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https://math.stackexchange.com/questions/775552/what-does-the-derivative-of-area-with-respect-to-length-signify | 1,726,108,798,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00756.warc.gz | 350,267,218 | 42,489 | # What does the derivative of area with respect to length signify?
Suppose that we have a square sheet of edge length $L$. Its area $A=L^2$.
Differentiating $A$ w.r.t. L, we get
$$\dfrac{dA}{dL}=2L$$
I do understand what it means to differentiate, graphically, it gives you the slope of the tangent at a point on the graph. But now, when I think of what differentiating means in the context of Area and length, it doesn't make any sense to me at all. What does $2L$ signify?
• Rate of change of area with respect to length. Commented Apr 30, 2014 at 11:53
Try to draw a square $ABCD$ with side equal to $L$. Now draw a slightly bigger square $AB'C'D'$ with side length $L+\Delta L$ (such that $DD'=BB'=\Delta L$). Now look at the $\Gamma$-like shape cut from $AB'C'D'$ by $ABCD$, you can split it into three parts: two thin rectangles $L\times \Delta L$ and one small square $\Delta L\times \Delta L$.
Now the derivative is in quite simpified terms "the difference of value of the function over the change of argument", so basically when you increase the side length by $\Delta L$, then the surface increases by $2L\Delta L$ and a negligeble term $(\Delta L)^2$.
One can also say that $2L$ signifies the permeter of the part of the square that got inflated.
• Did you mean Perimeter? Commented Apr 30, 2014 at 11:55
• I understood most of it...But to understand it completely, I think I need to know what it means for $2x$ to be the derivative of $x^2$. Commented Apr 30, 2014 at 12:00
• Is $2L$ the change in Area for unit change in length? Commented Apr 30, 2014 at 12:01
• No, but for an infinitely small $\Delta L$ the change in area is $2L\Delta L$. The $2L$ signifies how much the area would change relative to an infinitely small change in length. Imagine the graph for $x^2$ - for a certain $x$, if you'd go an infinitely small bit to the left or right, the slope would be $2x$, that's what it means. If the length becomes a very little bit bigger or smaller, it also changes the area by $2L *$ a very little bit. Commented Apr 30, 2014 at 13:59
• So you could write $\Delta A = 2 L \Delta L + (\Delta L)^2$. Dividing by $\Delta L$ gives $\frac{\Delta A}{\Delta L} = 2L + \Delta L$. If you now make $\Delta L$ very, very small (mathematically, take the limit $\Delta L \to 0$), you will get $2L$ on the right hand side. The left hand side is the definition of $\frac{dA}{dL}$. Commented Apr 30, 2014 at 14:07
Consider this picture:
Here, the green square is the square of area $A=L^2$ and red line is its increase.
When you increase the length $L$ by $dL$, the area $A$ gets increased by $2LdL$. So, to answer your question, significance of $2L$ is that it is the length of the red line on the picture ($dL$ is its width).
• This may be confusing since the area of the red part is actually 2LdL + dL^2... Commented Apr 30, 2014 at 16:33
• @Anh Indeed, but $(dL)^2$ part becomes infinitely smaller than $2LdL$ in the limit when $dL$ becomes infinitesimal, so it can be safely ignored. It matters for the second derivative though. Commented Apr 30, 2014 at 17:01
• @Danijel, you should add that to your answer, to make it more complete. Still, +1 for the picture Commented May 1, 2014 at 12:38
Thinking of the derivative graphically as the slope of the tangent is just one way to understand the meaning of the derivative. It's the most common, because it's how the derivative is motivated in most introductory calculus courses. But the meaning and value of the idea of a derivative is much deeper. The derivative measures the rate at which something changes. That's worth thinking about before you start with graphs and formulas. Here are some examples.
Suppose you're driving. Then the distance you've traveled changes as time goes by. If you're driving along at a constant 30 miles per hour then the distance increases by 30 miles for each hour of travel. The derivative of the distance is the rate: 30 miles per hour.
That's an easy example because the rate of travel is constant. Calculus was invented to handle situations where the rate is itself changing. For example, if you start from a red light and accelerate up to the legal speed limit of 30 miles per hour then your speed is changing. The derivative of the speed is the rate at which you're speeding up - the acceleration. You might measure that in (miles per hour) per second.
In economics, the number of customers for your product depends on the price you charge. When you raise the price, fewer people will buy from you. The derivative of the number of customers is the rate at which you lose them, measured in (customers lost) per (dollar increase in price). In this case the derivative is negative.
Populations change over time. For microorganisms you might choose to measure time in hours. Then the derivative of the population is the number of new organisms per hour. Then things get interesting, because the number of new organisms per hour depends on the population - the more organisms you have, the more of them there are to reproduce. So the derivative of the population, measured in new organisms per hour, is the product of the number of organisms and the birth rate. That means the derivative of the population (as time goes on) is proportional to the population. That leads to exponential growth.
You can describe the derivative of a graph of the function y = f(x) the same way. Here the height y changes as the value of x changes. The steeper the graph (at any particular point) the larger the change in y for any particular small change in x. The rate at which y changes is the derivative. You have to think only about small changes in x since the graph is a curve, whose steepness varies from place to place. As long as the change in x is small, the curve nearly matches the tangent, whose slope is just the rate of change you care about. (It's taken mathematicians centuries of work to make precise sense of the idea expressed roughly as "if you change x by just an infinitesimal amount then the curve and the tangent are the same".)
Now think about the question you asked. The area of a square depends on the length of its side. The derivative measures the rate at which the area changes when the side changes, measured in units like (square centimeters of area) per (centimeter of side). @TZakrevskiy 's answer above explains why that's just twice the side length. Here's an analogous question: explain why when you grow a circle of radius r the area changes at the rate 2 pi r.
I wish there were more time and more incentive to spend time in calculus classes on these ideas, rather than rushing to the rules and formulas for derivatives (and integrals).
• So derivative gives us the change in something "per what"? Commented Apr 30, 2014 at 15:33
• @shauryagupta Yes. The "what" in "per what" depends on how you frame your question - it's the independent variable. You could think about how the amount of gas you use depends on how far you travel, or about how it depends on the speed at which you travel (auto engine efficiency falls off at really high speeds). You'd have different derivatives in these examples. Commented Apr 30, 2014 at 15:59
• So if I differentiated amount of gas w.r.t. distance, I would get the change per how much of distance? Commented Apr 30, 2014 at 16:05
• @shauryagupta Yes. You would get the rate of change of gas used, in units (gallons of gas) per (miles traveled). In the other problem you would get the rate in (gallons of gas used) per (speed at which you travel). At high speeds an increase in the speed would cause an increase in the gas used (positive derivative). At low speeds an increase in the speed would cause a decrease in gas used (negative derivative). The derivative would be zero at the optimal speed. Commented Apr 30, 2014 at 16:16
• @shauryagupta Indeed it is. That's what makes calculus hard, and why it took so long for mathematicians to invent it. Commented Apr 30, 2014 at 16:19 | 1,998 | 7,948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.932565 |
https://mathworkorange.com/lagrange-polynomials/trinomials/solving-a-formula-for-a.html | 1,712,968,828,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00658.warc.gz | 357,589,590 | 12,874 | # Your Math Help is on the Way!
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greim decxs
Registered: 02.04.2006
From: Hollywood, Florida. "Sunshine State"
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Posted: Wednesday 08th of Aug 16:53 Thanks a million for the elaborate information. We will surely try this out. Hope we get our assignments finished with the help of Algebrator. If we have any technical clarifications with respect to its usage , we would definitely come back to you again. | 931 | 3,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.903616 |
https://www.jiskha.com/questions/1070829/a-ramp-leads-up-to-a-building-the-top-of-the-ramp-is-4-feet-above-the-ground-and-the | 1,585,830,101,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00440.warc.gz | 1,018,740,149 | 5,368 | # Algebra
A ramp leads up to a building. The top of the ramp is 4 feet above the ground, and the bottom is 14 feet from the building. what is the length of the ramp?
I think the answer is 18 ft.
choices are:
A. 18.6ft
B. 18ft
C. 14.6ft
D. 10ft
1. 👍 0
2. 👎 0
3. 👁 244
1. Nope.
This makes a right triangle. Use the Pythagorean Theorem to find the hypotenuse.
1. 👍 0
2. 👎 0
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# You are given the sample mean and the population standard deviation. Use this information to construct...
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of
78
dates, the mean record high daily temperature in a certain city has a mean of
85.43 degrees°F.
Assume the population standard deviation is
13.72 degrees°F.
The 90% confidence interval is
(nothing,nothing).
(Round to two decimal places as needed.)
The 95% confidence interval is
(nothing,nothing).
(Round to two decimal places as needed.)
Which interval is wider? Choose the correct answer below.
The 90% confidence interval
The 95% confidence interval
Interpret the results.
A.You can be certain that the mean record high temperature was within the 90% confidence interval for approximately
70
of the
78
days, and was within the 95% confidence interval for approximately
74
of the
78
days.
B.
You can be certain that the population mean record high temperature is either between the lower bounds of the 90% and 95% confidence intervals or the upper bounds of the 90% and 95% confidence intervals.
C.
You can be 90% confident that the population mean record high temperature is outside the bounds of the 90% confidence interval, and 95% confident for the95% interval.
D.
You can be 90% confident that the population mean record high temperature is between the bounds of the 90% confidence interval, and 95% confident for the95% interval.
Which interval is wider?
You can be 90% confident that the population mean record high temperature is between the bounds of the 90% confidence interval, and 95% confident for the 95% interval.
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April 11, 2016, 11:38 How to calcuate the thust on a wind turbine? #1 Senior Member Khamlaj Join Date: Nov 2010 Location: United States Posts: 198 Rep Power: 8 Hey All, I have had an issue figuring out the post-processing procedure required to find the thrust coefficient on a turbine blade. I thought that some of you might have dealt with the same post-processing before. The parameters of the turbine are as follows; 1- The turbine diameter is 4.5 m. 2- Hub radius 0.1m 3- Thickness of the actuator disk 0.14m When I used plot over line across the center line of the wind turbine, I obtained the following figure; 10msVP.jpg My misinterpretation led me to use the following relation to obtain the thrust coefficient; C_{t} = (p_{1}-p_{2})/0.5*\rho*V_{inf}^2 But, I finally figure out that I need to integrate the static pressure along the actuator disk. I used plot over line along the blade at a point above the hub till the tip of the blade. The coordinates are; {0.0624, 0, 0} and {0.0625, 3, 0} The file of this post-processing data.xlsx I have got the magnitude volume force as well as the position of each one. However, I don't exactly know what I do next. Which values should we take for the post-processing since I ran the case till 4000 iteration. Should I take the last iteration for the calculations of the thrust coefficient? What values do I need to use in order to find the the thrust? Has anybody experienced the same issue? Any guidance to achieve thrust... Best,
April 21, 2016, 16:49 Answer! #2 Senior Member Khamlaj Join Date: Nov 2010 Location: United States Posts: 198 Rep Power: 8 If anybody is working on the same problem and having issues, contact me. The problem has been solved. Thank you,
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## Mathterpieces: The Art of Problem-Solving
##### Australian Curriculum: Description
Foundation Year – Subitise small collections of objects (ACMNA003);
Year 1 – Represent and solve simple addition and subtraction problems using a range of strategies including counting on, partitioning, and rearranging parts (ACMNA015);
Year 2 – Describe, continue, and create number patterns resulting from performing addition or subtraction (ACMNA035).
##### Teaching ideas
Students investigate other situations involving subitising and counting on with and without distractors in routine and non-routine situations. Students could write the number sentences represented by art in this book (representational to abstract).
## 17 Kings and 42 Elephants
##### Australian Curriculum: Description
Subitise small collections of objects (ACMNA003)
##### Teaching ideas
Counting on with and without distractors in routine and non-routine environments
## Matherpieces: The Art of Problem-Solving
##### Australian Curriculum: Description
Subitise small collections of objects (ACMNA003); Represent and solve simple addition and subtraction problems using a range of strategies including counting on, partitioning and rearranging parts (ACMNA015); Describe, continue, and create number patterns resulting from performing addition or subtraction
##### Teaching ideas
Students investigate other situations involing subitising and counting on with and without distractors in routine and non-routine situations. Students could write the number sentences represented by art in this book (representational to abstract).
## When the King rides by
##### Australian Curriculum: Description
F. Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); Year 2 – Describe patterns with numbers and identify missing elements (ACMNA035)
##### Teaching ideas
Explore other mediums with patterns e.g. wrapping paper and fabric. Show students how patterns can help to predict --> expect --> plan.
## How Many Snails?
##### Australian Curriculum: Description
F. Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); Year 2 – Describe patterns with numbers and identify missing elements (ACMNA035)
##### Teaching ideas
Explore other mediums with patterns e.g. wrapping paper and fabric. Show students how patterns can help to predict --> expect --> plan.
## Each Orange Had 8 Slices
##### Australian Curriculum: Description
F.Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); F.Y – Represent practical situations to model addition and sharing (ACMNA004); Year 1 – Represent and solve simple addition and subtraction problems using a range of strategies including counting on, partitioning and rearranging parts (ACMNA015); Year 3 – Recall multiplication facts of two, three, five and ten and related division facts (ACMNA056)
##### Teaching ideas
Build on the concept of counting on, partitioning and part-part-whole with other object sequences.
## 12 Ways to Get to 11
##### Australian Curriculum: Description
F.Y – Establish understanding of the language and processes of counting by naming numbers in sequences, initially to and from 20, moving from any starting point (ACMNA001); F.Y – Represent practical situations to model addition and sharing (ACMNA004); Year 1 – Represent and solve simple addition and subtraction problems using a range of strategies including counting on, partitioning and rearranging parts (ACMNA015)
##### Teaching ideas
Build on the concept of counting on, partitioning and part-part-whole with other object sequences. | 1,090 | 5,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-35 | longest | en | 0.692163 |
https://plainmath.net/8306/quadratic-quadratic-intercept-quadratic-compression-approaches-approaches | 1,656,252,021,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103269583.13/warc/CC-MAIN-20220626131545-20220626161545-00556.warc.gz | 523,965,317 | 12,496 | # A quadratic has an axis of symmetry of x=m, and the quadratic hax an x-intercept at (m,0). The quadratic has a vertical compression as a approaches infinty y approaches negative infiniy. What is a possible equation of the quadratic?
A quadratic has an axis of symmetry of x=m, and the quadratic hax an x-intercept at (m,0). The quadratic has a vertical compression as a approaches infinty y approaches negative infiniy. What is a possible equation of the quadratic?
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The quadratic has an x-intercept at (w,0).
Its axis of symmetry is x=w,so it can be taken as $y=k{\left(x-w\right)}^{2}$ where k is constant.
From given information the quadratic has a vertical stretch and as x approaches $\mathrm{\infty }$ and y approaches $-\mathrm{\infty }$ so must have,k<0
So let us take k=−2.
Hence the possible equation of the quadratic becomes,$y=-2{\left(x-w\right)}^{2}$ | 282 | 1,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.923178 |
https://www.numbersaplenty.com/133330678 | 1,685,768,520,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00772.warc.gz | 1,013,918,818 | 3,237 | Search a number
133330678 = 21323593779
BaseRepresentation
bin1111111001001…
…11011011110110
3100021212220022021
413330213123312
5233113040203
621121423354
73206202115
oct774473366
9307786267
10133330678
1169297348
123879ab5a
13218137c0
14139c9c7c
15ba8a5bd
hex7f276f6
133330678 has 32 divisors (see below), whose sum is σ = 228614400. Its totient is φ = 57848736.
The previous prime is 133330663. The next prime is 133330679. The reversal of 133330678 is 876033331.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (133330679) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 33393 + ... + 37171.
It is an arithmetic number, because the mean of its divisors is an integer number (7144200).
Almost surely, 2133330678 is an apocalyptic number.
133330678 is a deficient number, since it is larger than the sum of its proper divisors (95283722).
133330678 is a wasteful number, since it uses less digits than its factorization.
133330678 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3876.
The product of its (nonzero) digits is 27216, while the sum is 34.
The square root of 133330678 is about 11546.8904039139. The cubic root of 133330678 is about 510.8695635616.
The spelling of 133330678 in words is "one hundred thirty-three million, three hundred thirty thousand, six hundred seventy-eight". | 467 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-23 | latest | en | 0.82714 |
https://nbviewer.jupyter.org/github/jklymak/Phy411/blob/master/lectures/Lecture-08-Sampling-Theorem.ipynb | 1,632,319,246,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00483.warc.gz | 463,300,264 | 315,810 | In [1]:
from IPython.core.display import HTML
def css_styling():
css_styling()
Out[1]:
Jody Klymak
# Week 8: Sampling Theorem and interpolation¶
We have not talked yet about the process of digitizing our data, though we have assumed it for all the computer examples so far. Digitization is usually carried out using an analog-to-digital converter on a preconditioned signal. You may have encountered ADCs in your electronics class, and we won't discuss how they work here. Preconditioning of the signal, however, is important to achieve a good representation of the signal because of aliasing of high frequencies can contaminate the signals we are interested in. The typical preconditioning step is to apply a low-pass filter to the data and then sampling fast enough to capture the filtered signal. Modern applications actually have very fast ADCs (GHz), that are then digitally filtered to return low-passed signals which are subsequently decimated for storage.
We also need to consider what happens when there are gaps in our time series, which can happen due to instrument failure, changing instruments etc. The approaches for this are probably review for most of you, and include linear interpoloation, nearest-neighbour interpolation, and cubic-spline interpolation.
## Sampling Theorem¶
The sampling theorem states that a continuous signal with limited frequency bandwidth can be exactly represented as discrete time series with dense enough time sampling. This means that the discretization is lossless, not just at the times of the discretization, but at all the times between.
Formally, let $x(t)$ be our continuous time series. If the continuous Fourier transform of $x(t)$ is $X(f)$, then $x(t)$ is bandlimited if $\left|X(f)\right|=0$ for $\left|f\right|>B$, for some frequency $B$, which we call the band limit.
Lets consider a signal that is white noise and has a peak at 31.5 Hz.
In [4]:
import numpy as np
import matplotlib.pyplot as plt
import scipy.signal as signal
import matplotlib.mlab as mlab
%matplotlib nbagg
N=1e6
dt = 1e-2
# make random noise.
x0 = np.random.randn(N)
# add a 31.5 Hz signal to x
t = np.arange(0,N*dt,dt)
x0 = x0+1.*np.cos(t*2*np.pi*31.5)
#Bandwidth = 2 Hz
B=2.
# filter to make Bandlimited signal
fp=B*0.7
fs = B*0.9
fNyq=1/2./dt
n,fn=signal.ellipord(fp/fNyq, fs/fNyq,.1, 60.)
b,a=signal.ellip(n,.1,60.,fn)
xBL = signal.filtfilt(b,a,x0)
fig,ax=plt.subplots(2,1,figsize=(5,5))
ax[0].plot(t,x0,color='0.5',label='full-spectrum signal')
ax[0].plot(t,xBL,color='k',label='band-limited signal')
ax[0].set_xlim([0,30]);ax[0].set_xlabel('time [s]');ax[0].set_ylabel('x [V]');ax[0].legend(fontsize='small',loc=0)
nfft=10240
pxx,f=mlab.psd(x0,NFFT=nfft,Fs=1./dt,noverlap=nfft/2.,window=mlab.window_hanning)
pxBL,f=mlab.psd(xBL,NFFT=nfft,Fs=1./dt,noverlap=nfft/2.,window=mlab.window_hanning)
ax[1].loglog(f,pxx,color='0.5')
ax[1].loglog(f,pxBL,color='k')
plt.axvline(x=B,linestyle='--',color='k');ax[1].text(B,1,' B=2 Hz');ax[1].set_ylim([1e-8,1e2]);ax[1].set_xlabel('f [Hz]');ax[1].set_ylabel('$G_{xx} [V^2 Hz^{-1}]$')
plt.tight_layout()
/Users/jklymak/anaconda/lib/python2.7/site-packages/ipykernel/__main__.py:10: DeprecationWarning: using a non-integer number instead of an integer will result in an error in the future
Here, for display purposes we have approximated a continuous signal with a signal that has a sampling frequency of 100 Hz (I can't plot a continuous signal!). Two signals are shown, one that is not bandlimited (to 50 Hz, anyways), and a second that is bandlimited at $B=2\ Hz$.
The point of the sampling theorem is that you would need to sample at about 4 Hz to capture all of the bandlimited signal, whereas you could not capture all of the full-sepctrum signal. Intuitively, that hopefully makes sense:
In [5]:
fig,ax=plt.subplots(1,1,figsize=(5,3))
ax.plot(t,x0,color='0.5',label='full-spectrum signal');ax.plot(t,xBL,color='k',label='band-limited signal')
ax.plot(t[::25],x0[::25],'d',color='0.5')
ax.plot(t[::25],xBL[::25],'d',color='k')
ax.set_xlim([10,12]);ax.set_xlabel('time [s]');ax.set_ylabel('x [V]');ax.legend(fontsize='small',loc=0)
Out[5]:
<matplotlib.legend.Legend at 0x108d5af90>
The grey diamonds do not begin to capture the variability of the grey line, whereas the black diamonds seem to do a reasonable job of the variability of the black line.
#### Sampling Theorem¶
States that if $x(t)$ is a bandlimited signal such that $\left|X(f)\right|=0$ for $\left| f\right|>B$, then $x(t)$ can be represented in full by a discrete time series $x_n$ sampled every $\Delta t$ if $\frac{1}{\Delta t} \geq 2B$, and in particular we can reconstruct $x(t)$ as:
$$x(t) = \sum_{n=-\infty}^{\infty} x(n\Delta t) \frac{\sin \left(\pi \left( \frac{t}{\Delta t}- n\right) \right)}{\pi\left(\frac{t}{\Delta t}-n\right)}$$
#### Proof¶
The proof is relatively straight forward. We note that because $X(f)$ is zero for $\left| f\right| >B$ then it can be expanded as a discrete Fourier series:
$$X(f) = \sum_{n=-\infty}^{\infty} c_n \mathrm{e}^{-j\frac{ 2\pi f n}{B}}$$
where $c_n$ is given by
\begin{align} c_n =& \frac{1}{2B} \int_{-B}^B X(f)\ \mathrm{e}^{+j2\pi f\frac{ n }{2B}}\ \mathrm{d}f\\ =& \frac{1}{2B}x\left(n/2B\right) \end{align}
So, this gives us the ingredients for $x(t)$:
\begin{align} x(t) &= \int_{-\infty}^{\infty} X(f) \mathrm{e}^{j2\pi f t}\ \mathrm{d}f\\ & = \frac{1}{2B} \int_{-B}^{B} \mathrm{e}^{j2\pi f t}\ \sum_{n=-\infty}^{\infty} x(n/2B) \mathrm{e}^{-j\frac{2\pi f n}{B}} \mathrm{d}f\\ & = \sum_{n=-\infty}^{\infty} x_n \int_{-B}^B \mathrm{e}^{j2\pi f \left(t-n/2B\right)}\ \mathrm{d}f\\ & = \sum_{n=-\infty}^{\infty} x_n \frac{\sin \left(\pi \left(2Bt -n \right) \right)}{\pi \left(2Bt-n\right)} \end{align}
where whe have defined $x_n = x(n/2B) = x(n\Delta t)$, and $\Delta t = 1/2B$ is the sample spacing.
Note that for $t=n/2B$, $x(t)=x_n$. For times in between the sample points, $t\neq n/2B$, we need an infinite sum of all the other points in the discrete time series to recover the true value at $x(t)$.
Of course a limitation of real data is that we do not have infinite data, so any estimate at $t\neq n/2B$ is going to be slightly imprecise. However, the sinc function rolls off quite quickly, so in practice this is not too much of a problem. Lets consider this using the example above. Here we subsample the bandlimitted time series by a factor of 25 and compare the reconstruction to the original time series.
In [8]:
# decimated time series:
tn = t[::25]
xn = xBL[::25]
Ns = [10,100,1000]
fig,ax=plt.subplots(1,1)
for N in Ns:
n = np.arange(N)
# reconstruct the full time series just from the subsampled data xn:
xreco=1.*xBL[:N*25] # trim the last N*25 data points...
for i in range(N*25):
xreco[i] = np.sum(xn[:N]*np.sinc(2*B*t[i]-n))
ax.plot(t[:N*25],(xreco-xBL[:N*25])/np.std(xBL),label='%d'%N)
plt.xlim([0,50])
plt.xlabel('t [s]')
plt.ylabel('$(x_{recon}-x(t))/std(x)$')
plt.ylim([-1.,1.])
plt.legend(fontsize='small')
Out[8]:
<matplotlib.legend.Legend at 0x11a7e0450>
So here we see the effect of having only a finite number of samples to reconstruct your data from; there is a bad edge effect (because you do not have the data from the negaitve side of the infinite sum), and then improvement in the estimate towards the center of the sample. Obviously the more data you have and the more away from the edges of the time series you are the better the approximation.
## Aliasing¶
An important flipside to the Sampling Theorem is that if you do have significant energy at frequencies greater than $f_S$, your sampling frequency, then your subsampled signal will contain that variance, but aliased to one of the resolved frequencies. This can be a huge problem:
In [10]:
xl=x0 # +np.cos(2*pi*7.001*t)
xna = xl[::25]
tna = t[::25]
pxna,fn2=mlab.psd(xna,Fs=1/dt/25,NFFT=nfft/25)
pxnB,fn=mlab.psd(xl,Fs=1/dt,NFFT=nfft)
fig,ax=plt.subplots(1,1)
ax.loglog(fn2,pxna,color='0.5',label='subsampled')
ax.loglog(fn,pxnB,color='k',label='raw')
ax.legend(fontsize='small',loc=0)
ax.set_xlabel('f [Hz]')
ax.set_ylabel('$G_{xx} [V^2 Hz^{-1}]$')
print 'Variance of subsampled: %1.3f'%np.var(xna)
print 'Variance of raw: %1.3f'% np.var(xl)
Variance of subsampled: 1.485
Variance of raw: 1.498
So we can see that the peak at 31.5 Hz is aliased back to about 0.1 Hz! Perhaps even worse, the high freqeuncy variance ahove 2 Hz is all wrapped back to the low frequencies. This shouldn't be surprising in light of Parseval's theorem - the integral under the two curves is the variance of the signal, and subsampling does not reduce the variance.
So, if we have variance at a frequency $f>f_{Nyq}$ where does it get aliased to? We can consider a frequency for the signal $f_0$ and determine its modulus with the Nyquist frequency:
$$f_0=mf_{Nyq} + \delta f$$
then
\begin{align} \cos\left( 2\pi f_0 t_n\right) &= \cos\left( 2\pi f_0 n\Delta t\right)\\ &= \cos\left( 2\pi m f_{Nyq} n\Delta t + 2\pi \delta f \Delta t \right)\\ &= \cos\left( \pi m n + 2\pi \delta f \Delta t \right)\\ &= \begin{cases} \cos\left(2\pi \delta f n \Delta t\right) & m\ \text{even}\\ \cos\left(\frac{2\pi m n \Delta t}{2\Delta t}+ \delta f n \Delta t\right) & m\ \text{odd}\\ \end{cases}\\ &= \begin{cases} \cos\left(2\pi \delta f \Delta t\right) & m\ \text{even}\\ \cos\left(\frac{2\pi n \Delta t}{2\Delta t}+ \delta f n \Delta t\right) & m\ \text{odd}\\ \end{cases}\\ &= \begin{cases} \cos\left(2\pi t_n \delta f \right) & m\ \text{even}\\ \cos\left(2\pi t_n \left(f_N-\delta f \right)\right) & m\ \text{odd}\\ \end{cases} \end{align}
so the aliased frequency, $f_A$, is given by
$$f_A =\begin{cases} \delta f & m\ \text{even}\\ f_N-\delta f & m\ \text{odd}\\ \end{cases}$$
So, for the case above, the spike at $f_0=31.5 Hz$ and a Nyquist frequency of $f_{Nyq}=2 Hz$, $m=15$, and $\delta f=1.5 Hz$, so $f_A=2-1.5\ Hz=0.5\ Hz$.
Note that we can get a bit confused if $f_0=mf_{Nyq}$ because $\delta f=0 Hz$ so $f_A=0$ or $f_A = f_{Nyq}$, therefore the mean or the highest frequency is affected, not the interior of the spectrum.
### Preconditioning¶
The solution to aliasing in practical applications is to make sure you apply an anti-aliasing filter to the analog signal before digitizing it. This is just a low-pass filter with its cut-off frequency significantly lower than the Nyquist frequency. This is typical in ADC's.
## Dealing with data gaps: Interpolation¶
We have seen in the weather data time series that there are freqeunt data gaps. We have been ignoring them, but that actually introduces phase distortion to your time series as you are removing data. Typically it is better to interpolate over the bad data.
It also happens sometimes that data is not collected at regular time (or space) intervals. For instance, some manual intervention is needed, and a technician runs the data when they can.
There are different methods to interpolate, which you are probably familair with:
### Nearest Neighbour Interpolation¶
This relatively trivial: If we have data $x_i$ collected at times $t_i$, where $t_i$ are not necessarily evenly spaced, then $x(t)=x_j$, for the $j$ where $|t-t_j|$ is the minimum over all $j$. This makes stair-stepped data and will have discontinuities between data points.
First, lets set up an example. x is our signal, which is then sampled at times t2 to make x2
In [37]:
import scipy.signal as signal
np.random.seed(1345)
N = 100000
T = 1000. # s.
t = np.arange(0,T,T/N)
fN = N/T/2.
b,a=signal.ellip(5,0.5,60.,0.75/fN)
x = signal.lfilter(b,a,(np.random.randn(N+1)))[:-1]
dt = np.random.rand(N)*0.75+0.75
t2 = np.cumsum(dt)
t2 = t2[t2<t[-1]]
x2 = np.interp(t2,t,x)
# t2 is unevenly spaced
y = x[range(25,N,50)]
ty = t[range(25,N,50)]
#y is evenly spaced at 2 Hz.
ygap=1.*y
tygap=1.*ty
ygap=np.delete(ygap,np.arange(30,60))
tygap=np.delete(tygap,np.arange(30,60))
fig,ax=plt.subplots()
ax.plot(t,x)
ax.plot(t2,x2,'d')
ax.plot(ty,y,'o')
ax.set_xlim((0,20)); ax.set_ylabel('x(t)');ax.set_xlabel('t [s]')
ax.legend(('x','unevenly spaced','evenly spaced 2 Hz'))
Out[37]:
<matplotlib.legend.Legend at 0x124092c50>
Here $x$ is a band-limited time series that we then downsample to 2 Hz (red dots). This is what we would like to retrieve. Now imagine we had only collected data at the green dots. Then we need to interpolate to the even 2-Hz time series.
In [38]:
yi=0.*ty
for n in range(0,np.shape(ty)[0]):
ind=np.argmin(abs(ty[n]-t2))
yi[n]=x2[ind]
nfft=64*4
Gyy,f = mlab.psd(y,nfft,Fs=2,window=mlab.window_hanning,noverlap=nfft/2)
Gyiy,f = mlab.psd(yi,nfft,Fs=2,window=mlab.window_hanning,noverlap=nfft/2)
Gxx,f0 = mlab.psd(x,nfft*50,Fs=100,window=mlab.window_hanning,noverlap=nfft/2)
Gneigh=Gyiy
fig,ax=plt.subplots(2,1,figsize=(6,6))
ax[0].plot(t,x,label='Orig')
ax[0].plot(ty,y,label='Ideal Interp')
ax[0].plot(ty,yi,label='Nearest')
ax[0].set_xlabel(r'$t \ \mathrm{[s]}$')
ax[0].set_ylabel(r'$x\ \mathrm{[V]}$')
ax[0].set_xlim((0,20))
ax[1].loglog(f0,Gxx,f,Gyy,f,Gyiy)
plt.xlim((1e-2,3.))
plt.ylim((1e-5,1e-1))
plt.xlabel(r'$f\ \mathrm{[Hz]}$');
plt.ylabel(r'$G_{xx} \ \mathrm{[V^2 Hz^{-1}]}$')
ax[1].legend(('Original fully resolved','Original even samples','nearest-neighbourhood interpolated from uneven'),loc=3,fontsize='small')
Out[38]:
<matplotlib.legend.Legend at 0x12484a590>
So, nearest-neighbour is not too bad in this case until the very highest frequencies where there is some error. Of course, the level of error depends on how far apart the data is in the uneven time series.
### Linear interpolation¶
Linear interpolation is simply interpolating onto a line formed by the two data points bracketing the time you are interested in getting a value at. So if you want to know the value of $x(t)$ where $t_j<t<t_{j+1}$, then
$$x(t) = x_j+\left(t-t_j\right)\frac{x_{j+1}-x_j}{t_{j+1}-t_j}$$
In [47]:
yi=0.*ty
for n in range(0,np.shape(ty)[0]):
if (ty[n]>t2[0]) & (ty[n]<t2[-1]):
ind=np.where(t2<ty[n])[0][-1]
yi[n]=x2[ind]+(ty[n]-t2[ind])*(x2[ind+1]-x2[ind])/(t2[ind+1]-t2[ind])
#yi=interp(ty,t2,x2)
nfft=64*4
Gyy,f = mlab.psd(y,nfft,Fs=2)
Gyiy,f = mlab.psd(yi,nfft,Fs=2)
Gxx,f0 = mlab.psd(x,nfft*50,Fs=100)
Gneigh=Gyiy
fig,ax=plt.subplots(2,1,figsize=(6,5))
ax[0].plot(t,x,label='Orig')
ax[0].plot(ty,y,label='Ideal Interp')
ax[0].plot(ty,yi,label='Nearest')
ax[0].set_xlabel(r'$t \ \mathrm{[s]}$')
ax[0].set_ylabel(r'$x\ \mathrm{[V]}$')
ax[0].set_xlim((0,20))
ax[1].loglog(f0,Gxx,label='Ideal')
ax[1].loglog(f,Gyy,label='Evenly sampled')
ax[1].loglog(f,Gyiy,label='Linearly interpolated')
plt.xlim((1e-2,3.))
plt.ylim((1e-5,1e-1))
plt.xlabel(r'$f\ \mathrm{[Hz]}$');
plt.ylabel(r'$G_{xx} \ \mathrm{[V^2 Hz^{-1}]}$')
ax[1].legend(loc=3,fontsize='small')
Out[47]:
<matplotlib.legend.Legend at 0x1285882d0>
This has less of a bias at low frequencies than the nearest-neighbour method. It has some roll off at high frequencies which is not too surprising given that a straightline has as little variance as possible and the data gaps cause us to lose variance.
Note that the code above has checks to make sure we are not off the ends of the time series. This is extrapolation, and should be dealt with carefully. There are a couple of common choices for what to do with if we want to extrapolate:
1. Linearly extrapolate: i.e. $x(t) = x_0+\left(t-t_0\right)\frac{x_{1}-x_0}{t_{1}-t_0}$.
2. Nearest neighbour: $x(t)=x_0$.
Obviously, the further out you extrapolate, the worse the extrapolation. Hopefully it doesn't change your signal, and is usually a bad idea in most situations.
### Cubic Spline Interpolation¶
Linear interpolation makes the data be continuous in time between the data points. Cubic spline takes this one step further and fits a cubic spline to keep the first and second derivatives continuous at the data points. This requires a universal fit, since the derivatives depend on the data non-locally.
So suppose, we have $N+1$ data points, $n=0,1,..N$, then we have $N$ segments between the data points, $i=0,1,...,N-1$. We want the fit $q(t)$ to go through the data and for the derivatives to be continuous at the data points. Let $q_i(t)$ be the fit between $t_i$ and $t_{i+1}$:
$$q_i(t)=a_i + b_i(t-t_i)+c_i(t-t_i)^2+d_i(t-t_i)^3.$$
We need to fit the $4N$ constants to complete the spline fit. To do this we choose matching conditions at the boundaries of the fit:
\begin{align} q_i(t_i)&=x_i \ \ \ \ \ \ \text{for}\ i=0...N-1\\ q_i(t_{i+1})&=x_{i+1}\ \ \ \ \text{for}\ i=0...N-1\\ q_i'(t_{i+1})&=q_{i+1}'(t_{i+1})\ \ \ \ \text{for}\ i=0...N-2\\ q_i''(t_{i+1})&=q_{i+1}''(t_{i+1})\ \ \ \ \text{for}\ i=0...N-2\\ \end{align}
This is $4N-2$ equations, and $q_0(t)$ and $q_{N-1}(t)$ require two boundary conditions. Different choices are possible, but a simple one is to set the second derivative to zero:
\begin{align} q_0''(t_0)&=0\\ q_{N-1}''(t_{N})&=0\\ \end{align}
We can easily eliminate $N$ equations, because the first condition means that $a_i=x_i$ for $i=0,...,N-1$. Defining $\Delta_i = t_{i+1}-t_i$ and solving for the other conditions gives:
\begin{align} x_i +\Delta_ib_i +\Delta_i^2c_i + \Delta_i^3d_i&=x_{i+1} \ \ \ \ \ \ \text{for}\ i=0...N-1\\ b_i+2\Delta_ic_i+3\Delta_i^2d_i &=b_{i+1}\ \ \ \ \text{for}\ i=0...N-2\\ 2c_i+6d_i\Delta_i&=2c_{i+1}\ \ \ \ \text{for}\ i=0...N-2\\ c_0&=0\\ 2c_{N-1}+6d_{N-1}\Delta_{N-1}&=0 \end{align}
With some patience we can eliminate $b_i$ and $d_i$ and get:
$$\Delta_i c_i + 2 (\Delta_i+\Delta_{i+1})c_{i+1}+\Delta_{i+1}c_{i+2}=3\left[\frac{x_{i+2}-x_{i+1}}{\Delta_{i+1}} - \frac{x_{i+1}-x_{i}}{\Delta_i}\right] \ \ \ \ \text{for}\ i=0,...,N-2$$
and $c_N=c_0=0$ giving us $N+1$ equations. Note that we need the $N+1$ equations because there is a $c_{i+2}$ term in the equation above.
$$\begin{pmatrix} 1 & 0 & 0 & \cdots & \cdots & \cdots & \cdots\\ \Delta_0 & 2\left(\Delta_0+\Delta_1\right) & \Delta_1 & \cdots & \cdots & \cdots & \cdots\\ \cdots & \cdots & \ddots & \cdots & \cdots & \cdots & \cdots\\ \cdots & \cdots & \cdots & \Delta_{N-3} & 2\left(\Delta_{N-3}+\Delta_{N-2} \right) & \Delta_{N-2} & 0\\ \cdots & \cdots & \cdots &\cdots & \Delta_{N-2} & 2\left(\Delta_{N-2}+\Delta_{N-1} \right) & \Delta_{N-1}\\ \cdots & \cdots & \cdots & \cdots & \cdots & 0 & 1 \end{pmatrix} \begin{pmatrix} c_0\\ c_1\\ c_2\\ \vdots\\ c_{N-2}\\ c_{N-1}\\ c_{N} \end{pmatrix} = 3\begin{pmatrix} 0\\ \frac{dx_1}{\Delta_1}- \frac{dx_0}{\Delta_0}\\ \frac{dx_2}{\Delta_2}- \frac{dx_1}{\Delta_1}\\ \vdots\\ \frac{dx_{N-1}}{\Delta_{N-1}}- \frac{dx_{N-2}}{\Delta_{N-2}}\\ 0 \end{pmatrix}$$
This an $(N+1) \times (N+1)$ matrix, and solving for $c_i$ is theoretically easy, but computationally challenging if $N$ is large. Fortunately, it is also a sparse matrix (i.e. most of the entries are zero) and there are effcient techniques for solving these. It is also possible to fit in chunks, and match boundary conditions across the chunks, which I believe is the algorithm usually used by packages.
In [52]:
from scipy.interpolate import interp1d
ff=interp1d(t2,x2,kind='cubic')
tyn=ty[ty>t2[0]]
tyn=tyn[tyn<t2[-1]]
yi=ff(tyn)
Gyy,f = mlab.psd(y,nfft,Fs=2)
Gyiy,f = mlab.psd(yi,nfft,Fs=2)
Gxx,f0 = mlab.psd(x,nfft*50,Fs=100)
Gneigh=Gyiy
fig,ax=plt.subplots(2,1,figsize=(5,5))
ax[0].plot(t,x,label='Orig')
ax[0].plot(ty,y,label='Ideal Interp')
ax[0].plot(tyn,yi,label='Nearest')
ax[0].set_xlabel(r'$t \ \mathrm{[s]}$')
ax[0].set_ylabel(r'$x\ \mathrm{[V]}$')
ax[0].set_xlim((0,20))
ax[1].loglog(f0,Gxx,label='Ideal')
ax[1].loglog(f,Gyy,label='Evenly sampled')
ax[1].loglog(f,Gyiy,label='Cubic Spline')
plt.xlim((1e-2,3.))
plt.ylim((1e-7,1e-1))
plt.xlabel(r'$f\ \mathrm{[Hz]}$');
plt.ylabel(r'$G_{xx} \ \mathrm{[V^2 Hz^{-1}]}$')
ax[1].legend(loc=3,fontsize='small')
Out[52]:
<matplotlib.legend.Legend at 0x1214beb10>
Here we see that the cubic spline does quite well at this type of signal.
### Which interpolation to use?¶
I usually just use linear interpolation, largely because it will behave in a predictable way. Cubic splines have have difficulty with noisy data, or data that has discontinuity. However, the best advice is to try both linear and cubic spline and see which has the best response for the data you are trying to interpolate.
In [ ]: | 6,862 | 20,104 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-39 | latest | en | 0.924779 |
https://physics.stackexchange.com/questions/467936/the-law-of-radioactive-decay-explanation-of-a-formula | 1,718,555,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00136.warc.gz | 409,175,508 | 39,617 | # The law of radioactive decay: explanation of a formula
The law of radioactive decay can be expressed in terms of $$\,\tau=1/\lambda$$ (average life) as:
$$N(t)=N_0e^{-t/\tau}, \quad \tag{1}$$
Why deriving the (1) I have: $$$$N'(t)=N_0(1-e^{-\lambda t})\, ?$$$$
• Your second equation shouldn't be the starting point of the derivation. How did you get there?
– noah
Commented Mar 21, 2019 at 22:12
• I have some notes of a research that I'm elaborating. I have finded this without any linkage. I don't get it. If I derive the (1) I don't get the second one. Commented Mar 21, 2019 at 22:16
• en.wikipedia.org/wiki/Radioactive_decay#One-decay_process . 5 secs of Googling.
– Gert
Commented Mar 21, 2019 at 22:17
• @Gert Could I please have a better explanation than Wikipedia, simpler and more complete? Actually, I haven't thought about searching on the web. Commented Mar 21, 2019 at 22:20
• I don't think you'll find anything simpler (or more complete). It really is a very simple problem, you know?
– Gert
Commented Mar 21, 2019 at 22:23
It comes from solving the differential equation
$$\frac{dN}{dt} = -\lambda N(t).$$
This equation comes from observations of the number of decay events $$N(t)$$. It's found through experiment that the rate of decay over a given time interval is proportional to the number of events recorded during that time. You can arrive at this conclusion by plotting the rate vs the number of events on a log log plot and finding that it is linear.
Formally, this is a differential equation. But solving it is really just a fact which you know already.
Which function $$N(t)$$ can you take the derivative of and get itself back times a constant?
The answer is exponentials, and so the solution to this equation is
$$N(t) = N(0) e^{-\lambda t}.$$
Edit: I should also note that you took the derivative incorrectly. The correct derivative is
$$N'(t) = \frac{d}{dt} N_0 e^{-\lambda t} = - \lambda N_0 e^{-\lambda t}$$
• "the rate of decay is proportional to the the rate of these events" That doesn't make a whole lot of sense. By your own formula (which is correct) the decay rate is proportional to the number of remaining atoms, macroscopically speaking at least.
– Gert
Commented Mar 21, 2019 at 23:31
• oops.. made a typo thanks Commented Mar 21, 2019 at 23:32
• @InertialObserver The derivative that I was founded is the same as your :-). My notes are probably wrong. In fact from the Wikipedia link provided by Gert: $N_{B}=N_{A_0}-N_A=N_{A_0}(1-e^{-\lambda t})$. Thank you very much for your answer. Commented Mar 22, 2019 at 10:38 | 733 | 2,578 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.937766 |
https://blog.dil.com.bd/2021/04/details-estimation-of-beam-according-to-real-project-drawing-excel-manual/ | 1,670,275,785,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00303.warc.gz | 162,158,116 | 28,398 | ### BEAM WORK:
Calculation of Floor Beam FB-1
General Information
• Clear length =55.75’
• Width=10 inch
• Depth=15 inch
• Consists of 3-20mm diameter straight bars at top & 2-16 mm extra top bar 3-20mm diameter straight bars at bottom & 2-16 mm extra bottom bar
• Diameter of stirrup is 10 mm spaced at 7inch center to center.
• Clear Cover to reinforcement provided is 1.5 inch.
Reinforcement Calculation:
Now we will calculate the length of reinforcement based on shapes of reinforcement required for reinforced concrete beam FB-1
Main Bar shape of is as shown below:
Length of Main Bar
= 55.75’– (2 x 1.5)/12 + (2 x 60d) (20mm=0.065ft)
= 55.5’+ (2x60x0.065)
=63.3’
Length of 20mm 63.3×6 =379.8 rft
Nos of Lapping
= 379.8/40 =9.49nos
Lap Length of Main Bar
=9x50x0.065 =29.25 ft
Total Length of Bar= 379.8+29.25 =409.05 rft
Weight = .75×409.05 =306.78 kg
Length of bar Extra Top Bar
:
L/4 of Corner portion of FB-1 =16’11”/4 = 4.23’
L/4 of mid portion of FB-1 =9’8”/4 = 2.41
So Total Ext.Top bar Length=(4×4.2)+(2×2.41)) =16.8+4.82=22’
Length of bar Extra bottom Bar:
L/7 of Corner Beam =16’11”/7=2.41’
L/7 of Mid Beam =9’8’11”/7=1.38’
So Total Ext.Top bar Length=((2×2.41×5)+( 5×1.38)) =31’
Length of total ExtraBar = 2x (31+22)=106’
RCC Beam Concrete Work
Now we will calculate the Now we will calculate Concrete Work required for FB-1
Nos= 2
LENGTH (ft) = 55.75′
B1= 10”
B2= 15”
X-AREA =(10×15)/144= 1.04 sft
Volume= 2×55.75×1.04 =116.14 cft
Wet Volume =1.54×116.14= 178.855 cft
For 1:2:4 ratio
Cement = (1/7)x178.855 =25.55 cft
Sand = (2/7)x178.855 = 51.10 cft
Brick Chips = (4/7)x178.855 = 102.20 cft
Similarly you can calculate for other’s Beam Reinforcement & Concrete Work | 654 | 1,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-49 | longest | en | 0.730002 |
http://mathoverflow.net/feeds/user/1508 | 1,369,142,453,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700014987/warc/CC-MAIN-20130516102654-00061-ip-10-60-113-184.ec2.internal.warc.gz | 176,879,417 | 21,860 | User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:20:56Z http://mathoverflow.net/feeds/user/1508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130623/fano-plane-drawings-embedding-pg2-2-into-the-real-plane/130648#130648 Answer by auniket for Fano plane drawings: embedding PG(2,2) into the real plane auniket 2013-05-14T23:28:12Z 2013-05-15T01:41:09Z <p>Does this one work? </p> <p><img src="http://i44.tinypic.com/28jjns9.jpg" alt="Fano"></p> http://mathoverflow.net/questions/129673/interpretation-of-multiplicity-of-a-point Interpretation of multiplicity of a point auniket 2013-05-04T19:51:42Z 2013-05-05T03:02:09Z <p>Let $x$ be a (closed) point on an algebraic variety $X$ (of dimension $n$) defined over an algebraically closed field $k$. What is the multiplicity $mult_x(X)$, and how to compute it?</p> <p>While having a hard time recently to compute the multiplicity of some surface singularities, I thought it might be useful to have a list of equivalent definitions. These are the ones I know of:</p> <p>Notations: let $m_x$ be the maximal ideal of $x$ at $X$. For definitions 1 to 3 below assume (a neighborhood of $x$ in) $X$ is embedded in a projective space $\mathbb{P}^N(\mathbb{k})$. </p> <p>Geometric Definitions:</p> <ol> <li><p>(Copying from Mumford, Algebraic Geometry I: This is valid only in the case $k = \mathbb{C}$.) For every linear subspace $L$ of dimension $N - n$ such that $x$ is a component of $L \cap X$, define a number $i(x;L \cap X)$ as follows: it is the unique number such that for every sufficiently small neighborhood $U$ of $x$ (in the classical topology), there is a neighborhood $U'$ of $L$ (in the space of $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$) such that if $L' \in U'$ and $L'$ intersects $X$ transversally, then $i(x;L \cap X) = |L' \cap X \cap U|$. Then $mult_x(X)$ is the minimum of $i(x;L \cap X)$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(\mathbb{C})$ for which $x$ is an isolated point of $L \cap X$.</p></li> <li><p>(Corollary of (1), but holds also over positive characteristic - or at least so I think. For my problem this turned out to be the right definition to use) For every linear subspace $L$ of dimension $N - n$ such that $L \cap X$ is discrete, let $s(x;L \cap X)$ be the number of points of intersection (counted with intersection multiplicity) of $L \cap X$ other than $x$. Then $mult_x(X) = \min\lbrace \deg(X) - s(x;L \cap X)\rbrace$ as $L$ runs over all $N-n$ dimensional linear subspaces of $\mathbb{P}^N(k)$ for which $L \cap X$ is discrete. </p></li> <li><p>(From Ramanujam's "On a geometric interpretation of multiplicity") Take a proper birational map $\phi: Y \to X$ such that the pull back of the maximal ideal of $P$ defines an effective Cartier divisor $D$ on $Y$. Then $mult_x(X) = (-1)^{n-1}D^n$.</p></li> </ol> <p>Algebraic Definitions:</p> <ol> <li><p>$mult_x(X)$ is $(n-1)!$ times the leading coefficient of the Hilbert–Samuel polynomial of $m_x$.</p></li> <li><p>If $X$ is a hypersurface in a neighbourhood of $x$ defined by a single equation $f$, then $mult_x(X)$ is the integer $q$ such that $f \in m_x^q \setminus m_x^{q+1}$.</p></li> </ol> <p>What other definitions are out there?</p> http://mathoverflow.net/questions/121186/jacobian-polynomial/121206#121206 Answer by auniket for jacobian polynomial auniket 2013-02-08T15:41:19Z 2013-02-08T16:24:04Z <p>Edit: The statement in the next paragraph is wrong! I misunderstood the result of Kaliman: it says that given $(f,g)$ as in the question, there is a polynomial automorphism $\phi$ of $\mathbb{C}^2$ such that each fiber of $\phi \circ (f,g): \mathbb{C}^2 \to \mathbb{C}^2$ is irreducible. So I would assume it is still hard to give a positive answer to the question, but clearly what I wrote below is false. </p> <p>I would assume the question is quite difficult, since a positive answer would imply the Jacobian conjecture by <a href="http://www.ams.org/journals/proc/1993-117-01/S0002-9939-1993-1106179-7/S0002-9939-1993-1106179-7.pdf" rel="nofollow"> this</a> result of Kaliman. </p> http://mathoverflow.net/questions/119702/normality-condition-on-graded-algebra/119707#119707 Answer by auniket for Normality condition on graded algebra auniket 2013-01-23T22:56:57Z 2013-01-24T17:40:44Z <p>Hi Isac, One simple criterion can be given which is analogous to Rees' valuations corresponding to ideals. Set $R := \mathbb{C}[x,y]$, define $\nu: R \to \mathbb{N} \cup \infty$ as $\nu(f) := \max\lbrace k: f \in m_k\rbrace$. Then I believe $A$ is normal if $\nu(f^k) = k\nu(f)$ for all $f \in R$. For other criteria, I would look into Kei-ichi Watanabe's articles.</p> <p>Edit: Here is a general approach.</p> <p>Claim 1: $A$ is integrally closed iff every <i> homogeneous </i> (with respect to the grading of $A$) element in $\mathbb{C}[x,y,t]$ (i.e. an element of the form $f(x,y)t^k$) which is integral over $A$ is in $A$.</p> <p>By Claim 1 the integral equation of $ft^k$ (for $f \in R$) over $A$ be of the form $z^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $g_e \in m_{ek}$. This proves Claim 2 below.</p> <p>Claim 2: $A$ is integrally closed iff $\bar m_k := m_k$ for all $k \geq 1$, where</p> <p>$\bar m_k := \lbrace f \in m_k: f^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $d \geq 0$ and $g_1, \ldots, g_d \in m_{ek}\rbrace$.</p> <p>In the special case that $m_k$ is a monomial ideal for each $k$, it suffices to prove the integral condition only for monomials in $m_k$. And in the special special case that $m_k = m^k$ for some monomial ideal $m$, it follows that your claim is true, i.e. $A$ is integrally closed iff the support of $m^k$ contains all the monomials in the cone spanned by its monomials.</p> http://mathoverflow.net/questions/117666/contracting-rational-curves-on-surfaces-and-getting-something-non-algebraic Contracting rational curves on surfaces and getting something non-algebraic auniket 2012-12-30T20:41:58Z 2012-12-30T20:41:58Z <p>Recently I posted an "announcement" on arxiv where I said something to the effect of "this is the first example we know where contracting (a tree of) rational curves from a non-singular algebraic surface (over $\mathbb{C}$) leads to a (normal) <i> non-algebraic </i> surface." Now that I am writing up the actual paper, I thought that may be I should broaden my knowledge base! So I ask: is there some example of this sort already known?</p> <p>Remarks: </p> <ol> <li><p>Grauert constructed (in this article: <a href="http://www.ams.org/mathscinet/pdf/137127.pdf" rel="nofollow"> mathscinet link</a>, <a href="http://link.springer.com/article/10.1007/BF01441136" rel="nofollow"> springerlink</a>) such non-algebraic surfaces by blowing down curves of genus $\geq 2$ (in Section 4.8, Example d) and remarked that he did not know if it is possible (to construct non-algebraic normal surfaces) from blowing down tori. </p></li> <li><p>An example of Nagata described in a <a href="http://mathoverflow.net/questions/79215/pathologies-of-analytic-non-algebraic-varieties" rel="nofollow"> previous question </a> of mine shows that it is indeed possible with tori. </p></li> <li><p>In the second paragraph of the first page of <a href="http://www.jstor.org/stable/2372985" rel="nofollow"> this article</a> (<a href="http://www.ams.org/mathscinet/pdf/146182.pdf" rel="nofollow">mathscinet link</a>) Artin mentions an example of Hironaka that shows that "in general there are no numerical criteria equivalent with (algebraic) contractibility of a given curve." Does anyone know what is this example?</p></li> <li><p>Happy New Year!</p></li> </ol> http://mathoverflow.net/questions/103657/degree-of-a-variety-is-well-defined/103675#103675 Answer by auniket for Degree of a variety is well-defined auniket 2012-08-01T09:45:02Z 2012-08-01T09:45:02Z <p>Hi Kiumars, this does not answer your question, since the proof works only when the field is $\mathbb{C}$, but when I was learning it, the most accessible and understandable proof for this case was that of Theorem 5.1 of Mumford's "Algebraic Geometry I: Complex Projective Varieties".</p> http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-pr Geometric interpretation of the exact sequence for the cotangent bundle of the projective space auniket 2012-06-02T04:44:47Z 2012-06-02T19:47:39Z <p>Edit: As Dan Petersen pointed out, this question is a duplicate of a <a href="http://mathoverflow.net/questions/5211/" rel="nofollow"> previous one</a>. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).</p> <p>I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field): $$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$ The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$. </p> <p>But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?</p> http://mathoverflow.net/questions/88750/functions-satisfying-one-one-iff-onto/88789#88789 Answer by auniket for functions satisfying "one-one iff onto" auniket 2012-02-18T01:15:04Z 2012-02-18T01:15:04Z <p>This addresses the "broader scope" of the question and possibly the comments of Uday on Donu's answer: An injective morphism from an affine algebraic variety over an algebraically closed field to itself is also surjective. Moreover, probably even more surprising is the fact that in the case that the field has characteristic zero (and of course algebraically closed), an injective endomorphism is actually a polynomial automorphism (that is the inverse is also a polynomial map!). See e.g. Chapter 4 of van den Essen's "Polynomial Automorphisms" for proofs of both these statements. Also from the same book: the map $x \mapsto x^3$ from $\mathbb{Q} \to \mathbb{Q}$ shows the necessity of algebraic closedness of the field, and the Frobenius automorphism $x \mapsto x^p$ of an algebraically closed field of characteristic $p > 0$ shows that the second statement is false for positive characteristic. Also, note that both statements are automatically true for proper varieties.</p> http://mathoverflow.net/questions/88788/terminology-for-the-image-of-the-diagonal-embedding Terminology for the image of the diagonal embedding. auniket 2012-02-18T00:45:44Z 2012-02-18T00:45:44Z <p>Let $X$ be a topological space equipped with maps into two spaces $\bar X_1$ and $\bar X_2$. Is there a standard notation/terminology for the closure $\bar X$ in $\bar X_1 \times \bar X_2$ of the <i> diagonal map </i> of $X$?</p> <p>In my case $X$ is an affine algebraic surface (in fact just $\mathbb{C}^2$) which is isomorphic to Zariski open subsets of complete surfaces $\bar X_1$ and $\bar X_2$ and the maps $X \to \bar X_j$'s are the corresponding embeddings.</p> <p>In a paper I wrote, I used the notation "birational join" for $\bar X$ following <a href="http://www.jstor.org/stable/1971467" rel="nofollow"> Spivakovsky, </a> but the referee does not like it. (S)He suggested something like the "fiber product" $\bar X_1 \times_X \bar X_2$, but that would require the arrows $X \to \bar X_j$ to be reversed. Similarly the cofiber product'' requires the arrows $\bar X \to \bar X_j$ to be reversed.</p> <p>Any suggestions would be much appreciated. Thanks!</p> http://mathoverflow.net/questions/80368/algebraic-curve-cannot-suddenly-end/80384#80384 Answer by auniket for Algebraic curve cannot suddenly end auniket 2011-11-08T13:14:34Z 2011-11-08T13:36:54Z <p>Following the idea of Felipe Voloch, I try to give a simple proof based on Puiseux series expansion. Let $C$ be a real algebraic curve at the origin. Look at the Puiseux series expansion (say in terms of $x$) of $C$ near $O$. By assumption one of the branches (over $\mathbb{C}$), call it $C_1$, has the form $$y = a_1x^{r_1} + a_2x^{r_2} + \cdots \quad\quad\quad (1)$$ for $a_i \in \mathbb{R}$. Let $q$ be the least common multiple of the denominators of $r_i$'s. If $q$ is odd, then the branch expands to <i> both </i> sides of the origin and therefore $C_1$ does not end abruptly. So assume $q$ is even. Let $\zeta := e^{2\pi i/q}$. For each $j$, $1 \leq j \leq q$, the complex curve corresponding to $C_1$ has a Puiseux expansion of the form $y = \sum_i a_i \zeta^{jp_i}x^{r_i}$, where $p_i = qr_i$. In particular, taking $j =q/2$ (so that $\zeta^j = -1$), we see that the complex curve corresponding to $C_1$ has an expansion of the form $$y = \sum_i a_i (-1)^{p_i}x^{r_i}. \quad\quad\quad (2)$$ It follows by the minimality of assumption on $q$ that there is $i$ such that $a_i\neq 0$ and $p_i$ is <i> odd </i>, and consequently, $(1)$ and $(2)$ give <i> different </i> real curves, and it follows that $C_1$ does not end abruptly.</p> <p>PS: The above arguments only show that $C_1$ has at least two end points on the boundary of a small enough disk centered at $O$. But it can not have more than two, because for all $j \not\in \lbrace q/2, q\rbrace$, $\zeta^j$ is non-real, so the corresponding parametrization does not give any real points.</p> http://mathoverflow.net/questions/79215/pathologies-of-analytic-non-algebraic-varieties Pathologies of analytic (non-algebraic) varieties. auniket 2011-10-26T23:46:30Z 2011-10-27T10:19:59Z <p><b> Note: </b> By an "analytic non-algebraic" surface below I mean a two dimensional compact analytic variety $X$ (over $\mathbb{C}$) which is not an algebraic variety.</p> <p>A property of Nagata's example (see the end of the post for the construction) of a non-algebraic normal analytic surface $X$ is the following: </p> <p>($\star$) $\quad$ There is a point $P$ on $X$ such that every (compact) algebraic curve $C$ on $X$ passes through $P$. </p> <p>In a paper I am writing I also constructed (to my surprise) some examples of non-algebraic normal analytic surfaces which have this peculiar property. </p> <p><b> Questions:</b> Is this sort of behaviour "normal" for such surfaces? Or, more precisely, if an analytic surface does not satisfy ($\star$), is it necessarily algebraic? How about for higher dimensions?</p> <p><b> Nagata's Construction (following Bădescu's book on surfaces):</b> Start with a smooth plane cubic $C$ and a point $P$ on $\mathbb{P}^2$ such that $P - O$ is not a torsion point (where $O$ is any of the inflection points of $C$) on $C$. Let $X_1$ be the blow up of $\mathbb{P}^2$ at $P$, and for each $i \geq 1$, let $X_{i+1}$ be the blow up of $X_i$ at the point of intersection of the strict transform of $C$ and the exceptional divisor on $X_i$. Each blow up decreases the self-intersection number of the strict transform $C_i$ of $C$ by $1$, so that on $X_{10}$ the self-intersection number of $C_{10}$ is $-1$. $X$ is the blow down of $X_{10}$ along $C_{10}$. By some theorems of Grauer and Artin, $X$ is a normal analytic surface.</p> http://mathoverflow.net/questions/78305/ample-divisors-on-projective-surfaces Ample divisors on projective surfaces auniket 2011-10-17T03:11:48Z 2011-10-18T19:44:55Z <p>Question: If $X$ is a projective surface and $U$ is an open affine subset of $X$, then is it true that $X \setminus U$ is the support of an (effective) ample divisor on $X$?</p> <p>Background: I was reading Goodman's paper <a href="http://www.jstor.org/stable/1970814" rel="nofollow"> "Affine open subsets of algebraic varieties and ample divisors"</a> which considers this same question for general varieties. Here is what I understand so far:</p> <ol> <li><p>If $\dim X = 1$, then the answer to the question is always affirmative. Indeed, if $X$ is a complete curve and $S$ is any finite set of points on $X$, then there is an effective ample Cartier divisor on $X$ which has support $S$ (this is Proposition 5 of the paper, and a straightforward application of the <a href="http://en.wikipedia.org/wiki/Ample_line_bundle#Intersection_theory" rel="nofollow"> Nakai-Moishezon criterion </a> of ampleness).</p></li> <li><p>For $\dim X = 2$, Theorem 2 of the paper states that the answer is positive if each point of $X\setminus U$ is factorial (i.e. its local ring is a UFD). Actually he proves it only assuming that $X$ is complete (i.e. a priori not necessarily projective) and as a corollary he proves Zariski's theorem that if all the singularities of a complete surface $X$ are contained in an affine open subset, then $X$ is projective. </p></li> <li><p>He presents two examples (of Hironaka and Zariski) where $X$ is a non-singular projective $3$-folds, but $X\setminus U$ is not the support of any ample divisor. </p></li> <li><p>In general he proves (in Theorem 1) that if $X$ is complete then a Zariski open subset $U$ of $X$ is affine iff the complement of (the isomorphic image of $U$) in a blow-up $X'$ of $X$ along a closed subscheme $F$ not meeting $U$ is the support of an effective ample Cartier divisor on $X'$.</p></li> <li><p>For $\dim X \geq 3$, he gives a criterion (in Theorem 3) for when the answer to the question is positive.</p></li> </ol> <p>As far as I can see, he does not mention anything about the status of the question (i.e. whether if there is a counter-example or not) for general projective surfaces. Therefore I ask it here. I would expect the answer to be negative, but can not think of any examples. For me particularly interesting would be the case when $X$ is normal. </p> <p><b> Edit: </b> As the example of Jason Starr in the comment shows: The answer is <i> negative </i> even for normal surfaces (see my comments for an attempt of proof). I wonder what happens if $X$ is rational. In any event, I would gladly accept Jason's answer if he writes one. (And I would also greatly appreciate any answer/remark about the rational case.)</p> http://mathoverflow.net/questions/76640/is-md-very-ample-if-d-is-ample/76873#76873 Answer by auniket for Is [mD] very ample if D is ample? auniket 2011-09-30T18:28:55Z 2011-10-01T14:15:37Z <p><del> At first a question: If I understand correctly, then the rounding down operation depends on your choice of basis for the Néron-Severi group, right? </del> </p> <p><del> So I am assuming you fix a basis $D_1, \ldots, D_k$ of $NS(X) \otimes_\mathbb{Z} \mathbb{R}$ and for each $D := \sum_j r_jD_j$, you define $[mD] := \sum [mr_j]D_j$. </del></p> <p><del> If this is true, then doesn't your assertion follow from the following geometric fact? </del></p> <p><del> Let $C$ be a full dimensional cone in $\mathbb{R}^k$ and $K$ be the standard cube of length $2$ in $\mathbb{R}^k$ centered at the origin, i.e. </del></p> <p><del> $K := \lbrace\sum_{j=1}^k s_je_j: -1 \leq s_j \leq 1$ for all $j$, $1 \leq j \leq k \rbrace$, </del></p> <p><del> where $e_1, \ldots, e_k$ are unit vectors along the axes. If $v$ belongs to the interior of a full dimensional cone $C$ in $\mathbb{R}^k$, then $mv + K$ also lies in the interior of $C$ for all sufficiently large $m$. </del></p> <p><del> If as your basis you choose ample divisors, then $K$ can be replaced by a cube of length one. </del></p> <p><b> Edit 3: </b> This is my 3rd attempt to give an elementary proof. It is essentially the same proof as in Edits 1 and 2, but with some corrections, and hopefully will be clearer. I hope you see that the idea is very simple and geometrically almost obvious. If it seems complicated, then the fault is in my exposition. </p> <p><b> Set Up: </b> Let $D_1, \ldots, D_k$ be ample divisors and $D := \sum_j r_jD_j$ for positive real numbers $r_1, \ldots, r_k$. Also, let $D_j = \sum_{i=1}^N a_{ji} C_i$, for irreducible divisors $C_i$ and integers $a_{ji}$. We want to show that $[mD]$ is very ample for large $m$.</p> <p>In the proof we will use the following fact about finite sums of integral points in a lattice:</p> <p><b> Lemma: </b> Let $v_1, \ldots, v_k \in \mathbb{Z}^N$ such that $\mathbb{Z}$-span of $v_j$'s equals $\mathbb{Z}^N$. Let $P$ be the convex hull (over $\mathbb{R}$) of $\lbrace 0, v_1, \ldots, v_k \rbrace$. Then there exists a positive real number $c$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that the (Euclidean) distance of $v$ from both the origin and the boundary of $nP$ is greater than $c$, then $v$ is in fact an non-negative integral linear combination of $v_1, \ldots, v_k$. </p> <p>The above statement (actually a more precise formulation of it) is due to Khovanskii. The proof is very elementary and beautiful, and is in Proposition 2 of <a href="http://www.springerlink.com/index/H71U3U6QR1727466.pdf" rel="nofollow"> this article.</a></p> <p>Here starts the proof:</p> <p>Step 1: Without loss of generality we may assume that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Indeed, it follows from <a href="http://en.wikipedia.org/wiki/Ample_line_bundle#Intersection_theory" rel="nofollow"> Kleiman's criterion, </a> and finite dimensionality of $N_1(X)$ that for every $m \gg 1$ and $\epsilon := (\epsilon_1, \ldots, \epsilon_N) \in \lbrace 1, 0, -1 \rbrace^N$, $D_{m,\epsilon} := mD_1 + \sum_{i=1}^N\epsilon_i C_i$ is ample. Choosing different values of $\epsilon$ and $m$ and adding $D_{m,\epsilon}$'s to the collection of $D_j$'s, we may ensure that $\mathbb{Z}$-span of $D_j$'s equals the $\mathbb{Z}$-span of $C_i$'s. Moreover, and this is essential, choosing $D_{m,\epsilon}$'s to be sufficiently close to the ray generated by $D_1$, we may ensure that $D$ still lies in the <i> interior </i> of the cone generated by $D_j$'s, i.e. $D = \sum_{j=1}^k r_jD_j$ with each $r_j$ being a <i> positive </i> real number. </p> <p>Step 2: For each $j$, $1 \leq j \leq k$, let $v_j := (a_{j1}, \ldots, a_{jN}) \in \mathbb{R}^N$, i.e. $v_j$ is the "coordinate" vector of $D_j$ for each $j$ (and therefore $v_j \in \mathbb{Z}^N$ for each $j$). Adding some big multiples of $D_j$'s to the existing collection of $D_j$'s if necessary, we may assume that $v := \sum r_j v_j$ is in the <i> interior </i> of the convex hull $P$ of $0, v_1, \ldots, v_k$. </p> <p>Step 3: For each $j$, $1 \leq j \leq k$, there exists a positive integer $m_j$ such that $mD_j$ is very ample for all $m \geq m_j$. Indeed, there is $l_j, n_j$ such that $n_jD_j$ is very ample and $mD_j$ is globally generated for all $m \geq l_j$. Setting $m_j := l_j + n_j$ does the job (due to Exercise II.7.5(d) of Hartshorne).