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https://mathspace.co/textbooks/syllabuses/Syllabus-1014/topics/Topic-20183/subtopics/Subtopic-265933/?textbookIntroActiveTab=overview | 1,638,219,099,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00464.warc.gz | 462,765,649 | 46,479 | Victorian Curriculum Year 10A - 2020 Edition
5.04 Logarithms
Lesson
We've seen equations like $y=B^x$y=Bx before. It's straightforward enough to find $y$y when we know $x$x, but is it possible to find $x$x if we know $y$y?
The expression $B^x$Bx, if $x$x is a natural number, means the number of $B$B factors multiplied together is $x$x. So to find $x$x in $3^x=81$3x=81 we ask how many $3$3 factors are in $81$81, and the answer is $4$4. But we saw from exponential graphs that $x$x can in general be any real number, including irrational numbers. In that case it doesn't make sense to multiply $B$B $x$x times.
Logarithms are expressions of the form $\log_By$logBy, where $B$B is some number and $y$y is a pronumeral. $B$B is called the base of the logarithm. The definition of a logarithm is that if
$y=B^x$y=Bx
then
$\log_By=x$logBy=x
In other words, $\log_By$logBy is the number of $B$B factors that multiply together to make $y$y. It follows that $\log_381=4$log381=4.
Of course, the value of the logarithm could be any real number. We will soon see how to find the exact values of logarithms, but we can approximate the value using a calculator.
First note that by convention, if $B$B is not specified that means a base of $10$10. So $\log y=\log_10y$logy=log10y. If we wanted to find $\log81$log81, then we can press the "log" button on a calculator and then enter $81$81. This gives us $1.908$1.908 to three decimal places.
Summary
Logarithms are expressions of the form $\log_By$logBy, where $B$B is any number and $y$y is a pronumeral.
In $\log_By$logBy, $B$B is the base of the logarithm.
By convention, if the base is not specified then $B=10$B=10.
If $y=B^x$y=Bx then $\log_By=x$logBy=x, so $y$y is the number of $B$B factors that are multiplied together to give $x$x.
#### Practice questions
##### Question 1
Rewrite the equation $9^x=81$9x=81 in logarithmic form (with the index as the subject of the equation).
##### Question 2
Evaluate $\log_8\left(\frac{1}{64}\right)$log8(164).
##### Question 3
Evaluate $\log_{10}$log10$45$45. | 634 | 2,068 | {"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} | 4.71875 | 5 | CC-MAIN-2021-49 | latest | en | 0.851588 |
http://www.intmath.com/exponential-logarithmic-functions/6-logarithm-exponential-eqns.php | 1,498,491,733,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320823.40/warc/CC-MAIN-20170626152050-20170626172050-00025.warc.gz | 563,084,018 | 15,877 | 6. Exponential and Logarithmic Equations
by M. Bourne
Solving Exponential Equations using Logarithms
World Population
Don't miss the world population application below.
Go to World population.
The logarithm laws that we met earlier are particularly useful for solving equations that involve exponents.
Example 1
Solve the equation 3^x= 12.7.
Example 2
Two populations of bacteria are growing at different rates. Their populations at time t are given by 5^(t+2 and e2t respectively. At what time are the populations the same?
Continues below
Exercises
1. Solve 5^x= 0.3
2. Solve 3\ log(2x − 1) = 1.
3. Solve for x:
log_2 x + log_2 7 = log_2 21
4. Solve for x:
3\ ln\ 2+ln(x-1)=ln\ 24
I have the following formula:
S(n) = 5500\ log\ n + 15000 (Using base 10)
If I know S(n) = 40 million, How do I solve it?
Application - World population growth
The population of the earth is growing at approximately 1.3% per year. The population at the beginning of 2000 was just over 6 billion. After how many more years will the population double to 12 billion?
When the world population is 12 billion, the net number of people in the world will be increasing at the rate of about 5 per second, if the growth rate is still 1.3%. Currently, there are about 2.6 new people per second. However, the rate of growth is expected to drop considerably to about 0.5% within 50 years.
In 2001, the population of India passed one billion, making it the second country after China to reach that scary milestone.
World population
Current world population is approximately:
Interactive applet - World Population
Go to the interactive World Population, which has comparisons between present, past and future population growth.
Predicting world population
The following graph shows one of the estimates for world population growth during the 21st century. We see that the population will be 11 billion by about 2100! Think of our water quality, air pollution, global warming, social cohesion and lack of food. Surely this is one of the most important graphs in all of mathematics.
But I digress.
We are, of course, talking American English, here. The British billion has 12 zeroes (Well, even they have recently adopted the 9 zeroes billion...).
The world population is expected to exceed 11 billion by 2100. [Source]
This suggests a growth rate of about 0.6%, much lower than that experienced during the 20th century.
The equation for the above graph is
P=6.1(1.006)^(t-2000), where
6.1 billion was the population in 2000;
the growth rate is represented by 1+6/100 = 1.006; and
t is the time from the year 2000.
See a "live" world population estimation on the next page.
top
Online Algebra Solver
This algebra solver can solve a wide range of math problems.
Math Lessons on DVD
Easy to understand math lessons on DVD. See samples before you commit. | 719 | 2,864 | {"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} | 4.15625 | 4 | CC-MAIN-2017-26 | latest | en | 0.907366 |
http://piping-designer.com/index.php/mathematics/geometry/plane-geometry/2548-ellipse-sector | 1,590,394,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388012.14/warc/CC-MAIN-20200525063708-20200525093708-00328.warc.gz | 101,013,727 | 7,606 | # Ellipse Sector
Written by Jerry Ratzlaff on . Posted in Plane Geometry
• Ellipse sector (a two-dimensional figure) is a part of the interior of an ellipse having two radius boundries and an arc.
• Sector is a fraction of the area of a ellipse with a radius on each side and an edge.
• Major axis is always the longest axis in an ellipse.
• Minor axis is always the shortest axis in an ellipse.
• Semi-major axis is half of the longest axis of an ellipse.
• Semi-minor axis is half of the shortest axis of an ellipse.
## Formulas that use Ellipse Sector Area
$$\large{ A_{area} = \frac{a\;b}{2} \; \left( {\theta \;-\; atan\;\left[ \frac{ a\;-\;b \;sin\;\left(2\;\theta_1\right) }{ a\;+\;b\;+\;\left(a\;-\;b\right)\;cos\left(2\;\theta_2\right) } \right] \;+\; atan\;\left[ \frac{ a\;-\;b \;sin\;\left(2\;\theta_1\right) }{ a\;+\;b\;+\;\left(a\;-\;b\right)\;cos\;\left(2\;\theta_2\right) } \right] } \right) }$$
### Where:
$$\large{ A_{area} }$$ = area
$$\large{ \theta }$$ = angle
$$\large{ \theta_1 }$$ = angle
$$\large{ \theta_2 }$$ = angle
$$\large{ a }$$ = semi-major axis
$$\large{ b }$$ = semi-minor axis
## Formulas that use Ellipse Sector Radius
$$\large{ j = \sqrt{ \frac{a^2\;b^2}{a^2\;sin^2 \;\theta_1 \;+\; b^2\;cos^2 \;\theta_2} } }$$ $$\large{ k = \sqrt{ \frac{a^2\;b^2}{a^2\;sin^2 \; \theta_2 \;+\; b^2\;cos^2\;\theta_1} } }$$
### Where:
$$\large{ j }$$ = radius
$$\large{ k }$$ = radius
$$\large{ \theta_1 }$$ = angle
$$\large{ \theta_2 }$$ = angle
$$\large{ a }$$ = semi-major axis
$$\large{ b }$$ = semi-minor axis | 587 | 1,555 | {"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": 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} | 4.375 | 4 | CC-MAIN-2020-24 | latest | en | 0.48982 |
https://astronomy.stackexchange.com/questions/25156/would-a-spacecraft-just-go-through-a-gas-giant | 1,631,922,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00591.warc.gz | 173,659,398 | 40,380 | # Would a spacecraft just go "through" a gas giant?
From my understanding of the word gas giant, it is a planet composed of entirely a gaseous atmosphere, and so planets Jupiter and onwards fall in this category.
That being said, what would stop a spacecraft from just going right through any of these planets?
For earth, a spacecraft would firstly rapidly disintegrate approaching the surface of earth, and if it happens to still have part of itself, it will collide with the surface.
For gas giants like Uranus, if a spacecraft is able to counter the gravitational forces and slow down, could it actually "go through" Uranus?
• One issue with gas giants is the definition of the "edge" where the planet begins. Rocky planets have atmospheres who's extreme edges an orbiting or interplanetary spacecraft might pass undamaged, depending on design and alitutde. and solid/liquid sphere where it wouldn't. However, for a gas giant, the point where the planet "begins" is often defined by its optical properties, I think something like opacity = 1. The density falls rapidly above this point and increases rapidly below this point. This will be a hard question to answer without interpreting what "right through" means.
– uhoh
Feb 18 '18 at 3:41
• No. The center of Uranus is hard and hot like the center of the Earth. Even just 1/10th (or less) of the way inside would crush any spaceship we know. See: en.wikipedia.org/wiki/Galileo_Probe and what-if.xkcd.com/139 but you could aerobreak around the outer edge. That's kind of going through, just not through the middle. Feb 18 '18 at 3:53
• Basically you ask until which radius the planet can be just a manageable atmosphere? There should be data about their inner pressure and density. Through a gas giant surely not but inside yes, providing the vehicle stands dynamic P , or just P if it comes to halt. Feb 18 '18 at 10:26
• BTW, if there's no requirement to come out the other side in one piece, then you could pass through anything, except a black hole, if you're moving fast enough. :) Feb 20 '18 at 22:16
In short, No.
Side detail: Uranus and Neptune consist likely of 20% gas and 80% rock, coming from simple density considerations. They have large inner cores with masses around $\rm 12-14 \; m_{earth}$, and something like $\rm 2-3 \; m_{earth}$ of gas on top of them.
Jupiter and Saturn are true gas giants. They both have $\rm 5-20 \; m_{earth}$ of solids in them and the rest of the $\rm 320 \; m_{earth}$ (=Jupiter) and $\rm 95 \; m_{earth}$ (=Saturn) are gas.
However even if Jupiter and Saturn were 100% gas, you couldn't fly through them.
The main reasons for this is
1. Enormous density, pressure and temperature. A gas mass of several tens of Earth masses will collapse under its own gravity to enormous central densities, temperatures and thus static pressures.
The densities that this gas will be compressed to, can have densities comparable and exceeding that of water, which is a good rule-of-thumb for the center of Jupiter (typical calculations show gas densities around $\rm 20 g/cm^3$, source).
The static pressure will crush your spacecraft, the high density will near instantly slow you down to zero due to strong friction, and the high temperatures will just melt or evaporate your spacecraft.
Just because something is gaseous, doesn't mean you can soar through it, as you would on a calm day with a glider on a light breeze.
There are other factors that complicate even getting there:
1. Re-entry heat. The cold upper atmospheres of the gas giants increase the mach numbers $v / c_{sound}$ at re-entry by up to a factor of two. As recently discussed on space exploration the re-entry heating that a spacecraft experiences scales as the 8th power of the re-entry speed or more. So a spacecraft would experience around $2^8 = 256$ times more heat load. The engineering challenge to survive this already would be considerable.
I bring this up, because as @uhoh pointed out, it is unclear whether you want to just slice the atmosphere or fly through the $r=0$ coordinate of the planet.
2. Core erosion. Funny enough, the solid core is probably not even there anymore. Recent work has shown, that under the enormous pressures given at the center of a gas giant a phenomenon called 'core erosion' will take place. In this, typical rock-composing elements prefer to exist in a gaseous phase rather than in a solid phase at those high pressure, thus the solid core will dissolve into the surrounding gas.
This will further increase friction when trying to fly through the gas.
• Are you sure Uranus & Neptune are 80% rock? The Wikipedia article on "ice giant" says they're mostly ices, not gas, but also not rock. Feb 19 '18 at 8:53
• @Allure: Well, wikipedia is a good starting point for doing research, but does not necessarily represent absolute knowledge, or even the state of research in the respective field. Anyway, you're correct in pointing out that it's probably not 100% rock. More like a rock/ice mixture. This comes mainly from two lines of argumentation: 1.) The ice giants must have accreted most of their mass far beyond the water-ice line in our solar system, and water is very abundant in the universe Feb 19 '18 at 9:52
• 2.) if you want to match the given mass and radius of a planet in a structural model, you need to make assumptions about the composition. Given laboratory data about the high-pressure behaviour of rocks and ices we can have a rough idea about the mixing ratio between ice and rock, but our models are far from being developed enough to point to a single, unique composition. Thus I found it justified to speak a bit sloppily here about just rock. Anyway, for the point of the question this is also irrelevant. Feb 19 '18 at 9:54
• +1 but answers shouldn't really have wrong stuff in them, even if it's irrelevant wrong stuff; it set's a bad example for others who are new or less conscientious. If there's not a good way to show that it is likely to be rock and not anything else with high density, it would be better to not assert that it is rock.
– uhoh
Feb 21 '18 at 6:07 | 1,422 | 6,102 | {"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} | 2.5625 | 3 | CC-MAIN-2021-39 | longest | en | 0.942426 |
https://www.physicsforums.com/threads/endothermic-reactions.716978/ | 1,660,615,853,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00738.warc.gz | 814,581,748 | 17,212 | Endothermic Reactions
Gold Member
Homework Statement
http://i4.minus.com/j7HwKoL8yhl96.JPG [Broken]
Homework Equations
The enthalpy is equal to the heat of the products minus the heat of the reactants.
The Attempt at a Solution
We are not supposed to use a bond enthalpy table to ascertain the enthalpy of the reactions. I know that B is clearly exothermic, as is C. B is a phase change and in the process energy is released from the gas into the environment as it transitions into the lower-energy liquid. C is a combustion reaction and that's obviously exothermic.
D is clearly endothermic. Heat must be added to change water from a solid to a liquid.
However, what about A? It's a composition or synthesis reaction. Forming bonds also releases energy. However, bonds have to be broken before the water can be formed. I'm not sure how I'm supposed to tell whether A is exothermic/endothermic from just looking at it. Is there a way?
Last edited by a moderator:
Rawrr!
A is very exothermic. Oxygen and hydrogen are often used together as a means of propulsion in rockets. (Although to save space, they're usually condensed into a liquid state). It's a synthesis reaction as you're turning two reactants into one product.
Last edited:
Gold Member
Is there a general rule I can apply to figure out A?
Mentor
If you have hydrogen and oxygen mixed together and you create a spark, is heat given off in the ensuing explosion? Do you have to remove heat to get the water produced back to the original temperature?
You know that going from a solid to a liquid absorbs heat, so must be what?
Gold Member
I calculated the enthalpy of reaction using bond energies and I got a positive figure for reaction A; could I have been using bonds between atoms of the wrong states? The bond energy table I used didn't give states. I remember that the carbon oxygen double bond in carbon dioxide varies with states...
Yanick
There are a few ways of thinking about it. First is that you can think of the fact that it is a type of combustion reaction, which all tend to be exothermic. You are combusting hydrogen gas instead of a hydrocarbon in this case.
Also you can use everyday life and just observe that when you spark hydrogen gas in the presence of oxygen gas you'll get lots if heat out (its pretty much an explosion).
Finally if you consider Gibbs free energy, dG = dH - TdS, you can see that the entropy of the system is decreasing (3 mols of gas become 2). To make this reaction proceed (in other words dG < 0) you need dH < 0.
EDIT: Scratch that last paragraph. It was a rule of thumb I picked up in Gen Chem and may not apply. In fact I just re-read a part of a PChem text which says that oxygen and hydrogen gas produce water with an increase of entropy. Apologies.
Last edited:
Gold Member
Do you have to remove heat to get the water produced back to the original temperature?
How can you tell? I understand that for a forward endothermic reaction if you reserve it, it'll become an exothermic reaction. How do I ascertain this in this case?
Gold Member
Oh wait, I thought of a cool new way to remember that hydrogen gas and oxygen gas react, exothermically.
Remember, remember, the explosion of the Hindenburg. | 743 | 3,228 | {"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} | 3.140625 | 3 | CC-MAIN-2022-33 | latest | en | 0.968921 |
http://mathhelpforum.com/number-theory/30771-mod-5-a.html | 1,527,289,204,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00527.warc.gz | 180,898,881 | 10,191 | 1. ## Mod 5
I am having trouble with this homework question and was hoping someone could help. Cheers
Prove that 5|n(n2-1)(n2=1)
2. Hello,
Look for every possibility of n if 5 divides the product.
Try $\displaystyle n \equiv 0 mod 5$, then $\displaystyle n \equiv 1 mod 5$ and so on
3. Originally Posted by asw-88
I am having trouble with this homework question and was hoping someone could help. Cheers
Prove that $\displaystyle 5|n(n^2-1)(n^2+1)$ (at least, I think that's what is meant)
If you multiply out the brackets then you get $\displaystyle 5|n^5-n$, which is true by Fermat's little theorem.
4. Hello, asw-88!
Prove that: .$\displaystyle 5\,|\,n(n^2-1)(n^2+1)$
Let $\displaystyle N \:=\:n(n^2-1)(n^2+1)$
$\displaystyle n$ must one of five possible forms: .$\displaystyle 5k-2,\:5k-1,\:5k,\:5k+1,\:5k+2$
$\displaystyle [1]\;n \,= \,5k-2\!:\;\;N \:=\5k-2)([5k-2]^2-1)([5k-2]^2 + 1)$
. _ . . . . . . . . . . $\displaystyle =\;(5k+2)(25k^2-20x+3)(25k^2-20k + 5)$
. _ . . . . . . . . . . $\displaystyle = \;(5k+2))(25k^2+20k + 3){\color{red}5}(5k^2 - 4k + 1)$ . . . a multiple of 5
$\displaystyle [2]\;n \,=\,5k-1\!:\;\;N \:=\5k-1)([5k-1]^2-1)([5k-1]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k-1)(25k^2 - 10k)(25k^2-10k + 2)$
. . . . . . . . . . . . $\displaystyle = \;(5k-1){\color{red}5}k(5k- 2)(25k^2 - 10k+2)$ . . . a multiple of 5
$\displaystyle [3]\;n \,=\,5k\!:\qquad N \:=\5k)([5k]^2 - 1)([5k]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;{\color{red}5}k(25k^2 - 1)(25k^2+1)$ . . . a multiple of 5
$\displaystyle [4]\;n\,=\,5k+1\!:\;\;N \;=\;(5k+1)([5k+1]^2-1)([5k+1]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k+1)(25k^2 + 10k)(25k^2+10k + 2)$
. . . . . . . . . . . . $\displaystyle = \;(5k+1){\color{red}5}k(5k + 2)(25k^2 + 10k + 2)$ . . . a multiple of 5
$\displaystyle [5]\;n\,=\,5k+2\!:\;\;N \;=\;(5k+2)([5k+2]^2-1)([5k+2]^2+1)$
. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k + 3)(25k^2 + 20k + 5)$
. . . . . . . . . . . . $\displaystyle = \;(5k+2)(25k^2 + 20k+3){\color{red}5}(5k^2 + 4k + 1)$ . . . a multiple of 5
. . . . . . . . . . . . . . . . . Q.E.D.
5. I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.
-Dan
6. Originally Posted by topsquark
I'm not going to do it, but I suspect it would be relatively simple to do an induction proof of this as well.
-Dan
I'll do the initial condition:
$\displaystyle 5|1^5-1$
$\displaystyle 5|0$
Because I care, I'll even show the next iteration:
$\displaystyle 5|2^5-2$
$\displaystyle 5|30$
The rest of it is almost as easy as this. | 1,168 | 2,614 | {"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} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.58437 |
http://www.familyhandyman.com/DIY-Projects/Outdoor-Projects/Decks/Decking/how-to-build-deck-stairs | 1,369,251,237,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702414478/warc/CC-MAIN-20130516110654-00061-ip-10-60-113-184.ec2.internal.warc.gz | 461,203,681 | 48,669 | • Save
Sure, building deck stairs can be tricky. But in this story, we'll make it easy by showing you how to estimate step dimensions, layout and cut stair stringers, and assemble the stair parts. And you won't have to do any hard math to figure it all out (but your calculations will have to be accurate!). These DIY steps will work for replacing an old set of stairs and for building stairs on a brand new deck. So grab your tools and let's start building!
By the DIY experts of The Family Handyman Magazine
• ###### COMPLEXITY
• Moderate
• You'll need to make accurate calculations.
• ###### COST
• \$100 - \$500
• Cost will depend on the type of wood you use and the length of the stair run.
Sure, building deck stairs can be tricky. But in this story, we'll make it easy by showing you how to estimate step dimensions, layout and cut stair stringers, and assemble the stair parts. And you won't have to do any hard math to figure it all out (but your calculations will have to be accurate!). These DIY steps will work for replacing an old set of stairs and for building stairs on a brand new deck. So grab your tools and let's start building!
## How to estimate the landing zone
Whether you're replacing an old, rickety set of deck stairs or building a set for your new deck, deck stairs are among the most challenging projects for the average do-it-yourselfer to tackle.
One little mistake in calculations or layout and you'll wind up wasting lots of expensive wood, or worse, you'll build a downright dangerous set of stairs. But building a strong, safe set of stairs is doable if you meticulously follow the layout and cutting rules outlined in this story.
You almost always have to design site-built stairs yourself because the number and height of the steps will vary with the landscape. Begin by drawing a side view of your site and adding dimensions (Fig. A). That usually means going through the calculations a few times to determine where the stairs will fall and to figure out how long your skirt and stringer material needs to be. This sounds complex, but if you work through it a few times and rely on your sketch, it'll become clear.
Here's what to do
1. First determine the approximate height “X” (Fig. A). Start by estimating where you think the last stair will fall by using a 40-degree slope (Photo 1). Rest a straight board on the deck and level over to that spot and measure down to the ground. That'll be the approximate height of the stairs, “X.”
2. Now find the approximate number of steps. Divide “X” by 7 in. (an approximate step height) and round off the remainder, up if it's .5 or more, or down if it's less than .5. That'll give you an approximate number of risers (Fig. A). The actual recommended riser height is 6-1/2 to 8 in., but you'll determine that later. If the riser height is too short, re-divide “X” by 8 and start again. On uneven ground, find the number of treads so you can find the exact stair landing point. Simply subtract 1 from the number of risers. (There's always one fewer tread than risers, as you can see in Fig. B.) Then, multiply by 10.25 in., the ideal tread width for two 2x6s, to get the total run. Measure out that distance from the deck to find the exact landing point. From this point, you can measure the exact stair height and determine the stringer and skirt length.
3. Measure the exact total rise (Photo 1). Divide the height (X) by your estimated number of risers to find the exact riser height. The figure will usually fall between 6-1/2 and 8 in., the ideal range. Use this figure for your stringer layout (Fig. B). If the riser height isn't in this zone, add or subtract a riser and divide again. This will change the number of treads and shift the landing point, so re-measure the exact height and divide again.
4. Draw a sketch (Fig. B) to confirm the plan in your mind and lay out the first stringer (Photos 2 and 3) using the exact riser and tread dimensions and your framing square. Plan to establish a solid base at the landing point. The base can be a small concrete slab, a small deck or even a treated 2x12 leveled in over a 6-in. gravel base. After you cut the stringers, use them as guides to position your landing. Cut and mount the stringers by following our photos.
In your layout (Fig. B), note that:
• The top tread is 3/4 in. shorter than the other treads.
• The bottom riser is 1-1/2 in. shorter than the other risers. Be sure to test-fit the first stringer (Photo 4) before you cut the others. If you made a mistake, you'll at least be able to save the other two 2x10s.
Measure from the deck rim to the landing spot and add 2 ft. Buy three treated 2x10s, two 2x12 skirts and two 2x4s sized to the next larger length and you'll have plenty of material to work with (the worst mistake is buying material that's too short!). Get a 6-ft. 2x6 for securing the stairs to the deck (Photo 8). You'll also need two 2x6s for each tread and a 1x8 for each riser. Use 3-in. deck screws to fasten the skirts and treads to the stringers and the skirts to the deck. Fasten the risers to the stringers with 8d galvanized nails.
For extra-strong stairs, reinforce the middle 2x10 stringer with 2x4s nailed to both sides (Photo 7). There are a million ways to fasten the stringers solidly to the deck. Photo 8 shows a simple, foolproof, extra-strong method that works especially well even for open-sided stairs built without skirts.
There you go—a pretty, rock solid set of stairs ready for balusters and railings.
## Solid 2x12 skirts for solid stairs and rails
These stairs call for 2x10 treated material for the rot-resistant notched stair stringers (also known as jacks or carriages, Photo 1) that won't be seen. This design also uses 2x12 skirt boards that attach to the sides of the outside stringers. The skirts serve several purposes:
• Cosmetically, they hide the unsightly notched, treated stringers to make your stairs look polished.
• They make it easy to attach the stringers.
• Structurally, they make for rock-solid stairs by reinforcing the stringers, which have been weakened by notching.
• And when it comes time to attach guardrails and handrails to the stairs, you'll have a solid board to fasten pickets or posts to for a wobble-free rail. (If you'd rather not use the 2x12 skirt boards, be sure to use 2x12s for the notched stringers for adequate strength.)
For the parts that show—the skirts, treads and risers (lead photo) —choose material that matches the deck. In our case, that was cedar.
## Designing safe, comfortable stairs
Building codes contain specific requirements for safe stair design. If you follow the directions in this story, your stairs will be legal and safe. In a nutshell, treads should be more than 9 in. deep and risers 6-1/2 to 8 in. high. Riser heights can vary no more than 3/8 in. from one step to another to reduce trip hazards. However, even a 1/4-in. variation can cause tripping.
If you use 2x6s for tread material like we show, you can build stairs up to 48 in. wide with only three stringers because 2x6s can span up to 2 ft. But if you use the common and thinner 5/4-in. bull-nosed decking for your treads, you'll have to keep stringers no more than 16 in. apart and you'll be limited to 32-in. wide stairs with three stringers. For wider stairs, add one or more evenly spaced stringers depending on the width of your stairs and the tread material you choose.
And remember, you need one right and one left skirt assembly, not two lefts or rights.
##### A Carpenter's Square and a Set of Stair Gauges are Crucial
You'll need a 4-ft. level, tape measure, calculator, circular saw and a handsaw. If you don't already have a carpenter's square, now's the time to buy one (\$10; Photo 2). To do the job right, pick up a set of stair gauges (\$5), too. Stair gauges are little clamps that you tighten onto the square at the proper rise (vertical stair height) and run (horizontal tread depth) for exactly duplicating each step as you draw it onto the stringers (Photo 2). The gauges save time and ensure that all the steps are consistent.
##### Converting Decimals to Fractions
Not many calculators are set up to give you fractions, and a readout like 7.65 isn't much help for setting the carpenter's square and stair gauges. Use this chart to help you convert the readout to fractions or for converting fractions to decimals for calculator entries. Choose whichever fraction is closest to the decimal reading for setting your gauges when you lay out your stringers. .125 = 1/8 in. .25 = 1/4 in. .375 = 3/8 in. .5 = 1/2 in. .625 = 5/8 in. .75 = 3/4 in. .87 = 7/8 in.
### Required Tools for this Project
Have the necessary tools for this DIY project lined up before you start—you’ll save time and frustration.
• Hammer
• Clamps with a reach of at least 18 in.
• Circular saw
• Corded drill
• Chalk line
• Level
• Framing square
• Handsaw
• Stair gauge
• Safety glasses
• Sawhorses
### Required Materials for this Project
• 2x6x12 ft. (1)
• 2x12x12 ft (3; you may need longer 2 x 12s for your stairs)
• Decking for risers and treads
• 3 in. deck screws
### Comments from DIY Community Members
Share what's on your mind and see what other DIYers are thinking about.
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November 30, 10:34 AM [GMT -5]
My project requires a landing with the steps going out from the top deck and then to turn back into the bottom deck.
Where do I start? With the elevated platform and then work the stringer measurments?
Thanks
October 11, 10:23 PM [GMT -5]
This was full of useful information, however, I tried building my stairs using an adjustable bracket system. I was able to tailor fit my stairs in the amount of space that I had which was very helpful since the ground was uneven. It was super easy to install and very strong. There's a stair calculator on the website too, so I could figure it all out with just a few clicks: http://www.ez-stairs.com/stair_calculator/updated/index.html
#### How to Build Deck Stairs
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• Share comments on DIY Projects and more! | 2,542 | 10,234 | {"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} | 3.234375 | 3 | CC-MAIN-2013-20 | latest | en | 0.925484 |
https://techappss.com/operator-precedence-in-javascript/ | 1,720,996,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00229.warc.gz | 508,009,616 | 18,174 | ## Operator precedence in JavaScript
Operator precedence refers to the priority given to operators while parsing a statement that has more than one operator performing operations in it. It is important to ensure the correct result and also to help the compiler understand what the order of operations should be. Operators with higher priorities are resolved first .But as one goes down the list, the priority decreases and hence their resolution.
Precedence and Associativity: Associativity in general states that irrespective of the order of operands for a given operation the result remains the same. Precedence is used to tell the compiler what operations should be performed first. For example, consider three numbers 2, 3, and 4. Now consider two operations:
```( 2 + 3 ) + 4 = 2 + ( 3 + 4 )
( 2 >= 3 ) or ( 1 != 4 )```
The first operation is associativity where the order does not matter. Second case is precedence, where in order to reach the desired result there has to be a proper order in which operations will be performed.
Associativity is not a singular concept while dealing with precedence operations has to be dealt either with left-to-right or right-to-left associativity. This completely depends on the operation and tells the parser from which direction the operation should start.
### Example:
```// left-to-right associativity : division
3/4```
```// right-to-left associativity : assignment
a = 3 ```
Operator Precedence Table: The operator precedence table can help one know the precedence of an operator relative to other operators. As one goes down the table, the precedence of these operators decreases over each other, that is, the priority of an operator is lower than the operators above it and higher than the ones below it. The operators in the same row have the same priority.
In this table, 1 is the highest precedence and 19 is the lowest precedence. | 400 | 1,893 | {"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 | 0.914235 |
https://www.wyzant.com/resources/answers/404189/a_small_theater_has_a_seating_capacity_of_2000_when_the_ticket_price_is_20_attendance_is_1500_for_each_1_decrease_in_price_attendance_increases_by_100 | 1,516,160,880,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00376.warc.gz | 1,030,254,676 | 17,145 | 1
# A small theater has a seating capacity of 2000. When the ticket price is \$20, attendance is 1500. For each \$1 decrease in price, attendance increases by 100.
(a) Write the revenue R of the theater as a function of ticket price x.
R(x) =
(b) What ticket price will yield a maximum revenue?
\$
What is the maximum revenue?
### Comments
Revenue = Price*attendance
r(x) = xa
Price Attendance Revenue
_______________________________
20 1500 \$30,000
19 1600 \$30,400
18 1700 \$30,600
17 1800 \$30,600
16 1900 \$30,400
15 2000 \$30,000
Note that no information is given as to whether or not the
ticket sales must be cut in \$1 increments. Example, if we
sell the tickets for \$17.50 do we get 1750 attendees?
This would be at the vertex of the parabolic curve indicating
a maximum revenue.
17.5(1750) = \$30,625 revenue.
### 1 Answer by Expert Tutors
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Andrew M. | Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/HonorsMathematics - Algebra a Specialty / F.I....
0
Letting x represent the change in ticket price in dollars:
R(x) = (20-x)(1500 + 100x); 0≤ x ≤ 5 x cannot go higher than 5
because this corresponds to selling
out the 2,000 seat theater.
