:
• 1B represents singles, which are worth 1 base each.
• 2B signifies doubles, granting the batter 2 bases.
• 3B stands for triples, adding 3 bases to the player’s total.
• HR denotes home runs, the ultimate hit that awards 4 bases.
• AB corresponds to at-bats, the number of times the player has faced a pitcher.
Bringing the Formula to Life
To illustrate this calculation, let’s take a look at a hypothetical example:
• Player A has 50 singles (1B), 20 doubles (2B), 5 triples (3B), and 15 home runs (HR) over 200 at-bats (AB).
• Plugging these numbers into the formula:
(50 + 20x2 + 5x3 + 15x4) / 200 = Slugging Percentage
• Calculating the total bases first:
(50 + 40 + 15 + 60) / 200 = 165 / 200 = 0.825
• Player A’s slugging percentage stands at an impressive 0.825.
Why Is Slugging Percentage Important?
Slugging percentage is a key metric for evaluating a player’s power and offensive contributions.
It provides a deeper understanding of how effectively a player can hit for extra bases, showcasing their ability to produce runs and impact the game offensively.
mastering the calculation of slugging percentage is crucial for both fans and analysts alike, offering valuable insights into a player’s performance beyond simple batting average.
Stay tuned for the next section, where we delve into real-world examples and the significance of slugging percentage in professional baseball.
How is Baseball Slugging Percentage Calculated?
So you want to dive into the world of baseball metrics and understand how slugging percentage is calculated?
Look no further!
In this section, I’ll walk you through a step-by-step guide on calculating slugging percentage like a pro.
Understanding Slugging Percentage
Before we get into the nitty-gritty of the calculation process, let’s ensure we’re on the same page regarding what slugging percentage actually represents.
Slugging percentage is a statistic used in baseball to measure the power of a hitter.
It provides insight into how well a batter hits for extra bases, such as doubles, triples, and home runs, as opposed to just singles.
The Formula for Slugging Percentage
To calculate slugging percentage, you will need to follow this simple formula:
Slugging Percentage = (Total Bases) / (At Bats)
Let’s break this down further:
• Total Bases: Add up the number of bases a batter reaches on hits. For example, a single equals 1 base, a double equals 2 bases, a triple equals 3 bases, and a home run equals 4 bases.
• At Bats: This refers to the number of times a batter comes up to the plate, excluding walks, hit-by-pitches, and sacrifices.
Example Calculation
Let’s walk through an example calculation to solidify your understanding:
Imagine a player has the following hits over a period:
– 10 singles
– 5 doubles
– 2 triples
– 3 home runs
Now, let’s calculate the slugging percentage for this player:
1. Calculate the total bases:
2. Singles: 10 * 1 = 10 bases
3. Doubles: 5 * 2 = 10 bases
4. Triples: 2 * 3 = 6 bases
5. Home runs: 3 * 4 = 12 bases
6. Add up the total bases: 10 + 10 + 6 + 12 = 38 bases
7. Determine the number of at-bats: Let’s say the player had 50 at-bats.
8. Apply the formula:
9. Slugging Percentage = Total Bases (38) / At Bats (50)
10. Slugging Percentage = 0.76 or 76%
Key Takeaways
By understanding and calculating slugging percentage, you gain valuable insights into a player’s power at the plate.
It goes beyond simple batting average and gives a more comprehensive view of a batter’s performance.
So, now that you’ve mastered the art of calculating slugging percentage, you’re on your way to becoming a baseball analytics aficionado!
Practice your calculations with different player stats to deepen your understanding of this essential metric.
Happy number crunching!
The Importance of Slugging Percentage in Baseball
When it comes to evaluating a baseball player’s performance, one key metric that stands out is the slugging percentage.
This metric is not only crucial for assessing individual player performance but also plays a significant role in determining overall team success.
Let’s delve into why slugging percentage is essential for player evaluation and team performance.
Understanding Slugging Percentage
Slugging percentage is a statistic used in baseball to measure the power and hitting ability of a player.
It takes into account the number of bases a player earns with their hits.
The formula to calculate slugging percentage is simple: divide the total number of bases a player has earned by the total number of at-bats.
For example, if a player has 10 total bases and 20 at-bats, their slugging percentage would be .500.
Assessing Player Performance
Slugging percentage provides valuable insights into a player’s ability to hit for extra bases.
A high slugging percentage indicates that a player is not only making contact with the ball but also driving it deep into the outfield, potentially leading to more runs scored.
Coaches and scouts often use slugging percentage to identify players with power-hitting potential.
Team Performance Impact
In addition to evaluating individual players, slugging percentage also influences team performance.
Teams with higher overall slugging percentages tend to score more runs and win games.
By having players who can consistently hit for power, teams increase their chances of driving in baserunners and ultimately crossing home plate.
Real-Life Examples
To understand the impact of slugging percentage further, let’s look at some real-life examples.
In the 2020 MLB season, players like Fernando Tatis Jr. and Luke Voit led their respective leagues in slugging percentage.
Their ability to consistently produce extra-base hits not only boosted their individual stats but also contributed to their team’s success on the field.
slugging percentage serves as a vital metric in the world of baseball, both for evaluating individual player performance and impacting overall team success.
By understanding and leveraging this statistic, players, coaches, and fans can gain deeper insights into the power-hitting abilities of athletes and the potential for offensive dominance on the
Final Thoughts
Understanding how baseball slugging percentage is calculated provides valuable insights into a player’s power at the plate and their ability to make impactful hits during a game.
By following the step-by-step guide we’ve outlined, you now have the knowledge to calculate slugging percentage like a pro.
Remember, a high slugging percentage signifies a player’s capability to contribute significantly to their team’s offense by hitting for extra bases.
Next time you watch a baseball game, pay attention to slugging percentage to gain a deeper appreciation for a player’s performance.
Keep crunching those numbers and uncover the hidden power of every swing! | {"url":"https://sporttasty.com/how-is-baseball-slugging-percentage-calculated/","timestamp":"2024-11-13T03:12:26Z","content_type":"text/html","content_length":"84761","record_id":"<urn:uuid:cbd9579a-6d33-4701-b312-7c588e598bec>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00837.warc.gz"} |
Ch. 11 Key Concepts - Intermediate Algebra | OpenStax
Key Concepts
11.1 Distance and Midpoint Formulas; Circles
• Distance Formula: The distance d between the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
• Midpoint Formula: The midpoint of the line segment whose endpoints are the two points $(x1,y1)(x1,y1)$ and $(x2,y2)(x2,y2)$ is
To find the midpoint of a line segment, we find the average of the x-coordinates and the average of the y-coordinates of the endpoints.
• Circle: A circle is all points in a plane that are a fixed distance from a fixed point in the plane. The given point is called the center, $(h,k),(h,k),$ and the fixed distance is called the
radius, r, of the circle.
• Standard Form of the Equation a Circle: The standard form of the equation of a circle with center, $(h,k),(h,k),$ and radius, r, is
• General Form of the Equation of a Circle: The general form of the equation of a circle is
11.2 Parabolas
• Parabola: A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of
the parabola.