</p> <p>Step 4: There exists a positive integer $m_0$ such that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of non-negative integers $s_1, \ldots, s_k$. Indeed, set $m_0 := \max \lbrace m_1, \ldots, m_k \rbrace$ and apply the same exercise of Hartshorne.</p> <p>Step 5: Let $v, v_1, \ldots, v_k$ and $P$ be as in Step 2. Let $c$ be the constant we get from applying Khovanskii's lemma to $v_1, \ldots, v_k$. Let $v_0 := m_0(v_1 + \cdots + v_k)$, where $m_0$ is as in Step 4. Since $v$ is in the interior of $P$, it follows that if $m$ is sufficiently large, then $[mv] - v_0$ is in the interior of $mP$ and the distance of $[mv] - v_0$ from the origin and the boundary of $mP$ is bigger than $c$. Therefore, Khovanskii's lemma implies that $[mv] - v_0 = \sum a_j v_j$ for non-negative integers $a_j$. Consequently, if $m$ is sufficiently large, then</p> <p>$$[mD] = m_0(D_1 + \cdots + D_0) + \sum a_j D_j$$ </p> <p>for non-negative integers $a_1, \ldots, a_k$. Step 4 then tells that $[mD]$ is very ample.</p> http://mathoverflow.net/questions/35514/pair-of-curves-joining-opposite-corners-of-a-square-must-intersect-proof/76896#76896 Answer by auniket for Pair of curves joining opposite corners of a square must intersect---proof? auniket 2011-10-01T00:08:35Z 2011-10-01T00:15:19Z <p>How about the following, using the <i> Nested Intervals Theorem </i> (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:</p> <p>We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$). </p> <p>If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = \lbrace t\rbrace$ for some $t \in [0,1]$. Then we must have $c_1(t) = c_2(t)$.</p> http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/76827#76827 Answer by auniket for Examples of seemingly elementary problems that are hard to solve? auniket 2011-09-30T06:45:24Z 2011-09-30T06:45:24Z <p>The Casas Alvero conjecture: Let $f \in \mathbb{C}[x]$ be a monic polynomial of degree $n$. Suppose that for each $k = 1, \ldots, n-1$, there is a common root of $f$ and $f^{(k)}$. Then $f = (x-a)^n$ for some $a \in \mathbb{C}$. It is known only for the case that $n$ is a prime power or two times a prime power (see for example, <a href="http://www.win.tue.nl/~jdraisma/talks/casasalverotalk.pdf" rel="nofollow"> this</a>). At some point I thought I proved it :-)</p> http://mathoverflow.net/questions/71952/do-the-elementary-properties-of-mixed-volume-characterize-it-uniquely/71967#71967 Answer by auniket for Do the elementary properties of mixed volume characterize it uniquely? auniket 2011-08-03T05:39:20Z 2011-08-03T05:39:20Z <p>I think the first three properties do indeed characterize mixed volume. For example, in two dimensions they imply that</p> <p>$V(A_1, A_2) = \frac{1}{2}(V(A_1 + A_2, A_1 + A_2) - V(A_1, A_1) - V(A_2,A_2))$ <br> $= \frac{1}{2}(Vol(A_1 + A_2) - Vol(A_1) - Vol(A_2)),$</p> <p>which gives the formula of mixed volume in terms of volume. You can perform the same trick to get in 3 dimensions:</p> <p>$V(A_1,A_2, A_3) = \frac{1}{6}(Vol(A_1+A_2+A_3) - Vol(A_1+A_2) - Vol(A_2+A_3)$ <br> $- Vol(A_3+A_1) + Vol(A_1) + Vol(A_2) + Vol(A_3))$</p> <p>In general I believe you get something like:</p> <p>$V(A_1, \ldots,A_n) = \frac{1}{n!}(Vol(A_1 + \cdots + A_n) - \sum_{i=1}^n Vol(A_1 + \cdots \hat A_i + \cdots + A_n)$ <br> $+ \cdots +(-1)^{n-1}\sum_{i=1}^n Vol(A_i))$</p> <p>I learned of this from Bernstein's paper that contains his famous result that the number of solutions in $(\mathbb{C}^*)^n$ of $n$ generic Laurent polynomials is precisely the mixed volume of their Newton polytopes.</p> http://mathoverflow.net/questions/70143/what-is-the-fan-of-the-toric-blow-up-of-mathbbp3-along-the-union-of-two-int/70324#70324 Answer by auniket for What is the fan of the toric blow-up of $\mathbb{P}^3$ along the union of two intersecting lines? auniket 2011-07-14T13:39:27Z 2011-07-14T13:39:27Z <p>To find the polytope associated to a toric variety directly you have to realize the variety as the closure of a map from the torus. In this case at least, it is not too hard to get such a description. Let the homogeneous coordinates of $\mathbb{P}^3$ be $[w:x:y:z]$ and the two lines be $C_1 := \lbrace w = x = 0 \rbrace$ and $C_2 := \lbrace w = y = 0 \rbrace$. Then the blow-up $B$ of $\mathbb{P}^3$ along $C_1 \cup C_2$ is the closure of <br> $\lbrace ([w:x:y:z], [w^2:wx:wy: wz: xy]) : [w:x:y:z] \in \mathbb{P}^3 \setminus (C_1 \cup C_2) \rbrace$ <br> in $\mathbb{P}^3 \times \mathbb{P}^4$. If we identify $\mathbb{C}^3$ with $\mathbb{P}^3 \setminus V(w)$, and write $X, Y, Z$ respectively for $x/w, y/w, z/w$, then $\mathbb{C}^3$ is embedded in $B$ via the map <br> $(X, Y, Z) \mapsto ([1:X:Y:Z), (1: X: Y: Z: XY])$ <br> Composing with the Segre embedding (and getting rid of duplicate coordinates), we get <br> $(X, Y, Z) \mapsto [1: X : Y: Z: X^2: XY: XZ: Y^2: YZ: Z^2: X^2Y: XY^2: XYZ]$ <br> Therefore the polytope is the convex hull of the exponents of these monomials. I believe its vertices are (0,0,0), (2,0,0), (0,2,0), (2, 1, 0), (1, 2, 0), (0, 0, 2) and (1,1,1). </p> <p>PS: There is a detail to be filled: blow-ups along singular subvarieties are not in general normal, so a priori $B$ might not be a normal toric variety (i.e. the polytope is associated not to $B$ but the normalization of $B$). But as David shows in his answer (and probably proved for a general torus invariant subspaces in the article he mentions in the comment), $B$ is indeed normal.</p> http://mathoverflow.net/questions/62627/intersection-of-curves-on-projective-toric-surface-and-some-enumerative-questions/62707#62707 Answer by auniket for Intersection of curves on projective toric surface and some enumerative questions auniket 2011-04-23T05:49:07Z 2011-04-23T05:49:07Z <p>The answer to your question B is yes <i> and </i> no :) You see: replacing $P$ by $kP$ for any $k \geq 1$ gives rise to the same toric surface (the latter is precisely the $k$-uple Veronese embedding of the former). And, the defining polynomial of any curve will fit (up to translation) in $kP$ for a sufficiently large $P$.</p> <p>For the phenomenon in your first question, I can give an answer in the case that the underlying field $\mathbb{k}$ is algebraically closed. Here it goes: let $f$ be the Laurent polynomial defining the curve $C$ and $S$ is an edge of the Newton Polygon $N$ of $f$. Let $\nu := (p, q)$ be an <i> outward pointing </i> (with respect to $N$) normal to $S$. For simplicity assume $q > 0$. Then there is a branch of $C$ with <i> degree-wise Puisuex series </i> of the form: $\gamma(t) = (t^p, \sum_{k = 0}^\infty a_k t^{q_k})$, where $q = q_0 > q_1 > \cdots$ are rational numbers with bounded denominators. Now let $\psi_P: X_P \to \mathbb{P}^N$ (where $N := |P \cap \mathbb{Z}^n| - 1$) be the embedding of the toric variety defined by the monomials in $P$, i.e. the restriction of $\psi_P$ to $(\mathbb{k}^*)^n$ is given by: $\psi_P(x) := [x^{\alpha_0}: \cdots : x^{\alpha_N}]$, where $P \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots \alpha_N \rbrace$. Let $Q$ be the face (i.e. edge or vertex) of $P$ such that $\nu$ is an outer normal to $Q$. W.l.o.g. assume that $Q \cap \mathbb{Z}^n = \lbrace \alpha_0, \ldots, \alpha_q \rbrace$, $q < N$. Then precisely the first $q+1$ coordinates of $x := \lim_{t \to \infty} \psi_P(\gamma(t))$ are non-zero, i.e. $x$ belongs to the subvariety of $X_P$ determined by $Q$.</p> <p>Remark: Usually in the books on toric varieties, <i> inner normals </i> are used instead of outer normals. That gives rise to a usual Puiseux series (the exponents being <i> increasing </i>, as opposed to the one in the preceding paragraph). But then the point at infinity (on the curve) is approached as $t \to 0$. I prefer that one approaches the point at infinity as $t$ approaches infinity as well.</p> http://mathoverflow.net/questions/59018/equations-defining-a-subvariety/59616#59616 Answer by auniket for equations defining a subvariety auniket 2011-03-25T22:03:42Z 2011-03-26T19:31:01Z <p>Note that $T := \tilde \phi(f^{-1}(Y))$ is a proper Zariski closed subset of $S$. Therefore, for generic $v_1, \ldots, v_s \in V$, $Z(v_1) \cap \cdots \cap Z(v_s) \cap T = \emptyset$. Consequently, in this case $P \cap Y = \emptyset$ and your question boils down to whether $v_1, \ldots, v_s$ generate the ideal of $Y$ (in a neighborhood of $Y$ in $X$), which should in general be false. Am I missing something?</p> <p>For example, let $X := Z(x_0^2x_2 - x_1^3) \subseteq \mathbb{P}^2$. The singular set of $X$ is $Y := \lbrace(0:0:1)\rbrace$ (with respect to homogeneous coordinates $(x_0: x_1: x_2)$ of $\mathbb{P}^2$). Let $L$ (resp. $V$) be the linear system with basis $x_0, x_1, x_2$ (resp. $x_0, x_1$). Then $S = \mathbb{P}^1$ and $T = \lbrace(0:1)\rbrace$. Therefore, if we take $v_1 := a_0x_0 + a_1x_1$ with $a_1 \neq 0$, then $Z(v_1) \cap T = \emptyset$ and consequently $P \cap Y = \emptyset$. Let $U$ be the affine neighborhood of $Y$ in $\mathbb{P}^2$ with coordinates $u_0 := x_0/x_2$ and $u_1 := x_1/x_2$. Then ideal of $Y$ on $U \cap X$ is $\mathcal{I} := \langle u_0, u_1 \rangle$ and the ideal generated by $v_1$ is $\mathcal{J} := \langle a_0u_0 + a_1u_1 \rangle$. Since the ideal in $\mathbb{C}[x,y]$ generated by $a_0u_0 + a_1u_1$ and $u_0^2 - u_1^3$ does <i> not </i> equal the ideal generated by $u_0$ and $u_1$, it follows that $\mathcal{I} \neq \mathcal{J}$. </p> <p><b> Edit: </b> The heuristics in the first paragraph remains valid in the case that $X$ is normal. Below I give an explicit example where $X$ is a normal surface. I don't know anything about secant varieties to comment about the validity of the statement in that case.</p> <p>Let $X$ be the weighted projective space $\mathbb{P}^2(1, 1, 2)$. We view $X$ as the toric surface corresponding to the polygon $\mathcal{P}$ which is the triangle in $\mathbb{R}^2$ with vertices $(0,0)$, $(2,0)$ and $(0,4)$. Let me draw $\mathcal{P} \cap \mathbb{Z}^2$.</p> <pre> | | | | | x-o-o-o-o- | | | | | x-o-o-o-o- | | | | | x-x-o-o-o- | | | | | x-x-o-o-o- | | | | | x-x-x-o-o- </pre> <p>Here I marked the integral points which belong to $\mathcal{P}$ by 'x' and the others by 'o' (the coordinates of the point at the bottom-left corner being $(0,0)$). Since $|\mathcal{P} \cap \mathbb{Z}^2| = 9$, it follows that $X$ is isomorphic to a subvariety of $\mathbb{P}^8$. Denote the homogeneous coordinates of $\mathbb{P}^8$ by $z_\alpha$ for all $\alpha \in \mathcal{P} \cap \mathbb{Z}^2$. Then the equations of $X$ in $\mathbb{P}^8$ determined by relations between $x_1^{\alpha_1}x_2^{\alpha_2}$ for all $\alpha := (\alpha_1, \alpha_2) \in \mathcal{P} \cap \mathbb{Z}^2$. </p> <p>Let $L$ be the linear system with basis $\lbrace z_\alpha \rbrace$ and $V$ be the subspace of $L$ with basis $\lbrace z_\alpha : \alpha \neq (2,0) \rbrace$. Then you can check that $Y := BS(V)$ (as a set) consists of the only singular point of $X$ and the blow-up $\tilde X$ of $X$ along $Y$ is non-singular. Moreover, $f^{-1}(Y)$ is a curve. Finally, $S$ is the toric surface corresponding to the polygon</p> <pre> | | | | | x-o-o-o-o- | | | | | x-o-o-o-o- Q := | | | | | x-x-o-o-o- | | | | | x-x-o-o-o- | | | | | x-x-o-o-o- </pre> <p>It follows that $S$ is non-singular, and $\dim (\tilde \phi(f^{-1}(Y))) \leq 1$. Therefore, for generic $v_1, v_2 \in V$, $Z(v_1) \cap Z(v_2) \cap \tilde \phi(f^{-1}(Y)) = \emptyset$, and consequently, $P \cap Y = \emptyset$. We claim that there is a neighborhood $U$ of $Y$ such that the ideal of $Y$ on $U$ can not be generated by $2$ elements. </p> <p>Indeed, let $U := X \setminus Z(z_{(2,0)})$. Then $U \cong \text{Spec}~ \mathbb{C}[x^{-1}, x^{-1}y, x^{-1}y^2] \cong \text{Spec}~ (\mathbb{C}[u,v,w]/\langle uw - v^2 \rangle)$ and $Y = Z(u,v,w) \subseteq U$. Since $uw - v^2$ is a homogeneous polynomial of degree $2$, the ideal generated by $u$, $v$ and $w$ in $\mathbb{C}[u,v,w]$ does not equal the ideal generated by $uw-v^2$, $g_1$ and $g_2$ for all $g_1, g_2 \in \mathbb{C}[u,v,w]$. This proves the claim and completes the counter example. </p> http://mathoverflow.net/questions/51180/are-there-non-projective-normal-surfaces-which-are-rational Are there non-projective normal surfaces which are rational? auniket 2011-01-05T06:26:43Z 2011-01-06T20:51:27Z <p>Every non-singular complete surface is projective. On the other hand, there are non-projective complete surfaces (see e.g. Excercise II.7.13 of Hartshorne) - and there are such examples where the surface is also normal (see e.g. <a href="http://reh.math.uni-duesseldorf.de/~schroeer/publications_pdf/on_non_proj.pdf" rel="nofollow"> this </a>). All the examples I have seen of complete normal non-projective surfaces are non-rational. Hence the question: are there (complete) rational non-projective normal surfaces?</p> <p>Edit: I just saw <a href="http://mathoverflow.net/questions/3624/nonprojective-surface" rel="nofollow"> a previous question </a> which asked for examples of normal non-projective varieties. So I guess this is a sub-question of that one.</p> http://mathoverflow.net/questions/47783/can-a-curve-intersect-a-given-curve-only-at-given-points Can a curve intersect a given curve only at given points? auniket 2010-11-30T11:21:54Z 2010-12-01T01:47:28Z <p>Clearly the question in the title has a positive answer for analytic (or smooth, or continuous ...) curves, but what about the algebraic category? More specifically, given an irreducible polynomial curve $X \subseteq \mathbb{CP}^2$ and points $P_1, \ldots, P_k$ on $X$, when can we find another curve $Y$ (defined by a polynomial) such that the $Y$ intersects $X$ only at $P_1, \ldots, P_k$? </p> <p>I find the question to be nontrivial even for $k = 1$. Here are some observations for $k = 1$ case:</p> <ol> <li><p>If $P$ is a point on $X$ with multiplicity $\deg X - 1$, then a tangent of $X$ through $P$ intersects $X$ only at $P$ (by Bezout's theorem).</p></li> <li><p>If $X$ is a rational curve and $X \setminus {P} \cong \mathbb{C}$, then there is a curve $Y$ such that $X \cap Y = {P}$.</p></li> <li><p>Let $X$ be a non-singular cubic. Give it a group structure such that the origin is an inflection point. Then for all $P \in X$, there exists $Y$ such that $Y \cap X = {P}$ iff $P$ is a torsion point in the group.</p></li> </ol> <p>If $X$ (of degree $d$) is non-singular at $P$, then the most direct approach for finding a $Y$ of degree $e$ intersecting $X$ only at $P$ seemed to blow it up $de$ times and look for the conditions under which $Y$ goes through each of the points on $X$ in the $i$-th infinitesimal neighborhood of $P$, $0 \leq i \leq de - 1$. But the conditions on the coefficients of the polynomial defining $Y$ did not appear very tractable.</p> <p><i> Edit: </i> I would like to make a correction to observation 3. This is what I know about a non-singular cubic curve $X$: If $P$ is an inflection point, then there is a curve $Y$ such that $Y \cap X = P$ (take $Y$ to be the tangent of $X$ at $P$). If $P$ is a non-torsion point (for the group structure on $X$ for which the origin is an inflection point), then there is no such $Y$. I don't know what happens for torsion points. </p> http://mathoverflow.net/questions/47428/cm-for-radical-ideal/47430#47430 Answer by auniket for CM for radical ideal auniket 2010-11-26T14:28:02Z 2010-11-26T14:28:02Z <p>Yes. From Eisenbud's Commutative Algebra: a ring $S$ is Cohen-Macaulay iff all the maximal ideal $m$ of $S$ satisfies codim($m$) = depth($m$). Now, the maximal ideals of $R/J$ are the same as $R/I$ and their depths and codimensions are the same as well.</p> http://mathoverflow.net/questions/37118/any-implemented-algorithm-to-compute-the-closure-of-an-affine-variety-in-a-produc Any implemented algorithm to compute the closure of an affine variety in a product of projective spaces? auniket 2010-08-30T06:16:57Z 2010-10-20T23:58:51Z <p>Let $I$ be an ideal of $k[x_1, \ldots, x_m, y_1, \ldots, y_n]$, $k$ being a field. Does any of the computer algebra systems implement any algorithm to calculate the generators of the 'bi-homogenization' $\tilde I$ of $I$ with respect to $x$ and $y$ variables? </p> <p>(Recall that the 'bi-homogenization' of a polynomial $f = \sum a_{\alpha, \beta} x^\alpha y^\beta$ is by definition $\tilde f := \sum a_{\alpha, \beta} x^\alpha y^\beta x_0^{d - |\alpha|} y_0^{e- |\beta|}$, where $x_0$ and $y_0$ are two new variables, $d := \deg_x(f)$ and $e := \deg_y(f)$. Then $\tilde I :=${$\tilde f: f \in I$}.)</p> <p>My motivation is geometric: to find the closure $\overset{-}{V}$ of a subvariety $V$ of $k^{m+n}$ in $\mathbb{P}^m \times \mathbb{P}^n$. Of course I could as well calculate the Segre embedding of $\overset{-}{V}$ in $\mathbb{P}^{mn + m +n}$, but I would like to have something computationally less expensive.</p> <p>I can think of an algorithm which involves introducing $n$ (or $m$, whichever is the smaller) new variables $t_1, \ldots, t_n$ and computing the monomial basis of an ideal $J$ in $k[x,y,t]$, where $J$ is to be constructed from $I$. But I was wondering if someone had already implemented some (possibly better) algorithm which would do this job. </p> http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/39937#39937 Answer by auniket for Jokes in the sense of Littlewood: examples? auniket 2010-09-25T08:16:22Z 2010-09-25T08:16:22Z <p>You can't prove something exists just by computing the probability of its existence - right? The first application of the <a href="http://en.wikipedia.org/wiki/Probabilistic_method" rel="nofollow"> "probabilistic arguments"</a> in Combinatorics I have encountered was <a href="http://books.google.com/books?id=KvQr9l0wgf8C&lpg=PP1&dq=proofs%20from%20the%20book&pg=PA82#v=onepage&q&f=false" rel="nofollow"> this;</a> took me a long time to get it. </p> http://mathoverflow.net/questions/37827/tensor-product-of-a-line-bundle-with-a-large-multiple-of-another-positive-line-bu Tensor product of a line bundle with a large multiple of another positive line bundle also positive? auniket 2010-09-05T21:31:25Z 2010-09-08T10:11:11Z <p>Let $X$ be a complex manifold and $\mathcal{L}$ be a positive line bundle on $X$. If $E$ is any other line bundle on $X$, then is it true that for all sufficiently large $m$, $\mathcal{L}^m \otimes E$ is also positive?</p> <p>When $X$ is compact, the answer is positive, and it follows by a standard compactness argument if you start with the definition that $\mathcal{L}$ is positive iff the Chern class $\omega$ of $\mathcal{L}$ satisfies: $\omega(x; v, Iv) > 0$ for all $x \in X$ and $v \in T_{\mathbb{R}, x}(X)$ (the real tangent space of $X$ at $x$) and $I: T_{\mathbb{R}, x}(X) \to T_{\mathbb{R}, x}(X)$ is the map induced by multiplication by $i$. </p> <p>So my real question is: is the above question true when $X$ is not compact? What if $X$ is an affine algebraic variety?</p> http://mathoverflow.net/questions/37082/appropriate-journal-to-publish-a-determinantal-inequality Appropriate journal to publish a determinantal inequality auniket 2010-08-29T20:41:56Z 2010-08-30T11:17:02Z <p>I have recently made the following observation:</p> <p>Let $v_i := (v_{i1}, v_{i2})$, $1 \leq i \leq k$, be <strike> non-zero </strike> <i> positive </i> elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m_{ij} := \max${$v_{ik}/v_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$. </p> <p>The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)? </p> <p>It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated. </p> <p><b> Edit: </b> Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.</p> http://mathoverflow.net/questions/27660/applications-of-compactness/27831#27831 Answer by auniket for Applications of compactness auniket 2010-06-11T16:02:28Z 2010-06-11T16:02:28Z <p><a href="http://books.google.ca/books?id=CbmaGqCRbhUC&printsec=frontcover&dq=fulton+toric+varieties&source=bl&ots=WGvezXIZHk&sig=NTCwwnU1ntXMwEH0Uo4Nz8qhYj8&hl=en&ei=o1gSTM2qFcP7lweJqPGPCA&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCEQ6AEwAg#v=snippet&q=gordon%27s&f=false" rel="nofollow">Gordan's lemma</a> is another application of "compact + discrete => finite", but it is one of the basic building blocks of the theory of toric varieties. <br></p> <p>In some sense the theory of division by polynomials (i.e. the Gröbner basis theory) is a manifestation of compactness, e.g. <a href="http://books.google.ca/books?id=E3sKOHH3990C&dq=ideals,+varieties,+and+algorithms&printsec=frontcover&source=bn&hl=en&ei=2VoSTKahKcOclgfUlvDpBw&sa=X&oi=book_result&ct=result&resnum=4&ved=0CCUQ6AEwAw#v=onepage&q=dickson%27s%20&f=false" rel="nofollow">Dickson's lemma</a> can be stated and proved as an application of compactness, see e.g. <a href="http://en.wikipedia.org/wiki/Dickson%27s_lemma" rel="nofollow"> the wikipedia entry</a>. I guess this is an example of the principle mentioned in the answer of Michael Greinecker.</p> http://mathoverflow.net/questions/27755/knuths-intuition-that-goldbach-might-be-unprovable/27761#27761 Answer by auniket for Knuth's intuition that Goldbach might be unprovable auniket 2010-06-11T01:39:48Z 2010-06-11T01:39:48Z <p>Again from combinatorics: <a href="http://en.wikipedia.org/wiki/Paris%E2%80%93Harrington_theorem" rel="nofollow"> The strengthened finite Ramsey theorem </a>. It was proved by Paris and Harrington that it is true, but not provable in Peano arithmetic. Wikipedia says: "This was the first "natural" example of a true statement about the integers that could be stated in the language of arithmetic, but not proved in Peano arithmetic". </p> http://mathoverflow.net/questions/24034/can-cantor-set-be-the-zero-set-of-a-continuous-function Can Cantor set be the zero set of a continuous function? auniket 2010-05-09T17:57:46Z 2010-05-11T07:11:38Z <p>More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?</p> <p>About two days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.</p> http://mathoverflow.net/questions/24059/is-the-absolutely-continuous-image-of-a-nowhere-dense-set-is-also-nowhere-dense Is the absolutely continuous image of a nowhere dense set is also nowhere dense? auniket 2010-05-10T04:27:16Z 2010-05-10T05:28:49Z <p>Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be an <i> <a href="http://en.wikipedia.org/wiki/Absolute_continuity" rel="nofollow">absolutely continuous</a> </i> map. Does $f$ map a nowhere dense subset of $[a, b]$ to a nowhere dense set?</p> <p>Remarks:</p> <ol> <li><p>The answer is "no" if $f$ is only assumed to be continuous and almost everywhere differentiable, e.g. take the <a href="http://en.wikipedia.org/wiki/Cantor_function" rel="nofollow"> Cantor function </a>. </p></li> <li><p>If $f$ is assumed to be $C^1$, then the answer is yes - a nice proof can be found at <a href="http://sci.tech-archive.net/Archive/sci.math.research/2005-02/0171.html" rel="nofollow">this page.</a> Essentially the same proof works if it is assumed that $f$ is not differentiable at at most countably many points. <b> Edit: </b> I would like to retract the previous sentence. Now I don't see why it should be true. <br></p></li> </ol> http://mathoverflow.net/questions/128951/reference-request-samuels-multiplicity-and-degree Comment by 2013-05-06T19:56:01Z 2013-05-06T19:56:01Z Hi Oleg, the comment of Steven Landsburg on this question: <a href="http://mathoverflow.net/questions/129673/interpretation-of-multiplicity-of-a-point" rel="nofollow" title="interpretation of multiplicity of a point">mathoverflow.net/questions/129673/…</a> implies that Samuel's book should have a reference to at least the first question of yours. http://mathoverflow.net/questions/129673/interpretation-of-multiplicity-of-a-point Comment by 2013-05-05T12:38:04Z 2013-05-05T12:38:04Z @Steven: Thanks! http://mathoverflow.net/questions/123375/contracting-a-curve-of-negative-self-intersection-on-a-surface Comment by 2013-04-27T01:07:02Z 2013-04-27T01:07:02Z Hi Philip, this article (of mine) gives a necessary and sufficient criterion for algebraicity in a special case: <a href="http://arxiv.org/abs/1301.0126" rel="nofollow">arxiv.org/abs/1301.0126</a> PS: I myself am interested in your Question 1, and I don't know of any other reference other than Grauert's original article, which is in German and therefore I can't read :( http://mathoverflow.net/questions/121817/a-question-on-counting-non-leading-monomials Comment by 2013-02-14T20:28:24Z 2013-02-14T20:28:24Z Is $a_{I,m}$ by definition $\lim_{k\to \infty} \sigma_{I,m}(k)/kh_I(k)$? If not, what is it? http://mathoverflow.net/questions/119702/normality-condition-on-graded-algebra/119707#119707 Comment by 2013-01-24T17:44:43Z 2013-01-24T17:44:43Z Hi Isac, A look at Chapter 5 of <a href="http://people.reed.edu/~iswanson/book/SwansonHuneke.pdf" rel="nofollow">people.reed.edu/~iswanson/book/SwansonHuneke.pdf</a> might help. http://mathoverflow.net/questions/117666/contracting-rational-curves-on-surfaces-and-getting-something-non-algebraic Comment by 2012-12-31T13:33:34Z 2012-12-31T13:33:34Z @Jason: OK, I admit it wasn't such a good choice of words :) Would you prefer if I change "something non-algebraic" to "non-algebraic surfaces"? http://mathoverflow.net/questions/117666/contracting-rational-curves-on-surfaces-and-getting-something-non-algebraic Comment by 2012-12-30T21:41:36Z 2012-12-30T21:41:36Z @Angelo: Thanks! But I knew of these (perhaps should have mentioned them in the question) and they do not contain (and as far as I can see, do not shed any light on the construction of) any such examples. http://mathoverflow.net/questions/116094/d-points-on-a-curve-which-are-in-the-base-locus-of-a-pencil-of-planes/116099#116099 Comment by 2012-12-12T20:26:24Z 2012-12-12T20:26:24Z Hi Francesco, the projection from a generic $p \in L$ may not be birational! The correct answer seems to be that either $L$ is a component of $C$, or there are hyperplanes $H_1, \ldots, H_m$ containing $L$ and curves $C_j \subseteq H_j$ of degree $d_j$ such that $1 \leq m \leq n$, $d_1 + \cdots d_m = d$ and $C = C_1 \cup \cdots \cup C_m$. http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers/22385#22385 Comment by 2012-12-09T18:29:42Z 2012-12-09T18:29:42Z Oops! I just voted up and destroyed the magic of 27! I am really sorry, now if I downvote it becomes 26! http://mathoverflow.net/questions/112232/typical-dimension-of-partial-derivatives/112265#112265 Comment by 2012-11-13T23:44:54Z 2012-11-13T23:44:54Z Oops! I can see that I was stupid. I will cross the answer out until I have something more useful to say. http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-pr/98646#98646 Comment by 2012-06-02T19:38:28Z 2012-06-02T19:38:28Z Thanks! This is pretty close to what I wanted. http://mathoverflow.net/questions/92338/isolated-solutions-of-a-polynomial-system Comment by 2012-03-27T05:49:33Z 2012-03-27T05:49:33Z Yeah! Careless mistake ... http://mathoverflow.net/questions/92338/isolated-solutions-of-a-polynomial-system Comment by 2012-03-27T05:32:36Z 2012-03-27T05:32:36Z $F$ is proper in a neighborhood of $F^{-1}(c)$ means that there is an open set $V$ containing $c$ such that $F$ restricted to $F^{-1}(V)$ is proper (i.e. for every compact subset $Z$ of $V$, $F^{-1}(Z)$ is also proper). E.g. $F := \mathbb{C}^2 \to \mathbb{C}^2$ be defined by $u = x^2y - x + y$ and $v = xy$ (where $(u,v)$ are the coordinates in the 'target'). Then $F$ is not proper at $F^{-1}(c)$ for every $c$ on the line $v = 1$. http://mathoverflow.net/questions/92338/isolated-solutions-of-a-polynomial-system Comment by 2012-03-27T04:52:32Z 2012-03-27T04:52:32Z Answer to 1) should be always true, I think. However, for 2), you need some sort of properness condition. More precisely, if $F$ is not proper in a neighborhood of $F^{-1}(c)$, then there is a curve $\gamma(t)$ such that as $t$ goes to infinity, $\gamma(t)$ goes to infinity and $F(\gamma(t))$ goes to $c$. Since there can be only finitely many isolated solutions of $F(x) = c$,your condition (2) will be violated. http://mathoverflow.net/questions/91351/properties-of-result-when-pre-and-post-multiplying-a-matrix-by-another-matrix Comment by 2012-03-16T06:23:36Z 2012-03-16T06:23:36Z No - any $m\times n$ matrix can be expressed as $BAB$ for some $m\times n$ matrix $B$ and $n\times m$ matrix $A$. In fact for each $B$ you can choose $A$ so that $BAB=B$ (to see it note that you can express any $B$ as $EJF$, where $E,F$ are invertible and $J$ is a block matrix with 4 blocks such that the left most block is a $p \times p$ identity matrix for some $p\leq\min(m,n)$ and the other blocks are zero matrices. Consequently $BAB=EJA'JF$, where $A'=FAE$. Choosing $A$ so that $A'$ has the "same" form as $J$ (of course the zero blocks have to have different dimensions) gives the result. | 17,633 | 54,397 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2013-20 | latest | en | 0.722891 |
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# Find the refractive index of a plastic block.
Extracts from this document...
Introduction
Name Jonathan Tam Class: 13A
Yew Chung International School Physics
Aim: Find the refractive index of a plastic block
Method:
I will use two methods to find the refractive index of plastic. The first one is to use the critical angle, and the other one is to use Snell’s Law.
Critical Angle method:
1. Set up the apparatus as shown. Put a piece of white paper under the apparatus. Turn off lights from the surroundings.
2. Put a black slid into the light box so that only a thin ray will be shone.
3. Turn on the light box. Adjust the angle of incidence so that there are no refracted rays.
Middle
to get refractive index of plastic.Repeat step 3 to 6 at least 7 times to get sufficient readings. Use a different angle of incidence each time.Compare the values obtained using the two methods.
Data Collection:
Critical angle method:
Critical Angle (± 0.5°) 47 48 47
Snell’s Law method:
Angle of Incidence (± 0.5°) Angle of Refraction (± 0.5°) 32 24 15 11 45 32 70 42 53 36 85 48 28 21
Conclusion
is 1.
0.530 0.407 0.259 0.191 0.707 0.530 0.940 0.669 0.799 0.588 0.996 0.743 0.469 0.358
The refractive index of plastic using Snell’s Law is: 1.344 ± 0.06
Conclusion and Evaluation:
From the two experiments, the refractive index of plastic is coherent to be between 1.344 to 1.36.
There are many systematic errors in these experiments, which came from the apparatus itself. These include:
• protractor
• light box
The protractor has a very high uncertainty of ± 0.5°. This has generated a high percentage error in our results (from 0.6% up to 3.3%). When we take sin for both angles and add the two percentage errors (or uncertainties) together, it has generated a whole range of refractive indexes.
However these results are acceptable because they both fall within the range of uncertainties (i.e. no contradictions). Therefore these experiments, although carries a high percentage error, can still be considered as valid.
This student written piece of work is one of many that can be found in our GCSE Waves section.
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# Related GCSE Waves essays
1. ## Investigation:To find the refractive index of cooking oil.
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https://webtoolsmate.com/cpm-calculator | 1,722,658,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00323.warc.gz | 503,433,096 | 111,303 | # CPM Calculator
\$
## What Is A CPM Calculator? Why Should You Use It?
Although you have been running your advertising campaigns quite effectively, how will you determine whether they are successful?
CPM is what you should refer to. You may use a CPM calculator to determine how well your ad campaign is doing and whether you've made any progress since you launched it.
This article will provide a complete guide to the CPM calculator and beneficial information about the tool. Let's read on and learn how to optimize your advertising budget!
## What Is A CPM Calculator?
CPM stands for Cost Per Mile. "Mile" means 1,000 in French, so Cost Per Mile is Cost Per Thousand Impressions.
CPM is a crucial advertising statistic that reveals how much advertisers pay publishers for every 1,000 ad impressions.
You need a CPM calculator to monitor the costs associated with your digital marketing campaigns that run on Facebook Ads, LinkedIn Ads, and Google AdWords.
The first stage in creating your advertising campaign is to establish a budget. Let's use a hypothetical \$10,000 expenditure as an example to help you comprehend the formula.
### The target number of impressions
The next step is determining how many impressions your campaign should generate. Have a look at a movement whose objective is 200,000 impressions.
### CPM formula
We calculate the CPM by dividing the total cost of your campaign by your desired impressions and multiplying the result by 1,000.
So the CPM formula is: CPM = Total Amount Spent / Total Impressions x 1,000
With the expenditure of 10,000 and the expected number of impressions is 200,000, we will have a CPM of 50.
By dividing \$10,000 by 200,000 impressions, you may spend \$10,000 on a project and get 200,000 impressions. The result is 50 when you multiply that value by 1,000.
It implies that for a project with a \$10,000 budget, you would spend \$50 per 1,000 impressions.
## Benefits Of Using Online CPM Calculators
You can use the formula we mentioned above to calculate your CPM model. But things may get more complicated since you will continuously work with up to thousands of impressions and adjust your advertising budgets.
We highly recommend online CPM calculators to simplify this task for you. Here are some reasons you want to opt for an impression calculator.
### Free
You can find web-based CPM calculators without having to download or install them. More interestingly, the tools are free.
Whenever you want to check your CPM model, access the website, insert the relevant values, and the cost per mile will appear.
### Security
Online marketers will love such tools because they are free of risks. They don't ask you to register or sign in, so you will work anonymously. The tools can't track or steal your IP as a result.
### User-friendly
Measuring how much advertisers pay for their impressions is super easy on the site. You need to tell them your total budget for the campaign and the number of images you have, and there you go. The tools will do everything automatically.
Moreover, some calculators use an advanced mode to compare and analyze the costs of two campaigns.
### No spam
Some websites allow ads, which may lead to suspicious pages and hack your account. Fortunately, you won't encounter any ads on online CPM calculators. Hence, your activities will be smooth and safe.
## How To Put CPM Into Use?
The CPM allows advertisers to analyze, compare, and improve their campaigns based on the reported data. You can use the results for multiple purposes, such as:
### Calculate the estimated cost of an ad campaign
The one who sells ad space establishes a CPM rate. Yet, you may calculate the cost of your marketing campaign using this formula for a specific amount of ad impressions, as we mentioned in the previous section.
### Calculate potential audience with your budget
You may decide whether the prospective audience is worthwhile if you have a defined rate and spending limit.
The formula is: Potential Audience = (Total Cost x 1,000) / CPM
For example, with a CPM of 10, a \$50,000 budget may give 5,000,000 impressions.
### Sell your space
If you run a website and intend to earn money from advertising, you can calculate the CPM based on the traffic to your website, and the sum of money advertisers are ready to pay to access that audience.
In digital marketing, a tool like Google Analytics can perform this step automatically. It will value your advertising space and sell it to those who place the highest bids.
### Maximize ad cost-benefit
CPM rates are handy for businesses and marketers trying to target the widest audience for the lowest advertising expense. Due to its flexibility, CPM may compare prices across multiple media and locations.
Of course, other elements like demographic data and ad accessibility will affect a marketing campaign's effectiveness. But for a starting point for cost analysis, CPM works best.
## What Is A Good CPM?
To determine a good CPM for your ad campaigns, you need to consider multiple factors, including:
• Previous advertising performance
• Industry-relevant ad data
It needs time to compile enough data to analyze the effectiveness of your ads, as is the case with most online advertising.
You may gather more data and refine your advertising strategies by conducting numerous online ads using multiple marketing tactics.
For CPM, lower isn't always better. A lower rate leads to lesser-quality traffic and maybe worse conversion rates.
On the other hand, a high CPM does not necessarily translate into leads or conversions of good quality. Online marketers can consider multiple variables when planning their online ads, including CPM.
## Factors That Affect CPM
Various elements impact the value of CPM. The following are some of the most critical aspects to consider while assessing CPM:
### User location
The region where you run your ads plays a vital role. Ads displayed in English-speaking, high-GDP countries often have a higher CPM than those done in other areas.
Larger formats can bring in a higher CPM than smaller ones. Advertisers recommend these sizes for ads run on different devices:
• Leaderboard: 720 x 90
• Half-page size: 300 x 600
• Mobile banner: 320 x 100
• Square: 300 x 250
Above-the-fold advertisements attract more people and generate higher CTR (Click Through Rate). As a result, their reach will be more impressive than small ads in the center or at the end of the content.
### Precious ad effectiveness
A webpage with a track record of delivering conversions and a high ROI might demand a higher cost. Publishers might request a higher CPM if they can demonstrate the value of their ad space.
### Page depth
Marketers will pay more to appear on page three of a newspaper than on page thirty. Similarly, it would be best if you spent more to occur early in the session than later in your user journey.
## Frequently Asked Questions
### 1. How to improve CPM?
Here are some proven ways that can enhance the CPM value:
• Create high-quality content
• Improve floors
• Team up with the demand side
• Experiment with different ad units
• Add new units
• Focus on SEO
### 2. How do you calculate CPM on a calculator?
To calculate the CPM formula, divide the ad campaign's total budget by the number of impressions, then multiply it by 1,000.
### 3. Why do we calculate CPM?
The CPM can benefit your business because it helps you increase sales conversion rates. Advertisements with high conversion rates often perform well when the advertiser pays per thousand views.
Moreover, optimizing CPM is an excellent way to create or raise the value of your business and make it stand out.
### 4. Is higher CPM better?
No. A high CPM rate often indicates that your campaign is underperforming and you need improvement to increase the number of impressions for your ads.
### 5. What is the average CPM?
The average CPM is the sum of all the CPMs for your different ads. Facebook and Instagram's average CPM for social media advertising in 2021 is approximately \$9, whereas this figure for Twitter and Linkedin is roughly \$6.50.
## Conclusion
CPM reveals the cost per thousand impressions, affecting the advertiser's ability to attract new leads and increase sales. This critical metric can also help you learn about your audience and value your ad space.
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## Geodesics on spheres are great circles
How does one prove that on $S^n$ (with the standard connection) any geodesic between two fixed points is part of a great circle?
For the special case of $S^2$ I tried an naive approach of just writing down the geodesic equations (by writing the Euler-Lagrange equations of the length function) and solving them to gain some insights but even if the equations are solvable I can't see how to show that they are great circles. (the solutions are some pretty complicated functions which don't give me much insight)
I checked the article on Great Circles on Wolfram Mathworld for a coordinate geometry approach to it but that article looked quite cryptic to me!
One knows that on compact semi-simple lie groups any one-parameter subgroup generates a geodesic and $S^n$ is the quotient of 2 compact semi-simple lie groups $SO(n+1)/SO(n)$. Is this line of thought useful for this question?
================================================================================= After some of the responses came let me put in "a" way of seeing the above for $S^2$ (wonder if it is correct). If $\theta$ and $\phi$ are the standard coordinates on $S^2$ then the equations for the curve are
$$\ddot{\theta} = \dot{\phi}^2 sin(\theta)cos(\theta)$$ $$\dot{\phi}sin^2{\theta} = k$$
where $k$ is some constant set by the initial data of the curve.
Now given the initial point I can choose my coordinate system such that the the initial data looks like $\dot{\phi}=0$, $\theta = \text{some constant}$, $\dot{\theta}=\text{some constant}$, $\phi = \text{some constant}$. Then the differential equations tell me that the $k=0$ and the only way it can happen for times is by having ,
$$\dot{\phi} = 0$$
Which clearly gives me a longitude in this coordinate system. Hence the geodesic equation gives as a solution a great circle.
Surely not an elegant proof like Bar's reference.
But I hope this is correct.
{As a friend of mine pointed out that this set of coordinates is motivated by the fact that the way the "energy" of the curve is being parametrized the z-component of the angular momentum is conserved which is in fact my second Euler-Lagrange equations}
-
By and large you save yourself a lot of time and effort by avoiding coordinates. Yes, your argument in coordinates works too but notice that you had to pick a very special coordinate system to make the argument. – Ryan Budney Jan 18 2010 at 18:45
In fact, the best co-ordinates to use are just the standard Cartesian co-ordinates on R^{n+1}. The Euler-Lagrange equations are much simpler to understand. Also, they work in any dimension. I find using spherical co-ordinates in higher dimensions very painful. – Deane Yang Jan 19 2010 at 19:16
Although Jose has made the essentially the same point, I just want to elaborate (this is really just a comment, but I always run out of room in the comment box).
What nobody else has mentioned explicitly is that you should have trouble solving the Euler-Lagrange equation for the length functional. The Euler-Lagrange equation is a second order ODE, but a highly degenerate one. And you know this before you even start. Why? Well, suppose you have a solution. Then if you reparameterize that curve using any arbitrary parameterization (i.e., any monotone function of the original parameter), the newly parameterized curve is still a solution to the Euler-Lagrange equation. That means that the ODE has an infinite dimensional space of solutions and it is nothing like any ODE we learned about in our ODE courses or textbooks.
A trick is needed to get around this, namely to use the so-called energy functional $E[\gamma] = \int_0^1 |\gamma'(t)|^2 dt$ (which is not invariant under reparameterization of the curve) instead of the length functional (which is). The Holder inequality shows that a minimum of the energy functional is necessarily a minimum of the length functional that is parameterized by a constant times arclength, i.e. a constant speed geodesic.
The Euler-Lagrange equation for the energy functional is a nice nondegenerate 2nd order ODE that can be handled by standard ODE techniques and theorems.
As for the standard sphere, there are many different ways to solve for the geodesics. To review the ways already suggested in other answers:
1) I recommend that you first do it without the machinery of Riemannian geometry and using only the Euclidean structure of $R^{n+1}$. Using the discussion above, you should be able to show that a curve on the unit sphere is a constant speed geodesic if and only if its acceleration vector is always normal to the sphere. You should then be able to work out the solutions to this ODE. The suggestion that you assume one point is the north pole and the other lies in a co-ordinate plane is a good one and makes the ODE easy to solve.
2) The other way is to do it all intrinsically. Here, I recommend using stereographic co-ordinates and assuming one point is the origin in those co-ordinates. Again, everything becomes very easy in that situation.
3) And the third way is to view the sphere as a homogeneous space and use formulas for that situation. I don't remember the details myself, but I learned them from the book by Cheeger and Ebin.
I recommend that you work through all 3 different ways, as well as any other way you can find.
As others have noted, the calculations for geodesics on hyperbolic space are identical, except that you are working with a "unit sphere" in Minkowski instead of Euclidean space. There is even a notion of stereographic projection (but onto what?). This is also fun to work out carefully.
Finally, I do want to note that after you work this all out and have it all in your head, it's a really beautiful picture and story. And if you find the right angle, it's all very simple, so you can work out the details yourself and not rely on reading a book line-by-line or having someone else show you all the details. Try to get the essential ideas and necessary tricks (like using the energy functional) from books, lectures, or teachers, but try to work everything else out from scratch (i.e., minimal reliance on theorems you can't prove yourself).
-
There are direct arguments as well -- many textbooks have standardized arguments that the shortest curve in Euclidean space connecting two points is a straight line. The primary tool is the triangle inequality. You could do the same on the sphere, using the sphere's intrinsic metric. Alternatively, there are cute proofs using the Cauchy-Crofton theorem for spherical, euclidean and hyperbolic geometry. – Ryan Budney Jan 18 2010 at 18:52
Agreed! I tend to shortchange direct geometric arguments, but they are definitely worth learning, too. – Deane Yang Jan 18 2010 at 19:46
Its very gratifying to get back such detailed expository answers! Thanks a lot. I was doing the Euler-Lagrange equations on the function $L = \dot{\theta}^2 + sin^2(\theta)\dot{\phi}^2$. Isn't that the same as doing it on the energy functional as you have suggested since the Energy Functional as you state is integral of my $L$. Your suggestion of calculation 1 is something I had done earlier. But how does that help in proving great circles are geodesics? Is there any canonical way of parameterizing the great circles as unit speed curves? (same was the spirit of Jose's answer) – Anirbit Jan 19 2010 at 6:04
I like to get my differential geometry students to prove great circles are length-minimizing using the Cauchy-Crofton theorem. That's kind of over-the-top but it's an entertaining argument. It also works in Euclidean and hyperbolic geometry, like the symmetry argument of Mariano's or, oh, that's the same as my comment to Jose's reply, too. – Ryan Budney Jan 19 2010 at 6:42
It is enough to show that the geodesics through a specific point are all of that form, and we can do this just for the north pole $N=(1,0,\dots,0)$, as the isometry group of the sphere is transitive. Moreover, we need only consider one unit tangent vector at $N$, for the stabilizer of $N$ in the isometry group of the sphere acts transitively on the unit vectors in the tangent space at $N$. So let $v=(0,1,0,\dots,0)$ be the initial speed of a geodesic $\gamma$ starting at $N$. Since the map $(x_1,\dots,x_n)\mapsto (x_1,x_2,-x_3,\dots,-x_n)$ preserves both the point and the vector, the geodesic $\gamma$ must be also preserved by it. It follows immediately that the curve is contained in a great circle.
-
A simple argument that shows that great circles are geodesics is that if you parametrise them in such a way that they have unit speed, their acceleration is normal to the surface. To show that all geodesics are great circles, just use uniqueness of the initial value problem for a geodesic after noticing that through every point and in any direction there is a great circle.
-
Another argument like this that I like is that the sphere has an isometric involution that fixes a given great circle and acts as the antipodal map on the normal bundle to the great circle. So the great circle can't have any normal curvature (as normal curvature is preserved by an isometry), so it's a geodesic. – Ryan Budney Jan 18 2010 at 18:46
For $S^2$, I would consider taking two points on the equator, and seeing if that is easier to solve. Since you can use isometries to move two arbitrary points onto the equator, that should prove it for $S^2$. For $S^n$ the same should also work.
It's the same trick that is used to find the geodesics in the upper half-plane with the hyperbolic metric, really. You compute the geodesics between two points lying on a vertical line, then use isometries to find the rest of them.
-
Actually it is pretty easy even to solve the Euler-Lagrange equations in the case of S^n. Please check example 7.3 in the following reference, where the sphere S^n is parameterized by unit vectors in R^(n+1).
-
I suppose you mean example 7.13 ? – Anirbit Jan 18 2010 at 16:50
This is Jose's argument. – Ryan Budney Jan 18 2010 at 18:42 | 2,417 | 10,167 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2013-20 | latest | en | 0.894453 |
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# Let $f\left( x \right) = {\sin ^4}x + {\cos ^4}x.$ Then f is an increasing function in the interval:$a.{\text{ }}\left[ {\dfrac{{5\pi }}{8},{\text{ }}\dfrac{{3\pi }}{4}} \right] \\ b.{\text{ }}\left[ {\dfrac{\pi }{2},{\text{ }}\dfrac{{5\pi }}{8}} \right] \\ c.{\text{ }}\left[ {\dfrac{\pi }{4},{\text{ }}\dfrac{\pi }{2}} \right] \\ d.{\text{ }}\left[ {0,{\text{ }}\dfrac{\pi }{4}} \right] \\$
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Hint: Check the graph of first derivative of the given function
Given equation is $f\left( x \right) = {\sin ^4}x + {\cos ^4}x.................\left( 1 \right)$
We know the function is increasing if its differentiation is greater than or equal to zero.
I.e.$f'\left( x \right) \geqslant 0$ so, differentiate equation 1 w.r.t.$x$
$\Rightarrow f'\left( x \right) = 4{\sin ^3}x\dfrac{d}{{dx}}\sin x + 4{\cos ^3}x\dfrac{d}{{dx}}\cos x \\ \Rightarrow f'\left( x \right) = 4{\sin ^3}x\left( {\cos x} \right) + 4{\cos ^3}x\left( { - \sin x} \right) \\ \Rightarrow f'\left( x \right) = 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right) \\$
As we know$2\sin x\cos x = \sin 2x$, and${\cos ^2}x - {\sin ^2}x = \cos 2x$, so apply this
$\Rightarrow f'\left( x \right) = - 2\sin 2x\cos 2x = - \sin 4x$
But for increasing function $f'\left( x \right) \geqslant 0$
$\Rightarrow - \sin 4x \geqslant 0 \\ \Rightarrow \sin 4x \leqslant 0 \\$
As we know $\sin x$is zero at $\left( {0,{\text{ }}\pi ,{\text{ }}2\pi } \right),$in the interval between $\left[ {0,2\pi } \right]$
So, in $\sin x$graph $\sin x$is less than or equal to zero in between $\left[ {\pi ,2\pi } \right]$
$\Rightarrow 4x \in \left[ {\pi ,2\pi } \right] \\ \Rightarrow x \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right] \\$
Hence, option $c$ is correct.
Note: - In such a type of question the key concept we have to remember is that for increasing function the differentiation of function w.r.t. the variable is always greater than or equal to zero, then simplify this we will get the required answer and the required answer is the shaded region in the figure. | 815 | 2,183 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-50 | latest | en | 0.539173 |
https://www.teacherspayteachers.com/Product/Multi-Digit-Multiplication-Game-for-3rd-4th-5th-6th-Grade-2488320 | 1,537,685,160,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159160.59/warc/CC-MAIN-20180923055928-20180923080328-00110.warc.gz | 870,917,461 | 26,922 | # Multi Digit Multiplication Game for 3rd, 4th, 5th, 6th Grade
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Product Description
Multi Digit Multiplication: These multi digit multiplication cootie catchers are a great way for students to have fun while they practice their skills with multi digit multiplication. How to Play and Assembly Instructions are included.
Multi Digit multiplication Cootie Catchers Contents:
There are 2 cootie catchers in this product, each one having 8 problems for a total of 16 problems involving multi digit multiplication. The first cootie catcher has 3 digit numbers multiplied by 2 digit numbers. The second cootie catcher has 4 digit numbers multiplied by 2 digit numbers. Step-by-Step answers are included.
Important: If you enjoyed this product, check out my other Math Cootie Catchers:
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\$2.25 | 1,515 | 5,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-39 | latest | en | 0.827457 |
https://brainly.com/question/276033 | 1,484,899,664,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280801.0/warc/CC-MAIN-20170116095120-00503-ip-10-171-10-70.ec2.internal.warc.gz | 796,906,350 | 8,738 | 2015-01-27T21:47:35-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Pick any number. Multiply the 7 and the 11 both by it,
and you'll have an equivalent ratio.
Examples:
(by 2) . . . 14 to 22
(by 6) . . . 42 to 66
(by 10). . . 70 to 110
(by a million) . . . 7 million to 11 million | 157 | 556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | latest | en | 0.93473 |
http://www.miltonline.com/2013/12/27/maximally-even-library/ | 1,611,743,087,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00556.warc.gz | 150,954,193 | 15,952 | # Maximally Even Library
UPDATE- Since the publication of Brad Osborn’s Kid Algebra (2014), I’m going to switch to his category of Euclidean rhythms (in their 4 types) to describe the patterns below. In summary, Euclidean rhythms (ER) are rhythms in which k onsets in n divisions are as similar as possible, which essentially means that they will only differ by at most one subdivision each. So in ER the groups are as similar as possible, but the term maximally even we will reserve for ER rhythms where the smaller note groups are as separated as much as possible. For example, (2,2,3,3) and (2,3,2,3) are both ER, but only the latter is maximally even.
This is a library of all the maximally even (including strictly even) rhythms for 2-7 rhythmic onsets within 6,8,12 and 16 beat cycles.
Maximal evenness (M.E.) describes a rhythm which is as evenly spread out as possible given both a number or events (rhythmic onsets), and a number of available slots (beats). Strict evenness (marked with a º) is a subset of M.E. and occurs when the hits are equally spaced. M.E. rhythms are intrinsic to much music making in a wide range of cultures from Sub-saharan Africa, South America to EDM and much in between.
The parenthesised number shows the number of displacements (or ‘rotations’) available for the rhythm in the beat-cycle, and allows for starting on rests. When the number of rotations equals the number of beats in the cycle this is marked with an * and represents maximally independence (MI – a common trait of African timelines and clave patterns). Note that 5,6 and 7 in 12 also represents maximally even pentatonic, hexatonic and heptatonic scale sets e.g. 3,3,2,2,2 represents all the modes of the major pentatonic as well as a 5 in 12 set of ME rhythms. As another example 2,2,1,2,2,1,2 (a rotation of 2,2,2,1,2,2,1) represents both the African standard time-line and the Mixolydian mode. Enjoy.
(Visited 225 times, 1 visits today)
Social tagging: > > > > > > > > > > | 512 | 1,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-04 | latest | en | 0.94022 |
https://convertingcolors.com/hex-color-00E5DD.html | 1,539,677,867,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510415.29/warc/CC-MAIN-20181016072114-20181016093614-00544.warc.gz | 661,120,984 | 8,279 | # Converting Colors
## Color Conversions
Get Conversions and Harmonies as JSON or XML.
## Color Details
The Hex color 00E5DD can be considered as a dark color, the websafe version is hex 33FFFF. A complement of this color would be FFAABE and the grayscale version is 9F9F9F.
A 20% lighter version of the original color is 71FFFF and 00ADA6 is the 20% darker color. If you saturate the color by 10% you get 00E5DD and if you desaturate by 10% it is 17E5DE.
### Distribution
These gradients show how the Hex color 00E5DD changes by changing the brightness by 10 percent. The first row shows a change by +10% for each color and the second row -10%.
00E5DD 4AFFFA 71FFFF 93FFFF B4FFFF D4FFFF F4FFFF FFFFFF
00E5DD 00C8C1 00ADA6 00918C 007772 005D5A 004443 002E2C 001118 000000
Similar to the brightness gradients but the following show a change of the Hex color 00E5DD by changing the saturation by 10% instead.
00E5DD 17E5DE 2EE5DF 45E5DF 5CE5E0 73E5E1 89E5E2 A0E5E3 B7E5E3 CEE5E4
## Color Harmonies
### Analogous
The analogous color harmony consists of three colors that are next the each other on the color wheel.
6EE3AF 00E5DD 00E2FF
The triadic color harmony groups three colors that are evenly spaced from another and form a triangle on the color wheel.
00E5DD E7BCFF FFC179
### Complementary
The complementary color scheme is a pair of colors which are on the opposite of each other on the color wheel.
00E5DD FFAABE
### Split Complementary
Split-complementary colors differ from the complementary color scheme. The scheme consists of three colors, the original color and two neighbours of the complement color.
FFB295 00E5DD FFAEEB
### Square
The square scheme is like the rectangle color scheme, but the four colors are evenly spaced on the color wheel.
### Rectangle
The rectangle color scheme consists of four colors that form a rectangle on the color wheel.
00E5DD 1FDDFF FFAABE FFBC80
### Sweetspot
The sweetspot groups the original color and five complimentary colors.
00E5DD B3FFFC 0BE500 52807E 000000 808080
## Color Images
A selection of SVG images using the hex version #00E5DD of the current color.
## Color Preview
### White Background
This preview shows how the Hex color 00E5DD looks on a white background.
### Black Background
This preview shows how the Hex color 00E5DD looks on a black background.
### Hex 00E5DD Background
This preview shows how black text looks on a background with the Hex color 00E5DD.
This preview shows how white text looks on a background with the Hex color 00E5DD.
## Color CSS
The css property to change the color of a text to Hex 00E5DD is called "color". The color property can be set on classes, ids or directly on the html element `.text, #text, p{ color:#00E5DD }`.
This example shows how text in the color #00E5DD looks like.
If you want to add a text shadow in that color use `.shadow{ text-shadow: 4px 4px 2px #00E5DD }`.
Here you see how black text with a 4 pixel #00E5DD colored shadow looks like.
### Border
This example shows the color as border, it can be applied via the css property "border" or "border-color".
The css property to change the border of a element to Hex 00E5DD is called "border". The border property can be set on classes, ids or directly on the html element `.border, #border, table{ border:4px solid #00E5DD }`.
If only the border color should be changed `.border{ border-color:#00E5DD }` can be used.
Here you see how a box with a 4 pixel #00E5DD colored shadow looks like.
If you want to add a box shadow in that color use `.boxshadow{ -moz-box-shadow:4px 4px 4px 4px #00E5DD; -webkit-box-shadow:4px 4px 4px 4px #00E5DD; box-shadow:4px 4px 4px 4px #00E5DD }`.
### Background
The css property to change the background color of a element to Hex 00E5DD is called "background". The background property can be set on classes, ids or directly on the html element `.background, #background, body{ background:#00E5DD }`.
If only the background color should be changed `.background{ background-color:#00E5DD }` can be used.
This example shows the color as background, it is applied via the css property "background".
To optimize and compress your css code you can use our online css compressor and optimizer based on csstidy. | 1,188 | 4,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.776093 |
https://thedevnews.com/producing-a-desk-from-stream-output/ | 1,721,843,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00821.warc.gz | 487,510,950 | 48,861 | Wednesday, July 24, 2024
HomeC ProgrammingProducing a Desk from Stream Output
# Producing a Desk from Stream Output
This month’s Train looked for and output the primary 100 cyclops numbers. However as an alternative of spewing them out in a protracted column, my resolution set them in a desk. The values marched throughout the display in neat rows. This trick is relatively straightforward to perform, however in my resolution I needed some flexibility with the column quantity.
I took code from the related a part of my Train resolution and rebuilt it right here to output sequential values:
### 2022_09_10-Lesson.c
```
#embrace <stdio.h>
int essential()
{
const int columns = 9;
const int values = 100;
int x,y,depend;
x = 0;
depend = values;
whereas( depend )
{
for( y=0; y<columns; y++ )
{
printf("%3d",(x*columns)+y );
count--;
if( !depend )
break;
if( y<columns-1)
putchar('t');
}
x++;
putchar('n');
}
return(0);
}```
The variety of `columns` is ready as a relentless, as is the `values` depend. I needed the desk output to deal with a variable variety of columns, but additionally to take care of a state of affairs when the final row doesn’t have a worth for every column.
The nested for loop handles the values throughout a row. The important thing to outputting the values sequentially is discovered within the printf() assertion: `(x*columns)+y)` This expression bases every row’s values on variable `x`, which is the row depend, occasions the whole variety of columns. Including the `y` worth completes the duty, as every worth is output sequentially, left-to-right, top-down, as proven in Determine 1.
Determine 1. The output displaying 100 values in 9 columns.
The expression `x*columns` increments the beginning (first column) worth by the worth of columns: 0, 9, 18, and so forth. The remainder of the row then falls into place based mostly on the worth of variable `y`.
This code is constructed in order that the column quantity might be adjusted with out requiring the loops to be rebuilt. If I set the worth of fixed `columns` to 5, the output adjustments as proven in Determine 2.
Determine 2. Desk output when the `columns` fixed is ready to 5.
When the rows and columns are even, as proven in Determine 2, the final row is full. However in Determine 1, you see just one merchandise on the ultimate row. The code that handles that is an if choice after the printf() assertion outputs the worth:
`if( !depend ) break;`
Variable `depend` is decremented after output. If its worth is zero, values are exhausted and the row can finish: The for loop stops and a newline is output. As a result of `depend` is zero, the outer whereas loop stops as properly.
Desk output raises the issue degree over simply spewing out every quantity on a line by itself. This methodology, rows throughout, is less complicated than coding a desk the place the values are set in columns marching left-to-right. I cowl such a desk output in subsequent week’s Lesson.
RELATED ARTICLES | 708 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.860347 |
http://forums.wolfram.com/mathgroup/archive/2001/Aug/msg00162.html | 1,716,684,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00729.warc.gz | 11,912,423 | 7,843 | Input in batch script
• To: mathgroup at smc.vnet.net
• Subject: [mg30296] Input in batch script
• From: Carlo Palmisano <palmisano at to.infn.it>
• Date: Wed, 8 Aug 2001 01:33:53 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
Question (partially Mathematica, partially Linux):
We have to pass two values (one integer and one string)
to a mathematica foo.m running in batch.
The integer is simply used in a sum;
the string has to be concatenated into a file name (Output)
^^^^^^^^^^^^
Let's imagine to have the following trivial program "foo.m"
which requires two inputs (one integer and one string)
from the keyboard
BUT has to run in **batch** (Linux with Mathematica 4.1).
*** foo.m begins ****
Clear["Global`*"]
a = 1;
number = Input["Enter number, e.g., 99 : "];
sumnumber = a + number;
sumnumber >> foo1.out
string = Input["Enter the string, e.g., 1212_800 : "]
string >>> foo1.out
m1 = MemoryInUse[]
sol = NDSolve[{y''[x] - y'[x] + y[x] ==0,y[0]==1,y'[0]=1},y[x],{x,0,3}]
m2 = MemoryInUse[]
m3 = (m2-m1) >>> foo???????.out
(*We have to replace foo???????.out with foo1212_800.out
where 1212_800 is read from the following batchfoo.com *)
**** foo.m end ***********
The relevant batchfoo.com (executable)
contains
math <<end1 >>foo.out
<<foo.m
99
"1212_800"
end1
The trouble lies in the line
m3 = (m2-m1) >>> foo??????.out
What shall we have to do in order to get
foo1212_800.out ?
Can you tell us how to modify the foo.m and the script ?
Several trials with, e.g., ToString failed.
Carlo Palmisano
University of Turin
via P. Giuria 1 - 10125 Torino - Italy
e-mail: palmisan at to.infn.it
```
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• Next by thread: Re: Input in batch script | 573 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.781654 |
http://mathhelpforum.com/algebra/13361-solve-systems-substitution.html | 1,480,994,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541876.62/warc/CC-MAIN-20161202170901-00200-ip-10-31-129-80.ec2.internal.warc.gz | 161,264,265 | 10,048 | # Thread: Solve systems by substitution
1. ## Solve systems by substitution
These substitution equation is really frustrating me what is the purpose of these equation I don't see how they are part of every day life.
Anyway any help on this equation is very helpful.
8x – 4y = 16
y = 2x – 4
2. Originally Posted by Patience
These substitution equation is really frustrating me what is the purpose of these equation I don't see how they are part of every day life.
Stop complaining. You should see the math I do. I purposely choose it so that I can have no meaning in real life. Think of it as abstract expressionism.