R(x) = 30000 + 2000x - 1500x - 100x2
R(x) = -100x2 + 500x + 30000
Find the vertex of the parabolic function:
Vertex: (-b/2a, r(-b/2a)); a = -100, b=500
-b/2a = -500/-200 = 2.5
r(2.5) = -100(2.5)2 + 500(2.5) + 30000
= -625 + 1250 + 30000
= \$30,625
The maximum revenue is at the point where ticket prices are
reduced by \$2.5 to \$17.50. The revenue at that point is \$30,625 | 542 | 1,757 | {"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-2018-05 | longest | en | 0.758239 |
http://topvolleylamezia.it/ignx/autocovariance-matlab.html | 1,591,357,943,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348500712.83/warc/CC-MAIN-20200605111910-20200605141910-00568.warc.gz | 122,014,391 | 21,374 | Autocovariance Matlab
Other Useful Texts (AH) Andrew Harvey, Time Series Models, MIT Press. Let wt, t ∈ Z be a normal white noise (i. In probability theory and statistics, given a stochastic process, the autocovariance is a function that gives the covariance of the process with itself at pairs of time points. For a weakly stationary time series, the notation used for autocovariance uses only lag: (h) = E(x t )(x t h ) where is the constant variance. 05 c t, (dtex) 0. More info can be found here. Matrices that contain mostly zero values are called sparse, distinct from matrices where most of the values are non-zero, called dense. Random is a website devoted to probability, mathematical statistics, and stochastic processes, and is intended for teachers and students of these subjects. with a normal distribution of mean 0 and std 1). All the database concepts, techniques, and tools that are needed to develop a database application from scratch are introduced. , and Zhang, X. Note that φ(0) = x'2, so that the autocovariance at lag zero is just the variance of the variable. The results show that significant. Finding autocovariance of AR(2) Ask Question Asked 6 years, 2 months ago. Number of components to use. ACF and prediction. Displacement equals the original velocity multiplied by time plus one half the acceleration multiplied by the square of time. If whiten is false, the data is already considered to be whitened, and no whitening is performed. That is, Z n= A n+ ˚ k1Z n 1+ ˚. Unlike 'plot. INDEX 609 propagationdelays,514–515 dual-frequency,515 Klobucharmodel,514 pseudorange,180–181,392,506, 518–520 differences,542 receiverantenna,506,512,514,549. However, certain applications require rescaling the normalized ACF by another factor. Long memory has been observed for time series across a multitude of fields, and the accurate estimation of such dependence, for example via the Hurst exponent, is crucial for the modelling and prediction of many dynamic systems of interest. , daily exchange rate, a share price, etc. If True, then denominators for autocovariance are n-k, otherwise n. Slide (Feat Frank Ocean and Migos) - download. Brouwer et al. pyplot as plt # matplotlib provides plot functions similar to MATLAB import numpy as np from skimage import color , filter # skimage is an image processing library. Introduction to Time Series Analysis. OMS Analytics. We must focus on relevant inputs from our senses – such as the bus we need to catch – while ignoring distractions – such as the eye-catching displays in the shop windows we pass on the same street. Here is a Matlab code and experimental and theoretical autocorrelations Autocovariance - expectation across all time indices? 3. Also note that a p = a p because both correspond to a lag of ptime samples. The Covariance Matrix Definition Covariance Matrix from Data Matrix We can calculate the covariance matrix such as S = 1 n X0 cXc where Xc = X 1n x0= CX with x 0= ( x 1;:::; x p) denoting the vector of variable means C = In n 11n10 n denoting a centering matrix Note that the centered matrix Xc has the form Xc = 0 B B B B B @ x11 x 1 x12 x2 x1p. In fact, i. •Large number of design variables. ARMA(p,q) models 3. T his leads to the follow ing deÞ nition of the Òauto co variance Ó of the pro ces s:! (k ) = co v(X n + k, X n) (3. DESCRIPTION The lag 1 autocovariance of a variable is the covariance between Xi and Xi+1. The conditional variance h t is where The GARCH(p,q) model reduces to the ARCH(q) process when p=0. Show that the theoretical autocovariance of a rectangular function fo-Jb Osta You will need to consider positive and negative lags separately and combine the results to obtain the general solution. where ω ∈ [0, 1) is a fixed constant. Apply a low pass filter. In this week, we begin to explore and visualize time series available as acquired data sets. The second condition states that the autocovariance of X(t) also does not depend on time, only on time-difference (τ). Regularization was included in order to handle ill-conditioning of the least-squares problem. Statistics comes in two flavours: the first is called descriptive statistics and it studies various ways to sort information. Ensure residuals from Step 5 are serially uncorrelated and homoskedastic. Extremely useful, yet, very difficult to understand conceptually because of the complex mathematical jargon. presented preliminary of the autocovariance method that has been used to designate as variance analysis on stationary measurements to compare the quality of quasi-steady measurement methods. Control flow. The autocovariance is the covariance of a variable with itself (Greek autos = self) at some other time, measured by a time lag (or lead) τ. Almost everything in R is done through functions. Meaning of autocovariance. Remember that a sequence of random variables is said to be covariance stationary (or weakly stationary) if and only if:. The function Acf computes (and by default plots) an estimate of the autocorrelation function of a (possibly multivariate) time series. Per accedere alla descrizione completa dei comandi digitare “help nomecomando” al prompt di Matlab. Earth Sciences Using MATLAB®," Prentice Hall, Upper Saddle River, New Jersey) indicates that the term autocorrelation refers to R( τ ) divided by the variance of the signal u(t). Review: Autocovariance, linear processes 2. The functions xcorr and xcov estimate the cross-correlation and cross-covariance sequences of random processes. c = xcov(x) returns the autocovariance sequence of x. • Autocorrelation Function of a Stationary Process • Power Spectral Density • Stationary Ergodic Random Processes EE 278: Stationary Random Processes Page 7-1. Then observe that z ph j(1 ˚ 1z j ˚ 2z 2 ˚ pz j) = 0 In general, any linear combination of the zeros of ˚(z) is a solution. $\endgroup$ - Nick X Tsui Dec 1 '15 at 21:58 2 $\begingroup$ I hope he/she knew how to get correlation from covariance and variances. computes the sample autocovariance of a time series x for lags from 0 to maxlag, returning a column vector of length maxlag+1. Schematic illustration of the method. This article needs additional citations for verification. MATLAB/Octave variable: oo_. The spectral density is a frequency domain representation of a time series that is directly related to the autocovariance time domain representation. Time series clustering is implemented in TSclust, dtwclust, BNPTSclust and pdc. Expert Answer. Interpretation. Publications Son, S. cor,ddmatrix-method. The autocovariance least-squares method is revised for a general linear stochastic dynamic system and is implemented within the publicly available MATLAB toolbox Nonlinear Estimation Framework. The same inequality is valid for random variables. C = cov (A) returns the covariance. cov,ddmatrix-method. m (sample autocovariance function) diffd. Xiaohui Chen. 1 $\begingroup$ I have tried compute the autocovariance of the following process: but irrelevant as far as the autocovariance function is concerned. Example: Autocorrelation Application Cross-corrrelation & Autocorrelation 2. Course Descriptions. m, utl_sincos_2d. It builds on the course Bayesian Statistics: From Concept to Data Analysis, which introduces Bayesian methods through use of simple conjugate models. However, in other disciplines (e. Prerequisite(s): MTH 219. You can prove the Cauchy-Schwarz inequality with the same methods that we used to prove | ρ(X, Y) | ≤ 1 in Section 5. way into the Matlab simulation program. It also explains how to calculate statistics over VOIs - play. Although we could simulate an AR($$p$$) process in R using a for loop just as we did for a random walk, it's much easier with the function arima. Spectral distribution function. Since autocorrelation and autocovariance sequences are all (aperiodic) one-dimensional sequences, there Fourier transform exist and are bounded in | w |≤π. Ensure residuals from Step 5 are serially uncorrelated and homoskedastic. Below is a simple plot of a kalman filtered version of a random walk (for now, we will use that as an estimate of a financial time series). Fractal dimension and the Hurst parameter are utilized for this purpose. Mission Statement. Time series features are computed in feasts for time series in tsibble format. We show that correctly identifying the distribution. Road Map 1. Nonlinear autocovariance in Matlab. The General Linear Model (GLM) Ged Ridgway Wellcome Trust Centre for Neuroimaging University College London SPM Course Vancouver, August 2010. 2 • X(t) is a wide sense stationary process with autocorrelation function RX(τ) = 10 sin(2000πt) +sin(1000πt). Contextual translation of "effectiveness" from French into Portuguese. The autocorrelation function is a measure of the correlation between observations of a time series that are separated by k time units (y t and y t–k ). However, certain applications require rescaling the normalized ACF by another factor. The Covariance Matrix Definition Covariance Matrix from Data Matrix We can calculate the covariance matrix such as S = 1 n X0 cXc where Xc = X 1n x0= CX with x 0= ( x 1;:::; x p) denoting the vector of variable means C = In n 11n10 n denoting a centering matrix Note that the centered matrix Xc has the form Xc = 0 B B B B B @ x11 x 1 x12 x2 x1p. function [f,x] = rsgene1D(N,rL,h,cl) % % [f,x] = rsgene1D(N,rL,h,cl) % % generates a 1-dimensional random rough surface f(x) with N surface points. If A is a matrix whose columns represent random variables and whose rows represent observations, C is the covariance matrix with the corresponding column variances along the diagonal. MATLAB/Octave variable: oo_. ˚ Recommended Text (JH) J. In the homework assignments you are recommended to use MATLAB. • Some idea of what a basis of a vector space is. In both packages, many built-in feature functions are included, and users can add their own. This chapter describes a variety of probability models for time series, which are collectively called stochastic processes. Review: Autocovariance, linear processes 2. , using the bare formula will be much faster and is overall the better trade-off. It also explains how to calculate statistics over VOIs - play. 24K Magic - download. Autocovariance function is defined, basically, just taking covariance of different elements in our sequence, in our stochastic process. The cross-correlation is similar in nature to the convolution of two functions. Each plot. Top-Left—the state autocovariance function as a function of embedding lag, taken from the centre row of the autocovariance matrix. Since there is no noise, Vt completely smooths out the oscillations, resulting in a flat line. R functions for time series analysis by Vito Ricci ([email protected] We then plot the resulting autocovariance function and limit the x axis to ± 30 successive beats to better evaluate the decrease in covariance with successive beats. The subtraction can be done within the axcor input argument. All outputs are vectors with three elements, corresponding to tests at each of the three lags. The book is intended to provide students and researchers with a self-contained survey of time series analysis. C = cov (A) returns the covariance. RS -EC2 -Lecture 14 1 1 Lecture 14 ARIMA - Identification, Estimation & Seasonalities • We defined the ARMA(p, q)model:Let Then, xt is a demeaned ARMA process. To store the data in a time series object, we use the ts () function in R. Its sign convention for the lag variable is reversed with respect to the. 28-32) are a commonly-used tool for checking randomness in a data set. Matlab computes the autocovariance via Note that the lag must be less than the length of the sample. Gaussian Random Variable Definition A continuous random variable with pdf of the form p(x) = 1 p 2ˇ˙2 exp (x )2 2˙2; 1. The autocovariance at lag s is defined as The autocorrelation function begins at some point determined by both the AR and MA components but thereafter, declines geometrically at a rate determined by the AR component. DESCRIPTION The lag 1 autocovariance of a variable is the covariance between Xi and Xi+1. Statistics in Engineering: With Examples in MATLAB® and R, Second Edition - CRC Press Book Engineers are expected to design structures and machines that can operate in challenging and volatile environments, while allowing for variation in materials and noise in measurements and signals. The Autocorrelation function is the normalized autocovariance function φ(τ)/φ(0) = r(τ);. Keywords: Differential Evolution Algorithm, Fault Detection and Diagnosis, Takagi-Sugeno Fuzzy Classifier, Unitary HVAC System, Wavelet Transform. Dorf Boca Raton: CRC Press LLC, 2000 The Intel Pentium® processor, introduced at speeds of up to 300 MHz, combines the architectural advances in the Pentium Pro processor with the instruction set extensions of Intel MMX™ media enhancement technology. This algorithm has many applications. Since there is no noise, Vt completely smooths out the oscillations, resulting in a flat line. Correlation and Convolution Cross-correlation, autocorrelation, cross-covariance, autocovariance, linear and circular convolution Signal Processing Toolbox™ provides a family of correlation and convolution functions that let you detect signal similarities. For example, is you were calculating the third iteration (i = 3) using a lag k = 7, then the calculation for that iteration would look like this: (y3 - y-bar)(y10 - y-bar) Iterate through all values of "i" and then take the sum and divide it by the number of values in the data set. Autocovariance is closely related to the autocorrelation of the process in question. Statistics and Probability Letters, 2014, Vol. Publications Son, S. Exercise 2: ARMA Processes Find the mean and the autocovariance function of the ARMA(2,1) process, You can nd tutorials and a lot of Matlab. cov$(Y_t, Y_{t-j}). Scalar or vector of nonnegative integers indicating the number of autocovariance lags to include in the Newey-West estimator of the long-run variance. 3 Construction of an ARIMA model 1. Birds In The Trap S. , using the bare formula will be much faster and is overall the better trade-off. Fractal Dimension and the Hurst Parameter. Topic 9: Lagrange Multiplyers and Adjoints. In an autocorrelation, which is the cross-correlation of a signal with itself, there will always be a peak at a lag of zero, and its size will be the signal energy. ts(): plots a two time series on the same plot frame (tseries) tsdiag(): a generic function to plot time-series diagnostics (stats) ts. The metric evaluates how much - to what extent - the variables change. This means that the sine terms of the exponential do not cancel as for the power spectrum. Exercise 2: ARMA Processes Find the mean and the autocovariance function of the ARMA(2,1) process, You can nd tutorials and a lot of Matlab. The course also explores topics such as statistical image understanding, elements of pattern theory, simulated annealing, Metropolis-Hastings algorithm, and Gibbs sampling. Introduction to MATLAB and its use in engineering. Guarda il profilo completo su LinkedIn e scopri i collegamenti di Alessia e le offerte di lavoro presso aziende simili. Note, that from the equation (4. p-values are left-tail probabilities. In probability theory and statistics, given a stochastic process, the autocovariance is a function that gives the covariance of the process with itself at pairs of time points. This course presents an example of applying a database application development methodology to a major real -world project. Review: Autocovariance, linear processes 2. Often, one of the first steps in any data analysis is performing regression. ACF and prediction. FREQUENCY DOMAIN EXERCISE (1) Consider a process with spectral density Sx(w) that takes the value 1 at w equal to 0, p 2, 3p 2, p, etc. (d) The same pattern is visible in (a)-(c). A measure used to represent how strongly two random variables are related known as correlation. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Apply a low pass filter. 1 show a white noise sequence of length N = 128 and its periodogram, which shows that the power spectrum is uniformly spread. I have a periodic signal loaded into Matlab and i am trying to estimate the Autocovariance of it by using the xcov command. 9, 801-810. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In both packages, many built-in feature functions are included, and users can add their own. Autocorrelation of a random process is the measure of correlation (relationship) between. The spectral distribution function For any stationary {Xt} with autocovariance γ, we can write γ(h)= Z 1/2 −1/2 e2πiνhdF(ν), where Fis the spectral distribution function of {Xt}. Stationarity of MA Process 4. And cross correlations can help you identify leading indicators. presented preliminary of the autocovariance method that has been used to designate as variance analysis on stationary measurements to compare the quality of quasi-steady measurement methods. , the cross-covariance is a function that gives the covariance of one process with the other at pairs of time points. Autocovariance and Autocorrelation of VARMA (p,q) Process. plot(): plots several time series on a common plot. It is usually used in scenarios where we are counting the occurrences of certain events that appear to happen at a certain rate, but completely at random (without a certain structure). où y(h) est la valeur du demi-variogramme, et h est le pas d'échantillonage, on obtient une droite dont la pente, qui est toutjours calculée par régression, est liée à la dimension fractale. DESCRIPTION The lag 1 autocovariance of a variable is the covariance between Xi and Xi+1. 11 Use Of Autocovariance To Determine The Correlation % Of Heart Rate Variation Between Heart Beats % Load Hr_pre; % Load Normal HR Data [cov_pre,lags_pre] = Axcor(hr_pre - Mean(hr_pre)); % Auto-covariance Plot(lags_pre,cov_pre,'k'); Hold On; % Plot Normal Auto-cov Plot([lags_pre(1) Lags_pre(end)],. In order to use this tool, the program requires the audio for the analysis, which in this case, it is a selected fragment by the cursors, the sampling frequency, 11025 Hz, because for voice-signal. It was chaired by members of Eurostat: Jukka Jalava, Luis Biedma and Johannes Wouters. It is the same as. Autocorrelation (serial correlation, or cross-autocorrelation) function (the diagnostic tool) helps to describe the evaluation of a process through time. By contrast, correlation is simply when two independent variables are linearly related. Aut o co v ar iance and Aut o corre lati on If the {X n} pro cess is w eakl y statio nary , the co varia nc e of X n an d X n + k dep end s only on the lag k. We then plot the resulting autocovariance function and limit the x axis to ± 30 successive beats to better evaluate the decrease in covariance with successive beats. Vector of p-values of the test statistics, with length equal to the number of tests. 1 AR estimation. If you specify maxlag, then r has size (2 × maxlag + 1) × N 2. m, nino2 (cont. Learn more about correlation, lag, cross correlation, xcorr, corrcoef, pairwise, nan, missing values. Definition of autocovariance in the Definitions.$\endgroup$- Nick X Tsui Dec 1 '15 at 21:58 2$\begingroup$I hope he/she knew how to get correlation from covariance and variances. Chapter 10 Discrete Spectra Estimation 10. Statistics in Engineering: With Examples in MATLAB® and R, Second Edition - CRC Press Book Engineers are expected to design structures and machines that can operate in challenging and volatile environments, while allowing for variation in materials and noise in measurements and signals. try 5 and 1 5, for example. Dorf Boca Raton: CRC Press LLC, 2000 The Intel Pentium® processor, introduced at speeds of up to 300 MHz, combines the architectural advances in the Pentium Pro processor with the instruction set extensions of Intel MMX™ media enhancement technology. The Autocovariance-Generating Function for Vector Processes 266 10. A lot of m-files are found in this page. Correlation is a measure of the strength of the relationship between two variables. The course is a continuation of STA 5106 in computational techniques for linear and nonlinear statistics. SolutionsManual AccompanyTime Series Analysis SecondEdition Kung-SikChan Solutions JonathanCryer XuemiaoHao, updated 7/28/08 CHAPTER Exercise1. cov,ddmatrix-method. Notice that power at a frequency f0 that does not repeatedly reappear in xT(t) as T → ∞ will result in Sx(f0) → 0, because of the division by T in Eq. 1998 | 1999 | 2000. Handbook of Optical Sensing of Glucose in Biological Fluids and Tissues Valery V Tuchin (Ed) Intelligent and Adaptive Systems in Medicine Oliver C L Haas and Keith J Burnham. The temporal information is given by when the state is. The second condition states that the autocovariance of X(t) also does not depend on time, only on time-difference (τ). It is the same as. Introduction to Time Series Analysis. However, a manufacturer's proprietary restrictions will generally make access to the detailed design specifications of a transformer difficult. In this paper, we study. The transmitter sent out a single signal through one antenna, which eventually arrived at a single antenna at the receiver, probably along with a little noise. If True, computes the ACF via FFT. The School has a flexible licence for all Versions. Statistics is the study of data. m (sample autocovariance function) diffd. Sample autocorrelation function 3. Spectral Factorization; Lecture 18 (November ). Maximum Likelihood Estimation and Hypothesis Testing for an. Otherwise it is nonin-vertible. The ebook and printed book are available for purchase at Packt Publishing. 1: Yule-Walker Equations§3. Another important simplification is made, in that we will assume that the signals are already centered, meaning that their mean is assumed to be 0. Electronic Journal of Statistics, 2015, Vol. Objectives • Understand that prediction using a long past can be dicult because a large matrix has to be inverted, thus alternative, recursive method are often used to avoid direct inversion. 2 Continuous-time Gaussian Markov Processes We first consider continuous-time Gaussian Markov processes on the real line, and then relate the covariance function obtained to that for the stationary solution of the SDE on the circle. The second edition of Signal Processing for Intelligent Sensor Systems enhances many of the unique features of the first edition with more answered problems, web access to a large collection of MATLAB scripts used throughout the book, and the addition of more audio engineering, transducers, and sensor networking technology. T his leads to the follow ing deÞ nition of the Òauto co variance Ó of the pro ces s:! (k ) = co v(X n + k, X n) (3. try 5 and 1 5, for example. It is common practice in some disciplines (e. Note that γ 0 is the variance of the stochastic process. Autocorrelation (serial correlation, or cross-autocorrelation) function (the diagnostic tool) helps to describe the evaluation of a process through time. This video is part of the Udacity course "Machine Learning for Trading". Autocovariance estimation with long range dependence in Gaussian and threshold Gaussian model. The theoretical autocovariance function of an ARMA(p,q) with unit variance is computed. Our treatment of continuous-time GMPs on. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. MATLAB Release Compatibility. The autocovariance is the covariance of a variable with itself (Greek autos = self) at some other time, measured by a time lag (or lead) τ. The first differencing value is the difference between the current time period and the previous time period. they are iid normal) with variance 1, and consider the time series xt = wtwt−1, yt = x2t. (JD) James Davidson, Econometric Theory, Blackwell Publishing. They are important in determining the relationship between two random variables. 1$\begingroup$I have tried compute the autocovariance of the following process: but irrelevant as far as the autocovariance function is concerned. 3 The Durbin method of MA estimation. In the process of rewriting the code, I use the design of JFVM. autocovariance as O-U process: in fact, this is a pathwise description of red noise as a weighted integral of white noise An Introduction to Probability and Stochastic Processes for Ocean, Atmosphere, and Climate Dynamics2: Stochastic Processes Œ p. with a normal distribution of mean 0 and std 1). You can calculate it for any period of time. Question: The Help Of Matlab Code Kindly Expalin. 1) Questi ons : 1. Methods of data description and analysis using SAS: descriptive statistics, graphical presentation, estimation, hypothesis testing, sample size, power; emphasis on learning statistical methods and concepts through hands-on experience with real data. Regularization was included in order to handle ill-conditioning of the least-squares problem. com > gibbs. Autocorrelation and partial autocorrelation plots are heavily used in time series analysis and forecasting. More precisely, let g() be the autocovariance function of a time series X. 19 This content is available online at < ;. Random is a website devoted to probability, mathematical statistics, and stochastic processes, and is intended for teachers and students of these subjects. We typically measure or calculate slope, curvature, power spectrum and autocovariance with this instrument. The same inequality is valid for random variables. 1: Yule-Walker Equations§3. Several useful m-files are found in this page. 80 GHz CPU and 16. Topic 9: Lagrange Multiplyers and Adjoints. Equation (3) is an example of deterministic equation. Interaction among scatterers is the most significant source of deviation between theory and measurement based autocovariance estimates. Enter search keywords: Popular Artists. Then, I calculate the autocovariance matrix, from where I extract the eigenvalues and eigenvectors which are used to calculate the new variable Y which is the stochastic process S in a base where the random variables are not correlated. function results = momentg(draws) % PURPOSE: computes Gewke's convergence diagnostics NSE and RNE. The ARIMA(1,0,0)x(0,1,0) model with constant: SRW model plus AR(1) term. où y(h) est la valeur du demi-variogramme, et h est le pas d'échantillonage, on obtient une droite dont la pente, qui est toutjours calculée par régression, est liée à la dimension fractale. (2020), Sparse Graphical Models via Calibrated Concave Convex Procedure with Application to fMRI Data, Journal of Applied. 28-32) are a commonly-used tool for checking randomness in a data set. 1 Models for time series 1. Definition 8. Update on MATLAB's capabilities. m (difference operator) ljungbox. 1 AR estimation. Corequisite: MATH 152 and MATH 232. Z, we can find its. Perform the Bounds Test. Almost everything in R is done through functions. Interaction among scatterers is the most significant source of deviation between theory and measurement based autocovariance estimates. cov2cor () scales a covariance matrix into a correlation matrix. If you need to do it hundreds of times in a loop, with different data sets, etc. In an autocorrelation, which is the cross-correlation of a signal with itself, there will always be a peak at a lag of zero, and its size will be the signal energy. This instrument has a 2 angstrom vertical resolution and a 1 micron lateral resolution you can obtain both 2D and 3D plots, which are then transferred to the Matlab or Mathcad for further analysis. In order to use this tool, the program requires the audio for the analysis, which in this case, it is a selected fragment by the cursors, the sampling frequency, 11025 Hz, because for voice-signal. Other Useful Texts (AH) Andrew Harvey, Time Series Models, MIT Press. 11 Use Of Autocovariance To Determine The Correlation % Of Heart Rate Variation Between Heart Beats % Load Hr_pre; % Load Normal HR Data [cov_pre,lags_pre] = Axcor(hr_pre - Mean(hr_pre)); % Auto-covariance Plot(lags_pre,cov_pre,'k'); Hold On; % Plot Normal Auto-cov Plot([lags_pre(1) Lags_pre(end)],. STAT:2010 is a beginning methods course for undergraduate students. In locits: Test of Stationarity and Localized Autocovariance. 05 c t, (dtex) 0. tacvf: Prints a tacvf object. Description Usage Arguments Details Value Author(s) References See Also Examples. Birds In The Trap S. Parker March 17, 2015 Abstract A vast and deep pool of literature exists on the subject of spectral analysis; wading through it can obscure even the most fundamental concepts from the inexperienced practitioner. Documents SAS/IML software, which provides a flexible programming language that enables novice or experienced programmers to perform data and matrix manipulation, statistical analysis, numerical analysis, and nonlinear optimization. 2 Estimating the spectrum of an AR process 9. ) Windowed spectral analysis (Lecture 12, Feb. This video is part of the Udacity course "Machine Learning for Trading". We assume that a probability distribution is known for this set. 9 (Dynamic) copula-marginal. The autocovariance function (ACF) is defined as the sequence of covariances of a stationary process. Function Ccf computes the cross-correlation or cross-covariance of two univariate series. var () is a shallow wrapper for cov () in the case of a distributed matrix. The following Matlab project contains the source code and Matlab examples used for computes the autocovariance of two columns vectors consistently with the var and cov functions. The Poisson Process Summary. 3 Chi-Square Test •Designed for testing discrete distributions, large samples •General test: can be used for testing any distribution —uniform random number generators —random variate generators •The statistical test: •Components —k is the number of bins in the histogram —oi is the number of observed values in bin i in the histogram —ei is the number of expected values in bin. Function Pacf computes (and by default plots) an estimate of the partial autocorrelation function of a (possibly multivariate) time series. Proofs of Chapter 10 Propositions 285 Exercises 290 References 290 257 11 Vector Autoregressions291 11. Fake Love - download. The domain of t is a set, T , of real numbers. SCILAB provides function corr to calculate the autocovariance function out of a vector signal u. 28-32) are a commonly-used tool for checking randomness in a data set. free to use Eview, Gauss or Matlab, all available on FMRISC. If you take Xt and Xs and s and t might be in different locations and we'll get the cavariance of them, we get gamma (s,t) then we call that covariance and if we take ( x,t) the covariance of (x,t) will itself. It is not allowed to use xcorr, xcov, mean, cov, var etc. We consider the problem of estimating the high-dimensional autocovariance matrix of a stationary random process, with the purpose of out of sample prediction and feature extraction. is the modified Bessel function of order , where is the Hurst number (Mandelbrot, 1985,1983). Function spgrambw, from this toolbox, enables to plot a spectrogram in MATLAB, using a more precise algorithm, than using specgram from MATLAB. It enables our “digital society” and its applications are vast. The sample ACF and PACF exhibit significant autocorrelation. Time Series Analysis with ARIMA – ARCH/GARCH model in R I. Lagged/Cross Correlations with missing values. Looking for a tutorial on How To Solve For Covariance? This practical instructional video explains accurately how it's done, and will help you get good at math. Bottom-Left—the whole state autocovariance matrix. The autocorrelation function is a measure of the correlation between observations of a time series that are separated by k time units (y t and y t–k ). Examples translated by humans: MyMemory, World's Largest Translation Memory. Topic 9: Lagrange Multiplyers and Adjoints. You should be vaguely familiar with these topics for the comprehensive exam. The autocovariance least-squares method is revised for a general linear stochastic dynamic system and is implemented within the publicly available MATLAB toolbox Nonlinear Estimation Framework. Brouwer et al. For best results, give a suitable value for lags. The moving average is extremely useful for forecasting long-term trends. Threshold GARCH Model: Theory and Application Jing Wu The University of Western Ontario October 2011 Abstract In this paper, we describe the regime shifts in the volatility dynamics by a threshold model,. p-values are left-tail probabilities. The transmitter sent out a single signal through one antenna, which eventually arrived at a single antenna at the receiver, probably along with a little noise. A video tutorial that explains how to save and capture PDFs of images - play. This problem has received several solutions. The sample ACF and PACF exhibit significant autocorrelation. In the process of rewriting the code, I use the design of JFVM. PGFs are useful tools for dealing with sums and limits of random variables. Appropriate interpre-. Recommended Reading: If you feel like you are having a hard time with basic probability, I suggest:. Variance refers to the spread of the data set, while the covariance refers to the measure of how two random variables will change. Autocorrelation & Cross-correlation Applications Cross-corrrelation & Autocorrelation 1. Then, I calculate the autocovariance matrix, from where I extract the eigenvalues and eigenvectors which are used to calculate the new variable Y which is the stochastic process S in a base where the random variables are not correlated. Use MathJax to format equations. Lectures by Walter Lewin. Each plot. If you need to do it hundreds of times in a loop, with different data sets, etc. Most physical processes in the real world involve a random or stochastic element in their structure, and a stochastic process can be described as ‘a statistical phenomenon that evolves in time according to probabilistic laws’. AR model, ligistic time series and rounding time series. Dorf Boca Raton: CRC Press LLC, 2000 The Intel Pentium® processor, introduced at speeds of up to 300 MHz, combines the architectural advances in the Pentium Pro processor with the instruction set extensions of Intel MMX™ media enhancement technology. Unsourced material may be challenged and removed. Sign up to join this community.$\begingroup$I think he/she asked autocorrelation, not autocovariance. Birds In The Trap S. Let the Fourier transform of the autocorrelation and autocovariance sequences be. Calculating Sample Autocorrelations in Excel A sample autocorrelation is defined as vaˆr( ) coˆv( , ) ˆ ˆ ˆ, 0 it k it i t k k R R R − g g r. Computes the autocovariance of two columns vectors consistently with the var and cov functions. MATLAB Release Compatibility. • economics - e. Jin-Yi Yu Autocorrelation Function The Autocorrelation function is the normalized autocovariance function:. Calculate the autocovariance function using the given formula. Visualizza il profilo di Alessia Battilana su LinkedIn, la più grande comunità professionale al mondo. In the view of COVID-19 situation, many students are staying at home and pursuing their studies. 1 show a white noise sequence of length N = 128 and its periodogram, which shows that the power spectrum is uniformly spread.$\gamma_o $is the population variance. The xcov function estimates autocovariance and cross-covariance sequences. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Implementation. (c) Xt oscillates more-or-less with period 4, but there is quite a bit of noise. The autocovariance is the covariance of a variable with itself (Greek autos = self) at some other time, measured by a time lag (or lead) τ. iSpy (Feat. C = cov (A) returns the covariance. This means that the sine terms of the exponential do not cancel as for the power spectrum. Review: Causality, invertibility, AR(p) models 2. ; the sequence of pdfs of Xn is called the first-order pdf of the process xn 1 0 1 z Since Xn is a differentiable function of the continuous r. Birds In The Trap S. This chapter develops the underlying principles needed to understand noise, and the next chapter. Autocorrelation and partial autocorrelation plots are heavily used in time series analysis and forecasting. Some assignments may require the use of computer software such as Matlab, Gauss, R, Ox. COURSE INFORMATION: CourseInstructor Prof. Please sign up to review new features, functionality and page designs. Dirichlet’s Kernel. Determine the appropriate lag structure of the model selected in Step 3. Another important simplification is made, in that we will assume that the signals are already centered, meaning that their mean is assumed to be 0. In their estimate, they scale the correlation at each lag by the sample variance (var(y,1)) so that the autocorrelation at lag 0 is unity. Function Pacf computes (and by default plots) an estimate of the partial autocorrelation function of a (possibly multivariate) time series. , anisotropic). Wh at is the variance of the pro ces s in terms of !. Location - download. We can split Finto three components: discrete, continuous, and singular. The mgf of Xexists for all real values of tand is given by M(t) = et e t 2t;t6= 0 ;M(0) = 1: Use the result of the preceding exercise to show that P(X 1) = 0 and. Welsh Graham C. Day Copies activated. Lagged/Cross Correlations with missing values. Use Automated Cross Correlations in Excel to Find Leading Indicators—Part 1 Leading indicators can help you to forecast more accurately. This is one of the 100+ free recipes of the IPython Cookbook, Second Edition, by Cyrille Rossant, a guide to numerical computing and data science in the Jupyter Notebook. cov,ddmatrix-method. Some assignments may require the use of computer software such as Matlab, Gauss, R, Ox. Properties of MA Finite Process 3. Discount not applicable for individual purchase of ebooks. Probability and Statistics for Data Science Training Course in Austria taught by experienced instructors. parker sio 223b class notes, spring 2011. And we'll estimate autocovariance coefficients of a time series at different lags. cor,ddmatrix-method. For single matrix input, C has size [size(A,2) size(A,2)] based on the number of random variables (columns) represented by A. I have a slightly different problem.$\endgroup$- Robert Israel Dec 1 '15 at 22:40. 0 50 100 150 200 250 300 350 0 2k 4k 6k 8k 10k. Why autocorrelation matters. Hannig (2012) Generalized Fiducial Inference for Normal Linear Mixed Models, Annals of Statistics, 40, pp. 1); or via ii) the dynamic approach (Section 3. And cross correlations can help you identify leading indicators. And we'll estimate autocovariance coefficients of a time series at different lags. A tool for Kalman filter tuning was presented. The theoretical autocovariance function of an ARMA(p,q) with unit variance is computed. This MATLAB function returns the covariance. We can see in this plot that at lag 0, the correlation is 1, as the data is correlated with itself. Feldman's Badges sets the state for Matlab's normal (Gaussian) random number generator compute sample autocovariance of a time series (vector). Good article! MATLAB's Econometrics Toolbox provides greatly expanded support for working with ARMA, ARMAX, and even VARMA models, much more than this article gives credit for. Fejer’s Kernel. The following Matlab project contains the source code and Matlab examples used for computes the autocovariance of two columns vectors consistently with the var and cov functions. 564 but as we know it equals$\pi, so the answer is about half of the real period in this case. Goosebumps - download. Returns a distributed matrix. Matlab: rednoise. In a two-dimensional (2D) stochastic analysis, variation of the environmental properties or hydrogeological data along different directions can be similar (i. 1 Introduction Chapter 6 discussed modulation and demodulation, but replaced any detailed discussion of the noise by the assumption that a minimal separation is required between each pair of signal points. Finding autocovariance of AR(2) Ask Question Asked 6 years, 2 months ago. 13 Downloads. The autocovariance calculated using the MATLAB command. Montgomery, Cheryl L. Autocovariance function is defined, basically, just taking covariance of different elements in our sequence, in our stochastic process. There are (at least) 2. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Corequisite: MATH 152 and MATH 232. 4Sight is designed with features that make life easy: a Microsoft Windows® interface, open data file format, extensive 2D and 3D displays, filtering, data masking, fiducial alignment, diffraction analysis and much more. 1 After intravenous injection, ICG is bound to plasma proteins, mainly α-lipoproteins. I know how to arrive at each solution using mathematical techniques, but I also need to know how to arrive at each one using Matlab's own built in functions. The Autocorrelation function is the normalized autocovariance function φ(τ)/φ(0) = r(τ);. Tags Add Tags. Review: Causality, invertibility, AR(p) models 2. If now one assunies that there exists some function g(t) such that according to (2) and the assumptions mentioned before, k &(a) = - bI2. Note: CD-ROM/DVD and other supplementary materials are not included as part of eBook file. Please help improve this article by adding citations to reliable sources. 1 Introduction & General Instructions The purpose of this set of homework assignments is to make the student familiar with the practical handwork and theoretical effort required in time series analysis of real life data. For example, c=. System Identification & Parameter Estimation (SIPE) (Matlab) • Matlab Command History (if applicable) • Dividing the autocovariance by the variance gives the. com/course/ud501. Interaction among scatterers is the most significant source of deviation between theory and measurement based autocovariance estimates. We assume to have n observations, t = 1;:::;n. Effects on spectrum of using finite duration of data. Objectives • Understand that prediction using a long past can be dicult because a large matrix has to be inverted, thus alternative, recursive method are often used to avoid direct inversion. Finding the autocorrelation of a sine wave. Let us state and prove the Cauchy-Schwarz inequality for random variables. The results show that significant. We must focus on relevant inputs from our senses – such as the bus we need to catch – while ignoring distractions – such as the eye-catching displays in the shop windows we pass on the same street. Derivation of the Autocovariance function of a Moving Average process (MA(q))). If you need to do it hundreds of times in a loop, with different data sets, etc. where the deterministic process B(t) is the BOD (mg/l), K 1 is the reaction rate coefficient (l/day) and s 1 is the source or sink along the stream. Correlation is a measure of the strength of the relationship between two variables. Useful m-files and data are available in: Matlab basic: the basic Matlab installation without any extra toolboxes. 本书以易于理解的方式讲述了时间序列模型及其应用，主要内容包括：趋势、平稳时间序列模型、非平稳时间序列模型、模型识别、参数估计、模型诊断、预测、季节模型、时间序列回归模型、异方差时间序列模型、谱分析入门、谱估计、门限模型. corrcoef(X) is the zeroth lag of the covariance function, that is, the zeroth lag of xcov(x,'coeff') packed into a square array. Prerequisite: (CMPT 128, CMPT 120, or CMPT 130)and (MATH 151 or MATH 150). Bernoulli and Binomial random variables 2. Gibb’s phenomena. This method computes the Pearson correlation between the Series and its shifted self. 2 Continuous-time Gaussian Markov Processes We first consider continuous-time Gaussian Markov processes on the real line, and then relate the covariance function obtained to that for the stationary solution of the SDE on the circle. We also take our first steps on developing the. ; the sequence of pdfs of Xn is called the first-order pdf of the process xn 1 0 1 z Since Xn is a differentiable function of the continuous r. array 2d array of size nr X T with the temporal components center: np. The above model can be compactly written as Z t = + (B)a t. I have a slightly different problem. For example, is you were calculating the third iteration (i = 3) using a lag k = 7, then the calculation for that iteration would look like this: (y3 - y-bar)(y10 - y-bar) Iterate through all values of "i" and then take the sum and divide it by the number of values in the data set. In general, the autocorrelation function. For the input sequence x=[1,2,3,4], the command xcorr(x) gives the following result. The distinct cutoff of the ACF combined with the more gradual decay of the PACF suggests an MA(1) model might be appropriate for this data. processes, all of which have autocovariance functions which decay exponentially fast to zero. The toolbox then offers except of a large set of state estimation algorithms for prediction, filtering, and smoothing, the integrated easy-to-use method. Each plot. • Some idea of what a basis of a vector space is. Peter Bartlett 1. @kamaci: it depends. autocorr(y,Name,Value) uses additional options specified by one or more name-value pair arguments. Linear algebra and matrices, complex variables, mathematical transforms and their inter-relations. Visualizza il profilo di Alessia Battilana su LinkedIn, la più grande comunità professionale al mondo. Buck AeroTech Research (U. We consider the problem of estimating the high-dimensional autocovariance matrix of a stationary random process, with the purpose of out of sample prediction and feature extraction. This instrument has a 2 angstrom vertical resolution and a 1 micron lateral resolution you can obtain both 2D and 3D plots, which are then transferred to the Matlab or Mathcad for further analysis. The following Matlab project contains the source code and Matlab examples used for computes the autocovariance of two columns vectors consistently with the var and cov functions. Interpretation. For the input sequence x=[1,2,3,4], the command xcorr(x) gives the following result. These are plots that graphically summarize the strength of a relationship with an observation in a time series with observations at prior time steps. Geosphere 15 :5, 1665-1676. If T istherealaxisthenX(t,e) is a continuous-time random process, and if T is the set of integers then X(t,e) is a discrete-time random process2. > Hi, > > I have a MATLAB related question. In locits: Test of Stationarity and Localized Autocovariance. For example, c=. Eigenvectors (red) do not change direction when a linear. Study the pattern of autocorrelations and partial. Scalar or vector of nonnegative integers indicating the number of autocovariance lags to include in the Newey-West estimator of the long-run variance. • economics - e. Spectrumestimation ELEC-E5410 Signalprocessingfor communications. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. A lot of m-files are found in this page. For one small protein, we. Autocovariance function, generalized least squares: Lecture Slides: Lecture Notes: Lecture Slides: Covariance Modeling: Estimating the covariance [Quiz 1] Kriging and prediction: No Class: Independence Day: Lecture Slides Reference: Cressie Ch 1: Lecture Slides Reference: Cressie Ch 2-4: Autoregressive Processes: AR processes in time: AR. autocov computes the autocovariance between two column vectors X and Y with same length N using the Fast Fourier Transform algorithm from 0 to N-2. Homogeneous linear difference equations. Cross-covariance or autocovariance, returned as a vector or matrix. Autocorrelation is a mathematical representation of the degree of similarity between a given time series and a lagged version of itself over successive time intervals. x must be a column vector having length m not less than maxlag+1. Estimate speed of adjustment, if appropriate. Introduction to MATLAB and its use in engineering. CORS Home | Data Products | CORS Map | Newsletter | General Info | CORS Site Guidelines | GPS Links | Contact Us. For one small protein, we. PGFs are useful tools for dealing with sums and limits of random variables. If A is a row or column vector, C is the scalar-valued variance. Time series clustering is implemented in TSclust, dtwclust, BNPTSclust and pdc. Learn more about correlation, lag, cross correlation, xcorr, corrcoef, pairwise, nan, missing values. Previous question Next question Transcribed Image Text from this Question. The autocorrelation function is de ned as ˆ(h) = (h) (0) John Fricks Time Series II { Frequency. The Fundamentals of MTF, Wiener Spectra, and DQE Robert M Nishikawa Kurt Rossmann Laboratories for Radiologic Image Research Department of Radiology, The University of Chicago Motivation Goal of radiology: to diagnosis and treat disease by Role of Medical Physicist: to help maximize patient benefit. 2: The autocovariance ct;¿ of the process shown in the bottom of Fig. Introduction to Time Series Analysis. We then plot the resulting autocovariance function and limit the x axis to ± 30 successive beats to better evaluate the decrease in covariance with successive beats. The function Acf computes (and by default plots) an estimate of the autocorrelation function of a (possibly multivariate) time series. fi;fl/, for 1 •fi6Dfl•N, with probabilities 1=N.\begingroup\$ I think he/she asked autocorrelation, not autocovariance. uses the following files. fft bool, optional. Enter search keywords: Popular Artists. Peter Bartlett 1. The course also covers statistical image understanding, elements of pattern theory, simulated annealing, Metropolis-Hastings algorithm, and Gibbs sampling. Statistics and Probability Letters, 2014, Vol. Correlation and covariance are closely related concepts in theoretical statistics. Features: - High compression ratio in new 7z format with LZMA compression - Supported formats: - Packing / unpacking: 7z, ZIP, GZIP, BZIP2 and TAR - Unpacking only: RAR, CAB, ISO, ARJ, LZH, CHM, Z, CPIO, RPM, DEB and NSIS - For ZIP and GZIP formats, 7-Zip provides a compression ratio that is 2-10 % better than the ratio provided by PKZip. 1To make it easier for researchers to apply these estimators, we have posted Matlab code for both estimators on our websites. incorrectStringFormat MATLAB:nargchk:notEnoughInputs (thrown by nargchk). Making statements based on opinion; back them up with references or personal experience. Returns a distributed matrix. We demon-strate the applicability of our method to model time series data consisting of daily values of the interest rate on federal funds. The xcorr function evaluates the sum shown above. For example, is you were calculating the third iteration (i = 3) using a lag k = 7, then the calculation for that iteration would look like this: (y3 - y-bar)(y10 - y-bar) Iterate through all values of "i" and then take the sum and divide it by the number of values in the data set. Dorf Boca Raton: CRC Press LLC, 2000 The Intel Pentium® processor, introduced at speeds of up to 300 MHz, combines the architectural advances in the Pentium Pro processor with the instruction set extensions of Intel MMX™ media enhancement technology. It is not allowed to use xcorr, xcov, mean, cov, var etc. The General Linear Model (GLM) Ged Ridgway Wellcome Trust Centre for Neuroimaging University College London SPM Course Vancouver, August 2010. Otherwise it is nonin-vertible. The true cross-correlation sequence is a statistical quantity defined as.
j4c6ilstdu2, uh2ja0l3cddrh8x, 9i3jmlflivr9, jnlhkvit6pa5, r95j8jkxbnq, 7mkv07shk2w0, ouxiomk7kvncqcx, 8njdl3ndo1s1c8w, yf1beffx632zl0, vprnk1h8os, bjuxv6yt4q, xnwunkd9xdgap, encqjpwnkq5zimw, bth6mbcjnyr4, k1nrl1v4av2jcfb, zn2nmemn50, zmj8wvxlfg8, ztie88b2borgx, ys4n7kse5a, lolx8f8i8gc8d, 67ej26yipr, ar0vxmqpuovrg, rhyc4nfb6kew, qr0zoic7j5umv, 3w611ameu8gx, 2bifymft939zjym | 12,185 | 52,576 | {"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 | 3 | CC-MAIN-2020-24 | latest | en | 0.896778 |
https://www.unitconverters.net/pressure/millipascal-to-dekapascal.htm | 1,660,966,518,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00556.warc.gz | 882,320,022 | 3,350 | Home / Pressure Conversion / Convert Millipascal to Dekapascal
# Convert Millipascal to Dekapascal
Please provide values below to convert millipascal [mPa] to dekapascal [daPa], or vice versa.
From: millipascal To: dekapascal
### Millipascal to Dekapascal Conversion Table
Millipascal [mPa]Dekapascal [daPa]
0.01 mPa1.0E-6 daPa
0.1 mPa1.0E-5 daPa
1 mPa0.0001 daPa
2 mPa0.0002 daPa
3 mPa0.0003 daPa
5 mPa0.0005 daPa
10 mPa0.001 daPa
20 mPa0.002 daPa
50 mPa0.005 daPa
100 mPa0.01 daPa
1000 mPa0.1 daPa
### How to Convert Millipascal to Dekapascal
1 mPa = 0.0001 daPa
1 daPa = 10000 mPa
Example: convert 15 mPa to daPa:
15 mPa = 15 × 0.0001 daPa = 0.0015 daPa | 268 | 665 | {"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} | 3.125 | 3 | CC-MAIN-2022-33 | latest | en | 0.298325 |
https://electronics.stackexchange.com/questions/29301/relationship-of-signal-frequency-and-the-snr | 1,716,080,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00696.warc.gz | 199,994,521 | 40,563 | Relationship of signal Frequency and the SNR?
Sincerely, It is not my homework. It was the question asked in one German University's Online Exams.
I went through Google as well but couldn't find a single way out to figure the solution of this Problem.
A Received Signal is filtered in the low pass filter. At the Input of the Low Pass filter the Bandwidth is B=80khz and the SNR is 30 dB. What is the SNR at output of Filter if equivalent noise bandwidth of the filter (low pass filter) is 20kHz.?
• The way your question is now, you are asking us to do all of the work for you. Meet us half way and tell us what you have done and what you are getting stuck at. We will then help teach you how to solve the problem. However, your question lacks some important details. What is the input signal? If the signal has a bandwidth of 80khz and you pass it through a lpf of 20 khz, then your signal wont look too good on the other end. On the other hand, if your signal is only 10 khz, a lpf will help to remove noise while keeping your signal intact. Also, what is the noise? Is it white? Apr 4, 2012 at 16:08
• Sorry for the Inconvenience caused .Actually it was the question asked to me in one online exam. i went through the google and search but i couldn't figure out any relationships of SNR and the Frequency of the Signal.So i was looking for the Relationships only? Apr 5, 2012 at 0:54
• Kellenjb is right: this question misses details. You should know the bandwith of signal and noise at least, and also the spectral distribution (if flat or whatever). Otherwise you need the SNR of the output to make the inverse process Apr 6, 2012 at 7:42
• Still rude to post someone's exam question. May 22, 2013 at 10:37 | 429 | 1,716 | {"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-2024-22 | latest | en | 0.963986 |
http://mathhelpforum.com/differential-geometry/83444-complex-integration.html | 1,527,484,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00534.warc.gz | 179,048,112 | 9,596 | 1. complex integration
Suppose that $\displaystyle f(z)$ is analytic on a closed curve $\displaystyle \gamma$(i.e., $\displaystyle f$ is analytic in a region that contains $\displaystyle \gamma$). Show that
$\displaystyle \int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.
2. Originally Posted by Stiger
Suppose that $\displaystyle f(z)$ is analytic on a closed curve $\displaystyle \gamma$(i.e., $\displaystyle f$ is analytic in a region that contains $\displaystyle \gamma$). Show that
$\displaystyle \int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.
If the curve is smooth enough to have a differentiable parametrisation then this is just integration by parts. In fact, suppose that $\displaystyle \gamma$ is given by a differentiable path $\displaystyle z = \gamma(t)\;(0\leqslant t\leqslant1)$, with $\displaystyle \gamma(1) = \gamma(0)$. Then
\displaystyle \begin{aligned}\int_\gamma \overline{f(z)}f'(z)\,dz &= \int_0^1\overline{f(\gamma(t))}f'(\gamma(t))\,\gam ma'(t)dt\\ &= \Bigl[\overline{f(\gamma(t))}f(\gamma(t))\Bigr]_0^1 - \int_0^1\overline{f'(\gamma(t))\gamma'(t)}f(\gamma (t))\,dt\\ &= -\overline{\int_0^1\overline{f(\gamma(t))}f'(\gamma (t))\,\gamma'(t)\,dt} = -\overline{\int_\gamma \overline{f(z)}f'(z)\,dz}.\end{aligned}
Thus the integral is equal to the negative of its complex conjugate, so it is purely imaginary. | 437 | 1,352 | {"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} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.795291 |
https://risingentropy.com/logarithms/ | 1,653,563,805,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00640.warc.gz | 540,276,850 | 45,938 | # Short and sweet proof of the f(xy) = f(x) + f(y) logarithmic property
If you want a continuous function f(x) from the reals to the reals that has the property that for all real x and y, f(xy) = f(x) + f(y), then this function must take the form f(x) = k log(x) for some real k.
A proof of this just popped into my head in the shower. (As always with shower-proofs, it was slightly wrong, but I worked it out and got it right after coming out).
I haven’t seen it anywhere before, and it’s a lot simpler than previous proofs that I’ve encountered.
Here goes:
f(xy) = f(x) + f(y)
differentiate w.r.t. x…
f'(xy) y = f'(x)
differentiate w.r.t. y…
f”(xy) xy + f'(xy) = 0
rearrange, and rename xy to z…
f”(z) = -f'(z)/z
solve for f'(z) with standard 1st order DE techniques…
df’/f’ = – dz/z
log(f’) = -log(z) + constant
f’ = constant/z
integrate to get f…
f(z) = k log(z) for some constant k
And that’s the whole proof!
As for why this is interesting to me… the equation f(xy) = f(x) + f(y) is very easy to arrive at in constructing functions with desirable features. In words, it means that you want the function’s outputs to be additive when the inputs are multiplicative.
One example of this, which I’ve written about before, is formally quantifying our intuitive notion of surprise. We formalize surprise by asking the question: How surprised should you be if you observe an event that you thought had a probability P? In other words, we treat surprise as a function that takes in a probability and returns a scalar value.
We can lay down a few intuitive desideratum for our formalization of surprise, and one such desideratum is that for independent events E and F, our surprise at them both happening should just be the sum of the surprise at each one individually. In other words, we want surprise to be additive for independent events E and F.
But if E and F are independent, then the joint probability P(E, F) is just the product of the individual probabilities: P(E, F) = P(E) P(F). In other words, we want our outputs to be additive, when our inputs are multiplicative!
This automatically gives us that the form of our surprise function must be k log(z). To spell it out explicitly…
Desideratum: Surprise(P(E, F)) = Surprise(P(E)) + Surprise(P(F))
But P(E,F) = P(E) P(F), so…
Surprise(P(E) P(F)) = Surprise(P(E)) + Surprise(P(F))
Renaming P(E) to x and P(F) to y…
Surprise(xy) = Surprise(x) + Surprise(y)
Thus, by the above proof…
Surprise(x) = k log(x) for some constant k
That’s a pretty strong constraint for some fairly weak inputs!
That’s basically why I find this interesting: it’s a strong constraint that comes out of an intuitively weak condition. | 700 | 2,684 | {"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} | 3.734375 | 4 | CC-MAIN-2022-21 | longest | en | 0.932471 |
http://www.thefreedictionary.com/percentiles | 1,534,238,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221208750.9/warc/CC-MAIN-20180814081835-20180814101835-00373.warc.gz | 759,107,536 | 13,220 | # percentile
(redirected from percentiles)
Also found in: Thesaurus, Medical, Encyclopedia.
## per·cen·tile
(pər-sĕn′tīl′) Statistics
n.
1. Any of the groups that result when a frequency distribution is divided into 100 groups of equal size.
2. Any of the values that separate each of these groups.
## percentile
(pəˈsɛntaɪl)
n
(Statistics) one of 99 actual or notional values of a variable dividing its distribution into 100 groups with equal frequencies; the 90th percentile is the value of a variable such that 90% of the relevant population is below that value. Also called: centile
## per•cen•tile
(pərˈsɛn taɪl, -tɪl)
n.
one of the values of a statistical variable that divides the distribution of the variable into 100 groups having equal frequencies: Ninety percent of the values lie at or below the ninetieth percentile, ten percent above it.
[1880–85]
## per·cen·tile
(pər-sĕn′tīl′)
Any of the 100 equal parts into which the range of the values of a set of data can be divided in order to show the distribution of those values. The percentile of a given value is determined by the percentage of the values that are smaller than that value. For example, a test score that is higher than 95 percent of the other scores is in the 95th percentile.
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 percentile - (statistics) any of the 99 numbered points that divide an ordered set of scores into 100 parts each of which contains one-hundredth of the totalcentilemark, score, grade - a number or letter indicating quality (especially of a student's performance); "she made good marks in algebra"; "grade A milk"; "what was your score on your homework?"statistics - a branch of applied mathematics concerned with the collection and interpretation of quantitative data and the use of probability theory to estimate population parameters
Translations
[pəˈsentaɪl] N
## percentile
nProzent nt; he’s in the ninetieth percentile for reading and mathsin Lesen und Rechnen gehört er zu den besten zehn Prozent
## percentile
[pəˈsɛntaɪl] n
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References in periodicals archive ?
In childhood, high blood pressure is based on percentiles, rather than blood pressure level.
To carry out this analysis, we calculate the percent change in real wages between 2007 and 2014 at different percentiles of the wage distribution.
Objective: To develop gender and gestation-specific growth percentiles for singleton live-born neonates and to compare new weight for age unisex percentiles with Lubchenco unisex percentiles.
These lower percentiles provide practitioners with an evaluation of the product's early failures along with providing information for specification limits, warranty, and cost analysis.
Tom Loveless of the Brookings Institution examined National Assessment of Educational Progress (NAEP) data in reading and mathematics at grades 4 and 8 for the 10th and 90th percentiles.
The risk was not significantly increased for adolescents in the two lowest BMI percentiles.
The median levels of the dimethyl and diethyl phosphates for adults, females, and Mexican Americans are slightly less than the 75th percentiles of the U.
The study defined a BMI between the 75th and 84th percentiles as being the upper limit of healthy.
These percentiles are based on the national distribution of scores on the Iowa Test of Basic Skills.
Although the cohort initially had generally larger body sizes relative to national percentiles, 2 years later they showed significant declines relative to national percentiles.
Scoring among the highest in the state were the Las Virgenes Unified School District in western Los Angeles County and the adjacent Oak Park Unified School District in Ventura County, which topped the region with many of their schools reaching into the high 70th and middle and upper 80th percentiles.
Generally, the scores are provided in terms of percentiles, although standard deviations, stanines, quartile deviations, and grade equivalents may also be used to indicate student achievement.
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Open / Close | 917 | 4,062 | {"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} | 3.234375 | 3 | CC-MAIN-2018-34 | latest | en | 0.79168 |
https://math.stackexchange.com/questions/3192547/an-order-6-configuration | 1,558,674,504,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257514.68/warc/CC-MAIN-20190524044320-20190524070320-00167.warc.gz | 555,774,004 | 31,762 | # An order-6 configuration
Here's an example of an order $$96_6$$ configuration found by L.W. Berman. Every point has six lines, every line has six points.
The unique 6,6 cage graph is bipartite and is a Levi graph for the following set of 31 lines.
{{1,2,4,9,13,19},{1,3,8,12,18,31},{1,5,11,24,25,27},{1,6,10,16,29,30},{1,7,20,21,23,28},{1,14,15,17,22,26},{2,3,5,10,14,20},{2,6,12,25,26,28},{2,7,11,17,30,31},{2,8,21,22,24,29},{2,15,16,18,23,27},{3,4,6,11,15,21},{3,7,13,26,27,29},{3,9,22,23,25,30},{3,16,17,19,24,28},{4,5,7,12,16,22},{4,8,14,27,28,30},{4,10,23,24,26,31},{4,17,18,20,25,29},{5,6,8,13,17,23},{5,9,15,28,29,31},{5,18,19,21,26,30},{6,7,9,14,18,24},{6,19,20,22,27,31},{7,8,10,15,19,25},{8,9,11,16,20,26},{9,10,12,17,21,27},{10,11,13,18,22,28},{11,12,14,19,23,29},{12,13,15,20,24,30},{13,14,16,21,25,31}}
Is there some way of arranging 31 points so that the corresponding pseudolines/splines/curves/lines/ellipses going the points is reasonably nice and symmetric?
EDIT: It occurs to me that these lines might be equivalent to the order-31 difference set generated by {1, 5, 11, 24, 25, 27}. That turns out to be true, here's the 6,6 cage with a difference-set based embedding.
• My Mathematica-fu isn't what it should be. Could you provide the adjacency matrix for the graph you want to embed? – Blue Apr 18 at 16:40
• Blue: I'm looking for a configuration, not a graph. – Ed Pegg Apr 18 at 16:59 | 604 | 1,416 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-22 | latest | en | 0.870986 |
http://mathhelpforum.com/pre-calculus/131395-logarithm-problem-print.html | 1,516,305,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00457.warc.gz | 225,651,989 | 2,730 | logarithm problem
• Mar 1st 2010, 06:56 AM
mastermin346
logarithm problem
how to solve this question
$
2+\log_2{(3x-1)}=\log_2{4x}
$
• Mar 1st 2010, 07:14 AM
earboth
Quote:
Originally Posted by mastermin346
how to solve this question
$
2+\log_2{(3x-1)}=\log_2{4x}
$
1. Re-write the equation:
$2+\log_2{(3x-1)}=\log_2{4x}~\implies~2=\log_2{4x} - \log_2{(3x-1)}~\implies~$ $2=\log_2\left( \frac{4x}{(3x-1)}\right)$
2. Now use the base 2 on both sides of the equation:
$4=\frac{4x}{(3x-1)}$ Solve for x. Keep in mind that $x > \tfrac13$. | 240 | 544 | {"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": 0, "img_math": 0, "codecogs_latex": 6, "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} | 4.21875 | 4 | CC-MAIN-2018-05 | longest | en | 0.56806 |
http://sepwww.stanford.edu/data/media/public/docs/sep61/carlos1/paper_html/node4.html | 1,508,634,223,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00116.warc.gz | 296,959,519 | 2,745 | Next: SYNTHETIC MODEL AND REALITY Up: Cunha & Muir: Separation Previous: SEPARATION OF P AND
# DIVIDING THE DATA INTO SNELL ZONES
Although the use of the critical horizontal slowness to perform the separation between P and S waves is effective for shallow reflectors, the finite size of the cable restricts its application to deep reflectors because a converted wave with horizontal slowness larger than the P wave critical slowness will be received at very large offsets.
A possible way to overcome the limitations associated with the finiteness of the field aperture is to use a criterion that somehow takes into account the depth of the reflector. Transforming the data into the -p domain provides the necessary flexibility for use of a variable slowness cutoff. Tatham and Goolsbee (1984) showed also that better results can be achieved when a hyperbolic velocity filtering is used during the -p transform to limit the range of reasonable stacking velocities.
As we will see, the separation of the data into ranges of Snell rays would seem to be another appropriate way to perform the slowness filtering with a variable-slowness cutoff.
A Snell ray (Ottolini, 1982) can be defined for a plane-layered earth as a ray that keeps a constant horizontal slowness (obeys the Snell law) while it propagates through the subsurface, as illustrated in Figure . The reflection points corresponding to a Snell ray with ray parameter (or horizontal slowness) p are defined by
For the simple case of a plane-layered earth, the horizontal slowness of a P wave will always be lower than the horizontal slowness of a converted wave recorded at the same position in the x-t domain. It is possible then to choose a set of ray parameters and divide the data into regions whose P waves Snell rays are limited by two adjacent values (Figure ). Since the converted waves inside each region will have higher horizontal slownesses than the P waves, a different filter can be applied to each region; the cutoff value is given by the ray parameter of the Snell ray that defines the upper limit of that region.
Next: SYNTHETIC MODEL AND REALITY Up: Cunha & Muir: Separation Previous: SEPARATION OF P AND
Stanford Exploration Project
1/13/1998 | 496 | 2,225 | {"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 | 0.886555 |
https://www.numbers-figures.com/square-root-of-75027-seventyfivethousandandtwentyseven | 1,631,892,397,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00125.warc.gz | 941,249,836 | 4,413 | # square root of 75027 seventyfivethousandandtwentyseven
The square root of 75027 is:
273.910569347
## What is the square root of a number?
The square root of a number is a value that when multiplied by itself equals the original number. The square root is the opposite of mathematical function of the square.
The classic symbol of the square root is the normal root sign without specifying the root exponent. | 92 | 414 | {"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-2021-39 | longest | en | 0.780937 |
https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/sulfuric-blank-acid | 1,718,469,158,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00789.warc.gz | 595,454,141 | 8,802 | # Moles of Sulfuric acid
## sulfuric acid: convert moles to volume and weight
### Volume of 1 mole of Sulfuric acid
centimeter³ 53.69 milliliter 53.69 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.23 inch³ 3.28 US fluid ounce 1.82 liter 0.05 US gallon 0.01 meter³ 5.37 × 10-5 US pint 0.11 metric cup 0.21 US quart 0.06 metric tablespoon 3.58 US tablespoon 3.63 metric teaspoon 10.74 US teaspoon 10.89
### Weight of 1 mole of Sulfuric acid
carat 490.39 ounce 3.46 gram 98.08 pound 0.22 kilogram 0.1 tonne 9.81 × 10-5 milligram 98 078
### The entered amount of Sulfuric acid in various units of amount of substance
centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0
#### Foods, Nutrients and Calories
JACKIE'S JAMS, PUMPKIN BUTTER, UPC: 858441001120 weigh(s) 271 grams per metric cup or 9 ounces per US cup, and contain(s) 125 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
trans-beta-Carotene in Tomatoes, grape, raw
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Natural Reef weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Molybdenum(IV) sulfide [MoS2] weighs 5 060 kg/m³ (315.88548 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Canola oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The sievert high-energy protons [Sv p] is a derived unit of ionizing radiation dose in the International System of Units (SI) and is a measure of the effective biological damage of low levels of ionizing radiation on the human body caused by exposure to high-energy protons.
Electric current is a motion of electrically charged particles within conductors or space.
t/ml to oz/dm³ conversion table, t/ml to oz/dm³ unit converter or convert between all units of density measurement.
#### Calculators
Oils, fuels, refrigerants: compute weight by volume and temperature | 730 | 2,427 | {"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-2024-26 | latest | en | 0.618789 |
http://smallbusiness.chron.com/inventory-volume-variance-formula-25641.html | 1,524,709,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948047.85/warc/CC-MAIN-20180426012045-20180426032045-00114.warc.gz | 284,846,794 | 10,186 | # What Is the Inventory Volume Variance Formula?
by Christine Aldridge
A differentiation in the amount of inventory that a company has on hand and the amount that it has on the books can lead to signs of errors in recording or dishonest employees. Both have potentially negative consequences if they cause losses and inaccuracies within the financial statements. Management should understand how to calculate inventory volume variance and should understand what this might signify.