Vertical Parabolas
General form Standard form
$y=ax2+bx+cy=ax2+bx+c$ $y=a(x−h)2+ky=a(x−h)2+k$
Orientation $a>0a>0$ up; $a<0a<0$ down $a>0a>0$ up; $a<0a<0$ down
Axis of symmetry $x=−b2ax=−b2a$ $x=hx=h$
Vertex Substitute $x=−b2ax=−b2a$ and $(h,k)(h,k)$
solve for y.
y- intercept Let $x=0x=0$ Let $x=0x=0$
x-intercepts Let $y=0y=0$ Let $y=0y=0$
• How to graph vertical parabolas $(y=ax2+bx+c(y=ax2+bx+c$ or $f(x)=a(x−h)2+k)f(x)=a(x−h)2+k)$ using properties.
1. Step 1. Determine whether the parabola opens upward or downward.
2. Step 2. Find the axis of symmetry.
3. Step 3. Find the vertex.
4. Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
5. Step 5. Find the x-intercepts.
6. Step 6. Graph the parabola.
Horizontal Parabolas
General form Standard form
$x=ay2+by+cx=ay2+by+c$ $x=a(y−k)2+hx=a(y−k)2+h$
Orientation $a>0a>0$ right; $a<0a<0$ left $a>0a>0$ right; $a<0a<0$ left
Axis of symmetry $y=−b2ay=−b2a$ $y=ky=k$
Vertex Substitute $y=−b2ay=−b2a$ and $(h,k)(h,k)$
solve for x.
y-intercepts Let $x=0x=0$ Let $x=0x=0$
x-intercept Let $y=0y=0$ Let $y=0y=0$
• How to graph horizontal parabolas $(x=ay2+by+c(x=ay2+by+c$ or $x=a(y−k)2+h)x=a(y−k)2+h)$ using properties.
1. Step 1. Determine whether the parabola opens to the left or to the right.
2. Step 2. Find the axis of symmetry.
3. Step 3. Find the vertex.
4. Step 4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
5. Step 5. Find the y-intercepts.
6. Step 6. Graph the parabola.
11.3 Ellipses
• Ellipse: An ellipse is all points in a plane where the sum of the distances from two fixed points is constant. Each of the fixed points is called a focus of the ellipse.
If we draw a line through the foci intersects the ellipse in two points—each is called a vertex of the ellipse.
The segment connecting the vertices is called the major axis.
The midpoint of the segment is called the center of the ellipse.
A segment perpendicular to the major axis that passes through the center and intersects the ellipse in two points is called the minor axis.
• Standard Form of the Equation an Ellipse with Center $(0,0):(0,0):$ The standard form of the equation of an ellipse with center $(0,0),(0,0),$ is
The x-intercepts are $(a,0)(a,0)$ and $(−a,0).(−a,0).$
The y-intercepts are $(0,b)(0,b)$ and $(0,−b).(0,−b).$
• How to an Ellipse with Center $(0,0)(0,0)$
1. Step 1. Write the equation in standard form.
2. Step 2. Determine whether the major axis is horizontal or vertical.
3. Step 3. Find the endpoints of the major axis.
4. Step 4. Find the endpoints of the minor axis
5. Step 5. Sketch the ellipse.
• Standard Form of the Equation an Ellipse with Center $(h,k):(h,k):$ The standard form of the equation of an ellipse with center $(h,k),(h,k),$ is
When $a>b,a>b,$ the major axis is horizontal so the distance from the center to the vertex is a.
When $b>a,b>a,$ the major axis is vertical so the distance from the center to the vertex is b.
11.4 Hyperbolas
• Hyperbola: A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant.
Each of the fixed points is called a focus of the hyperbola.
The line through the foci, is called the transverse axis.
The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola.
The midpoint of the segment joining the foci is called the center of the hyperbola.
The line perpendicular to the transverse axis that passes through the center is called the conjugate axis.
Each piece of the graph is called a branch of the hyperbola.
Standard Forms of the Equation a Hyperbola with Center $(0,0)(0,0)$
$x2a2−y2b2=1x2a2−y2b2=1$ $y2a2−x2b2=1y2a2−x2b2=1$
Orientation Transverse axis on the x-axis. Transverse axis on the y-axis.
Opens left and right Opens up and down
Vertices $(−a,0),(−a,0),$$(a,0)(a,0)$ $(0,−a),(0,−a),$$(0,a)(0,a)$
x-intercepts $(−a,0),(−a,0),$$(a,0)(a,0)$ none
y-intercepts none $(0,−a)(0,−a)$, $(0,a)(0,a)$
Rectangle Use $(±a,0)(±a,0)$$(0,±b)(0,±b)$ Use $(0,±a)(0,±a)$$(±b,0)(±b,0)$
asymptotes $y=bax,y=bax,$$y=−baxy=−bax$ $y=abx,y=abx,$$y=−abxy=−abx$
• How to graph a hyperbola centered at $(0,0).(0,0).$
1. Step 1. Write the equation in standard form.
2. Step 2. Determine whether the transverse axis is horizontal or vertical.
3. Step 3. Find the vertices.
4. Step 4. Sketch the rectangle centered at the origin intersecting one axis at $±a±a$ and the other at $±b.±b.$
5. Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle.
6. Step 6. Draw the two branches of the hyperbola.
Standard Forms of the Equation a Hyperbola with Center $(h,k)(h,k)$
$(x−h)2a2−(y−k)2b2=1(x−h)2a2−(y−k)2b2=1$ $(y−k)2a2−(x−h)2b2=1(y−k)2a2−(x−h)2b2=1$
Orientation Transverse axis is horizontal. Transverse axis is vertical.
Opens left and right Opens up and down
Center $(h,k)(h,k)$ $(h,k)(h,k)$
Vertices a units to the left and right of the center a units above and below the center
Rectangle Use a units left/right of center Use a units above/below the center
b units above/below the center b units left/right of center
• How to graph a hyperbola centered at $(h,k).(h,k).$
1. Step 1. Write the equation in standard form.
2. Step 2. Determine whether the transverse axis is horizontal or vertical.
3. Step 3. Find the center and $a,b.a,b.$
4. Step 4. Sketch the rectangle centered at $(h,k)(h,k)$ using $a,b.a,b.$
5. Step 5. Sketch the asymptotes—the lines through the diagonals of the rectangle. Mark the vertices.
6. Step 6. Draw the two branches of the hyperbola.
Conic Characteristics of $x2-x2-$ and $y2-y2-$ terms Example
Parabola Either $x2x2$ OR $y2.y2.$ Only one variable is squared. $x=3y2−2y+1x=3y2−2y+1$
Circle $x2-x2-$ and $y2-y2-$ terms have the same coefficients $x2+y2=49x2+y2=49$
Ellipse $x2-x2-$ and $y2-y2-$ terms have the same sign, different coefficients $4x2+25y2=1004x2+25y2=100$
Hyperbola $x2-x2-$ and $y2-y2-$ terms have different signs, different coefficients $25y2−4x2=10025y2−4x2=100$
11.5 Solve Systems of Nonlinear Equations
• How to solve a system of nonlinear equations by graphing.