Anyway any help on this equation is very helpful.
8x – 4y = 16
y = 2x – 4
8x-4y=16
y=2x-4
Thus,
8x-4(2x-4)=16
8x-8x+16=16
16=16
Which is true.
That means there were infinitely many solutions to that equation.
Thus, if x=t (some number) then the second equation tells us that,
y=2t-4.
Thus,
x=t
y=2t-4.
For any number t. | 270 | 951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-50 | longest | en | 0.962794 |
https://www.airmilescalculator.com/distance/sbt-to-scw/ | 1,603,455,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00407.warc.gz | 616,331,536 | 7,695 | # Distance between Sabetta (SBT) and Syktyvkar (SCW)
Flight distance from Sabetta to Syktyvkar (Sabetta International Airport – Syktyvkar Airport) is 877 miles / 1411 kilometers / 762 nautical miles. Estimated flight time is 2 hours 9 minutes.
## Map of flight path from Sabetta to Syktyvkar.
Shortest flight path between Sabetta International Airport (SBT) and Syktyvkar Airport (SCW).
## How far is Syktyvkar from Sabetta?
There are several ways to calculate distances between Sabetta and Syktyvkar. Here are two common methods:
Vincenty's formula (applied above)
• 876.985 miles
• 1411.370 kilometers
• 762.079 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 874.091 miles
• 1406.713 kilometers
• 759.564 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Sabetta International Airport
City: Sabetta
Country: Russia
IATA Code: SBT
ICAO Code: USDA
Coordinates: 71°13′9″N, 72°3′7″E
B Syktyvkar Airport
City: Syktyvkar
Country: Russia
IATA Code: SCW
ICAO Code: UUYY
Coordinates: 61°38′49″N, 50°50′42″E
## Time difference and current local times
The time difference between Sabetta and Syktyvkar is 2 hours. Syktyvkar is 2 hours behind Sabetta.
+05
MSK
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 142 kg (312 pounds).
## Frequent Flyer Miles Calculator
Sabetta (SBT) → Syktyvkar (SCW).
Distance:
877
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
877
Round trip? | 499 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.813254 |
https://programmer.ink/think/understanding-and-summary-of-red-black-tree.html | 1,718,498,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00479.warc.gz | 444,842,776 | 10,162 | # Understanding and summary of red black tree
Posted by GeXus on Tue, 08 Feb 2022 15:36:53 +0100
Copyright notice: This article is the original article of CSDN blogger "Zhang Yanfeng ZYF", which follows the CC 4.0 BY-SA copyright agreement. For reprint, please attach the original source link and this notice.
catalogue
1, Basic understanding of red black tree
(1) Understanding of the basic definition of red black tree
(2) Understanding that red black tree is "approximate equilibrium"
1. Remove the red node from the red black tree and analyze the height of the red black tree containing the black node
2. Add back the red node and analyze the height change
(3) Comparison between red black tree and AVL tree:
2, Analysis on the basic idea of realizing red black tree
(1) Understand the operation of rotate left and rotate right
(2) Balance adjustment of insertion operation
Case 1: if the concerned node is a, its uncle node d is red
Case 2: if the concerned node is a, its uncle node d is black, and the concerned node a is the right child node of its parent node b
Case 3: if the concerned node is a, its uncle node d is black, and the concerned node a is the left child node of its parent node b
The specific codes above are as follows:
(3) Balance adjustment for delete operation
1. Preliminary adjustment for deleted nodes
Case 1: if the node to be deleted is a, it has only one child node b
If a has two non child nodes of c, it is the case that a wants to delete its two child nodes
Case 3: if node a is to be deleted, it has two non empty child nodes, and the successor node of node a is not a right child node
2. Make secondary adjustment for concerned nodes
Case 1: if the node of interest is a, its brother node c is red
Case 2: if the node of interest is a, its brother node c is black, and the left and right child nodes d and e of node c are black
Case 3: if the node of interest is a, its brother node c is black, the left child node d of c is red, and the right child node e of c is black
Case 4: if the brother node c of node a is black and the right child node of c is red
The above specific codes can be seen:
# 1, Basic understanding of red black tree
## (1) Understanding of the basic definition of red black tree
The English of red black tree is "red black tree", which is called R-B Tree for short. It is a kind of relaxed balanced binary search tree
The nodes in the red black tree are marked as black and red. In addition, a red black tree also needs to meet the following requirements:
• The root node is black;
• Each leaf node is a black empty node (NIL), that is, the leaf node does not store data (the black and empty leaf nodes are omitted in the figure);
• Any adjacent nodes cannot be red at the same time, that is, red nodes are separated by black nodes;
• Each node, all paths from the node to its leaf node, contain the same number of black nodes;
## (2) Understanding that red black tree is "approximate equilibrium"
The original intention of balancing binary search tree is to solve the problem of performance degradation caused by dynamic update of binary search tree. Therefore, the meaning of "balance" can be equivalent to no degradation of performance. "Approximate balance" is equivalent to that the performance will not degrade too seriously.
The height of an extremely balanced binary tree (full binary tree or complete binary tree) is about log2n, so if you want to prove that the red black tree is approximately balanced, you only need to analyze whether the height of the red black tree is stably approaching log2n.
### 1. Remove the red node from the red black tree and analyze the height of the red black tree containing the black node
After the red node is deleted, some nodes have no parent node. They will directly take the grandfather node (the parent node of the parent node) of these nodes as the parent node. Therefore, the previous binary tree has become a quadtree.
Take out some nodes from the quadtree and put them at the leaf node, and the quadtree becomes a complete binary tree. Therefore, the height of a quadtree with only black nodes is smaller than that of a complete binary tree with the same number of nodes.
The height of the complete binary tree is similar to log2n. The height of the quad "black tree" here is lower than that of the complete binary tree, so the height of the "black tree" without red nodes will not exceed log2n.
### 2. Add back the red node and analyze the height change
In the red black tree, red nodes cannot be adjacent, that is, if there is a red node, there must be at least one black node to separate it from other red nodes.
The path containing the most black nodes in the red black tree will not exceed log2n, so after adding red nodes, the longest path will not exceed 2log2n, that is, the height of the red black tree is approximately 2log2n.
Therefore, the height of the red black tree is only twice that of the highly balanced AVL tree (log2n), and the performance does not decline much. The result derived in this way is not accurate enough. In fact, the performance of red black tree is better.
## (3) Comparison between red black tree and AVL tree:
• Although the time complexity of AVL tree is better than that of red black tree, the cpu is too fast for today's computers, and the performance difference can be ignored
• The insertion and deletion of red black tree is easier to control than AVL tree
• The overall performance of red black tree is slightly better than that of AVL tree (the rotation of red black tree is less than that of AVL tree)
# 2, Analysis on the basic idea of realizing red black tree
The balance process of the red black tree is very similar to the restoration of the Rubik's cube. The general process is: what kind of node arrangement we encounter, we will adjust it accordingly. As long as these fixed adjustment rules are followed, an unbalanced red black tree can be adjusted to balanced.
As mentioned above, among the four basic requirements that a qualified red black tree needs to meet, the third and fourth requirements may be destroyed in the process of inserting and deleting nodes, and "balance adjustment" is actually to restore the damaged third and fourth points. The specific analysis is as follows:
## (1) Understand the operation of rotate left and rotate right
Left rotation is the left rotation around a node. A, b and r in the figure represent subtrees and can be empty.
Specific code implementation:
```
/**
* Function Description: left-hand and right-hand need to be balanced
*
* @author yanfengzhang
* @date 2020-05-27 14:57
*/
private void rotateLeft(Entry<K, V> p) {
if (p !=
null) {
/*Get the right child node of the root node */
Entry<K, V> r = p.right;
/*Assign a value to the left node of the right child node of the root node*/
p.right = r.left;
if (r.left !=
null)
/*Assign the value of the root node to the currently disconnected following node*/ {
r.left.parent = p;
}
/*r In the future, it will become a new root node. p.parent is the root, making it a new following node */
r.parent = p.parent;
if (p.parent ==
null) {
root = r;
}
/*If p is a left child, let him still be a left child*/
else
if (p.parent.left == p) {
p.parent.left = r;
}
else {
p.parent.right = r;
}
/*Finally, the current exchanged follow-up value*/
r.left = p;
p.parent = r;
}
}
```
Right rotation is the right rotation around a node. A, b and r in the figure represent subtrees and can be empty.
Specific code implementation:
```
/**
* Function Description: right hand code
*
* @author yanfengzhang
* @date 2020-05-27 14:58
*/
private void rotateRight(Entry<K, V> p) {
if (p !=
null) {
Entry<K, V> l = p.left;
p.left = l.right;
if (l.right !=
null) {
l.right.parent = p;
}
l.parent = p.parent;
if (p.parent ==
null) {
root = l;
}
else
if (p.parent.right == p) {
p.parent.right = l;
}
else {
p.parent.left = l;
}
l.right = p;
p.parent = l;
}
}
```
## (2) Balance adjustment of insertion operation
The red black tree stipulates that the inserted node must be red. Moreover, the newly inserted nodes in the binary search tree are placed on the leaf nodes.
There are two special cases about the balance adjustment of the insertion operation:
• If the parent node of the inserted node is black, we don't have to do anything. It still meets the definition of red black tree.
• If the inserted node is the root node, we can directly change its color and turn it into black.
In addition, other situations will violate the definition of red black tree and need to be adjusted. The adjustment process includes two basic operations: left and right rotation and color change.
The balance adjustment process of red black tree is an iterative process. The node being processed is called the concerned node. The nodes of interest will constantly change with the continuous iterative processing. The first node of interest is the newly inserted node. After a new node is inserted, if the balance of the red black tree is broken, there are generally three situations:
Note: we only need to constantly adjust according to the characteristics of each situation, so that the red black tree can continue to meet the definition, that is, to maintain balance. In order to simplify the description, the brother node of the parent node is called the uncle node, and the parent node of the parent node is called the grandfather node.
### Case 1: if the concerned node is a, its uncle node d is red
The specific operations are as follows: set the color of parent node b and uncle node d of node a to black; Set the color of the grandfather node c of the node of interest a to red; The concerned node becomes the grandfather node c of a; Skip to case 2 or case 3.
### Case 2: if the concerned node is a, its uncle node d is black, and the concerned node a is the right child node of its parent node b
The specific operations are as follows: the concerned node becomes the parent node b of node a; Rotate left around the new focus node b; Skip to case three.
### Case 3: if the concerned node is a, its uncle node d is black, and the concerned node a is the left child node of its parent node b
The specific operations are as follows: rotate right around the grandfather node c of the concerned node a; Exchange the colors of parent node b and brother node c of node a, and the adjustment is completed.
### The specific codes above are as follows:
```
/**
* Function Description: insert a node
*
* @author yanfengzhang
* @date 2020-05-27 15:07
*/
private void insert(RBTreeNode<T> node) {
int cmp;
RBTreeNode<T> root =
this.rootNode;
RBTreeNode<T> parent =
null;
/*Under which parent node is the location node added*/
while (
null != root) {
parent = root;
cmp = node.key.compareTo(root.key);
if (cmp <
0) {
root = root.left;
}
else {
root = root.right;
}
}
node.parent = parent;
/*It means that there is no node at present, so the newly added node is the root node*/
if (
null == parent) {
this.rootNode = node;
}
else {
//Find the location of the new node under the current parent node
cmp = node.key.compareTo(parent.key);
if (cmp <
0) {
parent.left = node;
}
else {
parent.right = node;
}
}
/*Set the color of the inserted node to red*/
node.color = COLOR_RED;
/*Fixed to red black tree*/
insertFixUp(node);
}
/**
* Function Description: red black tree insertion correction
*
* @author yanfengzhang
* @date 2020-05-27 15:07
*/
private void insertFixUp(RBTreeNode<T> node) {
RBTreeNode<T> parent, gparent;
/*The parent node of the node exists and is red*/
while (((parent = getParent(node)) !=
null) && isRed(parent)) {
gparent = getParent(parent);
/*What if its grandfather node is empty? What if the parent node is the left child of the grandfather node*/
if (parent == gparent.left) {
RBTreeNode<T> uncle = gparent.right;
if ((
null != uncle) && isRed(uncle)) {
setColorBlack(uncle);
setColorBlack(parent);
setColorRed(gparent);
node = gparent;
continue;
}
if (parent.right == node) {
RBTreeNode<T> tmp;
leftRotate(parent);
tmp = parent;
parent = node;
node = tmp;
}
setColorBlack(parent);
setColorRed(gparent);
rightRotate(gparent);
}
else {
RBTreeNode<T> uncle = gparent.left;
if ((
null != uncle) && isRed(uncle)) {
setColorBlack(uncle);
setColorBlack(parent);
setColorRed(gparent);
node = gparent;
continue;
}
if (parent.left == node) {
RBTreeNode<T> tmp;
rightRotate(parent);
tmp = parent;
parent = node;
node = tmp;
}
setColorBlack(parent);
setColorRed(gparent);
leftRotate(gparent);
}
}
setColorBlack(
this.rootNode);
}
```
## (3) Balance adjustment for delete operation
The balance adjustment of deletion operation is divided into two steps:
The first step is to make preliminary adjustments for deleting nodes. The preliminary adjustment only ensures that the whole red black tree still meets the requirements of the last definition after a node is deleted, that is, all paths from the node to its leaf nodes contain the same number of black nodes;
The second step is to make secondary adjustment for the concerned node to make it meet the third definition of red black tree, that is, there are no two adjacent red nodes.
### 1. Preliminary adjustment for deleted nodes
In order to ensure that some nodes in the "black red tree" are defined as "black" or "black", the last node in the "red tree" will only meet the requirements of "black". If a node is marked as "black black", it should be counted as two black nodes when calculating the number of black nodes.
Note: if a node can be either red or black, it is represented by half red and half black in the figure. If a node is "red black" or "black black", a small black dot in the upper left corner is used to represent additional black.
### Case 1: if the node to be deleted is a, it has only one child node b
The specific operations are as follows: delete node A and replace node b with node A. This part of the operation is the same as that of ordinary binary search tree; Node a can only be black and node b can only be red. Other situations do not meet the definition of red black tree. In this case, we change node b to black; After adjustment, no secondary adjustment is required.
### If a has two non child nodes of c, it is the case that a wants to delete its two child nodes
The specific operation is as follows: if the successor node of node a is the right child node c, the right child node c must have no left child tree. We delete node A and replace node c with node a. This part of the operation is no different from the ordinary binary search tree deletion operation; Then set the color of node c to the same color as node a; If node c is black, in order not to violate the last definition of the red black tree, we add a black to the right child node d of node c. at this time, node d becomes "red black" or "black black"; At this time, the concerned node becomes node d, and the second step of adjustment will be done for the concerned node.
### Case 3: if node a is to be deleted, it has two non empty child nodes, and the successor node of node a is not a right child node
The specific operations are as follows: find the successor node D and delete it. Refer to CASE 1 for the process of deleting the successor node D; Replace node a with subsequent node D; Set the color of node d to the same color as node a; If node D is black, in order not to violate the last definition of the red black tree, we add a black to the right child node c of node D. at this time, node c becomes "red black" or "black black"; At this time, the concerned node becomes node c, and the second step of adjustment will be done for the concerned node.
### 2. Make secondary adjustment for concerned nodes
After the initial adjustment, the concerned node becomes a "red black" or "black black" node. For this concerned node, the secondary adjustment is carried out in four cases.
Note: the secondary adjustment is to ensure that there are no adjacent red nodes in the red black tree.
### Case 1: if the node of interest is a, its brother node c is red
Specific operation: rotate left around the parent node b of the concerned node a; The parent node b and the grandfather node c of the concerned node a exchange colors; The concerned nodes remain unchanged; Continue to select the appropriate rules from the four cases to adjust.
### Case 2: if the node of interest is a, its brother node c is black, and the left and right child nodes d and e of node c are black
Specific operation: change the color of brother node c of node a to red; Remove a black from the concerned node A. at this time, node a is pure red or black; Add a black to the parent node b of node a, and node b will become "red black" or "black black"; The concerned node changes from a to its parent node b; Continue to select the rules from the four cases to adjust.
### Case 3: if the node of interest is a, its brother node c is black, the left child node d of c is red, and the right child node e of c is black
Specific operation: rotate right around the brother node c of the concerned node a; Node c and node d exchange colors; The concerned nodes remain unchanged; Jump to CASE 4 and continue adjustment.
### Case 4: if the brother node c of node a is black and the right child node of c is red
Specific operation: rotate left around the parent node b of the concerned node a; Set the color of the brother node c of the node of interest a to the same color as the parent node b of the node of interest a; Set the color of the parent node a to black; Remove a black from the concerned node a, and node a becomes pure red or black; Set the uncle node e of the concerned node a to black; End of adjustment.
### The above specific codes can be seen:
```
/**
* Function Description: delete node
*
* @author yanfengzhang
* @date 2020-05-27 15:11
*/
private void remove(RBTreeNode<T> node) {
RBTreeNode<T> child, parent;
boolean color;
/*The left and right children of the deleted node are not empty*/
if ((
null != node.left) && (
null != node.right)) {
/*Get the successor node of the deleted node*/
RBTreeNode<T> replace = node;
replace = replace.right;
while (
null != replace.left) {
replace = replace.left;
}
/*node Node is not a root node*/
if (
null != getParent(node)) {
/*node Is the left node*/
if (getParent(node).left == node) {
getParent(node).left = replace;
}
else {
getParent(node).right = replace;
}
}
else {
this.rootNode = replace;
}
child = replace.right;
parent = getParent(replace);
color = getColor(replace);
if (parent == node) {
parent = replace;
}
else {
if (
null != child) {
setParent(child, parent);
}
parent.left = child;
replace.right = node.right;
setParent(node.right, replace);
}
replace.parent = node.parent;
replace.color = node.color;
replace.left = node.left;
node.left.parent = replace;
if (color == COLOR_BLACK) {
removeFixUp(child, parent);
}
node =
null;
return;
}
if (
null != node.left) {
child = node.left;
}
else {
child = node.right;
}
parent = node.parent;
color = node.color;
if (
null != child) {
child.parent = parent;
}
if (
null != parent) {
if (parent.left == node) {
parent.left = child;
}
else {
parent.right = child;
}
}
else {
this.rootNode = child;
}
if (color == COLOR_BLACK) {
removeFixUp(child, parent);
}
node =
null;
}
/**
* Function Description: delete repair
*
* @author yanfengzhang
* @date 2020-05-27 15:11
*/
private void removeFixUp(RBTreeNode<T> node, RBTreeNode<T> parent) {
RBTreeNode<T> other;
/*node It is not empty and black, and it is not the root node*/
while ((
null == node || isBlack(node)) && (node !=
this.rootNode)) {
/*node Is the left child of the parent node*/
if (node == parent.left) {
/*Get its right child*/
other = parent.right;
/*node The sibling node of the node is red*/
if (isRed(other)) {
setColorBlack(other);
setColorRed(parent);
leftRotate(parent);
other = parent.right;
}
/*node The sibling node of the node is black, and the two child nodes of the sibling node are also black*/
if ((other.left ==
null || isBlack(other.left)) &&
(other.right ==
null || isBlack(other.right))) {
setColorRed(other);
node = parent;
parent = getParent(node);
}
else {
/*node The sibling node of the node is black, and the right child of the sibling node is red*/
if (
null == other.right || isBlack(other.right)) {
setColorBlack(other.left);
setColorRed(other);
rightRotate(other);
other = parent.right;
}
/*node The sibling node of the node is black, and the right child of the sibling node is red, and the left child is any color*/
setColor(other, getColor(parent));
setColorBlack(parent);
setColorBlack(other.right);
leftRotate(parent);
node =
this.rootNode;
break;
}
}
else {
other = parent.left;
if (isRed(other)) {
setColorBlack(other);
setColorRed(parent);
rightRotate(parent);
other = parent.left;
}
if ((
null == other.left || isBlack(other.left)) &&
(
null == other.right || isBlack(other.right))) {
setColorRed(other);
node = parent;
parent = getParent(node);
}
else {
if (
null == other.left || isBlack(other.left)) {
setColorBlack(other.right);
setColorRed(other);
leftRotate(other);
other = parent.left;
}
setColor(other, getColor(parent));
setColorBlack(parent);
setColorBlack(other.left);
rightRotate(parent);
node =
this.rootNode;
break;
}
}
}
if (node !=
null) {
setColorBlack(node);
}
}
```
# References and links:
2. The beauty of data structure and algorithm, Wang Zheng (former Google Engineer), geek time, 2019
Topics: Algorithm data structure | 5,060 | 21,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.922447 |
https://nursingassignment.org/cartesian-datavector-homework/ | 1,643,041,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304572.73/warc/CC-MAIN-20220124155118-20220124185118-00566.warc.gz | 486,565,527 | 18,929 | June 21, 2017
###### Who are the oligarchs – How are the oligarchs Economic
June 21, 2017
Cartesian Data/Vector homework
Setup
You will need the class files:
~/cs162/baseCode/classes/incl/complex0.h
~/cs162/baseCode/classes/src/complex0.cpp
and the Homework Exercise files:
~/cs162/lu07/7_assign_code/hw02/incl/fns02.h
~/cs162/lu07/7_assign_code/hw02/src/fns02.cpp
~/cs162/lu07/7_assign_code/hw02/src/drvr02.cpp
Problem Statement
Define a complex class so that the following function can use it with correct results:
#include <iostream>
using namespace std;
#include “../incl/complex0.h” //to avoid confusion with complex.h
void demoComplex()
{
complex a( 3.0, 4.0 ); // initialize to (3, 4i)
complex c;
cout << “Enter a complex number (q to quit):n” ;
while( cin >> c )
{
cout << “c is ” << c << ‘n’ ;
cout << “complex conjugate is ” << ~c << ‘n’ ;
cout << “a is ” << a << ‘n’ ;
cout << “a + c is ” << a + c << ‘n’ ;
cout << “a – c is ” << a – c << ‘n’ ;
cout << “a * c is ” << a * c << ‘n’ ;
cout << “2 * c is ” << 2 * c << ‘n’ ;
cout << “c * 2 is ” << c * 2 << ‘n’ ;
cout << “Enter a complex number (q to quit):n” ;
}
cout << “Done!n” ;
}
Solving the Problem
General Approach
Declare the class in the class header file. Specify the data members needed, the constructors, inspectors, mutators, and facilitators indicated by the problem statement and the requirements of the demonstration function. Declare operator overloads as needed to satisfy the needs of the demonstration function. Overloads should be non-member auxiliary functions, not member or friend functions. Declare public member functions needed by the overload functions.
Define the member functions, testing each as you define it. Use the ~/cs162/lu07/7_assign_code/hw02/src/drvr02.cpp file for this, as well as executing the demonstration function.)
Declare and define the demonstration function as specified in the Problem Statement.
Specific Approach
Note that you have to overload the << and >> operators for this class. Use const wherever warranted – specifically with inspector functions, and Complex objects passed as reference parameters.
Provide overloads for both n * c and c * n
A sample run of the demonstration function might produce this output:
Enter a complex number (q to quit):
real: 10
imaginary: 12
c is (12, 12i)
complex conjugate is (10, -12i)
a is (3, 4i)
a + c is (13, 16i)
a – c is (-7, -8i)
a * c is (-18, 76i)
2 * c is (20, 24i)
c * 2 is (20, 24i)
Enter a complex number (q to quit):
real: q
Done!
Setup
You will need the class files:
~/cs162/baseCode/classes/incl/myvector.h
~/cs162/baseCode/classes/src/myvector.cpp
and the Homework files
~/cs162/lu07/7_assign_code/hw01/incl/fns01.h
~/cs162/lu07/7_assign_code/hw01/src/fns01.cpp
~/cs162/lu07/7_assign_code/hw01/src/drvr01.cpp
Problem Statement
Modify the Vector class header and implementation files so that the magnitude and angle are no longer stored as data components. Instead, they should be calculated on demand when the magval() and angval() methods are called.
Solving the Problem
General approach:
Leave the public interface unchanged (the same public methods with the same arguments) but alter the private section, including the private methods and their method implementations. Test the modified version with a third version of the randwalk() function, which should be left unchanged other than using the new version of the Vector class, because the public interface of the Vector class is unchanged.
Specific approach:
You will find class definitions here that you built in the Topic A Exploration Exercises.
Create a third version of the Vector class, Vector3.
Create a new version of the randwalk() function, randwalk3(), copying the function definition you find in the Topic A Exploration Three Exercise 1 file.
Modify the declaration for Vector3 to remove the polar coordinate data members, modify the prototypes of the private methods, and change the inline definitions of the inspectors for the now-deleted polar coordinate values.
The public inspectors for polar coordinate values must now invoke the revised facilitators to get the values they compute from x-y coordinate values. These were coded as inline definitions, and can stay that way – they will still have just one statement each.
Modify the definitions for Vector3 to update the private methods.
The private methods that were originally mutators for the polar coordinate data members will need to become facilitators that compute the polar coordinate equivalents of the x-y coordinates that are stored in the class. They should be constant functions that do not modify data member values, and return double values. The existing method definitions do the correct computations, but will need to return their results rather than setting the now-deleted polar coordinate data members.
The private methods that are mutators for the x-y coordinate data members can no longer read from the polar coordinate data members, but must receive polar coordinate values as method parameters. The mutator definitions do the correct computations, and will just need to be modified to work with their new input parameters.
While no public interface changes should be made, three of the public methods will need definition changes: the two constructors and the reset() method.
There are no more polar coordinate data members to set in any of the three.
If the input parameters are provided in polar coordinate form, they still need to be converted to x-y coordinate equivalent values before being stored in the data members. All the necessary pieces are there in the existing code, needing just slight modifications to create local variables and to pass those to the mutator functions.
Modify the randwalk3() function to declare variables of the Vector3 class. This is the only modification needed.
Add a call to randwalk3() to the Homework Exercise code file. The first two versions can still be tested through the Topic A project.
both are considered 1 assaignment. | 1,407 | 5,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.76393 |
https://papertowriters.com/mini-case-1-estimating-required-rate-of-return-interest-rate-using-risk-premiums-5-points-re/ | 1,669,946,437,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00424.warc.gz | 492,174,368 | 17,456 | # MINI-CASE 1: ESTIMATING REQUIRED RATE OF RETURN (Interest Rate) USING RISK PREMIUMS . [5 points] (Re
MINI-CASE 1: ESTIMATING REQUIRED RATE OF RETURN (Interest Rate) USING RISK PREMIUMS. [5 points]
(References: See essentially Schedule of Classes Units 1, 3 and 6) Preliminary Remarks:
The purpose of this mini-case is to demonstrate how the required rate of return on a fixed-income security can be determined, using the risk premium perspective. The mini-case brings together, in a simple model, several of the various risks associated with fixed-income securities. From a corporation perspective, the nominal interest rate to be determined in this mini-case can serve as a basis for estimating the firm’s before-tax cost of debt. At the issue of the security, if the nominal interest rate is set as the coupon rate that is equal to the yield to maturity (so that the price equals the par value), then one could interpret the obtained rate as a discount rate. From the perspective of investors (buyers of the security), the rate could be interpreted as the investor’s nominal required rate of return. The Mini-Case Assignment: Your company, Binghamton Truck, Inc., is about to offer a new issue of corporate bonds to the investing marketplace. You have been asked by your CFO to provide a reasonable estimate of the nominal interest rate (nominal yield), Rd, for a new issue of Aaa-rated bonds to be offered by Binghamton Truck. Some agreed-upon procedures related to generating estimates for key variables in the relevant equation, Rd = R*rf + IRP + DRP + MP + LP, are as follows:
The current (mid-2008) financial market environment is considered representative of the prospective tone of the market near the time of offering the new bonds to the investing public. This means that current interest rates will be used as benchmarks for some of the variable estimates. All estimates will be rounded off to hundredths of a percent; thus, 6.288 becomes 6.29 percent. The real risk-free rate of interest, R*rf, is the difference between the calculated average yield on 3-month Treasury bills and the inflation rate. The inflation-risk premium, IRP, is the rate of inflation expected to occur over the life of the bond under consideration. The default-risk premium, DRP, is estimated by the difference between the average yield on Aaa-rated bonds and 30-year Treasury bonds. The maturity premium, MP, is estimated by the difference between the calculated average yield on 30-year Treasury bonds and 3-month Treasury bills. Binghamton Truck’ bonds will be traded on the New York Exchange for Bonds, so the liquidity premium, LP, will be slight. It will be greater than zero; however, because the secondary market for the firm’s bonds is more uncertain than that of some other truck producers, it is estimated at 3 basis points.
Note: A basis point is one one-hundredth of 1 percent. (E.g., 1 basis point = 0.01%; 25 basis points = 0.25%) Based on your research, the mid-2008 estimates of the representative interest and inflation rates are as follows: (1) 3-Month T-Bills = 4.89% (2) 30-Year T-Bonds = 5.38% (use this as proxy for 20-year T-Bonds) (3) Aaa-Rated Corporate Bonds = 6.24% (4) Inflation Rate = 3.60%. Visit online Federal Reserve Bank of St. Louis (Google “Federal Reserve Bank of St. Louis FRED”) and update the above data with the most recently available rates for each of the above fixed income securities and for the inflation rate.
Required Task: Complete the Solution Table below, which is presented in form of a formula required to determine Rd. Place your answers (values) in the cells below the variables in the second row, and show your calculations below the Table, where applicable, of how you obtainned the value for each of the variables. Similarly, use your most recent collected rates (May 2019; see source below) to complete the third row of the worksheet below. Briefly comment on the differences between the two results (i.e. results obtained from above old data versus results obtained from recent data you collected).
Fill-In Table for Mini-Case 1 (show your work below the table, as appropriate):
R*rf
+ IRP
+ DRP
+ MRP
+ LRP
=
Rd
Using “old” data in the case above
Show in this space (add page as necessary) any calculations you performed to find any of the terms in the above table.
_________________________________
* Source of data for part II of Mini-Case 1:
FRED – ECONOMIC DATA, by The Federal Reserve Bank of St. Louis
https://fred.stlouisfed.org/categories
Copy and Paste the above URL into your browser, then look for the following data, under either “Prices” or “Interest Rates.” The “identifier” like TB3MS can be seen when you download the series (download option is provided in the top right corner of the data window of interest. When you click on “Download”, you will be given options for the format, like Excel, etc. I suggest choosing “Excel” if you decide to download the series.) Inflation Rate: PCETRIM12M159SFRBDAL
PCETRIM12M159SFRBDAL Trimmed Mean PCE Inflation Rate, Percent Change from Year Ago, Monthly,
Adjusted Three (3)-Month Treasury Bill [look under Treasury Bills]
TB3MS
3-Month Treasury Bill: Secondary Market Rate, Percent, Monthly, Not Seasonally Adjusted 30-Year Treasury Constant Maturity Rate [look under Treasury Constant Maturity]
DGS30
30-Year Treasury Constant Maturity Rate, Percent, Daily, Not Seasonally Adjusted 20-Year Treasury Constant Maturity Rate [look under Treasury Constant Maturity]
DGS20
20-Year Treasury Constant Maturity Rate, Percent, Daily, Not Seasonally Adjusted Moody’s Seasoned Aaa Corporate Bond Yield [look under Corporate Bonds]
AAA
Moody's Seasoned Aaa Corporate Bond Yield, Percent, Monthly, Not Seasonally Adjusted | 1,332 | 5,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.912632 |
https://experienceleaguecommunities.adobe.com/t5/adobe-livecycle-questions/percentage-field-format/td-p/152925 | 1,716,674,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00208.warc.gz | 209,542,331 | 39,104 | Expand my Community achievements bar.
# Adobe LiveCycle (Archived)
Enhance your AEM Assets & Boost Your Development: [AEM Gems | June 19, 2024] Improving the Developer Experience with New APIs and Events
SOLVED
## Percentage field format
Level 1
Hi,
I'm new to livecycle designer and I need your valuable help!
I'm making a form and I need to have some percentage fields, with data entered by filling the form.
These percentage fields should have format 000%. Ex. 005%, 035%, 100%
There should be an error message when the user tries to input a num greater than 100,
Also, I don't know if it could be done, to have a calculator script that summarizes all the input data and verify that the summary is aways 100. Ex. if field1 input=50, field2 input=30, field3 input=20, field4 input=0, field5 input=0, then it's ok!
1 Accepted Solution
Level 10
Hi,
the percentage fields often confuses users, as it internally calcultes with floating numbers between 0 and 1.
For example: If a users enters 0.77 it will display 77%.
When you want users to enter the same values that should represent the percentage value, then you can use a numeric pattern with a string for the %.
Num{999.8'%'}
To calculate a summary of fields you can use FormCalc ind the percentage fields calculate event.
var summary = Sum(field1, field2, field3, field4, field5)
if (summary gt 100) then
\$host.messageBox("Summary cannot be more that 100%. Please check your data.")
\$ = 0
else
\$ = summary
endif
Level 10
Hi,
the percentage fields often confuses users, as it internally calcultes with floating numbers between 0 and 1.
For example: If a users enters 0.77 it will display 77%.
When you want users to enter the same values that should represent the percentage value, then you can use a numeric pattern with a string for the %.
Num{999.8'%'}
To calculate a summary of fields you can use FormCalc ind the percentage fields calculate event.
var summary = Sum(field1, field2, field3, field4, field5)
if (summary gt 100) then
\$host.messageBox("Summary cannot be more that 100%. Please check your data.")
\$ = 0
else
\$ = summary
endif | 539 | 2,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.767193 |
http://openstudy.com/updates/4face2cfe4b059b524f87337 | 1,474,787,821,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660158.61/warc/CC-MAIN-20160924173740-00223-ip-10-143-35-109.ec2.internal.warc.gz | 195,967,065 | 21,441 | ## .Sam. 4 years ago $$\style{\text-shadow: 0 0 10px black;font-size:60px; } { \Huge \scr{ \color {white} {{\scr{\rlap{\color{red}{L }}{\color{white}{\;L} {{\scr{\rlap{\color{yellow}{a }}{\color{white}{\;a} } }}} {{\scr{\rlap{\color{lightgreen}{T }}{\color{green}{\;T} } }}} {{\scr{\rlap{\color{blue}{e }}{\color{lightblue}{\;e} } }}} {{\scr{\rlap{\color{pink}{X }}{\color{purple}{\;X} } }}} } }}}}}$$ $$\style{text-shadow: 0 0 1200px black;font-size:0px; } {\scr{F} \cal{U} \cal{N}!}$$
• This Question is Open
1. .Sam.
$i \hbar \frac{\partial}{\partial t}\Psi=\hat H \Psi$
2. .Sam.
ÁÆ}|¶¶
3. .Sam.
ДѬѯЎ҈әҸДԎ
4. .Sam.
$dy/dx, \operatorname{d}\!y/\operatorname{d}\!x, {dy \over dx}, {\operatorname{d}\!y\over\operatorname{d}\!x}, {\partial^2\over\partial x_1\partial x_2}y \prime, \backprime, f^\prime, f', f'', f^{(3)}, \dot y, \ddot y$
5. .Sam.
$\Huge \infty, \aleph, \complement, \backepsilon, \eth, \Finv, \hbar \Im, \imath, \jmath, \Bbbk, \ell, \mho, \wp, \Re, \circledS$
6. .Sam.
$\big\uparrow \Big\uparrow \bigg\uparrow \Bigg\uparrow \dots \Bigg\Downarrow \bigg\Downarrow \Big\Downarrow \big\Downarrow$
7. .Sam.
$\begin{\cases} 3x + 5y + z \\ 7x - 2y + 4z \\ -6x + 3y + 2z\end{\cases}$
8. .Sam.
$\Huge \cup, \Cup, \sqcup, \bigcup, \bigsqcup, \uplus, \biguplus$
9. .Sam.
$\doteq, \overset{\underset{\mathrm{\def}}{}}{=}, :=$
10. .Sam.
Example $$ax^2+bx+c=0$$ 123456
11. .Sam.
♜♞♝♛♚♝♞♜ ♟♟♟♟♟♟♟♟ ♙♙♙♙♙♙♙♙ ♖♘♗♕♔♗♘♖
12. TheViper
How to do this?
13. TheViper
@ParthKohli & @.Sam. plz tell:)
14. maheshmeghwal9
How?
15. ganeshie8
♜♞♝♛♚♝♞♜ ♟♟♟♟♟♟♟♟ ♙ ♙♙♙♙ ♙♙♙ ♖♘♗♕♔♗♘♖
16. .Sam.
€ £ Ұ ₴ \$ ₰ ¢ ₤ ¥ ₳ ₲ ₪ ₵ 元 ₣ ₱ ฿ ¤ ₡ ₮ ₭ ₩ ރ 円 ₢ ₥ ₫ ₦ zł ﷼ ₠ ₧ ₯ ₨ Kč र
17. .Sam.
♔ ♕ ♖ ♗ ♘ ♙ ♚ ♛ ♜ ♝ ♞ ♟ ♤ ♧ ♡ ♢ ♠ ♣ ♥ ♦ ⋆ ✢ ✣ ✤ ✥ ❋ ✦ ✧ ✩ ╰☆╮ ✪ ✫ ✬ ✭ ✮ ✯ ✰ ✡ ★ ✱ ✲ ✳ ✴ ❂ ✵ ✶ ✷ ✸ ✹ ✺ ✻ ✼ ❄ ❅ ❆ ❇ ❈ ❉ ❊
18. .Sam.
✖ ☢ ☣ ☤ ⚜ ♪ ♫ € ϟ
19. across
Someone needs to reply to @ganeshie8 accordingly!
20. .Sam.
$\Huge ϴρεηϟ☂ʊ∂⑂$
21. UnkleRhaukus
♜♞♝♛♚♝♞♜ ♟♟♟♟ ♟♟♟ ♟ ♙ ♙♙♙♙ ♙♙♙ ♖♘♗♕♔♗♘♖
22. ParthKohli
♜♞♝♛♚♝♞♜ ♟♟♟♟ ♟♟♟ ♗ ♟ ♙ ♙♙♙♙ ♙♙♙ ♖♘♗♕♔ ♘♖
23. maheshmeghwal9
♜♞♝♕♛♕♚♝♞♜ ♟♟♟♟ ♟♟♟ ♕ ♕ ♗ ♟ ♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕♕ ♕ ♙ ♕ ♙ ♙♙♙♙ ♙♙♙ ♖♘♗♕♔ ♘♖
24. .Sam.
javascript alert("booyan")
25. ParthKohli
javascript alert("booyan")
26. .Sam.
Test
27. .Sam.
$\begin{array}{|c|c|c|c|c|}\hline L&a&T&e&X\\ \hline 1&2&3&4&5\\ \hline\cup& \Cup& \sqcup& \bigcup& \bigsqcup\\\hline\frac{dy}{dx}&\oint&\int&\cong&\amalg\\\hline\nabla& \infty&\Re&\Im&\wp\\ \hline\end{array}$
28. TheViper
@.Sam. or @ParthKohli How did you write your question? plz tell :)
29. .Sam.
$\Huge \color{red}{\mathbb{R}} \color{Orange}{\mathbf{O}} \color{Yellow}{\boldsymbol{\Upsilon}} \color{Green}{\mathfrak{G}} \color{Blue}{\boldsymbol{\beta}} \color{Indigo}{\mathit{\Phi}} \color{Violet}{\mathfrak{V}}$
30. .Sam.
$\Huge \sideset{_1^2}{_3^4}\prod_a^b$ $\overbrace{ 1+2+\cdots+100 }^{5050}$ $\underbrace{ a+b+\cdots+z }_{26}$ $\Huge A \xleftarrow{n+\mu-1} B \xrightarrow[T]{n\pm i-1} C$ $\Huge \bigcap_{i=_1}^n E_i$ $\begin{bmatrix} 0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & 0\end{bmatrix}$
31. .Sam.
$\Large \Psi_{32 \pm 2}=\frac{a^{-3/2}}{162\pi^{1/2}}\frac{r^2}{a^2}e^{-r/3a}\sin^2 \theta e^{\pm 2 i \phi}$ $\Large C_V =9NK_B \frac{T^3}{\theta ^3 _D} \int\limits_0^{\theta _D /T} \frac{x^4e^x}{(e^x-1)^2}dx$ $$x = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4} } } }$$ $$\frac{ \begin{array}[b]{r} \left( x_1 x_2 \right)\\ \times \left( x'_1 x'_2 \right) \end{array} }{ \left( y_1y_2y_3y_4 \right) }$$ $\Huge {\scriptscriptstyle \int\limits f^{-1}(x-x_a)\,dx}$ $\iiiint\limits_{A}$ $\mathbf{A} = \begin{pmatrix} \dfrac{\varphi \cdot X_{n, 1}} {\varphi_{1} \times \varepsilon_{1}} & (x + \varepsilon_{2})^{2} & \cdots & (x + \varepsilon_{n - 1})^{n - 1} & (x + \varepsilon_{n})^{n}\\ \dfrac{\varphi \cdot X_{n, 1}} {\varphi_{2} \times \varepsilon_{1}} & \dfrac{\varphi \cdot X_{n, 2}} {\varphi_{2} \times \varepsilon_{2}} & \cdots & (x + \varepsilon_{n - 1})^{n - 1} & (x + \varepsilon_{n})^{n}\\ \dots &\dots &\dots &\dots &\dots\\ \dfrac{\varphi \cdot X_{n, 1}} {\varphi_{n} \times \varepsilon_{1}} & \dfrac{\varphi \cdot X_{n, 2}} {\varphi_{n} \times \varepsilon_{2}} & \cdots & \dfrac{\varphi \cdot X_{n, n - 1}} {\varphi_{n} \times \varepsilon_{n - 1}} & \dfrac{\varphi\cdot X_{n, n}} {\varphi_{n} \times \varepsilon_{n}} \end{pmatrix} + \mathbf{I}_{n}$
32. .Sam.
$\left( \bigvee (\, s_{i} \mid i \in I \,) \right)^{c} = \bigwedge (\, s_{i}^{c} \mid i \in I \,)$ $f(x) \overset{ \text{def} }{=} x^{2} - 1$ $\int_{\mathcal{D}} | \overline{\partial u} |^{2} \Phi_{0}(z) e^{\alpha |z|^2} \geq c_{4} \alpha \int_{\mathcal{D}} |u|^{2} \Phi_{0} e^{\alpha |z|^{2}} + c_{5} \delta^{-2} \int_{A} |u|^{2} \Phi_{0} e^{\alpha |z|^{2}}$ $f(x)= \begin{cases} -x^{2}, &\text{if x < 0;}\\ \alpha + x, &\text{if 0 \leq x \leq 1;}\\ x^{2}, &\text{otherwise.} \end{cases}$
33. .Sam.
$\Huge ∯= E•dA=\frac{∑q_{enc}}{ϵ_o}\\ \\ \Huge ∯B•dA=0 \\ \\ \Huge ∮E•dl=-\frac{d \Phi_B}{dt}\\ \\ \Huge ∮B•dl=μ_o I+μ_o ϵ_o \frac{d \Phi_E}{dt}$
34. .Sam.
$\Large \theta \rho \epsilon \eta \delta \tau \upsilon \partial \Upsilon$
35. anonymous
beautiful performance, I don't know how to show a^(polynomial) with that polynomial is up. I always have only the first number is up, the leftover is in bottom :( $\a^(2x^2 +3)$
36. anonymous
@Sam
37. .Sam.
38. ParthKohli
This is the best LaTeX thread ever. haha
39. anonymous
$\a^{2x^2+3}$yes, thanks
40. anonymous
how about the \ in the front? it looks weird
41. ParthKohli
What is \a?
42. anonymous
@ParthKohli you answer me, I posted it before you did
43. ParthKohli
I mean, why did you use \a? What were you trying to do?
44. anonymous
don't know, just follow others. so take it out? $a^{2x^3 +3}$ yaaaa, it's right!!!
45. anonymous
ty
46. ParthKohli
Oh, so that's what you wanted to do! Remember: when trying to type out plain letters, don't place a \ before them.
47. anonymous
yes, sir
$$\Huge \color{red}{~~~~~~~~~~~*WINNING* }$$ $$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$ $$\Large \color{red}{~\:\:\:\:\:\:\mathbb Don’t\:\:\mathbb Forget\:\:\mathbb To~~\mathbb Fan~~\mathbb And~~\mathbb Best~~\mathbb The~~\mathbb Answers!}$$ $$\Huge \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\:\mathbb Welcome\:\:\mathbb To\:\:\mathbb Open\mathbb Study}$$ $$\large \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\mathbb Snuggie\mathbb Lad }$$ $$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$$$\large \color{orange}{*}$$$$\small \color{purple}{*}$$$$\Tiny \color{pink}{*}$$$$\tiny \color{violet}{*}$$$$\Tiny \color{pink}{*}$$$$\small \color{purple}{*}$$$$\large \color{orange}{*}$$$$\Large \color{yellow}{*}$$$$\LARGE \color{green}{*}$$$$\huge \color{blue}{*}$$$$\Huge \color{red}{*}$$$$\huge \color{blue}{*}$$$$\LARGE \color{green}{*}$$$$\Large \color{yellow}{*}$$
49. .Sam.
Now that's some great work right there :)
50. e.mccormick
Sam, your question post is quite puzzling.... Here is a sometimes useful bit: $\left[\begin{matrix} a & b & c \\ d & e & f \\ h & i & j \end{matrix}\right] \left[\begin{matrix} k & l & m \\ n & o & p \\ q & r & s \end{matrix}\right] =\\ \left[\begin{matrix} ak+bn+cq & al+bo+cr & am+bp+cs \\ dk+en+fq & dl+eo+fr & dm+ep+fs \\ hk+in+jq & hl+io+jr & hm+ip+js \end{matrix}\right]$ And for fun: At the {Math} $$\large \cap$$ {Philosophy}... $\Large{\sum}_{k_0=g^it_0}^{e^rG_0}D^ek_ar^{t^{es}} \doteq M^i_nd$...the Cartesian $$k_0g^it_0~e^rG_0$$ sum.
51. UnkleRhaukus
@e.mccormick (i get it!)
52. .Sam.
$$\text{I'm just testing}$$$$\huge \text{~}$$
53. .Sam.
$\mbox {\text}$
54. .Sam.
$A \hline B$
55. e.mccormick
hline is for use in tabular, but even tabular is not working here: $\begin{tabular}{lll} Header & Header & Header \\ \hline data & data & data \\ data & data & data \\ data & data & data \end{tabular}$ As opposed to using it on: http://www.codecogs.com/latex/eqneditor.php http://latex.codecogs.com/gif.latex?%5Cbg_white%20%5Cbegin%7Btabular%7D%7Blll%7D%20Header%20%26%20Header%20%26%20Header%20%5C%5C%20%5Chline%20data%20%26%20data%20%26%20data%20%5C%5C%20data%20%26%20data%20%26%20data%20%5C%5C%20data%20%26%20data%20%26%20data%20%5Cend%7Btabular%7D Also, for what it looks like what you were trying to do, what about:$A\underline{\quad}B$
56. poopsiedoodle
if snuggie would stop using mah sparkles, that'd be great.
57. mathslover
@e.mccormick , you can make tables in LaTeX on OS too : $$\quad \begin{array}{|c|c|c|} \hline a & b & c & d \\ \hline 1 & 2 & 3 & 4 \\ \hline \end{array}$$ (Just an example)
58. poopsiedoodle
$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$ $$\Huge\color{orchid}{\mathscr{~~~HUEHUEHUE}}$$ $$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$ $$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$`
59. anonymous
╭━━━┳━━━┳━━━┳━━━╮╱╱╭┳━━━┳━━╮ ┃╭━╮┃╭━╮┃╭━╮┣╮╭╮┃╱╱┃┃╭━╮┃╭╮┃ ┃┃╱╰┫┃╱┃┃┃╱┃┃┃┃┃┃╱╱┃┃┃╱┃┃╰╯╰╮ ┃┃╭━┫┃╱┃┃┃╱┃┃┃┃┃┃╭╮┃┃┃╱┃┃╭━╮┃ ┃╰┻━┃╰━╯┃╰━╯┣╯╰╯┃┃╰╯┃╰━╯┃╰━╯┃ ╰━━━┻━━━┻━━━┻━━━╯╰━━┻━━━┻━━━╯ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██▄██░░░░██▄██▄██▄██▄██▄██░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████░░░░░░░█████░░░░░░░█████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░██▄██▄██▄██▄██▄██░░░░██▄██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░█████░░░░░░░█████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░░░░░░░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░░░░ ░░░░░░░░░░░░░░░██████████████░░░░░░ ░░░░░░░░░░░░░░░██▄██▄██▄██░░░░░░░░░ ░░░░░░░░░░░░░░░██████████████░░░░░░ ░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░░░░ ░░░░░░░░░░░░░░░░░░░░░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
60. e.mccormick
poopsiedoodle vs Machida.... hmmm... poopsiedoodle has color...eventually... when it finally renders. Machida has speed and tons of it. Lightnng code advantage! Machida wins! Color is nice, but I like reading things NOW and not in minutes when the overworked MathJax catches up.
61. e.mccormick
今日はまちだサン
62. anonymous
actually im goin mad for learn all of this (autodidact), but its so fun :D ╱╭╮╭╮╱╱╱╱╱╱╱╭╮ ╭╯╰┫┃╱╱╱╱╱╱╱┃┃ ╰╮╭┫╰━┳━━┳━╮┃┃╭┳━━╮ ╱┃┃┃╭╮┃╭╮┃╭╮┫╰╯┫━━┫ ╱┃╰┫┃┃┃╭╮┃┃┃┃╭╮╋━━┃ ╱╰━┻╯╰┻╯╰┻╯╰┻╯╰┻━━╯ ʞɔıɯɹoɔɔɯ˙ǝ
63. e.mccormick
$$\LaTeX$$ is not too bad. It is really powerful in the desktop version, and you can see that through some online sites that render full $$\LaTeX$$ wit PGF and TikZ, which lets you do accurate drawings and more.
64. e.mccormick
An example with PGF and TikZ https://www.writelatex.com/195052mwmzfz
65. poopsiedoodle
$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$ $$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$ $$\Huge\color{green}{\bf{~~~~~~~~~~~~~\ \ \ \ \ \ \ \ \ \ \ :(}}$$ $$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$$$\Huge\color{goldenrod}\star$$ $$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\LARGE\color{green}\ast$$$$\huge\color{blue}\ast$$
66. e.mccormick
All I see for the first minute....
67. .Sam.
$\begin{array}l\color{red}{\text{h}}\color{orange}{\text{a}}\color{yellow}{\text{p}}\color{green}{\text{p}}\color{blue}{\text{y}}\color{purple}{\text{ }}\color{purple}{\text{b}}\color{red}{\text{i}}\color{orange}{\text{r}}\color{yellow}{\text{t}}\color{green}{\text{h}}\color{blue}{\text{d}}\color{purple}{\text{a}}\color{purple}{\text{y}}\color{red}{\text{ }}\color{orange}{\text{♩}}\color{yellow}{\text{♪}}\color{green}{\text{♫}}\color{blue}{\text{♬}}\color{purple}{\text{♭}}\color{purple}{\text{♮}}\color{red}{\text{♯}}\color{orange}{\text{}}\end{array}$
68. blackops2luvr
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██▄██░░░░██▄██▄██▄██▄██▄██░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████░░░░░░░█████░░░░░░░█████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░██▄██▄██▄██▄██▄██░░░░██▄██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░█████░░░░░░░█████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░░░░░░░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░░░░ ░░░░░░░░░░░░░░░██████████████░░░░░░ ░░░░░░░░░░░░░░░██▄██▄██▄██░░░░░░░░░
69. NerdGamer_
$$\Huge *$$
70. NerdGamer_
╱╭╮╭╮╱╱╱╱╱╱╱╭╮ ╭╯╰┫┃╱╱╱╱╱╱╱┃┃ ╰╮╭┫╰━┳━━┳━╮┃┃╭┳━━╮ ╱┃┃┃╭╮┃╭╮┃╭╮┫╰╯┫━━┫ ╱┃╰┫┃┃┃╭╮┃┃┃┃╭╮╋━━┃ ╱╰━┻╯╰┻╯╰┻╯╰┻╯╰┻━━╯ ʞɔıɯɹoɔɔɯ˙ǝ
71. NerdGamer_
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██▄██░░░░██▄██▄██▄██▄██▄██░░░ ░░░████████░░░░█████████████████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████░░░░░░░█████░░░░░░░█████░░░ ░░░██▄██░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░██▄██▄██▄██▄██▄██░░░░██▄██▄██░░░ ░░░█████████████████░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░█████░░░░░░░█████░░░ ░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░ ░░░░░░░░░░░░░░░░░░░░░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░░░░ ░░░░░░░░░░░░░░░██████████████░░░░░░ ░░░░░░░░░░░░░░░██▄██▄██▄██░░░░░░░░░ ░░░░░░░░░░░░░░░██████████████░░░░░░ ░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░░░░ ░░░░░░░░░░░░░░░░░░░░░░░░████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░██▄██░░░ ░░░█████████████████████████████░░░ ░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░ ░░░█████████████████████████████░░░ ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
72. NerdGamer_
╔╗────╔════╗─╔═╗╔═╗ ║║────║╔╗╔╗║─╚╗╚╝╔╝ ║║──╔═╩╣║║╠╩═╗╚╗╔╝ ║║─╔╣╔╗║║║║║═╣╔╝╚╗ ║╚═╝║╔╗║║║║║═╬╝╔╗╚╗ ╚═══╩╝╚╝╚╝╚══╩═╝╚═╝Fun!
73. anonymous
$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$$$\huge\color{blue}\ast$$$$\LARGE\color{green}\ast$$$$\Large\color{purple}\ast$$$$\large\color{turquoise}\ast$$$$\normalsize\color{orchid}\ast$$$$\small\color{limegreen}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\tiny\color{orange}\ast$$$$\scriptsize\color{goldenrod}\ast$$$$\small\color{limegreen}\ast$$$$\normalsize\color{orchid}\ast$$$$\large\color{turquoise}\ast$$$$\Large\color{purple}\ast$$$$\huge\color{blue}\ast$$$$\Huge\color{red}\ast$$
74. Jaynator495
Peanut butter... sam must have felt a disturbence in the latorce... because she just logged on xD
75. Jaynator495
76. .Sammi.
$$\style{text-shadow: 0 0 5px black;font-size:40px; } { \sf \color {white} {{\tt\rlap{\color{white}{hey~there~sam~ಠᴥಠ }}{\color{white}{\;hey~there~sam}\color{white}{\hspace{7.5px}ಠᴥಠ }} }}}$$
77. .Sam.
$\style{\text-shadow: 0 0 10px black;font-size:50px; } { \sf \color {pink} {{\tt\rlap{\color{}{hey~there~sammi~ಠᴥಠ }}{\color{violet}{\;hey~there~sam}\color{purple}{\hspace{8px}ಠᴥಠ }} }}}$
78. .Sammi.
$$\style{text-shadow: 0 0 5px black;font-size:40px; } { \sf \color {white} {{\tt\rlap{\color{white}{\;hey~there~sam~ಠᴥಠ }}{\color{white}{hey~there~sam}\color{white}{\hspace{0.6px}~ಠᴥಠ }} }}}$$
79. .Sam.
$\style{\text-shadow: 0 0 10px black;font-size:50px; } { \sf \color {pink} {{\tt\rlap{\color{}{hey~there~sammi~ಠᴥಠ }}{\color{violet}{\;hey~there~sam}\color{violet}{\hspace{65px}ಠᴥಠ }} }}}$
80. .Sammi.
There you go ;)
81. .Sammi.
$$\style{text-shadow: 0 0 1200px black;font-size:40px; } { \sf \color {white} {{\tt\rlap{\color{white}{\;hey~there~sam~ಠᴥಠ }}{\color{white}{hey~there~sam}\color{white}{\hspace{0.6px}~ಠᴥಠ }} }}}$$ I think I did something wrong
82. .Sammi.
LOL So screwed up it isn't even showing anything!
83. .Sam.
$\style{\text-shadow: 0 0 10px black;font-size:50px; } { \bf \color {pink} {{\bf\rlap{\color{}{hey~there~sammi~ಠᴥಠ }}{\color{violet}{\;hey~there~sammi}\color{violet}{\hspace{10px}ಠᴥಠ }} }}}$
84. .Sammi.
WHOOT
85. .Sam.
$\style{\text-shadow: 0 0 10px black;font-size:60px; } { \Huge \mathscr{ \color {white} {{\mathscr{\rlap{\color{white}{WHOOT }}{\color{white}{\;WHOOT}} }}}}}$
86. .Sam.
$\style{\text-shadow: 0 0 10px black;font-size:60px; } { \Huge \mathscr{ \color {white} {{\mathscr{\rlap{\color{red}{L }}{\color{white}{\;L} {{\mathscr{\rlap{\color{yellow}{a }}{\color{white}{\;a} } }}} {{\mathscr{\rlap{\color{lightgreen}{T }}{\color{green}{\;T} } }}} {{\mathscr{\rlap{\color{blue}{e }}{\color{lightblue}{\;e} } }}} {{\mathscr{\rlap{\color{pink}{X }}{\color{purple}{\;X} } }}} } }}}}}$
87. Jaynator495
The stuff i have opened up :P
88. Jaynator495
89. Jaynator495
And sam :P
90. Jaynator495
Maybe now that ive changed the LaTeX subject forever you'll finally notice me ;)
91. Jaynator495
$$\style{text-shadow:none;background:black;padding:10pt;border:blue 2.5pt solid;cursor:url('//camo.githubusercontent.com/fb1f8165987c11d9d130146e9432bb46f596dfa4/68747470733a2f2f78736f636b6574732e6e65742f77702d636f6e74656e742f75706c6f6164732f323031352f31312f73706f6e6765626f622e706e67'), auto;}{\color{white}\sf\Huge Spongebob}$$
92. .Sam.
$\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 60px red, 0 0 50px blue; \color:black; font-size: 150px; font-weight:bold; padding:65px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{Wot?}}}$
93. .Sam.
$\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 50px violet, 0 0 80px red; \color:black; font-size: 40px; font-weight:bold; padding:30px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{Hey~there~Sammi~ಠᴥಠ}}}$
94. .Sam.
$\rlap{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 50px violet, 0 0 80px red; \color:black; font-size: 40px; font-weight:bold; padding:30px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{Hey~there~Sammi~ಠᴥಠ}}}}{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 5px violet, 0 0 0px pink; \color:black; font-size: 40px; font-weight:bold; padding:35px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {red} {{Hey~there~Sammi~ಠᴥಠ}}}}$
95. .Sam.
$\rlap{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 50px teal, 0 0 80px teal; \color:black; font-size: 40px; font-weight:bold; padding:30px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{♜♞♝♛♚♝♞♜ }}}}{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 5px indigo, 0 0 0px blue; \color:black; font-size: 40px; font-weight:bold; padding:35px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {blue} {{ }}}}$
96. .Sam.
$\rlap{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 50px pink, 0 0 90px red; \color:black; font-size: 40px; font-weight:bold; padding:32px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{❆~❇~❈~❉~❊ }}}}{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 5px indigo, 0 0 0px blue; \color:black; font-size: 41px; font-weight:bold; padding:30px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {red} {{❆~❇~❈~❉~❊ }}}}$
97. sammixboo
I like the snowflakes, pretty pretty
98. .Sam.
Yeah they do look pretty
99. Jaynator495
I was the first on to ever use the \style command in latex... ITS IN MY FREAKING TUTORIAL :/
100. poopsiedoodle
congrats?
101. Jaynator495
Ugh, what does a guy gotta do to get noticed around here... :/ @Whitemonsterbunny17
102. poopsiedoodle
not complain about not getting noticed?
103. Dqswag
$$\rlap{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 50px blue, 0 0 80px orange; \color:black; font-size: 40px; font-weight:bold; padding:30px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {white} {{I~see~you~ಠᴥಠ}}}}{\style{ fontFamily:segoe; background:rgba(0, 0, 0, 0.); //outline:rgba(0, 255, 0, 0.4) solid 5px; \text-shadow: 1px 1px 20px , 0 0 5px violet, 0 0 0px blue; \color:black; font-size: 40px; font-weight:bold; padding:35px; border:rgba(0, 255, 0, 0.6) solid 0px; } { \normalsize\cal \color {blue} {{ I~see~you~ಠᴥಠ }}}}$$ | 15,050 | 32,572 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-40 | longest | en | 0.270624 |
http://www.teos-10.org/pubs/gsw/html/gsw_IPV_vs_fNsquared_ratio.html | 1,679,394,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00731.warc.gz | 94,558,289 | 3,238 | # gsw_IPV_vs_fNsquared_ratio
Ratio of the vertical gradient of potential
density (with reference pressure, p_ref), to
potential density (75-term equation)
## USAGE:
[IPV_vs_fNsquared_ratio, p_mid] =
gsw_IPV_vs_fNsquared_ratio(SA,CT,p,p_ref)
## DESCRIPTION:
Calculates the ratio of the vertical gradient of potential density to
the vertical gradient of locally-referenced potential density. This
ratio is also the ratio of the planetary Isopycnal Potential
Vorticity (IPV) to f times N^2, hence the name for this variable,
IPV_vs_fNsquared_ratio (see Eqn. (3.20.17) of IOC et al. (2010)).
The reference sea pressure of the potential density surface must have
a constant value.
IPV_vs_fNsquared_ratio is evaluated at the mid pressure between
the individual data points in the vertical. This function uses the
computationally-efficient 75-term expression for specific volume in
terms of SA, CT and p (Roquet et al., 2015).
Note that the 75-term equation has been fitted in a restricted range of
parameter space, and is most accurate inside the "oceanographic funnel"
described in McDougall et al. (2003). The GSW library function
"gsw_infunnel(SA,CT,p)" is avaialble to be used if one wants to test if
some of one's data lies outside this "funnel".
## INPUT:
SA = Absolute Salinity [ g/kg ]
CT = Conservative Temperature [ deg C ]
p = sea pressure [ dbar ]
( i.e. absolute pressure - 10.1325 dbar )
p_ref = reference sea pressure of the potential density surface
[ dbar ]
SA & CT need to have the same dimensions.
p & p_ref may have dimensions 1x1 or 1xN or MxN, where SA & CT are MxN.
## OUTPUT:
IPV_vs_fNsquared_ratio
= The ratio of the vertical gradient of potential
density referenced to pr, to the vertical gradient
of locally-referenced potential density.
IPV_vs_fNsquared_ratio is ouput on the same
vertical (M-1)xN grid as p_mid.
IPV_vs_fNsquared_ratio is dimensionless
[ unitless ]
p_mid = mid pressure between the individual points of the
p grid. That is, p_mid is on a (M-1)xN grid.
p_mid has units of dbar. [ dbar ]
## EXAMPLE:
SA = [34.7118; 34.8915; 35.0256; 34.8472; 34.7366; 34.7324;]
CT = [28.8099; 28.4392; 22.7862; 10.2262; 6.8272; 4.3236;]
p = [ 10; 50; 125; 250; 600; 1000;]
p_ref = 0
[IPV_vs_fNsquared_ratio, p_mid] = ...
gsw_IPV_vs_fNsquared_ratio(SA,CT,p,p_ref)
IPV_vs_fNsquared_ratio =
0.999742244888022
0.996939883468178
0.986141997098021
0.931595598713477
0.861224354872028
p_mid =
1.0e+002 *
0.300000000000000
0.875000000000000
1.875000000000000
4.250000000000000
8.000000000000000
## AUTHOR:
Trevor McDougall and Paul Barker [ help@teos-10.org ]
## VERSION NUMBER:
3.05 (3rd June, 2016)
## REFERENCES:
IOC, SCOR and IAPSO, 2010: The international thermodynamic equation of
seawater - 2010: Calculation and use of thermodynamic properties.
Intergovernmental Oceanographic Commission, Manuals and Guides No. 56,
UNESCO (English), 196 pp. Available from the TEOS-10 web site.
See Eqn. (3.20.5) of this TEOS-10 Manual.
McDougall, T.J., D.R. Jackett, D.G. Wright and R. Feistel, 2003:
Accurate and computationally efficient algorithms for potential
temperature and density of seawater. J. Atmosph. Ocean. Tech., 20,
pp. 730-741.
Roquet, F., G. Madec, T.J. McDougall, P.M. Barker, 2015: Accurate
polynomial expressions for the density and specifc volume of seawater
using the TEOS-10 standard. Ocean Modelling.
The software is available from http://www.TEOS-10.org | 1,072 | 3,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.832851 |
https://www.tutorialspoint.com/write-a-golang-program-to-find-the-sum-of-digits-for-a-given-number | 1,685,413,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00454.warc.gz | 1,150,409,760 | 9,999 | # Write a Golang program to find the sum of digits for a given number
## Examples
• num = 125 => 1 + 2 + 5 = 8
• num = 101 => 1 + 0 + 1 = 2
• num = 151 => 1 + 5 + 1 = 7
## Approach to solve this problem
Step 1: Define a function that accepts numbers(num); type is int.
Step 2: Start a true loop until num becomes 0 and define res:=0.
Step 3: Find modulo and add to res.
Step 4: Divide num by 10.
Step 5: Return res.
## Program
Live Demo
package main
import "fmt"
func findDigitSum(num int) int {
res := 0
for num>0 {
res += num % 10
num /= 10
}
return res
}
func main(){
fmt.Println(findDigitSum(168))
fmt.Println(findDigitSum(576))
fmt.Println(findDigitSum(12345))
}
## Output
15
18
15 | 231 | 700 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-23 | latest | en | 0.450837 |
https://ibacnet.org/quadratic-equations-worksheet-igcse/ | 1,569,156,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00297.warc.gz | 486,573,900 | 30,410 | In Free Printable Worksheets216 views
4.11 / 5 ( 192votes )
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Author: Oksana Kerner
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 1,297 | 6,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-39 | latest | en | 0.936353 |
https://elitetermpapers.com/samples/true-false-write-39-t-39-if-the-statement-is-true-and-39-f-39-if-the-statement-is-fa-4302238-2/ | 1,709,573,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00330.warc.gz | 218,802,965 | 18,711 | # true false write 39 t 39 if the statement is true and 39 f 39 if the statement is fa 4302238
TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false. 1) A variance is the difference between the actual result and a budgeted amount.
1) _______ 2) Variances and flexible budgets help managers gain insights into why actual results differ from planned performance.
2) _______ 3) A static budget is a budget that can be changed or altered after it is developed.
3) _______ 4) The static-budget variance can be subdivided into the flexible-budget variance and the sales-volume
variance.
4) _______ 5) A flexible budget is a budget that is developed using budgeted revenue or cost amounts and is not adjusted at the end of the budgeted period.
5) _______ 6) A flexible budget enables managers to compute a richer set of variances than a static budget does.
6) _______ 7) “Determine the actual quantity of the revenue driver,” is one step in the development of a flexible budget.
7) _______ 8) The sales-volume variance is the difference between the flexible-budget amount and the static budget amount; unit selling prices, unit variable costs, and fixed costs are held constant.
8) _______ 9) The flexible budget variance is the difference between the actual results and the flexible-budget amount for the actual levels of the revenue and cost drivers.
9) _______ 10) The flexible budget variance pertaining to revenues is also called the variance of operating income.