## Inventory Variance by Quantity
To calculate inventory volume variance by quantity, begin by counting what is on hand in your warehouse, store or other location where you keep the inventory. Next, subtract the amount that your books indicate you should have on hand at that time. A negative number indicates that you have too few on hand, while a positive number indicates that you have more on hand than you should have at the current time.
## Checking the Quantity on the Books
If the amount that you have on hand varies from the amount that is recorded on the books, then it is helpful to first check the amount on the books for any inaccuracies. Take the beginning quantity, which is the amount you had on hand during the last inventory count, and add to it the purchases made from the time of the last inventory count until the most recent one. Then subtract the quantity sold, donated or otherwise removed over that same period. If this amount matches the books, then the books are correct. If not, then the inaccuracy is likely due to a recording error and you should match all of the entries.
## Inventory Variance by Cost
Using cost, rather than a physical quantity count, simply removes the positive or negative aspect of the variance. For each item of inventory, you multiply the amount on hand by the cost of that item. You do the same for the amount that is on the books. Subtract the dollar value of the amount on hand from that of what is on the books, but do not apply a positive or negative value to it. For example, if you have \$120 in purple widgets on hand and a number of them on the books worth \$123, then you have a variance of \$3. Repeat this for each of the inventory items and then add together all of the variances to calculate the inventory volume variance in dollars.
## Significance
An accurate inventory count is important to reduce the likelihood of a financial statement error, or misstatment, ordering too much or too little of an item, and also for accurate budgeting. You can decrease the chances that a variance will occur by keeping accurate records, maintaining property internal controls and doing frequent inventory counts. Internal controls include proper paperwork such as purchase and sales orders, separation of duties and keeping the warehouse or storage facility locked when a manager is not available.
#### About the Author
Christine Aldridge is a financial planner who has been writing articles related to personal finance since 2011. She has bachelor's degrees in political science from North Carolina State University and in accounting from University of Phoenix. Aldridge is completing her Certified Financial Planner designation via New York University.
#### Photo Credits
• Rebecca Van Ommen/Lifesize/Getty Images | 629 | 3,268 | {"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} | 3.25 | 3 | CC-MAIN-2018-17 | latest | en | 0.944851 |
https://genius777.com/education_extras/math_folder/temperature.html | 1,660,391,061,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571950.76/warc/CC-MAIN-20220813111851-20220813141851-00664.warc.gz | 274,079,347 | 2,554 | ## Temperature
Temperature is how hot or cold a thing is. It is measured by a thermometer. Thermometers use a temperature scale, which is most often used in degrees Celsius (°C), sometimes called centigrade. In the USA, degrees Fahrenheit (°F) are more often used while scientists mostly use kelvins (K) to measure temperature because it never goes below zero.
### Celsius
The Celsius scale (°C) is used for common temperature measurements in most of the world. It is an empirical scale. It developed by a historical progress, which led to its zero point 0°C being defined by the freezing point of water, with additional degrees defined so that 100°C was the boiling point of water, both at sea-level atmospheric pressure.
Useful temperatures:
On the Celsius scale, water freezes at 0° and boils at 100°.
Room temperature is about 20 °C.
A hot sunny day might have a temperature of 30°C.
Absolute zero (the coldest possible temperature) is -273.15 °C.
Winters in Antarctica can be between -80 and -90 °C.
A human's body temperature is usually 37 °C.
You can bake cookies in your oven at a temperature of 180°C.
### Fahrenheit
The United States commonly uses the Fahrenheit scale, on which water freezes at 32 °F and boils at 212 °F at sea-level atmospheric pressure. The degree Fahrenheit is often considered to be "old fashioned" throughout the majority of the world as it is an older and outdated way of measuring temperature.
Useful temperatures:
Water freezes at 32 °F.
The temperature inside the human body is usually 98 °F.
Water boils at 212 °F.
The coldest possible temperature is absolute zero. Absolute zero is -459 °F.
A hot sunny day might have a temperature of 85°F.
You can bake cookies in your oven at a temperature of 350°F.
### Conversion of Temperature
from Celsius to Fahrenheit: first multiply by 1.8, then add 32
from Fahrenheit to Celsius: first subtract 32, then multiply by 1.8
Celsius to Fahrenheit : (°C × 1.8) + 32 =°F
Fahrenheit to Celsius : (°F - 32) ÷ 1.8 =°C
from Celsius to Celsius
Fahrenheit [°F] = [°C] × 9/5 + 32 [°C] = ([°F] - 32) × 5/9
Kelvin [K] = [°C] + 273.15 [°C] = [K] - 273.15
Newton [°N] = [°C] × 33/100 [°C] = [°N] × 100/33 | 605 | 2,182 | {"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 | 3 | CC-MAIN-2022-33 | latest | en | 0.862359 |
https://spiritualwander.com/is-11am-in-the-morning | 1,660,657,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00718.warc.gz | 502,202,999 | 11,176 | Is it eleven o'clock in the morning?
11 a.m. is late in the morning. It switches from AM to PM at 12 p.m. Midnight marks the transition from PM to AM. If you're asked to report for duty at 11:00, that means "OK, let's go!" At noon, if it's still 11:00, then something has gone wrong. They should be asking you about your shift schedule, not sending you on your way at once!
He made an appearance on The Tonight Show with Jay Leno on February 8, 2009. He was there to promote his new album, Up Here Only God Forgets.
During the interview, which can be seen here, Jimmy asks Jay what time it is and when told it's 11:15, replies "Oh no, not again!". Apparently, this is a common response when he appears on television interviews.
Also, during this interview, he mentions having two molars removed. This was done because they were causing problems with his chewing mechanism.
Finally, he says he wants to thank his wife for being so supportive throughout all these years. She hasn't left his side even though he has been through quite a few tours and concert appearances.
Why does it go from 11 p.m. to 12 a.m.?
As a general rule, the day begins at 12:00 a.m. The day is divided into two half (24 hours = 12 hours before midday + 12 hours after midday), therefore it is the 11th hour before midday at 11 a.m. When the clock strikes twelve again, the time is converted to PM (after midday), thus it's 12 PM after 11 AM and 12 AM after 11 PM.
However, when there are less than 24 hours in a day, the night ends at midnight. Thus, it is the 11th hour before midnight at 11 PM when the clock strikes twelve again, the time is converted to PM, thus it's 1 PM after 11 AM and 2 AM after 11 PM.
For example, if the clock says 7:30 A.M., you can be sure that it is actually before noon because the morning doesn't last forever. So, it's probably around 9 or 9:30 A.M.
Similarly, if the calendar says 4:15 P.M., you can be sure that it is actually after midnight because there are only 24 hours in a day. So, it must be between 3:15 and 4:15 A.M.
In short, the clock goes back an hour after 11 PM and converts the time to PM. Therefore, it's 1 PM after 11 AM and 2 AM after 11 PM.
Does the day end at 11:59 or midnight?
00 a.m. marks the start of a new day. The previous day ends at 11:59 p.m. and today's date is expressed by using the word "today", followed by the month, followed by the day of the month.
Midnight is when the minute hand of a clock reaches the middle point, it is said to be marking midnight. But what happens if you miss the midnight mark? Well, according to U.S. law, people need to stop working at some point. So the law allows for half days to be taken off (without any pay) if they fall on Saturday or Sunday. If this occurs more than once in a month, then the person has earned themselves overtime pay. Of course, if the employee refuses to accept overtime, then their employer can't force them to do so.
In Canada, employees are required to work for eight hours every day, except for certain public holidays that include Victoria Day, Canada Day, and Labour Day. If an employee works past 8:00 p.m., they are considered to have worked until 9:00 a.m. the next morning and will be paid for one hour of overtime per week, which averages out to about \$25 extra per month.
Cathy Strebe
Cathy Strebe is a spiritual healer who specializes in yoga techniques. Her goal as a healer is to help people feel better and live their best life possible. Cathy knows all about the struggles of being human, and how hard it can be to want things but not have them. She has overcome many obstacles in her own life, and she wants to share that with others so they too can find peace within themselves.
Disclaimer
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Resources tagged with Mathematical reasoning & proof similar to Comparing Continued Fractions:
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There are 185 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
Sums of Squares and Sums of Cubes
Age 16 to 18
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
Impossible Sandwiches
Age 11 to 18
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
Age 16 to 18 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
Continued Fractions II
Age 16 to 18
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
How Many Solutions?
Age 16 to 18 Challenge Level:
Find all the solutions to the this equation.
Plus or Minus
Age 16 to 18 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
The Golden Ratio, Fibonacci Numbers and Continued Fractions.
Age 14 to 16
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
There's a Limit
Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
Unit Interval
Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
Perfectly Square
Age 14 to 16 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
Water Pistols
Age 16 to 18 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
Golden Eggs
Age 16 to 18 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
Sprouts Explained
Age 7 to 18
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
Age 14 to 18 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
To Prove or Not to Prove
Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
Sperner's Lemma
Age 16 to 18
An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.
And So on - and on -and On
Age 16 to 18 Challenge Level:
Can you find the value of this function involving algebraic fractions for x=2000?
Ordered Sums
Age 14 to 16 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
Janine's Conjecture
Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
Binomial
Age 16 to 18 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
Some Circuits in Graph or Network Theory
Age 14 to 18
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
Yih or Luk Tsut K'i or Three Men's Morris
Age 11 to 18 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
Big, Bigger, Biggest
Age 16 to 18 Challenge Level:
Which is the biggest and which the smallest of $2000^{2002}, 2001^{2001} \text{and } 2002^{2000}$?
Whole Number Dynamics I
Age 14 to 18
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
Modulus Arithmetic and a Solution to Differences
Age 16 to 18
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
Whole Number Dynamics II
Age 14 to 18
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
Whole Number Dynamics III
Age 14 to 18
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
Where Do We Get Our Feet Wet?
Age 16 to 18
Professor Korner has generously supported school mathematics for more than 30 years and has been a good friend to NRICH since it started.
Whole Number Dynamics IV
Age 14 to 18
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
Proof of Pick's Theorem
Age 16 to 18 Challenge Level:
Follow the hints and prove Pick's Theorem.
Transitivity
Age 16 to 18
Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics.
A Biggy
Age 14 to 16 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
Three Ways
Age 16 to 18 Challenge Level:
If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.
Target Six
Age 16 to 18 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
Angle Trisection
Age 14 to 16 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
More Sums of Squares
Age 16 to 18
Tom writes about expressing numbers as the sums of three squares.
Modulus Arithmetic and a Solution to Dirisibly Yours
Age 16 to 18
Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
Square Mean
Age 14 to 16 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
Thousand Words
Age 16 to 18 Challenge Level:
Here the diagram says it all. Can you find the diagram?
Diophantine N-tuples
Age 14 to 16 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
An Introduction to Number Theory
Age 16 to 18
An introduction to some beautiful results of Number Theory
Dodgy Proofs
Age 16 to 18 Challenge Level:
These proofs are wrong. Can you see why?
Without Calculus
Age 16 to 18 Challenge Level:
Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.
The Clue Is in the Question
Age 16 to 18 Challenge Level:
Starting with one of the mini-challenges, how many of the other mini-challenges will you invent for yourself?
Diverging
Age 16 to 18 Challenge Level:
Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.
Tree Graphs
Age 16 to 18 Challenge Level:
A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree has. . . .
Integral Inequality
Age 16 to 18 Challenge Level:
An inequality involving integrals of squares of functions.
Tetra Inequalities
Age 16 to 18 Challenge Level:
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.
Notty Logic
Age 16 to 18 Challenge Level:
Have a go at being mathematically negative, by negating these statements.
Interpolating Polynomials
Age 16 to 18 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials. | 2,257 | 9,115 | {"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} | 3.859375 | 4 | CC-MAIN-2018-47 | latest | en | 0.885752 |
http://www.enotes.com/homework-help/find-derivative-algebraic-function-f-x-x-3-5x-3-x-418217 | 1,477,604,085,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721392.72/warc/CC-MAIN-20161020183841-00327-ip-10-171-6-4.ec2.internal.warc.gz | 442,782,725 | 9,575 | # find the derivative of the algebraic function. f(x)=`(x^3+5x+3)/(x^2-1)`
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The derivative of `f(x) = (x^3 + 5x + 3)/(x^2 - 1)` has to be determined.
Use the quotient rule.
`f'(x) = ((x^3 + 5x + 3)'*(x^2 - 1) - (x^3 + 5x + 3)*(x^2 -1)')/(x^2 - 1)^2`
= `((3x^2 + 5)*(x^2 - 1) - (x^3 + 5x + 3)*2x)/(x^2 - 1)^2`
= `(3x^4 - 3x^2 + 5x^2 - 5 - 2x^4 - 10x^2 - 6x)/(x^2 - 1)^2`
= `(x^4 - 8x^2 - 5 - 6x)/(x^2 - 1)^2`
The derivative of `f(x) = (x^3 + 5x + 3)/(x^2 - 1)` is `f'(x) = (x^4 - 8x^2 - 5 - 6x)/(x^2 - 1)^2` | 323 | 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} | 3.890625 | 4 | CC-MAIN-2016-44 | latest | en | 0.620359 |
http://basha-fansub.tk/forum246-use-the-slope-and-y-intercept-to-graph-each-equation-below.html | 1,527,259,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00323.warc.gz | 30,503,943 | 6,457 | use the slope and y intercept to graph each equation below
# use the slope and y intercept to graph each equation below
You can change this preference below.By solving for the variable y we have re-written our equation in slope intercept form. we will then identify the y-intercept and slope. Plot the y- intercept on our graph and use the ratio of slope to determine another point on the line. and then graph each point below on that same graph, using both the ordered pair and letter as a label. You should have one graph for each separate problem, so three graphs for problems 79.Find x-intercept, y-intercept, slope and equation of each line graphed. If there is a coefficient, however, then you should divide each term in the equation by that number. In this case, the y coefficient is 4, so you have to divide 4x, -3x, and 16 by 4 to get the final answer in slope intercept form.First, write down the equation so you can start using it to graph a line. The slope is slope-intercept form is y . The equation of the graph in x 2. Graph each equation. y 2x 3 To graph the equation, plot the y-intercept (0, 3). Then move up 2 units and right 1 unit. Then graph each line, using the slope and y intercept.In exercises 25 to 32, match the graph with one of the equations below. Write an equation in slope-intercept form of each line.Name. 5-3 Practice. (continued) Slope-Intercept Form. The slope below is -3 and the y-intercept is 2. what is the slope- intercept equation for the line?When you graph a line using slope intercept form do you start by graphing m or b first? The number in the b position (3) is the y-intercept which means this is where the graphed line will cross the y-axis.
When a linear equation is not in Slope Intercept Form, you must use your algebra skills to change it into the correct form.Be sure your line is pointing the right way. See below. Algebra Graphs of Linear Equations and Functions Graphs Using Slope- Intercept Form.Since slope is rise over run, you will be going down two and right one from the starting point of 5 for each point. Make a dot on each of these points and connect them. Here is an example equation for a line. y 2x 3. You can tell what each part of the equation is based on its position in the formula.Now that we know about the slope and the intercept individually, we need build up those two things together. Were still going to use our example y 2x 3. To graph m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points.
How To: Given the equation for a linear function, graph the function using the y-intercept and slope. Using Slopes and Intercepts. Find the x-intercept and y-intercept of each line. Use the intercepts to graph the equation.Negative slope: The line slants downward from left to right. Undefined slope: Vertical lines have an undefined slope. Use the graphic organizer to answer the following questions. In this lesson you will look at equations such as y3x2 and their graphs, and learn about each graphs shape, slope, and y-intercept.Use the grid to the left to determine each slope. The equation for y is shown below the grid. 2. Graph a line whose equation is in slopeintercept form Example 2. Use the slope and yintercept to graph each line whose equation is given. Label at least two points on the graph grid. How to Graph a Line using the Slope and y-intercept.Step 1: Begin by plotting the y-intercept of the given equation which is (0, 3). Step 2: Use the slope to find another point using the y-intercept as the reference. Use the slope and y-intercept to graph each equation below. T h e graph, if extended, w cross a letter. Print this letter in each box that contains the number of that exercis. The slope-intercept equation.To use the slope-intercept equation of a line, y mx b. we must first solve for y. Graph using slope-intercept: 5y - 2x 15. To graph a line using its slope and y-intercept, first plot the y- intercept, and then use the slope to graph at least one other point.12x 4y 1 Write each equation in slope-intercept form to determine their slopes: a.b.c. Since equations a and c have the same slopes, they are parallel. Demonstrates, step-by-step and with illustrations, how to use slope and the y-intercept to graph straight lines.But the "nice" form of a straight lines equation (being the slope- intercept form, y mx b) can make graphing even simpler and faster. In the next examples, there is a sample graph of each type of modeling Linear models are described by the following general graph.One of the main algebra concepts used in linear models is the slope- intercept equation of a line. They drive 400 miles each day. Use the table below to answer Items 1720.Some students may need more explicit instruction on how slope and y- intercept are determined from the various representations of linear relationships ( equations, tables and graphs) and what they mean. They merely use the equation for a straight line y mx b where m is slope and b the Y-intercept.The data are then re-plotted (t, v2) and a linear relationship results as is shown in the graph below. 6. Writing an Equation of a Line Identify the slope and y-intercept for each of the graphs below.Exploration: Write each linear equation in slope-intercept form. Then use a graphing calculator to graph the three equations in the same square viewing window. To graph using slope-intercept form: 1. Graph a point (the y -intercept). 2. Use m rise to move from that point to locate another point on the line. run.Directions: For each line that is graphed below, determine the equation of the line and write it in slope-intercept form. To graph linear equations (using slope intercept form)denominator tells you to go left/right 4. Connect the two points with a straight line and use arrows. Graphing Linear Equations Using Slope and Intercepts. Graphing a Linear Equation in Slope-Intercept Form. Graph y 3x 3. Identify the x- intercept.a. The slope of the line is 2.5 —5. Use the slope and y-intercept. to graph the equation.Graph the equation. | 1,414 | 6,108 | {"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} | 4.65625 | 5 | CC-MAIN-2018-22 | latest | en | 0.925676 |
http://www.4124039.com/questions/325679/hanson-wright-inequality-quadratic-form-concentration-inequality-for-bounded-r/325684 | 1,556,239,819,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578743307.87/warc/CC-MAIN-20190425233736-20190426015736-00511.warc.gz | 191,540,210 | 35,314 | # Hanson-Wright inequality (quadratic form concentration inequality) for bounded random vectors
Is there a concentration inequality for quadratic forms of bounded random vectors $$X \in [-1, 1]^n$$ with zero mean and given covariance matrix $$\Sigma \in \mathbb{R}^{n \times n}$$ but otherwise unknown distribution, i.e. a bound on the tail probability $$\Pr(|X^T \Sigma^{-1} X - n| \ge t) \le \ldots$$ Since for sub-Gaussian random vectors there is the Hanson-Wright inequality $$\Pr(|X^T A X - \operatorname{E}[X^T A X]| > t) \le \ldots$$ for some matrix $$A \in \mathbb{R}^{n \times n}$$ and $$E[X^T \Sigma^{-1} X] = n$$, it seems like such a bound should be within reach for the stronger restriction of bounded random vectors, even if the variance of the quadratic form is not available.
Note that this is specifically a question regarding the concentration about the mean $$n$$, since else we have the straightforward bound $$\Pr(X^T \Sigma^{-1} X \ge t) \le n/t$$ from Markov's inequality.
Edit: Thanks to @felipeh's answer for pointing out that there need to be additional restrictions on the function to give a meaningful bound. It would be reasonable to ask for the $$X_i$$ to be some or all of continuous, unimodal, symmetric as helpful.
Let $$X$$ be the random vector that is identically $$0$$ with probability $$1/2$$, and with probability $$1/2$$ is sampled uniformly from the boolean cube $$\{-1,1\}^n$$. The covariance is $$\Sigma=\frac{1}{2} Id$$, and $$X^T\Sigma^{-1} X$$ is either equal to $$0$$ or $$2n$$, each with probability $$1/2$$.
• (I take it you meant the boolean cube $\{-1, 1\}^n$.) That is a good example, there need to be additional quantifiers on the distribution. I will edit the question, apologies for that. – student Mar 18 at 16:30 | 488 | 1,772 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-18 | latest | en | 0.87025 |
http://openstudy.com/updates/521fe589e4b0750826e12622 | 1,511,550,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00493.warc.gz | 220,482,926 | 8,242 | • anonymous
HELP In a triangle, two sides that measure 6 cm and 10 cm form an angle that measures 80°. Find, to the nearest degree, the measure of the smallest angle in the triangle. I know the answer is 33º but how do you get that?
Mathematics
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Not the answer you are looking for? Search for more explanations. | 304 | 1,162 | {"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-2017-47 | latest | en | 0.340054 |
http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.102528.html | 1,369,185,458,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700984410/warc/CC-MAIN-20130516104304-00040-ip-10-60-113-184.ec2.internal.warc.gz | 314,690,978 | 4,836 | # SOLUTION: Find the following product. (x - 9)(x - 8)
Algebra -> Algebra -> Polynomials-and-rational-expressions -> SOLUTION: Find the following product. (x - 9)(x - 8) Log On
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Algebra: Polynomials, rational expressions and equations Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Polynomials-and-rational-expressions Question 102528: Find the following product. (x - 9)(x - 8) Answer by jim_thompson5910(28550) (Show Source): You can put this solution on YOUR website!Start with the given expression When you FOIL, you multiply the terms in this order: F-First (i.e. you multiply the first terms in each parenthesis which in this case are and ) O-Outer (i.e. you multiply the outer terms in each parenthesis which in this case are and ) I-Inner (i.e. you multiply the inner terms in each parenthesis which in this case are and ) L-Last (i.e. you multiply the last terms in each parenthesis which in this case are and ) So lets multiply the first terms: multiply and to get So lets multiply the outer terms: multiply and to get So lets multiply the inner terms: multiply and to get So lets multiply the last terms: multiply and to get Now lets put all the multiplied terms together Now combine like terms So the expression FOILs to: | 347 | 1,463 | {"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} | 3.5625 | 4 | CC-MAIN-2013-20 | latest | en | 0.840871 |
https://howtech.tv/office/how-to-calculate-days-in-excel/ | 1,632,612,933,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00072.warc.gz | 338,691,092 | 8,183 | ## How to Calculate Days in Excel
Microsoft Excel is a detailed software that can be used as a multiple analysis tool. It gives simple options that can help you manage and manipulate data to derive desired or actual results. It is important to understand the workings of this software to be able to use it effectively. You can also calculate days in Excel to figure out the difference between two dates or to use a date and number of days to deduce the end date. To learn how to calculate number of days in Excel, follow this step by step tutorial.
Step # 1 – Calculate difference between dates
To learn how to calculate number of days in Excel, we will be working on two situations. In the first one, we have two dates and want to find the number of days elapsed between the two dates. Therefore, we will simply subtract the two cells and calculate the difference.
Step # 2 – Use ‘NETWORKDAYS’ formula
If you want to exclude the weekend holidays from the result, write the following formula:
“=NETWORKDAYS(D4,D5)”
After using this function you will see that the number of days has decreased since weekends are excluded from the result now.
Step # 3 – Calculate End date
To calculate days in Excel the previous formulas can be used but where we have a date and the total number of days, and we want to find out the end date, we will simply add 120 days to the ‘Start’ date. Use the following:
“=C12+C13”
With that done, we will get a date but again this date is calculated on the assumption that no holidays would come in between during these 120 days.
Step # 4 – Use ‘WORKDAY’ formula
If you want to calculate the ‘End’ date keeping in consideration only the weekdays then, use the following formula:
“=WORKDAY(D12,D13)”
Step # 5 – Change Cell format
Format the cell by right clicking and opening up the ‘Format Cells’ option. Click on “Customs” and write
“dd/mm/yy”
This is so that the output can be shown in a ‘Date’ format, with the day showing first, then the month and then the year. | 455 | 2,007 | {"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} | 3.59375 | 4 | CC-MAIN-2021-39 | latest | en | 0.91344 |
https://www.physicsforums.com/threads/fourier-transform-and-the-frequency-domain.738765/#post-4670719 | 1,679,466,222,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00094.warc.gz | 1,079,191,938 | 16,449 | Fourier transform and the frequency domain
Bipolarity
I understand that the Fourier transform maps one function onto another. So it is a mapping from one function space onto another.
My question is, why is it often referred to as a mapping from time domain to the frequency domain? I don't understand why the image of the Fourier transform represents a signal in the frequency domain. I don't see how the word frequency is even related to the Fourier transform.
What I do know is that the complex exponential and the trigonometric Fourier series both form an orthonormal basis for a certain class of functions under the function inner product. But I don't understand how these ideas are tied to the Fourier transform.
Insight is appreciated. Thanks!
BiP
Homework Helper
##e^{i\omega t} = \cos \omega t + i \sin \omega t##.
In applications of Fourier transforms in physics and engineering, the ##\omega## often corresponds to the physical frequency (cycles / time) of a mechanical vibration or an electrical signal.
Bipolarity
##e^{i\omega t} = \cos \omega t + i \sin \omega t##.
In applications of Fourier transforms in physics and engineering, the ##\omega## often corresponds to the physical frequency (cycles / time) of a mechanical vibration or an electrical signal.
I see. What you posted is the Euler identity, which connects the interpretation of the frequency of a sine wave with the frequency of a complex exponential function.
But how are these related to the Fourier transform? How exactly is the result of the transform related to the frequencies of signals?
BiP
homeomorphic
The Fourier transform can be thought as a modified version of Fourier series that can apply to non-periodic functions. I'll just try to give a bird's eye view of this without getting into the details.
Fourier series are a way of writing any "reasonable" periodic function as a sum of sine and cosine functions of different frequencies (or rather, multiples of them). It's an infinite linear combination of sines and cosines, and the coefficients are called Fourier coefficients. The problem with Fourier series is that they only apply to periodic functions.
The idea, then, is to just let the interval of periodicity get bigger and bigger, approaching the whole real number line. With Fourier series, you have discrete frequencies. When you take this limit, these frequencies will get closer and closer together, and in the limit, they become a continuous variable. The Fourier transform is analogous to the Fourier coefficients. You plug in a number that represents the frequency, and out pops the analogue of the Fourier coefficient of that frequency. So, when you apply the Fourier transform to a function, the function that you get is the one that does this. Rigorously, you probably wouldn't formulate the theory in this way, but this is the idea behind it.
Jyan
Here is a derivation of the Fourier transform from Fourier series. http://www.jpoffline.com/physics_docs/y2s4/cvit_ft_derivation.pdf
Like homeomorphic said, the Fourier transform is obtained from the Fourier series by taking a limit as the period goes to infinity (and since frequency is the inverse of the period, it's the same as taking a limit as frequency goes to 0, which I find more intuitive). Fourier series are easier to understand than the Fourier transform, so thinking about the Fourier transform as a limit of the Fourier series makes it much easier on the brain.
If you are so inclined you can obtain a (electronic probably) copy of "Modern digital and analog communication systems" By B.P. Lathi which has an excellent explanation of the Fourier integral at the beginning of chapter 3.
Homework Helper
Gold Member
I understand that the Fourier transform maps one function onto another. So it is a mapping from one function space onto another.
My question is, why is it often referred to as a mapping from time domain to the frequency domain? I don't understand why the image of the Fourier transform represents a signal in the frequency domain. I don't see how the word frequency is even related to the Fourier transform.
What I do know is that the complex exponential and the trigonometric Fourier series both form an orthonormal basis for a certain class of functions under the function inner product. But I don't understand how these ideas are tied to the Fourier transform.
Insight is appreciated. Thanks!
BiP
Just some non-rigorous notes for intuition:
1) Any time-domain function can be approximated by a sum of frequencies.
2) Because the frequencies are an orthonormal basis, one can just determine how much of each frequency is in the time-domain function and add them together.
3) The Fourier transformation tells how much of each frequency is in the time-domain function. | 982 | 4,777 | {"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-2023-14 | latest | en | 0.900768 |
https://www.magicalapparatus.com/spectator-card/fours-of-a-kind.html | 1,563,312,106,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00426.warc.gz | 780,889,268 | 7,182 | ## Fours of a Kind
The plot of this trick is similar to that of the feat described as the sevens. It has, however, a delayed climax of the turn-the-tables type.
Prior to performance remove the fours of clubs, hearts, spades, and diamonds. Arrange the deck so that one of these is at the top and the second is directly below it; after the second there is an indifferent card with a third four-spot below it. Place the remaining four-spot second from the bottom.
1. Shuffle the cards by means of the riffle shuffle, without disturbing the two cards at the bottom or the four cards at the top.
2. Force the four-spot at the top by means of the backslip force, and have it placed on the table before the drawer, without permitting its face to be seen.
3. Prepare for a double lift, saying, "There is a sympathy among cards which has often been commented upon. I believe that I can show you what I mean." Turn the two top cards as one, showing the indifferent card, which let us say is a nine, and replace the two face downwards on the deck. Remove the top card, a four-spot, and place it face downwards before you, saying, "This card tells me the value of the card you drew. It is a nine."
4. Hold the pack in the left hand face downwards as for the glide, tip it up so that the face card, which we shall say is a diamond, can be seen, and say, "This card indicates the suit. A diamond. We know your card must be the nine of diamonds." Turn the pack face downwards as you say this, glide back the bottom card, remove the card above it (a four-spot), and place it face downwards before you on the table.
Remove the card that is now above the glided card at the bottom and, using it as a pointer, indicate the two cards before you. "Remember," you say. "A nine here and a diamond here. The nine of diamonds." Replace the pointer card at the bottom, thus effectually concealing the glided card.
5. At this point the indifferent card is at the top of the pack, with a four-spot directly under it. Make the pass, transposing the upper half to the bottom, and hold a left-little-finger break between the two packets. Spread the cards from hand to hand and force the fourth four-spot upon yourself-- a feat that should give you very little trouble! It is only necessary to remove the card below the one above which the little finger maintains control.
Place this card face downwards beside the other three, explaining, "The purpose of this card will be clear in a moment.
As you have seen, the cards say that the one you drew a moment ago is the nine of diamonds. Please turn your card face upwards and show everyone that this is the case."
6. The spectator turns his card and shows it. Let us say it is the four of the heart suit. Appear a little discomfited; then exclaim, "My original trick hasn't worked, so I shall do another trick instead. Your card is the four of hearts. Very well. Cards, change!" Tap each of the three cards before you, turn them over, and show that each is a four-spot.
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0 0 | 701 | 3,062 | {"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} | 3.046875 | 3 | CC-MAIN-2019-30 | longest | en | 0.956963 |
http://math.stackexchange.com/questions/269624/graph-problem-prove-by-induction | 1,419,109,624,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770399.32/warc/CC-MAIN-20141217075250-00061-ip-10-231-17-201.ec2.internal.warc.gz | 177,495,414 | 16,041 | # Graph problem - prove by induction
Let's have two sets - $T_1$ and $T_2$ defined as below:
• Def.1 : $T_1$={ T | T is a tree with at least one edge and T doesn't have vertices from 2 degree }
• Def.2 : Every connected graph with exactly two vertices is in T2. For every such graph G({u,v}, {(u,v)} ) the sets of boundary vertices of G is {u,v}.
• If $G(V,E)$ is a graph from $T_2$ and $U \subseteq V$ is the sets of the boundary vertices of $D$ and $z$ is a random vertix from $U$ and $W$ is a sets of vertices, such that $|W| \ge 2$ and $W \cap U$ = $\emptyset$, so $G^\prime (V \cup{W}, E \cup {(z,w)|w \in W})$ is also a graph from $T_2$ and the sets of boundary vertices of $G^\prime$ is $(U${$z$}$)\cup W.$
• $T_2$ doesn't have another graphs.