1. Step 1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Step 2. Graph the first equation.
3. Step 3. Graph the second equation on the same rectangular coordinate system.
4. Step 4. Determine whether the graphs intersect.
5. Step 5. Identify the points of intersection.
6. Step 6. Check that each ordered pair is a solution to both original equations.
• How to solve a system of nonlinear equations by substitution.
1. Step 1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Step 2. Solve one of the equations for either variable.
3. Step 3. Substitute the expression from Step 2 into the other equation.
4. Step 4. Solve the resulting equation.
5. Step 5. Substitute each solution in Step 4 into one of the original equations to find the other variable.
6. Step 6. Write each solution as an ordered pair.
7. Step 7. Check that each ordered pair is a solution to both original equations.
• How to solve a system of equations by elimination.
1. Step 1. Identify the graph of each equation. Sketch the possible options for intersection.
2. Step 2. Write both equations in standard form.
3. Step 3. Make the coefficients of one variable opposites.
Decide which variable you will eliminate.
Multiply one or both equations so that the coefficients of that variable are opposites.
4. Step 4. Add the equations resulting from Step 3 to eliminate one variable.
5. Step 5. Solve for the remaining variable.
6. Step 6. Substitute each solution from Step 5 into one of the original equations. Then solve for the other variable.
7. Step 7. Write each solution as an ordered pair.
8. Step 8. Check that each ordered pair is a solution to both original equations. | {"url":"https://openstax.org/books/intermediate-algebra/pages/11-key-concepts","timestamp":"2024-11-03T05:39:34Z","content_type":"text/html","content_length":"423535","record_id":"<urn:uuid:7aa3681a-ac2a-447b-aa08-6c1115596210>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00368.warc.gz"} |
Nicos Georgiou
University of Sussex University of Sussex
Order of fluctuations for the discrete Hammersley process.
Probability Seminar
27th April 2018, 3:30 pm – 4:30 pm
Main Maths Building, SM4
We discuss the order of the variance on a lattice analogue of the Hammersley process, for which the environment on each site has independent, Bernoulli distributed values.
The last passage time is the maximum number of Bernoulli points that can be collected on a piecewise linear path, where each segment has strictly positive but finite slope.
For this model the shape function exhibits two flat edges and we study the order of the variance in directions that fall in the flat edge, in directions that approximate the edge of the flat edge,
and in directions in the strictly concave section of the shape for the i.i.d. model and for the associated equilibrium model with boundaries. This is joint work with Janosch Ortmann and Federico | {"url":"https://www.bristolmathsresearch.org/seminar/nicos-georgiou/","timestamp":"2024-11-10T13:10:31Z","content_type":"text/html","content_length":"54391","record_id":"<urn:uuid:0e909d48-cd7d-48c5-8a5f-eab12cd2624c>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00648.warc.gz"} |
A Secure Method for Data Storage and Transmission in Sustainable Cloud Computing
1 College of Computer Science and Technology, Xi’an University of Science and Technology, Xi’an, Shaanxi, 710054, China
2 Department of Computer Science (RCET Campus), University of Engineering and Technology, Lahore, 39161, Pakistan
3 Department of Information Sciences, Division of Science & Technology, University of Education, Lahore, 54700, Pakistan
4 Department of Computer Science, COMSATS University, Islamabad, Lahore, 54700, Pakistan
* Corresponding Author: Muhammad Usman Sana. Email:
Computers, Materials & Continua 2023, 75(2), 2741-2757. https://doi.org/10.32604/cmc.2023.036093
Received 16 September 2022; Accepted 06 January 2023; Issue published 31 March 2023
Cloud computing is a technology that provides secure storage space for the customer’s massive data and gives them the facility to retrieve and transmit their data efficiently through a secure network
in which encryption and decryption algorithms are being deployed. In cloud computation, data processing, storage, and transmission can be done through laptops and mobile devices. Data Storing in
cloud facilities is expanding each day and data is the most significant asset of clients. The important concern with the transmission of information to the cloud is security because there is no
perceivability of the client’s data. They have to be dependent on cloud service providers for assurance of the platform’s security. Data security and privacy issues reduce the progression of cloud
computing and add complexity. Nowadays; most of the data that is stored on cloud servers is in the form of images and photographs, which is a very confidential form of data that requires secured
transmission. In this research work, a public key cryptosystem is being implemented to store, retrieve and transmit information in cloud computation through a modified Rivest-Shamir-Adleman (RSA)
algorithm for the encryption and decryption of data. The implementation of a modified RSA algorithm results guaranteed the security of data in the cloud environment. To enhance the user data security
level, a neural network is used for user authentication and recognition. Moreover; the proposed technique develops the performance of detection as a loss function of the bounding box. The Faster
Region-Based Convolutional Neural Network (Faster R-CNN) gets trained on images to identify authorized users with an accuracy of 99.9% on training.
Nowadays, cloud computing is the most rapidly growing technology. It has tremendous advantages as it is faster in processing data, consumes less power, is low in cost, and has ample storage capacity.
But despite these advantages, cloud computing faces some challenges regarding user security and privacy. An essential aspect of improving the quality of service is the security of data in cloud
computing from cloud service providers [1]. Cloud computing is efficient, more reliable, and easy to use. It provides ample storage to their customers to save their data without having any virus
threat. It also provides secure data transmission between different equipment [2]. The availability, confidentiality, and data integrity are significant security concerns associated with cloud
computing. Therefore; for the confidentiality of the data, non-repudiation, authorization, and authentication of people who use the data are very important so that the unapproved parties are not able
to get or see any confidential data. Preserving data accurately and controlling the network devices and data from legal access is vital to keep the data integrity and to ensure valid and absolute
information. Availability of data is a big issue as it is stored in different locations in the cloud. The data availability whenever a client requests and the guarantee of data being available is
vital. In the public model of the cloud, authentication is required when the client retrieves their specific data. Authorization is needed to find out whether a particular individual has the
authority or not to perform an action on others’ data. Non-repudiation; is required to verify that a reliable person cannot withdraw after performing a job [3].
A lot of organizations adopted cloud computing, and hence as a consequence of loads of digital resources, security dilemmas are rising [4,5]. It is predictable recently that security in cloud
computing has become a considerable research focus [6–8]. Since the data is susceptible to intruder attacks, the key concern to the consumers is to find a secure cloud platform [9,10].
Numerous solutions have been provided by researchers, and several researchers are still working on them to find out more accurate and efficient solutions. The main objective of this research is to
look for a solution by applying the cryptographic technique for achieving data security in the cloud environment. Cryptography is one of the main approaches for the confidential transmission and
storage of data in the presence of a third party. Keeping information or data protected by changing the raw data into a configuration that is not easily readable is the technique of cryptography [11
]. Algorithms of cryptography with essential management techniques are particularly hopeful methods to achieve security and privacy in cloud computing [12,13]. When transferred and stored, data must
be encrypted constantly. If properly implemented this approach, data will never be easily understandable even if a different occupier can access the data. To keep non-repudiation and authentication
from unauthorized persons, data integrity and confidentiality are the foremost necessities of cryptography.