10) ______ 11) An input-price variance is the difference between actual quantity of input used and the budgeted quantity of input that should have been used, multiplied by the budgeted price.
1. Start by sharing the instructions of your paper with us
2. And then follow the progressive flow.
3. Have an issue, chat with us now
Regards,
Cathy, CS. | 448 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.786066 |
https://thedartmooryarncompany.com/attractions/your-question-how-many-litres-are-in-a-us-gallon-uk.html | 1,675,809,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00830.warc.gz | 569,079,312 | 18,750 | # Your question: How many Litres are in a US gallon UK?
Contents
## What is a US gallon in UK Litres?
The U.S. liquid gallon is defined as 231 cubic inches and equates to approximately 3.785 litres. One imperial gallon is equivalent to approximately 1.2 U.S. liquid gallons.
## Why are US gallons different to UK gallons?
The Americans had adopted a system where a gallon was comprised of 231 cubic inches of water. As a result, the U.S. gallon is 83.3 per cent of the Imperial gallon; put it another way, the Imperial gallon is about one-fifth or 20 per cent greater in volume than the American gallon.
## What is 10 US gallons in UK Litres?
US Gallons to Litres
US GALLONS IMPERIAL GALLONS LITRES
10 8.327 37.854
11 9.159 41.64
12 9.992 45.425
13 10.825 49.21
## Does the UK use gallons or liters?
the imperial gallon (imp gal), defined as 4.54609 litres, which is or was used in the United Kingdom, Canada, and some Caribbean countries; the US gallon (US gal) defined as 231 cubic inches (exactly 3.785411784 L), which is used in the US and some Latin American and Caribbean countries; and.
## What does the word Litres mean?
liter, litre, l, cubic decimeter, cubic decimetrenoun. a metric unit of capacity, formerly defined as the volume of one kilogram of pure water under standard conditions; now equal to 1,000 cubic centimeters (or approximately 1.75 pints)
THIS IS FUN: Quick Answer: What is the largest cemetery in UK?
## How much does 1000l of water weigh?
One litre of water has a mass of almost exactly one kilogram when measured at its maximal density, which occurs at about 4 °C. Similarly: one millilitre (1 mL) of water has a mass of about 1 g; 1,000 litres of water has a mass of about 1,000 kg (1 tonne).
## What is a litre of water?
litre (l), also spelled liter, unit of volume in the metric system, equal to one cubic decimetre (0.001 cubic metre). From 1901 to 1964 the litre was defined as the volume of one kilogram of pure water at 4 °C (39.2 °F) and standard atmospheric pressure; in 1964 the original, present value was reinstated.
## Is mpg US or imperial?
How Do I Convert MPG to L/100km? Miles and gallons are imperial measurements that are used in the United States. You’re unlikely to see these measures in Europe or other parts of the world. Instead, European countries use liters per 100 kilometers, or l/100km, to measure fuel efficiency.
## Is a US gallon the same as a Canadian gallon?
A canadian gallon is roughly 4.5 liters and a US gallon is 3.8 liters. No matter how you measure it, it’s gonna be expensive compared to what we once knew. In case you prefer to compare a Canadian gallon to a US gallon, a Canadian gallon equals 1.2 (1.2009) US gallons. | 712 | 2,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.906976 |
https://istopdeath.com/solve-for-x-27x92-x/ | 1,670,391,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711150.61/warc/CC-MAIN-20221207053157-20221207083157-00578.warc.gz | 350,836,758 | 15,941 | Solve for x 27^x=9^(2-x)
Create equivalent expressions in the equation that all have equal bases.
Since the bases are the same, then two expressions are only equal if the exponents are also equal.
Solve for .
Simplify .
Apply the distributive property.
Multiply.
Multiply by .
Multiply by .
Move all terms containing to the left side of the equation.
Add to both sides of the equation.
Divide each term by and simplify.
Divide each term in by .
Cancel the common factor of .
Cancel the common factor.
Divide by .
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Solve for x 27^x=9^(2-x) | 148 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-49 | latest | en | 0.908504 |
https://www.thestudentroom.co.uk/showthread.php?t=1166390&page=8 | 1,540,214,486,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515041.68/warc/CC-MAIN-20181022113301-20181022134801-00100.warc.gz | 1,097,373,121 | 46,618 | # OCR 2010 A2 Biology Unit 2 - Control, Genome and Environment watch
1. (Original post by Tinkerbelle ♥)
No same, I couldn't even tell you what it's for. All I can tell you is it involves maths. And complicated looking symbols.
Chi square basiaclly is based on Null hypothesis , in which the result you get from chi square (a value) tells if the hypothesis can be true or not for a particular experiment,mathematically its sigma (O-E)^2/E
where O: is the observed allele frequency,E: Expected frequency.
EDIT: made that in rush, would have posted more info but kinda busy atm.I will do once i get the time.
2. (Original post by ibysaiyan)
Chi square basiaclly is based on Null hypothesis , in which the result you get from chi square (a value) tells if the hypothesis can be true or not for a particular experiment,mathematically its sigma (O-E)^2/E
where O: is the observed allele frequency,E: Expected frequency.
EDIT: made that in rush, would have posted more info but kinda busy atm.I will do once i get the time.
Ahh thank you!! So is it's purpose kind of like Spearmans Rho? I'm going to go over it tomorrow Doesn't sound as scary as I was imagining, though!
3. Hey, I've got a couple of questions, rep for the best answer
1) Explain how cloning an animal can help save an endangered species of mammal?
2) State three ways of setting up a gene bank for the species being cloned?
4. hey guys havnt posted in a while but i think i'm in love with lac operon :Love:
5. (Original post by ILikeTheGame)
hey guys havnt posted in a while but i think i'm in love with lac operon :Love:
Hey, why whats with it?
6. (Original post by ibysaiyan)
Hey, why whats with it?
Because of switching off the offswitch XD - I assume lol
7. (Original post by win2kpro)
Because of switching off the offswitch XD - I assume lol
wait what switch lol? promoter xD?
8. So.. how is it going guys? =}
9. (Original post by ibysaiyan)
wait what switch lol? promoter xD?
Don't you mean operator? The promoter is where the RNA polymerase sits, the operator is where the inhibiter sits/binds to.
10. (Original post by xXxBaby-BooxXx)
Don't you mean operator? The promoter is where the RNA polymerase sits, the operator is where the inhibiter sits/binds to.
11. what exactly is an operator??
and can someone please tell me what stuff from AS is unit 2 module 1 buildin upon?? im confused here
haven't we learned about protein synsthesis last year?
12. (Original post by Remarqable M)
what exactly is an operator??
and can someone please tell me what stuff from AS is unit 2 module 1 buildin upon?? im confused here
haven't we learned about protein synsthesis last year?
The Operator is just a region/length of DNA where the inhibitor binds to. It's next to the structural genes (Z and Y) that code for lactose permease and beta-galactosidase.
And the protein synthesis we learnt about was mainly protein structure etc.
It builds upon biological molecules (structure of DNA and proteins) organelles (eg ribosomes) mitosis a small amount, selective breeding, dominant/recessive alleles, variation, kingdoom/domain/species and natural/artificial selection.
So it builds upon a lot. Hope that helps
13. (Original post by Remarqable M)
what exactly is an operator??
and can someone please tell me what stuff from AS is unit 2 module 1 buildin upon?? im confused here
haven't we learned about protein synsthesis last year?
Hey! welcome back, for unit 2 module one remember the basics meh whats transcription,translation,sickle cell anemia,apoptosis and operon region.I guess thats about it.Not much there... oh also homologous genes.
14. (Original post by xXxBaby-BooxXx)
The Operator is just a region/length of DNA where the inhibitor binds to. It's next to the structural genes (Z and Y) that code for lactose permease and beta-galactosidase.
And the protein synthesis we learnt about was mainly protein structure etc.
It builds upon biological molecules (structure of DNA and proteins) organelles (eg ribosomes) mitosis a small amount, selective breeding, dominant/recessive alleles, variation, kingdoom/domain/species and natural/artificial selection.
So it builds upon a lot. Hope that helps
Thank you. This is very useful and good luck with gettin an A* im glad you're part of this thread
15. (Original post by Remarqable M)
Thank you. This is very useful and good luck with gettin an A* im glad you're part of this thread
Oh I'm not getting an A* in...well anything tbh. I'm on a B in biology at the moment, I'd like an A, but even that's fairly unrealistic. Stupid HSW. I wish exams were just factual recall, I'm not too bad at that.
But thanks
16. (Original post by xXxBaby-BooxXx)
Oh I'm not getting an A* in...well anything tbh. I'm on a B in biology at the moment, I'd like an A, but even that's fairly unrealistic. Stupid HSW. I wish exams were just factual recall, I'm not too bad at that.
But thanks
Nah we will get there xD i dont think i will be getting an A* due to my last years practicals but meh an A would do .btw. remarq how is maths going =)?
Edit: I need some rest had a long session at the gym =/ .See you all later.
17. (Original post by ibysaiyan)
Nah we will get there xD i dont think i will be getting an A* due to my last years practicals but meh an A would do .btw. remarq how is maths going =)?
Edit: I need some rest had a long session at the gym =/ .See you all later.
not too bad i like everything in c4 except vectors.
18. (Original post by Remarqable M)
not too bad i like everything in c4 except vectors.
Hmm we can perhaps make one(thread) based on it :P?i wonder... anyways out for now.
19. (Original post by Remarqable M)
not too bad i like everything in c4 except vectors.
Urrrgh same. I hate vectors
20. Hey all.. how is it going ?
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### What is the intuition behind Expected Utility Theorem?
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### Proof that $U(\sum_{n=1}^{N}{p_nL_n})=\sum_{n}^{N}{p_nU(L_n)}$
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### How to Represent as a Payoff Matrix
I'm trying to represent the following as a pay-off matrix. I have 100 dollars to invest in one agricultural stocks with a choice of apples, pears or grapes. Return on investment relies on whether ...
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1 vote
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### Comparing & contrasting decision problems and normal games
I am trying to compare and contrast between decision problems and normal games. Are there any key concepts I should know? Any help would be greatly appreciated.
• 23
1 vote
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### Choice under Uncertainty: Relation between the certainty equivalent and the coefficient of relative risk aversion
That's my question! Propositon 6.C.4 (MWG (1995)): The following conditions for a Bernoulli utility function $u(\cdot)$ on amounts of money are equivalent: (i) $r_R(x,u)$ is decreasing in $x$. (iii) ...
63 views
### Uncertainty and Pareto efficient policies
There are two economic agents $i\in \{1,2\}$ with state dependent utility $u_{is}=-(x-b_{is})^2$ where $x\in R$ and $b_{is}\in R$ is bliss point of $i$ in state $s\in\{1,2\}$. Assume $b_{1s}\lt b_{2s}$...
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### Epstein zin and resolution of uncertainty
I'm reading Simon Gilchrist's notes here. I understood everything until and including page 14, where it reads $$\frac{W_h^{1-\rho} + W_l^{1-\rho}}{2} \geq \frac{c_h^{1-\rho} + c_l^{1-\rho}}{2}$$ and ...
• 131
54 views
### Proof of certainty equivalence
The questions say two lotteries are denoted L1 = (0.3,0.7,0.0) and L2= (0.9,0.0,0.1) and denote c(L1) and c(L2) the certainty equivalents of those two lotteries. Then prove L1 ≻ L2 if and only if c(L1)...
613 views
### Can the Certainty Equivalent be negative?
I am questioning if the CE of a lottery can be negative? For me it doesn't make much sense by definition. I encountered this problem on the following exercise: Imagine a case where we have a lottery(...
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### Lotteries = probability distribution?
Are "lotteries" in the model for choice under uncertainty not just probability distributions?
• 155
1 vote
39 views
### Term for risk AND ambiguity
This question is related to References for particular definitions of risk and uncertainty, which offers an excellent description of risk and uncertainty. Just as a recap: Knight (1921) described risk ...
• 105
1 vote
415 views
### Can we compare risks of lotteries?
I understand the concepts of risk-aversion, risk neutrality and risk-attraction. I wonder if it possible to compare risks between two lotteries without giving the utility function. For instance, Let ...
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### Intertemporal choice with possibility of death
Here is the setup: Suppose that there is an individual who lives up to two periods. He lives with absolute certainty during period $1$, and during this period his sub-utility function is given by: ...
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### How to prove the relationship between the expected value of a lottery and its certainty equivalent?
Utility function $u(x)$ is monotonic. I want to prove that $u(x)$ exhibits risk aversion if and only if for all lottery $F$: $E(x) \geq CE(F,u)$ (CE is certainty equivalent). (Definition of $CE$: the ...
• 372
1 vote
214 views
This is definition of Radner Equilibrium from Microeconomic Theory (Mas-Collel, Whinston and Green - Third Edition). I'm confused about two conditions: $\sum_k q_k \cdot z_{ki} \le 0$ (in yellow) The ...
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### Existence of optimal strategy in a choice problem with uncertainty and information structure
Consider a decision maker choosing an action, $y$, from the finiteset $\mathcal{Y}$, possibly without having complete information about the state of the world. More precisely, let $V$ be a ...
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### Robust predictions in single-agent decision problem with uncertainty
I would like your help to better understand the possibility of using the notion of Bayes Correlated Equilibrium (BCE) in a single-agent decision problem with uncertainty to make predictions on optimal ...
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### Optimal strategy in a single-agent choice problem under uncertainty
Consider the following single-agent choice problem under uncertainty. Let $V$ be the state of the world with support $\mathcal{V}$ and probability distribution $P_V\in \Delta(\mathcal{v})$. First, ...
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1 vote
48 views
Suppose that $q$ is a k-tuple vector of prices for the k assets whose quantities are given by the k-tuple $\theta$. I have just read that in the Radner Sequential Trade Equilibrium (not sure if this ...
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### Dominated lotteries in CPE
I have been looking into expectation-based loss aversion following Kőszegi-Rabin (2005, 2007). In particular, I find their choice-acclimating personal equilibrium (CPE) interesting, but it has a ...
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### Existence of 'best' and 'worst' lottery
How can 'the best and worst lotteries exist when the set of outcome is finite and the rational preference relation satisfies independence axiom' be proven?
1 vote
147 views
### Archimedean but not mixture continuous
In the context of preferences on a set of lotteries on a finite set $X$, what is an example of a preference that is independent, Archimedean but not mixture continuous? I know the mixture continuous ...
• 193
154 views
### Why Certainity Eqivalence in PIH only holds for quadratic utilities
In my current macro economics course, it has been stated that there is certainty equivalence in the random walk permanent income hypothesis ("which implies individuals act as if future consumption was ...
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328 views
### Question on uncertainity
Please imagine that Nicole is uncertain of her future wealth. Her wealth in the bad state of the world is zero. Her wealth in the good state is $w>0$. Each state is initially equally likely. ...
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### Indiference between two lotteries
Suppose that a binary relation satisfies only: Independence axiom: $L≿L′⟺α\circ L+(1−α)\circ L′′≿α\circ L′+(1−α) \circ L′′$ Reduction to simple lotteries: For all $g$, $g~g'$, $g'$ is the simple ...
329 views
### Uncertainty in an unfair gamble
If a risk averse person is given the option of a certain amount of 2000 or playing a lottery game giving him 10000 with 25% probability, and 500 with 75%, then what would he do ?. The bet here is not ...
• 133
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### Profit maximization under uncertainity
I have a seller say S and I have a buyer say B. Buyer’s willing to pay is equal to x which is private information. But Seller believe that it falls in the range [0,x1]. Seller’s belief distribution is ...
1k views
### What exactly is certainty equivalence in the context of DSGE models?
I keep reading about certainty equivalence in the context of DSGE models. I understand that it has something to do with "Getting rid of the expectation operator", but I'm not entirely sure? ...
• 803
59 views
### A growth model with regime switch
I have a very general question. I am reading this paper : http://www.webmeets.com/files/papers/eaere/2015/177/Discounting-HelsinkiBlind.pdf There is a catastrophic event probability and after the ...
• 2,095
990 views
### References for particular definitions of risk and uncertainty
I have some doubts about risk vs. uncertainty. I have read the thread "What is the difference between risk, uncertainty and ambiguity" and have skimmed through Knight's "Risk, Uncertainty, and Profit" ...
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204 views
• 1,053
400 views
### Inc Linear Transformation of Bernoulli Utility
According to MWG Proposition 6.B.2, it illustrates that the expected utility form is preserved only by increasing linear transformation. What is the significance of this proposition? The part I ...
• 1,053
1 vote
75 views
### Consequentialist View of Risk
In MWG, the authors introduce the consequentialist view of risk by assuming for any risky alternative, only the reduced lottery over final outcome matters to decision maker. From philosophical view, ...
• 1,053 | 2,637 | 10,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-27 | longest | en | 0.867228 |
http://spmath90509.blogspot.com/2009/10/scribe-post-for-october-16-2009.html | 1,503,510,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00424.warc.gz | 380,065,439 | 14,311 | ### Scribe Post for October 16, 2009
Saturday, October 17, 2009
Hey, 9-05. Today in class, we had a math test. It was mostly on surface area and bit on symmetry too. Today I will be showing you how to do questions on the test we had. One thing is, that you really need to know your formulas for all the shapes and also know when to use them.
These question are examples of what you need to know in order to answers the questions right. But they are not same as the test. The reason for this is that not everybody has taken the test yet and I don't have the test in front of me so I don't remember the exact measurements that were on the test.
These are some examples:
To figure the cylinder you need to know the formula for a cylinder.
Which is:
H=15
S.A. = 2πr²+2πrh
= 2π(3²)+2π(3)(15)
= 2π(9)+2π(45)
= 2(28.27)+2(141.37)
= (56.54)+(282.74)
= 339.28cm²
This is the formula for a rectangular prism
2(lw)+2(lh)+2(hw)
2(4)(2)+2(4)(1)+2(1)(2)
2(8)+2(4)+2(2)
16+8+4
=28
The homework is on the grade 9 homepage.
Melanee 9-06 said...
i know i'm not from 9-05 or anything.. i just like to comment.. good job on that post...:)
Lissa 9-05 said...
Good job on your scribe post! (: I like your use of colour.
NickyD905 said...
Very nice scribe Brendan. One little thing, there may be just a bit too much color. Yes color is nice but, you can easily have too much of a good thing. Also, there's two little picture, things... You may want to fix that. Still, great scribe Brendan!
melanie905 said...
Nicky, what do you mean too much colour? Haha, you should check out my posts. Well anyways, great job Brendan! | 486 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-34 | longest | en | 0.953547 |
https://kennyspeeds.com/service-and-maintenance/why-most-of-the-induction-motors-are-delta-connected.html | 1,652,758,200,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00521.warc.gz | 437,181,922 | 19,027 | # Why most of the induction motors are Delta connected?
Contents
For this reason, Induction Motor is started with Star connection to reduce the starting current by reducing the voltage across windings. … So in running condition, delta connection is used to develop high torque by providing the high voltage across windings.
## Is induction motor connected in Star or Delta?
In star delta starting an induction motor is connected in through a star connection throughout the starting period. Then once the motor reaches the required speed, the motor is connected in through a delta connection. A star delta starter will start a motor with a star connected stator winding.
## Why LT motors are Delta connected?
Starting torque should be large, as motors are of small capacity and hence Stator should be connected in Delta to have more current and hence more starting torque.
## Why is delta connection used?
Applications and Uses of Delta Connection
Delta connection is used for the running of induction motor, as it helps to get more speed and torque. … Delta connection is used in high voltage overhead transmission where insulation is not required but the flow of current is an important matter.
## Which is better Delta or star connection?
Star and Delta Connections are the two types of connections in a 3 – phase circuits. A Star Connection is a 4 – wire system and a Delta Connection is a 3 – wire system.
Comparison between Star and Delta Connections.
Star Connection (Y or Wye) Delta Connection (Δ)
Line Current and Phase Current are same. Line current is root three times the phase current.
## Which connection is better for motor Star or Delta?
The speeds of Star connected motors are slow as they receive 1/√3 voltage. The speeds of Delta connected motors are high because each phase gets the total of line voltage. … In Delta Connection, Motor receives highest Power output. In Star Connection, the phase voltage is low as 1/√3 of the line voltage.
## Is Delta high or low voltage?
If leads are numbered 1-6, the winding can usually be connected wye or delta. On machines rated for two voltages, the wye connection is for the high voltage; the delta connection is for the low voltage.
## Why HV motors are star connected?
In the high voltage motor, the current is often small, and the insulation grade of the motor is required to be higher, so the insulation of the motor with star connection is better and more economical.
## Can we run Delta motor in Star?
The answer to your question is no. One end of each star winding is connected internally to the neutral point. Since the neutral connection is internal, it cannot be opened to make a delta.
THIS IS EXCITING: Frequent question: Why does engine symbol on dashboard mean?
## What is Delta and Y connection?
Delta connection is used for shorter distances, whereas wye connection is used for power transmission networks for longer distances. Delta was primarily used at small industrial facilities that had a relatively large (240 VAC) motor load but only a small need for convenience outlets and lighting.
## Why delta connection has no neutral?
No there isnt. Delta connection is a 3 phase connection, which simply means that 3 wires (single phase) are connected such that their disimmilar ends are connected together. Being connected end to end, there isnt one common point as opposed to star connection.
## What is Delta delta connection?
A delta connection is a connection used in a three-phase electrical system in which three elements in series form a triangle, the supply being input and output at the three junctions.
## How do I know if my motor is delta or star?
A star only motor has four terminals, three phase terminals and a neutral. A delta only motor has three phase terminals only. A motor designed to be operated either star or delta has each end of each winding brought out, so you have six terminals to work with. | 791 | 3,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.929714 |
https://www.riddlesandanswers.com/v/232409/calculate-the-number-of-degrees-between-the-hour-hand-and-the-minute-hand-of-a-clock-nondigital-th/ | 1,610,964,041,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00234.warc.gz | 973,364,823 | 22,358 | # THE NON DIGITAL CLOCK RIDDLE
#### Trending Tags
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## The Non Digital Clock Riddle
Calculate the number of degrees between the hour hand and the minute hand of a clock (nondigital) that reads 3:15.
Hint:
The hour hand will have moved one-fourth of an hour; therefore, there will be 7.5 degrees between the two hands.
Did you answer this riddle correctly?
YES NO
Solved: 70% | 106 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.865695 |
https://www.education.com/resources/math-word-problems/CCSS-Math-Content-4/ | 1,656,516,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00223.warc.gz | 793,285,075 | 38,397 | # Search 4th Grade Math Word Problem Educational Resources
74 filtered results
74 filtered results
Math Word Problems
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Summer Word Problems
Worksheet
Summer Word Problems
Get your mind going with these super summer math word problems!
Math
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Math Mixed Review Part 1: Flying Through Fourth Grade
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Math Mixed Review Part 1: Flying Through Fourth Grade
From multi-digit subtraction to identifying prime numbers, assess each student’s mastery of a variety of fourth-grade math skills.
Math
Worksheet
Mixed Word Problems
Worksheet
Mixed Word Problems
Your fourth grader will hardly realize they are learning as they complete some entertaining mixed word problems to help out with birthday party logistics.
Math
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Autumn Word Problems
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Autumn Word Problems
Welcome the season with autumn-themed word problems. Your student will have the opportunity to practice addition, subtraction, multiplication, and division.
Math
Worksheet
Number Crunchers: Operations Practice
Workbook
Number Crunchers: Operations Practice
This workbook is packed with worksheets that let kids sharpen math skills by practicing the four basic math operations as well as factoring.
Math
Workbook
Measuring Cup Musings
Worksheet
Measuring Cup Musings
Students use a diagram of a measuring cup to answer six questions about measurement conversions between cups and ounces.
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Elevations: Multi-Step Word Problems
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Elevations: Multi-Step Word Problems
Students will solve these easy multi-step word problems to help them out. They'll use the chart provided to solve the addition and subtraction word problems.
Math
Worksheet
Ninja Multiplication
Workbook
Ninja Multiplication
Take a crack at multiplication from a different angle! This book supplies a series of sheets looking at multiplying numbers with flashcards, lattice grids and times tables.
Math
Workbook
Wild Word Problems: Multiplication
Worksheet
Wild Word Problems: Multiplication
These multi-digit multiplication word problems draw inspiration from the animal kingdom.
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Worksheet
Spring Word Problems
Worksheet
Spring Word Problems
Spring into the season with these math word problems!
Math
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Classroom Math: Multiplication Word Problems
Worksheet
Classroom Math: Multiplication Word Problems
Your fourth grader will be challenged to complete five word problems using multiplication with two-digit numbers.
Math
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Division Word Problems
Worksheet
Division Word Problems
Fourth graders will gain additional practice in strengthening their math skills with this worksheet featuring division word problems.
Math
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Math Skills: Word Problems
Worksheet
Math Skills: Word Problems
Test your fourth grader's command of general math skills by challenging him to complete a handful of word problems.
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Word Problems: Division
Worksheet
Word Problems: Division
Use this worksheet to practice long division with one-digit divisors.
Math
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Blake’s Afternoon at Sunnyvale Cinemas: Multi-Step Word Problems
Worksheet
Blake’s Afternoon at Sunnyvale Cinemas: Multi-Step Word Problems
Practice solving multi-step word problems in the context of real-world situations with this engaging math worksheet!
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Cool Cupcake Word Problems
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Cool Cupcake Word Problems
Kids answer sweet two-step word problems in this worksheet.
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Divide Them Up
Workbook
Divide Them Up
This workbook gets kids ready to be expert dividers, challenging them with whole-number quotients, remainders and mathematical word problems.
Math
Workbook
Multiplication: Word Problems (Part One)
Worksheet
Multiplication: Word Problems (Part One)
Solve word problems using one of the following strategies: draw an array, draw equal groups, skip count forward, repeated addition, or multiplication sentences.
Math
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Multiplication Word Problems: Money, Money, Money!
Worksheet
Multiplication Word Problems: Money, Money, Money!
Solve the money-themed word problems using multiplication. Don't forget to check your answers using the answer sheet.
Math
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Math Review Part 2: Fun with Multiplication and Division
Worksheet
Math Review Part 2: Fun with Multiplication and Division
Check students’ understanding of various multiplication and division strategies taught in fourth grade.
Math
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Math Review Part 3: Geometry Galore!
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Math Review Part 3: Geometry Galore!
Use this year-end assessment to check students’ grasp of key fourth grade geometry concepts.
Math
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Word Problem Assessment: Flying Through Fourth Grade
Worksheet
Word Problem Assessment: Flying Through Fourth Grade
In this assessment, your students will flex their math muscles as they use all four operations to solve a series of word problems.
Math
Worksheet
Divide in Real Life
Worksheet
Divide in Real Life
This math exercise allows your students to use the information and objects around as they apply their division skills in real-life scenarios.
Math
Worksheet | 989 | 4,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-27 | latest | en | 0.86468 |
http://math.stackexchange.com/questions/108478/if-a-normal-k-f-has-no-intermediate-extensions-then-k-f-is-prime | 1,466,857,305,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783393146.70/warc/CC-MAIN-20160624154953-00009-ip-10-164-35-72.ec2.internal.warc.gz | 198,525,770 | 18,142 | # If a normal $K/F$ has no intermediate extensions, then $[K : F]$ is prime
Let $K$ be a finite normal extension of $F$ such that there are no proper intermediate extensions of $K/F$. Show that $[K:F]$ is prime. Give a conterexample if $K$ is not normal over $F$.
-
Do you know some about Galois correspondance ? – Lierre Feb 12 '12 at 13:10
@Lierre: the Galois correspondence is not necessary here. You can prove it just starting from the definition of $[K:F]$. – Damian Sobota Feb 12 '12 at 13:32
@DamianSobota: Maybe (although I don't see any immediate solution without it) ! But if it is an exercise, we should guess what is the expected proof. That's why I asked. – Lierre Feb 12 '12 at 13:37
@Damian: really? This is quite amazing: could you please elaborate? – Georges Elencwajg Feb 12 '12 at 17:19
1) If $K/F$ is Galois and there is no strictly intermediate extension, Galois theory tells us that $Gal(K/F)$ has no non-trivial subgroup and thus is of order a prime $p$. Hence $K/F=p$ is prime too.
2) Let's exhibit for every $n\gt 1$ a field extension $K/F$ of degree $[K:F]=n$ without any strictly intermediate subfield.
a) Take any Galois extension $L/F$ with Galois group the full symmetric group $S_n$.
[This is easy to find: for example, take $k(T_1,...,T_n)/k(s_1,...,s_n)$ where the $T_1,...,T_n$ are indeterminates over an arbitrary field $k$ and the $s_i$'s are the elementatary symmetric in these indeterminates. There are examples with $F=\mathbb Q$ too, but that is more difficult]
b) Take $K=L^{ S_{n-1}}$, the fixed field under the subgroup $S_{n-1}\subset S_n$.
Since there is no subgroup strictly between $S_{n-1}$ and $S_n$, Galois theory implies that there is no strictly intermediate field between $F$ and $K$.
Edit
Answer 1) remains true under br69's weaker hypothesis that the extension only be finite and normal (but not necessarily Galois):
1') If the finite normal extension $K/F$ has no strictly intermediate extension, then $[K:F]$ is prime
Proof:
We start from the tower $F\subset K_{sep }\subset K$. The no-intermediate-field hypothesis ensures that one of the following two possibilities holds:
i) $K_{sep }= K$. Then the extension $F\subset K$ is Galois and we are back to the Galois case.
ii) $K_{sep }= F$. Then the extension $F\subset K$ is purely inseparable, hence we are in characteristic $p\gt 0$ and $[K:F]=p^r$.
Now for every $b\in K\setminus F$ there exists some power $b^{p^s}=c$ with $c\notin F$ but $c^p\in F$ .
The inclusions $F\subsetneq F(c)\subset K$ and the no-intermediate-field hypothesis force $F(c)=K$ and thus $[K:F]=[F(c):F]=p$ as desired.
-
Can you give me ab example where extension has composite order and still has no sub-extensions? – Swapnil Tripathi Oct 6 '14 at 22:20
@Swapnil: for any composite $n$ part 2) of the answer gives an example. – Georges Elencwajg Oct 7 '14 at 10:54 | 855 | 2,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-26 | latest | en | 0.854254 |
https://www.techiesin.com/how-many-ounces-is-50g/amp/ | 1,653,698,458,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00449.warc.gz | 1,189,642,378 | 19,588 | Techies In
# How many ounces is 50 g
## How many ounces = 50 gram ?
To convert 50g to oz divide the mass in grams by 28.349523125. The formula for 50 grams in ounces is [oz] = [50] / 28.349523125.
Thus we get (all results rounded to 2 decimal places):
50g to oz = 1.76oz
50 g in ounces = 1.76 ounces
50 grams to ounces = 1.76 ounces
## What is an Ounce?
The ounce is the designation for several different mass, weight or volume units and is derived almost unchanged from the uncia, an ancient Roman unit of measurement.
The avoirdupois ounce is 28.35 grams, and the fine and apothecary ounce is 31.103 grams.
As a unit of volume, the measurement of a fluid ounce is 1/16 pint or 29.57 millilitres in the US Customary System and 1/20 pint or 28.41 millilitres in the British Imperial System.
## What is a Gram?
The gram (SI unit symbol g) is a unit of mass in the metric system.
Initially defined in 1795 as “the absolute precise weight of a volume of pure water equal to one-hundredth of a cubic meter [1 cm3] and at the melting temperature of ice”, the actual temperature (~0 °C) was later changed to 4 °C, the temperature of the maximum density of water.
However, in the late 19 century, there were efforts to convert the base unit to kilogram and the gram into a derived unit.
Also, in 1960, the new International System of Units defined a gram as one-thousandth of a kilogram (i.e. a gram is one × 10−3 kg).
### What are the grams of 1 ounce?
There are about approx 28 grams in an ounce.
### What are the grams of 3.5 oz?
100g = 3.5274 ounces
### How do you convert 50 grams into ounces?
50 grams = 1.764 ounces
Formula: multiply the value in grams by the conversion factor ‘0.0352739619496’.
So, 50 grams = 50 × 0.0352739619496 = 1.76369809748 ounces.
50 grams as a usable fraction or an integer in ounces:
3/4 ounces (-0.78% smaller)
These are alternative values for 50 grams in ounces.
They represent a fraction of an integer close to the exact value (12, 14, 34 etc.). The approximation error, if any, is to the right of the value.
Knowing how to convert between different weight units can be an outstanding skill in the kitchen.
For example, you can use recipes created in the metric system, where weight is in grams, a style more familiar to American chefs, who typically measure ingredients in ounces of weight.
Another type of conversion occurs when you want to change the total quantity of the recipe, for example, to make half or double.
It can get even more complicated when you’re doing both types of conversions: changing from metric to imperial measurements (or vice versa) and changing totals.
Getting the proper conversions can make or break your bottom line.
## Imperial and Metric Equivalents
As long as you understand the basic equivalences, no fancy math skills must perform simple imperial to metric conversions.
You can use as many flows or tables as you like, or you can use a simple online conversion app.
Handy conversion apps are also available for smartphones.
### Rudimentary conversions include:
Ounces Pounds Grams Kilograms 1 1/16 28 0.028 4 1/4 113 0.113 8 1/2 227 0.227 16 1 454 0.454
Key Equivalencies:
• 1 ounce = 28 grams
• 1 pound = 16 ounces
• 1 pound = approximately 1/2 kilogram
• 1 kilogram = 1,000 grams
• 1 kilogram = 2.2 pounds
For example, if you have to convert a recipe developed in Europe to quantities more familiar to US cooks, converting grams to ounces is simply a matter of dividing the grams by 28.
## Ounces of weight vs Ounces of volume
It is important to remember that the term “ounce” is used both as a unit of weight and volume.
Dry ingredients like beans, flour, and sugar are often measured in ounces by weight, while liquid and other wet ingredients measure in volume or fluid ounces.
On packaged products, weight ounces we list as ‘NET WT OZ’, where WT means weight, while volume ounces are “NET OZ FL”, where FL stands for liquid.
If an ingredient requires FL OZ, don’t mistake measuring it by weight on a scale. Instead, use a graduated measuring cup marked in fluid ounces.
The gram (g) was the fundamental unit of mass in the 19-century centimetre-gram-second (CGS) system of units.
The CGS system coexisted with the meter-kilogram-second (MKS) units, first proposed in 1901 for much of the 20th century.
Still, the gram was replaced by the kilogram as the basic unit of mass when the MKS system became Chosen for SI base units in 1960.
### Applications
The gram is now the most widely used unit of measurement for non-liquid ingredients in cooking and grocery shopping around the world. Liquid ingredients can be measured by volume instead of mass.
Many regulations and legal requirements for the nutritional value labelling require the product’s relative contents per 100 g. The resulting number can also be read as a percentage.
## Gram Conversion Factors
1 gram (g) = 15.4323583529 grains (gr)
1 grain (gr) = 0.06479891 grams (g)
100 grams (g) = 3.527396195 ounces
1 gram (g) = 5 carats (ct)
1 avoirdupois ounce = 28.349523125 grams (g)
1 gram = 8.98755179 × 1013 joules (J) (by mass-energy equivalence)
1 troy ounce = 31.1034768 grams (g)
One-eleventh gram = 1 “eleventh gram” = 10−11 grams in the historic quadrant eleventh gram second (QES) system, also known as the hebdometer eleventh gram second (HUS) system (HUS system)
500 grams (g) = 1 Jin in Chinese units.
### Ounces to Pounds to Tons
Since your almond cookie recipe calls for two pounds of sliced almonds and doesn’t list ounces, you’ll need to convert or switch from one unit of measure to another.
Since we are going from a larger team to a smaller unit of action, we need to multiply.
One pound, also written as “lb,” equals 16 ounces, so multiply the number of pounds necessary by the number of ounces in a pound. You need 32 ounces to equal two pounds.
Some objects are so massive that ounces or pounds won’t do. Think of things like an elephant, a car, or a tub of almond cookies made in a factory.
In such cases, you need to measure the weight of the objects in tons.
One ton, also written as “T”, equals 2,000 pounds. To go from ounces to tons, you would need to do two conversions, first converting ounces to pounds and then converting pounds to tons by multiplying that value by 2,000.
If you were trying to go from a smaller unit like ounces to a larger unit like pounds, you would have to divide to convert the measurements. (32 ounces/16 ounces in a pound = 2 pounds)
## FAQ’s
You have just arrived at the FAQ section of our 50g per ounce, which includes some of the most frequently asked questions:
How much does 50g weigh?
Fifty grams is 1.7 ounces, which makes a CD case a good candidate for weighing around 50 grams.
How many ounces are there in 50 grams?
Ounce = [50] / 28.349523125.How many ounces in 50g?
How many ounces are there in 50 grams?
In 50 g there are 1.76 ounces
How many ounces are there in 100 grams?
There are 3.5274 ounces in 100 grams.
If you read all of our information, chances are you know all the answers on how to convert 50 grams to ounces.
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May 16, 2022 | 2,071 | 8,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.875452 |
https://www.kth.se/student/kurser/kurs/FSF3940 | 1,627,546,335,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00355.warc.gz | 892,447,874 | 14,530 | # FSF3940 Probability 7.5 credits
### Choose semester and course offering
Choose semester and course offering to see information from the correct course syllabus and course offering.
Headings with content from the Course syllabus FSF3940 (Spring 2019–) are denoted with an asterisk ( )
## Content and learning outcomes
### Course contents
Probability is the mathematical theory for studying randomness and is one of the fundamental subjects in applied mathematics.For a rigorous treatment of probability, the measure theoretic approach is a vast improvement over the arguments usually presented in undergraduate courses. This course gives an introduction to measure theoretic probability and covers topics such as the strong law of large numbers, the central limit theorem, conditional expectations, and martingales. It is expected, but not required, that students have had some exposure to measure theory prior to taking this course. Students will practice by studying applications and solving problems related to the theory.
### Intended learning outcomes
After completing the course students are expected to
• explain the foundations of probability in the language of measure theory,
• state the strong law of large numbers and give an outline of its proof
• have a basic knowledge of other 0-1 laws in probability
• have a working knowledge of weak convergence, characteristic functions, and the central limit theorem,
• give examples and applications of the strong law of large numbers and the central limit theorem,
• explain the concepts of recurrence and transience of random walks,
• explain the content of the Radon-Nikodym theorem and give an outline of its proof,
• explain the concept of conditional expectation, its properties and applications
• give an introduction to discrete time martingales and the martingale convergence theorem
• give examples and applications of martingales
• be able to solve basic problems related to the theory
### Course disposition
The course will consist of roughly bi-weekly discussion meetings (not standard lectures) where students present and discuss the material as well as some weekly exercises. The topic for each meeting is given below. Any of the books mentioned above can be used. The last two books are quite technical and perhaps not as accessible as the first six.
1. Measur and Integration (sigma-field, measure, integration)
2. Random variables, Expected Value, Independence
3. The Law of Large Numbers (Borel-Cantelli, 0-1 Laws, Applications)
4. The Central Limit Theorem (Generating Functions, Weak Convergence, Applications)
5. Random Walks (Recurrence, Transience)
6. Conditional Expectation (Radon-Nikodym Theorem, existence, properties)
7. Martingales (Martingale Convergence, Applications)
## Literature and preparations
### Specific prerequisites
A Master’s degree in mathematics, applied mathematics or related field including at least 30 ECTS in mathematics. Recommended courses are SF2940 Probability theory and SF2743 Advanced real analysis I.
### Recommended prerequisites
No information inserted
### Equipment
No information inserted
### Literature
1. Rick Durrett, Probability: Theory and Examples, 4th Edition, Cambridge Series in Statistical and Probabilistic Mathematics, 2010. ISBN 9780521765398
2. Jean Jacod and Philip Protter: Probability Essentials, Corrected Second Printing, Springer Verlag, 2004. ISBN 3-540-43871-8
3. Sid Resnick, A Probability Path, Birkhäuser Boston, 5th printing, 2005.
4. Allan Gut, Probability: A Graduate Course, Springer, 2005.
5. David Pollard, A User’s Guide to Measure Theoretic Probability, Cambridge University Press, 2002.
6. Patrick Billingsley, Probability and Measure, 3rd Edition, Wiley.
7. Daniel Stroock, Probability: An Analytic View, 2nd Edition, Cambridge University Press, 2011.
8. Olav Kallenberg, Foundations of Modern Probability, 2nd Edition, Springer, 2002.
## Examination and completion
If the course is discontinued, students may request to be examined during the following two academic years.
P, F
### Examination
• HEM1 - Home assignments, 3,5 hp, betygsskala: P, F
• TENM - Oral exam, 4,0 hp, betygsskala: P, F
Based on recommendation from KTH’s coordinator for disabilities, the examiner will decide how to adapt an examination for students with documented disability.
The examiner may apply another examination format when re-examining individual students.
The examination will be done as a combination of homework and oral exam.
### Other requirements for final grade
Hemwork and oral exam
### Opportunity to complete the requirements via supplementary examination
No information inserted
### Opportunity to raise an approved grade via renewed examination
No information inserted
### Ethical approach
• All members of a group are responsible for the group's work.
• In any assessment, every student shall honestly disclose any help received and sources used.
• In an oral assessment, every student shall be able to present and answer questions about the entire assignment and solution.
## Further information
### Course web
Further information about the course can be found on the Course web at the link below. Information on the Course web will later be moved to this site.
Course web FSF3940
SCI/Mathematics
### Main field of study
No information inserted
Third cycle | 1,135 | 5,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-31 | latest | en | 0.916009 |
https://www.experts-exchange.com/questions/27030952/Convert-datediff-calculation-to-decimal.html | 1,513,533,060,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597295.74/warc/CC-MAIN-20171217171653-20171217193653-00551.warc.gz | 740,164,525 | 30,939 | Solved
# Convert datediff calculation to decimal
Posted on 2011-05-11
Medium Priority
482 Views
Hi,
How would I convert this query to return a result in decimal:
After_Shift = datediff(minute, TimeAfterShiftStart, MAX(ClockOut)) / 60.0
0
Question by:NerishaB
LVL 15
Expert Comment
ID: 35736092
DATEDIFF(second, start_date, end_date) / 60.0 / 60.0
0
LVL 3
Accepted Solution
Krtyknm earned 1500 total points
ID: 35736104
Use CONVERT or CAST to convert it to decimal.
SELECT CONVERT(DECIMAL(5,1),datediff(minute, GETDATE()-1, GETDATE()) / 60)
SELECT CAST(datediff(minute, GETDATE()-1, GETDATE()) / 60 AS DECIMAL(5,1))
0
LVL 21
Expert Comment
ID: 35736107
It also give in decimal
0
Author Comment
ID: 35736132
@ Krtyknm: The result I am getting is something like : 10582011.001
The result should be : 0.011
0
Author Comment
ID: 35736261
What I actually need to do is find a way to use both the sum and the max in the same datediff function, like so:
TotalWorkHours = CAST(sum(datediff(minute, ClockIn, MAX( ClockOut)) / 60.0) as decimal(4,2)),
I get the following error though:
"Cannot perform an aggregate function on an expression containing an aggregate or a subquery."
Any help?
0
LVL 9
Expert Comment
ID: 35736373
Can you share your table schema with sample input and sample output?
0
LVL 3
Expert Comment
ID: 35736651
Hi NerishaB,
Can you explain brefily about your query like what are the fields and their datatype.
Thanks,
Karthik
0
Author Comment
ID: 35736759
The query is very large, see attached
; with emp_clock as
(
Select Personel_ID = Pers.Personel_ID,
FullName = Pers.[Name] + ' ' + Pers.Surname,
RowNo = row_number() over (partition by Pers.Personel_ID order by RC.ClockTime),
ClockDate = dateadd(day, datediff(day, 0, RC.ClockTime), 0),
DayAdj = case when Default_Shift_End < Default_Shift_Start then 1 else 0 end,
ClockTime = RC.ClockTime,
InOrOut = Dir.[Name],
ShiftName = ShiftRules.[Name],
ShiftStartTime = convert(varchar(10), ShiftRules.Default_Shift_Start, 108),
ShiftEndTime = convert(varchar(10), ShiftRules.Default_Shift_End, 108),
ShiftStart= dateadd(day, datediff(day, 0, RC.ClockTime), 0) + ShiftRules.Default_Shift_Start,
ShiftEnd = dateadd(day, datediff(day, 0, RC.ClockTime), 0) + ShiftRules.Default_Shift_End + case when ShiftRules.Default_Shift_End < ShiftRules.Default_Shift_Start then 1 else 0 end, BreakName= SRBreak.[Name], BreakStartTime = convert(varchar(10), SRBreak.SR_Break_StartTime, 108),
BreakEndTime = convert(varchar(10), SRBreak.SR_Break_EndTime, 108), BreakStart = dateadd(day, datediff(day, 0, RC.ClockTime), 0) + SRBreak.SR_Break_StartTime,
BreakEnd = dateadd(day, datediff(day, 0, RC.ClockTime), 0) + SRBreak.SR_Break_EndTime+ case when SRBreak.SR_Break_EndTime < SRBreak.SR_Break_StartTime then 1 else 0 end
FROM Personnel Pers
INNER JOIN RawClocks RC ON Pers.Personel_ID = RC.Person_ID
INNER JOIN Direction Dir ON RC.Direction_ID = Dir.Direction_ID
INNER JOIN Emp_ShiftGroup EmpShift ON Pers.Personel_ID = EmpShift.Employee_ID
INNER JOIN Rules_ShiftRulesGroup ShiftRules ON ShiftRules.ShiftRulesGroup_ID = EmpShift.ShiftRulesGroup_ID
INNER JOIN Rules_SRDay SRDay ON ShiftRules.ShiftRulesGroup_ID = SRDay.ShiftRulesGroup_ID And srday.cycledayno= EmpShift.StartDate_CycleDayNo + (day(RC.ClockTime - EmpShift.EmpSR_StartDate) % ShiftRules.SRShiftCycle_Days)-1
LEFT OUTER JOIN Rules_SRBreak SRBreak ON SRBreak.SRDay_ID = SRDay.SRDay_ID
),
timelog as
(
select Personel_ID = coalesce(ci.Personel_ID, co.Personel_ID),
FullName = coalesce(ci.FullName, co.FullName),
ClockIn= coalesce(ci.ClockTime, co.ClockDate - co.DayAdj + co.ShiftStartTime),
ClockOut = coalesce(co.ClockTime, ci.ClockDate + ci.DayAdj + ci.ShiftEndTime),
ShiftName = coalesce(ci.ShiftName, co.ShiftName),
ShiftStartTime = coalesce(ci.ShiftStartTime,co.ShiftStartTime),
ShiftEndTime = coalesce(ci.ShiftEndTime, co.ShiftEndTime),
ShiftStart = coalesce(ci.ShiftStart, co.ShiftStart),
ShiftEnd = coalesce(ci.ShiftEnd, co.ShiftEnd),
BreakName = coalesce(ci.BreakName, co.BreakName),
BreakStartTime = coalesce(ci.BreakStartTime, co.BreakStartTime),
BreakEndTime = coalesce(ci.BreakEndTime, co.BreakEndTime),
BreakStart= coalesce(ci.BreakStart, co.BreakStart),
BreakEnd = coalesce(ci.BreakEnd, co.BreakEnd)
from emp_clock ci
full outer join emp_clock con ci.Personel_ID = co.Personel_ID
and ci.RowNo = co.RowNo - 1 and co.ClockTime < dateadd(day, 1, ci.ClockDate) + ci.ShiftStartTime
where not (ci.InOrOut is null and co.InOrOut = 'IN')
and not (ci.InOrOut = 'OUT' and co.InOrOut is null)
and (ci.InOrOut = 'IN' or ci.InOrOut is null)
),
detail as
(
select Personel_ID, FullName,
ClockIn,
ClockOut,
TimeBeforeShiftStart = case when ClockIn < ShiftStart then ClockIn end,
TimeBeforeShiftEnd = case when ClockIn < ShiftStart then ShiftStartTime end,
TimeAfterShiftStart = case when ClockOut > ShiftEnd
then ShiftEnd end,
TimeAfterShiftEnd = case when ClockOut > ShiftEnd
then ClockOut
end,
TimeBreakStart = case when ClockIn between BreakStart and BreakEnd or ClockOut between BreakStart and BreakEnd
then case when ClockIn > BreakStart then ClockIn else BreakStart end end,
TimeBreakEnd = case when ClockIn between BreakStart and BreakEnd or ClockOut between BreakStart and BreakEnd
then case when ClockOut < BreakEnd then ClockOut else BreakEnd end
end
from timelog
),
summary as
(
select Personel_ID,
ClockDate = dateadd(day, datediff(day, 0, ClockIn), 0),
WorkHours = sum(datediff(minute, ClockIn, ClockOut) / 60.0),
BefShift = cast(isnull(sum(datediff(minute, TimeBeforeShiftStart, TimeBeforeShiftEnd)), 0) / 60.0 as decimal(10,2)),
AfterShift = isnull(sum(datediff(minute, TimeAfterShiftStart, TimeAfterShiftEnd)), 0) / 60.0,
WorkDuringBreak = isnull(sum(datediff(minute, TimeBreakStart, TimeBreakEnd) / 60.0), 0)
from detail
group by Personel_ID, dateadd(day, datediff(day, 0, ClockIn), 0)
)
select *
from summary
order by Personel_ID
0
Author Closing Comment
ID: 35830200
Thanks, your comment helped.
0
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# Stat219midterm_f08_sol - Math136/Stat219 Fall 2008 Midterm...
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Math136/Stat219 Fall 2008 Midterm Examination Friday, October 24, 2008, 11:00am - 12:30pm Write your name and sign the Honor code in the blue books provided. You have 90 minutes to solve all questions, each worth points as marked (maximum of 50). Complete reasoning is required for full credit. You may cite lecture notes and homework sets, as needed, stating precisely the result you use, why and how it applies. You may consult the following materials while taking the exam: 1. Stat219/Math136 Lecture notes, Fall 2008 version (the required text) 2. Kevin Ross’s Lecture slides posted in Coursework (Fall 2008 only) 3. Homework problems, sample exams, and solutions posted in Coursework (Fall 2008 only) 4. Your own graded homework papers 5. Your own notes, handwritten or typed Use of any other material is prohibited and constitutes a violation of the Honor Code. This includes, but is not limited to: other texts (including optional and recommended texts), photocopying of texts or notes, materials from previous sections of Stat219/Math136, the internet, programming formulas or other results in a calculator or computer, consultation with anyone during the exam (except for the Teaching Assistants or the Instructor). 1. (3 Points each) On a probability space (Ω , F , P ), let Y be a random variable with E ( Y 2 ) < and G ⊆ F be a σ -field. Define Var( Y |G ) = E ( Y 2 |G ) ( E ( Y |G ) 2 ) . Show the following. (Note: you must give a proof. Merely citing Exercise 2.3.7 will receive no credit.) a) Show that if Y is G -measurable then Var( Y |G ) = 0 almost surely. 1
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ANS. Since Y is G -measurable, Var( Y |G ) = E ( Y 2 |G ) ( E ( Y |G ) 2 ) = Y 2 ( Y ) 2 = 0 . b) Show that Var( Y ) = E (Var( Y |G )) + Var( E ( Y |G )) . ANS. Using the tower property and linearity of CE gives E (Var( Y |G )) = E [ E ( Y 2 |G ) ( E ( Y |G ) 2 )] = E ( Y 2 ) E [ E ( Y |G ) 2 ] . The definition of (unconditional) variance and the tower property implies Var( E ( Y |G )) = E [( E ( Y |G )) 2 ] ( E [ E ( Y |G )]) 2 = E [( E ( Y |G )) 2 ] ( E [ Y ]) 2 . Adding the above two equations yields E (Var( Y |G )) + Var( E ( Y |G )) = E ( Y 2 ) ( E ( Y )) 2 = Var( Y ) . c) Suppose that Y is G -measurable and X is a random variable on (Ω , F , P ) with E ( X 2 ) < . Show that Var( XY |G ) = Y 2 Var( X |G ) . ANS. Since Y is G -measurable, taking out what is known yields Var( XY |G ) = E (( XY ) 2 |G ) ( E ( XY |G ) 2 ) = Y 2 E ( X 2 |G ) ( Y E ( X |G )) 2 = Y 2 Var( X |G ) . 2. (3 Points each) Consider the probability space (Ω , F , P ), where Ω = (0 , 1); F is the Borel σ -field on (0 , 1), that is, F = σ ( { ( a, b ) : 0 < a < b < 1 } ); and P is the uniform probability measure. For n = 1 , 2 , . . . define X n ( ω ) = 2 n I A n ( ω ), ω Ω, where A n = parenleftbigg 1 2 1 2 n , 1 2 + 1 2 n parenrightbigg .
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NextVector Toolbox
... to the first element that differs. If the vector has no successor, next* returns [ ] (empty).If the vectors have repeated elements then next* returns the next distinct vector. Char inputs are allowed.As these functions are intended ...
Matlab
Min Max filter
It can eprform the following multidimensional array filtering: running filters in 1D and 2D, filters for image processing applications (erosion/dilatation), 3D and more.The MEX engine uses an algorithms that requires no more than three comparisons per element and per dimension in all configurations. Requirements: · MATLAB 7.8 · MEX ...
Matlab | 467 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.80325 |
http://wiki.zcubes.com/index.php?title=Manuals/calci/NUMSCORE&oldid=211742 | 1,653,005,006,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00136.warc.gz | 60,667,924 | 6,559 | # Manuals/calci/NUMSCORE
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
NUMSCORE (Limit,Start)
• and are any real numbers.
## Description
• This function shows numbers increasing by 1 with in the given limit.
• In , is the maximum number which is not included in the list.
• is the minimum number to start the sequence.
• Numbers are arithmetical value which are containing positive numbers,negative numbers and including zero.
• So this function is including the minimum number, but not including the maximum number.
• Also, when limit and start both are in Integers, function will return the result in Integers.
• Suppose start number is in decimal, then the function will return the result in decimal, increasing with 1.
## Examples
1. NUMSCORE(19,10) = 10 11 12 13 14 15 16 17 18
2. NUMSCORE(32,22.3) = 22.3 23.3 24.3 25.3 26.3 27.3 28.3 29.3 30.3 31.3
3. NUMSCORE(4,-11) = -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3
Numbers | 294 | 963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-21 | latest | en | 0.83169 |
https://www.weegy.com/?ConversationId=9467A25C&Link=i&ModeType=2 | 1,660,347,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00193.warc.gz | 912,671,455 | 11,973 | Disunity among Indian rulers helped the East India Company. True False
Question
Updated 5/25/2019 12:02:23 PM
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User: Disunity among Indian rulers helped the East India Company. True False
Weegy: True
User: Reincarnation is the concept of duties and obligations that Hindus must follow. True False
Question
Updated 5/25/2019 12:02:23 PM
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Flagged by emdjay23 [5/25/2019 12:02:23 PM]
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Reincarnation is the concept of duties and obligations that Hindus must follow. FALSE.
Questions asked by the same visitor
simply the equation 4^-2
Question
Updated 6/27/2016 9:10:46 AM
4^-2
= 1/4^2
= 1/16
simplify this equation -5^-2
Question
Updated 4/5/2015 12:55:57 PM
-5^-2 = 1/-5^2 = 1/25 = 0.04
Confirmed by jeifunk [4/5/2015 8:22:15 PM]
Variation in human skin color is a result of a. polygenic inheritance b. codominance. c. multiple alleles. d. intermediate inheritance.
Weegy: Variation in human skin color is a result of polygenic inheritance. (More)
Question
Updated 21 days ago|7/22/2022 11:51:53 AM
(15a^3)(-3a)
Weegy: (15a^3)(-3a) = -45a^4 (More)
Question
Solve the system by elimination. 4x + y = 8 -3x - y = 0
Question
Updated 9/10/2017 3:20:56 PM
Using the elimination method:
4x + y = 8
-3x - y = 0
(4x + y = 8) * 3
12x + 3y= 24
(-3x - y = 0) * 4
-12x - 4y = 0
(12x + 3y = 24) + (-12x - 4y = 0)
-y = 24
y = 24
4x + (-24) = 8
4x - 24 = 8
4x = 8 + 24
4x = 32
x = 32/4
x = 8
The solution set is: (8, -24)
36,403,277
';
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,173 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-33 | latest | en | 0.893175 |
https://www.teacherspayteachers.com/Product/French-Math-Problem-of-the-Week-Patterning-Les-suites-et-lalgebre-1697040 | 1,487,707,394,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170823.55/warc/CC-MAIN-20170219104610-00532-ip-10-171-10-108.ec2.internal.warc.gz | 899,171,187 | 26,250 | Total:
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# French Math Problem of the Week - Patterning (Les suites et l'algèbre)
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THIS PACKAGE IS PART OF A BUNDLE THAT INCLUDES ALL 5 MATH STRANDS. SAVE 20% WITH THE BUNDLE. CLICK TO SEE A PREVIEW OF THE GRADE 3 PROBLEM OF THE WEEK BUNDLE.
This specific French Problem of the Week package includes 12 Math word problems related to Patterning and Algebra (Les suites et l'algèbre). Each problem includes a little graphic organizer to help students organize their information, ideas and calculations. The file also includes a cover page for the unit.
The problem of the week is a great addition to your Math program and very manageable given that it only takes place once a week or as often as you choose. Students really enjoy the challenge of completing the problem. I usually reward students who get the answer correct with a Dojo Point.
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These problems are geared towards Grade 3 students, although they could also work in Grade 2 and 4 all depending on the level of your students.
This unit is also available for GRADE 2
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\$3.75 | 713 | 2,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-09 | longest | en | 0.8318 |
http://stackoverflow.com/questions/5029672/why-is-the-function-prototype-inside-a-different-function-block/5029785 | 1,419,698,713,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552343.99/warc/CC-MAIN-20141224185912-00089-ip-10-231-17-201.ec2.internal.warc.gz | 92,062,035 | 17,936 | # Why is the function prototype inside a different function block?
I am trying to understand C, by going through K&R. I have trouble understanding this code for two functions found in the book:
``````void qsort(int v[], int left, int right){
int i, last;
void swap(int v[], int i, int j);
if (left >= right)
return;
swap(v, left, (left+right)/2);
last = left;
for ( i = left+1; i<=right; i++)
if (v[i]<v[left])
swap(v,++last, i);
swap(v,left,last);
qsort(v,left,last-1);
qsort(v,last+1,right);
}
void swap(int v[], int i, int j){
int temp;
temp = v[i];
v[i] = v[j];
v[j] = temp;
}
``````
These two function perform a quicksort on a given array. In the main function I created an int array and called qsort. It compiled fine and ran fine. My question is, why is the prototype for swap() put in the function qsort() and not before main()?
-
It can be done both ways. I suppose this is done to put the prototype in a scope, not that it matters. – Stefan Dragnev Feb 17 '11 at 13:48
The prototype should be added before the actual function is used for first time. In this case, I do not think its a general practice to have prototype in `qsort()` function, however, it still serves the purpose. The prototype for `swap()` could also be added before `main()` too, don't think it will make a difference.
This is incorrect. The prototype for `swap()` could not be added in `main()`, as it must be declared before it's used in `qsort()`. It is a strange place to put the prototype, but the only other place it could be legally put would be before `swap()`. – Bradley Swain Feb 17 '11 at 14:54
You write a function prototype so that the compiler knows that function exists, and can use it. `swap()` is used inside `qsort()`, so it must appear before the line it is used. In this case, the `swap()` prototype is declared inside the `qsort()` function, but it could as well be declared before the function itself. Or you could define `swap()` before `qsort()` and remove the prototype. | 520 | 1,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2014-52 | latest | en | 0.898804 |
https://www.evwind.es/2018/06/20/evaluation-method-for-the-impact-of-wind-power-fluctuation-on-power-system-quality/63743 | 1,713,203,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00500.warc.gz | 719,822,749 | 28,253 | # Evaluation method for the impact of wind power fluctuation on power system quality
Abrupt changes of wind power generation output are a source of severe damage to power systems. Researchers at Kyoto University developed a stochastic modeling method that enables to evaluate the impact of such phenomena. The feature of the method lies in its significant computational effectiveness in comparison to standard Monte Carlo simulation, and its applicability to analysis and synthesis of various systems subject to extremum outliers.
The left figure is an example of time series of fluctuation of power generation outputs. The right figure shows the histogram of the frequency of corresponding power fluctuation. When we can assume mild fluctuation as illustrated by the red line, modeling with normal distribution is acceptable. However, wind power generation fluctuation contains abrupt changes as illustrated by the blue line. Consequently, the corresponding histogram has slowly-decaying heavy tails, which represents the extremal outliers. Credit: Japan Science and Technology Agency (JST)
Introduction of wind power generation into the electric power system is proceeding actively, mainly in the United States and Europe, and is expected to continue in Japan. However, upon the implementation, it is crucial to deal with prediction uncertainty of output fluctuation. The fluctuation of wind power generation is usually small, but it becomes extremely large due to the occurrence of gusts and turbulence at a non-negligible frequency. Such extreme outliers have been regarded as a source of severe damage to power systems.
To cope with such a fluctuation of wind power generation, the goal setting such as “absolutely keep the frequency fluctuation within 0.2 Hz” would be unattainable or would result in an overly conservative design. Therefore, the probabilistic goal setting such as “keep the frequency fluctuation within 0.2 Hz with 99.7 percent or more” is indispensable.
Probabilistic uncertainty is evaluated statistically, commonly by assuming that it obeys normal distribution for its mathematical processability. The output outliers in wind power are, however, more frequent than represented by normal distribution. Even if a complicated simulator can be constructed without assuming normal distribution, it is not realistic to investigate the statistical property by Monte Carlo simulation. This is because the required number of samples explodes before sufficiently many extreme outliers occur.
An evaluation was developed for the impact of power on power system quality. The method first builds probabilistic models assuming the stable distribution (an extension of the normal distribution) on the uncertainty. Then, instead of using the model as a simulator to generate data samples, we compute the statistical properties directly from parameters in the model. The important feature is 1. the influence of extreme outliers can be properly considered, 2. model can be determined easily from actual data, and 3. computation cost is very low. The method was proved to be valid through its application to frequency deviation estimation based on actual power system data.
This newly proposed probabilistic evaluation method enables us to quantitatively evaluate the power system risk caused by the occurrence of extremally abrupt changes of . Countermeasures based on the evaluation would contribute to improvement of the reliability and economic efficiency of the electric system. It should be also noted that the proposed method is applicable to analysis and synthesis of various systems which have extreme outliers.
Kenji Kashima et al. Stable Process Approach to Analysis of Systems under Heavy-tailed Noise: Modeling and Stochastic Linearization, IEEE Transactions on Automatic Control (2018). DOI: 10.1109/TAC.2018.2842145 | 704 | 3,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.946283 |
https://edurev.in/courses/627_Mathematics--Maths--Class-9-Docs--Videos--Tests?chapter=22104&utm_source=footer_class9&utm_medium=edurev_website&utm_campaign=footer | 1,686,332,013,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00739.warc.gz | 272,829,423 | 89,409 | Mathematics (Maths) Class 9
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# Mathematics (Maths) Class 9
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## About CBSE Mathematics (Maths) Class 9
EduRev's Mathematics (Maths) Class 9 Course is a comprehensive program designed for students of class 9. This course covers all the essential topics of mathematics, including algebra, geometry, trigonometry, and statistics. It provides a clear understanding of various mathematical concepts, making it easier for students to grasp complex topics. The course is designed by expert educators and includes interactive video lectures, practice quizzes, and solved examples to help students excel in their exams. With this course, students can enhance their mathematical skills and achieve academic success.
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Table of Content
Mathematics (Maths) Class 9 CBSE Syllabus 2023-2024
## Class 9 Mathematics Syllabus
• Number System: Understanding of irrational numbers, real numbers, and their properties. Representation of numbers on the number line, and conversion of decimals into fractions.
• Polynomials: Introduction to algebraic expressions, monomials, binomials, and trinomials. Factorization of polynomials, and finding common factors.
• Coordinate Geometry: Introduction to Cartesian plane, plotting of points, and finding distance between two points. Understanding of slopes, equations of lines, and their intersections.
• Linear Equations in Two Variables: Solving linear equations in two variables, and finding their solutions graphically.
• Introduction to Euclid's Geometry: Understanding of Euclid's axioms, postulates, and theorems.
• Lines & Angles: Understanding of lines and their properties, angles and their types, and parallel lines.
• Triangles: Introduction to triangles, their types, and properties. Understanding of congruence and similarity of triangles, and Pythagoras theorem.
• Circles: Understanding of circles, their properties, and theorems related to them.
• Heron's Formula: Formula for finding the area of a triangle using its sides.
• Surface Area & Volumes: Calculation of surface area and volumes of different geometric shapes.
• Statistics: Basic concepts of statistics, representation of data in tables and graphs, and calculation of mean, mode, and median.
• Area of Parallelograms & Triangles: Calculation of the area of parallelograms and triangles using base and height.
• Construction: Basic constructions using ruler and compass, and understanding of constructions related to triangles.
• Probability: Basic concepts of probability, and calculation of probability of simple events.
• CBSE Sample Question Papers: Practice papers based on the CBSE syllabus for Class 9 Mathematics.
• RD Sharma Solutions: Detailed solutions for the RD Sharma textbook for Class 9 Mathematics.
• RS Aggarwal Solutions: Detailed solutions for the RS Aggarwal textbook for Class 9 Mathematics.
• NCERT Textbooks & Solutions: Textbooks and solutions provided by NCERT for Class 9 Mathematics.
• Short & Long Question Answers: Comprehensive question and answer sets for the Class 9 Mathematics syllabus.
• NCERT Exemplar: Additional questions for practice and understanding of the Class 9 Mathematics syllabus.
This course is helpful for the following exams: Class 9
How to Prepare Mathematics (Maths) Class 9?
Preparing for Mathematics (Maths) Class 9 can be a daunting task for many students. However, with the right approach and mindset, it can become an enjoyable and rewarding experience. Here are some tips to help you prepare for your Class 9 Maths course offered by EduRev.
1. Understand the Syllabus:
It is essential to have a clear understanding of the syllabus to prepare effectively for your Maths Class 9 course. Familiarize yourself with the topics and concepts that will be covered in the course. Make a list of all the topics and subtopics and prioritize them according to your level of understanding.
2. Get the Right Study Material:
Having the right study material is crucial to preparation. The EduRev Maths Class 9 course offers comprehensive study material that includes videos, notes, and practice questions. Make use of these resources to build a strong foundation in Maths.
3. Practice Regularly:
Maths is a subject that requires consistent practice. Dedicate a fixed amount of time every day to practice Maths problems. Start with the basics and gradually move on to more complex problems. Practicing regularly will help you build your problem-solving skills and boost your confidence.
4. Seek Help When Needed:
If you find yourself struggling with a particular topic, don't hesitate to seek help. You can reach out to your teachers or classmates for assistance. You can also make use of the EduRev Maths Class 9 course's doubt-solving feature to clear any doubts you may have.
5. Take Mock Tests:
Taking mock tests is an excellent way to prepare for exams. It will help you identify your strengths and weaknesses and give you an idea of your progress. The EduRev Maths Class 9 course offers mock tests that simulate the actual exam and help you prepare for it.