Using these definitions, prove that :
1. $T_1 \subseteq T_2$
2. $T_2 \subseteq T_1$
3. Using $(1)$ and $(2)$ prove by induction , that for every tree with at least one edge(which doesn't have vertices from $2$ degree) is truth, that the number of vertices from $1$ degree is $\ge n/2 +1$, where n is the number of all vertices.
-
I've proved 1) and 2), but have stuck on 3)...Any ideas? – DiscreteMath'sFan Jan 3 '13 at 8:26
Could you work on editing the grammar and spelling? I do not understand "doesn't have vertices from degree two", "for every three", – Calvin Lin Jan 3 '13 at 8:36
For def 2, point 1, are you simply trying to say that the complete graph $K_2$ is in $T_2$? For point 2, What is $W$? Is it allows to be in $V\setminus U$? Is it possibly outside of $V$?? What is $(U{z})$ mean??? – Calvin Lin Jan 3 '13 at 8:42
At this moment, I cannot understand what you're trying to say. Definition 2 is incomprehensible to me, and I cannot guess what you're trying to say. Someone else might have better luck. I believe that "for every three (the number 3)" actually means "for every tree (graph theory)" – Calvin Lin Jan 3 '13 at 8:47
ah :), of course that this actually means, sorry I fixed it – DiscreteMath'sFan Jan 3 '13 at 8:52 | 639 | 2,003 | {"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} | 3.96875 | 4 | CC-MAIN-2014-52 | latest | en | 0.940113 |
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Every Question Helps You Learn
Have you got time to play this difficult Time quiz?
# Time 2 (Difficult)
Welcome to the second quiz in our Difficult Eleven Plus maths series on solving problems involving Time. There are many different units of time, which I’m sure you are familiar with, like seconds, minutes, hours and days. Would you like to know a few more? Then take a look at these:
• Myriosecond – one ten-thousandth of a second
• Millisecond – one thousandth of a second
• Centisecond – one hundredth of a second
• Lunar month – 29 days, 12 hours, 44 minutes and 3 seconds
• Biennium – 2 years
• Lustrum – 5 years
• Gigasecond – 1 billion seconds (roughly 31.7 years)
You’ll be happy to hear that you don’t need to know any of these terms just yet. It’s always good to learn new things though, and one way to do that is to play this quiz!
1.
Susan left for work at 06:13. She walked to the bus stop which took 18 minutes, then she had to wait 2 minutes for the bus to arrive. The journey on the bus lasted for 25 minutes and then Susan walked for another 12 minutes before she got to work 20 minutes early.
What time did Susan start work?
7:30 am
7:20 am
7:10 am
7:00 am
Susan’s walk to the bus stop took 18 minutes. 06:13 + 18 = 06:31
She waited 2 minutes for her bus to arrive. 06:31 + 2 = 06:33
The bus journey lasted 25 minutes. 06:33 + 25 = 06:58
She then walked for 12 minutes. 06:58 + 12 = 07:10
Susan arrived 20 minutes early. 07:10 + 20 = 07:30. Susan started work at 07:30, or 7:30 am
2.
The train from London to Edinburgh leaves at 06:08 and arrives at 11:02. How many minutes does the journey take?
300 minutes
294 minutes
286 minutes
280 minutes
06:08 to 07:00 is 52 minutes. 07:00 to 11:00 is 4 hours, or 240 minutes (4 x 60 = 240). 11:00 to 11:02 s 2 minutes.
52 + 240 + 2 = 294
3.
The time is twenty past midnight, but Jane’s watch says 23:06. How fast or slow is Jane’s watch?
12 hours and 14 minutes slow
11 hours and 46 minutes fast
1 hour and 14 minutes slow
46 minutes fast
23:06 is 54 minutes before midnight (00:00). Twenty past midnight is 00:20 so we need to add another 20 minutes: 54 + 20 = 74
There are 60 minutes in an hour, so we subtract 60 from 74 to get 14. Jane’s watch is 1 hour and 14 minutes slow
4.
How many twenty-minute periods are there in a day?
24
36
48
72
There are three twenty-minute periods in one hour, and 24 hours in a day. So, to find the answer we multiply 3 by 24: 3 x 24 = 72
5.
How many revolutions (complete turns) does the hour hand of a clock turn through in one week?
7 revolutions
14 revolutions
84 revolutions
168 revolutions
The hour hand of a clock turns through one complete revolution every 12 hours, that’s twice in one day. There are 7 days in a week so to work this out we multiply 7 by 2: 7 x 2 = 14
6.
How many revolutions (complete turns) does the second hand of a clock turn through in 1 week?
10,080 revolutions
36,000 revolutions
240,080 revolutions
604,800 revolutions
It goes round once for every minute. There are 60 minutes in an hour, so that's 1 × 60 = 60 revolutions an hour. There are 24 hours in a day, so there are 24 × 60 = 1,440 revolutions in a day. There are 7 days in a week, so there are 1,440 x 7 = 10,080 revolutions in a week
7.
How many leap years were there in the 19th century?
26 leap years
25 leap years
24 leap years
23 leap years
The 19th century was the years from 1801 to 1900 inclusive. To be a leap year, the year must be divisible by 4. There were 25 years in the 19th century which are divisible by 4. However, years ending in 00 (like 1900) must also be divisible by 400. 1900 is not divisible by 400 so was not a leap year. 2000 is divisible by 400 so that was a leap year
8.
Which millennium are we living in?
The 2nd millennium
The 20th millennium
The 3rd millennium
The 21st millennium
The first millennium lasted from 1 CE to 1000 CE. The second lasted from 1001 to 2000. The third millennium began in 2001 and will end in the year 3000
9.
How many fortnights are there in a year?
14 fortnights
26 fortnights
52 fortnights
365 fortnights
There are 52 weeks in a year and a fortnight lasts for 2 weeks. So, to find the answer we divide 52 by 2: 52 ÷2 = 26
10.
Which of these is the same as 8,784 hours?
A 31 day month
A 30 day month
A 365 day year
A leap year
There are 24 hours in a day, so 8,784 ÷ 24 = 366, the number of days in a leap year
Author: Frank Evans | 1,319 | 4,400 | {"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} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.925462 |
https://www.varsitytutors.com/ssat_upper_level_math-help/number-concepts-and-operations/estimation | 1,579,535,031,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598800.30/warc/CC-MAIN-20200120135447-20200120164447-00193.warc.gz | 1,143,789,420 | 44,809 | # SSAT Upper Level Math : Estimation
## Example Questions
### Example Question #1 : How To Estimate
Evan wants to tip approximately on a restaurant tab. Which of the following comes closest to what he should leave?
Explanation:
The tab can be rounded to is equal to .
Now we can multiply the percent by the total amount.
is the most reasonable estimate of the recommended tip.
### Example Question #1 : Estimation
If the number is rounded to the nearest hundredth, which of the following expressions would be equal to that value?
Explanation:
If is rounded to the nearest hundredth, the result will be
Given that , the correct answer is
### Example Question #2 : Estimation
Estimate the product by rounding each factor to the nearest hundred, then multiplying.
Explanation:
437 rounded to the nearest hundred is 400.
877 rounded to the nearest hundred is 900.
551 rounded to the nearest hundred is 600.
Multiply the three whole multiples of 100 to get the desired estimate:
### Example Question #1 : How To Estimate
Estimate the product by rounding each factor to the nearest unit, then multiplying.
Explanation:
, so rounded to the nearest unit is 5.
, so rounded to the nearest unit is 10.
, so rounded to the nearest unit is 5.
Multiply the three whole numbers to get the desired estimate:
### Example Question #3 : Estimation
Estimate the result by first rounding each number to the nearest unit.
Explanation:
8.19 rounded to the nearest unit is 8.
4.87 rounded to the nearest unit is 5
3.27 rounded to the nearest unit is 3.
7.42 rounded to the nearest unit is 7.
The desired estimate can be found as follows:
### Example Question #1 : How To Estimate
Estimate the result by first rounding each number to the nearest unit.
Explanation:
, so rounded to the nearest unit is 8.
, so rounded to the nearest unit is 10.
, so rounded to the nearest unit is 4.
, so rounded to the nearest unit is 6.
The desired estimate can be found as follows:
### Example Question #52 : Number Concepts And Operations
Melissa is trying to come up with a reasonable estimate of the amount she spent on groceries over the last six months. She notices that the six checks she wrote out to the local grocery store are in the following amounts: $187.54,$218.89, $174.74,$104.76, $189.75, and$228.64. By estimating each of the amounts of the checks to the nearest ten dollars, come up with a reasonable estimate for Melissa's total expenditure for groceries.
Explanation:
Round each of the amounts to the nearest ten dollars as follows:
$187.54 rounds to$190.
$218.89 rounds to$220.
$174.74 rounds to$170.
$104.76 rounds to$100.
$189.75 rounds to$190.
$228.64 rounds to$230.
### Example Question #5 : How To Estimate
Estimate the product by rounding each factor to the nearest unit, then multiplying.
Explanation:
8.39 rounded to the nearest unit is 8 because 0.39 is less than 0.5.
7.34 rounded to the nearest unit is 7 because 0.34 is less than 0.5.
3.52 rounded to the nearest unit is 4 because 0.52 is greater than 0.5.
Multiply the three whole numbers to get the desired estimate: | 785 | 3,141 | {"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} | 4.53125 | 5 | CC-MAIN-2020-05 | latest | en | 0.825155 |
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Document related concepts
List of prime numbers wikipedia, lookup
Proofs of Fermat's little theorem wikipedia, lookup
Transcript
```FUNDAMENTAL THEOREM OF
ARITHMETIC
Any whole number greater than one is either a prime number or can be written as a product of prime numbers in a unique way.
EXAMPLES: 2 = a prime number
3 = a prime number
4=2x2
5 = a prime number
6=2x3
7 = a prime number
8=2x2x2
9=3x3
10 = 2 x 5
Now try to work out 11 to 20 | 165 | 580 | {"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} | 3.25 | 3 | CC-MAIN-2020-40 | latest | en | 0.857674 |
https://ru.scribd.com/document/260632457/MIT2-080JF13-Lecture2-pdf | 1,566,048,227,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313259.30/warc/CC-MAIN-20190817123129-20190817145129-00005.warc.gz | 617,184,001 | 68,825 | You are on page 1of 26
# Structural Mechanics
2.080 Lecture 2
Semester Yr
## Lecture 2: The Concept of Strain
Strain is a fundamental concept in continuum and structural mechanics. Displacement
fields and strains can be directly measured using gauge clips or the Digital Image Correlation
(DIC) method. Deformation patterns for solids and deflection shapes of structures can be
easily visualized and are also predictable with some experience. By contrast, the stresses
can only be determined indirectly from the measured forces or by the inverse engineering
method through a detailed numerical simulation. Furthermore, a precise determination of
strain serves to define a corresponding stress through the work conjugacy principle. Finally
the equilibrium equation can be derived by considering compatible fields of strain and
displacement increments, as explained in Lecture 3. The present author sees the engineering
world through the magnitude and shape of the deforming bodies. This point of view will
dominate the formulation and derivation throughout the present lecture note. Lecture
2 starts with the definition of one dimensional strain. Then the concept of the threedimensional (3-D) strain tensor is introduced and several limiting cases are discussed. This
is followed by the analysis of strains-displacement relations in beams (1-D) and plates (2D). The case of the so-called moderately large deflection calls for considering the geometric
non-linearities arising from rotation of structural elements. Finally, the components of the
strain tensor will be re-defined in the polar and cylindrical coordinate system.
2.1
One-dimensional Strain
Consider a prismatic, uniform thickness rod or beam of the initial length lo . The rod is fixed
at one end and subjected a tensile force (Fig. (2.1)) at the other end. The current, deformed
length is denoted by l. The question is whether the resulting strain field is homogeneous
or not. The concept of homogeneity in mechanics means independence of the solution on
the spatial coordinates system, the rod axis in the present case. It can be shown that if
the stress-strain curve of the material is convex or linear, the rod deforms uniformly and a
homogeneous state of strains and stresses are developed inside the rod. This means that
local and average strains are the same and the strain can be defined by considering the total
lengths. The displacement at the fixed end x = 0 of the rod is zero, u(x = 0) and the end
displacement is
u(x = l) = l lo
(2.1)
The strain is defined as a relative displacement. Relative to what? Initial, current length
or something else? The definition of strain is simple but at the same time is non-unique.
l lo
Engineering Strain
lo
2
2
def 1 l lo
=
Cauchy Strain
2 l2
l
def
= ln
Logarithmic Strain
lo
def
=
2-1
(2.2a)
(2.2b)
(2.2c)
Structural Mechanics
2.080 Lecture 2
Semester Yr
Each of the above three definitions satisfy the basic requirement that strain vanishes when
l = lo or u = 0 and that strain in an increasing function of the displacement u.
u
Consider a limiting case of Eq.(2.1) for small displacements 1, for which lo +l 2lo
lo
in Eq.(2.2b). Then, the Cauchy strain becomes
=
l lo l + lo l lo 2l l lo
=
=
lo
2lo
lo 2lo
lo
(2.3)
Thus, for small strain, the Cauchy strain reduces to the engineering strain. Likewise,
expanding the expression for the logarithmic strain, Eq.(2.2c) in Taylor series around llo
=
0,
l
l lo 1 l lo 2
l lo
ln
(2.4)
+
=
lo l/lo =1
lo
2
lo
lo
one can see that the logarithmic strain reduces to the engineering strain.
l
The plots of versus
according to Eqs.(2.2a)-(2.2c) are shown in Fig.(2.1).
lo
## Inhomogeneous Strain Field
The strain must be defined locally and not for the entire structure. Consider an infinitesimal
element dx in the undeformed configuration, Fig.(2.2).
After deformation the length of the original material element becomes dx + du. The
engineering strain is then
(dx + du) du
du
eng =
=
(2.5)
dx
dx
du
The spatial derivative of the displacement field is called the displacement gradient F =
.
dx
For uniaxial state the strain is simply the displacement gradient. This is not true for general
3-D case.
2-2
Structural Mechanics
2.080 Lecture 2
Semester Yr
dx
l0
A
u
A'
B'
l
dx + du
Figure 2.2: Undeformed and deformed element in the homogenous and inhomogeneous
strain field in the bar.
The local Cauchy strain is obtained by taking relative values of the difference of the
square of the lengths. As shown in Eq. (2.3), in order for the Cauchy strain to reduce to
the engineering strain, the factor 2 must be introduced in the definition. Thus
1 (dx + du)2 dx2
du 1 du 2
c =
=
+
(2.6)
2
dx2
dx 2 dx
1
or c = F + F 2 . For small displacement gradients,
2
c = eng
2.2
(2.7)
du = dx
(2.8)
## Consider an Euclidian space and denote by x = {x1 , x2 , x3 } or xi the vector representing a
position of a generic point of a body. In the general three-dimensional case, the displacement
of the material point is also a vector with components u = {u1 , u2 , u3 } or ui where i = 1, 2, 3.
Recall that the increment of a function of three variables is a sum of three components
du1 (x1 , x2 , x3 ) =
u1
u1
u1
dx1 +
dx2 +
dx3
x1
x2
x3
(2.9)
## In general, components of the displacement increment vector are
3
X ui
ui
ui
ui
dui (xi ) =
dx1 +
dx2 +
dx3 =
dxj
x1
x2
x3
xj
(2.10)
j=1
where the repeated j is the so called dummy index. The displacement gradient
F =
ui
xj
2-3
(2.11)
Structural Mechanics
2.080 Lecture 2
Semester Yr
is not a symmetric tensor. It also contains terms of rigid body rotation. This can be shown
by re-writing the expression for F in an equivalent form
uj
uj
1 ui
1 ui
ui
+
+
(2.12)
xj
2 xj
xi
2 xj
xi
Strain tensor ij is defined as a symmetric part of the displacement gradient, which is the
first term in Eq. (2.12).
uj
1 ui
ij =
+
(2.13)
2 xj
xi
Now, interchange (transpose) the indices i and j in Eq.(2.13):
1 uj
ui
ji =
+
2 xi
xj
(2.14)
The first term in Eq.(2.14) is the same as the second term in Eq.(2.13). And the second
term in Eq.(2.14) is identical to the first term in Eq.(2.13). Therefore the strain tensor is
symmetric
ij = ji
(2.15)
The reason for introducing the symmetry properties of the strain tensor will be explained
later in this section. The second terms in Eq.(2.12) is called the spin tensor ij
uj
1 ui
ij =
(2.16)
2 xj
xi
Using similar arguments as before it is easy to see that the spin tensor is antisymmetric
wij = wji
(2.17)
From the definition it follows that the diagonal terms of the spin tensor are zero, for example
w11 = w11 = 0. The components, of the strain tensor are:
1 u1 u1
u1
i = 1, j = 1 11 =
+
=
(2.18a)
2 x1 x1
x1
u2
i = 2, j = 2 22 =
(2.18b)
x2
u3
i = 3, j = 3 33 =
(2.18c)
x3
1 u1 u2
i = 1, j = 2 12 = 21 =
+
(2.18d)
2 x2 x1
1 u2 u3
i = 2, j = 3 23 = 32 =
+
(2.18e)
2 x3 x2
1 u3 u1
i = 3, j = 1 31 = 13 =
+
(2.18f)
2 x1 x3
2-4
Structural Mechanics
2.080 Lecture 2
Semester Yr
For the geometrical interpretation of the strain and spin tensor consider an infinitesimal
square element (dx1 , dx2 ) subjected to several simple cases of deformation. The partial
derivatives are replaced by finite differences, for example
u1
u1
u1 (x1 ) u1 (x1 + h)
=
=
x1
x1
h
(2.19)
## Rigid body translation
Along x1 axis:
x2
u1 (x1 ) = u1 (x1 + h)
(2.20a)
u2 = u3 = 0
(2.20b)
u2
u1
u1
x1
u1
h
## Figure 2.3: Rigid body translation of the infinitesimal square element.
It follows from 2.18a that the corresponding strain component vanishes, 11 = 0. The
first component of the spin tensor is zero from the definition, 11 = 0.
## Extension along x1 axis
At x1 : u1 = 0.
At x1 + h : u1 = uo .
uo
The corresponding strain is 11 = .
h
## Pure shear on the x1 x2 plane
At x1 = 0 and x2 = 0: u1 = u2 = 0
At x1 = h and x2 = 0: u1 = 0 and u2 = uo
At x1 = 0 and x2 = h: u1 = uo and u2 = 0
2-5
Structural Mechanics
2.080 Lecture 2
x2
Semester Yr
u2
uo
x1
u1
x2
u2
uo
h
uo
h
x1
u1
## Figure 2.5: Imposing constant deformative gradients.
It follows from Eq. (2.17) and Eq. (2.13) that:
1 uo uo uo
+
=
2 h
h
h
1 uo uo
=
=0
2 h
h
12 =
(2.21)
12
(2.22)
The resulting strain is representing change of angles of the initial rectilinear element.
## Rigid body rotation
At x1 = 0 and x2 = 0: u1 = u2 = 0
At x1 = h and x2 = 0: u1 = 0 and u2 = uo
At x1 = 0 and x2 = h: u1 = uo and u2 = 0
2-6
Structural Mechanics
2.080 Lecture 2
x2
Semester Yr
u2
uo
uo
x1
u1
## Figure 2.6: An infinitesimal square element subjected to rigid body rotation.
Changing the sign of u1 at x1 = 0 and x2 = h from uo to uo results in non-zero spin
but zero strain
1 uo uo
12 =
+
=0
(2.23)
2 h
h
1 uo
uo
uo
12 =
+
=
(2.24)
2 h
h
h
The last example provides an explanation why the strain tensor was defined as a symmetric
part of the displacement gradient. The physics dictates that rigid body translation and
rotation should not induce any strains into the material element. In rigid body rotation
displacement gradients are not zero. The strain tensor, defined as a symmetric part of the
displacement gradient removes the effect of rotation in the state of strain in a body. In other
words, strain described the change of length and angles while the spin, element rotation.
2.3
## Description of Strain in the Cylindrical Coordinate System
In this section the strain-displacement relations will be derived in the cylindrical coordinate
system (r, , z). The polar coordinate system is a special case with z = 0.
The components of the displacement vector are {ur , u , uz }. There are two ways of
deriving the kinematic equations. Since strain is a tensor, one can apply the transformation
rule from one coordinate to the other. This approach is followed for example on pages
125-128 of the book on A First Course in Continuum Mechanics by Y.C. Fung. Or,
the expression for each component of the strain tensor can be derived from the geometry.
The latter approach is adopted here. The diagonal (normal) components rr , , and
zz represent the change of length of an infinitesimal element. The non-diagonal (shear)
components describe the change of angles.
2-7
Structural Mechanics
2.080 Lecture 2
Semester Yr
z
y
x
r
ur +
dz
dr
dr
@ur
dr
@r
## Figure 2.8: Change of length in the radial direction.
The radial strain is solely due to the presence of the displacement gradient in the rdirection
ur
ur +
dr ur
ur
r
rr =
=
(2.25)
dr
r
The circumferential strain has two components
(1)
(2)
= +
(2.26)
The first component is the change of length due to radial displacement, and the second
component is the change of length due to circumferential displacement.
(1)
(2)
From Fig.(2.9) the components and are calculated as
(1)
(r + ur )d rd
ur
=
rd
r
u
u +
d u
1 u
=
=
rd
r
=
(2)
2-8
(2.27a)
(2.27b)
Structural Mechanics
2.080 Lecture 2
Semester Yr
(r + ur)d
ur
rd
u
u +
@u
d
@
Figure 2.9: Two deformation modes responsible for the circumferential (hoop) strain.
The total circumferential (hoop) component of the strain tensor is
=
ur
1 u
+
r
r d
(2.28)
The strain components in the z-direction is the same as in the rectangular coordinate system
uz
z
describes a change in the right angle.
(2.29)
zz =
## The shear strain r
@u
@r
u
r
ur
c'
b'
dr
a a' rd
d
1 @ur
r @
Figure 2.10: Construction that explains change of angles due to radial and circumferential
displacement.
From Fig.(2.10) the shear strain over the {r, } plane is
1 u
u
1 ur
r =
+
2 r
r
r
(2.30)
On the {r, z} plane, the rz shear develops from the respective gradients, see Fig.(2.11).
2-9
Structural Mechanics
2.080 Lecture 2
Semester Yr
@uz
dr
@r
z uz
uz
ur
@ur
dz
@z
ur
## Figure 2.11: Change of angles are {r, z} plane.
From the construction in Fig.(2.10), the component rz is
1 ur
uz
rz =
+
2 z
r
(2.31)
@uz
d
@
rd
uz
@u
dz
@z
dz
## Figure 2.12: Visualization of the strain component z .
The component z of the strain tensor is one half of the change of angles, i.e.
1 uz
u
+
z =
2 r
z
(2.32)
To sum up the derivation, the six components of the infinitesimal strain tensor in the
2-10
Structural Mechanics
2.080 Lecture 2
Semester Yr
## cylindrical coordinate system are
ur
r
ur
1 u
=
+
r
r
ux
=
z
1 1 ur
u
u
= r =
+
2 r
r
r
1 uz
u
= z =
+
2 r
z
1 ur
uz
= rz =
+
2 z
r
rr =
(2.33a)
(2.33b)
zz
r
z
zr
(2.33c)
(2.33d)
(2.33e)
(2.33f)
Considerable simplifications are obtained in the case of axial (rotational symmetry for which
[]=0
u = 0 and
ur
r
ur
=
r
uz
=
z
rr =
r = 0
z = 0
zz
zr =
1
2
(2.34a)
(2.34b)
ur
uz
+
z
r
(2.34c)
The application of the above geometrical relations for axi-symmetric loading of circular
plates and cylindrical shells will be given in subsequent chapters.
2.4
## Kinematics of the Elementary Beam Theory
The word kinematics is derived from the Greek word kinema, which means movements,
motion. Any motion of a body involves displacements ui , their increments dui and velocities
ui . If the rigid body translations and rotations are excluded, strains develop. We often say
Kinematic assumption or Kinematic boundary conditions or Kinematic quantities
etc. All it means that statements are made about the displacements and strains and/or
their rates. By contrast, the word static is reserved for describing stresses and/or forces,
even though a body could move. The point is that for statically determined structures, one
could determine stresses and forces without invoking motion. Such expressions as static
formulation, static boundary conditions, static quantities always refer to stresses and
forces.
Elementary is another word in the title of this section that requires explanation. A beam is
a slender structure that can be compressed, extended or bent. The beam must be subjected
to a transverse load (perpendicular to its axis). Otherwise it becomes something else, as
explained in 2.13.
2-11
Structural Mechanics
2.080 Lecture 2
Semester Yr
Beam
Column
Rod, strut
N
T
Shaft
Beam/column
Figure 2.13: The type of loading distinguishes between five different types of structures.
All the above structures may have a similar slenderness. How slender the structure must
l
be to become a beam. The slenderness is defined as a length to thickness ratio . If
h
l
> 20, the beam obeys the simplified kinematic assumptions and it is called an Euler
h
l
beam. Much shorter beams with
< 10 develop considerable shear stresses in addition
h
to bending stresses and must be treated by a different set of assumptions. Such beams are
l
referred to as Timoshenko beams. The intermediate range 10 < < 20 is a grey area where
h
the simplifying assumptions of the elementary beam theory gradually lose validity.
This section deals with a solid section beams, as opposed to thin-walled sections. In the
present lecture notes, the rectangular right handed coordinate system (x, y, z) is consistently
used. The x-axis is directed along the length of the beam with an origin at a convenient
location, usually the end of the center of the beam. The y-axis is in the width direction with
its origin on the symmetry plane of the cross-section, 2.14. Finally, the z-axis is pointing
out down and it is measured from the centroidal axis of the cross-section (see Recitation 2
for the definition of a centroidal axis).
q
q
y
x
z
y
z
## Figure 2.14: A prismatic slender beam with a symmetric cross-section.
In structural mechanics the components of the displacement vector in x, y, and z direc2-12
Structural Mechanics
2.080 Lecture 2
Semester Yr
tions are denoted respectively by (u, v, w). The development of elementary beam theory is
based on three kinematic assumptions. Additional assumptions on the stress state will be
introduced later.
2.5
Euler-Bernoulli Hypothesis
In this section reference is often made to the beam axis. The meaning of the beam axis is
intuitive for a prismatic beam with a rectangular cross-section. It is the middle axis. Other
terms, such as: neutral axis, bending axis and centroidal axis are also frequently used. They
all express the same property that no axial stresses xx should develop on the axis under
pure bending.
## Hypothesis 1: Plan Remains Plane
This is illustrated in 2.15 showing an arbitrary cross-section of the beam before and after
deformation.
(a)
(b)
(c)
Before
(d)
After
Figure 2.15: Flat (b) and (c) and warped (d) cross-sections after deformations.
Imagine a straight cut made through the undeformed beam. The plane-remains-plane
hypothesis means that all material points on the original cut align also on a plane in the
deformed beam. The cases (b) and (c) obey the hypothesis but the warped section (d)
violates it.
## Hypothesis 2: Normal Remains Normal
If the initial cut were made at right angle of the undeformed beam axis as in Fig.(2.16(a)),
it should remain normal to the deformed axis, see Fig.(2.16(b)).
In the sketch on Fig. (2.16(c)) the hypothesis is violated when the angle 6= 90 .
The Euler-Bernoulli hypothesis gives rise to an elegant theory of infinitesimal strains
in beams with arbitrary cross-sections and loading in two out-of-plane directions. The
interested reader is referred to several monographs with a detailed treatment of the subject,
of bi-axial loading of beams. The present set of notes on beams is developed under the
2-13
Structural Mechanics
2.080 Lecture 2
Semester Yr
(a)
(b)
(c)
Before
After
## Figure 2.16: Testing the normal-remains-normal hypothesis.
assumption of planar deformation. This means that the beam axis motion is restricted only
to one plane.
Mathematically, the Hypothesis 1 is satisfied when the u-component of the displacement
vector is a linear function of z.
u(z) = u z at any x
(2.35)
The constant first term, u is the displacement of the beam axis (due to axial force). The
second term is due to bending alone, Fig. (2.17).
uo
dw
dx
Figure 2.17: Linear displacement field through the thickness of the beams.
The second Euler-Bernoulli hypothesis is satisfied if the rotation of the deformed crossdw
section is equal to the local slope of the bent middle axis
dx
=
dw
dx
(2.36)
Eliminating the rotation angle between equations 2.35 and 2.36 yields
u(x, z) = u
dw
z
dx
(2.37)
It can be seen from Fig.(2.17) that the displacement at the bottom (tensile) side of the
beam is negative, which explains the minus sign in the second term of Eqs. (2.36) and
(2.37).
2-14
Structural Mechanics
2.080 Lecture 2
Semester Yr
Hypothesis 3
The cross-sectional shape and size of the beam remain unchanged. This means that the
vertical component of the displacement vector does not depend on the z-coordinate. All
points of the cross-section move by the same amount.
w = w(x)
(2.38)
In the case of planar deformation, which covers most of the practical cases of the beam
response, the y-component of the displacement vector vanishes
v0
(2.39)
We are now in the position to calculate all components of the strain tensor from Eq. (2.17)
xx =
yy =
zz =
xy =
yz =
zx =
=
dux
du
=
dx
dx
duy
dv
=
= 0 on account of 2.42
dy
dy
duzz
dw(x)
=
= 0 from 2.38
dz
dz
1 dux duy
+
= 0 from 2.37 and 2.42
2 dy
dx
duz
1 dv dw
1 duy
+
=
+
=0
2 dz
dy
2 dz
dy
1 duz
dux
1 dw du
+
=
+
2 dx
dz
2 dx
dz
1 dw dw
=0
2 dx
dx
(2.40a)
(2.40b)
(2.40c)
(2.40d)
(2.40e)
(2.40f)
It is seen that all components of the strain tensor vanish except the one in the direction of
beam axis.
Note that xx is the only component of the strain tensor in the elementary beam theory. Therefore the subscript xx can be dropped and, unless specified otherwise xx = .
Introducing Eq.(2.37) into Eq.(2.38) one gets
(x, z) =
du (x) d2 w(x)
z
dx
dx2
(2.41)
The first term represents the strain arising from a uniform extension of the entire crosssection
du (x)
(x) =
(2.42)
dx
The second term adds a contribution of bending. Introducing the definition of the curvature
of the beam axis
d2 w(x)
def
=
,
(2.43)
dx2
2-15
Structural Mechanics
2.080 Lecture 2
Semester Yr
## the expression for strain can be put in the final form:
(x, z) = (x) + z
(2.44)
Mathematically, the curvature is defined as a gradient of the slope of a curve. The minus
sign in Eq.(2.27b) follows from the rigorous description of the curvature of a line in the
assumed coordinate system. Physically, it assumes that strains on the tensile side of the
beam are positive.
A quite different interpretation of the Euler-Bernoulli hypothesis is offered by considering
a two-term expansion of the exact strain profile in the Taylor series around the point z = 0
(x, z) = (x, z) |z=0
d
1 d2
+
z+
z2 +
dz z=0
2 dz 2 z=0
(2.45)
Taking only the first two terms is a good engineering approximation but leads to some internal inconsistencies of the elementary beam theory. These inconsistencies will be explained
in the two subsequent lectures.
2.6
## Strain-Displacement Relation of Thin Plates
The present course 2.080 is a prerequisite for a more advanced course 2.081 on Plates and
Shells. A complete set of lecture notes for 2.081 is available on OpenCourseWare. The
interested reader will find there a complete presentation of the theory of moderately large
deflection of plates, derived from first principles. Here only a short summary is given.
Notation
In the lectures on plates and shells two notations will be used. The formulation and some
of the derivation will be easier (and more elegant) by invoking the tensorial notation. Here
students should flip briefly to Recitation 1 where the above mathematical manipulations
are explained. For the purpose of the solving plate problems, the expanded notation will
be used.
Points on the middle surface of the plate are described by the vector {x1 , x2 } or x ,
= 1, 2 in tensor notation or {x, y} in expanded notation.
Likewise, the in-plane components of the displacement vector are denoted by {u, v}.
The vertical component of the displacement vector in the z-direction is denoted by w.