The concept of cryptography is centered on two standard terms: encryption and decryption. The process of conversion of data recognized using plain text aimed at an unreadable configuration,
identified as ciphertext, is encryption. Ciphertext cannot be understood by unauthorized persons. Converting data again into its previous form or making the understanding of encrypted information so
it can be understood or read by authorized persons is termed decryption [14,15]. For information security, there are numerous encryption algorithms extensively employed. For example, Amazon Simple
Storage Service (Amazon S3), 256-bit Advanced Encryption Standard (AES) is one of the most capable encryption algorithms. There are three types of cryptography algorithms: symmetric algorithms,
hashing, and asymmetric algorithms. Symmetric key cryptography is also known as the “Secret Key Encryption Algorithm” for encoding and decoding the private key [16].
In asymmetric key cryptography, symmetric for encoding and decoding, double exclusive keys are applied. The public key on the network is available to anyone. To encode data, the public key gets
employed. Data can only be decoded by the only private key. In keeping data secure, the private key is set aside secret. The benefit of applying asymmetric key encryption is that it offers an
improved distribution of scalability and key compared to symmetric systems. Some standard Asymmetric Key Algorithms are Diffie-Hellman, Elliptic Curve Cryptography (ECC), El Gamal, Digital Signature
Algorithm (DSA), and RSA: 1, one of the initial asymmetric cryptosystems RSA. In the cryptosystem, RSA is still the most used and employed. It is named after the three researchers who made this
system Ron Rivest, Len Adleman, and Adi Shamir. It includes two keys, a private and a public key. Information twisted with a public key can get translated only, with a private key. With its private
key engraving, a separate message, that’s recognized as a digital signature in this verification procedure. The server represents the authentication of the public key. Afterward that using the public
key server verifies then the digital signature is returned to the user [17].
Kumar et al. [18] proposed an efficient technique in a virtual environment related to the storage of cloud data focusing on problems and suggested techniques by using a public key cryptosystem to
offer data security and storage in the cloud by using the modified RSA algorithm concept to offer improved security in the cloud for data storage.
In terms of storage as well as computational complexities and solving these complexities in clouds. Ambika et al. [19] proposed the SKT-RSA technique based on the Secure Key Transmission (SKT). There
proposed technique is a distribution scheme of cluster keys that is tree based. And they also proposed the distribution of keys between the end user and certified authority. Gupta et al. [20] study
proposed RSA and Blowfish as two individual cryptographic calculations and examined the effect on speed and security, how they work, and when these two algorithms combine to form one hybrid algorithm
after suitable modification. In terms of encryption and decryption time, their study presented a comparative analysis of hybrid algorithms and individual algorithms. Dhamodaran et al. [21] modified
RSA, as on large blocks, RSA is largely byte-parallel and is computationally intensive. In a distributed environment System will perform parallel the encoding and decoding process and in terms of
execution time, the performance analysis confirms progress and maintains security.
In the cloud, to resolve the confidentiality trouble in transmitting sensitive data, the n-RSA encryption algorithm for a multi-level security model is proposed [22]. A security valuation is
conducted by the power cloud security center of the data to decide the security level. According to the security level of the data, then, for the n-RSA algorithm, select the suitable prime number. A
proposed method for power clouds is investigated with advantages and disadvantages. The safe transmission of power cloud data is shown in experimental results, and the proposed method can efficiently
develop flexibility and security. For securing the cloud, [23] proposed an innovative algorithm combining the RSA algorithm and Ciphertext Policy-Identity Attribute-Based Encryption (CP-IDABE). The
performance of the RSA-CP-IDABE algorithm for varying data sizes based on the time it takes for encryption, decryption, and execution is evaluated. Matched with the existing algorithms, the results
achieved by the proposed method show higher efficiency.
Seth et al. [24] proposed an integrating encryption technique in the cloud for secure data storage. The proposed method ensures secure and safe data transmission in the cloud using different data
fragmentation techniques and double encryption methods. Jiang et al. [25] proposed a model by using Faster R-CNN for the detection of faces. By using a dataset of WIDER face, they train a face
detection model in Faster R-CNN, and proposed a model of face detection in another work that is centered on the evolutionary Haar filter [26] set. There are 159,424 faces and 12,880 images in
training. They described experimented images of the WIDER dataset. By using deep learning, the authors presented a new face detection system, and attain advanced recognition performance on the
benchmark evaluation by a distinguished Face Detection Data Set and Benchmark (FDDB). Specifically, Sun et al. [27] developed the modern Faster R-CNN system by uniting several approaches, including
model pre-training, hard negative mining, feature concatenation, proper calibration of key parameters, and multi-scale training. Consequently, the suggested scheme gained advanced face detection
performance, and with regards to Receiver Operating Characteristic (ROC) curves on the FDDB benchmark evaluation, was graded as one of the best models.
Based on Faster R-CNN, Wu et al. [28] proposed a Different Scale Face Detector (DSFD), the novel network, while performing a Faster R-CNN in real-time, can advance the accuracy of face detection. A
multi-task effective region proposal network joined to attain the human face Region of Interest (ROI), Region Proposal Network (RPN) with improving face detection is developed. An anchor is
consistently formed on the top feature while setting the ROI as a limit, mapped by the multi-task RPN. And the anchor shared with the facial signs is mined with a human face scheme. The authors
proposed a Fast R-CNN network, which is parallel-type. The schemes are allotted to three parallel Fast R-CNN networks according to the different percentages of the images they cover. A range of
approaches is presented in the face detection network, comprising feature concatenation, feature pyramid, and multi-task learning. Related to advanced face detection approaches, for example,
HyperFace, UnitBox, and FastCNN, on standard benchmarks comprising Annotated Faces in the Wild (AFW), FDDB, WIDER FACE, and PASCAL face, the proposed DSFD technique attains good performance.
Faster R-CNN, in object detection applications, is a common technique. Based on Faster R-CNN, an improved model Face R-CNN is proposed in which a facial feature enhancement technique related to the
attention process is used. Syntax-guided network (SG-NET) combined with Face R-CNN to unite the produced image with the unique convolutional features that in the feature map improve the concentration
on the area of the face and in the case of large-scale obstruction, can efficiently attain face detection. Through the analysis by testing and training Face R-CNN on the dataset of Wider Face, the
experimental results show that the improved model has a further noticeable recognition outcome on blocking faces, and is greater than the Faster R-CNN accuracy rate average is 3.5% [29].
To recognize candidate face frames that are to be detected with partial occlusion. The Non-Maximum Suppression (NMS) technique based on Faster R-CNN practices a strict threshold. In multifaceted
scenes with fractional obstruction of the face and irregular lighting, the occurrence of mislaid and incorrect face detection is likely to happen. To solve this issue, Yan et al. [30] suggested a
novel face detection technique through CNN to extract facial features, and the Region Proposal Network (RPN) detected and generated a considerable amount of face candidate frames; by linear weighting
technique, the hard threshold of NMS is enhanced, and by the linearly weighted NMS face candidate frame is selected. On the FDDB dataset, comparison experiment results show, the newly proposed face
detection method has high detection robustness, and accuracy, and under partial occlusion and uneven lighting, can efficiently avoid missed detection and wrong detection of abundant faces.