In conclusion, preparing for Maths Class 9 requires dedication and effort. By following the above tips and utilizing the resources offered by the EduRev Maths Class 9 course, you can build a strong foundation in Maths and excel in your exams.
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Importance of Mathematics (Maths) Class 9
Importance of Mathematics (Maths) Class 9 Course for Class 9
Introduction
Mathematics is one of the most important subjects that a student studies in school. It is a subject that is used in our daily lives and is essential for many careers. The Class 9 Maths course is an important foundation for higher level studies. In this article, we will discuss the importance of the Class 9 Maths course and how it can benefit students.
Benefits of Class 9 Maths Course
The Class 9 Maths course is designed to help students develop a strong foundation in mathematics. It helps them understand the basic concepts and principles of mathematics, which are essential for higher level studies. Some of the key benefits of the Class 9 Maths course are:
1. Builds Problem-Solving Skills: The Class 9 Maths course helps students develop problem-solving skills. They learn how to solve complex problems using logical reasoning and mathematical concepts.
2. Enhances Analytical Skills: The Class 9 Maths course also helps students improve their analytical skills. They learn how to analyze and interpret data using mathematical tools and techniques.
3. Develops Critical Thinking: The Class 9 Maths course helps students develop critical thinking skills. They learn how to evaluate and analyze information, and make informed decisions.
4. Prepares for Higher Studies: The Class 9 Maths course is a foundation for higher level studies in mathematics and other related fields. It prepares students for the challenges they will face in higher studies.
Conclusion
In conclusion, the Class 9 Maths course is an important foundation for higher level studies. It helps students develop problem-solving skills, analytical skills, critical thinking skills, and prepares them for higher studies. It is essential for students to take the Class 9 Maths course seriously and put in the effort to learn and understand the concepts. With the right guidance and support, students can excel in mathematics and achieve their academic goals.
;
Mathematics (Maths) Class 9 FAQs
1. What is the syllabus of Maths for Class 9?
Ans. The Maths syllabus for Class 9 generally includes topics such as real numbers, polynomials, coordinate geometry, linear equations in two variables, Euclid's geometry, triangles, quadrilaterals, areas of parallelograms and triangles, circles, constructions, surface areas and volumes, statistics, and probability.
2. How can I score well in Maths in Class 9?
Ans. To score well in Maths in Class 9, you should focus on understanding the concepts thoroughly. Practice regularly, solve as many problems as possible, and revise regularly. You can also take help from your teachers or tutors if you face any difficulties in understanding the topics.
3. Is it necessary to solve previous year's question papers for M | 1,728 | 8,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.914405 |
https://documents.pub/document/ncctm-leadership-conference-october-24-2012-dpi-update-dpi-update.html | 1,679,913,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00295.warc.gz | 256,008,400 | 26,280 | Home > Documents > NCCTM Leadership Conference October 24, 2012 DPI UPDATE DPI UPDATE
NCCTM Leadership Conference October 24, 2012 DPI UPDATE DPI UPDATE
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NCCTM Leadership Conference October 24, 2012 DPI UPDATE DPI UPDATE. NC Assessment Schedule. 2012 – 2013 NC Assessments 2013 – 1014 NC Assessments 2014 – 2015 SBAC, Common Exams, and Math I EOC. COMMON EXAMS (MSLs). Local Option Math II and Local Option Math III (CCSS-M) - PowerPoint PPT Presentation
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NCCTM Leadership Conference October 24, 2012 DPI UPDATE
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• NCCTM Leadership ConferenceOctober 24, 2012DPI UPDATE
DPI UPDATE
• NC Assessment Schedule2012 2013 NC Assessments2013 1014 NC Assessments2014 2015 SBAC, Common Exams, and Math I EOC
• COMMON EXAMS(MSLs)Local Option Math II and Local Option Math III (CCSS-M)Geometry and Algebra II (2003 SCoS and CCSS-M)Geometry and Algebra II (CCSS-M)Integrated Mathematics III (CCSS-M)AFM, Discrete, and Pre-Cal (2003 SCoS)
• Resources RebornLessons for Learning On the Wiki
• http://www.ncdpi.wikispaces.net
• Three Shifts
Focus, Coherence, Rigor
Student Achievement Partners
• Research
NC EducatorsCCSS ProgressionsSmarter Balanced Assessment ConsortiumNC Department of Public Instruction
• Accessible Mathematics10 Instructional Shifts That Raise Student Achievement
• Steven Leinwand
• It's Instruction Stupid"
• Making the Case for K-5
• Making the Case for K-5
If a student leaves second grade not knowing math, he leaves 12th grade not knowing math.
Robert R.
• What Do We Know?
Mathematics is the gate keeper!
• What Do We Know? If you have a problem with Algebra in your schools, you have to fix it in K-4. Kathy Richardson
• What Do We Know?Using 6 longitudinal data sets,.the results show that early math skills have the greatest predictive power (of future academic success), followed by reading and attention skills. Developmental Psychology 2007, Vol. 43
• Research
Students are shown this number. Teacher points to the 6 and says, Can you show me this many?16
• Research
The teacher points to the 1 in the tens place and asks, Can you show me this many?16
• Research
By third grade nearly half the students still do not get this concept of place value..16
• More research - It gets worse!
A number contains 18 tens, 2 hundreds, and 4 ones. What is that number?
• More research - It gets worse!
A number contains 18 tens, 2 hundreds, and 4 ones. What is that number?
1824
• More research - It gets worse!
A number contains 18 tens, 2 hundreds, and 4 ones. What is that number?
1824 2184
• More research - It gets worse!
A number contains 18 tens, 2 hundreds, and 4 ones. What is that number?
1824 2184 218.4
• More research - It gets worse!
A number contains 18 tens, 2 hundreds, and 4 ones. What is that number?
1824 2184 218.4 384
• And worse
35 x25 175 70 245
Deborah Ball
• And worse
35 x25 255 800 1055
Deborah Ball
• 8 + 4 = [ ] + 5Thinking Mathematically: Integrating Arithmetic & Algebra in Elementary School Carpenter, Franke, & LeviHeinemann, 2003
• 8 + 4 = [ ] + 5
Percent Responding with AnswersGrade7121712 & 171st - 2nd3rd - 4th5th - 6th
• 8 + 4 = [ ] + 5
Percent Responding with AnswersGrade7121712 & 171st - 2nd5581383rd - 4th5th - 6th
• 8 + 4 = [ ] + 5
Percent Responding with AnswersGrade7121712 & 171st - 2nd5581383rd - 4th94925105th - 6th
• 8 + 4 = [ ] + 5
Percent Responding with AnswersGrade7121712 & 171st - 2nd5581383rd - 4th94925105th - 6th276212
• How about Fractions?Estimate the answer to 12/13 + 7/8.A. 1B. 2C. 19D. 21
• How about Fractions?Estimate the answer to 12/13 + 7/8.A. 1B. 2C. 19D. 21 Only 24% of 13 year olds answered correctly. Equal numbers of students chose the other answers. NAEP
• Teaching for Understanding
Lets do some Math!
• In Conclusion..
• DPI Contact Informationhttp://www.ncdpi.wikispaces.net
Kitty RutherfordElementary Mathematics [email protected] MaynorSecondary Mathematics [email protected]
Barbara BissellK 12 Mathematics Section [email protected] HartK-12 Program [email protected]
This is the website for mathematics. We are posting presentations here (). The site also contains links to most requested documents. You do not have to join to be able to access the materials.
www.ncdpi.wikispaces.net*Focus Task*The Major Work of the Grade has been identified for North Carolina based on information provided by NC educators, NCDPI, and SBAC.
*The Major Work identifies the primary focus for each grade level, at the cluster level. The supporting/additional work is very important- but while not the focus of that grade, it supports the major work.* 2012 Karen A. Blase and Dean L. Fixsen* 2012 Karen A. Blase and Dean L. Fixsen**Did you get all that?Good news! He wrote a book, Accessible Mathematics. Well discuss some of those instructional shifts he talked about in this resource after the break.
*Grayson Wheatlys research with 5,000 middle schoolers were given the following task:This task was posed to 5,000 5-8 graders.Some students gave this answer. Others knew the tens had to be in the middle, so..Many gave this answer, knowing about decimals, and that you could only have 3 digits if a number was in the hundreds.Around 50% of the middle school students gave the correct answer.
*Grayson Wheatlys research with 5,000 middle schoolers were given the following task:This task was posed to 5,000 5-8 graders.Some students gave this answer. Others knew the tens had to be in the middle, so..Many gave this answer, knowing about decimals, and that you could only have 3 digits if a number was in the hundreds.Around 50% of the middle school students gave the correct answer.
*Grayson Wheatlys research with 5,000 middle schoolers were given the following task:This task was posed to 5,000 5-8 graders.Some students gave this answer. Others knew the tens had to be in the middle, so..Many gave this answer, knowing about decimals, and that you could only have 3 digits if a number was in the hundreds.Around 50% of the middle school students gave the correct answer.
*Grayson Wheatlys research with 5,000 middle schoolers were given the following task:This task was posed to 5,000 5-8 graders.Some students gave this answer. Others knew the tens had to be in the middle, so..Many gave this answer, knowing about decimals, and that you could only have 3 digits if a number was in the hundreds.Around 50% of the middle school students gave the correct answer.
*Deborah Ball has done extensive research on the mathematics teachers need to understand to teach math well. Here are 2 multiplication problems with wrong answers. There are no mistakes in basic facts.
This is the result of teaching procedures without understanding.*Deborah Ball has done extensive research on the mathematics teachers need to understand to teach math well. Here are 2 multiplication problems with wrong answers. There are no mistakes in basic facts.
Recommended | 1,864 | 7,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.767629 |
https://www.bartleby.com/essay/The-Effect-Of-Concentration-On-Osmosis-FCWJSWCAYMV | 1,669,561,422,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00408.warc.gz | 732,643,528 | 8,723 | # The Effect Of Concentration On Osmosis
Decent Essays
Discussion 1-Describe the pattern of results by the graph you have drawn. There was no change in length at 0%, the potato cylinder stayed 40 mm, however at 3% the potato decreased by 24mm to become 36 mm. At 5%, the potato’s length decreased by 3 mm to become 37 mm in length. Therefore, there is a negative correlation in the graph. At 0%, there was an increase of 12.32% in the potato cylinders mass but at 2%, the potato cylinders mass decreased to -31.02%. At 5%, the potato cylinders mass decreased -26.65%. There is a negative correlation in the graph. 2-Discuss the reliability if the results when compared to other groups. Use examples. When compared to other groups, the results of the weight change were mildly reliable. For instance, the 2 % solution percentage results for the groups were -32.35%, -30.7% and -30%, these are all similar and so are reliable however n some results, like the 1 5 solution, group 1 had -17.98% and group 3 has -17.7% but group 2 had -32.5%. So, because of these differences the results are not reliable.…show more content…
Systematic errors are errors within the equipment such as measuring cylinders which may not be exact or scales which could be damaged and so give incorrect weights. Random errors are those made by human errors such as having parallax errors distorting measurements, temperatures in the room affecting results or slow reaction times when stopping a stop watch or placing things in solutions. Systematic and random errors can be reduced by many repeats of the whole experiment and then averaging the results and also, decreasing as many variables as possible by using precise | 387 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-49 | latest | en | 0.968578 |
http://www.sage.unsw.edu.au/currentstudents/ug/projects/Vonk/INS%20overview.htm | 1,505,860,409,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686043.16/warc/CC-MAIN-20170919221032-20170920001032-00648.warc.gz | 558,489,295 | 4,425 | Gravity modelling is an integral part of INS. INS can only operate in situations in which the gravitational field is known, or known to be insignificant. The accurate measurement of gravity is therefore a limiting factor to the accuracy of INS. Gravity is composed of two parts, gravitation and centripetal acceleration. Gravitation is defined by Newton’s Law of Gravitation, which relates the attraction of two masses by their weight and separation. Centripetal acceleration is the rotational acceleration that occurs because of the rotation of the Earth. The combination of the two components gives the local gravity vector to which a ‘plumb bob’ would align itself if held above the Earth.
Before an INS can operate it must first determine its attitude, position and velocity. Gimballed INS require that the axes of the platform are parallel to the navigation coordinates, while for a strapdown INS the alignment involves the calculation of the initial values of the coordinate transformation from the sensor coordinates to navigation coordinates. There are four common methods employed for INS alignment, they include:
· Optical Alignment – This takes two forms, local alignment using ground based systems such as coordinated survey marks and a theodolite or space based alignment using a star tracker, which is primarily used for alignment in space.
· Transfer Alignment – this is achieved using velocity matching with an aligned and operating INS. This is commonly used in military vehicles to align a slave INS in a missile from the master INS in the vehicle. This requires that the onboard INS be in working order and the manoeuvres of the vehicle are suitable for the transfer of alignment.
· GPS-aided Alignment – this uses position matching with GPS to estimate the alignment variables. It can be done without any specific movements but does take time for the navigation solution to settle to acceptable levels.
· Gyrocompassing Alignment – this is achieved using the sensed direction of the vertical while stationary. Latitude is determined through the angle between the earth rotation vector and the horizontal. The system is incapable of calculating longitude and this has to be input into the system.
All inertial navigation systems exhibit similar qualities regardless of their configuration, gimballed or strapdown, although both systems have their nuances. The system provides information about the velocity, position and attitude, all of which degrade over time. The system does bound the effect of some errors through methods such as Schuler tuning, but others such as gyro drift cause an unavoidable continuous error growth.
The quality of the system is defined by its error growth per unit of time. The quality of the system is directly related to the quality of the components. There are many applications for which inertial navigation systems are used, both short and long duration. Short term applications generally do not require sustained performance and consequently use systems that minimise the error growth for a short period after which is not considered. In these situations the quality of the components is not paramount and cheaper alternatives can be used. Longer duration applications require the use of precise instrumentation but continuos error growth is still unavoidable. The point at which the solution becomes unusable is determined by the requirements of the application and the instrumentation. The ability to determine this point is only capable through the known or estimated error growth of the system.
The system is also influenced by gravity, the system has to apply a value of gravity for each calculation and the inability to perfectly model the effect of gravity causes errors that continually propagate in the vertical channel. This can transfer to horizontal measurements however the effect is bounded in these channels, unlike the vertical channel.
The system is almost self-reliant, but it is dependent upon the input of data to determine its initial position, even through gyrocompassing the system cannot determine longitude. The system does provide a continuous data stream once operational and achieves this at a high output rate. The system is autonomous once operational and therefore is incapable of being jammed or detected in most circumstances. | 785 | 4,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-39 | latest | en | 0.915161 |
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-3-section-3-4-volume-and-area-exercise-page-147/45 | 1,720,950,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00400.warc.gz | 707,223,532 | 12,594 | ## Elementary Technical Mathematics
0.085 $km^2$
$85,000m^2=85,000m^2\times(\frac{1km}{1000m})^2=85,000m^2\times\frac{1km^2}{1,000,000m^2}=\frac{85,000}{1,000,000}km^2=0.085km^2$ | 92 | 179 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-30 | latest | en | 0.458194 |
http://gogeometry.com/geometry/tangent_line_to_a_circle_theorems_problems_high_school_college_index_17.html | 1,558,551,571,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256948.48/warc/CC-MAIN-20190522183240-20190522205240-00507.warc.gz | 85,431,904 | 4,344 | # Circle Tangent Line, Theorems and Problems - Table of Content, Page 17
Geometry Problem 259.
Equilateral Triangle, Incircle, Tangency Points, Side, Distances, Squares,
Tangent lines to circle.
Geometry Problem 258.
Equilateral Triangle, Circumcircle, Point, Vertices, Side, Distances, Squares,
Tangent lines to circle.
Geometry Problem 248.
Napoleon's Theorem III. Inner and outer Napoleon triangles, Area,
Tangent lines to circle.
Geometry Problem 247.
Napoleon's Theorem II. Internal Equilateral triangles. Inner Napoleon triangle,
Tangent lines to circle.
Geometry Problem 246.
Napoleon's Theorem I. External Equilateral triangles. Outer Napoleon triangle,
Tangent lines to circle.
Geometry Problem 209.
Tangent line to circle.
Geometry Problem 208.
Triangle, Excircles, Angles, 360 degrees,
Tangent line to circle.
Geometry Problem 207.
Tangent lines to circle.
Problem 206.
Area of a Right Triangle, Inradius, and Exradius relative to the hypotenuse,
Tangent lines to circle.
Geometry Problem 205.
Right Triangle Area, Exradii relatives to legs or catheti,
Tangent lines to circle.
Geometry Problem 204.
Tangent line to circle.
Geometry Problem 203.
Tangent lines to circle.
Geometry Problem 202.
Right Triangle, Incircle, Excircles relatives to catheti, Points of Tangency, Exradius, Semiperimeter,
Tangent line to circle.
Geometry Problem 201.
Right Triangle, Excircles, Points of Tangency, Exradius, Semiperimeter,
Tangent lines to circle.
Geometry Problem 197.
Geometry Problem 187.
Right Triangle, Altitude, Incenters, Circles, Angles, Tangent line to circle.
Geometry Problem 190.
Tangent circles, Chord, Perpendicular, Distance, Tangent line to circle.
Geometry Problem 200.
RightTriangle, Incircle, Excircles, Points of Tangency, Inradius,
Tangent lines to circle.
Geometry Problem 196.
Geometry Problem 195. | 444 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-22 | latest | en | 0.684589 |
https://www.formulaconversion.com/formulaconversioncalculator.php?convert=yardsperhour_to_millimeterspersecond | 1,508,528,274,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824325.29/warc/CC-MAIN-20171020192317-20171020212317-00579.warc.gz | 961,143,488 | 16,223 | # Yards/hour to millimeters/second (yd/hr to mm/s) Metric conversion calculator
Welcome to our yards/hour to millimeters/second (yd/hr to mm/s) conversion calculator. You can enter a value in either the yards/hour or millimeters/second input fields. For an understanding of the conversion process, we include step by step and direct conversion formulas. If you'd like to perform a different conversion, just select between the listed Speed units in the 'Select between other Speed units' tab below or use the search bar above. Tip: Use the swap button to switch from converting yards/hour to millimeters/second to millimeters/second to yards/hour.
## millimeters/second (mm/s)
(not bookmarks)
Swap
< == >
1 yd/hr = 0.254 mm/s 1 mm/s = 3.93700787 yd/hr
Algebraic Steps / Dimensional Analysis Formula
yd/hr * 1 yd/s 3600 yd/hr * 3 ft/s 1 yd/s * 1 m/s 3.2808 ft/s * 1000 mm/s1 m/s = mm/s
Direct Conversion Formula
yd/hr * 1 mm/s3.93700787 yd/hr = mm/s
feet/hour 0 feet/second 0 kilometers/hour 0 kilometers/second 0 knots 0 meters/hour 0 meters/second 0 miles/hour 0 miles/second 0
If you would like to switch between Speed units, select from the tables below
centimeters/hour centimeters/second decimeters/hour decimeters/second dekameters/hour dekameters/second feet/hour feet/second gigameters/hour gigameters/second hectometers/hour hectometers/second kilometers/hour kilometers/second knots megameters/hour megameters/second meters/hour meters/second micrometers/hour micrometers/second miles/hour miles/second millimeters/hour millimeters/second yards/hour yards/second < == > centimeters/hour centimeters/second decimeters/hour decimeters/second dekameters/hour dekameters/second feet/hour feet/second gigameters/hour gigameters/second hectometers/hour hectometers/second kilometers/hour kilometers/second knots megameters/hour megameters/second meters/hour meters/second micrometers/hour micrometers/second miles/hour miles/second millimeters/hour millimeters/second yards/hour yards/second
Active Users | 542 | 2,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-43 | latest | en | 0.687508 |
https://scientific-calculator.org/how-to-calculate-mortgage-payment/ | 1,696,261,982,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511000.99/warc/CC-MAIN-20231002132844-20231002162844-00053.warc.gz | 549,781,146 | 13,799 | # How to calculate mortgage payment?
To calculate your mortgage payment, you will need to know the following information:
1. The amount of the loan: This is the total amount you are borrowing to purchase your home.
2. The interest rate: This is the percentage of the loan amount that you will pay in interest.
3. The loan term: This is the length of time you have to pay back the loan, usually expressed in years.
4. The frequency of payments: Most mortgages require payments to be made monthly.
Once you have this information, you can use the following formula to calculate your mortgage payment:
M = P[r(1+r)^n]/[(1+r)^n-1]
Where:
M is the mortgage payment P is the loan amount r is the monthly interest rate (the annual interest rate divided by 12) n is the number of payments (the loan term in years multiplied by 12)
For example, let’s say you are borrowing \$200,000 at an annual interest rate of 3.5% for a loan term of 30 years (360 payments). Your monthly interest rate would be 3.5%/12 = 0.2917%, and your mortgage payment would be:
M = \$200,000[0.002917(1+0.002917)^360]/[(1+0.002917)^360-1]
M = \$898.09
This means your monthly mortgage payment would be approximately \$898.09.
Note that this is just a rough estimate of your mortgage payment, and the actual payment may vary depending on additional factors such as taxes, insurance, and any fees associated with your loan.
## How to calculate mortgage interest
To calculate the mortgage interest on a loan, you will need to know the following information:
1. The amount of the loan: This is the total amount you are borrowing to purchase your home.
2. The interest rate: This is the percentage of the loan amount that you will pay in interest.
3. The loan term: This is the length of time you have to pay back the loan, usually expressed in years.
4. The frequency of payments: Most mortgages require payments to be made monthly.
To calculate the total mortgage interest you will pay over the life of the loan, you can use the following formula:
Total Interest = P x (r/12) x n
Where:
P is the loan amount r is the annual interest rate n is the number of payments (the loan term in years multiplied by 12)
For example, let’s say you are borrowing \$200,000 at an annual interest rate of 3.5% for a loan term of 30 years (360 payments). Your total mortgage interest would be:
Total Interest = \$200,000 x (0.035/12) x 360
Total Interest = \$70,800
This means you will pay approximately \$70,800 in mortgage interest over the life of the loan.
Note that this is just a rough estimate of the mortgage interest you will pay, and the actual amount may vary depending on additional factors such as taxes, insurance, and any fees associated with your loan.
## What is mortgage ?
A mortgage is a loan that is used to finance the purchase of a home. When you take out a mortgage, you agree to pay back the loan over a set period of time, usually 15 or 30 years, with interest. The interest is a percentage of the loan amount that you pay to the lender for the privilege of borrowing the money.
To get a mortgage, you typically need to provide the lender with information about your credit history, employment, and income. The lender will use this information to determine whether you are a good risk for a loan and, if so, what interest rate to charge you.
There are many types of mortgages available, including fixed-rate mortgages, adjustable-rate mortgages, and government-backed mortgages such as FHA loans and VA loans. Each type of mortgage has its own set of terms and conditions, so it’s important to shop around and compare different options before deciding on a mortgage. | 832 | 3,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-40 | longest | en | 0.96372 |
https://grf-labs.github.io/grf/articles/categorical_inputs.html | 1,722,757,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00259.warc.gz | 234,548,027 | 5,226 | In this example, we will illustrate several ways of dealing with categorical variables when using grf.
One of the approaches below relies on grf’s sister package sufrep. Let’s install and load it first.
library(grf)
install.packages("https://github.com/grf-labs/sufrep/blob/master/sufrep_0.1.0.tar.gz?raw=true", repos = NULL, type = "source")
#> Installing package into '/tmp/RtmpRA3IOm/temp_libpath195759d6d3af'
#> (as 'lib' is unspecified)
library(sufrep)
Let’s pretend we would like to estimate mileage per gallon (mpg) from number of cylinders (cyl), quarter-mile time (qsec), and car brand name (brand, created below).
# Create a categorical column with brand name
df <- within(mtcars, {
# E.g. 'Mazda RX4' --> 'Mazda'
brand <- factor(sapply(rownames(mtcars), function(x) strsplit(x, " ")[[1]][1]))
})
x <- c("cyl", "qsec") # Continuous variables
g <- c("brand") # Categorical variable
#> cyl qsec brand
#> Mazda RX4 6 16.46 Mazda
#> Mazda RX4 Wag 6 17.02 Mazda
#> Datsun 710 4 18.61 Datsun
#> Hornet 4 Drive 6 19.44 Hornet
#> Hornet Sportabout 8 17.02 Hornet
#> Valiant 6 20.22 Valiant
This code would raise an error, because data is not numerical.
# rf <- regression_forest(X=df[c(x, g)], Y=df$mpg) We can consider three approaches here. • Simply assign integers to each category (convert ‘AMC’ to 1, ‘Cadillac’ to 2, etc.) • One-hot encode the categories (as many binary columns as there are categories) • Use a sufficient representation of the category. Here we will use the means method from the sufrep package. The last method involves substituting the brand column by averages of the continuous columns cyl and qsec, grouped by category. If you are curious about why that works, or would like to know more about sufficient representations, please check out our sufrep paper (ArXiv). # Solution 1: Transform variable into numbers X1 <- within(df[c(x, g)], brand <- as.numeric(brand)) rf1 <- regression_forest(X1, df$mpg)
# Solution 2: One-hot encoding
X2 <- model.matrix(~ 0 + ., df[c(x, g)])
rf2 <- regression_forest(X2, df$mpg) # Solution 3: 'Means' encoding using the 'sufrep' package encoder <- make_encoder(df[x], df$brand, method="means")
X3 <- encoder(df[x], df$brand) #> [1] 22 2 rf3 <- regression_forest(X3, df$mpg)
Different approaches can yield different forest performance.
mse1 <- mean(rf1$debiased.error) mse2 <- mean(rf2$debiased.error)
mse3 <- mean(rf3\$debiased.error)
print("MSE when representing categorical variables as...")
#> [1] "MSE when representing categorical variables as..."
print(paste0("Integers: ", mse1))
#> [1] "Integers: 15.6597259578249"
print(paste0("One-hot vectors: ", mse2))
#> [1] "One-hot vectors: 14.2686431201772"
print(paste0("'Means' encoding [sufrep]: ", mse3))
#> [1] "'Means' encoding [sufrep]: 14.7096993292856" | 839 | 2,862 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.726049 |
https://www.teacherspayteachers.com/Product/New-Series-Math-Bowl-Percent-of-Change-IncreaseDec-Station-Activity-3276938 | 1,508,492,773,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00651.warc.gz | 1,003,329,221 | 23,065 | Total:
\$0.00
# [New Series] Math Bowl | Percent of Change: % Increase/Dec. | Station Activity
Common Core Standards
Product Rating
4.0
1 rating
File Type
PDF (Acrobat) Document File
1 MB|8 pages
Share
Product Description
Math Bowl is a new series that I have created to help keep your students "on a roll" in math! (Cheesy, I know!) In this activity, students compete against each other either individually or with a partner as they work to solve word problems. The objective of this game is for students to achieve a “strike” (knock down all of the bowling pins). The best part about this resource is that it is self correcting!
This resource includes 10 Percent of Change task cards that students must work through. Answers are provided on the bowling pins for students to cross out (aka "knock down") as they solve each problem. The first team that achieves a strike is the winner.
Topics covered:
Percent Increase, Percent decrease, Markup, & markdown
What's Included?
2 Game Boards
TEKS Aligned:
7.4D (R) The student is expected to solve problems involving ratios, rates, and percents, including multi‐step problems involving percent increase and percent decrease, and financial literacy problems.
Total Pages
8 pages
Included
Teaching Duration
45 minutes
Report this Resource
\$2.50
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\$2.50 | 336 | 1,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-43 | latest | en | 0.933533 |
https://www.esaral.com/q/how-many-bricks-each-of-size-25-cm-13-5-cm-6-cm-86529 | 1,726,306,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00774.warc.gz | 679,368,821 | 11,419 | # How many bricks, each of size 25 cm × 13.5 cm × 6 cm,
Question:
How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?
Solution:
Volume of the brick $=25 \times 13.5 \times 6=2025 \mathrm{~cm}^{3}$
Volume of the wall $=800 \times 540 \times 33=14256000 \mathrm{~cm}^{3}$
Total number of bricks $=\frac{\text { Volume of the wall }}{\text { Volume of each brick }}=\frac{14256000}{2025}=7040$ bricks | 168 | 476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.705036 |
http://physicsclasses.online/what-is-vectors/ | 1,696,186,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00611.warc.gz | 31,066,722 | 119,318 | Categories
# What do you mean by the term vector quantity ?
## What do you mean by the term vector quantity? what are the uses of vector quantity? Explain different types of vector.
VECTORS-
Physical quantity having both magnitude and direction called vector quantity . And the physical quantity having magnitude only called scalar quantity . E.X.- suppose a man move 10 km , now in that case I am discussing about its distance only i.e. it has magnitude only but there is no any information about its direction hence it is a scalar quantity, if we say man is moving 10 km towards East(or any other direction ) so we can say we are discussing about its displacement i.e. it has both magnitude and direction. A vector quantity is represented as an arrow on its top ex;- ? ⃗
Vectors can be divided into two ways –
• Polar vector- a vector which has a starting point or a point of application. E. X .- displacement , force etc.
• (ii) Axial vector – Vectors which represent rotational effects and acts along the axis of rotation in accordance with right hand screw rule.
## 2,661 replies on “What do you mean by the term vector quantity?”
[…] the addition of two vectors – For the addition of two vectors represent these two vectors by arrowed lines using the same scale . Displace the second vector as […]
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3
In this category, I find counting to be effective at disproving conjectures. If it did have a left adjoint $F$, then there would be a bijection $$\hom(FA, B) \cong \hom(A, X \times B)$$ so we would have $$b^{|FA|} = (|X| b)^{|A|}$$ for every finite cardinal $b$. Supposing $|X| > 0$, we would need $$|FA| = |A| (1 + \log_b |X|)$$ for all ...
-1
This is only difficult if you get confused by the following facts that are stupidly written in all the textbooks A limit of a particular diagram is a right Kan extension A functor assigning limits to diagrams is a right adjoint but a left Kan extension. What happens if you apply a right adjoint to a limit is that you still have a limit, but if you apply ...
0
(I misread your question and wrote the answer about limits; but everything below holds with colimiting cocones in place of limiting cones). Given a functor category $[\mathcal E,\mathcal C]$, a diagram $\mathcal D\xrightarrow{J}[\mathcal E,\mathcal C]$, and a limiting cone $F\overset{\phi_A}\Rightarrow JA$ over $J$, it is not in general true that for any ...
3
I see that this question has already some good answer. Nonetheless allow me to give a personal perspective on the matter. The definition of currying and partial application are the following: currying is an operation that takes a function of two (or maybe more argument) and return a function-valued function partial application is an operation that takes a ...
4
No, it doesn't unless $X$ is a singleton. The very first condition to check for a functor to have a left adjoint is that it should preserve limits (such as products, equalizers...). But clearly in general, if $X$ has at least two element, $$X \times (Y \times Z) \not\cong (X \times Y) \times (X \times Z),$$ and so the functor doesn't preserve products, thus ...
3
Good question. No, it does not have a left adjoint. One of the most important properties of adjoints is that right adjoints commute with limits (including products and the terminal object), while left adjoints commute with colimits (including coproducts and the initial object). For example, since $\emptyset$ is an initial object and $X \times -$ is a left ...
1
Not really if you are considering pure category theory (as the tag suggests): $$\operatorname{Hom}(X \times Y, Z) \cong \operatorname{Hom}(X, \operatorname{Hom}(Y,Z)).$$ As usual when you have a canonical natural isomorphism, in most context you can replace one object with the other without changing anything. The notions of "returning" and "evaluating" and ...
-1
As far as I can tell, the only real difference that is ever noted is that partial application combines two steps into one: (1) first it curries a function $f$ and then (2) evaluates the resulting curried function $\tilde{f}$ with a given value. But then, at least in my unimportant opinion, that shouldn't be called "partial application", since one would ...
1
The answer to your question is kind of: strongest and weakest topologies are certainly terminal and initial objects in certain categories, but they are required to satisfy an additional property. What I'm writing is partially taken from the Joy of Cats which is freely available online, and partially taken from the nlab. Explicitly, the category of ...
2
Yes, the final topology is the final object in the category of topologies on the given set making the given functions continuous, which is the poset whose objects are such topologies and whose maps are inclusions of sets, and dually for the initial topology. These words weren't necessarily invented to agree with each other, though. Arbitrary limits and ...
0
Well I found that the problem is with the definition of $\kappa$-accessibility and many resources just don't care about this notational problem and take it for granted! Here we go: A $\mathtt {Set}$-functor $T:\mathtt {Set} \to \mathtt {Set}$ is defined to be $\kappa$-accessible for a regular cardinal $\kappa$ iff for all sets $X$ and all $x\in TX$ there ...
0
The category of elements of $\mathcal Set\xrightarrow{F}\mathcal C$ is also a comma category $(\{*\}\downarrow F)$, i.e. a kind of weak pullback of $\mathbf 1\xrightarrow{\{*\}}\mathcal Set\xleftarrow{F}\mathcal C$, in the sense that we have pair of functors $\mathbf 1\leftarrow(\{*\}\downarrow F)\xrightarrow{\Pi_F}\mathcal C$ together with a natural ...
2
It might be amusing to see how this works in the context of covering spaces. If $G$ is discrete, a $G$-bundle is specified by a representation $\rho: \pi_1(B) \to G$; this representation is determined by picking a basepoint in the total space $x \in E$, and sending a loop to the unique $g$ such that if you lift the loop to start at $x$, it ends at $gx$. This ...
4
To directly answer your question: Andrej meant "monomorphism". Since the objects of a topos don't typically consist of (raw, unstructured) sets, the usual notions of injectivity and surjectivity don't apply. But for convenience, people still talk about "injections" and "surjections" in those contexts (and mean monomorphisms and epimorphisms, respectively). ...
1
I have missed that $A$ and $Z$ are not objects of ${\sf C}_A$ but only objects of ${\sf C}$, hence there is no circularity in the definition.
2
For (1), you are right. For (2): if $S \to B$ is a subobject, and $f: A \to B$ is any morphism, then we can form (because $\mathscr B$ is finitely complete) the pullback $$\require{AMScd} \begin{CD} f^{-1}(S) @>>> S\\ @VVV @VVV \\ A @>{f}>> B. \end{CD}$$ Because monomorphisms are stable under pullback, $f^{-1}(S) \to A$ is a subobject. ...
5
You trivialize $E\times_B E\to E$ using exactly the nullhomotopy of $f\circ p:E\to BG$. For this you need to know that principal $G$-bundles over $X$, up to isomorphism, are in bijection with maps $X\to BG$, up to homotopy, and that given $f:X\to BG$ classifying $q:Z\to X$ and a map $g:Y\to X,f\circ g$ classifies $Z\times_X Y\to Y$. The first fact is proven ...
2
I'll write only maps that are actual functions to avoid confusion about contravariance. A map $f:Y\to P(X)$ corresponds to a map $f':X\to P(Y)$ by $f'(x)=\{y:x\in f(y)\}$. Then $f''(y)=\{x:y \in f'(x)\}=\{x:y\in \{\hat y: x\in f(\hat y)\}\}=\{x:x\in f(y)\}=f(y)$. Switching $X$ and $Y$, we see that the prime operation is indeed bijective. I don't understand ...
2
The point you're missing is that in a general topos, the internal logic is not necessary Boolean, so $\neg$ does not necessarily give complements. Topological spaces are a good source of examples for intuitionistic logic; in a lattice of open sets, $\neg$ gives the exterior of an open set. Usually, the complement of the set is not open, and thus does not ...
1
Regarding the second question, the center crops up in a different way: If $G$ is a group viewed as a category and $1_G$ is the identity functor, then $Z(G)$ is the group of natural transformations $1_G \to 1_G$. It can be useful to extend this to more general categories: for an arbitrary category $\mathbf{C}$, we can define $Z(\mathbf{C}) = \hom(1_\mathbf{... 3 There is a sense in which they always coincide if$\prod_{j\in J}$is understood to be an internal product--that is, a right adjoint to the functor$J^*:\mathbf{C}/1\to\mathbf{C}/J$given by pullback along the unique morphism$J\to 1$. The intuition for thinking of such a right adjoint as a product is that such a pullback is like taking an object in$\mathbf{...
2
What you're saying seems confused. First, you say from your previous question you have concluded $\mathrm{Hom}_{\mathbf{Set}}(J,X)=X^J\cong\prod_{j\in J}X$ – do you not see that this is exactly the product (really power) and exponential coinciding in $\mathbf{Set}$? Second, you say that in the category of topological spaces, the product (really power) and ...
1
$\require{AMScd}$I don't quite understand the question, but basically, I think you're trying to define a particular double category. Objects. Sets Arrows. Relations Proarrows. Relations Squares. We assume that each square has at most one filler, and that it has a filler iff the condition $$\quad(a,a')\in\alpha\wedge(b,b')\in\beta\implies\... 2 Such a characterisation of 'relations between relations' is not possible. As a counterexample, let A,B,A',B',\alpha,\beta be whatever you want them to be, let R = \varnothing, let R' = A' \times B'. Then: (a',b') \in R' for all a' \in A' and b' \in B', so the statement on the right-hand side of your \Leftrightarrow symbol is true for all a,... 3 Well, it is quite trivial: both groups consist of a single object, so the functor can only map the first object to the second. For two arrows g,h in the first group, and a functor f to the second group, functoriality means that f(gh) = f(g)f(h). But this is precisely the definition of a group homomorphism. 1 Interesting questions. Let start be saying that your observations on the subcategories of groups are correct: if you exclude the empty category all the subcategories of a group are exactly the submonoids of the group (by the way there is a subcategory that is full: the group itself). About the posets, a subcategory of a poset-category P is a poset Q ... 2 Jean Goubeault-Larrecq has written a detailed textbook-level introduction to "Lawvere" metric spaces from a topological point of view in Chapter 6 of his book Non-Hausdorff Topology and Domain Theory. He defines a hemi-metric space to be a set X equipped with a set-function d\colon X\times X\to[0,\infty] such that d(x,x)=0 d(x,y)\leq d(x,z)+d(z,y) ... 1 I'll try to answer your questions in order. I think you're right, it should be enough to prove that K(f+g)=K(f)+K(g), which is exactly the definition of additivity. Again I think that your argument is sound. He's being redundant because every semigroup-homomorphism between groups preserves identities and inverses. K has been defined as the pointwise ... 4 This is another example of "equivalence instead of equality". Monic maps "ought" not be characterized by f(x) = f(y) \implies x=y: the right characterization ought to be f(x) \cong f(y) \implies x \cong y. (note: the following was written around the symmetric version; where we further assume d(x,y) = d(y,x). Without this assumption the details are ... 1 No, because a map of abelian groups can preserve addition without preserving the identity and inverses, but this is not the case for subtraction. Thus you answer this yourself in 2: he's being redundant. A morphism of additive functors is an arbitrary natural transformation. Thus since K is an equalizer with respect to natural transformations out of any ... 5 I assume you are looking for some real world examples of pseudo-metric spaces (that correspond to what you call Lawvere-metric spaces). Well a family of examples is given by the \mathcal L^p-spaces (where p is real number in [1,\infty]). In general for (\Omega,\mathcal F, \mu) a measured space, we have the vector space \mathcal L^p(\Omega) whose ... 3 The answer is yes, both formulations are equivalent. Let's see the correspondence. Let's see that giving a contravariant functor F:C^{op}\to D is equivalent to giving another functor G:C\to D^{op}. Assume we know F. At the level of objects it's clear what to do: for any a\in Ob(C)=Ob(C^{op}) we define G(a):=F(a)\in Ob(D^{op})=Ob(D). Now, given ... 3 The point is that the f_{ij} are the entries in a matrix:$$\begin{pmatrix} f_{11}& f_{12}\\ f_{21} &f_{22} \end{pmatrix}.$$In fact, in the case when our category is the category of (left) R-modules for some ring R, this description tells us that a map R^n\rightarrow R^m (where we think of R^n as R\oplus\dots\oplus R with n copies of ... 2 If by all the \times you mean \times_k, then your diagram is not necessarily cartesian, as the example of X = Y = \text{Spec}(\mathbb{R}) and L = \mathbb{C} shows (use that \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}), so I assume the \times on the right (resp. left) denotes fiber product over k (resp. L). ... 2 For (1), you're right, but one can (and probably the author intended this) even strengthen this by saying that any four morphisms f_{ij}: A_i\to B_j give rise to a morphism f: A_1\oplus A_2\to B_1\oplus B_2. For (2)... you're also right, it's just that what you say stands in no contradiction with the statement in question: namely, when you consider p_i\... 3 The 'corresponds' just means that there is a correspondence between e.g. the class \text{ Ob}(C) of objects of C and the class [1,C] of all functors from 1 to C. Likewise for the other cases. In fact, you probably have already seen a similar statement in the special case where the category C has only identity morphisms: elements of a class ... 1 Hint: functions don't need to be surjective. A functor C \to D does not have to "represent" every object in D: there is a functor that maps a field K to the abelian group K^* of invertible elements of K, but not every abelian group is the group of invertible elements of a field. 1 Well, yes, one functor 1\to C determines exactly one object in C, but this can be any object. So, the functors 1\to C correspond to objects of C. 3 An Alexandrov space is canonically preordered by x\leq y when y is in all the opens x is in. This is clearly reflexive and transitive, but in general need not be antisymmetric: the latter holds exactly for T_0 spaces. A preordered set gives rise to an Alexandrov topology with open sets the upward closed sets. It's not too hard to see this ... 4 The category of affine schemes is nothing but the opposite category of commutative rings. So \mathrm{Spec}\, \mathbb Z is a terminal object in the category of affine schemes because \mathbb Z is an initial object in the category of commutative rings (the only morphism \mathbb Z \to R is the one sending 1 to 1_R...). 1 Even with the extra condition that \mathcal{A} is a Grothendieck category, it may still have no simple objects. I think the following is the easiest example I know. Let R be a (necessarily non-noetherian) commutative local ring with non-zero maximal ideal \mathfrak{m} satisfying \mathfrak{m}^2=\mathfrak{m}. Let \mathcal{C} be the category of R-... 1 The answer to your question is yes f is indeed the zero of the abelian group \mathscr C(A,B). Here is the proof. Since \mathscr C(0,0) has only one object it is the trivial group (the group with only the identity). We know that in a pre-additive category composition is bilinear so if r \in \mathscr C(A,0) and l \in \mathscr C(0,B) (the only ... 2 Yes: Deconstruct your f:A \to B as gh, where h:A \to 0 and g:0 \to B. As there is a unique map g:0 \to B, we must have g+g = g and hence, by bilinearity, gh + gh = (g+g)h = gh. But then$$f + f = gh + gh = gh = f,$$and cancelling f from both sides yields that f is the zero object in the Hom-set. 2 I think Lectures and Exercises on Functional Analysis by A. Y. Helemskii might be exactly what you're looking for. Quoting from the introduction: Perhaps the main idea is that our book is written from the categorical point of view. Everywhere we stress and comment on the categorical nature of the fundamental constructions and results (like the ... 4 From the comments, the original issue is solved and now you ask how to figure out the nature of this unique map. You can think of the map X \to \operatorname{Spec} \mathbb Z as a map, which sends a point to the characteristic of its residue field. 2 One possible way to "see" composition of arrows of the original category \mathcal{C} in \operatorname{Ar}(\mathcal{C}) is the following: Suppose f:x\to y and g:y\to z, let 1_x,1_y,1_z\in\operatorname{Ar}(\mathcal{C}) be the identity morphisms of x,y,z respectively, then f gives a morphism 1_f:1_x\to1_y in the obvious way, and similarly for ... 2 The arrow category is equipped with two functors s, t : \text{Ar}(C) \to C giving the source and target of an arrow. Composition is a functor$$\text{Ar}(C) \times_{\text{Ob}(C)} \text{Ar}(C) \to \text{Ar}(C) where the LHS is a (2-)pullback, with one of the maps $\text{Ar}(C) \to \text{Ob}(C)$ being source and one being target. This expresses precisely ...
4
Note that there exists a $G$-equivariant map $f:G/X\to G/Y$ iff some conjugate of $X$ is contained in $Y$, and there is a $G$-equivariant bijection iff some conjugate of $X$ is equal to $Y$. So the question is, if $gXg^{-1}\subseteq Y$ and $hXh^{-1}\supseteq Y$ for some $g,h\in G$, must $X$ be conjugate to $Y$? The answer is no. For instance, fix any ...
1
A map $\Phi:S\rightarrow \prod_{j\in J} E_j$ is the same thing as a family of maps $(\Phi_j:S\rightarrow E_j)_{j\in J}$, where $\Phi_j$ is just the $j$-th coordinate in $\prod_{j\in J} E_j$. So to give a map $f_0:S\rightarrow\prod_{i\in I}\prod_{\varphi\in M_i} G_{i,\phi}$, it is enough to give each "coordinates". For the coordinate corresponding to $(i,\... 1 An answer among others. A topology on$X$is obtained by the relation$r\subseteq X\times \mathcal P(X)$where$(a,S)\in r\iff a\in \overline S$. Then a "canonical morphism"$(X,r)\to (Y,s)$is a function$f:X\to Y$such that$x\in \overline{f^{-1}(T)}\implies f(x)\in\overline T\$, which means continuity. But of course it is possible to chose other ...
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Equivalent fractions can best be visualised by application a bar model. If you booty two identical confined and breach one in bisected and the added into sixths you will see that one bisected is absolutely the aforementioned as three sixths, i.e. ¹/₂ = ³/₆.
This works because back 1 is the numerator and 2 is denominator of a fraction, and both are assorted by 3, you get 3 and 6 (or three-sixths).
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#### Resources tagged with Visualising similar to The Path of the Dice:
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Visualising. Working systematically. Generalising. Compound transformations. Cubes. Dice. Tangram. Interactivities. Games. Practical Activity.
### Right or Left?
##### Stage: 2 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### The Path of the Dice
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A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
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Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
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Can you fit the tangram pieces into the outlines of the chairs?
### Construct-o-straws
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Make a cube out of straws and have a go at this practical challenge.
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How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### World of Tan 19 - Working Men
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Can you fit the tangram pieces into the outline of this shape. How would you describe it?
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Can you fit the tangram pieces into the outline of Little Ming playing the board game?
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Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
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Can you fit the tangram pieces into the outline of this telephone?
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##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
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Can you fit the tangram pieces into the outline of Mai Ling?
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Exchange the positions of the two sets of counters in the least possible number of moves
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Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### Taking Steps
##### Stage: 2 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### Midpoint Triangle
##### Stage: 2 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Endless Noughts and Crosses
##### Stage: 2 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### World of Tan 2 - Little Ming
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### Regular Rings 1
##### Stage: 2 Challenge Level:
Can you work out what shape is made by folding in this way? Why not create some patterns using this shape but in different sizes?
### Coin Cogs
##### Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
### Seeing Squares
##### Stage: 1 and 2 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Stage: 1 and 2
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### Move Those Halves
##### Stage: 2 Challenge Level:
For this task, you'll need an A4 sheet and two A5 transparent sheets. Decide on a way of arranging the A5 sheets on top of the A4 sheet and explore ...
### Counter Roundup
##### Stage: 2 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### Turning Cogs
##### Stage: 2 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same. | 1,974 | 8,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-47 | latest | en | 0.834301 |
https://in.mathworks.com/matlabcentral/cody/problems/42809-sorting-integers-by-their-digits-level-1/solutions/1704980 | 1,591,320,341,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492295.88/warc/CC-MAIN-20200604223445-20200605013445-00079.warc.gz | 388,531,731 | 15,619 | Cody
# Problem 42809. Sorting integers by their digits (Level 1)
Solution 1704980
Submitted on 9 Jan 2019 by Athi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
v = [14 3 268 14 210 1 80]; w_correct = [1 14 14 210 268 3 80]; assert(isequal(soort(v),w_correct))
2 Pass
v = [246 24680 2468 246 24 2 24680 24 2468 2]; w_correct = [2 2 24 24 246 246 2468 2468 24680 24680]; assert(isequal(soort(v),w_correct))
3 Pass
v = ones(1,100)*randi(100); w_correct = v; assert(isequal(soort(v),w_correct))
4 Pass
v = []; w_correct = []; assert(isequal(soort(v),w_correct)) | 247 | 691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-24 | latest | en | 0.529921 |
http://forums.codeguru.com/printthread.php?t=539027&pp=15&page=1 | 1,521,291,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645069.15/warc/CC-MAIN-20180317120247-20180317140247-00570.warc.gz | 117,247,619 | 3,126 | # Idiot/Noob HelpLine :) <----------HERE
• August 8th, 2013, 03:15 PM
percy121
Idiot/Noob HelpLine :) <----------HERE
Hey guys im rather new to programming and I have a newbie question for you:
I am making a program as a project, that's purpose is to tell whether a number is divisable by 1,2,3,4,5,6,7,8,9 and/or 10.
So far I have the graphic parts done like JOptionPanes and SystemOuts but I am stuck on the Calculation Parts. I need the program to separate the digits in a potentially 200 digit long number and "read" them. Here is an Example of one of the Equations i need to do:
To tell whether a number is divisable by 11 you must take the even number places (reading left to right: odd,even,odd,etc.), add them up, then add all the odd places, then take the sums, and subtract the evens from the odds.
So... Fairly specifically... how would I do this? (Please include all var's and associated scanners: Remember Im new at this)
Thanks,
JB ;)
• August 9th, 2013, 03:15 AM
LukeCodeBaker
Re: Idiot/Noob HelpLine :) <----------HERE
All you have to do is to use modulo operator. If you don't know what modulo operator is, see wiki http://en.wikipedia.org/wiki/Modulo_operation or simple example here: http://www.cprogramming.com/tutorial/modulus.html.
• August 9th, 2013, 02:46 PM
percy121
Re: Idiot/Noob HelpLine :) <----------HERE
Hey Luke aor anybody else... If I could have a java example of the modulo operator that would be great! Thanks! | 403 | 1,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-13 | longest | en | 0.875682 |
https://sofsource.com/algebra-resources/rational-equations/online-ti-83-plus-calculator.html | 1,552,959,790,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201882.11/warc/CC-MAIN-20190319012213-20190319034213-00039.warc.gz | 629,901,792 | 12,097 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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## NCERT Solutions Class 10th Maths Chapter 12 Areas related to Circles PDF Download Free 2018-19 Mathematics
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• Latest CBSE Sample Papers Class 12th 2018 | 641 | 2,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-13 | latest | en | 0.721189 |
http://www.jiskha.com/display.cgi?id=1192574754 | 1,493,203,061,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121267.21/warc/CC-MAIN-20170423031201-00191-ip-10-145-167-34.ec2.internal.warc.gz | 583,793,590 | 4,387 | # Jake
posted by on .
If 142.38 g of l water at 21.7 degrees C is placed into a syrofoam cup with 174.36 g of ice at -27.5 degrees C what would be the final temperature of the entire contents at equilibrium? If it is partially frozen, how many g of the ice has melted?
given values
cp water = 4.184 j/g degree c
cp ice = 2.114 j/g degree c
heat of fusion of ice is 335 J/g
so far, i have this:
(174.36g)(2.114g)(27.5 C)
= 10247 g to three sig figs = 10200 J
is that i calculate the energy required to raise the ice to 0 degrees C?
now i know that the next steps are to find the enrgy required to melt the ice and the energy reuired to cool the water down to 0 degrees C. how do i do those two calculations?
after that i did this to try and melt the ice
q = (mass l H20/molar mass H20)X(335 j/g) the 335 is the heat of fusion.
is that correct? if not, what am i supposed to do?
• Jake - ,
Your 10,247 J I get 10,136 J. Check my arithmetic. THEN, I would keep all the numbers and round at the end; otherwise, it may lead to rounding errors. I'll assume the 10,136 J is correct to move the ice from -27.5 to Oo C.
Then we need to calculate the energy required to cool the liquid water from 21.7 C to zero C. That is
142.38 x 4.184 x 21.7 = check me on this but I get about 13,000 J or so. You do the exact number.
So if you look at the numbers, we can move the ice to zero and move the lqiuid water to zero and have about 3,000 J left in the liquid water. That will melt some of the ice. How much?
q = mass x 335 J/g. Calculate the mass of ice that will melt.
I hope this helps. Check my thinking. Check my work. | 480 | 1,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-17 | latest | en | 0.924523 |
https://studyres.com/doc/466524/supply-and-demand--price-and-quantity-determination-in | 1,621,009,438,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00632.warc.gz | 549,874,499 | 9,184 | # Download Supply and Demand: Price and Quantity Determination in
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Transcript
The Economics
Department, UMR
Presents:
Supply and Demand: Price
and Quantity
Determination in
Competitive Markets
Starring
Changes in
Equilibrium
u
Featuring
uIncreases
in Demand, c.p.
uIncreases in Supply, c.p
uDecreases in Demand, c.p.
uDecreases in Supply, c.p.
uSimultaneous Changes
Changes in Equilibrium
u
u
Remember that Supply and Demand
are drawn under the ceteris paribus
assumption.
Any factors which cause Supply and/or
Demand to change will affect
equilibrium price and quantity.
Change in Demand
Demand will change for any of the
factors discussed previously.
u Remember PINTE (except for the price
of the good itself)
u For instance, let’s say the demand for
CDs increased due to an increase in
income
u
Increase in Demand
P
S
P*’
P*
E’
E
D’
D
0
Q* Q*’
Q/t
Change in Supply
Supply will change for any of the
factors discussed previously
u Remember PENT (except for the price
of the good itself)
u For instance, let’s say that the
government lowers taxes on CDs
u
Increase in Supply
P
S
S’
P*
E
P*’
E’
D
0
Q* Q*’
Q/t
Changes in Demand and
Supply
To determine the impact of both supply
and demand changing:
u First examine what happens to
equilibrium price and quantity when
just demand shifts.
u Second, examine what happens to
equilibrium price and quantity when
just supply changes
u Finally, add the two effects together.
Changes in Demand and Supply
u
u
u
u
u
u
When supply and demand move in the same
direction equilibrium price is ambiguous
When supply and demand move in opposite
directions equilibrium quantity is ambiguous
If P and Q both increase the dominant force
must have been an increase in D
If P and Q both decrease the dominant force
must have been an decrease in D
If P increases and Q decreases, the dominant
force must have been a decrease in S
If P decreases and Q increases the dominant
force must have been an increase in S
Explaining Changes in
Equilibrium Price and Quantity
P
Initial equilibrium
A
B
C
F
Another equilibrium
Q/t
Moving to quadrant B implies the dominate force was an
increase in demand. To quadrant C, the dominate force is
a decrease in demand. Moving to quadrants A or F
implies the dominate force was supply (decrease for A,
and increase for F)
Increase in Supply and
Demand: Q, )P ?
Q
P
S
S’
P*’
P*
D’
D Q/t
0
P
Q*
Q*’
)D > )S > 0
D’
S
S’
P*
P*’
0
D
Q* Q*’ Q/t
0 < )D < )S
Decrease in Supply and
Demand: Q, )P ?
Q
S’
P
S
P*
P*’
S
D’
P*’
P*
D
D’ Q/t
0
S’
P
Q*’ Q*
)D > )S > 0
D
0
Q*’Q*
Q/t
0 < )D < )S
Increase in Supply and
Decrease in Demand: P, )Q ?
Q
D
P
P*
S
S
S’
D’
S’
P*
P*’
P*’
D’
0
P
Q*’Q*
)D dominates, )Q < 0
Q/t
0
D
Q*Q*’ Q/t
)S dominates, )Q > 0
Decrease in Supply and
Increase in Demand: P, )Q ?
Q
P
D
D’
S’
S’
S
S
P*’
P*’
P*
0
P
Q*Q*’
)D dominates, )Q < 0
D’ P*
Q/t
0
D
Q*’Q*
Q/t
)S dominates, )Q > 0 | 977 | 3,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-21 | latest | en | 0.860945 |
https://www.molottery.com/powerball/understanding_chances.jsp | 1,660,341,925,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00790.warc.gz | 800,501,574 | 8,925 | # Understanding Powerball Chances
No matter how many people play Powerball, the chances of winning never change. The chances of winning are based on the amount of number combinations, not the number of players. Quick Pick tickets are generated randomly by the terminal at the retailer, and there is no central computer that is controlling ticket generation.
Many people ask us to explain how their chances of winning are determined. So, let's start at the beginning!
The Powerball game is played by selecting five numbers out of 69 choices (the numbers from 1 to 69), and by selecting one extra number from 1 to 26 (that's the Powerball). When it's time to draw the winning numbers, there are two machines tumbling the ball sets at the same time: one machine has the 69 white balls, and the other one has the 26 red "Powerballs." So the Powerball game is like holding two drawings at the same time; two independent events played simultaneously. Any chances of winning has to consider what happens to the number selections within one drum combined with the possibilities of what's happening in the other drum. That idea is really important to keep in mind. A winning set of numbers will be whatever five balls are selected out of the white-ball drum plus the one ball selected from the red-ball drum.
But the chances of winning look confusing because even at the smallest winning level (just getting the Powerball right), many people think the chances of winning that should be 1 in 26 (one correct choice out of a drum of 26 balls, right?). But we're not just playing that one drum; in order to say you "only matched the Powerball" that means you have to have missed all five of the white balls that were chosen in that other drum! The chances of doing that combined with your 1-in-26 shot of getting the Powerball actually come out to being 1 in 38.3. In other words, there is one chance in 38.3 that you'd pick the winning Powerball number and miss selecting any of the five winning white-ball numbers.
So when you hear a "chances of winning" calculation, it's describing the chance that you will have chosen the winning numbers, including the chances you chose numbers that didn't get drawn. Since there are several ways to win (from zero to all five of the white balls, with and without having the Powerball), that means people who calculate the chances of winning have to figure each level of winning or not winning in each of the two drums.
For example, in order to win the big jackpot, a winner must have chosen all five white balls correctly, as well as the Powerball number. To describe the chances of winning, you have to say what your possible chances are of choosing the winning balls against choosing any of the 64 non-winning balls in the one drum, combined with your chances of choosing the winning Powerball against choosing any of the 25 non-winning red balls in that drum. That's a lot of combinations - more than 292 million, in fact. But there's an easy formula that gets you there.
Where do you start? Any calculation of chances for a Lottery game involves three basic elements:
1. The total number of balls (numbers) you are choosing from, which is called the field or (f).
2. The number of balls the player selects on his\her play slip, which is called the pick or (p).
3. The number of correct matches between the player's picks and the numbers drawn out of the machine, which is called the match (m).
The mathematical formula looks like this:
Each part of that formula, which is in parentheses, is a "binomial coefficient," which is a calculation on how many ways there are to make different combinations. In our case, we're talking about how many combinations of lottery balls you could make with a set of 69 (or 26) balls. What that means is that for whatever numbers you have, the binomial can be rewritten as follows:
X! means "x factorial"; for example, 5! is 5 x 4 x 3 x 2 x 1 = 120.
Now we're ready to look at winning that jackpot!
To do that, we said you'd have to describe the chances of picking five numbers including 64 non-winning ones out of a total of 69. Then you combine that with the figure to describe your chances of picking one winning Powerball number, including choosing 25 non-winning numbers. So using our binomials and factorials, it looks like this:
Powerball chances of winning the jackpot:
(5 of 5 plus the Powerball)
So, 69 x 68 x 67 x 66 x 65 = 1,348,621,560 and 5 x 4 x 3 x 2 x 1 = 120; divide one by the other:
1,348,621,560 / 120 = 11,238,513 (chances of 11,238,513 to 1 of picking all 5 white balls out of 69)
And don't forget that now we have to combine the possibilities of choosing the Powerball, which we know is one in 26. That formula works just like the one above:
So, 11,238,513 / 1 x 26 / 1 = 292,201,338.
Now, that's just one win scenario: the big jackpot. But you can also win eight other prizes ranging from \$4 to \$1,000,000.
(In order to win a prize, your Powerball number must at least match the Powerball number drawn, or you must match at least three of the five white-ball numbers drawn.)
When you combine the chances of all those levels and all those combinations, it comes out to 1 in 24.87.
So each player has a 1-in-24.87 chance to win any of the prizes (jackpot, \$1,000,000, \$50,000, \$100, \$7, or \$4)!
We hope this helps to better explain the chances of winning the Powerball jackpot.
| About Us | Careers | Calendar | Play Responsibly | Where to Play | Game Rules
Visit Us On
Players must be 18 years or older to purchase Missouri Lottery tickets.
*In the event of a discrepancy, official winning numbers prevail over any numbers posted on this website. | 1,338 | 5,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-33 | latest | en | 0.970069 |
https://en.wikipedia.org/wiki/Talk:Angle_of_parallelism | 1,505,947,741,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687484.46/warc/CC-MAIN-20170920213425-20170920233425-00550.warc.gz | 661,243,280 | 14,149 | # Talk:Angle of parallelism
WikiProject Mathematics (Rated B+ class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
B+ Class
Mid Importance
Field: Geometry
## Bonola and Halsted
In 1912 Roberto Bonola published a textbook with Open Court in Chicago on the topic of non-Euclidean geometry. When Dover re-issued the book half a century later, it included work by G.B. Halsted that significantly improved coverage of the subject, particularly a translation of Lobachevski. It is that appendix that is cited for an English reference for the topic "angle of parallelism". Recently the Bonola text has been scanned into Archive.org. One can now find the link at the end of Non-Euclidean geometry. Since the text scanned was the original 1912 edition it does not include Halsted's translation of Lobachevski. Hence I have removed the link as inappropriate for this article.Rgdboer (talk) 22:45, 19 February 2008 (UTC)
The Halsted manuscript has become available Google Books so the old print source in the back of the Dover edition of Bonola is unnecessary.
The text is available by template:Geometrical Researches on the Theory of Parallels, p. 13, at Google Books.
Bonola's original edition did not include Halsted and the paging is unclear in editions that include the translation.Rgdboer (talk) 01:24, 10 February 2015 (UTC)
## Napier ?
The second of the equivalent descriptions of angle of parallelism is
${\displaystyle \tan({\tfrac {1}{2}}\Pi (a))=e^{-a}.}$
Here the e is the base of natural logarithms. The historic English reference cited is Halsted 1891, now on GB:
Geometrical Researches on the Theory of Parallels, p. 41, at Google Books.
Unfortunately Halsted considers unit of length in connection with selection of a base for logarithm, and in doing so refers to Napierian logarithm, a topic frequently confused with natural logarithm, but is in fact a rather different function. All reference to Napierian logarithm should be avoided; better references might be found to complete the thought Halsted meant to express before this misdirection.Rgdboer (talk) 02:04, 11 February 2015 (UTC) Correct date 1891.Rgdboer (talk) 02:07, 11 February 2015 (UTC)
Consequently, the following quotation from Halsted was removed:
...${\displaystyle \tan({\tfrac {1}{2}}\Pi (x))=\theta ^{-x}}$ where ${\displaystyle \theta }$ may be any arbitrary number, which is geater than unity, since ${\displaystyle \Pi (x)=0}$ for ${\displaystyle x=\infty }$.
Since the unit by which lines are measured are arbitrary , so we may also understand by ${\displaystyle \theta }$ the base of Napierian logarithms.
The quotation is found on page 41 of Halsted's translation of Lobachevsky.Rgdboer (talk) 02:21, 20 February 2015 (UTC)
## Negative angle
The following confusing contribution was removed:
By definition for a negative angle p:
${\displaystyle \Pi (p)+\Pi (-p)=\pi }$
Before the text was made consistent with a as variable segment length, some uses were p. It does not make sense to then call p an angle. Negative segment length doesn't make sense either. Perhaps a discussion here can clarify the idea, or a reference can be produced to justify this addition.Rgdboer (talk) 20:08, 7 March 2015 (UTC)
Reference: Halsteds translation of Lobachevsky's "Geometrische Untersuchungen zur Theory der Parallellinien" page 20-21, end of paragraph 23 (taken from Bonola):
Since we are wholy at liberty we will understand by the symbol ${\displaystyle \Pi (p)}$ when the line p is expressed by a negative number we will assume
${\displaystyle \Pi (p)+\Pi (-p)=\pi }$,
as equation which shall hold for all values of p, positive as well as negative and for p = 0.
Maybe my shortening of this was not really clear, or maybe it is later given another meaning
WillemienH (talk) 20:51, 9 March 2015 (UTC)
## Compass and straightedge construction
The following section was moved here for discussion. The third step assumes b and l intersect (with no justification). The section has no attribution, and the appeal to Lambert quadrilateral is strange because there is no link to this article there. Recommend finding feature of this angle of parallelism in that quadrilateral and documenting it before another effort like this:
The angle of parallelism between an point P and a line l can be constructed by:
• Construct line b through point P perpendicular to line l
• Construct line d through point P perpendicular to line b
• Point B is the intersection of line b and line l
• Construct circle a with center P going trough B
• Construct circle c with center B going trough P
• Point C is one of the intersections of line l and circle c
• construct line e through point C perpendicular to line d
• Points D and E are where line e intersects with circle a (D near C)
• The lines PD and PE are the limiting parallels trough P to l
• The angle ∠BPD is the angle of parallelism for segment PB.
The lines l, b, d and e are the sides of a Lambert quadrilateral with the non right angle at C.
Hyperbolic geometry is now a classical subject with numerous sources, some of which may support the idea sparking this contribution. — Rgdboer (talk) 03:16, 9 November 2015 (UTC)
The construction is from Bonola's "Non euclidean Geometry" on page 217, and is also in Coxeter's "Non euclidean Geometry" on page 204. I did simplify the construction a bit by fixing the distance BC to be equal PB. (which is not really needed, BC must be equal to PD and PE)
I don't undertstand your remark "The third step assumes b and l intersect (with no justification)", b is the line from point P perpendicular to line l. How can b and l be perpendicular but not intersecting? WillemienH (talk) 22:40, 11 November 2015 (UTC)
Good to see some references (will check them out soon). Also, now the construction of the Lambert quadrilateral makes sense to me (sorry about slowness). But still there is the question of relation to this article, Angle of parallelism. The construction belongs on the quadrilateral page. Now the important thing here is the question: Is ∠BPD the angle of parallelism for segment PB ? — Rgdboer (talk) 00:52, 12 November 2015 (UTC)
Yes, ∠BPD is the angle of parallelism for segment PB. (if we assume CD < CE ) It is about the construction of an angle of parallelism , so why should it on the page for the lambert quadrilateral? i just added that remark to make the construction more understandable. Long time I thought that ∠C was right while in fact it is acute, the ((unnamed) angle opposite ∠B is right. I updated the construction above with these remarks. Ps I cannot follow the proof of this construction, maybe you can add a bit about that as well. WillemienH (talk) 01:08, 12 November 2015 (UTC)
Have put the construction in the Quadrilateral article. Must rush now to check Coxeter before Library shuts. — Rgdboer (talk) 01:16, 12 November 2015 (UTC)
Coxeter is making use of Cayley-Klein metric theory in his "proof" of the construction of the parallels through a given point not on a reference line. Straightedge and compass constructions are part of Euclidean geometry and can be used for the model Coxeter gives as figure 10.4A on page 204. On the other hand, the hyperbolic plane is a curved surface where geodesics replace straight lines and the "geodesic-straightedge" is science fiction. Coxeter gives as references Baldus (1927) and Mohrmann (1930) for his construction of the parallels. They may be helpful. As for this article on Angle of parallelism, it is the function Π(x) that Lobachevski introduced that forms the subject. Nevertheless, the configuration in Coxeter's diagram is very instructive and may well have a place somewhere in the encyclopedia, perhaps in Beltrami-Klein model. But that model does not represent angles that agree with their Euclidean magnitudes (it is not conformal), whereas the Poincare half-plane model used in this article does have the conformal property. — Rgdboer (talk) 22:45, 12 November 2015 (UTC)
I would like to add the construction back to the article. the article is on the angle of parallelism in the hyperbolic plane , (not just on how to measure/ calculate it) and this is just a good way to construct it as it is independent of the model used.