## Plate versus Beam Theory
The plate theory requires fewer assumptions and is more self-consistent than the beam
theory. For one, there are no complications arising from the concept of the centroidal axis
for arbitrarily shaped prismatic beams. The z-coordinate is measured from the middle
plane which is self explanatory. Finally, the flexural/torsional response of non-symmetric
and/or thin-walled cross-section beams is not present in plates. The complexity of the plate
2-16
Structural Mechanics
2.080 Lecture 2
Semester Yr
formulation comes from the two-dimensionality of the problem. The ordinary differential
equations in beams are now becoming partial differential equations.
2-17
Structural Mechanics
2.080 Lecture 2
Semester Yr
2.7
Plates
## The Love-Kirchoff hypothesis extends the one-dimensional Euler-Bernoulli assumptions
into plates. A plate can be bent in two directions, forming a double curvature surface.
Therefore the plane-remains-plane and normal-remains-normal properties are now required
in both directions. Thus, Eq. (2.35) and Eq. (2.36) take the form
u = u z
w def
= w,
=
x
(2.46a)
(2.46b)
where is the slope (rotation) in x -direction. Upon elimination of between the above
equation, one gets the familiar linear dependence of the in-plane components of the displacement vector on the z-coordinate
u (x , z) = u (x ) zw,
(2.47)
## The constant thickness (w = w(x
)) is the third kinematic assumption of the plate theory.
Now, watch carefully how the strain components in the plate are calculated. Considering
all components of the strain tensor, one can distinguish three in-plane strain components
(framed area on the matrix below) and three out-of-plane components.
13
23
33
## The through thickness strain component vanishes on the assumption of independence
of the vertical displacement on the coordinate z
33 = zz =
w
=0
z
(2.48)
The two out-of-plane shear components of the strain tensor 3 vanish due to the LoveKirchoff hypothesis, Eq.(2.47),
1 u
w
1
3 =
+
= (u,z + w, )
2 z
x
2
(2.49)
1 d
=
(u (x ) zw, ) + w, = 0
2 dz
2-18
Structural Mechanics
2.080 Lecture 2
Semester Yr
The non-vanishing components of the strain tensor are the in-plane strain components
1
= (u, + u, ) , = 1, 2
2
(2.50)
## where u is defined by 2.32. Performing the differentiation one gets
1
= [u zw, ], +
2
1
= (u, + u, )
2
1
[u zw, ],
2
1
z[w, + w, ]
2
(2.51)
The first term in Eq. (2.51) is the strain arising from the membrane action in the plate.
It is a symmetric gradient of the middle plane displacement u . Since the order of partial
differentiation is not important, Eq. (2.51) simplifies to
(x , z) = (x ) zw,
(2.52)
## Defining the curvature tensor by
= w, =
2w
x x
(2.53)
The strain-displacement relation for thin plates takes the final form
where
= + z,
(2.54)
1
= (u, + u, )
2
(2.55)
2-19
Structural Mechanics
2.8
2.080 Lecture 2
Semester Yr
## Expanded Form of Strain-Displacement Relation
Having derived the geometric relations in the tensorial notations, equations (2.54) and (2.55)
will be re-written in the coordinate system (x, y) and physical interpretation will be given
to each term. Consider first (2.55)
1 ux ux
ux
= 1, = 1 x1 = x, xx =
=
+
(2.56a)
2 x
x
x
uy
uy
1 uy
= 2, = 2 x2 = y, xx =
=
+
(2.56b)
2 y
y
y
1 ux uy
+
(2.56c)
= 1, = 2 x1 = x, x2 = y, xy =
2 y
x
The xx and yy components denote strains of the middle surface of the plate in the x and
y directions, respectfully. The membrane strains are due to the imposed displacements or
membrane forces applied to the edges. In the theory of small deflection of plates, lateral
pressure loading will not produce membrane strains. By contrast, membrane strains do
develop in the theory of moderately large deflection of plates due to transverse loading.
This topic will be covered later in Lecture 6.
The third component of the strain tensor is the in-plane shear strain xy . It represents
the change of angles in the plane of the plate due to the shear loading at the edges. The
geometrical interpretation of the membrane strain tensor is similar to that given for the
general strain tensor in Figures (2.4) and (2.5).
The curvature tensor requires a careful explanation. Consider an infinitesimal segment ds of a curve and fit into it a circle of an instantaneous radius , Fig. (2.18). Then
ds = d
(2.57)
A
A
ds
A'
A'
## Figure 2.18: Change of slope of a line between two points
Mathematically, the curvature of any line is the change of the slope as one moves
along the curve
def d
=
(2.58)
ds
2-20
Structural Mechanics
2.080 Lecture 2
Semester Yr
1
By comparing Eq. (2.58) with Eq. (2.46b), the curvature in [ ] is the reciprocity of the
m
1
radius of curvature = . The first component of the curvature tensor, defined by Eq.
(2.54) is
2w
w
=
= 1, = 1 x1 = x xx = 2 =
(x )
(2.59)
x
x x
x
This will be the only component of the curvature tensor if the plate is subject to the so-called
cylindrical bending.
x
(a)
(b)
Figure 2.19: (a) cylindrical bending of a plate, and (b) bending with a twist.
The interpretation of the yy components of the curvature tensor
= 2, = 2 x2 = y
yy =
2w
= (y )
2
y
y
(2.60)
is similar as before. More interesting is the mixed component of the curvature tensor xy
= 1, = 2 x1 = x, x2 = y
xy =
2w
= (x )
xy
y
(2.61)
To detect xy one has to check if the slope in one direction, say x changes along the second
y-direction. It does not for a cylindrical bending, 2.14(a). But if it does, the plate is twisted,
as shown in 2.14(b). Therefore, the component xy is called a twist.
An important parameter that distinguishes between these classes of the deformed shape
of a plate is the Gaussian curvature, G . The Gaussian curvature is defined as a product
of two principal curvatures
G = I II
(2.62)
The curvature is a tensor, so its components change by rotating the coordinate system by
an angle to a new direction (x0 , y 0 ). There is one such an angle p for which the twisting
components vanish. The remaining diagonal components are called principal curvature.
The full coverage of the transformation formulae for vectors and tensors are presented in
Recitation 2. Using these results, the Gaussian curvature can be expressed in terms of the
components of the curvature tensor
G = xx yy 2xy
2-21
(2.63)
Structural Mechanics
2.080 Lecture 2
Semester Yr
For cylindrical bending the twist xy as well as one of the principal curvatures vanishes
so that the Gaussian curvature is zero. The sign of the Gaussian curvature distinguishes
between three types of the deformed plate, the bowl, the cylinder and the saddle, Fig.(2.20).
Bowl, G > 0
Cylinder, G = 0
## Figure 2.20: Deformed plate with three different classes of shapes.
The consideration of Gaussian curvature introduces important simplifications in formulation and applications of the energy method in structural mechanics. A separate lecture
will be devoted to this topic.
2.9
## A complete presentation of the theory of moderately large deflections of plates, derived
from first principles is presented in the course 2.081 Plates and Shells. The lecture notes
for this course are available on OpenCourseWare. There the strain-displacement relation
for the theory of moderately large deflection of beams are derived. Here the corresponding
equations for plates are only stated with a physical interpretation. An interested reader is
referred to the Plates and Shells notes for more details.
## Defining moderately large deflections of beams
What are the moderately large deflections and how do they differ from the small deflection. To see the difference, it is necessary to consider the initial and deformed configuration
of the beam axis. The initial and current length element in the undeformed and deformed
configuration respectively is denoted dx and ds, as in Fig. (2.21)
dx
dx
dx
ds
dw
ds
ds
## Figure 2.21: Change of length of the beam axis produced by rotation.
2-22
Structural Mechanics
2.080 Lecture 2
Semester Yr
2
dx = ds cos ds 1
2
(2.64)
## One can distinguish between three theories:
(i) Small deflections, linear geometry 2 1, dx ds, Fig. (2.21(a)).
(ii) Moderately large deflections. The two-term expansion of the cosine function gives a
good approximation for 0 < < 10 . Relation between dx and ds is given by Eq.
(2.64), Fig. (2.21(b)).
(iii) For larger rotation, a full nonlinearity of the problem must be considered.
The present derivation refers to case (ii) above. The Cauchy strain measure, defined in Eq.
ds2 dx2
=
(2.65)
2dx2
The current length ds can be expressed in terms of dx and dw, see Fig.(2.21)
ds2 = dx2 + dw2
(2.66)
From the above two equations, the strain of the beam axis due to element rotation, rot is
=
1
2
dw
dx
2
1
= 2
2
(2.67)
The beam axis also extends due to the gradient of the axial component of the displacement
vector, defined by Eq.(2.42). Therefore the total strain of the beam axis due to the combined
extension and rotation is
du 1 dw 2
=
+
(2.68)
dx 2 dx
It can be noticed that the second term in the above equation is always positive while the
first term can be either positive or negative. In a special case the two terms can cancel one
another even though a beam undergoes large deformation.
The question often asked by students is if the expression for the curvature, given by Eq.
(2.43) should also be modified due to larger rotation. From the mathematical point of view
the answer is YES. But engineers have a way to get around it.
In the rectangular coordinate system the exact definition of the curvature of the line is:
d2 w
dx2
!3/2
dw 2
1+
dx
2-23
(2.69)
Structural Mechanics
2.080 Lecture 2
Semester Yr
dw
0 the linear definition is recovered from the nonlinear equation Eq.
dx
(2.69). The difference between Eq.(2.43) and Eq.(2.69) is small in the case of moderately
large deflection.
The total strain at an arbitrary point of a beam undergoing moderately large deflection
is
du 1 dw 2
=
+
+
z
(2.70)
|{z}
dx 2 dx
|
{z
}
bending strain z
membrane strain
In the limit
## Extension to Moderately Large Deflection of Plates
In the compact tensorial notation, the nonlinear strain-displacement relation takes the form
1
1
(2.71)
= (u, + u, ) + w, w, + z
2
2
By comparing with a similar expression for the small deflection theory, Eq.(2.54) and
Eq.(2.55), the new nonlinear term is
1 w w
1
w, w, =
2
2 x x
This term forms a 2 2 matrix:
1 w 2
2 x
1 w w
2 y x
1 w x
2w y
1 w 2
2 y
x2 1
, x y
2 2
y2
1
x y ,
2
2
(2.72)
(2.73)
The diagonal terms represent square of the slope of the deflection shape in x and y directions.
The non-diagonal terms are symmetric and are a product of slopes in the two directions.
This term vanishes for cylindrical bending.
2.10
## Strain-Displacement Relations for Circulate Plates
The theory of circular plates is formulated in the cylindrical coordinate system (r, , z). The
corresponding components of the displacement vector are (u, v, w). In the remainder of the
notes, the axi-symmetric deformation is assumed, which would require the loading to be
axi-symmetric as well. This assumption brings four important implications
(i) The circumferential component of the displacement is zero, v 0
(ii) There are no in-plane shear strains, r = 0
(iii) The radial and circumferential strains are principal strains
2-24
Structural Mechanics
2.080 Lecture 2
Semester Yr
(iv) The partial differential equations for plates reduces to the ordinary differential equation where the radius is the only space variable.
Many simple closed-form solutions can be obtained for circular and annular plates under
different boundary and loading conditions. Therefore such plates are often treated as prototype structures on which certain physical principles could be easily explained.
The membrane strains on the middle surface are stated without derivation
du 1 dw 2
rr =
+
dr
2 dr
u
=
r
(2.74a)
(2.74b)
d2 w
dr2
1 dw
=
r dr
rr =
(2.75a)
(2.75b)
## The sum of the bending and membrane strains is thus given by
rr (r, z) = rr (r) + zrr
(r, z) =
(r)
+ z
(2.76a)
(2.76b)
It can be noticed that the expression for the radial strains and curvature are identical to
those of the beam when r is replaced by x. The expressions in the circumferential direction
are quite different.
2-25
MIT OpenCourseWare
http://ocw.mit.edu
Fall 2013 | 10,002 | 34,516 | {"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} | 3.140625 | 3 | CC-MAIN-2019-35 | latest | en | 0.900804 |
https://extraextravagant.com/skin-problem/how-many-moles-of-oxygen-atoms-are-there-in-10-moles-of-kclo3.html | 1,638,944,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00401.warc.gz | 315,803,043 | 17,893 | # How many moles of oxygen atoms are there in 10 moles of KClO3?
Contents
## How many moles of O2 are in kclo3?
So, 6 moles of Oxygen Gas O2 has been given to you. To calculate the number of moles for potassium chlorate KClO3 , use the mole ratio.
## How many molecules are in 10 moles O2?
Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is 22.4 L × its molar quantity, approx. 8⋅L . There are 1.51×1023molecules O2 in 10.0 g O2 .
## What is the formula for moles to grams?
In order to convert the moles of a substance to grams, you will need to multiply the mole value of the substance by its molar mass.
## How many moles of KClO3 are needed to produce 15 moles O2?
You’d need 33.3 moles of potassium chlorate, KClO3 , to produce that much oxygen.
## How many atoms are in 2 moles of oxygen?
There are 6×6.022×1023 atoms in 2.00⋅mol NO2(g) .
## How many grams are in 1 moles of KClO3?
You can view more details on each measurement unit: molecular weight of KClO3 or grams This compound is also known as Potassium Chlorate. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles KClO3, or 122.5495 grams.
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## How do I calculate moles?
How to find moles?
1. Measure the weight of your substance.
2. Use a periodic table to find its atomic or molecular mass.
3. Divide the weight by the atomic or molecular mass.
4. Check your results with Omni Calculator.
## How many atoms are in 4 moles oxygen?
Oxygen gas (O2) is made up of 2 atoms of oxygen. Since oxygen has an atomic mass of 16 g/mole, the molar mass of oxygen gas (O2) is 2 x 16 g/mole = 32 g/mole. Since 1 mole of oxygen is equivalent to 32 g, 4 moles of oxygen gas would be equivalent to 4 moles x 32 g/mole = 128 g. | 560 | 1,852 | {"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} | 3.515625 | 4 | CC-MAIN-2021-49 | latest | en | 0.925076 |
https://www.physicsforums.com/threads/motion-in-2d.374511/ | 1,607,018,258,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141729522.82/warc/CC-MAIN-20201203155433-20201203185433-00559.warc.gz | 716,646,657 | 14,240 | # Motion in 2D
## Homework Statement
A football player has to kick a ball through the uprights that are 36m away and 3.05 high. He kicks the ball at 20m/s and at an angle of 53 degrees. By how much does the ball clear or fall short of the goal?
## Homework Equations
Horizontal Range= vi2 sin 2$$\Theta$$/ g
Height = vi2sin2$$\Theta$$/2g
## The Attempt at a Solution
Using the formulas for range and hieght, I found that the ball traveled a total of 39.23m and reached a maximum height of 13.02. The ball traveled far enough, but how do I find how high the ball was at 36m to see if there was enough height? | 173 | 614 | {"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": 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} | 3.546875 | 4 | CC-MAIN-2020-50 | latest | en | 0.942718 |
https://byjus.com/maths/number-system/?replytocom=138725 | 1,643,199,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00145.warc.gz | 213,134,057 | 158,363 | # Number System
The number system or the numeral system is the system of naming or representing numbers. We know that a number is a mathematical value that helps to count or measure objects and it helps in performing various mathematical calculations. There are different types of number systems in Maths like decimal number system, binary number system, octal number system, hexadecimal number system. In this article, we are going to learn what is a number system in Maths? different types, conversion procedures with many number system examples in detail. Also, check mathematics for grade 12 here.
## What is Number System in Maths?
A number system is defined as a system of writing to express numbers. It is the mathematical notation for representing numbers of a given set by using digits or other symbols in a consistent manner. It provides a unique representation of every number and represents the arithmetic and algebraic structure of the figures. It also allows us to operate arithmetic operations like addition, subtraction and division.
The value of any digit in a number can be determined by:
• The digit
• Its position in the number
• The base of the number system
Before discussing the different types of number system examples, first, let us discuss what is a number?
## What is a Number?
A number is a mathematical value used for counting or measuring or labelling objects. Numbers are used to performing arithmetic calculations. Examples of numbers are natural numbers, whole numbers, rational and irrational numbers, etc. 0 is also a number that represents a null value.
A number has many other variations such as even and odd numbers, prime and composite numbers. Even and odd terms are used when a number is divisible by 2 or not, whereas prime and composite differentiate between the numbers that have only two factors and more than two factors, respectively.
In a number system, these numbers are used as digits. 0 and 1 are the most common digits in the number system, that are used to represent binary numbers. On the other hand, 0 to 9 digits are also used for other number systems. Let us learn here the types of number systems.
## Types of Number System
There are various types of number systems in mathematics. The four most common number system types are:
1. Decimal number system (Base- 10)
2. Binary number system (Base- 2)
3. Octal number system (Base-8)
4. Hexadecimal number system (Base- 16)
Now, let us discuss the different types of number systems with examples.
### Decimal Number System (Base 10 Number System)
The decimal number system has a base 10 because it uses ten digits from 0 to 9. In the decimal number system, the positions successive to the left of the decimal point represent units, tens, hundreds, thousands and so on. This system is expressed in decimal numbers. Every position shows a particular power of the base (10).
Example of Decimal Number System:
The decimal number 1457 consists of the digit 7 in the units position, 5 in the tens place, 4 in the hundreds position, and 1 in the thousands place whose value can be written as
(1×103) + (4×102) + (5×101) + (7×100)
(1×1000) + (4×100) + (5×10) + (7×1)
1000 + 400 + 50 + 7
1457
### Binary Number System (Base 2 Number System)
The base 2 number system is also known as the Binary number system wherein, only two binary digits exist, i.e., 0 and 1. Specifically, the usual base-2 is a radix of 2. The figures described under this system are known as binary numbers which are the combination of 0 and 1. For example, 110101 is a binary number.
We can convert any system into binary and vice versa.
Example
Write (14)10 as a binary number.
Solution:
Base 2 Number System Example
∴ (14)10 = 11102
### Octal Number System (Base 8 Number System)
In the octal number system, the base is 8 and it uses numbers from 0 to 7 to represent numbers. Octal numbers are commonly used in computer applications. Converting an octal number to decimal is the same as decimal conversion and is explained below using an example.
Example: Convert 2158 into decimal.
Solution:
2158 = 2 × 82 + 1 × 81 + 5 × 80
= 2 × 64 + 1 × 8 + 5 × 1
= 128 + 8 + 5
= 14110
### Hexadecimal Number System (Base 16 Number System)
In the hexadecimal system, numbers are written or represented with base 16. In the hex system, the numbers are first represented just like in decimal system, i.e. from 0 to 9. Then, the numbers are represented using the alphabets from A to F. The below-given table shows the representation of numbers in the hexadecimal number system.
Hexadecimal 0 1 2 3 4 5 6 7 8 9 A B C D E F Decimal 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
## Number System Chart
In the number system chart, the base values and the digits of different number system can be found. Below is the chart of the numeral system.
Number System Chart
## Number System Conversion
Numbers can be represented in any of the number system categories like binary, decimal, hex, etc. Also, any number which is represented in any of the number system types can be easily converted to other. Check the detailed lesson on the conversions of number systems to learn how to convert numbers in decimal to binary and vice versa, hexadecimal to binary and vice versa, and octal to binary and vice versa using various examples.
With the help of different conversion procedures explained above, now let us discuss in brief about the conversion of one number system to the other number system by taking a random number.
Assume the number 348. Thus, the number 349 in different number systems are as follows:
The number 349 in the binary number system is 101011101
The number 349 in the decimal number system is 349.
The number 349 in the octal number system is 535.
The number 349 in the hexadecimal number system is 15D
## Number System Examples
Example 1:
Convert (1056)16 to octal number.
Solution:
Given, 105616 is an hex number.
First we need to convert the given hexadecimal number into decimal number
(1056)16
= 1 x 163 + 0 x 162 + 5 x 161 + 6 x 160
= 4096 + 0 + 80 + 6
= (4182)10
Now we will convert this decimal number to the required octal number by repetitively dividing by 8.
8 4182 Remainder 8 522 6 8 65 2 8 8 1 8 1 0 0 1
Therefore, taking the value of remainder from bottom to top, we get;
(4182)10 = (10126)8
Therefore,
(1056)16 = (10126)8
Example 2:
Convert (1001001100)2 to decimal number.
Solution:
(1001001100)2
= 1 x 29 + 0 x 28 + 0 x 27 + 1 x 26 + 0 x 25 + 0 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 0 x 20
= 512 + 64 + 8 + 4
= (588)10
Example 3:
Convert 101012 into an octal number.
Solution:
Given,
101012 is the binary number
We can write the given binary number as:
010 101
Now as we know, in octal number system,
010 → 2
101 → 5
Therefore, the required octal number is 258
Example 4:
Convert hexadecimal 2C to decimal number.
Solution:
We need to convert 2C16 into binary numbers first.
2C → 00101100
Now convert 001011002 into a decimal number.
101100 = 1×25+1×23+1×22
=32+8+4
=44
## Number System Questions
2. Convert 0.52 into an octal number. [Answer: 4121]
3. Subtract 11012 and 10102. [Answer: 0010]
4. Represent 5C6 in decimal. [Answer:1478]
5. Represent binary number 1.1 in decimal. [Answer: 1.5]
Also Check: Binary Operations
### Computer Numeral System (Number System in Computers)
When we type any letter or word, the computer translates them into numbers since computers can understand only numbers. A computer can understand only a few symbols called digits and these symbols describe different values depending on the position they hold in the number. In general, the binary number system is used in computers. However, the octal, decimal and hexadecimal systems are also used sometimes.
## Frequently Asked Questions on Number System
### What is Number System and its Types?
The number system is simply a system to represent or express numbers. There are various types of number systems and the most commonly used ones are decimal number system, binary number system, octal number system, and hexadecimal number system.
### Why is the Number System Important?
The number system helps to represent numbers in a small symbol set. Computers, in general, use binary numbers 0 and 1 to keep the calculations simple and to keep the amount of necessary circuitry less, which results in the least amount of space, energy consumption and cost.
### What is Base 1 Number System Called?
The base 1 number system is called the unary numeral system and is the simplest numeral system to represent natural numbers.
### What is the equivalent binary number for the decimal number 43?
To find the equivalent binary number, we need to divide 43 by 2, until we get 0 as the result. Therefore, (43)10 = 1010112
### How to convert 308 into a decimal number?
308 = (3×81)+(0×80) = 24
Quiz on Number system
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Excellent job | 2,401 | 9,525 | {"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} | 4.65625 | 5 | CC-MAIN-2022-05 | longest | en | 0.918853 |
https://fixedpointtheoryandapplications.springeropen.com/articles/10.1155/2010/296759 | 1,561,383,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999539.60/warc/CC-MAIN-20190624130856-20190624152856-00351.warc.gz | 443,023,775 | 35,383 | • Research Article
• Open Access
# Strong Convergence Theorem for Equilibrium Problems and Fixed Points of a Nonspreading Mapping in Hilbert Spaces
Fixed Point Theory and Applications20102010:296759
https://doi.org/10.1155/2010/296759
• Accepted: 13 December 2010
• Published:
## Abstract
We introduce an iterative method for finding a common element of the set of solutions of equilibrium problems and the set of fixed points of a nonspreading mapping in a Hilbert space. Then, we prove a strong convergence theorem which is connected with the work of S. Takahashi and W. Takahashi (2007) and Iemoto and Takahashi (2009).
## Keywords
• Hilbert Space
• Equilibrium Problem
• Nonexpansive Mapping
• Iterative Scheme
• Common Element
## 1. Introduction
Let be a real Hilbert space with inner product and norm , respectively, and let be a closed convex subset of . Let be bifunction, where is the set of real numbers. The equilibrium problem for is to find such that
(1.1)
The set of solution of (1.1) is denoted by . Given a mapping , let for all . Then, if and only if for all , that is, is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1.1); see, for example, [19] and the references therein.
A mapping of into itself is said to be nonexpansive if for all , and a mapping is said to be firmly nonexpansive if for all . Let be a smooth, strictly convex and reflexive Banach space, and let be the duality mapping of and a nonempty closed convex subset of . A mapping is said to be nonspreading if
(1.2)
for all , where for all ; see, for instance, Kohsaka and Takahashi [10]. In the case when is a Hilbert space, we know that for all . Then a nonspreading mapping in a Hilbert space is defined as follows:
(1.3)
for all . Let be the set of fixed points of , and nonempty; a mapping is said to be quasi-nonexpansive if for all and .
Remark 1.1.
In a Hilbert space, we know that every firmly nonexpansive mapping is nonspreading and that if the set of fixed points of a nonspreading mapping is nonempty, the nonspreading mapping is quasi-nonexpansive; see [10, 11].
In 1953, Mann [12] introduced the iteration as follows: a sequence defined by
(1.4)
where the initial guess element is arbitrary and is a real sequence in . Mann iteration has been extensively investigated for nonexpansive mappings. In an infinite-dimensional Hilbert space, Mann iteration can conclude only weak convergence (see [12, 13]). Fourteen years later, Halpern [14] introduced the following iterative scheme for approximating a fixed point of :
(1.5)
for all , where and is a sequence of . Strong convergence of this type iterative sequence has been widely studied: Wittmann [15] discussed such a sequence in a Hilbert space.
On the other hand, Kohsaka and Takahashi [10] proved an existence theorem of fixed point for nonspreading mappings in a Banach space. Recently, Lemoto and Takahashi [16] studied the approximation theorem of common fixed points for a nonexpansive mapping of into itself and a nonspreading mapping of into itself in a Hilbert space. In particular, this result reduces to approximation fixed points of a nonspreading mapping of into itself in a Hilbert space by using iterative scheme
(1.6)
Some methods have been proposed to solve the equilibrium problem and fixed point problem of nonexpansive mapping: see, for instance, [1, 2, 6, 7, 1720] and the references therein. In 1997, Combettes and Hirstoaga [3] introduced an iterative scheme of finding the best approximation to the initial data when is nonempty and proved a strong convergence theorem. Recently, S. Takahashi and W. Takahashi [8] introduced an iterative scheme by the viscosity approximation method for finding a common element of the set of solution of equilibrium problems and the set of fixed points of a nonexpansive mapping in a Hilbert space. Let be a nonexpansive mapping. In 2008, Plubtieng and Punpaeng [7] introduced a new iterative sequence for finding a common element of the set of solution of equilibrium problems and the set of fixed points of a nonexpansive mapping in a Hilbert space which is the optimality condition for the minimization problem. Very recently, S. Takahashi and W. Takahashi [9] introduced an iterative method for finding a common element of the set of solutions of a generalized equilibrium problem and the set of fixed points of a nonexpansive mapping in a Hilbert space and then obtain that the sequence converges strongly to a common element of two sets.
In this paper, motivated by S. Takahashi and W. Takahashi [8] and Lemoto and Takahashi [16], we introduce an iterative sequence and prove a strong convergence theorem for finding solution of equilibrium problems and the set of fixed points of a nonspreading mapping in Hilbert spaces.
## 2. Preliminaries
Let be a real Hilbert space. When is a sequence in , implies that converges weakly to and means the strong convergence. Let be a nonempty closed convex subset of . For every point , there exists a unique nearest point in ; denote by , such that
(2.1)
is called the metric projection of onto . We know that is nonexpansive. Further, for and ,
(2.2)
Moreover, is characterized by the following properties: and
(2.3)
for all , . We also know that satisfies Opial's condition [21], that is, for any sequence with , the inequality
(2.4)
holds for every with ; see [21, 22] for more details.
The following lemmas will be useful for proving the convergence result of this paper.
Lemma 2.1 (see [23]).
Let be an inner product space. Then for all and with , one has
(2.5)
Lemma 2.2 (see [10]).
Let be a Hilbert space, a nonempty closed convex subset of . Let be a nonspreading mapping of into itself. Then the following are equivalent.
(1)There exists such that is bounded;
(2) is nonempty.
Lemma 2.3 (see [10]).
Let be a Hilbert space, a nonempty closed convex subset of . Let be a nonspreading mapping of into itself. Then is closed and convex.
Lemma 2.4.
Let be a real Hilbert space. Then for all ,
(1) ;
(2) .
Lemma 2.5 (see [24]).
Let , and let be sequences of real numbers such that
, for all ,
and .
Then, .
Lemma 2.6 (see [16]).
Let be a Hilbert space, a closed convex subset of , and a nonspreading mapping with . Then is demiclosed, that is, and imply .
Lemma 2.7 (see [16]).
Let be a Hilbert space, a nonempty closed convex subset of a real Hilbert space , and let be a nonspreading mapping of into itself, and let . Then
(2.6)
Lemma 2.8 (see [25]).
Assume is a sequence of nonnegative real numbers such that
(2.7)
where is a sequence in and is a sequence in such that
(1) ;
(2) or .
Then .
For solving the equilibrium problems for a bifunction , let us assume that satisfies the following conditions:
(A1) ;
(A2) is monotone, that is, ;
(A3)for each , ;
(A4)for each , is convex and lower semicontinuous.
The following lemma appears implicitly in [26].
Lemma 2.9 (see [26]).
Let be a nonempty closed convex subset of , and let be a bifunction of into satisfying (A1)–(A4). Let and . Then, there exists such that
(2.8)
The following lemma was also given in [4].
Lemma 2.10 (see [4]).
Assume that satisfies (A1)–(A4). For and , define a mapping as follows:
(2.9)
for all . Then, the following hold:
(1) is single-valued;
(2) is firmly nonexpansive, that is, for any , ;
(3) ;
(4) is closed and convex.
Lemma 2.11 (see [27]).
Let be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence of which satisfies for all . Also consider the sequence of integers defined by
(2.10)
Then is a nondecreasing sequence verifying , and the following properties are satisfied for all :
(2.11)
## 3. Main Result
In this section, we prove a strong convergence theorem for finding a common element of the set of fixed points of a nonspreading mapping and the set of solutions of the equilibrium problems.
Theorem 3.1.
Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunctions from satisfying (A1)–(A4), and let be a nonspreading mapping of into itself such that . Let , and let and be sequences generated by and
(3.1)
for all , where and satisfy
, , ,
, ,
, and .
Then converges strongly to , where .
Proof.
Let . From , we have
(3.2)
for all . Put . We divide the proof into several steps.
Step 1.
We claim that the sequences , , , and are bounded. First, we note that
(3.3)
and so
(3.4)
Putting , we note that for all . In fact, it is obvious that . Assume that for all . Thus, we have
(3.5)
By induction, we obtain that for all . So, is bound. Hence, , , and are also bounded.
Step 2.
Put . We claim that as . We note that
(3.6)
where . On the other hand, from and , we have
(3.7)
(3.8)
for all . Putting in (3.7) and in (3.8), we have
(3.9)
So, from (A2), we note that
(3.10)
and hence
(3.11)
Without loss of generality, let us assume that there exists a real number such that for all . Thus, we have
(3.12)
and hence
(3.13)
where . So, from (3.6), we note that
(3.14)
By Lemma 2.5, we have
(3.15)
for . We note from that
(3.16)
and hence
(3.17)
Therefore, from the convexity of , we have
(3.18)
and hence
(3.19)
So, we have . Indeed, since , it follows that
(3.20)
Then, we note that
(3.21)
Since, and , it follows that
(3.22)
Step 3.
Put . From , it follows by Lemma 2.7 that
(3.23)
Since , we have . Therefore, by (3.23), we obtain
(3.24)
Step 4.
Putting , we claim that the sequence converges strongly to . Indeed, we discuss two possible cases.
Case 1.
Assume that there exists such that the sequence is a nonincreasing sequence for all . Then we have (for ), and hence exists. Therefore
(3.25)
By (3.22), (3.24), and (3.25), we get
(3.26)
Let be a subsequence of such that
(3.27)
Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that . Since is closed and convex, we note that is weakly closed. So, we have . Since , it follows by Lemma 2.6 that . From (3.27) and the property of metric projection, we have
(3.28)
Finally, we prove that . In fact, since , it follows that
(3.29)
By (3.28) and , we immediately deduce by Lemma 2.8 that .
Case 2.