In cloud infrastructure, when a user sends the data, it is first transferred to the cloud server side, and after encryption, data is sent to the receiver. Data owners have no access or control over
the data once they send it, and can’t monitor and check the security status. All the problems related to user data security are handled or controlled at the cloud server. Data-centric security is an
approach in which users have the authority to control data and check the security aspects when the data is at the server end. The driving force behind modifying the RSA in the proposed research is
the secured and safe transmission of data through a secured transmission of key. Also, the appropriate authentication technique is required as this is missing in previous research. In previous
research for encryption and decryption, RSA Algorithm is used, in which the N key is the product of two prime numbers which can be identified, and someone can easily hack the entire system without
any difficulty by knowing the private key.
The main objectives of this research are:
• To enhance the security level of data that is stored on the cloud server.
• To develop a more advanced algorithm for encryption and decryption.
• A face recognition method based on Faster R-CNN was suggested to ensure that only the authorized person can access the keys.
Three famous mathematicians discovered the RSA algorithm for data encryption and decryption named, Ronald Rivest, Adi Shamir, and Leonard Adleman in, 1978. How efficiently an RSA algorithm works to
secure the user data relies on the complexity of the mathematical function, the more complex the function means that it is difficult to guess or solve by any other person. To give access authorized
users to send or receive information via the cloud RSA Algorithm is being implemented and through this technique, data is not retrieved by any attacker or authorized person. When a user saves their
data on the cloud, it first gets encrypted, and then saved on the cloud. When the user requires data that is being stored on a cloud, first, it sends a request for data retrieval to the cloud
provider, and then after proper identification and authorization of user data is sent to the specified user. In the RSA algorithm, each message is converted into integers. RSA works on public and
private keys. A private Key is only familiar to the authorized user and it is used to decrypt the data which is first encrypted by the cloud provider.
The whole process consists of three steps which are:
1. Key Generation:
Calculate the product of any two random numbers as
Euler function
Now take any number named e to estimate (e,φ(v))=1,and1<e<φ(n).
φ(v) multiplication inverse is named d, which also satisfies the following equation.
The keys (E, n) are public, while the keys (D, n) are private.
2. Encryption
M information is encrypted by the correspondent by using the following equation where C is encrypted ciphertext.
3. Decryption
Cipher C has been decrypted by the receiver, who receives the information.
RSA Algorithm is the most popular algorithm for the encryption and decryption of data in a more secure way. RSA helps in the identification of service providers and authentication over insecure
mediums of communication, and it’s challenging to hack the system and retrieve other persons’ information. RSA encryption cracking is as challenging as huge numbers factoring [31,32]. The RSA
algorithm comprises three steps:
1. To generate a public and private key
2. Encryption
3. Decryption
Two prime numbers, modular multiplicative inverses that are generated randomly, are selected as w and x.
Calculate n=w∗x
Calculate f(n)=(w−1)(x−1)
Choose number e, e<f(n).
Calculate D, such that module1 (modf(n)) and e<f(n).
Private key = {D,n}
Public key = {E,n}
Cipher text C = message E mod n
Plain text P = cipher text D mod n
We divide the platform into three steps
1. Public key1 + plaintext → ciphertext1
2. Public key2 + ciphertext1 → ciphertext 2
3. Private key + ciphertext1 → plain text
We propose a modified RSA Algorithm in which it is not easy to crack the system and get access to other users’ private data. Previously for encryption and decryption, RSA Algorithm is used, and the N
key, is the product of two prime numbers which can effortlessly be identified, and someone can easily hack the entire system by knowing the private key.
The following are the steps of a modified RSA System.
1. n1=p×q
2. n2=r×s
3. N=n1∗n2
4. φ(N)=(p−1)∗(q−1)∗(r−1)∗(s−1)
5. E=E1∗E2
6. After the calculation of E, we get two encryptions and one decryption key by just dividing the key by E1 and E2.
Fig. 1 presents a flow chart demonstration of the improved RSA algorithm. To calculate n and mod φ(N) four different prime numbers p*q*r*s are taken as the input of the random number (a, b) is
carefully chosen from the range 1<e<fφ(n) as the exponent of the public key. To utilize as the exponent of the private key multiplicative inverse modularity of the random numbers (w, x) is
considered. Encryption and decryption are completed using those exponents of the public and private keys.
In the modified RSA model, instead of two prime numbers, we have used four that are p, q, r, and s. E, f, and N are used to generate a public key where f is chosen randomly, and D, g, and N are used
to generate the private key in which g is randomly generated. In the previous RSA algorithm as N is the most complex number, and in this proposed model, if a person (hacker) knows the N value, then
they cannot trace the private key and public key because he cannot estimate the values of four prime numbers [33–36]. RSA keys are usually 1024 or 2048 bits long. However, researchers consider that
against all attacks, 1024-bit keys are no longer wholly protected. That is why the minimum key length of 2048 bits are used by some industries and the government.
Key size: 2048. Prime number p, q, r, s secret keys are given below.
P: J6NnsfaTBJoNcTE9MgyYssdAeuJI8oUjPgBWMS81w5jl/lqvqJLJcXF85zWha2RTllCsTnC9
R: NpM1vC4NZokOJSeNkvdlhyGhfkpak7v1nYWan6/P6U+iAK/fLXKBY25pk9T54pLqUF7
S: c4wWaEhdA1Qei+fizmE6+cew/45Xqc0ScBDHUAVnjKMygG7plT69aJt/1Jw6z4AhpSL89F
Table 1 presents the proposed key size as 2048 and time needed for encryption is 0.331, and the time required for decryption is 0.321 compared to other key sizes previously used. The function φ (n)
analyzes the number of elements in the specified dataset. Similarly, key sizes 2048 and 1024 with the same number of elements show less.
In Fig. 2 the comparison of encryption and decryption time of the proposed modified RSA the RSA previously used is given. The results recognized that to maintain equilibrium between security and
speed, the optimal or best key size is 2048 bits as compared to the remaining key size values, and it is visible that the time required for encrypting or decrypting is improved than the previous RSA
models with a similar number of elements. Allowing factorization of 1024 and 2048-bit keys exploit likely because of poor entropy presented by a usual mathematical arrangement of the prime factors
that also result in the revealing of keys with this usual structure.
4 Region-Based Convolutional Neural Network
For more security of data, a trained neural network is implemented to first recognize the person by capturing their image through the camera and then to identify whether it is an authorized person or
not by comparing their image with the images placed in the neural network database.
4.1 Faster R-CNN for Face Detection
For face recognition, we have implemented Faster R-CNN for the recognition and identification of users or clients to ensure that only the authorized person will be able to retrieve or send data
through a cloud environment, as shown in Fig. 3 Faster R-CNN has two main elements [37–39]. The first one is an RPN that is a fully connected network, generating regional proposals that are
additionally used as input. The second one is a Fast R-CNN detector which classifies every ROI [40]. As an input, the features of an image are taken from an RPN that produces rectangular object
proposals in the form of a set, and each proposal has the object’s score. To create region proposals directly in the network, the Faster R-CNN indicator enhances an RPN. The RPN utilizes anchor boxes
for the detection of the object. In the network, making regional proposals are better and faster in regulating the data.