Bonola gives some other proofs of the construction, but they all seem rather complicated. I need to study them in more detail myself. WillemienH (talk) 10:00, 13 November 2015 (UTC)
Saddle shape surface, suggestive of hyperbolic plane
Check out Compass-and-straightedge construction and see that it is presuming Euclidean geometry. There is no straightedge to be used for constructions in Lobachevsky's plane. — Rgdboer (talk) 02:37, 14 November 2015 (UTC) | 2,193 | 8,918 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-39 | latest | en | 0.936693 |
https://mail.python.org/pipermail/numpy-discussion/2015-December/074291.html | 1,632,211,013,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057199.49/warc/CC-MAIN-20210921070944-20210921100944-00350.warc.gz | 427,271,548 | 2,268 | # [Numpy-discussion] Q: Use of scipy.signal.bilinear
R Schumacher rays at blue-cove.com
Tue Dec 8 13:12:03 EST 2015
```Sorry - I'll join there.
- Ray
At 10:00 AM 12/8/2015, you wrote:
>On Tue, Dec 8, 2015 at 9:30 AM, R Schumacher
><<mailto:rays at blue-cove.com>rays at blue-cove.com> wrote:
>We have a function which describes a frequency
>response correction to piezo devices we use. To
>flatten the FFT, it is similar to:
>Cdis_t = .5
>N = 8192
>for n in range(8192):
>Â B3 = n * 2560 / N
>Â Fc(n) = 1 /
>((B3/((1/(Cdis_t*2*pi))**2+B3**2)**0.5)*(-0.01*log(B3) + 1.04145))
>
>In practice it really only matters for low frequencies.
>
>I suggested that we might be able to do a time
>domain correction as a forward-reverse FFT
>filter using the function, but another said it
>can also be applied in the time domain using a bilinear transform.
>So, can one use
><http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.signal.bilinear.html>http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.signal.bilinear.html
>and, how does one generate b,a from the given
>Fourrier domain flattening function?
>
>
>This should go to either
><mailto:scipy-user at scipy.org>scipy-user at scipy.org
> or <mailto:scipy-dev at scipy.org>scipy-dev at scipy.org
>
>Chuck
>_______________________________________________
>NumPy-Discussion mailing list
>NumPy-Discussion at scipy.org
>https://mail.scipy.org/mailman/listinfo/numpy-discussion
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``` | 501 | 1,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-39 | latest | en | 0.673817 |
https://dietetyka.net.pl/3117/21dUvv2/notes_critical_speed_formula_for_ball_mill_in_meters_.html | 1,632,740,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00570.warc.gz | 252,729,048 | 6,739 | #### Ball Mill Critical Speed Mineral Processing & Metallurgy
A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies. Use our online formula The mill speed is typically defined as the percent of the Theoretical
#### critical speed formula for ball mill
Calculate Critical Speed Ball Mill India. Calculate Critical Speed Ball Mill India The formula to calculate critical speed is given below n c 42305 sqtdd n c critical speed of the mill d mill diameter specified in meters d diameter of the ball in practice ball mills are driven at a speed of 5090 of the critical speed the factor being influenced by economic consideration.derivation for the
#### Mill Critical Speed Calculation
Effect of Mill Speed on the Energy Input In this experiment the overall motion of the assembly of 62 balls of two different sizes was studied. The mill was rotated at 50, 62, 75 and 90% of the critical speed. Six lifter bars of rectangular cross-section were used at equal spacing. The overall motion of the balls at the end of five revolutions is shown in Figure 4. As can be seen from the
#### Critical Speed Tumbling Mill Formula
Mill Critical Speed Formula Derivation Grinding The formula to calculate critical speed is given below N c = 42 305 /sqt(D d) N c = critical speed of the mill D = mill diameter specified in meters d = diameter of the ball In practice Ball Mills are driven at a speed of 50 90% of the critical speed the factor being influenced by economic
#### Critical Speed Of Ball Mill Formula Derivation
Ball mill critical speed derivation . Ball Mills Mine EngineerCom ball mill critical speed derivation,If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill The point where the mill becomes a centrifuge is called the Critical Speed, and dragThe second is air drag, which.
#### Ball Mill Operating Speed Mechanical Operations Solved
The critical speed of ball mill is given by, where R = radius of ball mill; r = radius of ball. For R = 1000 mm and r = 50 mm, n c = 30.7 rpm. But the mill is operated at a speed of 15 rpm. Therefore, the mill is operated at 100 x 15/30.7 = 48.86 % of critical speed.
#### Critical Speed Of A Ball Mill
Notes Critical Speed Formula For Ball Mill In Meters. The critical speed of a ball mill is the speed at which the balls just ata obtained by fitting a first-order decay equation, y a becx, using tableurve 2 d meters of 8 and 16 cm, were rotated with a erweka ar400 variable speed oller vr 1983 a note on energy-size reduction relationships in
#### to calculate critical speed of ball mill practical
how to calculate the critical speed of rotating drum to calculate critical speed of ball mill practical. there is a critical rotation speed above . formula for critical speed of a rotating mill,factory that .
#### Critical Speed Of Ball Mill Formula
Ball Mill Notes Critical Speed Formula For Ball Mill In Meters. The critical speed formula of ball mill 4229 is a horizontal cylindrical equipment, which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m, and the mill with diameter of
#### Ball Mill Critical Speed Mineral Processing & Metallurgy
A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a mill as speed varies. Use our online formula The mill speed is typically defined as the percent of the Theoretical
#### Critical Speed Of Ball Mill Formula Derivation
Ball mill critical speed derivation . Ball Mills Mine EngineerCom ball mill critical speed derivation,If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill The point where the mill becomes a centrifuge is called the Critical Speed, and dragThe second is air drag, which.
#### Critical Speed Equation Of A Ball Mill
Ball Mills Mine Engineer.Com. information on ball mills. A Ball Mill grinds material by rotating a cylinder with steel grinding balls, This formula calculates the critical speed of any ball mill.
#### Critical Speed Calculation Of Ball Mill
Ball Mill Critical Speed Mineral Processing . 2021-3-10 A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell.
#### how to calculate mill critcal speed
Ball Mill Critical Speed Mineral Processing & 2021-1-22 A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell.
#### Critical Speed Calculation Formula Of Ball Mill
Nov 18, 2013 formula for calculating the critical speed of a ball mill, ball mill speed? newbie questions 2 most if the actual speed of a 6 ft diameter ball mill Read more how to calculate critical speed of a ball mill YouTube
#### Critical Speed Of Ball Mill Formula
Ball Mill Notes Critical Speed Formula For Ball Mill In Meters. The critical speed formula of ball mill 4229 is a horizontal cylindrical equipment, which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m, and the mill with diameter of
#### to calculate critical speed of ball mill practical
how to calculate the critical speed of rotating drum to calculate critical speed of ball mill practical. there is a critical rotation speed above . formula for critical speed of a rotating mill,factory that .
#### Equation Formula Critical Speed Of The Ball Mill 2cmuc
Critical Speed Of A Ball Mill Formula . The formula to calculate critical speed is given below N c 42305 sqtDd N c critical speed of the mill D mill diameter specified in meters d diameter of the ball In practice Ball Mills are driven at a speed of 5090 of the critical speed the factor being influenced by economic consideration.
#### critical speed of ball mill pdf Pochiraju Industries Ltd
Mill Work Index which expresses the resistance of a material to ball milling. . from 40 ball mills show that the average critical speed is 0.73 (Morrell, 1996). Get Price 5-Ball Mill-8 Imimg
#### SAGMILLING.COM .:. Mill Critical Speed Determination
Result #1: This mill would need to spin at RPM to be at 100% critical speed. Result #2: This mill's measured RPM is % of critical speed. Calculation Backup: the formula used for Critical Speed is: N c =76.6(D-0.5) where, N c is the critical speed,in revolutions per minute,
#### Crush Plant Why Critical Speed Of Ball Mill Crusher
Why Critical Speed Of Ball Mill Is Less Than One How Much » defination of ball mill critical speed » critical speed of ball mill 76.6 » calculator for ball mill critical speed » critical speed formula for ball mill in meters
#### The working principle of ball mill Meetyou Carbide
Mainly the rotation speed of the ball mill, the size and number of the grinding body, the volume of the object to be polished, the grinding medium and the grinding time. Wherein D is the diameter of the mill barrel (meter). Let D = 0.5 m, then. This is the critical speed of the 180 litre wet mill currently used in cemented carbide production.
#### formulae for cement mill capacity
Cement Mill Capacity Calculation Formula. Cement mill capacity calculation formula cement mill critical mill speed formula kimsburgersNl jan raw mills usually operate at critical speed and cement mills at calculation of the critical mill speed g weight of a grinding ball in kg w angular velocity of the mill tube in radialsecond w n di inside mill diameter in meter effective mill diameter n
#### Critical Speed Of Ball Mill Formula Derivation
Ball mill critical speed derivation . Ball Mills Mine EngineerCom ball mill critical speed derivation,If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill The point where the mill becomes a centrifuge is called the Critical Speed, and dragThe second is air drag, which.
#### Critical Speed Calculation Of Ball Mill
Ball Mill Critical Speed Mineral Processing . 2021-3-10 A Ball Mill Critical Speed (actually ball, rod, AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell’s inside surface and no balls will fall from its position onto the shell.
#### Critical Speed Equation Of A Ball Mill
Ball Mills Mine Engineer.Com. information on ball mills. A Ball Mill grinds material by rotating a cylinder with steel grinding balls, This formula calculates the critical speed of any ball mill.
#### critical speed of a ball mill Pochiraju Industries Ltd
Ball Mill Operating Speed Mechanical Operations Solved Problems In a ball mill of diameter 2000 mm, 100 mm dia steel balls are being used for grinding. Presently, for the Calculations: The critical speed of ball mill is given by,. Get Price
#### SAGMILLING.COM .:. Mill Critical Speed Determination
Result #1: This mill would need to spin at RPM to be at 100% critical speed. Result #2: This mill's measured RPM is % of critical speed. Calculation Backup: the formula used for Critical Speed is: N c =76.6(D-0.5) where, N c is the critical speed,in revolutions per minute,
#### The working principle of ball mill Meetyou Carbide
Mainly the rotation speed of the ball mill, the size and number of the grinding body, the volume of the object to be polished, the grinding medium and the grinding time. Wherein D is the diameter of the mill barrel (meter). Let D = 0.5 m, then. This is the critical speed of the 180 litre wet mill currently used in cemented carbide production.
#### Critical Speed Calculation Formula Of Ball Mill
Nov 18, 2013 formula for calculating the critical speed of a ball mill, ball mill speed? newbie questions 2 most if the actual speed of a 6 ft diameter ball mill Read more how to calculate critical speed of a ball mill YouTube
#### Velocity Calculation In Ball Mills Crusher Mills, Cone
formula calculates the critical speed of a ball mill. Critical Speed Calculation Of Ball Mill Raymond Grinding Mill. CEMENT MILL FORMULAS MILL CRITICAL VELOCITY = 76 / (D)^1/2 MILL Ball Mill 1. n = C (A-B
#### critical speed of ball mill pdf Pochiraju Industries Ltd
Mill Work Index which expresses the resistance of a material to ball milling. . from 40 ball mills show that the average critical speed is 0.73 (Morrell, 1996). Get Price 5-Ball Mill-8 Imimg
#### formulae for cement mill capacity
Cement Mill Capacity Calculation Formula. Cement mill capacity calculation formula cement mill critical mill speed formula kimsburgersNl jan raw mills usually operate at critical speed and cement mills at calculation of the critical mill speed g weight of a grinding ball in kg w angular velocity of the mill tube in radialsecond w n di inside mill diameter in meter effective mill diameter n
#### critical speed of ball mills MC World
Critical speed ball mill calculation.The critical speed of the mill c is defined as the speed at which a single ball will just rod and ball mills in mular al and bhappu r b editors mineral processing plant design raw mills usually operate at 7274 critical speed and cement mills at 7476 32 calculation of the critical mill speed g weight of a.
#### prove of critical speed equtaion in ball mill
Equation Formula Critical Speed Of The Ball Mill 2cmuc. A ball mill critical speed actually ball rod ag or sag is the speed at which the centrifugal forces equal gravitational forces at the mill shells inside surface and no balls will fall from its position onto the shell the imagery below helps explain what goes on inside a mill as speed varies use our online formula the mill speed is
#### Why Critical Speed Of Ball Mill Is Less Than One
why operating speed of ball mill is less than its critical mill speed critical speed paul o. abbe . mill speed critical speed. mill speed. no matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar
#### Cement Mill Critical Speeed And Effect- EXODUS Mining machine
Critical Speed Of The Ball Mill Vdchari. Aug 23 2018 critical speed of the ball mill let a grinding ball of mass m is in motion in a mill of diameter d meters it is at a position making an angle a at the center called the angle of repose the force acting on the ball are | 2,798 | 12,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.888554 |
https://brainly.ph/question/322919 | 1,487,795,798,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171043.28/warc/CC-MAIN-20170219104611-00507-ip-10-171-10-108.ec2.internal.warc.gz | 686,912,614 | 9,868 | # The next term in the sequence A3, B9, C15, D21, E27 is
2
by tremaex
2016-06-21T21:42:30+08:00
The next term in the sequence A3,B9,D21,E27 is F33
2016-06-21T21:43:38+08:00
The next term in the sequence A3, B9, C15, D21, E27 is F33.
3 + 6 = 9
9 + 6 = 15
15 + 6 = 21
21 + 6 = 27
27 + 6 = 33 | 146 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-09 | latest | en | 0.82277 |
https://everythingwhat.com/can-ac-be-transformed | 1,723,167,980,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00678.warc.gz | 195,891,208 | 9,639 | science physics
# Can AC be transformed?
43
AC can be converted to and fromhighvoltages easily using transformers. AC is also capableofpowering electric motors. Motors and generators are the exactsamedevice, but motors convert electrical energy into mechanicalenergy(if the shaft on a motor is spun, a voltage is generated attheterminals!).
In this regard, how is AC current generated?
An alternating current is produced by anelectricgenerator. As the wire rotates in the magnetic field, thechangingstrength of the magnetic field through the wire produces aforcewhich drives the electric charges around the wire. Theforceinitially generates an electric current inonedirection along the wire.
Likewise, can a transistor converts DC to AC? A single transistor cannot convert DCtoAC. But if you use a combination of minimum 2transistorwith 2 capacitors you can do it. Thecircuit is known ashalf bridge inverter circuit. The antiparalleldiodes across thetransistor only necessary for inductiveload only to pumpthe current back to the source.
Thereof, why can transformers only use AC?
Why Transformers Only Works WithAlternatingCurrent The main coil is linked to an ACsupply.The altering current generates a changing magnetic field.Thismakes an alternating voltage in the minor coil. This makesanAC in the circuit associated to thesecondarycoil.
Why does AC current travel longer?
AC is (otherwise) preferred because it isfareasier to convert to a different voltage using a transformer.Thereason why AC is primarily used for longdistancetransmission is due to the fact that it is very easy toincreasethe voltage of AC with a transformer.
Professional
## What is AC signal?
An electrical signal is a voltage or currentwhichconveys information, usually it means a voltage. The term canbeused for any voltage or current in a circuit.AlternatingCurrent (AC) Alternating Current(AC)flows one way, then the other way, continuallyreversingdirection.
Professional
## Why AC current is used in homes?
Since high voltages are more efficient forsendingelectricity great distances, AC electricity has anadvantageover DC. This is because the high voltages from thepowerstation can be easily reduced to a safer voltage foruse inthe house. Changing voltages is done by the useof atransformer.
Professional
## How is current measured?
The SI unit of electric current is theampere,which is the flow of electric charge across a surface at therateof one coulomb per second. The ampere (symbol: A) is an SIbaseunit Electric current is measured using adevicecalled an ammeter.
Explainer
## How is AC converted to DC?
A rectifier is an electrical device thatconvertsalternating current (AC), which periodicallyreversesdirection, to direct current (DC), which flows inonly onedirection. The process is known as rectification, sinceit"straightens" the direction of current.
Explainer
## Is power dissipated negative?
You cannot have negative power in aresistor,unless the resistor is generating power likedissimilarjunctions in metal. But the positive power isalways greaterthan the negative power, due to losses. Butthe overall timeaveraged power will be positive, ie,supplied from thesource and dissipated in thecircuit.
Explainer
## Which is better AC or DC?
Why is DC better than AC? UnlikeAC,a direct current undergoes no switching. There are noperiods andthe current flows in a single direction with a steadyvoltage. Asalready mentioned, DC is prone to lose power asheat –a characteristic that Edison exploited to light thefirstbulb.
Pundit
## What is the current?
Current is a flow of electrical chargecarriers,usually electrons or electron-deficient atoms. The commonsymbolfor current is the uppercase letter I. In analternatingcurrent (AC), the flow of charge carriersreverses directionperiodically.
Pundit
## Is a sine wave AC or DC?
Pulsed DC is commonly produced fromAC(alternating current) by a half-waverectifier or afull-wave rectifier. Full waverectified ac ismore commonly known as Rectified AC.PDC has somecharacteristics of both alternating current(AC) anddirect current (DC) waveforms.
Pundit
## Why are transformers rated in KVA?
The Copper loss(I2R) occurs due to the flowofthe current in the transformer winding and the Iron orcoreloss occurs due to the voltage. These losses do not depend onthepower factor so that is why the transformer rating inKVA notKW. These are the Three Reasons Why Transformeris Ratedin KVA.
Pundit
## How do AC transformers work?
A transformer is an electrical apparatusdesignedto convert alternating current from one voltage to another.It canbe designed to "step up" or "step down" voltages and works onthemagnetic induction principle. A voltage is then induced intheother coil, called the secondary or output coil.
Pundit
## Why can't transformers use DC?
Transformers are used to transfer electricpowerfrom one end to another through the medium of Magnetic Field.Dueto constant flux there will be no emf induced in primaryorsecondary and hence transformer will be of nouse.So, you can safely say that Current Transformerscannot workwith DC Supply.
Teacher
## What is the difference between AC motor and AC generator?
Difference between the AC Motor and theDCMotor
In the AC motor, the source of power isACmains supply whereas in DC motor power is obtainedfrombatteries. In AC motors the armature is stationary andthemagnetic field rotates whereas in DC motors it isviceverse.
Teacher
## Why do we use transformers?
A transformer is a device that is usedtoeither raise or lower voltages and currents in anelectricalcircuit. In modern electrical distributionsystems,transformers are used to boost voltage levelsso asto decrease line losses during transmission.
Teacher
## What is rms value of current?
The RMS value of a set of values (oracontinuous-time waveform) is the square root of the arithmeticmeanof the squares of the values, or the square of thefunctionthat defines the continuous waveform. In physics, theRMScurrent is the "value of the direct currentthatdissipates power in a resistor."
Teacher
## What is a step up transformer?
A transformer that increases the voltagefromprimary to secondary (more secondary winding turns thanprimarywinding turns) is called a step-uptransformer.Conversely, a transformer designed to dojust the oppositeis called a step-downtransformer.
Reviewer
## What happens when DC is applied to transformer?
When a dc supply is provided tothetransformer primary no self induced emf will begenerated(no back emf). Therefore heavy current will flow inthetransformer primary winding which may result in burningdownthe transformer primary winding.
Reviewer
## Can DC current produce electromagnetic waves?
In a DC circuit operating with aconstantcurrent, the electrical charges, usually electrons,onlyexperience very tiny accelerations and very little energyisradiated as electromagnetic waves. Thecurrentcarrying charged particles in AC circuits arecontinuouslyaccelerating and always emitelectromagneticwaves.
Reviewer
## Can we change DC to AC?
Inverters can also be used with transformerstochange a certain DC input voltage into acompletelydifferent AC output voltage (either higher orlower) butthe output power must always be less thanthe inputpower: it follows from the conservation ofenergy that aninverter and transformer can't give outmorepower
Reviewer
## Which is dangerous AC or DC?
DC is more dangerous in one respect inthatit causes more electrolysis in the tissues thanAC.Realistically (statistically) speaking, AC ismoredangerous that DC. This comes from the factthat120/240v AC is the voltage that we are most likelytoencounter - which can kill us.
Co-Authored By:
10
24th April, 2020
67 | 1,673 | 7,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.841329 |
http://stsdas.stsci.edu/stsci_python_epydoc/nictools/nictools.SP_FirstDerivatives.DerivVar-class.html | 1,386,465,514,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163057146/warc/CC-MAIN-20131204131737-00085-ip-10-33-133-15.ec2.internal.warc.gz | 169,893,874 | 3,360 | [frames] | no frames]
# Class DerivVar
source code
Numerical variable with automatic derivatives of first order
Instance Methods
__init__(self, value, index=0, order=1) source code
`list`
__getitem__(self, order) Returns: a list of all derivatives of the given order source code
__repr__(self) source code
__str__(self) source code
__coerce__(self, other) source code
__cmp__(self, other) source code
__neg__(self) source code
__pos__(self) source code
__abs__(self) source code
__nonzero__(self) source code
__sub__(self, other) source code
__rsub__(self, other) source code
__mul__(self, other) source code
__rmul__(self, other) source code
__truediv__(self, other) source code
__div__(self, other) source code
__floordiv__(self, other) source code
__rdiv__(self, other) source code
__pow__(self, other, z=None) source code
__rpow__(self, other) source code
exp(self) source code
log(self) source code
log10(self) source code
sqrt(self) source code
sign(self) source code
sin(self) source code
cos(self) source code
tan(self) source code
sinh(self) source code
cosh(self) source code
tanh(self) source code
arcsin(self) source code
arccos(self) source code
arctan(self) source code
arctan2(self, other) source code
gamma(self) source code
Method Details
### __init__(self, value, index=0, order=1)(Constructor)
source code
Parameters:
• `value` (number) - the numerical value of the variable
• `index` (`int`) - the variable index, which serves to distinguish between variables and as an index for the derivative lists. Each explicitly created instance of DerivVar must have a unique index.
• `order` (`int`) - the derivative order, must be zero or one
Raises:
• `ValueError` - if order < 0 or order > 1
### __getitem__(self, order)(Indexing operator)
source code
Parameters:
• `order` (`int`) - derivative order
Returns: `list`
a list of all derivatives of the given order
Raises:
• `ValueError` - if order < 0 or order > 1
Generated by Epydoc 3.0.1 on Thu Jan 13 11:20:15 2011 http://epydoc.sourceforge.net | 616 | 2,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-48 | latest | en | 0.339405 |
https://number.academy/5008352 | 1,716,160,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00673.warc.gz | 388,860,072 | 11,224 | # Number 5008352 facts
The even number 5,008,352 is spelled 🔊, and written in words: five million, eight thousand, three hundred and fifty-two, approximately 5.0 million. The ordinal number 5008352nd is said 🔊 and written as: five million, eight thousand, three hundred and fifty-second. The meaning of the number 5008352 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 5008352. What is 5008352 in computer science, numerology, codes and images, writing and naming in other languages
## What is 5,008,352 in other units
The decimal (Arabic) number 5008352 converted to a Roman number is (M)(M)(M)(M)(M)(V)MMMCCCLII. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
5008352 seconds equals to 2 months, 1 day, 23 hours, 12 minutes, 32 seconds
5008352 minutes equals to 1 decade, 4 months, 6 days, 32 minutes
### Codes and images of the number 5008352
Number 5008352 morse code: ..... ----- ----- ---.. ...-- ..... ..---
Sign language for number 5008352:
Number 5008352 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 5008352 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 5008352 are 2 * 2 * 2 * 2 * 2 * 156511
The factors of 5008352 are 1, 2, 4, 8, 16, 32, 156511, 313022, 626044, 1252088, 2504176, 5008352.
Total factors 12.
Sum of factors 9860256 (4851904).
#### Powers
The second power of 50083522 is 25.083.589.755.904.
The third power of 50083523 is 125.627.446.921.161.310.208.
#### Roots
The square root √5008352 is 2237,934762.
The cube root of 35008352 is 171,092753.
#### Logarithms
The natural logarithm of No. ln 5008352 = loge 5008352 = 15,426617.
The logarithm to base 10 of No. log10 5008352 = 6,699695.
The Napierian logarithm of No. log1/e 5008352 = -15,426617.
### Trigonometric functions
The cosine of 5008352 is 0,990063.
The sine of 5008352 is -0,140626.
The tangent of 5008352 is -0,142038.
## Number 5008352 in Computer Science
Code typeCode value
5008352 Number of bytes4.8MB
Unix timeUnix time 5008352 is equal to Friday Feb. 27, 1970, 11:12:32 p.m. GMT
IPv4, IPv6Number 5008352 internet address in dotted format v4 0.76.107.224, v6 ::4c:6be0
5008352 Decimal = 10011000110101111100000 Binary
5008352 Decimal = 100102110011112 Ternary
5008352 Decimal = 23065740 Octal
5008352 Decimal = 4C6BE0 Hexadecimal (0x4c6be0 hex)
5008352 BASE64NTAwODM1Mg==
5008352 MD55132a6d7a6d1d1a265969788a93bbd3f
5008352 SHA15734ee092f344b08ed90ae39e827f552438570fc
5008352 SHA2241e8614c1e165e642ed111321f60e48d9e9566e2fbdfcc6a40bde3414
5008352 SHA256d7c609fae25212679c09eaac5ff1c65bdce8150c8fd9f0b3414f34ee20fafc44
5008352 SHA38487fc73c6dd88e507249cd7dc65722c25ce313b4c3ac928af68a9075ed6b07f600a5891eeafed1c5a595ebdccb324856c
More SHA codes related to the number 5008352 ...
If you know something interesting about the 5008352 number that you did not find on this page, do not hesitate to write us here.
## Numerology 5008352
### Character frequency in the number 5008352
Character (importance) frequency for numerology.
Character: Frequency: 5 2 0 2 8 1 3 1 2 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 5008352, the numbers 5+0+0+8+3+5+2 = 2+3 = 5 are added and the meaning of the number 5 is sought.
## № 5,008,352 in other languages
How to say or write the number five million, eight thousand, three hundred and fifty-two in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 5.008.352) cinco millones ocho mil trescientos cincuenta y dos German: 🔊 (Nummer 5.008.352) fünf Millionen achttausenddreihundertzweiundfünfzig French: 🔊 (nombre 5 008 352) cinq millions huit mille trois cent cinquante-deux Portuguese: 🔊 (número 5 008 352) cinco milhões e oito mil, trezentos e cinquenta e dois Hindi: 🔊 (संख्या 5 008 352) पचास लाख, आठ हज़ार, तीन सौ, बावन Chinese: 🔊 (数 5 008 352) 五百万八千三百五十二 Arabian: 🔊 (عدد 5,008,352) خمسة ملايين و ثمانية آلاف و ثلاثمائة و اثنان و خمسون Czech: 🔊 (číslo 5 008 352) pět milionů osm tisíc třista padesát dva Korean: 🔊 (번호 5,008,352) 오백만 팔천삼백오십이 Danish: 🔊 (nummer 5 008 352) fem millioner ottetusinde og trehundrede og tooghalvtreds Hebrew: (מספר 5,008,352) חמישה מיליון ושמונת אלפים שלוש מאות חמישים ושתיים Dutch: 🔊 (nummer 5 008 352) vijf miljoen achtduizenddriehonderdtweeënvijftig Japanese: 🔊 (数 5,008,352) 五百万八千三百五十二 Indonesian: 🔊 (jumlah 5.008.352) lima juta delapan ribu tiga ratus lima puluh dua Italian: 🔊 (numero 5 008 352) cinque milioni e ottomilatrecentocinquantadue Norwegian: 🔊 (nummer 5 008 352) fem million åtte tusen tre hundre og femtito Polish: 🔊 (liczba 5 008 352) pięć milionów osiem tysięcy trzysta pięćdziesiąt dwa Russian: 🔊 (номер 5 008 352) пять миллионов восемь тысяч триста пятьдесят два Turkish: 🔊 (numara 5,008,352) beşmilyonsekizbinüçyüzelliiki Thai: 🔊 (จำนวน 5 008 352) ห้าล้านแปดพันสามร้อยห้าสิบสอง Ukrainian: 🔊 (номер 5 008 352) п'ять мільйонів вісім тисяч триста п'ятдесят два Vietnamese: 🔊 (con số 5.008.352) năm triệu tám nghìn ba trăm năm mươi hai Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,054 | 5,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-22 | latest | en | 0.779588 |
https://marxistphilosophyofscience.com/tag/matriz/ | 1,679,926,251,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00407.warc.gz | 432,316,650 | 97,592 | ## Rank and trace are equal for a real symmetric idempotent matrix
Proposition. Let $latex mathbf{X} in mathbb{R}^{n times n}$ be a matrix that is symmetric ($latex mathbf{X}^top = mathbf{X}$) and idempotent ($latex mathbf{X}^2 = mathbf{X}$). Then the rank of $latex mathbf{X}$ is equal to the trace of $latex mathbf{X}$. In fact, they are both equal to the sum of the eigenvalues of $latex mathbf{X}$.
The proof is relatively straightforward. Since $latex mathbf{X}$ is real and symmetric, it is orthogonally diagonalizable, i.e. there is an orthogonal matrix $latex mathbf{U}$ ($latex mathbf{U}^top mathbf{U} = mathbf{I}$) and a diagonal matrix $latex mathbf{D}$ such that $latex mathbf{D} = mathbf{UXU}^top$ (see here for proof).
Since $latex mathbf{X}$ is idempotent,
latex begin{aligned} mathbf{X}^2 &= mathbf{X}, mathbf{U}^top mathbf{D}^2 mathbf{U} &= mathbf{U}^T mathbf{DU}, mathbf{D}^2 &= mathbf{D}. end{aligned}
Since $latex mathbf{D}$ is a diagonal matrix, it implies that the entries on the diagonal must be zeros or ones. Thus, the number of ones on the diagonal (which is $latex text{rank}(mathbf{D})… View original post 16 more words Advertisement ## Asymptotic distribution of the Pearson chi-square statistic Imagen tomada de ResearchGate. I recently learned of a fairly succinct proof for the asymptotic distribution of the Pearson chi-square statistic (from Chapter 9 of Reference 1), which I share below. First, the set-up: Assume that we have$latex n$independent trials, and each trial ends in one of$latex J$possible outcomes, which we label (without loss of generality) as$latex 1, 2, dots, J$. Assume that for each trial, the probability of the outcome being$latex j$is$latex p_j > 0$. Let$latex n_j$denote that number of trials that result in outcome$latex j$, so that$latex sum_{j=1}^J n_j = n$. Pearson’s$latex chi^2$-statistic is defined as$latex begin{aligned} chi^2 = sum_{text{cells}} dfrac{(text{obs} – text{exp})^2}{text{exp}} = sum_{j=1}^J dfrac{(n_j – np_j)^2}{np_j}. end{aligned}$Theorem. As$latex n rightarrow infty$,$latex chi^2 stackrel{d}{rightarrow} chi_{J-1}^2$, where$latex stackrel{d}{rightarrow}$denotes convergence in distribution. Before proving the theorem, we prove a lemma that we will… View original post 614 more words ## General chi-square tests Imagen tomada de Lifeder. In this previous post, I wrote about the asymptotic distribution of the Pearson$latex chi^2$statistic. Did you know that the Pearson$latex chi^2$statistic (and the related hypothesis test) is actually a special case of a general class of$latex chi^2$tests? In this post we describe the general$latex chi^2$test. The presentation follows that in Chapters 23 and 24 of Ferguson (1996) (Reference 1). I’m leaving out the proofs, which can be found in the reference. (Warning: This post is going to be pretty abstract! Nevertheless, I think it’s worth a post since I don’t think the idea is well-known.) Let’s define some quantities. Let$latex Z_1, Z_2, dots in mathbb{R}^d$be a sequence of random vectors whose distribution depends on a$latex k$-dimensional parameter$latex theta$which lies in a parameter space$latex Theta$.$latex Theta$is assumed to be a non-empty open subset… View original post 696 more words ## SOBRE LOS TENSORES: SU INTERPRETACIÓN CONCEPTUAL Como señala (Kaplan, 1985, pág. 297), cuando se introducen coordenadas curvilíneas los métodos matriciales ya no resultan adecuados para el análisis de las operaciones vectoriales fundamentales. El análisis deseado se puede llevar a cabo con la ayuda de las estructuras matemáticas conocidas como tensores. Los tensores son el resultado de un producto tensorial denotado como A⨂B. Un producto tensorial generaliza la noción de producto cartesiano o producto directo A × B y de suma directa A⨁B para espacios de coordenadas curvilíneas conocidos como variedades (como por ejemplo, las variedades pseudo-riemannianas bajo la cual está modelada la Teoría General de la Relatividad); lo anterior se afirma porque si se verifican las propiedades de un tensor u operador tensorial se podrá verificar que se comporta como una suma, pero su resultado (el espacio o conjunto generado) se comporta como una multiplicación. Esto está relacionado con poder generalizar nociones geométricas (que a nivel de matrices de datos tiene implicaciones en poder medir las longitudes entre los datos –y todo lo que eso implica, ni más ni menos que la base de las mediciones de todo tipo-), como por ejemplo la ortogonalidad entre vectores para una gama más general de superficies entre muchísimas otras cuestiones; de hecho, una variedad generaliza el concepto de superficie. En el lenguaje de programación R, un array multidimensional es un tensor, es decir, el resultado de un producto tensorial entre vectores, mientras que una matriz es resultado de un producto cartesiano entre vectores y es por ello que los primeros se pueden concebir geométricamente como un cubo n-dimensional o una estructura cúbica de medición con n-coordenadas, que además pueden ser curvilíneas. Una matriz es un tensor de dos dimensiones o coordenadas lineales. Un vector es una flecha que representa una cantidad con magnitud y dirección, en donde la longitud de la flecha es proporcional a la magnitud del vector y la orientación de la flecha revela la dirección del vector. También se puede representar con vectores otras cosas, como áreas y volúmenes. Para hacer esto, se debe hacer a la longitud del vector una magnitud proporcional a la magnitud del área a calcular y la dirección del vector debe ser ortogonal a la superficie o región de la cual se desea estimar el área o volumen. Los vectores base o vectores unitarios (cuando la base del espacio lineal es canónica, es decir, que cada vector que conforma la base está compuesto en su pertinente coordenada por la unidad y en el resto por ceros) tienen longitud 1. Estos vectores son los vectores directores del sistema de coordenadas (porque le dan dirección a cada uno de los ejes del plano, puesto que precisamente cada uno representa un eje). Para encontrar los componentes de un vector (en el caso de un sistema de tres coordenadas, el componente x, el componente y, el componente z) se proyecta el vector sobre el eje que corresponde al componente a encontrar, por ejemplo, si se desea encontrar el componente x del vector, la proyección se realiza sobre X. Entre mayor sea el ángulo entre un vector y un eje de referencia (X,Y,Z), menor será la magnitud del componente correspondiente a dicho eje (este componente, en este ejemplo, puede ser x, o z); el inverso también es cierto. La magnitud de cualquier vector dentro del plano real o complejo puede determinarse como combinación lineal de los vectores base con el campo de los reales o los complejos, respectivamente. Esto implica que la magnitud de un vector (y por consiguiente de los componentes dentro del mismo, al ser una estructura lineal) puede expresarse como determinada cantidad de vectores unitarios (de longitud 1) de los diferentes ejes de coordenadas, en donde cada componente del vector se expresará unívocamente en una cantidad determinada de vectores unitarios del eje correspondiente a dicho componente. Para generalizar los resultados anteriores a un vector de vectores A (que entre otras cosas permite agruparlos en una misma estructura matemática -por ello a nivel del programa R los arrays tienen contenido del mismo tipo y relacionado entre sí[1]-), se establece que dicho vector A tendrá los componentes A_X, A_Y, A_Z, que representan a los componentes X, Y y Z, respectivamente. Se requiere establecer un índice para cada vector (el índice es en este caso el subíndice) porque sólo existe un indicador direccional (es decir, un vector base) por componente (porque cada componente se corresponde con su respectivo eje). Esto es lo que hace a los vectores ser tensores de rango 1, que tienen un índice o un vector base por componente. Bajo la misma lógica, los escalares pueden ser considerados tensores de rango cero, porque los escalares no tienen ningún indicador direccional (son una cantidad con magnitud, pero sin sentido) y, por consiguiente, no necesitan índice. Los tensores son combinaciones entre componentes auxiliares de naturaleza diversa (parámetros, coeficientes, pendientes, que son en última instancia algún elemento de algún campo escalar o anillo) y componentes centrales (los miembros de la base del espacio vectorial o módulo, que expresan las variables fundamentales del sistema que se describe), que sirven para estimar de forma más robusta (en términos de precisión cuantitativa y especificidad cualitativa) las coordenadas de un sistema de referencia. El número de índices de cada tensor será igual al número de vectores base por componente (en el caso de los tensores, los componentes y los vectores base no tienen necesariamente una relación uno-a-uno, por lo que a un componente le puede corresponder más de un vector base o vector director del sistema de coordenadas). Considérense, por ejemplo, las fuerzas que actúan al interior de un objeto sólido cualquiera en un espacio de tres dimensiones. Este interior está segmentado en términos de superficies (que son regiones de dicho espacio a manera de planos) por los vectores base de tipo área X, Y, Z. Asúmase además que cada una de las fuerzas actúa en cada una de las regiones del espacio (esto no necesariamente es así, sólo se usa un ejemplo así para que sea más fácilmente capturable a la intuición; aunque lógico-formalmente sí es así, filosóficamente y en términos de las ciencias aplicadas no necesariamente). Lo anterior significa que, debido a la diferente dirección de los vectores base, la acción de dichas fuerzas tiene orientaciones diferentes según la región del espacio de la que se trate. Esto es así porque cada vector base tiene una dirección diferente (al menos si su dirección se estudia cuando está anclado al origen) y cada vector base determina la dirección de la acción de cada fuerza en la región del espacio que a dicho vector base le corresponde (una región -lo que de forma más general puede concebirse como una caracterización dentro de un sistema referencial- estudiada puede estar compuesta por subregiones bajo el efecto de fuerzas diferentes). Así, para poder caracterizar completamente las fuerzas que actúan dentro del objeto sólido (lo que equivale precisamente a caracterizar completamente al objeto sólido mismo -bajo las limitaciones que la teoría tiene frente a la práctica-), es necesario que cada fuerza pueda ser expresada en términos de todas las regiones del espacio en las que actúa (cada región se corresponde con un vector director o vector base), por lo que cada fuerza se debe vincular a la correspondiente cantidad regiones del sólido en las que actúa (se debe vincular a la correspondiente cantidad de vectores base a los que está asociada). Así, los tensores permiten caracterizar completamente todas las fuerzas posibles y todas las regiones posibles sobre las que actúan tales fuerzas. Los tensores permiten que todos los observadores en todos los sistemas de coordenadas de referencia (marco referencial, de ahora en adelante) puedan estar de acuerdo sobre las coordenadas establecidas. El acuerdo no consiste en un acuerdo sobre los vectores base (que pueden variar de un espacio a otro), tampoco en los componentes (que pueden variar según el campo escalar), sino en las combinaciones entre vectores base y componentes. La razón de lo anterior radica en que al aplicar una transformación sobre los vectores base (para pasar de un sistema referencial a otro de alguna forma equivalente), en el contexto de los tensores, la estructura algebraica resultante tendrá invariablemente una única dirección sin importar el marco referencial; por su parte, al transformar un componente se logran mantener las combinaciones entre componentes y vectores base para todos los observadores (i.e., para todos los marcos referenciales -cada observador está en un marco referencial-). Por tanto, los tensores expresan matemáticamente (i.e., lógico-formalmente) la unidad a nivel del fenómeno (social o natural) de las fuerzas contrarias entre sí que lo componen, así como también la tensión que implica la lucha de tales fuerzas por imponerse la una a la otra durante el proceso evolutivo del fenómeno estudiado. Como se señala en (Universidad de Granada, 2022), en el contexto de la estadística aplicada, un array es un tipo de dato estructurado que permite almacenar un conjunto de datos homogéneo, es decir, todos ellos del mismo tipo y relacionados. Cada uno de los elementos que componen un vector pueden ser de tipo simple como caracteres, entero o real, o de tipo compuesto o estructurado como son vectores, estructuras, listas. A los datos almacenados en un array se les denomina elementos; al número de elementos de un array se les denomina tamaño o rango del vector; este rango puede determinarse de forma equivalente, en el caso de arrays multidimensionales (tensores), a través del número de ejes. Para acceder a los elementos individuales de un array se emplea un índice que será un número entero no negativo que indicará la posición del elemento dentro del array. Para referirse a una posición particular o elemento dentro del array, se especifica el nombre del array y el número de posición del elemento particular dentro del mismo, el índice. Los arrays en gran parte se definen como las variables ordinarias, excepto en que cada array debe acompañarse de una especificación de tamaño (número de elementos). Para un array unidimensional, el tamaño se especifica con una expresión entera positiva encerrada entre paréntesis cuadrados. La expresión es normalmente una constante entera positiva. En suma, cada dimensión de un tensor/array multidimensional (que, al ser en sí mismo una estructura de datos con las propiedades usuales de los números, es también un espacio vectorial, específicamente un espacio euclidiano) está compuesta por un número de filas y columnas especificado. En la mayoría de los casos, los tensores se pueden considerar como matrices anidadas de valores que pueden tener cualquier número de dimensiones. Un tensor con una dimensión se puede considerar como un vector, un tensor con dos dimensiones como una matriz y un tensor con tres dimensiones se puede considerar como un paralelepípedo. El número de dimensiones que tiene un tensor se llama su rango y la longitud en cada dimensión describe su forma. El rango de un tensor es el número de índices necesarios para seleccionar de forma única cada elemento del tensor (TensorFlow, 2022). El rango también se conoce como “orden” o “grado”; como se señaló antes, otra forma de ver los tensores es como arrays multidimensionales (RStudio, 2022). Como señala (Weisstein, 2022), formalmente hablando el rango de un tensor es el número total de índices contravariantes y covariantes de un tensor, relativos a los vectores contravariantes y covariantes, respectivamente. El rango R de un tensor es independiente del número de dimensiones N del espacio subyacente en el que el tensor se localice. Adicionalmente, se señala en la documentación R sobre el paquete ‘tensor’, que el producto tensorial de dos arrays es teóricamente un producto exterior de tales arrays colapsados en extensiones específicas al sumar a lo largo de las diagonales apropiadas. Por ejemplo, un producto matricial es el producto tensorial a lo largo de la segunda extensión de la primera matriz y la primera extensión de la segunda matriz. En el modelo de datos multidimensional, los datos se organizan en una jerarquía que representa diferentes niveles de detalles. Un modelo multidimensional visualiza los datos en forma de cubo de datos. Un cubo de datos permite modelar y visualizar datos en múltiples dimensiones. Se define por dimensiones y hechos. Las dimensiones son las perspectivas o entidades sobre las cuales una organización mantiene registros. Por ejemplo, una tienda puede crear un almacén de datos de ventas para mantener registros de las ventas de la tienda para la dimensión de tiempo, artículo y ubicación. Estas dimensiones permiten registro para realizar un seguimiento de las cosas, por ejemplo, las ventas mensuales de artículos y las ubicaciones en las que se vendieron los artículos. Cada dimensión tiene una tabla relacionada con ella, llamada tabla dimensional, que describe la dimensión con más detalle. ### Referencias Fleisch, D. A. (2012). What’s a tensor? Recuperado el 26 de Marzo de 2022, de Dan Fleisch: https://www.youtube.com/watch?v=f5liqUk0ZTw geeksforgeeks. (26 de Marzo de 2022). Multidimensional Arrays in C / C++. Obtenido de geeksforgeeks.org: https://www.geeksforgeeks.org/multidimensional-arrays-c-cpp/ java T point. (Marzo de 25 de 2022). What is Multi-Dimensional Data Model? Obtenido de Data Warehouse: https://www.javatpoint.com/data-warehouse-what-is-multi-dimensional-data-model Kaplan, W. (1985). CÁLCULO AVANZADO. MÉXICO, D.F.: COMPAÑÍA EDITORIAL CONTINENTAL, S.A. DE C.V., MÉXICO. Patidar, P. (14 de Diciembre de 2019). Tensors — Representation of Data In Neural Networks. Obtenido de Medium: https://medium.com/mlait/tensors-representation-of-data-in-neural-networks-bbe8a711b93b Paul, S. (12 de Septiembre de 2018). Investigating Tensors with PyTorch. Obtenido de DataCamp: https://www.datacamp.com/community/tutorials/investigating-tensors-pytorch RStudio. (25 de Marzo de 2022). Tensors and operations. Obtenido de TensorFlow for R: https://tensorflow.rstudio.com/tutorials/advanced/customization/tensors-operations/ TensorFlow. (25 de Marzo de 2022). tf.rank. Obtenido de TensorFlow Core v2.8.0 : https://www.tensorflow.org/api_docs/python/tf/rank Universidad de Granada. (25 de Marzo de 2022). Arrays y cadenas. Obtenido de Departamento de Ciencias de la Computación e Inteligencia Artificial de la Universidad de Granada: https://ccia.ugr.es/~jfv/ed1/c/cdrom/cap5/f_cap52.htm. Weisstein, E. W. (25 de Marzo de 2022). Tensor Rank. Obtenido de MathWorld – A Wolfram Web Resource: https://mathworld.wolfram.com/TensorRank.html [1] Véase (Universidad de Granada, 2022). ## PROCESO DE SELECCIÓN DE VARIABLES EXPLICATIVAS EN MODELOS ESTADÍSTICOS ## ISADORE NABI # PROCESO DE SELECCIÓN DE VARIABLES EXPLICATIVAS ## Introducción: Sobre la necesidad de un proceso de selección de predictores Usualmente se tiene interés en explicar los datos de la forma más simple, lo cual en el contexto de la teoría de las probabilidades (especialmente en la teoría bayesiana de probabilidades) se conoce como el *principio de parsimonia*, el cual está inspirado en el principio filosófico conocido como *navaja de Ockham*, la cual establece que en igualdad de condiciones la explicación más simple suele ser la más probable. El principio de parsimonia adopta diferentes formas según el área de estudio del análisis inferencial en el que se encuentre un investigador. Por ejemplo, una parametrización parsimoniosa es aquella que usa el número óptimo de parámetros para explicar el conjunto de datos de los que se dispone, pero "parsimonia" también puede referirse a modelos de regresión parsimoniosos, es decir, modelos que utilizan como criterio de optimización emplear la mínima cantidad de coeficientes de regresión para explicar una respuesta condicional Y. El principio de parsimonia, los procesos matemáticos de optimización regidos por el criterio de alcanzar un mínimo y la navaja de Ockham son un mismo tipo de lógica aplicado en escalas de la existencia (que podríamos llamar en general "materia", como lo hace Landau en sus curso de física teórica) cualitativamente diferentes. La historia de la Filosofía demuestra que el único sistema que podría ser aplicado así exitosamente es el sistema hegeliano (lo que obedece a que parcialmente sigue la lógica de la existencia misma, como han demostrado Marx, Engels, Lenin, Levins, Lewontin y el mismo Hegel en su extensa obra). ¿Cómo es posible la vinculación en distintas escalas cualitativas de la realidad del principio de la navaja de Ockham? A que todas esas ideas responden a la escuela filosófica de Ockham, que era la escuela nominalista. Retomando lo que señalan (Rosental & Iudin. Diccionario Filosófico, Editorial Tecolut, 1971. p.341; véase https://www.filosofia.org/enc/ros/nom.htm), el nominalismo fue una corriente de la filosofía medieval que consideraba (ya es una escuela extinta) que los conceptos generales tan sólo son nombres de los objetos singulares. Los nominalistas afirmaban que sólo poseen existencia real las cosas en sí, con sus cualidades individuales (es decir, las generalizaciones para ellos no tenían valor gnoseológico en sí mismas sino como recurso gnoseológico). Los nominalistas van más allá, planteando que las generalizaciones no sólo no existen con independencia de los objetos particulares (esta afirmación en correcta, lo que no es correcto es pensar que lo inverso sí es cierto), sino que ni siquiera reflejan las propiedades y cualidades de las cosas. El nominalisto se hallaba indisolublemente vinculado a las tendencias materialistas, ya que reconocía la prioridad de la cosa y el carácter secundario del concepto. Por supuesto, las generalizaciones aunque menos reales que los objetos particulares (y de ahí la sujeción de la teoría a la práctica en un concepto que las une conocido en la teoría marxista como *praxis*) no deja por ello de ser real en cuanto busca ser una representación aproximada (a largo plazo cada vez más aproximada a medida se desarrollan las fuerzas productivas) de la estructura general (interna y externa, métrica y topológica) común que tienen tales fenómenos naturales o sociales. Marx señaló que el nominalismo fue la primera expresión del materialismo de la Edad Media. Con todo, los nominalistas no comprendían que los conceptos generales reflejan cualidades reales de cosas que existen objetivamente y que las cosas singulares no pueden separarse de lo general, pues lo contienen en sí mismas (y esto no tiene un carácter únicamente marxista, sino que incluso el célebre formalista David Hilbert señaló, según la célebre biógrafa de matemáticos Constance Reid que "The art of doing mathematics consists in finding that special case which contains all the germs of generality"). Así, el defecto fundamental de la navaja de Ockham es el no considerar algún conjunto de restricciones que complementen al criterio de selección de la explicación basado en que sea la idea más simple. Como se señala en https://www.wikiwand.com/en/Occam%27s_razor, "En química, la navaja de Occam es a menudo una heurística importante al desarrollar un modelo de mecanismo de reacción (...) Aunque es útil como heurística en el desarrollo de modelos de mecanismos de reacción, se ha demostrado que falla como criterio para seleccionar entre algunos modelos publicados seleccionados (...) En este contexto, el propio Einstein expresó cautela cuando formuló la Restricción de Einstein: "Difícilmente se puede negar que el objetivo supremo de toda teoría es hacer que los elementos básicos irreductibles sean tan simples y tan pocos como sea posible sin tener que renunciar a la representación adecuada de un dato único de experiencia"." La clave en la expresión anterior de Einstein es "sin tener que renunciar a...", lo que se cristaliza nítidamente en una frase que señala la fuente citada es atribuida a Einstein, pero no ha sido posible su verificación: "Todo debe mantenerse lo más simple posible, pero no lo más simple". Como se verifica en https://www.statisticshowto.com/parsimonious-model/, en general, existe un *trade-off* entre la bondad de ajuste de un modelo y la parsimonia: los modelos de baja parsimonia (es decir, modelos con muchos parámetros) tienden a tener un mejor ajuste que los modelos de alta parsimonia, por lo que es necesario buscar un equilibrio. La parsimonia estadística es deseada porque un mínimo de coeficientes de regresión implica un mínimo de variables y un mínimo de estos implica un mínimo de variables explicativas, lo que puede ser útil en casos de que exista colinealidad entre las variables explicativas, así como también permite ahorrar tiempo y dinero en lo relativo a la inversión de recursos destinada al estudio, aunque no necesariamente garantice que en general (considerando el impacto posterior de las decisiones tomadas con base en el estudio y otros factores) se ahorre tiempo y dinero. ## Modelos Jerárquicos Existen diferentes tipos de modelos jerárquicos. Los hay de diferente tipo, algunos más complejos que otros (complejidad a nivel teórico, matemático y computacional); ejemplos de tales modelos son las mixturas de probabilidad y se pueden estudiar en https://marxianstatistics.files.wordpress.com/2020/12/sobre-los-estimadores-de-bayes-el-analisis-de-grupos-y-las-mixturas-gaussianas-isadore-nabi.pdf. Aquí se tratará con modelos jerárquicos más simples, como los abordados en (Kutner, Nachtsheim, Neter & Li. p.294-305). Como señalan los autores referidos en la p.294., los modelos de regresión polinomial tienen dos tipos básicos de usos: 1. Cuando la verdadera función de respuesta curvilínea es de hecho una función polinomial. 2. Cuando la verdadera función de respuesta curvilínea es desconocida (o compleja), pero una función polinomial es una buena aproximación a la función verdadera. El segundo tipo de uso, donde la función polinomial se emplea como una aproximación cuando se desconoce la forma de la verdadera función de respuesta curvilínea, es muy común. Puede verse como un enfoque no paramétrico para obtener información sobre la forma de la función que modela la variable de respuesta. Un peligro principal en el uso de modelos de regresión polinomial es que las extrapolaciones pueden ser peligrosas con estos modelos, especialmente en aquellos con términos de orden superior, es decir, en aquellos cuyas potencias sean iguales o mayores a 2. Los modelos de regresión polinomial pueden proporcionar buenos ajustes para los datos disponibles, pero pueden girar en direcciones inesperadas cuando se extrapolan más allá del rango de los datos. Así, como señalan los autores referidos en la p.305, el uso de modelos polinomiales no está exento de inconvenientes. Estos modelos pueden ser más costosos en grados de libertad que los modelos no-lineales alternativos o los modelos lineales con variables transformadas. Otro inconveniente potencial es que puede existir multicolinealidad grave incluso cuando las variables predictoras están centradas. Una alternativa al uso de variables centradas en la regresión polinomial es usar polinomios ortogonales. Los polinomios ortogonales están no-correlacionados, puesto que la ortogonalidad de sus términos implica independencia lineal entre los mismos. Algunos paquetes de computadora usan polinomios ortogonales en sus rutinas de regresión polinomial y presentan los resultados ajustados finales en términos tanto de los polinomios ortogonales como de los polinomios originales. Los polinomios ortogonales se discuten en textos especializados como (Drapper & Smith, Applied Linear Regression). A veces, se ajusta una función de respuesta cuadrática con el fin de establecer la linealidad de la función de respuesta cuando no se dispone de observaciones repetidas para probar directamente la linealidad de la función de respuesta. ## Caso de Aplicación ### 1. Conversión de Matriz de Datos a Marco de Datos La base ´Vida.Rdata´ contiene datos para los 50 estados de los Estados Unidos. Estos datos son proporcionados por U.S. Bureau of the Census. Se busca establecer las relaciones que existen entre ciertas variables del Estado que se analice y la esperanza de vida. A continuación, se presenta una descripción de las variables que aparecen en la base en el orden en que aparecen: + **esper**: esperanza de vida en años (1969-71). + **pob**: población al 1 de Julio de 1975. + **ingre**: ingreso per capita (1974). + **analf**: porcentaje de la población analfabeta (1970). + **crim**: tasa de criminalidad por 100000 (1976). + **grad**: porcentaje de graduados de secundaria (1970). + **temp**: número promedio de días con temperatura mínima por debajo de los 32 grados (1931-1960) en la capital del estado. + **area**: extensión en millas cuadradas. Debe comenzarse leyendo el archivo de datos pertinente mediante la sintaxis load("Vida.Rdata"). Si se observa la estructura de la base de datos, se verifica que es simplemente una matriz. Por tanto, si se utiliza la sintaxis names(base) no se obtiene información alguna, mientras que si se trata de llamar a alguna de las variables por su nombre, como por ejemplo base$esper, R informa de un error y lo mismo ocurre si se usa attach(base). Esto sucede porque la estructura de datos invocada no está definida como un marco de datos o data.frame. Por ello, debe comenzarse por convertir dicha matriz de datos en un marco de datos o data.framey posteriormente puede verificarse si las sintaxis antes mencionadas son ahora operativas.
“{r}
setwd(“C:/Users/User/Desktop/Carpeta de Estudio/Mis Códigos en R”)
names(base)
base=data.frame(base)
names(base)
“
### 2. Obtención de todos los modelos posibles dado un determinado conjunto de variables dentro del marco de datos
Pueden obtenerse los $R^2$ ajustados de todos los modelos posibles con las 7 variables disponibles. Para hacerlo, puede construirse primero un objeto con todos los predictores y llamarlo **X** para posteriormente construir un objeto llamado **sel** aplicando la función leaps (perteneciente a la librería con el mismo nombre) de la siguiente forma: sel=leaps(x,y, method="adjr2"). Nótese que el objeto construido mediante la sintaxis leaps, es decir, **sel**, es una lista con 4 componentes cuyos nombres pueden obtenerse con la sintaxis names(sel). Así, puede llamarse a cada uno de tales componentes por separado usando el signo $, por ejemplo, sel$which. Antes de proceder a realizar los cálculos definidos antes, se estudiará a nivel general la sintaxis leaps.
La sintaxis leaps usa un algoritmo eficiente (parsimonioso) de ramificación y cota para realizar una búsqueda exhaustiva de los mejores subconjuntos de las variables contenidas en el marco de datos para predecir y realizar análisis de regresión lineal; este tipo de algoritmo, según https://www.wikiwand.com/en/Branch_and_bound, es un paradigma de diseño de algoritmos para problemas de optimización discreta y combinatoria, así como optimización matemática. Un algoritmo de ramificación y acotación consiste en una enumeración sistemática de soluciones candidatas mediante la búsqueda en el espacio de estados: se piensa que el conjunto de soluciones candidatas forma un árbol enraizado con el conjunto completo en la raíz; "si las cosas fuesen tal y como se presentan ante nuestros ojos, la ciencia entera sobraría" dijo Marx alguna vez. El algoritmo explora las ramas del árbol representado por los subconjuntos del conjunto de soluciones posibles al problema de optimización. Antes de enumerar las soluciones candidatas de una rama, el algoritmo sigue el siguiente proceso descarte de ramas: la rama se compara con los límites estimados superior e inferior de la solución óptima y se descarta (la rama en su conjunto) si no ella puede producir una solución mejor que la mejor encontrada hasta ahora por el algoritmo (véase https://cran.r-project.org/web/packages/leaps/leaps.pdf, p.1). Como se señala en la documentación antes citada, dado que el algoritmo devuelve el mejor modelo de cada tamaño (aquí se refiere a los modelos estadísticamente más robustos según un número de variables fijo que se considere) no importa si desea utilizar algún criterio de información (como el AIC, BIC, CIC o DIC). El algoritmo depende de una estimación eficiente de los límites superior e inferior de las regiones/ramas del espacio de búsqueda. Si no hay límites disponibles, el algoritmo degenera en una búsqueda exhaustiva.
A pesar de lo señalado relativo a que la búsqueda realiza por leaps es independiente de cualquier criterio de información utilizado, puede omitirse este hecho con la finalidad de que sea posible incorporar a esta práctica el estudio de los criterios de información. A continuación, se presenta una lista de los mejores modelos siguiendo el criterio de $R^2$ ajustado más alto, lo que se indica al interior de la sintaxis leaps mediante methods="adjr2".
“
{r}
attach(base)
library(leaps)
sel
names(sel)
sel$adjr2 sel$which
sel$label sel$size
“
Adicionalmente, es posible construir una matriz, almacenarla bajo el nombre **mat** con el contenido de las filas sel$which y sel$adjr2, agregando un contador para identificar cada modelo mediante la sintaxis cbind. La estructura de datos **mat** contiene todos los diferentes modelos de regresión lineal (a diferentes tamaños de los mismos) mediante la sintaxis leaps para la base de datos utilizada.
“{r}
k=nrow(sel$which) k mat=data.frame(cbind(n=1:k,sel$which,round(sel$adjr2,2))) mat head(mat[order(-mat$V9),],10)
“
Así, puede construirse un subconjunto de **mat** que contenga sólo los modelos cuyo coeficiente de determinación ajustado sea mayor o igual que 0.68.
“{r}
subcon=subset(mat,sel$adjr2>=0.68) head(subcon[order(-subcon$V9),],10)
“
Nótese que los cuatro modelos con el $R^2$ ajustado más alto son los modelos 28, 38, 39, y 40, cuyo tamaño oscila entre 4 o 5 variables explicativas; si se utiliza la sintaxis print es posible verificar que en las filas está el modelo como tal (si la variable se toma en consideración tiene asignado un "1", mientras que en caso contrario un "0"), mientras que en las columnas se localizan las posibles variables a utilizar.
Como se puede verificar en https://support.minitab.com/es-mx/minitab/18/help-and-how-to/modeling-statistics/regression/supporting-topics/goodness-of-fit-statistics/what-is-mallows-cp/, el Estadístico $C_p$ de Mallows sirve como ayuda para elegir entre múltiple modelos de regresión. Este estadístico ayuda a alcanzar un equilibrio importante con el número de predictores en el modelo. El $C_p$ de Mallows compara la precisión y el sesgo del modelo completo con modelos que incluyen un subconjunto de los predictores. Por lo general, deben buscarse modelos donde el valor del $C_p$ de Mallows sea pequeño y esté cercano al número de predictores del modelo más la constante $p$. Un valor pequeño del $C_p$ de Mallows indica que el modelo es relativamente preciso (tiene una varianza pequeña) para estimar los coeficientes de regresión verdaderos y pronosticar futuras respuestas. Un valor del $C_p$ de Mallows que esté cerca del número de predictores más la constante indica que, relativamente, el modelo no presenta sesgo en la estimación de los verdaderos coeficientes de regresión y el pronóstico de respuestas futuras. Los modelos con falta de ajuste y sesgo tienen valores de $C_p$ de Mallows más grandes que p. A continuación se presenta un ejemplo.
Así, para el ejemplo aquí utilizado (que responde a la base de datos antes especificada) puede obtenerse el estadístico $C_p$ de Mallows para todos los modelos posibles con las 7 variables disponibles. Para ello puede usarse la función leaps; nótese que no es necesario indicarle a R que obtenga el estadístico de Mallows mediante la sintaxis method=Cp puesto que este método es el establecido por defecto en la programación de R, por lo que en el escenario en que no se indique un "method" en específico el programa utilizará por defecto el criterio del estadístico de Mallows.
“{r}
sel = leaps(X,esper)
names(sel)
sel$Cp “ Complementariamente, puede construirse una nueva matriz **mat** que en lugar de los criterios sel$which y sel$adjr2 siga los criterios sel$which, sel$Cp y sel$size, agregando al igual que antes un contador para identificar cada modelo. Esto implicará la sobreeescritura de la matriz **mat**. Pueden seleccionarse con antelación únicamente las filas de **mat** que se corresponden con los modelos seleccionados en el punto anterior y comparar la columna del $C_p$ con la columna $size$ que corresponde al número de coeficientes (p). En cada caso puede determinarse si el modelo es sesgado o no, sin perder de vista que un modelo es sesgado según el estadístico de Mallows si $C_p>p$. De lo anterior se desprende que se está buscando un conjunto de modelos insesgados para los cuales se verifica la condición $C_p<p$ antes mencionada.
“{r}
mat=data.frame(cbind(1:k,sel$which,round(sel$Cp,2),sel$size)) colnames(mat)[9]<-“CP” colnames(mat)[10]<-“p” mat[c(28,38,39,40),] “ Como puede observarse, en todos los modelos arrojados por la sintaxis leaps cumplen con la condición antes especificada, por lo que es posible afirmar que, sobre todo respecto a los modelos 28, 38, 39 y 40, que son buenos candidatos para ser utilizados (los mejores modelos son los mismos cuatro que en el literal anterior). ### 4. Suma de Cuadrados Residuales de Predicción (PRESS) ####4.1. Aproximación Gráfica Como se señala en (https://pj.freefaculty.org/guides/stat/Regression/RegressionDiagnostics/OlsHatMatrix.pdf, p.9), la PRESS no es otra cosa que el error de estimación correspondiente a un valor particular de la variable condicional$Y$; la estimación de PRESS a veces es útil como una medida resumida de la capacidad de un modelo para predecir nuevas observaciones. Las líneas de comando presentadas a continuación expresan la configuración de la función personalizada plot.press, que es una función empírica que se aproxima gráficamente a los PRESS mediante el siguiente procedimiento: a) Crea un modelo solamente con la variable **ingre**. b) Toma el Estado i-ésimo y crea otro modelo basado en los demás Estados (excepto el i-ésimo). c) Grafica las dos líneas de regresión y marca la observación del Estado i-ésimo en rojo para que se observe como se diferencian las dos líneas a la altura del ingreso de ese Estado. d) Estima el promedio de la esperanza de vida para el i-ésimo Estado usando las dos ecuaciones. “{r} plot.press=function(i){ mod =lm(esper~ingre,base) mod1=lm(esper ~ ingre,base[-i,]) plot(base$ingre,base$esper,pch=18,xlab=”ingreso”,ylab=”esperanza”) points(base$ingre[i],base$esper[i],pch=18,col=2) abline(mod) abline(mod1,lty=2,col=2) abline(v=base$ingre[i],col=4,lty=2)
legend(3000,max(esper),c(“completo”,paste(“falta el “,i,sep=””)),col=c(1,2),lty=c(1,2),bty=”n”)
yi=predict(mod,data.frame(ingre=base$ingre[i])) yii=predict(mod1,data.frame(ingre=base$ingre[i]))
res=c(yi,yii)
names(res)=c(“y_i”,”y_i(i)”)
return(round(res,2))
}
“
Así, puede usarse la función plot.press con diferentes estados, por ejemplo, con Alaska (i=2) o algún otro.
“{r}
plot.press(2)
plot.press(15)
plot.press(10)
“
#### 4.2. Aproximación Inferencial vía Residuos Estandarizados
Como señala https://www.statisticshowto.com/what-is-a-standardized-residuals/, los residuos estandarizados permiten normalizar el conjunto de datos de estudio en el contexto del análisis de regresión y de la ejecución de pruebas de hipótesis chi-cuadrado $χ^2$. Un residuo estandarizado es una razón: la diferencia entre el valor observado y el valor esperado (condicional, a posteriori) sobre la desviación estándar del valor esperado en la prueba de chi-cuadrado.
Como se señala en https://online.stat.psu.edu/stat501/lesson/11/11.4, existen varias medidas para identificar valores extremos de X (observaciones de alto $leverage$ o $influencia$) y valores de Y inusuales (valores atípicos). Al intentar identificar valores atípicos, un problema que puede surgir es cuando existe un valor atípico potencial que influye en el modelo de regresión hasta tal punto que la función de regresión estimada se "arrastrada" hacia el valor atípico potencial, de modo que no se marca como un valor atípico utilizando el criterio usual de residuos estandarizados. Para abordar este problema, los residuos eliminados ofrecen un criterio alternativo para identificar valores atípicos. La idea básica de esto es eliminar las observaciones una a la vez, reajustando cada vez el modelo de regresión en las n – 1 observaciones restantes. Luego, se comparan los valores de respuesta observados con sus valores ajustados basados en los modelos con la i-ésima observación eliminada. Esto produce residuos eliminados (no estandarizados). La estandarización de los residuos eliminados produce residuos eliminados studentizados, como se verá teóricamente a continuación.