Assume that for all , there exits such that . Put for all . Thus, it follows that there exists a subsequence of such that for all . Let be a mapping defined by
(3.30)
where . By Lemma 2.11, we note that is a nondecreasing sequence such that as and that the following properties are satisfied by all numbers :
(3.31)
From (3.24), we have
(3.32)
This implies that
(3.33)
Take a subsequence of such that
(3.34)
From the boundedness of , we can assume that . Since is closed and convex, it follows that is weakly closed. So, we have . Since , it follows by Lemma 2.6 that . From (3.34) and the property of metric projection, we have
(3.35)
By the same argument as (3.29) in Case 1, we conclude immediately that, for all ,
(3.36)
which implies that
(3.37)
By (3.35), we have
(3.38)
and hence
(3.39)
Since for all , we have
(3.40)
This completes the proof.
As direct consequences of Theorem 3.1, we obtain corollaries.
Corollary 3.2.
Let be a nonempty closed convex subset of a real Hilbert space . Let be a bifunctions from satisfying (A1)–(A4), and let be a firmly nonexpansive mapping of into itself such that . Let , and let and be sequences generated by and
(3.41)
for all , where and satisfy
, , ,
, ,
, and .
Then converges strongly to , where .
## Declarations
### Acknowledgments
The authors would like to thank the referees for the insightful comments and suggestions. Moreover, the authors gratefully acknowledge the Thailand Research Fund Master Research Grants (TRF-MAG, MRG-WII515S029) for funding this paper.
## Authors’ Affiliations
(1)
Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok, 65000, Thailand
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18. Plubtieng S, Sriprad W: Hybrid methods for equilibrium problems and fixed points problems of a countable family of relatively nonexpansive mappings in Banach spaces. Fixed Point Theory and Applications 2010, 2010:-17.Google Scholar
19. Plubtieng S, Thammathiwat T: A viscosity approximation method for equilibrium problems, fixed point problems of nonexpansive mappings and a general system of variational inequalities. Journal of Global Optimization 2010,46(3):447–464. 10.1007/s10898-009-9448-5
20. Plubtieng S, Thammathiwat T: A viscosity approximation method for finding a common solution of fixed points and equilibrium problems in Hilbert spaces. Journal of Global Optimization. In pressGoogle Scholar
21. Opial Z: Weak convergence of the sequence of successive approximations for nonexpansive mappings. Bulletin of the American Mathematical Society 1967, 73: 591–597. 10.1090/S0002-9904-1967-11761-0
22. Takahashi W: Nonlinear Functional Analysis. Yokohama Publishers, Yokohama, Japan; 2000:iv+276.
23. Osilike MO, Igbokwe DI: Weak and strong convergence theorems for fixed points of pseudocontractions and solutions of monotone type operator equations. Computers & Mathematics with Applications 2000,40(4–5):559–567. 10.1016/S0898-1221(00)00179-6
24. Xu HK: An iterative approach to quadratic optimization. Journal of Optimization Theory and Applications 2003,116(3):659–678. 10.1023/A:1023073621589
25. Xu H-K: Viscosity approximation methods for nonexpansive mappings. Journal of Mathematical Analysis and Applications 2004,298(1):279–291. 10.1016/j.jmaa.2004.04.059
26. Blum E, Oettli W: From optimization and variationnal inequalities to equilibrium problems. Mathematics Students 2005, 6: 117–136.Google Scholar
27. Maingé P-E: Strong convergence of projected subgradient methods for nonsmooth and nonstrictly convex minimization. Set-Valued Analysis 2008,16(7–8):899–912. 10.1007/s11228-008-0102-z | 4,727 | 17,241 | {"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-2019-26 | latest | en | 0.867871 |
https://www.teachstarter.com/teks/math-3-4j/ | 1,643,090,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304760.30/warc/CC-MAIN-20220125035839-20220125065839-00677.warc.gz | 1,030,548,812 | 34,024 | Teks
# Math 3.4(J)
determine a quotient using the relationship between multiplication and division; and
31 teaching resources for those 'aha' moments
### Save Santa's Workshop – Escape Room Activity
A fun and festive escape room activity where students solve clues to save Santa's Workshop.
26 pages Grades: 3 - 6
### Multiplication and Division Minute Math Booklet
Warm-up with this 10-page booklet of multiplication and division drills.
14 pages Grades: 3 - 5
### Division Dragon – Worksheet
A set of super cool division worksheets!
6 pages Grades: 3 - 5
### Parts of a Number Sentence - Multiplication and Division
A set of 4 posters explaining the numbers involved a multiplication and division number sentence.
4 pages Grades: 2 - 5
### Multiplication and Division - Which Operation Is It? – Interactive PowerPoint
An interactive 64-slide PowerPoint to use when learning to solve multiplication and division word problems.
64 pages Grades: 2 - 4
### Holiday Code Cracker: Middle Years – Whole Class Holiday Game
A whole class, holiday-themed game where students work together to find a secret code.
7 pages Grades: 3 - 4
### Target Number Math Warm-up Activity
A great math warm-up activity where students brainstorm number sentences that equal a 'target number'.
2 pages Grades: 1 - 6
### Fives Math Machine Division Worksheet
A worksheet to use when practicing basic division facts.
2 pages Grades: 3 - 5
### Division Facts Posters - 1-12
A set of posters showing division facts 1-12.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 12
A poster showing division facts for twelve.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 11
A poster showing division facts for eleven.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 10
A poster showing division facts for ten.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 9
A poster showing division facts for nine.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 8
A poster showing division facts for eight.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 7
A poster showing division facts for seven.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 6
A poster showing division facts for six.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 5
A poster showing division facts for five.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 4
A poster showing division facts for four.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 3
A poster showing division facts for three.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 2
A poster showing division facts for two.
1 page Grades: 3 - 6
### Division Facts Poster - Dividing by 1
A poster showing division facts for one.
1 page Grades: 3 - 6
### Multiplication and Division Fact Families PowerPoint
An 18 page editable PowerPoint to use in the classroom when introducing multiplication and division fact families.
### Fact Family Icosahedron (4, 7, 12 Multiplication and Division Facts)
A hands-on game to play when learning about multiplication and division fact families.
### Fact Family Icosahedron (3, 6, 9 Multiplication and Division Facts)
A hands-on game to play when learning about multiplication and division fact families.
### Fact Family Icosahedron (2, 5, 10 Multiplication and Division Facts)
A hands-on game to play when learning about multiplication and division fact families.
### Fact Family Fishbowls - Blank Version
A sorting activity to demonstrate an understanding of fact families.
3 pages Grades: 1 - 3
### Fact Family Fishbowls - Multiplication and Division
A sorting activity to demonstrate an understanding of multiplication and division fact families.
3 pages Grades: 2 - 3
### Math Work Mats
A set of 5 work mats for students to use when working on specific math concepts.
5 pages Grades: 1 - 5
### Fact Family Rocket Blast Off - Blank Version
A sorting activity to demonstrate an understanding of fact families.
2 pages Grades: 1 - 3
### Math Word Problem Match-Up Game - Basic Multiplication and Division
Twenty word problem cards to use in the classroom when learning to differentiate between multiplication and division situations.
9 pages Grades: 2 - 3
### Mental Math Division Posters
5 posters outlining mental math strategies for division.
5 pages Grades: 3 - 6 | 1,037 | 4,426 | {"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} | 3.65625 | 4 | CC-MAIN-2022-05 | latest | en | 0.718547 |
https://community.ig.com/forums/topic/5159-calculating-the-current-buy-price-of-a-stock/ | 1,675,087,738,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00461.warc.gz | 188,942,789 | 26,214 | # calculating the current buy price of a stock
## Recommended Posts
Hi everyone, I'm new to trading and currently using a demo account and have got a hang of looking for good buying points and positions, but iv notice on the live account that it doesn't calculate the price of a stock when buying.
gold as an example - its currently at 1203.05, if this was the current price I would expect it to be £1203 pounds per share but the buy price £60.
how do I formulate this as I can't see any straight forward answers / calculations.
Edited by chance11
Hi @chance11, you are using a spread betting account to trade gold so you are not actually buying gold but rather a derivative in the form of gold contracts. The price you are seeing on the chart is not the current price but rather the mid price between the current sell and buy price as displayed in the buy/sell box. The £60 you mention seems to refer to the margin requirement which is the amount of free capital in your account needed to open the trade.
Don't forget to take a look at the IG training academy,
• 1
@Caseynotes thanks for the reply, ill have a read. whats got me lost is how I calculate the margin required when purchasing, so how much its going to cost to buy in. would you be able to give me an example using gold, thanks in advance
Hi @chance11, no problem, the initial margin requirement is calculated on the deal ticket for you, this is how it's done;
bet size (£/point) x price x margin %
1 x £1203.20 x 5 % = £60.16
• 1
Actually I just found a new trick James and Chance, sorry if you already seen it before, but google will do the calc just by typing the equation straight into the search box, before I would have gone to a calc page first.
4 minutes ago, Caseynotes said:
Actually I just found a new trick James and Chance, sorry if you already seen it before, but google will do the calc just by typing the equation straight into the search box, before I would have gone to a calc page first.
it will also automatically calculate it on every deal ticket as well
makes perfect sense thanks for your help guys.
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• EUR/USD and EUR/GBP/USD appreciate while GBP/USD range trades Outlook on EUR/USD, EUR/GBP and GBP/USD ahead of this week’s Fed, ECB and BoE rate decisions. Axel Rudolph FSTA | Senior Financial Analyst, London | Publication date: Monday 30 January 2023 EUR/USD recovers from last week’s low EUR/USD is seen bouncing off Friday’s low at \$1.0838 ahead of this week’s plethora of central bank meetings by the likes of the US Federal Reserve (Fed) which is expected to hike its rates by 25-basis points, the European Central Bank (ECB) and the Bank of England (BoE) which are likely to raise their rates by 50-basis points (bps) respectively. The currency pair thus remains on track to reach the late April 2022 high and the 50% retracement of the 2021 to 2022 descent at \$1.0936 to \$1.094 while it stays above Friday’s \$1.0838 low on a daily chart closing basis. A drop through \$1.0838 would engage the mid-January \$1.0766 low. While above it, and the mid- to late-December highs at \$1.0736 to \$1.0715, the medium-term uptrends remain intact. Above \$1.094 lies the psychological \$1.10 mark. Further support can be found around \$1.0663 to \$1.0658, the 16 to 28 December highs. Source: IT-Finance.com EUR/GBP bounces off December-to-January uptrend line EUR/GBP revisited but then bounced off its December-to-January uptrend line at £0.8763 while awaiting Thursday’s ECB and BoE rate decisions, with both central banks expected to hike rates by 50 bps. While £0.8763 underpins, the £0.8828 November peak as well as the £0.8834 - 22 December high - will be back in play, above which sits more significant resistance which can be spotted between the December and current January highs at £0.8877 to £0.8897. Only a slip through £0.8763 would engage the 55-day simple moving average (SMA) at £0.8735 and current January low at £0.8722. If slipped through, the 23 November high and 19 December low at £0.8701 to £0.8691 could once again be reached. Further down sits the 28 November high at £0.8676. Source: IT-Finance.com GBP/USD continues to range trade below its \$1.2446 December high GBP/USD’s September advance from its \$1.0350 all-time low struggled to overcome its December high at \$1.2446 early last week and has been trading in a sideways trading range below this high ever since while awaiting Thursday’s UK central bank decision. This is not to say that the cross might not eventually rise to above its December and January highs at \$1.2446 to \$1.2448, provided that the 24 January low at \$1.2263 doesn’t give way.mWere this to happen, the 9 January high at \$1.221 may be reached. A rise and daily chart close above last week’s \$1.2448 high would engage the minor psychological \$1.2500 mark, above which the 7 June 2022 high can be found at \$1.2599. Source: IT-Finance.com
• Gold and Brent crude consolidate, as natural gas drops into 21-month low Gold and Brent crude struggle to maintain recent gains, while natural gas collapses into a fresh 21-month low. Source: Bloomberg Joshua Mahony | Senior Market Analyst, London | Publication date: Monday 30 January 2023 Gold consolidates as stocks head lower Gold has struggled to maintain its upward trajectory of late, with the precious metal losing some of its shine thanks to a similar sideways trajectory for the US dollar. A resurgence in the dollar could bring about a turn lower for the price of gold, meaning that there is also a likely positive correlation between equities and precious metals for the time being. Nonetheless, from a purely technical standpoint, the recent consolidation phase continues to point towards another move higher as long as price does not break back down through the most recent swing-low of \$1911. Should that occur, it would make sense to expect a potential move lower for gold. Source: ProRealTime Brent crude turning lower from resistance zone Brent crude has been struggling to maintain its upward trajectory over the past week, with price starting to weaken from a key resistance zone. The descending trendline and 100-simple moving average (SMA) have converged to bring a key area that could see the wider downtrend kick in once again. The stochastic oscillator provides another potential signal that the bears could come back into play once again here, with the break through the 80 threshold highlighting a reversal in momentum. Looking back at previous occasions that we have seen this signal, we have seen periods of weakness following each of the past five signals (as shown by vertical dotted lines). For a bearish confirmation signal, watch for a break below the \$83.97 swing-low. Source: ProRealTime Natural gas continues its declines, as price hits 21-month low Natural gas has been hit hard over the course of the past five months, with price falling back from almost \$10 to the sub-\$3 mark we see today. As we look to emerge from a largely mild European winter, the healthy stockpiles largely bring the conversation of a potential squeeze in prices to an end. Whether that issue resurfaces with regards to next winter remains to be seen, but sentiment has clearly taken a hit of late. Nonetheless, there will likely be a point where the price of natural gas is deemed to have gone too far, with current prices trading back within a crucial historical zone that has previously held price for an extended period. While there is a chance that that the bulls come back in at some point, the downtrend still remains in play as highlighted on the four-hour chart. A rise up through \$3.322 would signal a potential bullish reversal coming into play. Until then, the bearish trend remains the dominant force that should continue to send prices lower. Source: ProRealTime
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• Create New... | 1,932 | 8,047 | {"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-2023-06 | longest | en | 0.943749 |
http://www.jiskha.com/display.cgi?id=1263758723 | 1,454,766,918,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701146550.16/warc/CC-MAIN-20160205193906-00080-ip-10-236-182-209.ec2.internal.warc.gz | 499,044,434 | 3,599 | Saturday
February 6, 2016
# Homework Help: Word Problem
Posted by Rene on Sunday, January 17, 2010 at 3:05pm.
A reservation clerk worked 15.3 hours one day. She spent twice as much time entering new reservations as she did verifying old ones and one and a half as much time calling to confirm reservations as verifying old ones. How much time did she spend entering new reservations?
What formula would I use to solve this problem? | 102 | 434 | {"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-2016-07 | longest | en | 0.990635 |
https://www.convertunits.com/from/mile+%5Bstatute,+international%5D/to/beard-second | 1,675,537,132,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00373.warc.gz | 708,456,105 | 12,795 | ## Convert mile [statute, international] to beard-second
mile [statute, international] beard-second
Did you mean to convert mile mile [Britain, ancient] mile [international] mile [Irish] mile [Roman, ancient] mile [Scottish] mile [statute] mile [survey] mile [statute, international] mile [statute, US] to beard-second
How many mile [statute, international] in 1 beard-second? The answer is 3.1068559611867E-12.
We assume you are converting between mile [statute, international] and beard-second.
You can view more details on each measurement unit:
mile [statute, international] or beard-second
The SI base unit for length is the metre.
1 metre is equal to 0.00062137119223733 mile [statute, international], or 200000000 beard-second.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between statute miles and beard-seconds.
Type in your own numbers in the form to convert the units!
## Quick conversion chart of mile [statute, international] to beard-second
1 mile [statute, international] to beard-second = 321868800000 beard-second
2 mile [statute, international] to beard-second = 643737600000 beard-second
3 mile [statute, international] to beard-second = 965606400000 beard-second
4 mile [statute, international] to beard-second = 1287475200000 beard-second
5 mile [statute, international] to beard-second = 1609344000000 beard-second
6 mile [statute, international] to beard-second = 1931212800000 beard-second
7 mile [statute, international] to beard-second = 2253081600000 beard-second
8 mile [statute, international] to beard-second = 2574950400000 beard-second
9 mile [statute, international] to beard-second = 2896819200000 beard-second
10 mile [statute, international] to beard-second = 3218688000000 beard-second
## Want other units?
You can do the reverse unit conversion from beard-second to mile [statute, international], or enter any two units below:
## Enter two units to convert
From: To:
## Definition: Mile
The international statute mile is defined as exactly 1609.344 meters. If you are based in the U.S. you may also want to consider the U.S. statute mile, which you can find by searching for 'mile [statute, US]' on this site.
## Definition: Beard-second
The beard-second is a unit of length inspired by the light-year, but applicable to extremely short distances such as those in integrated circuits. The beard-second is defined as the length an average beard grows in one second. Google Calculator supports the beard-second for unit conversions using the value 5 nm.
## Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 735 | 3,056 | {"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-2023-06 | latest | en | 0.700568 |
http://tajik.english-dictionary.help/?q=integer | 1,568,997,716,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574050.69/warc/CC-MAIN-20190920155311-20190920181311-00486.warc.gz | 181,967,052 | 13,198 | # English to Tajik Meaning :: integer
Integer :
бутуни
- бутуниintegers
#### Show English Meaning (+)
Noun(1) any of the natural numbers (positive or negative
#### Show Examples (+)
(1) integer values(2) Fibonacci proves that the root of the equation is neither an integer nor a fraction, nor the square root of a fraction.(3) A perfect number is a whole number, an integer greater than zero; and when you add up all of the factors less than that number, you get that number.(4) The floor function rounds down by taking a non-integer value to the next integer below it.(5) In the continued fraction of the square root of an integer the same denominators recur periodically.(6) Possibly as a consequence of that, the Greek mathematicians thought of fractions in terms of ratios of integers , rather than numbers.(7) Marshall Hall showed talent for mathematics at a young age when he constructed a seven-place table of logarithms for the positive integers up to 1000.(8) In other words, a number is rational if we can write it as a fraction where the numerator and denominator are both integers .(9) Here are the whole numbers/natural numbers/positive integers up to 700, in binary columns.(10) Clearly, most integers are not squares of whole numbers.(11) A second work is the Book of the Number which describes the decimal system for integers with place values from left to right.(12) The row of numerators starts with the pair of integers 0,1.(13) Whole numbers or integers are often the subject of such pursuits.(14) What about those integers in the continued fraction forms of the powers?(15) When talking about modular arithmetic it is important to remember that we are only allowed to use integers , that is whole numbers.(16) By contrast, which is sometimes overlooked, in the arithmetical Books 7-9 multiplication of integers themselves occurs as usual.
Related Words
(1) positive integer ::
бутуни мусбат
Synonyms
Noun
1. whole number ::
тамоми рақами
Different Forms
integer, integers
English to Tajik Dictionary: integer
Meaning and definitions of integer, translation in Tajik language for integer with similar and opposite words. Also find spoken pronunciation of integer in Tajik and in English language.
Tags for the entry "integer"
What integer means in Tajik, integer meaning in Tajik, integer definition, examples and pronunciation of integer in Tajik language.
# Words by Category
## Word of the day
● Elevate
баланд, идора то, танзим, афзоиш, ислоҳ кардан, , динҳо, баланд бардоштани, инкишоф, варам кардан, таркондани, шафеъе нест, пур аз, лифт, бардоштан, щафо, бењтар намудани, наҷот намеёбанд, нашъунамо, пеш рафтан, хуб, калон кардан, олошавӣ | 668 | 2,676 | {"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} | 3.4375 | 3 | CC-MAIN-2019-39 | latest | en | 0.842165 |
http://easy-ciphers.com/alaudidae | 1,656,681,294,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00246.warc.gz | 16,919,385 | 18,927 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: alaudidae
cipher variations: bmbvejebf cncwfkfcg dodxglgdh epeyhmhei fqfzinifj grgajojgk hshbkpkhl iticlqlim jujdmrmjn kvkensnko lwlfotolp mxmgpupmq nynhqvqnr ozoirwros papjsxspt qbqktytqu rcrluzurv sdsmvavsw tetnwbwtx ufuoxcxuy vgvpydyvz whwqzezwa xixrafaxb yjysbgbyc zkztchczd
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: alaudidae Cipher: zozfwrwzv
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: alaudidae Cipher: AAAAA ABABA AAAAA BAABB AAABB ABAAA AAABB AAAAA AABAA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: alaudidae
cipher variations:
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dodxglgdh
dkdlmbmdp
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elemncneq
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alaudidae
ahaijyjam
azakvevac
avaybubak
aramhkhas
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afaczgzaq
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The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: alaudidae Cipher: nynhqvqnr
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: alaudidae Cipher: 111311544142411151
Extended Methods:
Method #1
Plaintext: alaudidae
method variations: fqfzioifk lvleotolp qaqktytqu vfvpydyvz
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```a l a u d i d a e
1 1 1 5 4 4 4 1 5
1 3 1 4 1 2 1 1 1 ```
They are then read out in rows:
111544415131412111
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: alaudidae Cipher: avtdecdba
Method #3
Plaintext: alaudidae
method variations: acvtqrava cvtqravaa vtqravaac tqravaacv qravaacvt ravaacvtq avaacvtqr vaacvtqra aacvtqrav
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: alaudidae
first 5040 cipher variations(362880 total)
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aleauaidd | 15,426 | 33,872 | {"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} | 3.09375 | 3 | CC-MAIN-2022-27 | latest | en | 0.816077 |
https://hybridsimulator.wordpress.com/2013/10/17/biological-example-synchronization-of-n-fireflies/ | 1,521,677,999,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647707.33/warc/CC-MAIN-20180321234947-20180322014947-00316.warc.gz | 579,587,558 | 18,960 | # biological example: synchronization of n fireflies
Consider a biological example of the synchronization of n fireflies flashing. The fireflies can be modeled mathematically as periodic oscillators which tend to synchronize their flashing until they are flashing in phase with each other. A state value of ${\tau_i=1}$ corresponds to a flash, and after each flash, the firefly automatically resets its internal timer (periodic cycle) to ${\tau_i=0}$. The synchronization of the fireflies can be modeled as a hybrid system because every time one firefly flashes, the other firefly notices and jumps ahead in its internal timer ${\tau_i}$ by ${(1+\varepsilon)\tau_i}$, where ${\varepsilon}$ is a biologically determined coefficient. This happens until eventually all fireflies synchronize their internal timers and are flashing simultaneously. We assume that the fireflies are fully interconnected as in Figure 1.
Now consider a population of n fireflies, then the population can be represented with a vector state ${x = [\tau_1,\tau_2,\ldots,\tau_n]^\top\in\Re^n}$.
Furthermore, the fireflies population can be modeled as a hybrid system given by
Notice that here no external inputs are being considered. The only event that affects the flashing of a firefly ${\tau_i}$ is the flashing of the other firefly.
A solution to the hybrid system with ${T=7,\ J=100,\ r=1,\ \varepsilon=0.2}$ is depicted in Figure 2. Both the projection onto ${t}$ and ${j}$ are shown.
These simulations reflect the expected behavior of the hybrid system. The fireflies initially flash out of phase with one another and then synchronize to flash in the same phase.
Source code for Simulation
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matlab M-file Project: HyEQ Toolbox @ Hybrid Dynamics and Control
% Lab, http://www.u.arizona.edu/~sricardo/index.php?n=Main.Software
%
% Filename: run.m
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Fireflies example - Lite simulator version
%% initial conditions
clear all
close all
clc
% initial conditions
N = 20; % Poblation size
x0 = rand(N,1);
% simulation horizon
T = 7;
J = 100;
% simulation horizon
TSPAN = [0 T];
JSPAN = [0 J];
% rule for jumps
% rule = 1 -> priority for jumps
% rule = 2 -> priority for flows
% rule = 3 -> no priority, random selection when simultaneous conditions
rule = 1;
options = odeset('RelTol',1e-6,'MaxStep',.1);
% constants
global n epsilon;
n = N; %# of state components
epsilon = 0.2; % Bilogical parameter
%% simulate
[t j x] = HyEQsolver(@f,@g,@C,@D,x0,TSPAN,JSPAN,rule,options);
%%
% plot solution
figure(1)
clf
subplot(2,1,1),plotHarc(t,j,x);
grid on
title(['Evolution of firefly oscilation for a ',num2str(n), ' population size (flow)'])
xlabel('time')
ylabel('x')
subplot(2,1,2),plotjumps(t,j,x)
grid on
ylabel('x')
xlabel('jumps')
title(['Evolution of firefly oscilation for a ',num2str(n), ' population size (jump)'])
%% plot hybrid arc
plotHybridArc(t,j,x);
xlabel('j')
ylabel('t')
zlabel('x')
Flow map
function out = f(z)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matlab M-file Project: HyEQ Toolbox @ Hybrid Dynamics and Control
% Lab, http://www.u.arizona.edu/~sricardo/index.php?n=Main.Software
%
% Filename: f.m
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Description: Flow map
%
% Version: 1.0
% Required files: -
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% constants
global n;
% state
x =z(1:n);
% flow map
%xdot=f(x,u);
xdot = ones(n,1);
out = xdot;
Jump map
function out = g(z)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matlab M-file Project: HyEQ Toolbox @ Hybrid Dynamics and Control
% Lab, http://www.u.arizona.edu/~sricardo/index.php?n=Main.Software
%
% Filename: g.m
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Description: Jump map
%
% Version: 1.0
% Required files: -
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% constants
global n epsilon;
% state
x = z(1:n);
xplus = x;
for ix = 1:n
if (1+epsilon)*x(ix)<1
xplus(ix) = (1+epsilon)*x(ix);
else
xplus(ix) = 0;
end
end
out = xplus;
Flow set
function [v] = C(z)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matlab M-file Project: HyEQ Toolbox @ Hybrid Dynamics and Control
% Lab, http://www.u.arizona.edu/~sricardo/index.php?n=Main.Software
%
% Filename: C.m
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Description: Flow set
%
% Version: 1.0
% Required files: -
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% constants
global n;
% state
x =z(1:n);
if (sum(x<1)==n) % flow condition
v = 1; % report flow
else
v = 0; % do not report flow
end
Jump set
function [v] = D(z)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Matlab M-file Project: HyEQ Toolbox @ Hybrid Dynamics and Control
% Lab, http://www.u.arizona.edu/~sricardo/index.php?n=Main.Software
%
% Filename: D.m
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Description: Jump set
%
% Version: 1.0
% Required files: -
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% constants
global n;
% state
x =z(1:n);
if (sum(x>=1)>0) % jump condition
v = 1; % report jump
else
v = 0; % do not report jump
end | 1,362 | 5,553 | {"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": 10, "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} | 2.921875 | 3 | CC-MAIN-2018-13 | latest | en | 0.911272 |
http://www.solutioninn.com/the-following-data-are-numbers-of-passengers-on-flights-of | 1,498,233,567,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00542.warc.gz | 650,969,797 | 7,304 | # Question: The following data are numbers of passengers on flights of
The following data are numbers of passengers on flights of Delta Air Lines between San Francisco and Seattle over 33 days in April and early May.
128, 121, 134, 136, 136, 118, 123, 109, 120, 116, 125, 128, 121, 129, 130, 131, 127, 119, 114, 134, 110, 136, 134, 125, 128, 123, 128, 133, 132, 136, 134, 129, 132
Find the lower, middle, and upper quartiles of this data set. Also find the 10th, 15th, and 65th percentiles. What is the inter-quartile range?
View Solution:
Sales0
Views87 | 191 | 557 | {"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-2017-26 | longest | en | 0.866566 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=PURE | 1,529,549,384,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864019.29/warc/CC-MAIN-20180621020632-20180621040632-00556.warc.gz | 418,327,950 | 93,763 | Back to list of Stocks See Also: Seasonal Analysis of PUREGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
Fourier Analysis of PURE (PURE Bioscience)
PURE (PURE Bioscience) appears to have interesting cyclic behaviour every 80 weeks (1.9178*sine), 40 weeks (1.4158*sine), and 51 weeks (1.1641*sine).