In the RPN structure, from the convulsion layer, the input of RPN is a feature map. There are two data flows, the upper flow categorizes the boxes with negative or positive labels, and the lower flow
analyses the compensation of regression in the bounding box. At that time, both of the flows combined with a fully connected layer to produce and filter appropriate proposals. To produce region
proposals, we mapped a tiny sliding window that is the output of the recent pooled convolution layer on the map of convolution features. For a Faster R-CNN detector, a set of regional proposals is
provided as input. Every proposal is distributed and provided a feature map of fixed dimension among an ROI pooling layer and, mapped these features into a vector feature by Fully Connected Layers
(FCLs). In the classification process, these vector features are the inputs of the box classification layer (CLs) and box regression layer (Reg). In the Faster R-CNN model, the softmax classifier is
The main part of Faster R-CNN that is presented is extracted. For this difficulty, deep features are used. To extract the features, Visual Geometry Group (VGG) 16 architecture is used to train a net
dataset on an image. For human feature detection, a Faster R-CNN is used, as shown in Fig. 4.
The network architecture of VGG16 layers is described below in Fig. 5. VGG16 network has three fully connected layers and 15 convolution layers. At the end of the network, one softmax layer is
present. Sixty-four (64) filters are present in the first two layers of convolutions, 128 filters have been used in the 3rd, and 4th layers of convoluted, 256 filters have been used for 5th, and 6th
layers of convolution, and 512 filters have been used in the last five convolutions layers. Three layers that are fully connected have 4096, 4096, and 2622 neurons correspondingly. Maximum pooling of
2 × 2 is practical in every block.
4.3 Region of Interest Pooling
The regions have uninformed sizes proposed by the RPN. An ROI pooling layer is used to make them constant, which receives the regions projected by the RPN, and other confrontations make them of a
similar size. ROI pooling layer sends the proposal to the following fully connected layer after pooling them into the same size if their size is different for the classification of face and
regression of position adjustment. In short, the key goal of ROI pooling is to provide fully connected layers to the fixed-length output. The ROI pooling functionality is enlightened in Fig. 6.
4.4 Classification and Bounding Box Regressor
The key determination of a bounding box Regressor is to advance once the object has been classified and managed for the bounding box through support vector machine SVM or softmax classifier. Whether
or not the object does not correctly fit inside a bounding box that is generated, through the classifier adjusts the four offset values of the bounding box at that time the bounding box regressor is
employed to adjust that object accurately within a created bounding box. For classification, commonly, Softmax and Support Vector Machine (SVM) is applied to detect and classify multiple objects
within an image using the bounding box, and Faster R-CNN represents high accuracy results. ROI pooling is used by the fully connected layers, and the permanent length of the output is formed. In
fully connected layers, there are two pipelines. The first one is to predict the class present in the ROI box of the object, and the other one is to predict the object’s region of interest.
The recognition flow is shown in Fig. 6. Though the complete recognition procedure remains unaffected, based on R-CNN the subsequent developments have been made: 1) To unite the images into the fixed
size, it is no longer required by capturing and standardizing processes before, in an attempt to resolve the trouble of image misrepresentation and the loss of information initiated by regularization
processes; 2) Utilizing spatial pyramid pooling layer substitute the pooling layer of the preceding convolution layer. Thus, spatial pyramid pooling has the following benefits: 1) Spatial pyramid
pooling progression images of arbitrary aspect ratio, by significant scalable pooling layer, and arbitrary scale, make fixed size output, and through multi-scale enhance the robustness of the
extracted features; 2) The recurrent counting of convolution layer is effectually resolved and overall efficiency is improved Since directly from the complete feature mapping the features of entire
nominee regions are extracted.
In a pre-trained network, the convolution layers in the RPN are tracked by a 3 × 3 CNN layer. This communicates in the input image to map a receptive field or large spatial window (e.g., 228 × 228
for VGG16) at a center stride vector to a low-dimensional feature (e.g., 16 for VGG16). For regression and classification branches, two 1 × 1 convolutional layers are at that time included in all
spatial windows. To distribute with altered scales and aspect ratios of objects, the RPN anchors are presented. We devise to look out of the RPN modules as they share convolutional layers for the
complete arrangement. In the convolutional maps, an anchor is on every descending location in the middle of every spatial window. Every anchor is related to an aspect ratio and with a scale. After
the default situation, we utilize three aspect ratios (1:1, 1:2, and 2:1) and three scales (256-d, 4 * 256d, and 16 * 256d pixels) and, at every location leading to k = 9 anchors. Compared to an
anchor, every proposal is parameterized. Thus, we have at maximum conceivable offers for a map size of the convolutional feature. The similar features of every sliding location are utilized to
regress k proposals, as an alternative of extracting training, a single regressor, sets of features, and using Stochastic Gradient Descent (SGD) training of the RPN must be completed in an end-to-end
approach for both regression and classification purpose.
We take an image base dataset from https://www.kaggle.com/dataturks/face-detection-in-images and generate results in MATLAB of how accurate results or matching occurs for user data when a client
wants to access data by face recognition method. The proposed model hyperparameters with values are shown in Table 2.
For the duration of the training, the total loss begins at 2.413 and then progressively declines as the number of training steps rises at 50–60 k steps, ending in a loss of 0.390–0.104. For
regression, the commonly used loss function is in Mean Squared Error (MSE). The squared changes between predicted and true values are the loss in the mean managed data. MSE is complex to outliers and
specified instances with similar input feature standards, and their mean target value is optimal prediction. Peak Signal-to-Noise Ratio (PSNR) calculates in decibels between two images. This ratio is
utilized as a measurement of quality between the compressed and the original image, which improved the feature of the compressed image that developed the PSNR. The PSNR and MSE are utilized to
compare the quality of compression of the image PSNR signify the amount of the peak error. The lower the value of MSE, the lower the error.
The batch size is one, and the learning rate is in the range of 0.002 to 0.0002, as shown in Figs. 7a–7c. Fig. 7d shows the performance evaluation of the dataset by training on the proposed neural
network. For learning the algorithm, Stochastic Gradient Descent is used. To complete the Linux Ubuntu, the training acquired 5 h and 55 min on 16.04 virtual machines.
As shown in Fig. 8, on the training dataset, the Faster R-CNN model fits 99%. We are assured from the result of training that the proposed model in the testing dataset would fit above 90%.
The significant aspects of the proposed Faster R-CNN, compared to previous techniques are shown in Table 3.
RSA Algorithm plays a vital role to secure and save data and information efficiently in a sustainable cloud environment. It’s a modern, and well-developed method to secure medical data. The RSA
Algorithm method works far better than other data security algorithms. In our technique, we use a neural network for user identification and authentication and after the user, proper recognition RSA
algorithm technique is implemented for data transmission and retrieval. RSA protects, unauthorized users, and if the attacker intentionally or willingly gets access to cloud data, it does not have
the authority to decrypt it and extract information from it. Data protection in the cloud depends on how strong and efficient its encryption and decryption technique is.