Formalmente, es un resultado conocido del álgebra lineal que $y=Xβ+ε$, en donde $X_{n×p}$, $\hat{β}=(X'X)^{-1}X-y$ y $\hat{y}=X\hat{β}=X(X'X)^{-1}X'y=Hy$, donde $H=X(X'X)^{-1}X'$ es la matriz conocida como *matriz sombrero*. Los residuos son $e=y-\hat{y}=y-Hy=(I-H)y$. Adicionalmente, se sabe que la varianza poblacional $σ^2$ es desconocida y puede estimarse mediante la suma de cuadrados medios del error $MSE$. Así, los residuos pueden ser expresados mediante la ecuación $e_i^*=\frac{e_i}{\sqrt{MSE}}$ y se conocen como *residuos semistudentizados*. Puesto que la varianza de los residuos depende tanto de $σ^2$ como de $X$, la varianza estimada es $\hat{V}(e_i)=MSE(1-h_{ii})$, donde $h_{ii}$ es el $i$-ésimo elemento de la diagonal principal de la matriz sombrero. Así, los residuos estandarizados, también conocidos como *residuos internamente studentizados*, tienen la forma $r_i=\frac{e_i}{\sqrt{MSE(1-h_{ii})}}$. Sin embargo, se sabe que es imposible que un residuo individual y el MSE (que es la varianza del conjunto de residuos) no estén correlacionados (existe dependencia lineal) y, por consiguiente, es imposible que $r_i$ siga una distribución t de Student. Lo anterior representa un impedimento para realizar pruebas de significancia estadística de los coeficientes de regresión, puesto que la distribución t es junto con la F los dos tipos de distribución más utilizados (y no sólo en el contexto de regresión) para realizar pruebas de hipótesis, dentro de las cuales las pruebas de significancia de coeficientes son un tipo de ellas. La solución a la problemática antes descrita consiste en eliminar la $i$-ésima observación, ajustar la función de regresión a las $n-1$ observaciones restantes y luego obtener nuevas $\hat{y}$'s que pueden ser denotadas como $\hat{y}_{i(i)}$. La diferencia $d_i=y_i-\hat{y}_{i(i)}$ es llamada *residuo eliminado*. Una expresión equivalente que no requiere recomputación es: $d_i=\frac{e_i}{1-h_{ii}}$.
Los residuos eliminados expresados de la forma anterior son la base para encontrar los residuos conocidos como *residuos eliminados studentizados* o *resiudos studentizados externamente*, los cuales adoptan la forma $t_i=\frac{d_i}{\sqrt{{\frac{MSE}{1-h_{ii}}}}}\sim{\sf t_{n-p-1}}$ o $t_i=\frac{e_i}{\sqrt{{{MSE(1-h_{ii})}}}}\sim{\sf t_{n-p-1}}$; véase https://stats.stackexchange.com/questions/99717/whats-the-difference-between-standardization-and-studentization/99723.
En lo que a la estimación de los diferentes tipos de residuos se refiere, debe comenzarse por obtener las **influencias** o **leverage** del modelo usando hatvalues(mod); debe recordarse que las influencias son utilizadas para determinar que tanto impacto tiene una observación sobre los resultados de la regresión. Precisamente el análisis descriptivo anterior, en el que en una de las rectas de regresión (de las dos que aparecen en cada una de las cincuenta gráficas posibles) se omitía un Estado, tenía como finalidad verificar cuánto impactaba su ausencia (la del Estado sustraido) en la estimación realizada sobre la media condicional de $Y$. Al utilizar la sintaxis "mod=lm(esper~ingre,base)" se está planteando un modelo con la totalidad de Estados, del cual se calculan sus valores sombrero mediante la sintaxis h = hatvalues(mod), sus residuos mediante r=mod$res, se estima el residuo de un modelo en el que no se considera el Estado i-ésimo en el análisis (en este caso Alaska) mediante pred.r = r[2]/(1-h[2]) y, finalmente, la validez estadística de la estimación pred.r = r[2]/(1-h[2]) se determina contrastándola con respecto al resultado de restarle a la media estimada$\hat{Y}_2$(porque en este caso para Alaska, que ocupa la fila dos en la base de datos, que es una base de datos de corte transversal) la media estimada$\hat{Y}$del modelo que no considera al i-ésimo Estado (aquí es Alaska). “{r} mod=lm(esper~ingre,base) h = hatvalues(mod) r=mod$res
pred.r = r[2]/(1-h[2])
round(pred.r,2)
esper[2]-73.07
plot.press(2)
“
Finalmente, puede obtenerse la Suma de Cuadrados Residuales de Predicción $PRESS$ utilizando los residuos eliminados globales (no únicamente para el Estado de Alaska) mediante la siguiente ecuación: $$PRESS=\sum{( \frac{r_i}{1-h_i}} )^2$$.
“{r}
d=r/(1-h)
press=t(d)%*%d
round(press,2)
“
### 5. Comparación de Modelos vía $PRESS$
Es posible comparar el modelo que únicamente contempla la variable ingreso **ingre** con el que se obtiene en un modelo que contenga en su lugar la cantidad de población del Estado **pop** y su tasa de criminalidad **crim**. Esto con el fin de verificar cuál de los dos modelos es más sensible a valores extremos de X al realizar estimaciones de la media condicional $\hat{Y}$ de la variable *esperanza de vida*.
“{r}
mod=lm(esper~ingre,base)
r=mod$res h=hatvalues(mod) d=r/(1-h) press=t(d)%*%d round(press,2) “ “{r} mod2= lm(esper~pop+crim,base) r=mod2$res
h=hatvalues(mod2)
d=r/(1-h)
press=t(d)%*%d
round(press,2)
“
Se observa que el modelo mod es más sensible, puesto que su PRESS es más alto (89.32).
Debe decirse que la matriz "d" es conocida también como *matriz de Gramm*, por lo que su determinante es igual al producto de sí y su transpuesta, es decir, t(d)%*%d. Como se verifica en https://www.wikiwand.com/en/Gram_matrix, la matriz de Gramm cuyos elementos pertenecen a los reales tiene la característica de ser simétrica (matriz cuadrada que es igual a su transpuesta); la matriz de Gramm de cualquier base ortonormal (conjunto de vectores linealmente independientes que generan un espacio lineal -conocido como *span lineal*- denso dentro del espacio de referencia) es una matriz identidad.
El modelo anterior puede expandirse en predictores considerando ahora población **pop**, nivel de ingreso **ingre**, porcentaje de población analfebeta **analf** y la extensión en millas cuadradas **area** para explicar la esperanza de vida (medida en años).
“{r}
mod0= lm(esper~pop+ingre+analf+area,base)
r=mod0$res h=hatvalues(mod0) d=r/(1-h) press=t(d)%*%d round(press,2) “ El modelo mod0 es aún más sensible a los datos provistos por el Estado de Alaska que el modelo mod Así como se amplió la cantidad de variables en consideración al pasar del modelo mod al modelo mod0, también podría realizarse el procedimiento anterior para un modelo que considere la totalidad de las variables disponibles. Una forma para evitar escribir todas las variable en es usar un punto después de **~**, además de indicar de cuál base provienen los datos. De esta forma R entiende que debe considerar todas las variables de esa base como predictores, con excepción de la variable que se indica como respuesta. “{r} mod_comp= lm(esper~., base) r=mod_comp$res
h=hatvalues(mod_comp)
d=r/(1-h)
press=t(d)%*%d
round(press,2)
“
Como se verifica de las pruebas antes realizadas, el modelo completo mod_comp tiene una $PRESS$ menor (más bajo) que el modelo que utiliza 4 predictores (*i.e.*, mod0) para explicar la media condicional de la esperanza de vida, lo que indica menor *leverage* en relación al Estado de Alaska.
### 6. Construcción Escalonada de Modelos de Predicción
#### 6.1. Aspectos Teóricos Generales
Como se conoce de los cursos de álgebra lineal, el mecanismo de *eliminación gaussiana* o *reducción de por filas*, es un proceso secuencial de *operaciones elementales entre filas* realizadas sobre la correspondiente matriz de coeficientes con la finalidad de estimar el rango de la matriz, el determinante de una matriz cuadrada y la inversa de una matriz invertible, en cuanto este mecanismo prepara las condiciones para resolver el sistema de ecuaciones; sobre los orígenes históricos de este mecanismo debe decirse que, como se señala en https://en.wikipedia.org/wiki/Gaussian_elimination, casos particulares de este método se conocían descubiertos por matemáticos chinos (sin prueba formal) en el año 179 de la era común C.E. (que es una forma no-cristiana de expresar la era que inicia en el año en que se supone nació Jesucristo).
Los mecanismos matemáticos anteriores, utilizados en el procedimiento estadístico de selección de los predictores de la media condicional de alguna variable de respuesta, se conocen como *regresión escalonada*. Como se señala en https://en.wikipedia.org/wiki/Stepwise_regression, la regresión escalonada es un método de ajuste de modelos de regresión en el que la elección de las variables predictivas se realiza mediante un procedimiento automático (...) En cada paso, se considera una variable para sumar o restar del conjunto de variables explicativas basado en algún criterio preespecificado. Por lo general, esto toma la forma de una secuencia hacia adelante, hacia atrás o combinada de pruebas F o pruebas t. La práctica frecuente de ajustar el modelo final seleccionado seguido de reportar estimaciones e intervalos de confianza sin ajustarlos para tener en cuenta el proceso de construcción del modelo ha llevado a llamadas a dejar de usar la construcción escalonada de modelos por completo (...) o al menos asegurarse de que en el modelo la incertidumbre se refleja correctamente (...) Las alternativas incluyen otras técnicas de selección de modelos, como $R^2$ ajustado, ek criterio de información de Akaike, el criterio de información bayesiano, el $C_p$ de Mallows, la $PRESS$ o la *tasa de falso descubrimiento*.
La construcción escalonada de un modelo puede suscitarse fundamentalmente de tres maneras:
1.*Selección hacia adelante*, que implica comenzar sin variables en el modelo, comprobar lo que ocurre al adicionar cada variable utilizando un criterio de ajuste del modelo elegido, agregando la variable (si la hubiese) cuya inclusión permita la mejora estadísticamente más significativa del ajuste y repetir este proceso hasta ningún predictor mejore el modelo de manera estadísticamente significativa. Véase https://www.analyticsvidhya.com/blog/2021/04/forward-feature-selection-and-its-implementation/
2. *Eliminación hacia atrás*, que implica comenzar con todas las variables candidatas, probar la eliminación de cada variable utilizando un criterio de ajuste del modelo elegido, eliminar la variable (si la hubiese) cuya pérdida produce el deterioro más insignificante estadísticamente del ajuste del modelo, y repetir este proceso hasta que no se pueden eliminar más variables sin una pérdida de ajuste estadísticamente insignificante. Véase https://www.analyticsvidhya.com/blog/2021/04/backward-feature-elimination-and-its-implementation/?utm_source=blog&utm_medium=Forward_Feature_Elimination.
3. *Eliminación bidireccional*, una combinación de 1 y 2, probando en cada paso las variables que se incluirán o excluirán.
#### 6.2. Método de Eliminación Hacia Atrás en R
##### 6.2.1. Eliminación Hacia Atrás con Probabilidad F
Para eliminar variables secuencialmente se usa la función drop1, que proporciona el estadístico F correspondiente a la eliminación de una única variable explicativa; el estadístico F arrojado por esta sintaxis debe interpretarse como la probabilidad de materialización de la probabilidad de rechazar $H_0:β_1=B_2=⋯=B_i=0$ siendo esta verdadera. A causa de lo anterior, un valor F alto indica que la probabilidad de la materialización antes descrita es alta y, ante semejante riesgo, la decisión racional es fallar en rechazar $H_0$ sobre la significancia estadística nula global de los coeficientes de regresión. Fallar en rechazar $H_0$ implica que probabilísticamente hablando no existen consecuencias relevantes (a nivel de capacidad predictiva) si se elimina el modelo en cuestión, por lo que un F mayor que el nivel de significancia $α$ preestablecido (que es la probabilidad de cometer error tipo I, fijada por el investigador con base a la información histórica y a criterios de experto experimentado) significa que ese coeficiente de regresión no es estadísticamente relevante y puede eliminarse.
Puede escribirse el modelo completo (con los 7 predictores) y luego utilizar drop1(mod,test="F") para verificar cuál es la primera variable que se recomienda eliminar tras el proceso antes descrito. Como se adelantó, se deben eliminar aquellos predictores cuyo valor de probabilidad F sea más alto.
“{r}
mod3=lm(esper~., base)
moda=mod3
drop1(moda,test=”F”)
“
Si se comparan los resultados de la sintaxis drop1 con los de summary, se puede verificar que las probabilidades F y t coinciden. Esto sucede en este ejemplo porque no hay variables categóricas con más de 2 categorías; sin embargo, cuando se cuenta con variables categóricas con más de 2 categorías, no se debe usar summary porque en tal caso las probabilidades F y t no son equivalentes.
“{r}
summary(moda)
“
De los resultados anteriores se desprende que el primer predictor a ser eliminado es la variable **area**, pues tiene la probabilidad F más alta. Para materializar la eliminación se puede actualizar el modelo anterior mediante moda=update(moda,.~.-area).
“{r}
moda=update(moda,.~.-area)
drop1(moda,test=”F”)
“
Y así puede continuarse hasta que, por ejemplo, todas las probabilidades sean menores a 0.15 (o a algún valor$α$ preestablecido de la forma antes descrita).
“{r}
moda=update(moda,.~.-analf)
drop1(moda,test=”F”)
moda=update(moda,.~.-ingre)
drop1(moda,test=”F”)
“
Finalmente, se obtiene que el modelo sugerido contempla las variables **pop**, **crim**, **grad** y **temp**.
##### 6.2.2. Eliminación Hacia Atrás con AIC
Adicionalmente, en lugar de usar el criterio de la probabilidad F se pueden usar criterios de información. Para usar el criterio de Akaike (AIC) simplemente no se indica nada en test, pues el AIC es el criterio por defecto que utiliza drop1. En este caso, la columna de AIC indica el valor del AIC que se obtendría si se elimina esa variable. Puesto que el objetivo es aumentar el AIC (porque eso haría al predictor candidato de ser eliminado), entonces se elimina la variable que más disminuye el AIC, generando luego un nuevo modelo (con las variables que menos disminuyen el AIC) que se compara con el modelo anterior, y así sucesivamente, hasta que la eliminación de cualquier variable aumenta el AIC con respecto al modelo anterior en lugar de disminuirlo, puesto que esta es la señal que en términos de robustez estadística del modelo no es recomendable eliminar más predictores.
“{r}
moda=mod3
drop1(moda)
moda=update(moda,.~.-area)
drop1(moda)
moda=update(moda,.~.-analf)
drop1(moda)
moda=update(moda,.~.-ingre)
drop1(moda)
“
El procedimiento antes descrito se puede realizar de forma automática con la sintaxis step mediantestep(mod). Tras ello, puede almacenarse el resultado en una estructura de datos (aquí llamada "mod4"#") y aplicar summary sobre dicho objeto.
“{r}
mod4=step(mod3)
summary(mod4)
“
##### 6.2.2. Eliminación Hacia Atrás con BIC
###### 6.2.2.1. Aspectos Teóricos Relevantes del BIC
Si en lugar del criterio AIC se desease utilizar el criterio bayesiano de información (BIC) se debe indicar en la sintaxis step mediante k=log(n). Debe agregarse que, como se señala en (Bishop, Christopher M. Pattern Recognition and Machine Learning. 2006, p. 217), el criterio bayesiano de información penaliza la complejidad del modelo y es el criterio expuesto por Bishop en el lugar referido el que muestra la penalización que el BIC ejerce sobre la complejidad del modelo y que se conoce como *factor de Occam*.
“{r}
knitr::include_graphics(“FOTO4.JPG”)
“
Debe decirse sobre el factor de Occam que, como puede verificarse en [David J. Spiegelhalter, Nicola G. Best, Bradley P. Carlin & Angelika Van Der Linde. Bayesian measures of model complexity and fit. Journal of Royal Statistical Society, Series B (Statistical Methodology); https://rss.onlinelibrary.wiley.com/doi/pdf/10.1111/1467-9868.00353] y en (van der Linde, Angelika. A Bayesian view of model complexity. Statistica Neerlandica xx, year xx-xx, special issue: All Models Are Wrong...; https://statmodeling.stat.columbia.edu/wp-content/uploads/2013/08/snavdlmc.pdf), no existe una definición analítica para el mismo, *i.e.*, una definición que pueda ser sustentada lógicamente desde algún marco teórico en congruencia clara y directa con un marco matemático autodemostrable dentro de teoría de conjuntos ZF-C (Zermelo-Fraenkel con Axioma de Elección) que la modele.
Brooks (p. 616-18) plantea que la investigación (como casi toda buena investigación) deja preguntas abiertas, específicamente él señala que la ecuación 9 de la página 587 utiliza para calcular dicha complejidad el valor esperado, pero ¿por qué no la moda o la mediana?, ¿cuál es la justificación teórica de ello?, y de ello se deriva también ¿cómo se debe decidir entonces que el parámetro estimado debe ser la media, la moda o la mediana?, lo cual es relevante en cuanto podría conducir a diferencias importantes con el DIC; finalmente, ¿cómo se pueden ser comparables el análisis del modelo bajo el DIC con el análisis del modelo bajo las probabilidades posteriores (enfoque bayesiano) y por qué difieren?, ¿pueden ambas ser "correctas" de alguna manera significativa?
Por su parte, Jim Smith (p. 619-20) señala que no encontró errores técnicos (*i.e.*, matemáticos), pero que encontró cuatro problemas fundacionales. El primero que señala es que las implicaciones predictivas de todas las configuraciones del prior relativas a las variaciones en los ejemplos resueltos en la Sección 8 son increíbles (no en un sentido que podría considerarse positivo), puesto que según Smith no representan juicios de expertos cuidadosamente obtenidos, sino las opiniones de un usuario de software vacío. También señala que, al principio de la Sección 1, los autores afirman que quieren identificar modelos sucintos que parecen describir la información [¿acerca de valores de parámetros "verdaderos" incorrectos (ver Sección 2.2)?] en los datos con precisión, sin embargo, señala también que en un análisis bayesiano, la separación entre la información de los datos y el prior es artificial e inapropiada; señala que "Un análisis bayesiano en nombre de un experto en auditoría remota (Smith, 1996) podría requerir la selección de un prior que sea robusto dentro de una clase de creencias de diferentes expertos (por ejemplo, Pericchi y Walley (1991)). A veces, los prior predeterminados pueden justificarse para modelos simples. Incluso entonces, los modelos dentro de una clase de selección deben tener parametrizaciones compatibles: ver Moreno et al. (1998). Sin embargo, en los ejemplos en los que "el número de parámetros supera en número a las observaciones", afirman que sus enfoques de enfoque, es poco probable los prior predeterminados (por defecto) muestren alguna robustez (estadística). En particular, fuera del dominio de la estimación local vaga o de la estimación de la varianza de separación (discutida en la Sección 4), aparentemente los antecedentes por defecto pueden tener una fuerte influencia en las implicaciones del modelo y, por lo tanto, en la selección.", de lo cual se deriva una razonable insatisfacción ante la expresión la afirmación de los autores y autora sobre la baja probabilidad de que los prior muestren robustez.
Martyn Plummer (p. 621) señala lo que a su juicio son debilidades en la derivación heurística del DIC y de ello se deriva su señalamiento de sustento formal ;como señalan (Rosental & Iudin. Diccionario Filosófico. Editorial Tecolut, 1971. p. 215-216),
en términos históricos la palabra "heurística" proviene del griego εὑρίσκω, que significa "discuto". Es el arte de sostener una discusión y floreció sobre todo entre los sofistas de la antigua Grecia. Surgida como medio de buscar la verdad a través de la polémica, se escindió pronto en dialéctica y sofística. Sócrates, con su método, desarrolló la primera. En cambio, la sofística, tendiente sólo a alcanzar la victoria sobre el contrincante en la discusión, redujo la heurística a una suma de procedimientos que podían aplicarse con el mismo éxito tanto para demostrar una aseveración, cualquiera que fuese, como para refutarla. De ahí que ya Aristóteles no estableciera ninguna diferencia entre heurística y sofística. En la actualidad, al hablar de métodos heurísticos se hace referencia a una especie de atajos para las derivaciones rigurosas que implican mayor costo computacional, por lo que su carácter de verdad es siempre de corto plazo (provisional).
Mervyn Stone (p. 621) señala que la investigación de 2002 "bastante económico" en lo relativo a la *verdad fundamental* (véase https://marxianstatistics.files.wordpress.com/2020/12/sobre-los-estimadores-de-bayes-el-analisis-de-grupos-y-las-mixturas-gaussianas-isadore-nabi.pdf, p. 43-44), que si la sección 7.3 pudiera desarrollarse rigurosamente (puesto que le parece gnoseológicamente cuestionable el uso de $E_Y$), "(...) otra conexión (a través de la ecuación $(33)$) podría ser que $DIC ≈ −2A$. Pero, dado que la sección 7.3 invoca el supuesto de "buen modelo" y pequeños $|\hat{θ}-θ|$ para la expansión de la serie de Taylor (es decir, $n$ grande), tal conexión sería tan artificial como la de $A$ con el criterio de información de Akaike: ¿por qué no seguir con la forma prístina (hoy en día calculable) de $A$, que no necesita $n$ grande o verdad? , ¿y cuál acomoda la estimación de θ en el nivel de independencia de un modelo bayesiano jerárquico? Si la sensibilidad del logaritmo a probabilidades insignificantes es objetable, los bayesianos deberían estar felices de sustituirlo por una medida subjetivamente preferible de éxito predictivo." Es imposible cuestionar a Stone en cuanto a que, dado el enseñoramiento que en la teoría bayesiana de probabilidades tiene la escuela bayesiana subjetiva, el promedio del gremio bayesiano estaría filosóficamente satisfecha con renunciar a elementos objetivos (en este caso son requerimientos preestablecidos por la teoría del aprendizaje estadístico que condicionan la validez gnoseológica del modelo propuesto como un todo, como una muestra grande y/o una verdad fundamental) si representan un punto de discordia y pueden ser sustituidos por algún criterio de decisión que pueda ser determinado; que en paz descanse su alma https://www.ucl.ac.uk/statistics/sites/statistics/files/meryvn-stone-obituary.pdf.
Christian P. Robert y D. M. Titterington (p. 621) señalan que la estructura matemática planteada por los autores de la investigación para determinar la complejidad de un modelo desde la perspectiva bayesiana parecería hacer un uso duplicado (repetido en dos ocasiones) del conjunto de datos, la primera vez lo hacen para determinar la distribución posterior y la segunda para calcular la verosimilitud observada (o verosimilitud a priori, sin considerar información adicional). Este uso duplicado del conjunto de datos puede conducir a un sobreajuste del modelo; señalan que este tipo específico de problemática surgió antes en la investigación de (Aitkin, 1991).
Seguramente el invitado más célebre entre todos los que asistieron a este maravilloso coloquio académico fue John Nelder, padre de los modelos lineales generalizados. Antes de exponer su planteamiento, deben introducirse algunas cuestiones. En primer lugar, el *escape de amoníaco* en aplicaciones industriales es a lo que los autores se refieren (y se refirará Nelder) como *stack loss* (p. 609). En segundo lugar, la tabla 2 a la que se referirá Nelder es la siguiente:
“{r}
knitr::include_graphics(“TABLA2.JPG”)
“
Así, Nelder (p. 622) señala: "Mi colega, el profesor Lee, ha planteado algunos puntos generales que conectan el tema de este artículo con nuestro trabajo sobre modelos lineales generalizados jerárquicos basados en la probabilidad. Quiero plantear un punto específico y dos generales. (a) El profesor Dodge ha demostrado que, de las 21 observaciones en el conjunto de datos de pérdida de amoníaco, ¡solo cinco no han sido declaradas como valores atípicos por alguien! Sin embargo, existe un modelo simple en el que ninguna observación aparece como un valor atípico. Es un modelo lineal generalizado con distribución gamma, log-link y predictor lineal x2 + log.x1 / Å log.x3 /: Esto da las siguientes entradas para la Tabla 2 en el documento: 98.3 92.6 6.2 104.5 (estoy en deuda con el Dr. Best por calcularlos). Es claramente mejor que los modelos existentes usados en la Tabla 2. (b) Este ejemplo ilustra mi primer punto general. Creo que ha pasado el tiempo en que bastaba con asumir un vínculo de identidad para los modelos y permitir que la distribución solo cambiara. Deberíamos tomar como nuestro conjunto de modelos de línea base al menos la clase de modeloos lineales generalizados definida por distribución, enlace y predictor lineal, con la elección de escalas para las covariables en el caso del predictor lineal. (c) Mi segundo punto general es que, para mí, no hay suficiente verificación de modelos en el artículo (supongo que el uso de tales técnicas no va en contra de las reglas bayesianas). Por ejemplo, si un conjunto de efectos aleatorios es suficientemente grande en número y el modelo postula que están distribuidos normalmente, sus estimaciones deben graficarse para ver si se parecen a una muestra de tal
distribución. Si parecen, por ejemplo, fuertemente bimodales, entonces el modelo debe revisarse." Que en paz descanse su alma.
Anthony Atkinson (p. 622) señala que dirige su participación al contexto de la regresión, concluyendo que este criterio de selección de modelos (el BIC planteado por los autores, que es el estimado mediante la sintaxis de R) es un primer paso, que necesita ser complementado
mediante pruebas de diagnóstico y gráficos. Para finalizar plantea que "Estos ejemplos muestran que la búsqueda hacia adelante es una herramienta extremadamente poderosa para este propósito. También requiere muchos ajustes del modelo a subconjuntos de datos. ¿Puede combinarse con los apreciables cálculos de los métodos de Monte Carlo de la cadena de Markov de los autores?" Que en paz descanse su alma.
A.P. Dawid plantea que el artículo debería haberse titulado "Medidas de la complejidad y el ajuste del modelo bayesiano", ya que según él son los modelos, no las medidas, los que son bayesianos. Una vez que se han especificado los ingredientes de un problema, cualquier pregunta relevante tiene una respuesta bayesiana única. La metodología bayesiana debe centrarse en cuestiones de especificación o en formas de calcular o aproximar la respuesta. No se requiere nada más (...) Un lugar donde un bayesiano podría querer una medida de la complejidad del modelo es como un sustituto de p en la aproximación del criterio de información de Bayes a la probabilidad marginal, por ejemplo, para modelos jerárquicos. Pero en tales casos, la definición del tamaño de muestra $n$ puede ser tan problemática como la de la dimensión del modelo $p$. Lo que necesitamos es un mejor sustituto del término completo $p⋅log(n)$". En línea con la gnoseología marxiana, lo adecuado parecería ser considerar que tanto los modelos como las medidas son bayesianos (o de otra escuela de filosofía de las probabilidades).
Las participaciones restantes son no tanto relativas a cuestiones metodológicas como a cuestiones filosóficas-fundacionales de la teoría bayesiana de las probabilidades y de la teoría de las probabilidades en general (puesto que el DIC, que es un criterio de información presentado por los mismos autores que presentan el BIC, no es bayesiano debido a que es una generalización del AIC -que es frecuentista-); de hecho, la transición de cuestiones metodológicas a filosóficas-fundacionales se expresa en el planteamiento de Dawid, quien aunque aborda cuestiones metodológicas lo hace con base en la lógica filosófica de que los modelos y no las medidas son los que pueden ser (o no) bayesianos. Por supuesto, estas últimas son las participaciones más importantes, sin embargo, abordalas escapa a los límites de esta investigación, por lo que para tan importante tarea se dedicará indudablemente un trabajo posterior.
###### 6.2.2.2. Ejecución de la Eliminación Hacia Atrás con el BIC
“{r}
n = nrow(base)
mod5=step(mod3,k=log(n))
summary(mod5)
“
#### 6.3. Método de Selección Hacia Adelante en R
A propósito de lo señalado por Anthony Atkinson, para realizar un proceso de selección hacia adelante se puede usar la función add1 inciando con un modelo que no contenga ninguna variable e indicando en scope cuales son todas las variables disponibles. Ello se realiza de la siguiente forma: add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area).
“{r}
mod6 = lm(esper~1,base)
modb=mod6
add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area)
“
En este caso se escoge agregar la variable que disminuya más el AIC. En este caso es **crim**. Se actualiza el modelo y se continúa hasta que todas tengan un AIC más bajo que el anterior: modb=update(modb,.~.+crim).
“{r}
modb=update(modb,.~.+crim)
add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area)
add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area)
modb=update(modb,.~.+temp)
add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area)
modb=update(modb,.~.+pop)
add1(modb, scope=~pop + ingre + analf + crim + grad + temp + area)
“
De forma similar se puede usar step para indicar scope (además de indicar direction="forward") de la siguiente forma: step(mod6,direction="forward",scope=~pop + ingre + analf + crim + grad + temp + area). scope "define la gama de modelos examinados en la búsqueda por pasos. Debe ser una fórmula única o una lista que contenga los componentes superior e inferior, ambas fórmulas. Consulte los detalles sobre cómo especificar las fórmulas y cómo se utilizan." (véase https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/stepAIC.html).
En este caso, tiene la logica del modelo hacia adelante, se va ingresando las variables que reducen el AIC y luego quedan las que no estan en el modelo, osea las que incrementaria el AIC.
“{r}
mod7=step(mod6,direction=”forward”,scope=~pop + ingre + analf + crim + grad + temp + area)
summary(mod7)
“
## ASPECTOS TEÓRICOS GENERALES SOBRE LA MATRIZ DE DISEÑO ESTRUCTURAL
Como se señala en (Eppinger & Browning, 2012, págs. 2-4), la matriz de diseño estructural (DSM de ahora en adelante, por sus siglas en inglés) es una herramienta de modelado de redes que se utiliza para representar los elementos que componen un sistema y sus interacciones, destacando así la arquitectura del sistema (o estructura diseñada). DSM se adapta particularmente bien a aplicaciones en el desarrollo de sistemas de ingeniería complejos y, hasta la fecha, se ha utilizado principalmente en el área de gestión de ingeniería. Sin embargo, en el horizonte hay una gama mucho más amplia de aplicaciones de DSM que abordan problemas complejos en la gestión de la atención médica, los sistemas financieros, las políticas públicas, las ciencias naturales y los sistemas sociales. El DSM se representa como una matriz cuadrada N x N, que mapea las interacciones entre el conjunto de N elementos del sistema. DSM, una herramienta muy flexible, se ha utilizado para modelar muchos tipos de sistemas. Dependiendo del tipo de sistema que se modele, DSM puede representar varios tipos de arquitecturas. Por ejemplo, para modelar la arquitectura de un producto, los elementos de DSM serían los componentes del producto y las interacciones serían las interfaces entre los componentes (figura 1.1.a).
Para modelar la arquitectura de una organización, los elementos de DSM serían las personas o equipos de la organización, y las interacciones podrían ser comunicaciones entre las personas (figura l.1.b). Para modelar una arquitectura de proceso, los elementos del DSM serían las actividades en el proceso, y las interacciones serían los flujos de información y/o materiales entre ellos (figura l.l.c). Los modelos DSM de diferentes tipos de arquitecturas pueden incluso combinarse para representar cómo se relacionan los diferentes dominios del sistema dentro de un sistema más grande (figura l.l.d). Por tanto, el DSM es una herramienta genérica para modelar cualquier tipo de arquitectura de sistema. En comparación con otros métodos de modelado de redes, el principal beneficio de DSM es la naturaleza gráfica del formato de visualización de la matriz. La matriz proporciona una representación muy compacta, fácilmente escalable y legible de forma intuitiva de la arquitectura de un sistema. La figura l.3.a muestra un modelo DSM simple de un sistema con ocho elementos, junto con su representación gráfica dirigida equivalente (dígrafo) en la figura 1.3.b.
En comparación con otros métodos de modelado de redes, el principal beneficio de DSM es la naturaleza gráfica del formato de visualización de la matriz. La matriz proporciona una representación muy compacta, fácilmente escalable y legible de forma intuitiva de la arquitectura de un sistema. La figura l.3.a muestra un modelo DSM simple de un sistema con ocho elementos, junto con su representación equivalente como grafo dirigido (dígrafo) en la figura 1.3.b. En los estudios iniciales de DSM, a muchos les resulta fácil pensar que las celdas a lo largo de la diagonal de la matriz representan los elementos del sistema, análogos a los nodos en el modelo de dígrafo; sin embargo, es necesario mencionar que, para mantener el diagrama de matriz compacto, los nombres completos de los elementos a menudo se enumeran a la izquierda de las filas (y a veces también encima de las columnas) en lugar de en las celdas diagonales. También es fácil pensar que cada celda sobre la diagonal principal de la matriz puede tener entradas que ingresan desde sus lados izquierdo y derecho y salidas que salen desde arriba y abajo. Las fuentes y destinos de estas interacciones de entrada y salida se identifican mediante marcas en las celdas fuera de la diagonal (en la figura anterior expresadas con una letra X) análogas a los arcos direccionales en el modelo de dígrafo. Examinar cualquier fila de la matriz revela todas las entradas del elemento en esa fila (que son salidas de otros elementos).
Si se observa hacia abajo, cualquier columna de la matriz muestra todas las salidas del elemento en esa columna (que se convierten en entradas para otros elementos). En el ejemplo simple de DSM que se muestra en la figura 1.3.a, los ocho elementos del sistema están etiquetados de la A a la H, y hemos etiquetado tanto las filas como las columnas de la A a la H en consecuencia. Al leer la fila D, por ejemplo, vemos que el elemento D tiene entradas de los elementos A, B y F, representados por las marcas X en la fila D, columnas A, B y F. Al leer la columna F, vemos ese elemento F tiene salidas que van a los elementos B y D. Por lo tanto, la marca en la celda fuera de la diagonal [D, F] representa una interacción que es tanto una entrada como una salida dependiendo de si se toma la perspectiva de su proveedor (columna F) o su receptor (fila D). Es importante notar que muchos recursos de DSM usan la convención opuesta, la transposición de la matriz, con las entradas de un elemento mostradas en su columna y sus salidas mostradas en su fila. Las dos convenciones transmiten la misma información, y ambas se utilizan ampliamente debido a las diversas raíces de las herramientas basadas en matrices para los sistemas de modelado.
En este sentido, como se verifica en (IBM, 2021), en diversos escenarios aplicados puede existir más de una función discriminante[1], como se muestra a continuación.
En general, como se verifica en (Zhao & Maclean, 2000, pág. 841), el análisis discriminante canónico (CDA, por nombre en inglés) es una técnica multivariante que se puede utilizar para determinar las relaciones entre una variable categórica y un grupo de variables independientes. Uno de los propósitos principales de CDA es separar clases (poblaciones) en un espacio discriminante de menor dimensión. En este contexto es que cuando existe más de una función discriminante (cada una de estas puede verse como un modelo de regresión lineal), un asterisco (*) como en este caso (para el caso del programa SaaS) u otro símbolo denotará la mayor correlación absoluta de cada variable con una de las funciones canónicas. Dentro de cada función, estas variables marcadas se ordenan por el tamaño de la correlación. Para el caso de la tabla presentada en la figura anterior, su lectura debe realizarse de la siguiente manera:
1. “Nivel educativo” está más fuertemente correlacionado con la primera función y es la única variable más fuertemente correlacionada con esta función.
2. Años con empresa actual, “Edad” en años, “Ingresos del hogar” en miles, “Años” en la dirección actual, “Retirado” y “Sexo” están más fuertemente correlacionados con la segunda función, aunque “Sexo” y “Jubilación” están más débilmente correlacionados que los otros. Las demás variables marcan esta función como función de “estabilidad”.
3. “Número de personas en el hogar” y “Estado civil” están más fuertemente correlacionados con la tercera función discriminante, pero esta es una función sin utilidad, así que estos predictores son prácticamente inútiles.
### REFERENCIAS
Eppinger, S. D., & Browning, T. R. (2012). Design Structure Matrix Methods and Applications. Cambridge, Massachusetts: MIT Press.
IBM. (2021). Análisis discriminante. Obtenido de SPSS Statistics: https://www.ibm.com/docs/es/spss-statistics/version-missing?topic=features-discriminant-analysis
IBM. (2021). Matriz de estructura. Obtenido de SaaS: https://www.ibm.com/docs/es/spss-modeler/SaaS?topic=customers-structure-matrix
Wikipedia. (23 de Junio de 2021). Linear classifier. Obtenido de Statistical classification: https://en.wikipedia.org/wiki/Linear_classifier
Zhao, G., & Maclean, A. L. (2000). A Comparison of Canonical Discriminant Analysis and Principal Component Analysis for Spectral Transformation. Photogrammetric Engineering & Remote Sensing, 841-847. Obtenido de https://www.asprs.org/wp-content/uploads/pers/2000journal/july/2000_jul_841-847.pdf
[1] Como se verifica en (de la Fuente Fernández, pág. 1), un discriminante es cada una de las variables independientes con las que se cuenta. Además, como se verifica en (IBM, 2021), una función discriminante es aquella que, mediante las diferentes combinaciones lineales de las variables predictoras, busca realizar la mejor discriminación posible entre los grupos. No debe olvidarse que, como se señala en (Wikipedia, 2021), En el campo del aprendizaje automático, el objetivo de la clasificación estadística es utilizar las características de un objeto para identificar a qué clase (o grupo) pertenece.
## INTRODUCCIÓN A LOS ENSAYOS CLÍNICOS DESDE LA TEORÍA ESTADÍSTICA Y RSTUDIO: ASOCIACIÓN Y CORRELACIÓN DE PEARSON, SPEARMAN Y KENDALL
###Se utiliza para describir la distribución de una suma de variables aleatorias al cuadrado. También se utiliza para probar la bondad de ajuste de una distribución de datos, si las series de datos son independientes y para estimar las confianzas que rodean la varianza y la desviación estándar de una variable aleatoria de una distribución normal.
### COEFICIENTES DE CORRELACIÓN
###Coeficiente de Correlación de Pearson (prueba paramétrica): https://statistics.laerd.com/statistical-guides/pearson-correlation-coefficient-statistical-guide.php, https://www.wikiwand.com/en/Pearson_correlation_coefficient.
###Coeficiente de Correlación de Spearman (prueba no-paramétrica): https://statistics.laerd.com/statistical-guides/spearmans-rank-order-correlation-statistical-guide.php, https://www.wikiwand.com/en/Spearman%27s_rank_correlation_coefficient, https://www.statstutor.ac.uk/resources/uploaded/spearmans.pdf.
###Coeficiente de Correlación de Kendall (prueba no-paramétrica): https://www.statisticshowto.com/kendalls-tau/, https://towardsdatascience.com/kendall-rank-correlation-explained-dee01d99c535, https://personal.utdallas.edu/~herve/Abdi-KendallCorrelation2007-pretty.pdf, https://www.wikiwand.com/en/Kendall_rank_correlation_coefficient.
####Como se verifica en su forma más general [véase Jeremy M. G. Taylor, Kendall’s and Spearman’s Correlation Coefficient in the Presence of a Blocking Variable, (Biometrics, Vol. 43, No. 2 (Jun., 1987), pp.409-416), p. 409], en presencia de “empates”, conocidos también como “observaciones vinculadas” (del inglés “ties”, que, como se verifica en http://www.statistics4u.com/fundstat_eng/dd_ties.html, significa en el contexto de las estadísticas de clasificación de orden -rank order statistics- la existencia de dos o más observaciones que tienen el mismo valor, por lo que imposibilita la asignación de números de rango únicos), es preferible utilizar el coeficiente de correlación de Spearman rho porque su varianza posee una forma más simple (relacionado con el costo computacional, puesto que la investigación de Jeremy Taylor emplea como herramienta de estadística experimental la metodología Monte Carlo, lo que puede verificarse en https://pdodds.w3.uvm.edu/files/papers/others/1987/taylor1987a.pdf).
### RIESGO RELATIVO
####Como se verifica en https://www.wikiwand.com/en/Odds_ratio, el riesgo relativo (diferente a la razón éxito/fracaso y a la razón de momios) es la proporción de éxito de un evento (o de fracaso) en términos del total de ocurrencias (éxitos más fracasos).
### RAZÓN ÉXITO/FRACASO
####Es el cociente entre el número de veces que ocurre un evento y el número de veces en que no ocurre.
####INTERPRETACIÓN: Para interpretar la razón de ataque/no ataque de forma más intuitiva se debe multiplicar dicha razón Ψ (psi) por el número de decenas necesarias Ξ (Xi) para que la razón tenga un dígito d^*∈N a la izquierda del “punto decimal” (en este caso de aplicación hipotético Ξ=1000), resultando así un escalar real υ=Ψ*Ξ (donde υ es la letra griega ípsilon) con parte entera que se interpreta como “Por cada Ξ elementos de la población de referencia bajo la condición especificada (en este caso, que tomó aspirina o que tomó un placebo) estará presente la característica (u ocurrirá el evento, según sea el caso) en (d^*+h) ocasiones, en donde h es el infinitesimal a la derecha del punto decimal (llamado así porque separa no sólo los enteros de los infinitesimales, sino que a su derecha se encuentra la casilla correspondiente justamente a algún número decimal).
### RAZÓN DE MOMIOS
####DEFÍNICIÓN: Es una medida utilizada en estudios epidemiológicos transversales y de casos y controles, así como en los metaanálisis. En términos formales, se define como la posibilidad que una condición de salud o enfermedad se presente en un grupo de población frente al riesgo que ocurra en otro. En epidemiología, la comparación suele realizarse entre grupos humanos que presentan condiciones de vida similares, con la diferencia que uno se encuentra expuesto a un factor de riesgo (mi) mientras que el otro carece de esta característica (mo). Por lo tanto, la razón de momios o de posibilidades es una medida de tamaño de efecto.
####Nótese que es un concepto, evidentemente, de naturaleza frecuentista.
####La razón de momios es el cociente entre las razones de ocurrencia/no-ocurrencia de los tratamientos experimentales estudiados (una razón por cada uno de los dos tratamientos experimentales sujetos de comparación).
### TAMAÑO DEL EFECTO
####Defínase tamaño del efecto como cualquier medida realizada sobre algún conjunto de características (que puede ser de un elemento) relativas a cualquier fenómeno, que es utilizada para abordar una pregunta de interés, según (Kelly y Preacher 2012, 140). Tal y como ellos señalan, la definición es más que una combinación de “efecto” y “tamaño” porque depende explícitamente de la pregunta de investigación que se aborde. Ello significa que lo que separa a un tamaño de efecto de un estadístico de prueba (o estimador) es la orientación de su uso, si responde una pregunta de investigación en específico entonces el estadístico (o parámetro) se convierte en un “tamaño de efecto” y si sólo es parte de un proceso global de predicción entonces es un estadístico (o parámetro) a secas, i.e., su distinción o, expresado en otros términos, la identificación de cuándo un estadístico (o parámetro) se convierte en un tamaño de efecto, es una cuestión puramente epistemológica, no matemática. Lo anterior simplemente implica que, dependiendo del tipo de pregunta que se desee responder el investigador, un estadístico (o parámetro) será un tamaño de efecto o simplemente un estadístico (o parámetro) sin más.
setwd(“C:/Users/User/Desktop/Carpeta de Estudio/Maestría Profesional en Estadística/Semestre II-2021/Métodos, Regresión y Diseño de Experimentos/2/Laboratorios/Laboratorio 2”)
## ESTIMAR EL COEFICIENTE DE CORRELACIÓN DE PEARSON ENTRE TEMPERATURA Y PORCENTAJE DE CONVERSIÓN
###CÁLCULO MANUAL DE LA COVARIANZA
prom.temp = mean(temperatura)
prom.conversion = mean(porcentaje.conversion)
sd.temp = sd(temperatura)
sd.conversion = sd(porcentaje.conversion)
n = nrow(vinilacion)
covarianza = sum((temperatura-prom.temp)*(porcentaje.conversion-prom.conversion))/(n-1)
covarianza
###La covarianza es una medida para indicar el grado en el que dos variables aleatorias cambian en conjunto (véase https://www.mygreatlearning.com/blog/covariance-vs-correlation/#differencebetweencorrelationandcovariance).
###CÁLCULO DE LA COVARIANZA DE FORMA AUTOMATIZADA
cov(temperatura,porcentaje.conversion)
###CÁLCULO MANUAL DEL COEFICIENTE DE CORRELACIÓN DE PEARSON
###Véase https://www.wikiwand.com/en/Pearson_correlation_coefficient (9 de septiembre de 2021).
coef.correlacion = covarianza/(sd.temp*sd.conversion)
coef.correlacion
###CÁLCULO AUTOMATIZADO DEL COEFICIENTE DE CORRELACIÓN DE PEARSON
cor(temperatura,porcentaje.conversion) ###Salvo que se especifique lo contrario (como puede verificarse en la librería de R), el coeficiente de correlación calculado por defecto será el de Pearson, sin embargo, se puede calcular también el coeficiente de Kendall (escribiendo “kendall” en la casilla “method” de la sintaxis “cor”) o el de Spearman (escribiendo “spearman” en la casilla “method” de la sintaxis “cor”).
cor(presion,porcentaje.conversion)
###VÍNCULO, SIMILITUDES Y DIFERENCIAS ENTRE CORRELACIÓN Y COVARIANZA
###El coeficiente de correlación está íntimamente vinculado con la covarianza. La covarianza es una medida de correlación y el coeficiente de correlación es también una forma de medir la correlación (que difiere según sea de Pearson, Kendall o Spearman).
###La covarianza indica la dirección de la relación lineal entre variables, mientras que el coeficiente de correlación mide no sólo la dirección sino además la fuerza de esa relación lineal entre variables.
###La covarianza puede ir de menos infinito a más infinito, mientras que el coeficiente de correlación oscila entre -1 y 1.
###La covarianza se ve afectada por los cambios de escala: si todos los valores de una variable se multiplican por una constante y todos los valores de otra variable se multiplican por una constante similar o diferente, entonces se cambia la covarianza. La correlación no se ve influenciada por el cambio de escala.
###La covarianza asume las unidades del producto de las unidades de las dos variables. La correlación es adimensional, es decir, es una medida libre de unidades de la relación entre variables.
###La covarianza de dos variables dependientes mide cuánto en cantidad real (es decir, cm, kg, litros) en promedio covarían. La correlación de dos variables dependientes mide la proporción de cuánto varían en promedio estas variables entre sí.
###La covarianza es cero en el caso de variables independientes (si una variable se mueve y la otra no) porque entonces las variables no necesariamente se mueven juntas (por el supuesto de ortogonalidad entre los vectores, que expresa geométricamente su independencia lineal). Los movimientos independientes no contribuyen a la correlación total. Por tanto, las variables completamente independientes tienen una correlación cero.
## CREAR UNA MATRIZ DE CORRELACIONES DE PEARSON Y DE SPEARMAN
####La vinilación de los glucósidos se presenta cuando se les agrega acetileno a alta presión y alta temperatura, en presencia de una base para producir éteres de monovinil.
###Los productos de monovinil éter son útiles en varios procesos industriales de síntesis.
###Interesa determinar qué condiciones producen una conversión máxima de metil glucósidos para diversos isómeros de monovinil.
cor(vinilacion) ###Pearson
cor(vinilacion, method=”spearman”) ###Spearman
## CREAR UNA MATRIZ DE VARIANZAS Y COVARIANZAS (LOCALIZADAS ESTAS ÚLTIMAS EN LA DIAGONAL PRINCIPAL DE LA MATRIZ)
cov(vinilacion)
## GENERAR GRÁFICOS DE DISPERSIÓN
plot(temperatura,porcentaje.conversion)
plot(porcentaje.conversion~temperatura)
mod = lm(porcentaje.conversion~temperatura)
abline(mod,col=2)
###La sintaxis “lm” es usada para realizar ajuste de modelos lineales (es decir, ajustar un conjunto de datos a la curva dibujada por un modelo lineal -i.e., una línea recta-, lo cual -si es estadísticamente robusto- implica validar que el conjunto de datos en cuestión posee un patrón de comportamiento geométrico lineal).
###La sintaxis “lm” puede utilizar para el ajuste el método de los mínimos cuadrados ponderados o el método de mínimos cuadrados ordinarios, en función de si la opción “weights” se llena con un vector numérico o con “NULL”, respectivamente).
### La casilla “weights” de la sintaxis “lm” expresa las ponderaciones a utilizar para realizar el proceso de ajuste (si las ponderaciones son iguales para todas las observaciones, entonces el método de mínimos cuadrados ponderados se transforma en el método de mínimos cuadrados ordinarios). Estas ponderaciones son, en términos computacionales, aquellas que minimizan la suma ponderada de los errores al cuadrado.
###Las ponderaciones no nulas pueden user usadas para indicar diferentes varianzas (con los valores de las ponderaciones siendo inversamente proporcionales a la varianza); o, equivalentemente, cuando los elementos del vector de ponderaciones son enteros positivos w_i, en donde cada respuesta y_i es la media de las w_j unidades observacionales ponderadas (incluyendo el caso de que hay w_i observaciones iguales a y_i y los datos se han resumido).
###Sin embargo, en el último caso, observe que no se utiliza la variación dentro del grupo. Por lo tanto, la estimación sigma y los grados de libertad residuales pueden ser subóptimos; en el caso de pesos de replicación, incluso incorrecto. Por lo tanto, los errores estándar y las tablas de análisis de varianza deben tratarse con cuidado.
###La estimación sigma se refiere a la sintaxis “sigma” que estima la desviación estándar de los errores (véase https://stat.ethz.ch/R-manual/R-devel/library/stats/html/sigma.html).
###Si la variable de respuesta (o dependiente) es una matriz, un modelo lineal se ajusta por separado mediante mínimos cuadrados a cada columna de la matriz.
###Cabe mencionar que “formula” (la primera entrada de la sintaxis “lm”) tiene un término de intersección implícito (recuérdese que toda ecuación de regresión tiene un intercepto B_0, que puede ser nulo). Para eliminar dicho término, debe usarse y ~ x – 1 o y ~ 0 + x.
plot(presion~porcentaje.conversion)
mod = lm(presion~porcentaje.conversion) ###Ajuste a la recta antes mencionado y guardado bajo el nombre “mod”.
abline(mod,col=2) ###Es crear una línea color rojo (col=2) en la gráfica generada (con la función “mod”)
## REALIZAR PRUEBA DE HIPÓTESIS PARA EL COEFICIENTE DE CORRELACIÓN
###Para estar casi seguros (en relación al concepto de convergencia) Para asegurar que existe al menos una leve correlación entre dos variables (X,Y) se tiene que probar que el coeficiente de correlación poblacional (r) no es nulo.
###Para que la prueba de hipótesis tenga validez se debe verificar que la distribución de Y para cada X es normal y que sus valores han sido seleccionados aleatoriamente.
###Si se rechaza la hipótesis nula, no se asegura que haya una correlación muy alta.
###Si el valor p es menor que el nivel de significancia se rechaza la Ho de que el coeficiente de correlación entre Y y X es cero en términos de determinado nivel de significancia estadística.
###Evaluar la significancia estadística de un coeficiente de correlación puede contribuir a validar o refutar una investigación donde este se haya utilizado (siempre que se cuenten con los datos empleados en la investigación), por ejemplo, en el uso de modelos lineales de predicción.
###Se puede utilizar la distribución t con n-2 grados de libertad para probar la hipótesis.
###Como se observará a continuación, además de la forma estándar, también es posible calcular t como la diferencia entre el coeficiente de correlación.
###Si la probabilidad asociada a la hipótesis nula es casi cero, puede afirmarse a un nivel de confianza determinado de que la correlación es altamente significativa en términos estadísticos.
###FORMA MANUAL
ee = sqrt((1-coef.correlacion^2)/(n-2))
t.calculado = (coef.correlacion-0)/ee ###Aquí parece implicarse que el valor t puede calcularse como el cociente entre el coeficiente de correlación muestral menos el coeficiente de correlación poblacional sobre el error estándar de la media.
cor.test(temperatura,porcentaje.conversion) ###El valor del coeficiente de correlación que se ha estipulado (que es cero) debe encontrarse dentro del intervalo de confianza al nivel de probabilidad pertinente para aceptar Ho y, caso contrario, rechazarla.
cor.test(temperatura,presion)
###Como se señala en https://marxianstatistics.com/2021/09/05/analisis-teorico-de-la-funcion-cuantil-en-r-studio/, calcula el valor umbral x por debajo del cual se encuentran las observaciones sobre el fenómeno de estudio en una proporción P de las ocasiones (nótese aquí una definición frecuentista de probabilidad), incluyendo el umbral en cuestión.
qt(0.975,6)
### EJEMPLO DE APROXIMACIÓN COMPUTACIONAL DE LA DISTRIBUCIÓN t DE STUDENT A LA DISTRIBUCIÓN NORMAL
###El intervalo de confianza se calcula realizando la transformación-z de Fisher (tanto con la función automatizada de R como con la función personalizada elaborada) como a nivel teórico), la cual se utiliza porque cuando la transformación se aplica al coeficiente de correlación muestral, la distribución muestral de la variable resultante es aproximadamente normal, lo que implica que posee una varianza que es estable sobre diferentes valores de la correlación verdadera subyacente (puede ampliarse más en https://en.wikipedia.org/wiki/Fisher_transformation).
coef.correlacion+c(-1,1)*qt(0.975,6)*ee ###Intervalo de confianza para el estadístico de prueba sujeto de hipótesis (el coeficiente de correlación, en este caso) distribuido como una distribución t de Student.
coef.correlacion+c(-1,1)*qnorm(0.975)*ee ###Intervalo de confianza para el estadístico de prueba sujeto de hipótesis (el coeficiente de correlación, en este caso) distribuido normalmente.
## CASO DE APLICACIÓN HIPOTÉTICO
###En un estudio sobre el metabolismo de una especie salvaje, un biólogo obtuvo índices de actividad y datos sobre tasas metabólicas para 20 animales observados en cautiverio.
rm(list=ls()) ###Remover todos los objetos de la lista
###Coeficiente de Correlación de Pearson
###Se rechaza la hipótesis nula de que la correlación de Pearson es 0.
###Coeficiente de correlación de Spearman
(t.s=corr*(sqrt((n-2)/(1-(corr^2)))))
(gl=n-2)
(1-pt(t.s,gl))*2
###Se rechaza la hipótesis nula de que la correlación de Spearman es 0.
###Ambas oscilan entre -1 y 1. El signo negativo denota la relacion inversa entre ambas. La correlacion de Pearson mide la relación lineal entre dos variables (correlacion 0 es independencia lineal, que los vectores son ortogonales). La correlación de Pearson es para variables numérica de razón y tiene el supuesto de normalidad en la distribución de los valores de los datos. Cuando los supuestos son altamente violados, lo mejor es usar una medida de correlación no-paramétrica, específicamente el coeficiente de Spearman. Sobre el coeficiente de Spearman se puede decir lo mismo en relación a la asociación. Así, valores de 0 indican correlación 0, pero no asegura que por ser cero las variables sean independientes (no es concluyente).
### TABLAS DE CONTINGENCIA Y PRUEBA DE INDEPENDENCIA
###Una tabla de contingencia es un arreglo para representar simultáneamente las cantidades de individuos y sus porcentajes que se presentan en cada celda al cruzar dos variables categóricas.
###En algunos casos una de las variables puede funcionar como respuesta y la otra como factor, pero en otros casos sólo interesa la relación entre ambas sin intentar explicar la dirección de la relación.
###CASO DE APLICACIÓN HIPOTÉTICO
###Un estudio de ensayos clínicos trataba de probar si la ingesta regular de aspirina reduce la mortalidad por enfermedades cardiovasculares. Los participantes en el estudio tomaron una aspirina o un placebo cada dos días. El estudio se hizo de tal forma que nadie sabía qué pastilla estaba tomando. La respuesta es que si presenta o no ataque cardiaco (2 niveles),
rm(list=ls())
aspirina
str(aspirina)
attach(aspirina)
names(aspirina)
str(aspirina)
View(aspirina)
#### 1. Determinar las diferencias entre la proporción a la que ocurrió un ataque dependiendo de la pastilla que consumió. Identifique el porcentaje global en que presentó ataque y el porcentaje global en que no presentó.
e=tapply(aspirina$freq,list(ataque,pastilla),sum) ###Genera la estructura de la tabla con la que se trabajará (la base de datos organizada según el diseño experimental previamente realizado). prop.table(e,2) ###Riesgo Relativo columna. Para verificar esto, contrástese lo expuesto al inicio de este documento con la documentación CRAN [accesible mediante la sintaxis “?prop.table”] para más detalles. prop.table(e,1) ###Riesgo Relativo fila. Para verificar esto, contrástese lo expuesto al inicio de este documento con la documentación CRAN [accesible mediante la sintaxis “?prop.table”] para más detalles. (et=addmargins(e)) ###Tabla de contingencia. addmargins(prop.table(e)) ####Distribución porcentual completa. ###Si se asume que el tipo de pastilla no influye en el hecho de tener un ataque cardíaco, entonces, debería de haber igual porcentaje de ataques en la columna de médicos que tomaron aspirina que en la de los que tomaron placebo. ###Se obtiene el valor esperado de ataques y no ataques. ### Lo anterior se realiza bajo el supuesto de que hay un 1.3% de ataques en general y un 98.7% de no ataques. #### 2. Usando los valores observados y esperados, calcular el valor de Chi-Cuadrado para determinar si existe dependencia entre ataque y pastilla? ###Al aplicar la distribución Chi cuadrado, que es una distribución continua, para representar un fenómeno discreto, como el número de casos en cada unos de los supuestos de la tabla de 2*2, existe un ligero fallo en la aproximación a la realidad. En números grandes, esta desviación es muy escasa, y puede desecharse, pero cuando las cantidades esperadas en alguna de las celdas son números pequeños- en general se toma como límite el que tengan menos de cinco elementos- la desviación puede ser más importante. Para evitarlo, Yates propuso en 1934 una corrección de los métodos empleados para hallar el Chi cuadrado, que mejora la concordancia entre los resultados del cálculo y la distribución Chi cuadrado. En el articulo anterior, correspondiente a Chi cuadrado, el calculador expone, además de los resultados de Chi cuadrado, y las indicaciones para decidir, con arreglo a los límites de la distribución para cada uno de los errores alfa admitidos, el rechazar o no la hipótesis nula, una exposición de las frecuencias esperadas en cada una de las casillas de la tabla de contingencia, y la advertencia de que si alguna de ellas tiene un valor inferior a 5 debería emplearse la corrección de Yates. Fuente: https://www.samiuc.es/estadisticas-variables-binarias/valoracion-inicial-pruebas-diagnosticas/chi-cuadrado-correccion-yates/. ###Como se señala en [James E. Grizzle, Continuity Correction in the χ2-Test for 2 × 2 Tables, (The American Statistician, Oct., 1967, Vol. 21, No. 4 (Oct., 1967), pp. 28-32), p. 29-30], técnicamente hablando, la corrección de Yates hace que “(…) las probabilidades obtenidas bajo la distribución χ2 bajo la hipótesis nula converjan de forma más cercana con las probabilidades obtenidas bajo el supuesto de que el conjunto de datos fue generado por una muestra proveniente de la distribución hipergeométrica, i.e., generados bajo el supuesto que los dos márgenes de la tabla fueron fijados con antelación al muestreo.” ###Grizzle se refiere con “márgenes” a los totales de la tabla (véase https://www.tutorialspoint.com/how-to-create-a-contingency-table-with-sum-on-the-margins-from-an-r-data-frame). Además, la lógica de ello subyace en la misma definición matemática de la distribución hipergeométrica. Como se puede verificar en RStudio mediante la sintaxis “?rhyper”, la distribución hipergeométrica tiene la estructura matemática (distribución de probabilidad) p(x) = choose(m, x) choose(n, k-x)/choose(m+n, k), en donde m es el número de éxitos, n es el número de fracasos lo que ) y k es el tamaño de la muestra (tanto m, n y k son parámetros en función del conjunto de datos, evidentemente), con los primeros dos momentos definidos por E[X] = μ = k*p y la varianza se define como Var(X) = k p (1 – p) * (m+n-k)/(m+n-1). De lo anterior se deriva naturalmente que para realizar el análisis estocástico del fenómeno modelado con la distribución hipergeométrica es necesario conocer la cantidad de sujetos que representan los éxitos y los fracasos del experimento (en donde “éxito” y “fracaso” se define en función del planteamiento del experimento, lo cual a su vez obedece a múltiples factores) y ello implica que se debe conocer el total de los sujetos experimentales estudiados junto con su desglose en los términos binarios ya especificados. ###Lo mismo señalado por Grizzle se verifica (citando a Grizzle) en (Biometry, The Principles and Practice of Statistics in Biological Research, Robert E. Sokal & F. James Rohlf, Third Edition, p. 737), especificando que se vuelve innecesaria la corrección de Yates aún para muestras de 20 observaciones. ###Adicionalmente, merece mención el hecho que, como es sabido, la distribución binomial se utiliza con frecuencia para modelar el número de éxitos en una muestra de tamaño n extraída con reemplazo de una población de tamaño N. Sin embargo, si el muestreo se realiza sin reemplazo, las muestras extraídas no son independientes y, por lo tanto, la distribución resultante es una hipergeométrica; sin embargo, para N mucho más grande que n, la distribución binomial sigue siendo una buena aproximación y se usa ampliamente (véase https://www.wikiwand.com/en/Binomial_distribution). ###Grados de libertad correspondientes: número de filas menos 1 por número de columnas menos 1. ###Ho = Hay independencia entre el ataque y las pastillas. (tabla.freq<-xtabs(freq~ataque+pastilla, data=aspirina)) ###La tabla de frecuencias contiene tanto las frecuencias observadas como las esperadas. ###La frecuencia esperada es el conteo de observaciones que se espera en una celda, en promedio, si las variables son independientes. ###La frecuencia esperada de una variable se calcula como el producto entre el cociente [(Total de la Columna j)/(Total de Totales)]*(Total Fila i). ###PRUEBA CHI-CUADRADO AUTOMATIZADA (prueba.chi<-chisq.test(tabla.freq,correct=F) ) ###La sintaxis “chisq.test” sirve para realizar la prueba de Chi-Cuadrado en tablas de contingencia y para realizar pruebas de bondad de ajuste. names(prueba.chi) ###PRUEBA CHI-CUADRADO PASO A PASO (esperado<-prueba.chi$expected) ###valores esperados
###Si el valor p es mayor que el nivel de significancia se falla en rechazar Ho, si es menor se rechaza Ho.
###Se rechaza Ho con un nivel de significancia alfa de 0.05. Puesto que se tiene una probabilidad muy baja de cometer error tipo I, i.e., rechazar la hipótesis nula siendo falsa.
## SOBRE LA CREACIÓN Y DESTRUCCIÓN DE VALOR EN LOS SISTEMAS DE ECONOMÍA POLÍTICA CAPITALISTA EN PARTICULAR Y EN LOS SISTEMAS ECONÓMICOS EN GENERAL (BORRADOR)
### REFERENCIAS
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Zachariah, D. (Junio de 2006). Labour value and equalisation of profit rates: a multi-country study. Indian Development Review, 4, 1-20. | 31,300 | 116,290 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-14 | latest | en | 0.736451 |
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Question 51 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N 43,000
B
N 28,000
C
N13,000
D
N3,000
Explanation.
No official Explanation yet!
Question 52 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N 60,000
B
N75,000
C
N85,000
D
N185,000
Explanation.
No official Explanation yet!
Question 53 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N8m
B
N 11M
C
N12M
D
N15M
Explanation.
No official Explanation yet!
Question 54 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N9M
B
N11M
C
N13M
D
N16M
Explanation.
No official Explanation yet!
Question 55 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N190,000
B
N210,000
C
N220,000
D
N230,000
Explanation.
No official Explanation yet!
Question 56 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
LAST IN FIRST OUT
B
FIRST IN FIRST OUT
C
AVERAGE COST
D
WEIGHTED AVERAGE
Explanation.
No official Explanation yet!
Question 57 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N4.00
B
N5.00
C
N5.50
D
N6.00
Explanation.
No official Explanation yet!
Question 58 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N4200
B
N2700
C
N4500
D
N3900
Explanation.
No official Explanation yet!
Question 59 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N3900
B
N2500
C
N4100
D
N2700
Explanation.
No official Explanation yet!
Question 60 Principles of Accounts | JAMB/UTME (2007 | 2006 | 2005 | 2004 | 2003 | 2002 | 2001 | 2000)
A
N5475
B
N5725
C
N4400
D
N5000
Explanation.
No official Explanation yet!
Question Map | 827 | 2,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.593338 |
https://www.manualslib.com/manual/1570903/Mitsubishi-Electric-Gug-01sl-E.html?page=56 | 1,611,125,336,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00689.warc.gz | 868,078,973 | 34,035 | # Enthalpy [Kj/Kg(Da)]; Lgh-100Rvx-E + Gug-02Sl-E + Puhz-Zrp71; Return Air; (Lossnay Outlet Air) - Mitsubishi Electric GUG-01SL-E Technical Manual
Direct expansion coil unit for lossnay, lossnay return air and supply air temperature control
9
Model selection and capacity calculation
CHAPTER
Example 3-1 : Cooling for an office floor by SA temperature control
For details of each calculation, see Example 1-1.
STEP1. Calculation of required cooling capacity
I: Calculation of required ventilation air volume and selection of Lossnay unit
The floor is assumed same room with Example 1-1.
Therefore, required air volume is also same with Example 1-1.
Required fresh air volume
II : Calculation of cooling load to determine the required capacity
One of the main purposes of SA temperature control is to make up for ventilation load.
As written in Example 1-1, ventilation load is calculated as following under this conditions.
Ventilation load per unit area = ρ x n x V
Outdoor Air
Cooling
### Return Air
Required cooling capacity to make up for above ventilation load (not total cooling load)
III : Calculation of Lossnay energy recovery effect
Lossnay energy recovery effect can also be calculated by the same way with Example1-1.
Energy recovered by Lossnay
Example 1-1 condition
Dry-bulb temperature
Wet-bulb temperature
Absolute humidity
[kg/kg(DA)]
Relative humidity
Enthalpy
[kJ/kg(DA)]
STEP2. Selection of Dx-coil unit system
Select Dx-coil unit and outdoor unit as shown below.
LGH-100RVX-E
Temperature control feature
Supply air temperature control
STEP3-1. Calculation of Cooling capacity of Dx-coil unit
I : Read out characteristics from the specification sheet
(1) Calculation conditions
(A) Lossnay + Dx-coil System configuration
(B) Temperature control feature
(2) Characteristics read out from specifications
(D) Cooling capacity under specification condition
- Lossnay recovery
- Cooling capacity of Dx-coil unit
53
3
1000m
/h
x ( h
f
Dry Bulb Temp. Relative Humidity Wet Bulb Temp.
33 °C
26 °C
2
= 53.0 [W/m
] × 200 [m
Outdoor air
[˚C]
33
[˚C]
27
0.0201
[%]
63
84.6
Lossnay model
- h
)
190.8 kJ/m
O
R
63%
27 °C
50%
18.7 °C
2
] = 10.6 [kW]
7.5 kW
Return air
26
18.7
0.0105
50
52.9
LGH-100RVX-E + GUG-02SL-E + PUHZ-ZRP50
SA Temperature control
9.5 kW
4.2 kW
5.3 kW
LGH-100RVX-E
= 53.0 W/m
2
2
Enthalpy
Enthalpy Difference
84.6 kJ/kg(DA)
31.8 kJ/kg(DA)
52.8 kJ/kg(DA)
### (Lossnay outlet air)
21.4
0.0133
62.0
GUG-02SL-E
PUHZ-ZRP50
28
56
#### This manual is also suitable for:
Gug-03sl-eGug-02sl-e | 773 | 2,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.782156 |
https://stacks.math.columbia.edu/tag/05EP | 1,620,393,847,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988793.99/warc/CC-MAIN-20210507120655-20210507150655-00055.warc.gz | 569,309,245 | 6,171 | Lemma 15.88.13. Let $\varphi : R \to S$ be a flat ring map and $(f_1, \ldots , f_ t) = R$. Then $\text{Can}$ and $H^0$ are quasi-inverse equivalences of categories
$\text{Mod}_ R = \text{Glue}(R \to S, f_1, \ldots , f_ t)$
Proof. Consider an object $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$. By Algebra, Lemma 10.24.5 there exists a unique module $M$ and isomorphisms $M_{f_ i} \to M_ i$ which recover the glueing data $\alpha _{ij}$. Then both $M'$ and $M \otimes _ R S$ are $S$-modules which recover the modules $M_ i \otimes _ R S$ upon localizing at $f_ i$. Whence there is a canonical isomorphism $M \otimes _ R S \to M'$. This shows that $\mathbf{M}$ is in the essential image of $\text{Can}$. Combined with Lemma 15.88.11 the lemma follows. $\square$
There are also:
• 2 comment(s) on Section 15.88: Formal glueing of module categories
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 367 | 1,099 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-21 | latest | en | 0.604527 |
https://www.slideshare.net/RaviSekhar35/finite-element-analysis-rajendra-mpdf | 1,685,697,952,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00545.warc.gz | 1,092,503,088 | 72,833 | ### Finite Element Analysis Rajendra M.pdf
1. Applied Finite Element Analysis for Civil and Mechanical Engineering Applications A Conceptual Course Presented by Rajendra Machavaram Assistant Professor, AgFE, IIT Kharagpur for Faculty Development Program at Audisankara College of Engineering and Technology Gudur, Andhra Pradesh, India. Rajendra M. 1 AgFE, IIT KGP
2. Finite Element Method (Introduction & History) The Finite Element Method (FEM) is a numerical procedure for analyzing structures and continua. It was first applied to stress analysis and today it is applied to other problems, such as, heat transfer, fluid flow, lubrication, electric and magnetic fields. It is also used to design buildings, electric motors, heat engines, ships, airframes and space crafts. History: Fig. A coarse mesh of gear tooth. R. Courant (1943) described a piecewise polynomial solution for the torsion problem. The procedure was impractical at that time due to lack of digital computers. 1950s: Work in the aircraft industry introduced the FE method to practicing engineers 1960s: The name Finite Element was coined (R. W. Clough) 1963: The mathematical validity of the FE method was recognized and the method was expanded from its structural beginnings to include heat transfer, groundwater flow, magnetic fields and other areas 1970s: Large general purpose FE software began to appear 1980s: The software was available on microcomputers, complete with color graphics and pre and post processors (G. E. Smith) 1990 to Present: Application of FE using various softwares on different areas is started Rajendra M. 2 AgFE, IIT KGP
3. Finite Element Method (General Procedure) Step 1: Divide the structure or continuum into finite elements, called as mesh generation. Step 2: Formulate the properties of each element Step 3: Assemble the elements to obtain the finite element model of the structure Step 4: Apply the known loads (Initial conditions) Step 5: Apply the support conditions (Boundary conditions) Step 6: Solve the simultaneous linear algebraic equations to determine nodal DOF (Displacements/ temperatures). Step 7: Calculate the element properties, such as strains, stresses, heat flux etc. Rajendra M. 3 AgFE, IIT KGP
4. Finite Element Method (Terminology) Elements: Finite elements are fragments of the structure, these are triangular or quadrilateral areas on two dimensional space (OR) Tetrahedron or Hexahedron on three dimensional volumes. Nodes: Nodes are the connectors that fasten elements together, which appear on element boundaries. Degrees of Freedom (DOF): Number of independent parameters Shape Function: It is a polynomial function which defines the element field variables in terms of field variables of the nodes. Stiffness Matrix: It is an element characteristic matrix in structural mechanics. It relates nodal displacements to nodal forces. Conductivity Matrix: It is an element characteristic matrix in heat conduction. It relates nodal temperatures to nodal fluxes. Mass Matrix: It is an element characteristic matrix in structural dynamics. It relates nodal velocities to nodal fluxes. Rajendra M. 4 AgFE, IIT KGP
5. Finite Element Method (Problem Formulation) Finite Element Method Problem Formulation describes the procedure for determining the element characteristic matrix. There are three important ways to derive an element characteristic matrix. 1. Direct Method: It is based on physical reasoning. It is limited to very simple elements. 2. Variational Method: It is applicable to problems that can be stated by certain integral expressions such as the expression for potential energy. 3. Weighted Residual Methods: These are particularly suited to problems for which differential equations are known but no variational statement is available. a) Collocation Method: Impulse functions are selected as weighted functions. b) Subdomain Method: Each weighting function is selected as unity over a specific region. c) Galerkin’s Method: This uses the same functions for weights, that were used in the approximating equation. This approach is the basis of the finite element method for problems involving first-derivative terms (Potential or Kinetic Energy variation). d) Least Squares Method: It utilizes the residual as the weighting function and obtains a new error term. This error is minimized with respect to the unknown coefficients in the approximate solution. Rajendra M. 5 AgFE, IIT KGP
6. Finite Element Method (Problem Formulation) Constitutive Matrix Formulation (Three Dimensional Problem)–Classical Mechanics Consider a continuous three-dimensional (3D) elastic solid with a volume V and a surface area S. The solid can be loaded by body forces fb and surface forces fs in any distributed fashion in the volume of the solid. Since cij=cji, there are altogether 21 independent material constants for a fully anisotropic material Rajendra M. 6 AgFE, IIT KGP
7. Finite Element Method (Problem Formulation) Constitutive Matrix Formulation (Three Dimensional Problem)–Classical Mechanics Isotropic Material Isotropic Material – Plane Stress (Thin Shells) and Plane Strain (Thick Shells) Rajendra M. 7 AgFE, IIT KGP
8. Finite Element Method (Problem Formulation) Shape Function Formulation (Three Dimensional Problem) Consider an element with nd nodes at xi (i=1,2,..,nd), where xT = {x, y, z} for three dimensional problem. There should be nd shape functions for each displacement component for an element. Where uh is the approximation of the displacement component, pi(x) is the basis function of monomials in the space coordinates x, and αi is the coefficient for the monomial pi(x). Rajendra M. 8 AgFE, IIT KGP
9. Finite Element Method (Problem Formulation) Stiffness and Mass Matrix Formulation (Three Dimensional Problem) Strain Energy Kinetic Energy Work Done Dynamic Equilibrium Equation Static Equilibrium Equation Rajendra M. 9 AgFE, IIT KGP
10. Finite Element Method (Elements) Basic Element Shapes: The shapes, sizes, number and configuration of the elements have to be chosen carefully such that the original body or domain is simulated as closely as possible without increasing the computational effort needed for the solution. Mostly the choice of the type of element is dictated by the geometry of the body and the number of independent coordinates (DOF) necessary to describe the system. One Dimensional Elements (a) Bar (2 DOF at 2 nodes) and (b) Beam (4 DOF at 2 nodes) Two Dimensional Elements (a) Triangle (b) Rectangle (c) Quadrilateral (d) Parallelogram Three Dimensional Elements (a) Tetrahedron (b) Rectangular Prism (c) Hexahedron Rajendra M. 10 AgFE, IIT KGP
11. Finite Element Method (Bar 1D Element) Bar Element: Stiffness Matrix Formulation: Consider a uniform prismatic elastic bar of length ‘L’ with elastic modulus ‘E’ and cross-sectional area ‘A’. A node is located at each end with axially directed displacements. Direct Method: The stretch in the bar is given by And the forces is calculated as For displacement at node 1 due to force at node 2, and the displacement at node 2 due to force at node 1 is given by Rajendra M. 11 AgFE, IIT KGP
12. Finite Element Method (Bar 1D Element) Bar Element: Stiffness Matrix Formulation: Consider a uniform prismatic elastic bar of length ‘L’ with elastic modulus ‘E’ and cross-sectional area ‘A’. A node is located at each end with axially directed displacements. Variational Approach: Stiffness matrix of a element is given by Where B is the strain-displacement matrix, E is the material property matrix (Constitutive matrix) and dV is an increment of the element volume V. Rajendra M. 12 AgFE, IIT KGP
13. Finite Element Method (Truss 1D Element) The displacement within the element varies linearly, hence it is called linear element Truss Element: Stiffness Matrix Formulation: A truss is one of the simplest and most widely used structural members. It is a straight bar that is designed to take only axial forces, therefore it deforms only in its axial direction. It has one DOF at each node. Rajendra M. 13 AgFE, IIT KGP
14. Finite Element Method (Truss 1D Element) Truss Element: Stiffness Matrix Formulation: A truss is one of the simplest and most widely used structural members. It is a straight bar that is designed to take only axial forces, therefore it deforms only in its axial direction. Rajendra M. 14 AgFE, IIT KGP
15. Finite Element Method (Truss 1D Element) Truss Element: Stiffness Matrix Formulation: Considering de as element displacements in local coordinate system and De as element displacements in global coordinate system. If T is transformation matrix to transform the local coordinates to global coordinate system Rajendra M. 15 AgFE, IIT KGP
16. Finite Element Method (Truss 1D Element) Truss Element: Problems A structure is made of three planer truss members as shown in Fig. A vertical downward force of 1000 N is applied at node 2. The properties of each member is given in Table. Rajendra M. 16 AgFE, IIT KGP
17. Finite Element Method (Truss 1D Element) Truss Element: Problems Rajendra M. 17 AgFE, IIT KGP
18. Finite Element Method (Truss 1D Element) Truss Element: Problems Rajendra M. 18 AgFE, IIT KGP
19. Finite Element Method (Truss 1D Element) Truss Element: Problems Rajendra M. 19 AgFE, IIT KGP
20. Finite Element Method (Truss 1D Element) Truss Element: Problems Rajendra M. 20 AgFE, IIT KGP
21. Finite Element Method (Truss 1D Element) Truss Element: Problems (Exercise) A structure is made of three planer truss members as shown in Fig. A vertical downward force of 1000 N is applied as shown. All the truss members are of the same material (E=69.0 GPa) and with the same cross-sectional area of 0.01 m2. Determine the stress and strain induced in each bar. Rajendra M. 21 AgFE, IIT KGP
22. Finite Element Method (Beam 1D Element) Beam Element: Element Characteristic Matrix Formulation A beam is another simple but commonly used structural component. It is also geometrically a straight bar of an arbitrary cross-section, but it deforms only in directions perpendicular to its axis. Beams are subjected to transverse loading, including transverse forces and moments that result in transverse deformation. The stresses on the cross-section of a beam are the normal stress and shear stress. There are several theories for analyzing beam deflections. These are (i) A theory for thin beams: Euler-Bernoulli beam theory (ii) A theory for thick beams: Temoshenko beam theory In thin beam theory, the transverse planes of the beam before and after bending are always perpendicular to the beam axis. The shear stress is assumed to be negligible. Rajendra M. 22 AgFE, IIT KGP
23. Finite Element Method (Beam 1D Element) Beam Element: Element Characteristic Matrix Formulation In planar beam elements there are 2DOF at a node in its local coordinate system. These are translation (v) along y-axis and rotation about z-axis (θz) in xy plane. Therefore the beam element has a total of 4DOF. Consider a beam element of length l = 2a with nodes 1 and 2 at each end of the element. As there are 4DOF for a beam element, there should be four shape functions. Considering natural coordinate system, the natural coordinate system has its origin at the centre of the element, and the element is defined from -1 to +1. The relation between the natural coordinate system and the local coordinate system is The shape functions of beam element using natural coordinate system is The third order polynomial is chosen because there are four unknowns in the polynomial, which can be related to the four nodal DOFs in the beam element. Rajendra M. 23 AgFE, IIT KGP
24. Finite Element Method (Beam 1D Element) Beam Element: Element Characteristic Matrix Formulation The shape functions of beam element using natural coordinate system is Rajendra M. 24 AgFE, IIT KGP
25. Finite Element Method (Beam 1D Element) Beam Element: Element Characteristic Matrix Formulation The shape functions of beam element using natural coordinate system is Rajendra M. 25 AgFE, IIT KGP
26. Finite Element Method (Beam 1D Element) Beam Element: Element Characteristic Matrix Formulation The stiffness and mass matrices of beam element are Rajendra M. 26 AgFE, IIT KGP
27. Finite Element Method (Beam 1D Element) Beam Element: Problem A cantilever beam is fixed at one end and it has a uniform cross-sectional area as shown in Fig. A download load of P=1000 N applied at the free end. The beam is made of aluminium and its properties are given in Table. Determine the deflection at centre of the beam using FEA. The second moment of area of the cross-sectional area about the z-axis is given as Rajendra M. 27 AgFE, IIT KGP
28. Finite Element Method (Beam 1D Element) Beam Element: Problem Applying boundary conditions Rajendra M. 28 AgFE, IIT KGP
29. Finite Element Method (Beam 1D Element) Beam Element: Problem Deflection and rotation at centre of the beam ( ) are Solve the above problem, considering it as two element FEA model as Exercise Rajendra M. 29 AgFE, IIT KGP
30. Finite Element Method (Frame 1D Element) Frame Element: Characteristic Matrix Formulation (Three Dimensional Problem) A frame element is formulated to model a straight bar of an arbitrary cross-section, which can deform not only in the axial direction but also in the directions perpendicular to the axis of the bar. The bar is capable of carrying both axial and transverse forces, as well as moments. Therefore, a frame element is seen to posses the properties of both truss and beam elements. The frame element developed is also known in many commercial software packages as the General beam element or simply Beam element. Planar frame element has 3 DOF at each node, hence it has total 6 DOF. Spatial frame element has 6 DOF at each node, hence it has total 12 DOF. Rajendra M. 30 AgFE, IIT KGP
31. Finite Element Method (Frame 1D Element) Frame Element: Characteristic Matrix Formulation (Three Dimensional Problem) Spatial frame element has 6 DOF at each node, hence it has total 12 DOF. Rajendra M. 31 AgFE, IIT KGP
32. Finite Element Method (Frame 1D Element) Frame Element: Characteristic Matrix Formulation (Three Dimensional Problem) Spatial frame element has 6 DOF at each node, hence it has total 12 DOF. Based on global coordinate system Rajendra M. 32 AgFE, IIT KGP
33. Finite Element Method (Triangular 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Triangular element is the first type of element developed for 2D solids. The formulation is also simplest among all the 2D solid elements. It has been found that the linear triangular element is less accurate compared to linear quadrilateral elements. Triangular elements are normally used to generate the mesh of a 2D model involving complex geometry with acute corners. It has 2 DOF at each node and this element has 3 nodes and a total of 6 DOF. Area Coordinates Rajendra M. 33 AgFE, IIT KGP
34. Finite Element Method (Triangular 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Triangular element It has 2 DOF at each node and this element has 3 nodes and a total of 6 DOF. Eisenberg and Malvern (1973) Rajendra M. 34 AgFE, IIT KGP
35. Finite Element Method (Rectangular 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Rectangular element Triangular elements are less accurate than rectangular elements, but they are preferred due to meshing problem for complex geometry. Its shape function formulation is easy compared to triangular element. Rectangular element has 2 DOF at each node and it has four nodes, hence the element has a total of 8 DOF. The dimensions of the element is defined here as 2a×2b×h. A local natural coordinate system (ξ, η) with its origin located at the centre of the rectangular element is defined. Rajendra M. 35 AgFE, IIT KGP
36. Finite Element Method (Rectangular 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Rectangular element The stiffness and mass matrices of Rectangular element are Gauss Integration Rajendra M. 36 AgFE, IIT KGP
37. Finite Element Method (Quadrilateral 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Quadrilateral Element Quadrilateral element is more practical and useful element than rectangular and triangular elements because of its unparalleled edges. The Quadrilateral element has four nodes with 2 DOF at each node and a total of 8 DOF. However, there can be a problem for the integration of the mass and stiffness matrices for a quadrilateral element, because of the irregular shape of the integration domain. The Gauss integration scheme cannot be implemented directly with quadrilateral elements. Hence, key in the development of a quadrilateral element is the coordinate mapping from irregular shape of local coordinate system to square shape of natural coordinate system. Rajendra M. 37 AgFE, IIT KGP
38. Finite Element Method (Quadrilateral 2D Element) 2D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Linear Quadrilateral Element The stiffness and mass matrices are J is the Jacobian matrix Rajendra M. 38 AgFE, IIT KGP
39. Finite Element Method (Hexahedron 3D Element) 3D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Hexahedron Element Hexahedron element is a 3D element with eight nodes and six surfaces. It has 3 DOF for each node and the element has a total of 24 DOF. It is again useful to define a natural coordinate system (ξ, η, ζ) with its origin at the centre of its transformed cube, as this makes it easier to construct the shape functions and to evaluate the matrix integration. The characteristic matrix formulation is similar to that of Quadrilateral element. J is the Jacobian matrix Rajendra M. 39 AgFE, IIT KGP
40. Finite Element Method (Hexahedron 3D Element) 3D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Hexahedron Element Hexahedron element is a 3D element with eight nodes and six surfaces. It has 3 DOF for each node and the element has a total of 24 DOF. J is the Jacobian matrix Rajendra M. 40 AgFE, IIT KGP
41. Finite Element Method (Hexahedron 3D Element) 3D Element: Characteristic Matrix Formulation (Three Dimensional Problem) Hexahedron Element The stiffness and mass matrices are Rajendra M. 41 AgFE, IIT KGP
42. Finite Element Method (References) 1. Robert D. Cook. (1995). Finite Element Modeling for Stress Analysis. First Edition John Wiley & Sons, Inc. 2. Robert D. Cook, David S. Malkus and Michael E. Plesha. (1989). Concepts and Applications of Finite Element Analysis. Third Edition. John Wiley & Sons, Inc. 3. Singiresu S. Rao. (2004). The Finite Element Method in Engineering. Fourth Edition. Elsevier Science & Technology Books. 4. Larry J. Segerlind. (1984). Applied Finite Element Analysis. Second Edition. John Wiley & Sons, Inc. 5. G. L. Narasaiah. (2008). Finite Element Analysis. First Edition. B. S. Publications. 6. G. R. Liu and S. S. Quek. (2003). The Finite Element Method – A Practical Course. First Edition. Elsevier Science Ltd. 7. J. N. Reddy. (2006). An Introduction to the Finite Element Method. Third Edition. McGraw Hill, Inc. Thank You Rajendra M. 42 AgFE, IIT KGP | 4,531 | 19,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-23 | latest | en | 0.912615 |
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# What is law of mass action and explain it?
## What is law of mass action and explain it?
law of mass action, law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.
### Who discovered chemical equilibrium?
The development of thermodynamic theory of equilibria—in particular, equilibria of chemical reactions—owed much to J. W. Gibbs (1873-1878) and Le Chatelier (1885), who discovered the principle of displacement of equilibria under conditions of external change.
Who has given law of mass action?
Just over 150 years ago, on 15 March 1864, Peter Waage and Cato Guldberg (Figure 1) published a paper in which they propounded what has come to be known as the Law of Mass Action 1. In this article we review the history of its discovery and early applications in pharmacology.
What is the law of mass action MCAT?
The Law of Mass Action links the rate of a chemical reaction as proportional to the concentrations of the reactants and products in a chemical reaction. For a chemical reaction mixture that is in equilibrium, the ratio between the concentration of the reactants and products is constant.
## What is the importance of mass action law?
This law can be used to explain the behavior exhibited by solutions in dynamic equilibria. The law of mass action also suggests that the ratio of the reactant concentration and the product concentration is constant at a state of chemical equilibrium.
### Who discovered reversible reactions?
Berthollet
The concept of a reversible reaction was introduced by Berthollet in 1803, after he had observed the formation of sodium carbonate crystals at the edge of a salt lake (one of the natron lakes in Egypt, in limestone):
Who invented kinetics?
In 1864, Peter Waage and Cato Guldberg pioneered the development of chemical kinetics by formulating the law of mass action, which states that the speed of a chemical reaction is proportional to the quantity of the reacting substances.
What is law of mass action example?
For example, if the temperature in a system containing a mixture of ice and water is uniformly 273.15 K, the net amount of ice formed and melted will be zero. The amount of liquid water will also remain constant, if no vapor escapes from the system.
## What are the importance of law of mass action?
### Is kinetics on the MCAT?
The MCAT may test your ability to determine if a certain product is a kinetic or thermodynamic product. Thus, it will be important to be able to determine activation energy and compare these quantities.
Is law of mass action and equilibrium law same?
The law of mass action is concerned with the position of equilibrium in a reversible reaction i.e. at equilibrium both forward and back reactions are taking place at equal rates. The law of mass action says nothing about this rate but only tells us what the reactant and product concentrations will be at equilibrium. | 640 | 3,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | longest | en | 0.941703 |
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### Is a projective module of constant finite rank finitely generated?
If $R$ is a (commutative) ring and $P$ is a projective $R$-module, then every localization of $P$ at a prime of $R$ is free by Kaplansky's theorem, and has a well-defined rank. If these ranks are all ...
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### A weak version of Bass' conjecture
Let $A$ be a finitely generated $\mathbb{Z}$-algebra which is a UFD. Then (a special case of) the Bass conjecture states that $K_0(A)$ is a finitely generated abelian group. As far as I am aware, this ...
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### Metabolic vs stably metabolic
Let $A$ be a commutative ring with unit. A non-degenerate symmetric bilinear form $\phi$ on a finitely generated projective $A$-module $P$ is called metabolic if there is a direct summand $L$ of $P$ ...
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### Projective Modules/Algebras: decomposition of linear functions, and the rank formula
Let $A$ be a ring, $B$ a finite projective $A$-algebra, and $C$ a finite projective $B$-algebra. We can show that $C$ is also finite and projective when regarded as an $A$-algebra (by, for instance, ...
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### An analogue of the Bass-Quillen conjecture with power or Laurent series
The famous Quillen-Suslin theorem (formerly known as Serre's problem/conjecture) states that every projective module over $k[x_1,\dots, x_n]$ is free for $k$ a field. Replacing $k$ by a more general ...
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### Baer's criterion for projective modules
Let $R$ be a commutative ring. If necessary, assume that $R$ has any convenient properties you like. Is there some $R$-module $Q$ such that an $R$-module $P$ is projective if and only if ...
840 views
### Is every projective $\mathbf{Z}[x]$-module free?
Is every finitely generated projective $\mathbf{Z}[x]$-module free?
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### Projective Modules and their Determinants, Extended or not?
Let $A$ be a commutative noetherian ring, and let $P$ be a projective $A[T]$-module with constant rank $n$. Let $L$ be the determinant of $P$, $\wedge^n(P)$. We say that $P$ (resp. $L$) is extended ...
475 views
### When are two projective modules of equal rank isomorphic?
Let $R$ be a commutative ring and let $M,N$ be two finitely generated projective $R$-modules which have equal rank (not necessarily constant). What kind of general results are there concerning the ...
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### Commutator tensors and submodules
Let $k$ be a commutative ring with $1$, and let $B$ be a submodule of a $k$-module $A$. For every $n\in\mathbb N$ and every $k$-module $V$, let $K^n\left(V\right)$ denote the kernel of the canonical ...
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### Freeness of modules along ring homomorphisms
This question arises from my discussion with a Master student. It concerns with the following situation: let $\phi: R \to S$ be a homomorphism between Noetherian commutative rings. Suppose the ...
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### Example of a projective module which is not a direct sum of f.g. submodules?
This semester I am teaching a graduate course in commutative algebra, and I have been taking the occasion to try to look at the proofs of some the results in my basic source material (Matsumura, ...
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### Unimodular column property
Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property. I would like to know if there is a ring $R$ that doesn't ...
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### Nonfree projective module over a regular UFD?
What is the simplest example of a domain $R$ which is regular (in particular Noetherian) and factorial which admits a finitely generated projective module that is not free? In fact I'll be at least ...
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### Stably free module not finitely generated is free
Hi. I have read that stably free modules not finitely generated are free; this is proved in M.R. Gabel, stably free projectives over commutative rings, Thesis, Brandeis Univ., Waltham, MA 1972. But ...
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### Projective modules over semi-local rings
Let $R$ be a semi-local ring, and $M$ a finite projective $R$-module. If the localizations $M_m$ have the same rank for all maximal ideals $m$ of $R$ then $M$ is free.
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### Projective dimension of zero module
Is there any consensus on what the projective dimension of the zero module should be? Here are three statements one commonly encounters in textbooks, sometimes with or without the condition $M\neq 0$: ...
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### What is the insight of Quillen's proof that all projective modules over a polynomial ring are free?
One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free. It involves some of the most basic notions in ... | 1,256 | 4,711 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2014-35 | latest | en | 0.8556 |
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# Problem Using the Power (^) Operation in Mathematica
Posted 11 years ago
In am new to Mathematica and just learning how to use it.I have encountered a rather unusual problem, most operations, such as + - / * work fine, at least when I used the numpad to type them in. However, I cannot get the power operation to work consistently. Somtimes mathematica will resolve and expression such as 2^2, other times it will not. The problem also occurs with the alternatives, such as ctrl + ^ and power[2, 2]. I do not understand why it is working so inconsistently. Can you please provide me with help regarding this function?
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Posted 11 years ago
"power[2,2]" will not work because the word power must be capitalized. All built-in Mathematica expressions are capitalized. Can you give us an example of where the carrot operator "^" is not working for you? What happens when you try to use it?In general it should always work when given two numbers like:2^33.14^10etc.If you have not read the virtual book, this may help you become more famailiar with how to get started with Mathematica.
Posted 11 years ago
Actually I was capitalizing Power in the program, I just wrote it down wrong in the thread.More explicitly though, here is what happens.Sometimes a write:6^4 then enterThis will giveIn:= 4^6Out:= 4096Other times I will write6^4 then enterThis will then give absolutely nothing, it will just advance to the next line waiting for a text prompt.I hope that provides a better idea of what is going on.
Posted 11 years ago
Did you mean: " ... then Shift + Enter ... " each time when you wrote " ... then enter ... "Like you, I am only beginning to use Mathematica, but usually it requires Shift + Enter to evaluate an expression.But you probably already know this.Good Luck.
Posted 11 years ago
Actually I did not realize that, because I only had to press enter for Mathematica to evaluate most of the expressions I was using. But your right, by pressing shift + enter, it is now consistently evaluating these expressions.Thanks for the help.
Posted 11 years ago
Please see the virtual book for an overview of how to use Mathematica. You are using Wolfram|Alpha in Mathematica instead of using Mathematica directly. When you use Wolfram|Alpha in Mathematica, you will notice big orange equal signs to the left of your code. This means you aren't using Mathemtica directly, but using Woflram|Alpha. When you use Mathematica directly Ctrl+Enter is always needed to evaluate an expression.
Posted 11 years ago
Sean - thanks for the useful insights about the relationships between Mathematica and Wolfram|Alpha.Can you elaborate - for Mr Barbieri and me - on the difference between Shift+Enter and Ctrl+Enter?Shift+Enter seems to evaluate an expression, whereas Ctrl+Enter seems only to rewrite and return the expression in an alternate format.Thanks, Ed
Posted 11 years ago
Sean Clarke wrote:When you use Mathematica directly Ctrl+Enter is always needed to evaluate an expression.That has to be a typo -- Shift+Enter evaluates, while Ctrl+Enter does something else (inserts a new row when entering a table or a matrix). By the way, this is a more exhaustive listing of the various Keyboard Shortcuts.
Posted 11 years ago
Yes. Sorry. The "Ctrl" is a typo and should be "Shift". Sorry about that. | 793 | 3,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.936497 |
https://au.answers.yahoo.com/question/index?qid=20210228015427AAb5G7x | 1,618,451,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00163.warc.gz | 234,563,576 | 20,162 | Kuy asked in Science & MathematicsMathematics · 2 months ago
# What is (x^(1/5))^2?
Relevance
• 2 months ago
x^(2/5)
• ?
Lv 7
2 months ago
It means the fifth root of x squared. You can do that two ways:-
Take 32 as a convenient example for x.
You could square 32 giving 1024 and find the fifth root which is 4.
Or you could find the fifth root of 32 which is 2 and square that to give 4 that way.
If you just want the simplification of the algebra (x^(1/5))^2 = x^(2/5)
• 2 months ago
x^(2/5)
• 2 months ago
[x^(1/5)]^2
=x^(2/5) | 181 | 542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-17 | latest | en | 0.931655 |
https://senseis.xmp.net/?NotcherCode | 1,591,421,866,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00414.warc.gz | 511,792,476 | 3,777 | # Notcher Code
Difficulty: Intermediate Keywords: Life & Death
On this page we devise a coding system for notchers, in order to set up a systematic approach to their life and death status.
Suppose the notcher is black and the surrounding group white.
Notcher 112NS = nabxy
Below squared stones (): notch (n)
Circled stones (): legs (ab)
Cross marked (): hips (xy)
1. The notcher has a notch of n spaces.
2. The notcher has a left leg of length a and a right leg of length b. The legs are the (black) stones on the second line.
3. The notcher has a degree of weakness, x on the left hip and y on the right hip:
• The letter W means that the hip is white (weak or an (angle) wedge),
• N means that the stone on the hip is absent (neutral),
• while S means that it is black (solid or strong).
We put this information into a code nabxy, where n, a, b take numerical values, and x and y can be either W, N or S.
The status of the notcher assumes no outside liberties and no escape threat (e.g. a descent that threatens to live and to escape). That would have to be covered by a second status in parentheses or the like.
The code is leg-normal if
• a < b or
• a = b and x <= y (W < N < S: weaker before stronger)
It is hip-normal if
• x < y or
• x = y and a <= b
Use leg-normal if you put up a page. Use hip-normal if you search in the Table of Notchers (that gives tables of reasonable size). A flip might be necessary: an exchange of a and b as well as x and y.
See Table of Notchers for lots of examples.
Notcher Code last edited by RobertPauli on January 20, 2018 - 15:16 | 414 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-24 | longest | en | 0.895842 |
7.ksjo.pl | 1,561,403,694,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00228.warc.gz | 3,571,383 | 6,015 | ## Voltage meter diagram
By far the most commonly used meter in the United States is the form 2s meter. Here is a form 2s meter wiring diagram. I also want to offer some notes about the form 2s service here. One of the
questions that I often get is about how to wire a form 9s meter. I get this question so often I put up a form 9s meter wiring diagram. 7805 Voltage Regulator IC is a member of 78xx series of fixed linear voltage regulator ICs. Description, Pinout and PDF Datasheet of 7805 IC. A voltmeter is an instrument used for measuring
electrical potential difference between two points in an electric circuit. Analog voltmeters move a pointer across a scale in proportion to the voltage of the circuit; digital voltmeters give a numerical display of voltage by use of an analog to digital converter. A voltmeter in a circuit diagram is represented by the letter V in a circle. Multimeters can be a key tool in many of the electric tasks that you might need to do around the
house. They are used to measure current, voltage and resistance which … Installation Manual Three Phase Energy Meter For HXE310 CT & CTPT Meter Hexing Electrical Co., Ltd. www.hxgroup.cn [2013.3] The disadvantages are that a bias voltage is required because the gate threshold voltage can be anywhere between 2 and 4 volts for any power mosfet. Here this circuit diagram is for +12V regulated (fixed voltage) DC power supply.This power supply circuit diagram is ideal for an average current requirement of 1Amp.
This circuit is based on IC LM7812.It is a 3-terminal (+ve) voltage regulator IC. It has short circuit protection , thermal overload protection. An electricity meter, electric meter, electrical meter, or energy meter is a device that
measures the amount of electric energy consumed by a residence, a business, or an electrically powered device. Electric utilities use electric meters installed at customers' premises for billing purposes.They are typically calibrated in billing units, the most common one being the kilowatt hour (kWh). Here this circuit diagram is for +12V regulated (fixed voltage) DC power supply.This power supply circuit diagram is ideal for an average current requirement of 1Amp. This circuit is based on IC LM7812.It is a 3-terminal (+ve) voltage regulator IC. It has short circuit protection , thermal overload protection.
Rated 4.8 / 5 based on 500 reviews. | 530 | 2,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-26 | longest | en | 0.906018 |
https://www.teacherspayteachers.com/Product/Place-Value-Digital-Activities-and-Quiz-for-Google-Classroom-4696632 | 1,566,200,345,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00064.warc.gz | 999,704,338 | 22,355 | # Place Value Digital Activities and Quiz for Google Classroom™
Subject
Resource Type
Common Core Standards
Product Rating
4.0
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12 Slides + 10 Question Quiz
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Product Description
This hands-on digital activity provides step by step review lessons for place value. It's perfect to use as a whole group interactive whiteboard activity, or in guided math small groups. Students will love the interactive, movable pieces as they learn about reading large numbers, standard and word form, understanding the value of a digit, expanded form and expanded notation, and pictorial representation of numbers using place value disks and base 10 blocks.
Check out the PREVIEW to see a more detailed view!
This is a DIGITAL RESOURCE for Google Classroom™. You and your students will need a Google Driveaccount to use this resource.
Skills covered in this product include:
• word form
• standard form
• value of a digit
• expanded form
• expanded notation
• pictorial number representation
This product is aligned to:
TEKS 3.2A - compose and decompose numbers up to 100,000 as a sum of so many ten thousands, so many thousands, so many hundreds, so many tens, and so many ones using objects, pictorial models, and numbers, including expanded notation as appropriate.
CCSS 4.NBT.A.2 - Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form.
Included in this resource:
• Teacher Notes and Tips (1 page)
• 12 Google Slides™ with guided lessons and practice activities
• Printable Student Key Cards for students to check their own work (4 pages)
• 10 Question Quiz using Google Forms™
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 668 | 2,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-35 | latest | en | 0.728153 |
http://www.heiqigong.com/academy/lesson/valuating-variable-annuity-contracts.html | 1,582,779,636,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00389.warc.gz | 199,089,996 | 34,384 | Valuating Variable Annuity Contracts
Instructor: Bianca Ince
Bianca has FINRA Series 7, 63, SIE licenses and has licensing program at her firm for 5+ years.
Variable annuities are an attractive option for investors hoping to maximize their growth potential before retirement. During this lesson, we will review how the value of these contracts is determined.
Variable Annuity Phases
Carl is an investor thinking ahead. How can he make sure to maximize his growth potential before retirement?
Variable annuities are retirement vehicles in which the policyholder invests money for a period of time and then receives payments during retirement. The period in which Carl invests his premiums is called the accumulation phase. The annuitinization phase begins once he elects to start receiving payments.
During the accumulation phase, the premiums Carl makes are used to purchase accumulation units of the investments he chooses. The accumulation unit measures the value of a contribution to the investment selected. The number of accumulation units purchased, similar to shares of stock, will fluctuate with the market.
Therefore, during a downmarket, a contribution will purchase more shares of an investment than it would when market prices are up. The purpose of the accumulation phase is to boost the value of the separate account. Therefore, any dividends received are reinvested to purchase more accumulation units of that investment. Carl can elect to change investments according to his investment objectives and/or risk tolerance.
Calculating Accumulation Unit Value
So, how much ownership of an investment can Carl purchase with the premium amount? The accumulation unit value (AUV) can be calculated by dividing the total value of the investment by the number of accumulation units outstanding (owned).
Say the total value of Investment Pool XYZ is \$3 million, with 150k units outstanding:
• investment value / # units outstanding = AUV
• \$3,000,000 / 150,000 = \$20
Now the number of units a premium can purchase can be calculated by dividing the premium amount by AUV.
Carl should keep in mind that these figures will all fluctuate with the market. For example, say the market took a downturn, and the value of Investment XYZ decreased by half to \$1.5 million. The AUV would decrease by half as well to \$10, but the total units purchased with a \$500 premium would increase to 50 units. Since the AUV is 50% less, the same premium amount can purchase double the units.
Due to fluctuation within the market, the overall value of the separate account will consistently change. However, the value of the Carl's investments and the overall value of the separate account should also increase. This is the primary advantage variable annuities have over other types of annuity products.
Surrender Value
Variable annuities, like other retirement vehicles, are intended to be long-term investments. As a result, insurance companies deter annuitants from canceling their contracts with a surrender charge, typically a percentage of the cash value (the original amount paid for the contract).
During the surrender period, typically the first 10-15 years of the contract, Carl will receive only the cash surrender value if he cancels, which is the cash value after the surrender charge has been subtracted.
Say Carl decides to cancel his variable annuity contract during the surrender period. The cash value of the contract is \$20k, and the insurance company will access a 25% surrender charge upon cancellation. To calculate the surrender cash value he would receive:
First, determine the dollar amount of the surrender charge
\$20,000 * 25% = \$5,000. This is the dollar amount of surrender charge.
Next, subtract the amount of surrender charge from the cash value of the policy.
\$20,000 - \$5,000 = \$15,000. This is the surrender cash value annuitant will receive upon cancellation.
After the surrender period ends, the annuitant would not incur a surrender charge for canceling the policy. Therefore, the cash value amount and the surrender cash value would be the same.
Annuitization Phase
When Carl decides to begin receiving annuity payments, the annuitization phase begins. This could be when he retires or before (under certain circumstances). During the annuitization phase, the accumulation units in the separate account will be converted into annuity units. The annuity units will then be sold in the market depending on the annuity payout option selected.
If a fixed payout is selected, the amount of each payout is the same regardless of market performance. Variable payments, however, would change according to market fluctuation.
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https://www.ukessays.com/essays/engineering/computational-fluid-dynamic-analysis-of-race-car-wings-engineering-essay.php | 1,481,012,298,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00272-ip-10-31-129-80.ec2.internal.warc.gz | 1,053,379,404 | 19,781 | # Computational Fluid Dynamic Analysis Of Race Car Wings Engineering Essay
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This paper is to identify drag and lift coefficient of front and read wings of a race car with different angles of attack. Process simulation and analysis for both the model design was conducted with computer-aided drawing software and analyzed using software COSMOSFloworks to perform a turbulent simulation of the airflow on the front and rear wings of a race car with different angles of attack. Results are presented graphically, showing lift to drag ratio, lift (CL) and drag coefficients (CD) for the different cases. The aerofoils were tested at angles of attack ranging from 0°to 20°.The purpose of study is to find the suitable profile of aerofoil to generate the required lift force or drag force to the race car. The results were used in a simulation of a typical Formula SAE race car. It was shown that an angle of 16° below the horizontal seems to indicate stalling conditions except for the Formula One race car. Furthermore, the highest lift-to-drag ratio of front wing is found to be Formula Three race car with angles of attack of 0°meanwhile the highest lift-to-drag ratio of rear wing also found to be Formula One race car with angle of attack of 5°.
Keywords
CFD; Race Car; k-ε turbulent modeling; Aerofoil; COSMOS FloWorks 2008
## Overview
First racing cars were primarily designed to achieve high top speeds and the main goal was to minimize the air drag. But at high speeds, cars developed lift forces, which affected the stability. In order to improve their stability and handling, engineers mounted inverted wings proï¬les to generating negative lift [13]. In open-wheel racing series such as Formula One, a front wing is inverted to produce downforce, that is, negative lift, leading to an enhancement of traction and cornering ability of cars [14]. The main function of the front wing is to generate downforce on the front-axis. In interaction with the bargeboards and the aerodynamic at the rear of the car, it is crucial for the performance of the car[8].The front wing is operated in close proximity to a solid boundary, known as the ground effect regime, where different flow features are exhibited, compared with the free stream condition. Although wind tunnel testing remains a signiï¬cant tool for aerodynamic development, computational fluid dynamics(CFD) plays an important role because of its efï¬cient cost performance compared with wind tunnel testing , and the detailed flow information that is available[14]. Drag reduction is not commonly the main target of top race car aerodynamic optimisation but it is still an important factor for low power vehicles (F3, electric/solar cars)[4].
## Objectives
The objectives of this project is to design three types of both wings for race car and to analyze aerodynamic characteristics of the wings on a race car by using software Computational Fluid Dynamics (CFD) software. Then, coefficient of drag and coefficient of lift for all the wings will be evaluated to figure out which profile of aerofoil is the best in term of lift-to-drag ratio. Lastly, the selected profile of aerofoil will be improved by carry out the new design of wing for race car.
## Scope
In any investigation, there are always different approaches that can be taken. Numerical, analytical and experimental approaches all have their advantages and disadvantages. CFD initially appeared suited to this investigation owing to the large number of tests. The scope of this investigation is limited to a 2-D study shape to eliminate the vortex phenomena at the tip of aerofoil. In reality, the wings would be exposed to three dimensional (3-D) flows; however this would increase the number of variables to be explored greatly [7]. Race car typically use multi-element wings[2]. The multi-element design allows for much higher lift than is possible from a single element wing of similar dimensions. This investigation has been limited to the effects on a single element wing. A benefit of the single element is the simplicity of the design. The mechanism to actively control a single element wing would be a lot simpler than that for a multi-element, thus easier to implement in both FSAE and production vehicles. The angles of attack of wings this investigation is limited in the range of 0° to 20°.
## The Physical Model for Airfoil
A schematic configuration and a coordinate system of airfoils are shown in Fig.2.1a and Fig.2.1b.
Figure 2.1a- Airfoil and its components
Figure 2.1b- Airfoil nomenclature
From both figures above, the chord line is a line connecting leading a trailing edge. The chord length is the distance from the leading to the trailing edge, measured along the chord line. The camber is the maximum distance between mean camber line and chord line. The thickness is the distance between the upper and lower surfaces [13].The flow around an airfoil is assumed to be turbulent and steady state with incompressible fluid [14]. The amount of lift L produced by the airfoil, can be expressed in term of lift coefficient CL
(1)
Where V denotes the free stream velocity, Ï is fluid density and A is the airfoil area [13].
Besides that, the drag coefficient CD for vehicle body can define as: [6]
(2)
Where D is the drag and A is the airfoil area (A=b*c) [9], b and c are wingspan and chord length respectively [16].
## Ground Effect
The Wings as downforce generating aerodynamic devices appeared in the 1960s. By 1970, the rear wing was placed at the rear of the car, behind and above the rear wheels, and the front wing in front of the front wheels in ground effect. This basic arrangement of the front and rear wings has remained the same since then [15]. The main difference between wing application in aviation and car racing is that cars are in contact with the ground. Therefore, wing experiences some additional effects due to ground proximity [13]. With the airfoil close to the ground, the velocity on the upper surface is slower than for the airfoil in free air. On the lower surface of the wing, the velocity is higher for the airfoil in ground effect than the one in free air using the principle of continuity to maintain the same flow under the wing downstream as the wing upstream [3].
## Lift-to-Drag Ratio
(3)The lift-to-drag ratio is the amount of lift generated by a wing divided by the drag it creates by moving through the air.
## Numerical Method for Flow Analysis
The numerical model was set up and run using the COSMOSFloworks Computational Fluid Dynamics (CFD) code. Due To the assumption of isothermal flows and no heat transfer, the energy equation was not introduced [3]. The front wing is basically attached to the sides of the body of the race car, while the back wing is attached using a strut to the body of the car. These would normally introduce some 3-D effects into the flow patterns developed around these wings .It is expected, though, that the results of the 2-D simulation will be uniformly affected by these construction details. However, the relative results and trends for the different simulations are expected to stay relatively the same[3].Any end effects for either wings, such as vortex generation, are assumed to be relatively small due to the relatively small Reynolds number of the flow. The simulation here only concentrate on effect of the angle of attack on the front and rear wing. For conducting flow analysis in FloWorks we need to know the airspeed. Reynolds number and airspeed are related by following equation:[16]
(4)
Where Ï is fluid density, V is horizontal component of airspeed , c is chord length and is fluid viscosity.
## Methodology
This chapter explains the designs and simulations.
## Conceptual Design of Formula Mazda Race Car
With the airfoil curve dimension given [3], the modified airfoil curve coordinates was obtained from the software JAVAFOIL. A 3D front wing model was designed in SolidWorks with 381.85mm chord and 1400mm span is shown in Figure 3.1a.
## .
Figure 3.1a-Drawing of front wing (airfoil)
Figure 3.1b-Drawing of rear wing (airfoil)
The Figure 3.1b shows the drawing of rear wing designed by SolidWorks with the chord length is 452.14mm and span of the wing is 1300mm.
## Conceptual Design of Formula One Race Car
With the airfoil curve coordinates given [12]. A 3D front wing model was designed with some modification (The trailing edge given was not closed) in SolidWorks with 380.19mm chord and 1400mm span is shown in Figure 3.2a. The airfoil curve coordinates of rear wing is designed using JAVAFOIL with the information given [11].The profile of rear wing is based on NACA 2104 which shown in Figure 3.2b.
Figure 3.2a- Drawing of front wing (airfoil)
Figure 3.2b- Drawing of rear wing (airfoil)
## Conceptual Design of Formula Three Race Car
With the airfoil curve coordinates given [12]. A 3D front wing model was designed in SolidWorks with 380.04mm chord and 1400mm span is shown in Figure 3.3a. The airfoil curve coordinates of rear wing is designed using JAVAFOIL with the information given [11].The profile of rear wing is based on NACA 2312 which shown in Figure 3.3b.
Figure 3.3a- Drawing of front wing (airfoil)
Figure 3.3b- Drawing of rear wing (airfoil)
## Flow Analysis and Simulation in FloWorks
We considered using 2D flow analysis vs. 3D flow analysis over the surface of the wing model designed in SolidWorks[1]. In 3D flow analysis, the fluid dynamics computations are performed over the entire region of the solid, whereas, in 2D flow analysis the computations are performed in the cross sectional area of the solid. The various factors that came into our consideration for evaluating 2D flow analysis vs. 3D flow analysis were: time complexity and memory requirements for 3D computational analysis, practicalities involved in extrapolating 3D results from a series of 2D analysis, and the goal of developing a general method that could be used in subsequent research on other wing models [16]. The Reynolds number of 1.4 million was used for front wing of race car which is based on the aerofoil chord length of 0.38m, design speed of 58m/s, density of fluid is 1.2041kg/m3 and the fluid viscosity is 1.84e-5Pa.s. For the rear wing of race car, the Reynolds number of 1.7 million was used which is based on the aerofoil chord length of 0.45m.
## Finish
A model of the airfoil was created in SolidWorks and the part was then exported to CosmosFloWorks. First, the coordinates for the airfoil of race car were imported into SolidWorks in the form of text file. Next, the airfoil profile was extruded. A trimetric view of the finished wing section in SolidWorks is shown in Figure 3.5.
Figure 3.5- The finished model of the wing section in SolidWorks
Using the CosmosFloWorks wizard, the SI unit system was first chosen followed by the choice of the external analysis type option. Next, air was chosen as the default fluid. The size of the computational domain in the stream wise direction was 0.368 m in front of the leading edge and 0.585 m after the trailing edge. In order to get a reasonable calculation time, a 2D plane steady flow calculation was selected. A free stream velocity of 58 m/s, a wall surface roughness of 0 micrometer and a turbulence intensity of 0.5% were chosen for the settings following the CosmosFloWorks wizard [5]. Lastly, two global goals and 1 surface goal need to be specified. X-component of Force is defined as aerodynamic drag force meanwhile the Y-component of the force represents the lift force. The surface goal used as criterion for stopping the flow analysis process by selecting the lower surface of airfoil and parameter is average total pressure.
## Results & Discussions
The results of the computer simulations were only consider for free air without ground effect and compiled and plotted graphically. The lift coefficient and drag coefficient is plotted against the angles of attack. In the same manner, the lift-to-drag ratio was plotted against the angles of attack. These result maybe will revised later for improving the accuracy of the result.
## Simulation 1 - Formula Mazda Race Car
Table 4.1a- Simulation result for front wing
## L/D Ratio
0
807.422
101.695
0.678
0.085
7.940
4
1108.052
147.735
0.931
0.124
7.500
5
1268.124
174.324
1.065
0.146
7.275
8
1405.842
226.975
1.181
0.191
6.194
10
1627.157
279.311
1.367
0.235
5.826
12
1709.297
326.045
1.436
0.274
5.243
15
1806.083
396.969
1.517
0.333
4.550
16
1942.339
429.391
1.632
0.361
4.523
20
1719.125
509.954
1.444
0.428
3.371
Table 4.1b- Simulation result for rear wing
## L/D Ratio
0
441.399
69.959
0.408
0.065
6.309
4
685.813
101.324
0.633
0.094
6.769
5
760.236
106.275
0.702
0.098
7.153
8
887.205
151.066
0.819
0.140
5.873
10
1003.711
183.783
0.927
0.170
5.461
12
1084.839
218.815
1.002
0.202
4.958
15
1198.604
270.751
1.107
0.250
4.427
16
1223.307
278.042
1.130
0.257
4.400
20
1009.002
375.431
0.932
0.347
2.688
Figure 4.1a- Front wing variation of the CD and CL as a function of AOA.
Figure 4.1b- Rear wing variation of the CD and CL as a function of AOA
From the Table 4.1a and Table 4.1b, the simulation result shows that the airfoil is beginning to approach a "stall condition" at 16° angle of attack as the critical angle of attack was achieved [16]. At a certain angle of attack there is no longer a smooth flow of air over the lower surface of a wing. At this point the lift produced by the wing is no longer sufficient to support the weight of the race car and so the race car is said to be in a stall condition. This angle is called the critical angle of attack. As we go on increasing the angle of attack beyond this critical point there is a sudden drop in lift force (the wing is stalled)as shown in Figure 4.1a and Figure 4.1b. So we can say that for a given airfoil the maximum lift is produced at the critical angle of attack.
## Simulation 2 - Formula One Race Car
Table 4.2a- Simulation result for front wing
## L/D Ratio
0
255.179
48.621
0.236
0.045
5.248
4
447.884
53.484
0.414
0.049
8.374
5
500.979
61.587
0.463
0.057
8.135
8
669.044
83.715
0.618
0.077
7.992
10
742.107
108.065
0.685
0.100
6.867
12
755.574
122.088
0.698
0.113
6.189
15
981.181
174.258
0.906
0.161
5.631
16
1016.044
191.147
0.938
0.177
5.316
20
1172.827
251.666
1.083
0.232
4.660Table 4.2b- Simulation result for rear wing
## L/D Ratio
0
147.106
18.849
0.125
0.016
7.805
4
425.351
31.452
0.361
0.027
13.524
5
494.393
31.396
0.419
0.027
15.747
8
694.763
65.524
0.589
0.056
10.603
10
824.518
95.021
0.699
0.081
8.677
12
880.392
132.412
0.747
0.112
6.649
15
992.648
203.268
0.842
0.172
4.883
16
979.813
212.350
0.831
0.180
4.614
20
1033.540
333.587
0.877
0.283
3.098Figure 4.2a- Front wing variation of the CD and CL as a function of AOA.
Figure 4.2b- Rear wing variation of the CD and CL as a function of AOA.
From the Table 4.2a and Figure 4.2b, the lift force still increasing when the angle of attack increased. This phenomenon indicates that the critical attack of angle of both wings of formula one race car is more than 20° angle of attack. The critical angle is dependent upon the profile of the wing, its planform, its aspect ratio, and other factors, but is typically in the range of 8 to 20 degrees relative to the incoming wind for most subsonic airfoils. The graph for lift coefficient vs. angle of attack follows the same general shape for all airfoils, but the particular numbers will vary. The rear wing of Formula One race car with 5° angle of attack is achieve the highest lift-to-drag ratio among all the race cars. From the Figure 4.2a and Figure 4.2b show the airfoil is not yet begin to reach the "stall" condition where the angle of attack increases beyond a certain point such that the lift begins to decrease.
## Simulation 3 - Formula Three Race Car
Table 4.3a- Simulation result for front wing
## L/D Ratio
0
377.748
43.199
0.351
0.040
8.744
4
548.895
78.886
0.509
0.073
6.958
5
596.547
90.630
0.554
0.084
6.582
8
713.457
128.911
0.662
0.120
5.534
10
793.868
148.813
0.737
0.138
5.335
12
860.532
183.200
0.799
0.170
4.697
15
972.378
237.509
0.902
0.220
4.094
16
968.535
259.049
0.900
0.240
3.739
20
686.048
328.960
0.637
0.305
2.086 Table 4.3b- Simulation result for rear wing
## L/D Ratio
0
164.329
21.574
0.139
0.018
7.617
4
431.943
39.280
0.366
0.033
10.996
5
506.952
44.851
0.430
0.038
11.303
8
719.325
76.319
0.610
0.065
9.425
10
864.084
97.833
0.733
0.083
8.832
12
979.200
135.459
0.831
0.115
7.229
15
1043.234
186.148
0.885
0.158
5.604
16
1139.602
215.209
0.967
0.183
5.295
20
872.240
315.443
0.740
0.268
2.765
Figure 4.3a- Front wing variation of the CD and CL as a function of AOA
Figure 4.3b- Rear wing variation of the CD and CL as a function of AOA.
From the Table 4.3a and Table 4.3b, the simulation result shows that the airfoil is beginning to approach a "stall condition" at 16° angle of attack as the critical angle of attack was achieved. The process of flow detaching from the surface often happens instantaneously when the angle of attack is increased, making the loss of lift rather sudden and dangerous-this is called stall [10]. The plot of lift coefficient vs. angle of attack is significant to us because it gives the critical angle of attack (the angle of attack at which lift coefficient is maximum) for the particular airfoil as shown in Figure 4.3a and Figure 4.3b. For each aerofoil, the highly cambered side which features accelerated flow (and associated suction force) is called the 'suction side' of the aerofoil. This corresponds to the bottom side of the inverted aerofoil. The other, less cambered side features slower flow and, usually, a positive pressure force. This side is called the 'pressure side' and is the top side of the inverted aerofoil. the airfoil. This low velocity region extends downstream of the trailing edge into the wake of the airfoil. In addition, there is a marked decrease in CL by about 23%-32%, which may indicate that between 16° and 20° angle of attack there is a potential for a "stall" condition with the airfoil.
## Comparison of Simulation Result
Figure 4.4a- Lift-to-Drag Ratio for all race cars (front wing)
Figure 4.4b- Lift-to-Drag Ratio for all race cars (rear wing)
The Figure 4.4a and Figure 4.4b show the variation of the lift-to-drag ratio for various angles of attack of the wings. The highest lift-to-drag ratio for front wing occurs at 0â-¦ (Formula Three) meanwhile the highest lift-to-drag ratio for rear wing occurs at 5â-¦ (Formula One).
## Conclusion & Recommendation
After simulation of all the three airfoils, we could conclude that the front wing of Formula Three race car and the rear wing of Formula One race car, which was giving a lift-to-drag ratio of 8.744 and 15.747 respectively, will be the better choice which could be used for the airfoil for the race car. For both airfoils, the hydrodynamic performance of the foils is signiï¬cantly affected by the angle of attack. If the profiles tested are judged according to the performance, the wing profile of Formula One race car is found to be the optimum in term of lift-to-drag ratio.
## Recommendation
The lift-to-drag ratio can be improved by utilizes the aspect ratio and the cross sectional area of wing. As we know, the wing span and the chord length will influence the aspect ratio of the wing. Future work is suggested to perform a new design improvement by changing the aspect ratio of the wing for the selected wings of race car. Thus, the lift-to-drag ratio of the new design of the race car wings should be higher than the current design. | 5,306 | 20,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-50 | longest | en | 0.953885 |
https://www.horiba.com/scientific/products/particle-characterization/education/la-960/method-development/simplify-choosing-a-refractive-index-with-the-method-expert/?L=846 | 1,585,980,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370520039.50/warc/CC-MAIN-20200404042338-20200404072338-00017.warc.gz | 944,821,096 | 12,189 | ## Choosing a Refractive Index
The accuracy of your laser diffraction particle size measurement depends by varying degrees upon the accuracy of your choice of refractive index. How to identify good refractive index from bad is one of the most common questions (and most common headaches) of the typical laser diffraction user.
## Why is Refractive Index Important?
One of the key factors affecting the accuracy of many laser diffraction particle size measurements is the choice of refractive index (RI). Consisting of real and imaginary components the RI describes how light interacts with a material. The real component is often either listed in the software library, located through a literature or internet search, or can be directly measured. The imaginary component, or i term is a value between 0.0 and 10.0 that correlates to the degree of difference between a transparent and opaque particle. The i term is sometimes easy to deduce - for example samples that are transparent and spherical and therefore have an imaginary value of zero. For non-transparent particles the i term is a non-trivial selection and directly affects result accuracy. The LA-960 Method Expert software provides a structured, automated approach with expert advice to choose the optimum i term.
Download the White Paper about the revolutionary Method Expert software (You need to be logged in)
The LA-960 Method Expert: Guided, Automated Method Development Software
## Choosing the Optimum Imaginary Component
The raw data for the experiment is the scattered light measured on the instrument detectors. This scattered light is then used to calculate the particle size distribution (PSD). An error calculation called the R parameter quantifies the error when converting from the raw data to the PSD. This error calculation can be used to select the optimum i term; the lower the R parameter, the better the calculation (result). The i term generating the lowest R parameter is then used for all future measurements of this sample.
## Automation by Method Expert
The sample is measured once and then this raw data is used for optimization in the Method Expert. The real component of RI is kept fixed and the i term is varied as determined by the user – typically 0, 0.01, 0.1, 1, and 10. A screen shots from the wizard set up is shown below.
Click image to enlarge: Setup screen for the Imaginary RI Wizard
After the calculations are completed a summary report is generated the displays the following information as seen below:
• PSD for each calculation
• D90, D50, D10 vs. RI
• R parameter vs. RI
Click image to enlarge: Result screen for the Imaginary RI Wizard
For this and every test the Method Expert provides the user with Expert Advice containing guidance on how to make the best possible choice.
Click image to enlarge: Expert Advice for Imaginary RI Wizard
In this case the user should select 1.0 as the best i term for this sample and can choose from a number of options to report the results.
The same wizard can also be used to select the optimum real RI component while keeping the i term fixed. This approach works for the majority of samples, but the user must still also decide that the result generated meets expectations or correlates to another technique or product performance.
The user can then perform the other automated tests within the Method Expert software including structured tests to select the optimum: | 688 | 3,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-16 | latest | en | 0.844142 |
https://www.mrexcel.com/board/threads/formula-help-please.273093/ | 1,725,708,054,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00456.warc.gz | 865,662,965 | 22,761 | #### CwBoy1963
##### New Member
I hope I can explain this well enough, here goes......I have a spreadsheet divided out by different tabs and on one page it takes various data from these tabs and gives final numbers for a particular month. Here is one of my formulas I am having issues with: =IF('Inspections-Page 3'!\$E\$5:\$E\$69="Carson", COUNTIF('Inspections-Page 3'!\$G\$5:\$G\$69, "H"),0)
Here is what I am trying to get the formula to do, Find all of the cells labled Carson and if they have an 'H' in the corresponding cells then count these and put the number here. What ends up happening when I use the above formula is it ignores the IF statement and only does the COUNTIF of all of the 'H' which is not what I want. I think I am missing maybe a THEN statement or something else in the middle??? Can someone solve this for me? Thank you!!!
### Excel Facts
Easy bullets in Excel
If you have a numeric keypad, press Alt+7 on numeric keypad to type a bullet in Excel.
One way is with an array formula:
=SUM(('Inspections-Page 3'!\$E\$5:\$E\$69="Carson")*('Inspections-Page 3'!\$G\$5:\$G\$69="H"))
Confirmed with Control + Shift + Enter
I'll bet there are other ways to do this too. Regards
Thank you, but that did not work. I thought * was mutiply? How do you multiply words with out telling the formula that you are counting the words within the box and then making them into a number? I am confused!!
Hmmm...
If the Cell in Column 1 is Carson (TRUE) then the true will evaluate to 1, and if the Cell in Column 2 is H (TRUE) then the true will evaluate to 1, so:
1 * 1 = 1,
You entered the formula with Ctrl + Shift + Enter, correct? Not just enter.
Let me think here...
Edit: note, FALSE is 0, so anything times 0 is 0 ... hence, not counted.
Try:
=SUMPRODUCT(('Inspections-Page 3'!\$E\$5:\$E\$69="Carson")*('Inspections-Page 3'!\$E\$5:\$E\$69="H"))
This is not an array formula, so just Enter when you put the formula in.
Alex,
That did not work either..hmmmm. Can I send the spreadsheet to you possibly and you look at it? I did not do the spreadsheet myself, I am just working on it for a coworker. It was done in Excel 2002 and I moved it to my computer with Excel 2003 on Windows 2000. We work in the State of Nevada Fire Marshal Office and this spreadsheet tabulates statistics of HAZMAT and Fire & Life Safety Inspections done throughout the State. I have appointments all day today and tomorrow so if I dont get in touch with you until Monday, you know why. I appreciate any help you can render. Thank you for your efforts thus far.
CwBoy1963 - a quick and dirty solution to this would be to set up a helper column which concatenates Cols E and G like this -
Code:
``=E1&G1``
Then do your countif on this new column, counting instances of "CarsonH".
Alex's array formula does work. I built a test sheet to try it out, resluts in the photo:
I copied the text of his array formula, pasted it into a cell. Then instead of hitting enter or just clicking away, you must hit CTRL+SHIFT+ENTER. This gives you an array formula instead of a normal formula.
From your description of the application, I made a guess that your list will be various inspections for various counties in Nevada. I generated a list of counties at random in the range \$E\$5:\$E\$69, and a random list of made-up inspection types in \$G\$5:\$G\$69. I made sure Carson had at least one "H" type inspection. Then I applied the array formula to my invented list, and it worked. I then altered the array to look up the county and inspection type, again using the CTRL+SHIFT+ENTER key combination. I then copied across and down by dragging the corner of the cell like you would for any normal formula.
If you send me your email addres, I can send you the sheet.
Thank you, I will try it all out on Monday. Do you have to turn on the Shift+Ctrl+Enter for an array, it does not seem to matter if I hit just enter or the combo, it still comes back as an error.
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Go back | 1,318 | 5,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.909559 |
www.1sthealthinsurancequotes.com | 1,516,581,373,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00453.warc.gz | 390,905,571 | 5,344 | # What is an Actuary?
In a nutshell, the actuary is the person or group of persons who works with complicated formulas to determine what premium your life or health insurance company needs to charge.
Actuarial science is a complex mathematical science that dates back to 17th century when mathematicians and scientists in several European countries became fascinated by the scientific aspects of probability. Prior to this time, some groups existed that attempted to provide some form of life insurance, burial expense provision, or life annuities—an early attempt to create a retirement plan. Deciding what premium a person needed to pay was extremely difficult; most groups failed and fell into obscurity in just a few years.
Prior to the development of computers, actuaries had to use long formulas and pages of tables to calculate the likelihood of any single event happening to a specific group of people. While it was not possible to pinpoint a particular person for an event—such as death or catastrophic illness, the 20th century brought the study of probability, statistics, economics, and financial theory together to form a precise and accurate actuarial science.
The calculations of the actuaries thus allow an insurance company to predict, with surprising accuracy, the likelihood of the occurrence heart disease, diabetes, stroke, cancer, Alzheimer's Disease, and many other chronic ailments among a particular group of people who have something in common—such as living in a certain area, or having a family history of illness.
Today's actuaries have an even more challenging job than those of the 19th and 20th centuries due to cultural changes, a global interaction among people who at one time would have had no chance of contact, and the extreme variation in life styles. Fortunately for them, calculations that once would have taken months to work out can now be accomplished in seconds with the aid of computers and calculators.
The work of an actuary is beneficial to both the consumer and the company, although it may seem to be more for the protection of a company. The ability to accurately predict the likelihood of a particular health event occurring within a certain group (called cohort) of people is supposed to ensure that premiums will be assessed fairly based on the risk to the company. Companies do, however, have the ability to select their own target groups and establish premiums based on the actuarial tables specific to that group. Thus, a group of seniors in a particular geographic area will be more likely to have heart disease than a similar age group in a different geographic area. These differences lead to widely differing premiums in different parts of the country. If they chose a different grouping method—such as economic status or education without regard to age, for example—the necessary premiums would be radically different.
Modern companies use the finely tuned actuarial calculations to both avoid undesirable risk, determine the amount of money they must keep available for paying claims, and to estimate the profits that will be available to share holders. Consequently, fewer health insurance companies go bankrupt today, and few actuaries have to worry about being out of work. | 612 | 3,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-05 | latest | en | 0.963168 |
http://pvs.csl.sri.com/mail-archive/pvs-help/msg00612.html | 1,493,004,851,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118963.4/warc/CC-MAIN-20170423031158-00468-ip-10-145-167-34.ec2.internal.warc.gz | 312,630,963 | 2,414 | # Re: "<" on a subset of nat is well_founded
```Mauro Gargano wrote:
> Hi everybody,
>
> I want to do an induction proof starting from the index "i!1+2" so
> I used the induction scheme (induct "j" :name
> "wf_induction[upfrom(i!1+2),<]")
>
> but I cannot get rid of the subgoal in which PVS asks me to show that
> the "<" is wellfounded on the set upfrom(i!1 + 2):
>
> |-------
>
> [1] well_founded?[upfrom(i!1 + 2)]
>
> (restrict[[real, real], [upfrom(i!1 + 2), upfrom(i!1 + 2)],
>
> boolean]
>
> (<))
>
>
> How can I instantiate properly the wf_nat axiom?
>
>
>
>
> Mauro.
>
>
Hi,
To prove this subgoal, you can expand "well_founded?" and "restrict".
This gives:
|-------
{1} FORALL (p: pred[upfrom(2 + i!1)]):
(EXISTS (y: upfrom(2 + i!1)): p(y)) IMPLIES
(EXISTS (y: (p)): FORALL (x: (p)): (NOT (x < y)))
Then (skosimp*) will introduce predicate "p!1" and you can
finish the proof by instantiating the wf_nat axiom with
"{ n: nat | n>=2+i!1 AND p!1(n) }".
But there's an easier way of doing induction from "i!1+2",
by using the lemmas from theory "bounded_int_inductions" in
the prelude. They exist in two forms:
upfrom_induction: LEMMA
(pf(m) AND (FORALL jf: pf(jf) IMPLIES pf(jf + 1)))
IMPLIES (FORALL jf: pf(jf))
UPFROM_induction: LEMMA
(FORALL jf: (FORALL kf: kf < jf IMPLIES pf(kf)) IMPLIES pf(jf))
IMPLIES (FORALL jf: pf(jf))
You can do either (induct "j" :name "upfrom_induction[i!1+2]")
or (induct "j" :name "UPFROM_induction[i!1+2]") depending
on which form you want to use.
Note that (induct "j") is often all you need. PVS will
choose an induction lemma based on the type of "j".
For example, if "j" is of type "upfrom(i!1+2)"
then (induct "j") has the same effect as
(induct "j" :name "upfrom_induction[i!1+2]")
Bruno
``` | 624 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-17 | latest | en | 0.875517 |
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