This modified version of the RSA Algorithm is far better than the previously implemented RSA Algorithm for secure data storage and transmission. Only the authorized user has the authority to retrieve
or get information from the cloud, and if someone hacked the system, then they cannot decrypt the data. Through the implementation of a modified RSA algorithm, we claim and guarantee the security of
data in the cloud environment. For more security, we also implemented a trained neural network for person recognition. The Faster R-CNN gets trained on images to identify authorized users with an
accuracy of 99.9% on training and 93% on testing. Future research focuses on the implementation of the RSA cryptography algorithm efficiently and reliably with Artificial Neural Network (ANN)
technique. Additionally, improve the accuracy of training with low-resolution images in face detection.
Funding Statement: This work is supported by the Natural Science Basic Research Plan in Shaanxi Province of China (Program No. 2019JM-348).
Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.
1. S. E. Kafhali, I. E. Meir and M. Hanini, “Security threats, defense mechanisms, challenges, and future directions in cloud computing,” Archives of Computational Methods in Engineering, vol. 29,
no. 1, pp. 223–246, 2022. [Google Scholar]
2. S. Shahrin, A. Rosli, M. H. J. A. Hadi and H. Awang, “A theoretical framework of the secure environment of virtual reality application in tertiary tvet education using blockchain technology,”
Journal of Contemporary Social Science and Education Studies, vol. 1, no. 1, pp. 39–46, 2021. [Google Scholar]
3. M. N. Birje, P. S. Challagidad, R. Goudar and M. T. Tapale, “Cloud computing review: Concepts, technology, challenges and security,” International Journal of Cloud Computing, vol. 6, no. 1, pp.
32–57, 2017. [Google Scholar]
4. P. Srivastava and R. Khan, “A review paper on cloud computing,” International Journal of Advanced Research in Computer Science and Software Engineering, vol. 8, no. 6, pp. 17–20, 2018. [Google
5. I. M. Khalil, A. Khreishah and M. Azeem, “Cloud computing security: A survey,” Computers, vol. 3, no. 1, pp. 1–35, 2014. [Google Scholar]
6. K. -K. R. Choo, J. D. Ferrer and L. Zhang, “Cloud cryptography: Theory, practice and future research directions,” Future Generation Computer System, vol. 62, no. C, pp. 51–53, 2016. [Google
7. P. Samarati, S. D. C. D. Vimercati, S. Murugesan and I. Bojanova, Cloud Security: Issues and Concerns. Chichester, England: John Wiley & Sons, pp. 1–14, 2016. [Google Scholar]
8. M. U. Sana and Z. Li, “Efficiency aware scheduling techniques in cloud computing: A descriptive literature review,” PeerJ Computer Science, vol. 7, no. 24, pp. e509, 2021. [Google Scholar]
9. H. Wang, S. Wu, M. Chen and W. Wang, “Security protection between users and the mobile media cloud,” IEEE Communications Magazine, vol. 52, no. 3, pp. 73–79, 2014. [Google Scholar]
10. K. Hashizume, D. G. Rosado, E. F. Medina and E. B. Fernandez, “An analysis of security issues for cloud computing,” Journal of Internet Services and Applications, vol. 4, no. 1, pp. 1–13, 2013. [
Google Scholar]
11. D. R. Stinson and M. B. Paterson, Cryptography: Theory and Practice, 4th ed., NewYork, USA: Chapman and Hall/CRC Press, pp. 1–14, 2018. [Google Scholar]
12. M. K. Neha, “Enhanced security using a hybrid encryption algorithm,” International Journal of Innovative Research in Computer and Communication Engineering, vol. 4, no. 7, pp. 13001–13007, 2016.
[Google Scholar]
13. N. Chintawar, S. Gajare, S. Fatak, S. Shinde and G. Virkar, “Enhancing cloud data security using elliptic curve cryptography,” International Journal of Advanced Research in Computer and
Communication Engineering, vol. 5, no. 3, pp. 1–4, 2016. [Google Scholar]
14. B. Qin, H. Wang, Q. Wu, J. Liu and J. Domingo-Ferrer, “Simultaneous authentication and secrecy in identity-based data upload to cloud,” Cluster Computing, vol. 16, no. 4, pp. 845–859, 2013. [
Google Scholar]
15. M. U. Sana, Z. Li, F. Javaid, H. B. Liaqat and M. U. Ali, “Enhanced security in cloud computing using neural network and encryption,” IEEE Access, vol. 9, pp. 145785–145799, 2021. [Google Scholar
16. H. Delfs and H. Knebl, “Symmetric-key cryptography,” in Introduction to Cryptography, 3rd ed., Berlin, Heidelberg: Springer, pp. 11–48, 2015. [Google Scholar]
17. V. Agarwal, A. K. Kaushal and L. Chouhan, “A survey on cloud computing security issues and cryptographic techniques,” in Social Networking and Computational Intelligence, vol. 100. Singapore:
Springer, pp. 119–134, 2020. [Google Scholar]
18. Y. K. Kumar and R. M. Shafi, “An efficient and secure data storage in cloud computing using modified RSA public key cryptosystem,” International Journal of Electrical and Computer Engineering,
vol. 10, no. 1, pp. 530, 2020. [Google Scholar]
19. S. Ambika, S. Rajakumar and A. Anakath, “A novel RSA algorithm for secured key transmission in a centralized cloud environment,” International Journal of Communication Systems, vol. 33, no. 5,
pp. e4280, 2020. [Google Scholar]
20. A. Gupta, S. Gupta and N. Yadav, “Enhancement of security using B-RSA algorithm,” in Inventive Communication and Computational Technologies, vol. 89. Singapore: Springer, pp. 439–450, 2020. [
Google Scholar]
21. M. Dhamodaran, E. Punarselvam, S. D. Varshan, P. D. Kumar, C. Saravanan et al., “Security and privacy of sensitive data in cloud computing using RSA,” International Journal of Scientific Research
in Science and Technology, vol. 8, no. 2, pp. 657–661, 2021. [Google Scholar]
22. Y. Wang, Q. Ma, L. Li, T. Guan, Y. Geng et al., “An encryption method of power cloud data based on n-RSA,” in Advances in Intelligent Information Hiding and Multimedia Signal Processing, vol. 211
. Singapore: Springer, pp. 416–423, 2021. [Google Scholar]
23. S. Chandel, G. Yang and S. Chakravarty, “RSA-CP-IDABE: A secure framework for multi-user and multi-owner cloud environment,” Information-An International Interdisciplinary Journal, vol. 11, no. 8
, pp. 382, 2020. [Google Scholar]
24. B. Seth, S. Dalal, V. Jaglan, D. N. Le, S. Mohan et al., “Integrating encryption techniques for secure data storage in the cloud,” Transactions on Emerging Telecommunications Technologies, vol.
30, no. 4, pp. e4108, 2022. [Google Scholar]
25. H. Jiang and E. Learned-Miller, “Face detection with the faster R-CNN,” in 12th IEEE Int. Conf. on Automatic Face & Gesture Recognition, Washington, DC, USA, pp. 650–657, 2017. [Google Scholar]
26. M. Besnassi, N. Neggaz and A. Benyettou, “Face detection based on evolutionary Haar filter,” Pattern Analysis and Applications, vol. 23, no. 1, pp. 309–330, 2020. [Google Scholar]
27. X. Sun, P. Wu and S. C. Hi, “Face detection using deep learning: An improved faster RCNN approach,” Neurocomputing, vol. 299, no. 2, pp. 42–50, 2018. [Google Scholar]
28. W. Wu, Y. Yin, X. Wang and D. Xu, “Face detection with different scales based on faster R-CNN,” IEEE Transactions on Cybernetics, vol. 49, no. 11, pp. 4017–4028, 2019. [Google Scholar] [PubMed]
29. L. Hai and H. Guo, “Face detection with improved face R-CNN training method,” in 3rd Int. Conf. on Control and Computer Vision, Macau China, pp. 22–25, 2020. [Google Scholar]
30. H. Yan, X. Wang, Y. Liu, Y. Zhang and H. Li, “A new face detection method based on faster R-CNN,” Journal of Physics: Conference Series, vol. 1754, no. 1, pp. 012209, 2021. [Google Scholar]
31. A. V. N. Krishna, “A randomized cloud library security environment,” in Cloud Security: Concepts, Methodologies, Tools, and Applications, Hershey, PA, USA: IGI Global, pp. 1087–1107, 2019. [
Google Scholar]
32. R. Biswas, S. Bandyopadhyay and A. Banerjee, “A fast implementation of the RSA algorithm using the GNU MP library,” in National Workshop on Cryptography, IIIT-Calcutta, India, pp. 1–15, 2003. [
Google Scholar]
33. S. Sharma, P. Sharma and R. S. Dhakar, “RSA algorithm using modified subset sum cryptosystem,” in 2nd Int. Conf. on Computer and Communication Technology, Allahabad (UPIndia, pp. 457–461, 2011. [
Google Scholar]
34. A. Mouse, “Sensitivity of changing the RSA parameters on the complexity and performance of the algorithm,” Journal of Applied Sciences, vol. 5, no. 1, pp. 60–63, 2005. [Google Scholar]
35. M. A. Islam, M. A. Islam, N. Islam and B. Shabnam, “A modified and secured RSA public key cryptosystem based on n prime numbers,” Journal of Computer and Communications, vol. 6, no. 3, pp. 78–90,
2018. [Google Scholar]
36. M. Thangavel, P. Varalakshmi, M. Murrali and K. Nithya, “An enhanced and secured RSA key generation scheme (ESRKGS),” Journal of Information Security and Applications, vol. 20, no. 1, pp. 3–10,
2015. [Google Scholar]
37. S. Wan and S. Goudas, “Faster R-CNN for multi-class fruit detection using a robotic vision system,” Computer Networks, vol. 168, no. 12, pp. 107036, 2020. [Google Scholar]
38. W. Yang, Z. Li, C. Wang and J. Li, “A multi-task Faster R-CNN method for 3D vehicle detection based on a single image,” Applied Soft Computing, vol. 95, no. 2, pp. 106533, 2020. [Google Scholar]
39. M. Arman, M. Hasan, F. Sadia, A. K. Shakir, K. Sarker et al., “Detection and classification of road damage using R-CNN and faster R-CNN: A deep learning approach,” in Int. Conf. on Cyber Security
and Computer Science, Dhaka, Bangladesh, pp. 730–741, 2020. [Google Scholar]
40. Z. Huang, S. Watanabe, Y. Fujita, P. García, Y. Shao et al., “Speaker diarization with region proposal network,” in Int. Conf. on Acoustics, Speech and Signal Processing (ICASSP), Barcelona,
Spain, pp. 6514–6518, 2020. [Google Scholar]
41. R. Chauhan, K. K. Ghanshala and R. C. Joshi, “Convolutional neural network (CNN) for image detection and recognition,” in First Int. Conf. on Secure Cyber Computing and Communication (ICSCCC),
Jalandhar, India, pp. 278–282, 2018. [Google Scholar]
42. A. Saranya, K. Kottursamy, A. A. AlZubi and A. K. Bashir, “Analyzing fibrous tissue pattern in fibrous dysplasia bone images using deep R-CNN networks for segmentation,” Soft Computing, vol. 26,
no. 16, pp. 7519–7533, 2022. [Google Scholar] [PubMed]
43. S. C. Hsu, C. L. Huang and C. H. Chuang, “Vehicle detection using simplified, fast R-CNN,” in Int. Workshop on Advanced Image Technology (IWAIT), Chiang Mai, Thailand, pp. 1–3, 2018. [Google
Cite This Article
APA Style
Sana, M.U., Li, Z., Kiren, T., Liaqat, H.B., Naseem, S. et al. (2023). A secure method for data storage and transmission in sustainable cloud computing. Computers, Materials & Continua, 75(2),
2741-2757. https://doi.org/10.32604/cmc.2023.036093
Vancouver Style
Sana MU, Li Z, Kiren T, Liaqat HB, Naseem S, Saeed A. A secure method for data storage and transmission in sustainable cloud computing. Comput Mater Contin. 2023;75(2):2741-2757 https://doi.org/
IEEE Style
M.U. Sana, Z. Li, T. Kiren, H.B. Liaqat, S. Naseem, and A. Saeed, “A Secure Method for Data Storage and Transmission in Sustainable Cloud Computing,” Comput. Mater. Contin., vol. 75, no. 2, pp.
2741-2757, 2023. https://doi.org/10.32604/cmc.2023.036093
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Satisficing-Planning MetricTime Track
In this planning domain there are 40 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized. For the problems in this domain, the main challenge is
computing a valid plan with *good quality*, but for a fully-automated planner in some cases even finding any valid plan is hard. Computing the optimal solution in this domain is known NP-hard
In this planning domain there are 20 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized. For the problems in this domain, the main challenge is
computing a valid plan with *good quality*, but for a fully-automated planner in some cases even finding any valid plan is hard.
In this planning domain there are 30 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized. For the problems in this domain, the main challenge is
computing a valid plan with *good quality*, but for a fully-automated planner in many cases even finding any valid plan is hard.
In this planning domain there are 50 problems (named by numbers on the x-axis). The problems not indicated are solved by no planner. In the quality plot, the plan metric has to be minimized.
In this planning domain there are 40 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized. The problems not indicated are solved by no planner. For the
problems in this domain, the main challenge is computing a valid plan with *good quality*, but for a fully-automated planner in some cases even finding any valid plan is hard.
In this planning domain there are 30 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized.
In this planning domain there are 30 problems (named by numbers on the x-axis). In the quality plot, the plan metric has to be minimized. For the problems in this domain, the main challenge is
computing a valid plan with *good quality*, but for a fully-automated planner in some cases even finding any valid plan is hard. | {"url":"https://ipc06.icaps-conference.org/deterministic/plot-results/satisficing/MetricTime/index.html","timestamp":"2024-11-12T21:51:08Z","content_type":"text/html","content_length":"5045","record_id":"<urn:uuid:eaa7127f-6223-4d17-b226-f992f84729bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00210.warc.gz"} |