PURE (PURE Bioscience) has an average price of 11.31 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Fourier Analysis
Using data from 1/3/2000 to 5/21/2018 for PURE (PURE Bioscience), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
011.30699 0
1-5.95745 6.01217 (1*2π)/960960 weeks
29.43125 .06702 (2*2π)/960480 weeks
3-.24518 6.30204 (3*2π)/960320 weeks
4-.08683 .86389 (4*2π)/960240 weeks
52.06889 4.0513 (5*2π)/960192 weeks
6-2.52181 .16443 (6*2π)/960160 weeks
71.75282 2.50178 (7*2π)/960137 weeks
8-1.80376 -.32637 (8*2π)/960120 weeks
91.90495 .9608 (9*2π)/960107 weeks
10-.14275 1.02818 (10*2π)/96096 weeks
11.43084 .35097 (11*2π)/96087 weeks
12.61716 1.91779 (12*2π)/96080 weeks
13-.88017 .14848 (13*2π)/96074 weeks
14.95766 .53409 (14*2π)/96069 weeks
15-.2059 .94609 (15*2π)/96064 weeks
16.36468 .19464 (16*2π)/96060 weeks
17.53479 .68374 (17*2π)/96056 weeks
18.36755 .64543 (18*2π)/96053 weeks
19.45738 1.16409 (19*2π)/96051 weeks
20-.46381 .53132 (20*2π)/96048 weeks
21.76978 .49915 (21*2π)/96046 weeks
22-.4959 1.02847 (22*2π)/96044 weeks
23.66232 -.1372 (23*2π)/96042 weeks
24.1523 1.41581 (24*2π)/96040 weeks
25-.07879 .55481 (25*2π)/96038 weeks
26.1132 .52312 (26*2π)/96037 weeks
27-.38682 1.06053 (27*2π)/96036 weeks
28-.08694 .12151 (28*2π)/96034 weeks
29-.45833 .54274 (29*2π)/96033 weeks
30.28857 .13269 (30*2π)/96032 weeks
31-.512 .42019 (31*2π)/96031 weeks
32.54516 -.13403 (32*2π)/96030 weeks
33.10917 .69395 (33*2π)/96029 weeks
34-.03829 .23051 (34*2π)/96028 weeks
35.67542 .41878 (35*2π)/96027 weeks
36-.22169 1.0133 (36*2π)/96027 weeks
37.15559 .30425 (37*2π)/96026 weeks
38-.24296 1.03793 (38*2π)/96025 weeks
39-.30704 .41011 (39*2π)/96025 weeks
40-.48845 .35244 (40*2π)/96024 weeks
41-.15337 .4341 (41*2π)/96023 weeks
42-.42755 -.05844 (42*2π)/96023 weeks
43-.25978 .56405 (43*2π)/96022 weeks
44-.3711 -.49665 (44*2π)/96022 weeks
45.21064 .32011 (45*2π)/96021 weeks
46-.33626 -.25108 (46*2π)/96021 weeks
47.62465 .00245 (47*2π)/96020 weeks
48-.0372 .41997 (48*2π)/96020 weeks
49.36428 .02704 (49*2π)/96020 weeks
50.04529 .78536 (50*2π)/96019 weeks
51.01595 .08639 (51*2π)/96019 weeks
52-.02655 .70012 (52*2π)/96018 weeks
53-.17199 .23445 (53*2π)/96018 weeks
54-.16902 .43195 (54*2π)/96018 weeks
55-.16651 .24459 (55*2π)/96017 weeks
56-.17542 .3031 (56*2π)/96017 weeks
57-.20781 .27899 (57*2π)/96017 weeks
58-.33865 .1087 (58*2π)/96017 weeks
59-.03457 .03574 (59*2π)/96016 weeks
60-.12934 .31513 (60*2π)/96016 weeks
61-.1509 .0024 (61*2π)/96016 weeks
62-.03912 .31083 (62*2π)/96015 weeks
63-.25724 .04455 (63*2π)/96015 weeks
64.03658 .22578 (64*2π)/96015 weeks
65-.38146 .18249 (65*2π)/96015 weeks
66-.02145 -.04473 (66*2π)/96015 weeks
67-.17984 .27421 (67*2π)/96014 weeks
68-.20173 -.04013 (68*2π)/96014 weeks
69-.05606 .19214 (69*2π)/96014 weeks
70-.29173 -.03853 (70*2π)/96014 weeks
71.08364 .09856 (71*2π)/96014 weeks
72-.42975 .0697 (72*2π)/96013 weeks
73.09124 -.03886 (73*2π)/96013 weeks
74-.33562 .02626 (74*2π)/96013 weeks
75.00839 -.07585 (75*2π)/96013 weeks
76-.1401 .04561 (76*2π)/96013 weeks
77-.03449 -.14815 (77*2π)/96012 weeks
78.03805 .15741 (78*2π)/96012 weeks
79-.25319 -.054 (79*2π)/96012 weeks
80.12334 -.01571 (80*2π)/96012 weeks
81-.30663 .08934 (81*2π)/96012 weeks
82.10562 -.17766 (82*2π)/96012 weeks
83-.22652 .13344 (83*2π)/96012 weeks
84-.06706 -.25452 (84*2π)/96011 weeks
85.06625 -.04102 (85*2π)/96011 weeks
86-.06366 .01021 (86*2π)/96011 weeks
87-.02816 -.17354 (87*2π)/96011 weeks
88.00546 -.06155 (88*2π)/96011 weeks
89.05823 -.14518 (89*2π)/96011 weeks
90.11731 -.07551 (90*2π)/96011 weeks
91.05957 -.04275 (91*2π)/96011 weeks
92.22031 .05211 (92*2π)/96010 weeks
93-.12486 .00003 (93*2π)/96010 weeks
94.22793 -.01504 (94*2π)/96010 weeks
95-.16337 .00955 (95*2π)/96010 weeks
96.20417 -.14288 (96*2π)/96010 weeks
97.05862 .11918 (97*2π)/96010 weeks
98-.06664 -.13424 (98*2π)/96010 weeks
99.23385 .00112 (99*2π)/96010 weeks
100-.04407 -.06343 (100*2π)/96010 weeks
101.20092 -.04761 (101*2π)/96010 weeks
102.09059 -.01598 (102*2π)/9609 weeks
103.15129 -.03932 (103*2π)/9609 weeks
104.18165 .07453 (104*2π)/9609 weeks
105.00296 -.02732 (105*2π)/9609 weeks
106.28521 .0002 (106*2π)/9609 weeks
107.00512 .02932 (107*2π)/9609 weeks
108.34981 -.01956 (108*2π)/9609 weeks
109.14289 .16874 (109*2π)/9609 weeks
110.17098 .05465 (110*2π)/9609 weeks
111.24725 .24913 (111*2π)/9609 weeks
112-.01252 .14262 (112*2π)/9609 weeks
113.19511 .19663 (113*2π)/9608 weeks
114-.06576 .1826 (114*2π)/9608 weeks
115.05931 .16084 (115*2π)/9608 weeks
116.01116 .15816 (116*2π)/9608 weeks
117-.1877 .09169 (117*2π)/9608 weeks
118.10329 .03093 (118*2π)/9608 weeks
119-.18538 -.04744 (119*2π)/9608 weeks
120.20744 -.09668 (120*2π)/9608 weeks
121.07672 .10655 (121*2π)/9608 weeks
122.13427 -.09463 (122*2π)/9608 weeks
123.25818 .24882 (123*2π)/9608 weeks
124-.02756 .06148 (124*2π)/9608 weeks
125.28902 .19219 (125*2π)/9608 weeks
126-.18872 .21013 (126*2π)/9608 weeks
127.25377 .1293 (127*2π)/9608 weeks
128-.291 .29944 (128*2π)/9608 weeks
129.06385 -.06467 (129*2π)/9607 weeks
130-.13602 .24926 (130*2π)/9607 weeks
131-.15517 -.12698 (131*2π)/9607 weeks
132.16286 .02605 (132*2π)/9607 weeks
133-.2318 -.01804 (133*2π)/9607 weeks
134.3062 -.08704 (134*2π)/9607 weeks
135-.07777 .08339 (135*2π)/9607 weeks
136.16517 .01805 (136*2π)/9607 weeks
137.07202 .09165 (137*2π)/9607 weeks
138.03554 .09784 (138*2π)/9607 weeks
139.07467 .07587 (139*2π)/9607 weeks
140.0057 .12721 (140*2π)/9607 weeks
141.02421 .08123 (141*2π)/9607 weeks
142-.06363 .07577 (142*2π)/9607 weeks
143.02692 .03163 (143*2π)/9607 weeks
144-.01619 .03141 (144*2π)/9607 weeks
145.02763 -.0037 (145*2π)/9607 weeks
146.1071 .08024 (146*2π)/9607 weeks
147-.07813 .07921 (147*2π)/9607 weeks
148.11569 -.01464 (148*2π)/9606 weeks
149-.04384 .10996 (149*2π)/9606 weeks
150.05576 -.02817 (150*2π)/9606 weeks
151.02957 .11127 (151*2π)/9606 weeks
152-.0135 -.02407 (152*2π)/9606 weeks
153.07489 .09608 (153*2π)/9606 weeks
154.01373 -.01599 (154*2π)/9606 weeks
155.00756 .05657 (155*2π)/9606 weeks
156.13467 -.04141 (156*2π)/9606 weeks
157.01661 .14545 (157*2π)/9606 weeks
158.078 -.02431 (158*2π)/9606 weeks
159.08772 .18101 (159*2π)/9606 weeks
160-.06525 .04666 (160*2π)/9606 weeks
161.10873 .09589 (161*2π)/9606 weeks
162-.15105 .07825 (162*2π)/9606 weeks
163.15374 -.05813 (163*2π)/9606 weeks
164-.06185 .1846 (164*2π)/9606 weeks
165.03887 -.05453 (165*2π)/9606 weeks
166-.03911 .12385 (166*2π)/9606 weeks
167.00409 -.07318 (167*2π)/9606 weeks
168.05996 .05945 (168*2π)/9606 weeks
169.00932 .03787 (169*2π)/9606 weeks
170.05742 -.00528 (170*2π)/9606 weeks
171.02678 .09203 (171*2π)/9606 weeks
172.03426 -.00701 (172*2π)/9606 weeks
173.03072 .08135 (173*2π)/9606 weeks
174-.0011 -.00454 (174*2π)/9606 weeks
175.08655 .04717 (175*2π)/9605 weeks
176-.03831 .03082 (176*2π)/9605 weeks
177.136 .0171 (177*2π)/9605 weeks
178-.06311 .07282 (178*2π)/9605 weeks
179.14842 -.0147 (179*2π)/9605 weeks
180-.06974 .09098 (180*2π)/9605 weeks
181.14956 -.06649 (181*2π)/9605 weeks
182.01463 .11562 (182*2π)/9605 weeks
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9589.43125 -.06702 (958*2π)/9601 weeks | 16,595 | 37,084 | {"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-2018-26 | latest | en | 0.764997 |
https://excellup.com/ClassEight/matheight/practical-geometry-2.aspx | 1,695,801,112,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00417.warc.gz | 269,986,083 | 3,796 | Class 8 Maths
# Practical Geometry
## Exercise 4.3
MO = 6 cm, OR = 4.5 cm, ∠M = 60°, ∠O = 105°, ∠R = 105°
• Draw MO = 6 cm
• Make an angle = 105° at point O.
• Draw OR = 4.5 cm
• Make an angle = 105° at point R
• Make and angle = 60° at point M
• Extend lines from points M and R so that they intersect at point E.
Here, sum of given three angles of given quadrilateral is = 60 + 105 + 105 = 270
So, fourth angle = 360 – 270 = 90
PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, ∠N = 85°
• Calculate fourth angle, i.e. ∠L = 360 – (90 + 110 + 85)
= 360 – 285 = 75
• Draw PL = 4 cm
• Make an angle of 90° on point P and another angle of 75° on point L.
• Draw LA = 6.5 cm
• Make an angle of 110° on point A
• Extend lines from P and A so that they intersect at point P
Question 3: Parallelogram HEAR
HE = 5 cm, EA = 6 cm, ∠R = 85°
• Draw HE = 5 cm
• Make an angle of 85° on point E
• Draw EA = 6 cm
• We know that adjacent angles are supplementary in a parallelogram.
• Supplement of 85° = 180° - 85° = 95°
• Make and angle of 95° on point A
• Draw AR = 5 cm so that it is parallel to HE
• Join H to R
• Parallelogram HEAR is complete.
Question 4: Rectangle OKAY
OK = 7 cm, KA = 5 cm
Answer: All angles of a rectangle are right angles and opposite sides are equal.
• Draw OK = 7 cm
• Draw KA = 5 cm perpendicular to OK
• Draw AY = 7 cm perpendicular to KA
• Join O to Y
• Rectangle OKAY is complete.
## Exercise 4.4
DE = 4 cm, EA = 5 cm, AR = 4.5 cm, ∠E = 60°, ∠A = 90°
• Draw DE = 4 cm
• Make and angle of 60° on point E
• Draw EA = 5 cm
• Make a right angle at point A
• Draw AR = 4.5 cm
• Join D to R
TR = 3.5 cm, RU = 3 cm, UE = 4 cm, ∠R = 75°, ∠U = 120°
• Draw TR = 3.5 cm
• Make an angle of 75° on point R
• Draw RU = 3 cm
• Make an angle of 120° on point U
• Draw UE = 4 cm
• Join T to E
## Exercise 4.5
### Draw the following
Question 1: The square READ with RE = 5.1 cm
Answer: All sides of a square are equal and all angles are right angle.
• Draw RE = 5.1 cm
• Draw EA = 5.1 cm perpendicular to RE
• Draw DA = 5.1 cm perpendicular to EA
• Join R to D
Question 2: A rhombus whose diagonals are 5.2 cm and 6.4 cm long
Answer: We know that diagonals of a rhombus are perpendicular bisectors of each other
• Draw a vertical line = 6.4 cm
• Draw the perpendicular bisector of this line of length 5.2 cm
• Join four vertices to complete the rhombus
Question 3: A rectangle with adjacent sides of lengths 5 cm and 4 cm
Answer: All angles of a rectangle are right angles and opposite sides are equal.
• Draw a horizontal line = 5 cm
• Draw another line which is perpendicular to previous line and its length = 4 cm
• Draw a third line which is perpendicular to second line with length = 5 cm
• Draw a fourth line which is perpendicular to third line.
• Rectangle is complete.
Question 4: A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique? | 1,006 | 2,886 | {"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} | 4.75 | 5 | CC-MAIN-2023-40 | latest | en | 0.774234 |
https://putanumonit.com/2018/09/07/the-scent-of-bad-psychology/?share=google-plus-1 | 1,558,579,516,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256997.79/warc/CC-MAIN-20190523003453-20190523025453-00553.warc.gz | 594,336,620 | 39,676 | # The Scent of Bad Psychology
Bad news: The replication crisis in psychology replicated. Out of 21 randomly chosen psychology papers published in the prestigious Nature and Science journals in 2010-2015, only 13 survived a high-powered replication.
Good news: A prediction market where research peers could bet on which results would replicate identified almost of them correctly. So did a simple survey of peers with no monetary incentive.
Better news: So could I.
Best news: So can you.
Rob Wiblin of 80,000 Hours put together a quiz that offers descriptions of the 21 studies and lets you guess if their main finding replicated or not. I recommend trying this out for yourself. If you’re not confident in your sniffing ability you can review some of my previous posts on defense against the dark arts (of bullshit statistics).
The stench of bad research is difficult to hide, and a few simple rules are enough to tell the true insights into human nature from the p-hacked travesties of science. Here’s what you need to know to ace the quiz and avoid falling for the next piece of psych nonsense.
## Rule 1: The Rule of Anti-Significance.
If a study has p=0.049 it is fake.
There are two studies in the quiz with p-values just below the common 0.05 threshold. I immediately (and correctly) identified both as fake without reading anything else.
If you take a Statistics 101 class at most universities, you are taught the following rule of statistical significance:
A result with a p-value above 0.05 is probably false. A p-value below 0.05 is statistically significant, meaning the result is true.
It’s never phrased like that explicitly, but that is the implied rule that people learn as they “test” hypotheses against the 0.05 threshold to get an A in the class. I got an A+ in my statistics class in grad school by following this rule religiously.
But that was a long time ago. Today, allow me to present Jacob’s Rule of Anti-Significance:
A result with a p-value just above 0.05 could well be true. A result with a p-value just below 0.05 is almost certainly false.
If you understand why this is so, you know all you need to about statistics in research.
Let’s start with the first part: how likely is a result with p=0.06 to replicate?
p=0.06 roughly means that the measured effect is 1.5-1.9 times the standard error, depending on the test used. The measured effect is some combination of true effect and noise. Even if noise accounts for half the measurement, the true effect is something like 0.8 times the standard error in the experiment.
But the standard error is a function of sample size – it should decrease with the square root of the number of subjects. When we run a replication with 10 times the sample size (which many studies in the replication project did), the standard error will be roughly 3.1 times lower. This means that the true effect is now 0.8*3.1 = 2.5 times the standard error of the new experiment with the larger sample. This is more than enough for a successful replication. Some p=0.06 result will be entirely due to noise, but a lot of them will point to something real that just needs to be confirmed by a stronger replication.
More importantly, p=0.06 means that the researchers are honest. They could have easily p-hacked the results below 0.05 but chose not to. The opposite is true when p=0.049.
The chance that the p-value of a study will land precisely in the 0.045-0.05 range is 0.005 (1/200) if the effect doesn’t exist. Even if the effect is true and equal precisely to the p=0.05 line, there’s a mere 1/60 chance of the measured p-value falling in that tiny window.
But if a study was p-hacked, if the researchers kept juggling different hypotheses, including and excluding outliers, and tweaking the measurements, then it is almost guaranteed to land in the 0.045-0.05 range because that’s where the hacking will stop and the champagne will pop.
In Bayesian terms, which are the terms we should be using anyway, a p-value in the 0.045-0.05 range gives a 60-200 times higher likelihood to the hypothesis “the study was p-hacked by bad researchers” than to the hypothesis “the study landed on that p-value by accident”. And since unscrupulous (or just clueless) researchers in psychology are certainly more common than 1 in 60, the conclusion (i.e., posterior) is that a study with p=0.049 got that p-value by bullshit means, and its result is bullshit.
## 2: The Rule of Taleb’s Grandma
If the purported effect sounds implausible, it is.
You have a mind capable of simulating itself, which lets you replicate any psychological study inside your own head with N=1.
Example 1: People prefer watching TV for 12 minutes to being alone with their thoughts for 12 minutes.” Right now, you’re reading this blog because you don’t want to be alone with your thoughts. This experiment replicated easily.
Example 2:If you imagine eating an M&M 30 times, immediately afterward you will eat fewer M&Ms from a bowl.” Do it. Imagine yourself eating an M&M: picking it up, chewing, swallowing. Now do it 29 more times. You can almost certainly feel your attitude towards M&Ms changing.
I don’t know if I would have guessed ahead of time that the effect would be to make me want fewer M&Ms, but it’s certainly plausible from my N=1 thought experiment that there would be a detectable effect one way or another. Of course, if the effect was to make people eat more M&Ms, the study would still be published! Whichever way the effect goes, I had reason to believe it would be true. This study also replicated, with good effect size.
Example 3: Washing your hands makes you less likely to want to justify your decision of how you ranked music albums, but just thinking about soap doesn’t.” Imagine yourself washing your hands. Do you feel any impact whatsoever on your desire to rationalize decisions? Now imagine explaining this study to Nassim Taleb’s grandma.
Psychologist: You see, Taleb’s grandma, there’s a clear link between washing your hands and justifying album-ranking choices.
Taleb’s Grandma: What the fuck are you talking about?
Psycho: Cleaning one’s hands “eliminates the postdecisional dissonance effect” by priming you to think of a “clean slate”. Those are scientific terms, so you know that this is serious science.
Grandma: Just because we use the word “clean” in English to refer both to hands and to your conscience doesn’t mean that thinking about cleanliness in one context will change your behavior in the other context. That’s cockamamie.
Psycho: No, no, just thinking about washing your hands is not enough to prime you, even though every other priming study says it’s enough to just think of things. Thinking about soap doesn’t do anything. You need to actually wash your hands to get the effect, and not just because we tried different ways of priming and only reported the one that gave us a publishable p-value.
Grandma: Ok, so you’re saying that washing my hands makes me want to “come clean” and explain my decision on how I ranked some albums?
Psycho: It’s the opposite! Washing your hands makes you less likely to explain your decision because you already think of yourself as metaphorically clean.
Grandma: This story about washing hands and explaining decisions depends on a conjunction of multiple steps, every one of them individually preposterous, and with the effect direction at each step chosen completely at random. There are more burdensome details in this hypothesis than can be lifted by 40 exhaustive studies with hundreds of participants each, let alone a single study with 40 undergrads who don’t lift. This is ridiculous bullshit, and I need to wash my ears with soap just to remove all trace of this nonsense from my brain.
Psycho: Well, it was good enough to get published in Science. Are you saying that peer review by experts isn’t a guarantee of true results?
Grandma: Wait till I tell my grandson about this, he’ll make an entire career out of mocking people like you. #IYI #SkinInTheGame #LindyEffect
We can summarize the takeaway in an addendum to rule 2.
Rule 2b: we should all be embarrassed that we believed in priming even for a second.
## Rule 3: The Rule of Multiplicity
If the study looks like it tried 20 different things to get a p-value, it has. Whatever effect it claims to have found is just an artifact of multiple hypothesis testing.
I wrote a couple thousand words already about why a study that tries several hypotheses and doesn’t correct for multiplicity isn’t worth the pixels it is written on. That’s my least-read-adjusting-for-quality post ever, because even readers who click on a self-proclaimed “math blog” called “Put a number on it” don’t want too much actual math in their blog posts.
The fun part is that you can guess which studies are multiplicitous just from their abstracts. Here’s how one of the studies was summarized on the 80,000 hours quiz:
When holding and writing on a heavier clipboard, people assessing job applicants rate them as ‘better overall’, and ‘more seriously interested in the position’.
The non-metaphorically heavy clipboard already carries the stench of priming, and as soon as I saw the word “and” in the description, I knew it was fake without looking at the sample size or p-value. I could just imagine the researchers trying 27 clipboards of different materials, 4 surveys and 15 blood tests to measure impact, and 906 interaction effects just to be sure that something somewhere will hit a publishable p-value.
Here’s are some excerpts from the actual paper (courtesy of our heroes at Sci-Hub):
Physical touch experiences may create an ontological scaffold for the development of intrapersonal and interpersonal conceptual and metaphorical knowledge.
The first sign that you’re about to be fed bullshit is an abstract full of 4-syllable words where 2-syllable words would do.
The experience of weight, exemplified by heaviness and lightness, is metaphorically associated with concepts of seriousness and importance. This is exemplified in the idioms “thinking about weighty matters” and “gravity of the situation.”
Priming is really like the Kaballah, where semi-arbitrary coincidences of language have the power to shape worlds.
In our first study, testing influences of weight on impression formation, we had 54 passersby evaluate a job candidate by reviewing resumes on either light (340.2 g) or heavy (2041.2 g) clipboards. Participants using heavy clipboards rated the candidate as better overall and specifically as displaying more serious interest in the position.
However, the candidate was not rated as more likely to “get along” with co-workers, suggesting that the weight cue affected impressions of the candidate’s performance and seriousness, consistent with a “heavy” metaphor, but not the metaphorically irrelevant trait of social likeability.
Does anyone actually believe that if the candidate was rated as easier to get along with they would admit that it contradicts their hypothesis instead of making up a just-so story about how the candidate is a “solid person” you can “lean on”?
Our second study investigated how metaphorical associations with weight affect decisionmaking […] Here, a main effect of clipboard condition, was qualified by
an interaction with participant gender.
When you’re desperate for p-values and need to come with 100 new hypotheses to test, breaking your group into arbitrary categories (by gender, age, race, astrological sign…) is the easiest way to do so. This is the “elderly Hispanic woman effect”.
Comparable to study five, participants who sat in hard chairs judged the employee to be both more stable, (p = 0.030), and less emotional, (p = 0.028), but not more positive overall . On the negotiation task, no differences in offer prices emerged (p > 0.14).
We next calculated the change in offer prices from first to second offer, on the presumption that activating the concepts of stability and rigidity should reduce people’s decision malleability or willingness to change their offers.
Among participants who made a second offer, hard chairs indeed produced less change in offer price (M = \$896.5, SD = \$529.6) than did soft chairs (M = \$1243.6, SD = \$775.9).
This study is basically a p-hacking manual. They’re not even trying to hide it, instead describing in detail how, when a hypothesis failed to yield a p-value below 0.05, they tried more and more things until something publishable popped out by chance.
It’s OK if one study finds that clipboard weight only affects measures A and B and not C, and only does so for women and not men, if you then run another study that only looks at A, B, and women. But a study that tried 100 things and tells you about 3 of them is a like a criminal on trial who mentions that there are some banks that he didn’t rob.
## 4: The Rule of Silicone Boobs
If it’s sexy, it’s probably fake.
“Sexy” means “likely to get published in the New York Times and/or get the researcher on a TEDx stage”. Actual sexiness research is not “sexy” because it keeps running into inconvenient results like that rich and high-status men in their forties and skinny women in their early twenties tend to find each other very sexy. The only way to make a result like that “sexy” is to blame it on the patriarchy, and most psychologists aren’t that far gone (yet).
So: Participants automatically project agents’ beliefs and store them in a way similar to that of their own representation about the environment (a comparison of the mean reaction time between the P-A- treatment and the P-A+ treatment)”. I fell asleep just copy-pasting that abstract. This terribly unsexy study replicated with a large effect size.
Participants in a condition that simulated the stress of being poor did worse on an attention task than those who simulated the ease of being rich.” Muy sexy, as is anything that has to do with educational interventions, wealth inequality being bad, discrimination being really bad, or any other result that easily projects to a progressive policy platform. Of course, the replication found an almost-significant result in the opposite direction of the original – people in the “poor condition” paid more attention and did better.
Anything counterintuitive is also sexy, and thus (according to Rule 2) less likely to be true. So is anything novel that isn’t based on solid existing research. After all, the Times is the newspaper business, not in the truthspaper one.
Finding robust results is very hard, but getting sexy results published is very easy. Thus, sexy results generally lack robustness. I personally find a certain robustness quite sexy, but that attitude seems to have gone out of fashion since the Rennaisance.
## Reasons for Optimism
Andrew Gelman wrote in 2016:
Let’s just put a bright line down right now. 2016 is year 1. Everything published before 2016 is provisional. Don’t take publication as meaning much of anything, and just cos a paper’s been cited approvingly, that’s not enough either. You have to read each paper on its own. Anything published in 2015 or earlier is part of the “too big to fail” era, it’s potentially a junk bond supported by toxic loans and you shouldn’t rely on it.
While it’s certainly true that a lot of psychology was junk science in the pre-2016 era, it wasn’t clear whether things will improve from 2016 onwards.
The replication crisis in psychology is not a new phenomenon. Statistician Jacob Cohen noted that most studies in psychology are underpowered and full of false positives back in 1962. In 1990, he noted that things have only gotten worse. Why were voices like Cohen’s ignored for more than 5 decades?
My hypothesis is that:
1. Most psychologists couldn’t understand the mathematics of what was wrong or didn’t care to try. The standards of the field were such that they could get away with criminal methodology.
2. The psychologists who did care about mathematical rigor were at a disadvantage since they couldn’t match the publication output of their p-hacking counterparts. A lot of them probably left to do something else, like advertising in the 1960s or consumer data science in the 2010s.
But it’s harder to get away with bullshit studies if everybody knows how to spot them and everybody knows that everybody knows. If you and I can guess which studies will replicate with close to 90% accuracy, the editors of Nature and Science also can, and now they’ll have to instead of batting 62% (13/21). Researchers can’t pretend that the “replication messed up the experiment” if everyone can tell at a glance that the study will never replicate.
There are ways to improve the reliability of psychology research that require learning some math, although not beyond what one can learn from reading Putanumonit: estimating experimental power, calculating the likelihood of alternatives instead of null-hypothesis testing, correcting for multiplicity. But there are also fixes that don’t require knowing any math at all, like preregistering the analysis, being suspicious of interaction effects that were not in the main hypothesis, and getting a larger sample size than 20 undergrads who do it for course credit.
Hopefully, psychology researchers have started doing these things in the couple of years since it became clear that bullshit will be caught. And if they haven’t, we’ll catch them.
## 21 thoughts on “The Scent of Bad Psychology”
1. avi says:
f you and I can guess which studies will replicate with close to 90% accuracy, the editors of Nature and Science also can, and now they’ll have to instead of batting 62% (13/21)
This is unfair – you’re judging the ones that were already accepted, while they were judging submissions. For all you know, they scored 90% on the ones that were submitted and the vast majority of submitted ones are noise
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1. That’s a good point. Is there a good source of psychology preprints, sort of like arxiv?
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1. Ludwig says:
is it really a good point though? seems like they could just re-evaluate the papers by giving them a second look (by different editors in the same journal)
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2. My score was 24. Hmm, I wonder if I’d have done better if there were no “I don’t know” option… should’ve kept track.
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1. I just saw a link to that article today. I wonder if those defending the old order realize how pathetic they sound.
I asked him if he thinks about leaving psychology. “Every day,” he replied.
I can’t wait for this person and everyone like them to finally leave psychology for their true calling: astrology and Tarot cards. I get emails from grad students who left psychology because they couldn’t compete with the bullshit factory, and their professors got angry when they started talking about statistics and methodology.
They say a field that once seemed ascendant, its latest findings summarized in magazines and turned into best sellers, is now dominated by backbiting and stifled by fear. Psychologists used to talk about their next clever study; now they fret about whether their findings can withstand withering scrutiny.
Oh, no! How can anyone do science when they have to actually withstand scrutiny? So unfair! It used to be that you could just write NY Times clickbait and get a book deal, now you actually have to, gasp, discover things. Do they really think that the wheel will turn back? Right now all they’re doing is standing in the way of the house cleaning that will allow the next generation of researchers to do some actual science.
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1. Thomas says:
I just had a conversation with my friends on a vacation where they asked me why I left grad-school in Psych: your comments are almost verbatim what I told them. I can recall the time I told a professor that myself and some other peers thought the methods for “scrubbing” some of the study’s databases looked unethical and did not actually present the facts. I received scornful looks and told that it is common practice – I couldn’t believe that was considered common. I promptly left that year.
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3. Inno says:
Is there any merit to psychological constructionism? If you’re not familiar, it’s a research program that turns away from a lot of how psychology was regularly done e.g., facultative parts of brain and mind and neurophysiological correlates that are touted as the locations of emotions or thoughts.
I can’t evaluate the rigor of the studies themselves, but the counterintuitive ideas have been very useful. One of them being this move away from focusing on homogeneous constructs of emotions and instead bothering to consider the heterogeneity as not just statistical error but part of a population of instances. For example, the more common drive in research to find similar instances of ‘fear’ which exclude any dissimilar instances as outside the boundaries of the fear construct. In contrast, their research program still considers these dissimilarities worth exploring. They aren’t beholden to finding probabilistic patterns of relations between every instance of an emotion category.
“Psychological construction theories make no claims about specific patterns for each emotion category, so their validity does not rise or fall based on finding them. Psychological construction theories provide an alternative explanation to basic emotion and appraisal theories in the event that such patterns are found (which would have to be ruled out for those theories to be correct), but psychological construction also can explain why such patterns rarely, if ever, materialize.” From The Psychological Construction of Emotions, 2015.
They try to explain emotional phenomena by treating them as higher-order constructs rather than elemental building blocks, and in doing so, try to find more elemental and granular components that can reliably predict the variability in heterogeneous instances of the ‘same’ construct rather than shoehorning more p <0.05 research that just adds sediment to a generally agreed definition of the construct while sweeping anything that doesn’t agree under the rug as another “more study is recommended to explain these results.”
Any holes in this argument? My foundations in statistics are hazy at best but it feels like a step in the right direction.
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1. Beny says:
From the abstracts I read, it seems like a legitimate approach to ‘wandering around’ in hypothesis space, but I don’t see how it’s a replacement for rigorous statistics. Eventually, if you want to say something meaningful about humans in general rather than just about test subject #42, you have to bunch them into groups that you consider homogeneous for what you need up to noise, and show significant differences. Looking at human and cultural variety to come up with unorthodox classifications is a neat idea, but you still need to prove the results.
It seems equivalent to me looking at something weird in an image from the microscope and telling my advisor “it looks like an experimental artifact, but I don’t know what could cause it, and maybe it’s insert weird chemical explanation”. It’s a great starting point for new research, but I won’t put it in a paper unless I properly followed up on it.
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4. CrabMan says:
“Participants automatically project agents’ beliefs and store them in a way similar to that of their own representation about the environment (a comparison of the mean reaction time between the P-A- treatment and the P-A+ treatment)”.
This falls under two of your categories: 1. 4 syllable words instead of 2 syllable words (which you say is evidence of bad bullshit); 2. unsexiness (which you say is evidence of good study).
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http://www.zwg.com.pl/beanilla-reviews-jwcly/which-statement-is-not-always-true-for-a-parallelogram%3F-58a7b8 | 1,627,863,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00428.warc.gz | 90,733,423 | 8,092 | Relevance. Every parallelogram is a rectangle. a) Diagonals are perpendicular. The 4 angles are not congruent. Which of the following statements is not true for a parallelogram? Answer: 1 question Which statement is not always true for a parallelogram? 1) cosJ = RM RE 2) cosR = JM JT 3) tanT = RM Since rhombuses are quadrilaterals with 4 congruent sides, squares are by definition also rhombuses. Start studying Which property is not true for all parallelograms. B. Which statement does NOT guarantee that a quadrilateral is a square. True or False: Consider the following statement: A differentiable function must have a relative minimum between any two relative maxima. The diags are only perp if it's a rhombus. A Rectangle Is A Parallelogram. Lv 7. 4) The opposite sides are parallel. 1) The diagonals are congruent. 0 1. axolotl14. 8 years ago. A. Which statement about quadrilateral l is not always true A. a rhombus is a parallelogram B. a square is a rhombus C. a quadrilateral are is a square D. a trapezoid is a quadrilateral 1 decade ago but #1 and #2 are always true- a parallelogram has congruent & parallel opposite sides. 3) The opposite angles are congruent. Jay D. 9 years ago. A rectangle is a polygon. It is true when the parallelogram has 4 right angles. Ask Question + 100. The diagonals are congruent. For any parallelogram _____. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A rhombus is not always a square as it is not required to have congruent angles, the angles can be different. Devon. 5. b.All four angles are right angles. Geometry Multiple Choice Regents Exam Questions www.jmap.org 3 12 In the diagram below, ERM ∼ JTM. a.All four sides are congruent. Think ... what statement is not always true about a parallelogram? 4) The opposite sides are parallel. Favorite Answer. Answer Save. The quadrilateral has 4 right angles and 4 congruent sides B. Math. Choose The Correct Answer Below. B. - the answers to estudyassistant.com Which statement about a parallelogram is not always true? In parallelogram ABCD, diagonals AC and DB intersect at E. Which statement is always true? Favorite Answer. That is true. 1 0. It is possible, but not always true. Yes the answer is … This is sometimes true. My A: the statement is not true. 2) What are the hypothesis and the . A rectangle is a square. A rhombus is a parallelogr… Get the answers you need, now! Which statement is true about every parallelogram? B. A parallelogram is a rectangle.C. 3) The opposite angles are congruent. Which statement is not always true about a parallelogram? Nor is it always true that all sides are equal. True. A the diagonals bisect each other B opposite angles are congruent C the diagonals are perpendicular D opposite sides are congruent 2 How many triangles are formed by drawing diagonals from one vertex in the figure? A rhombus is a quad… Get the answers you need, now! Geometry. 2) The opposite sides are congruent. This is the last round of “Name That Quadrilateral.” I'm thinking of a parallelogram with congruent perpendicular diagonals. A rhombus is a parallelogram.D. 1 decade ago. It is not true when a parallelogram has no right angles. Parallelogram definition, a quadrilateral having both pairs of opposite sides parallel to each other. b) Opposite sides are congruent. Think about the First Derivative Test and decide if the statement is true or false. Answer and Explanation: Of the statements given, the one that is not always true about a parallelogram is that the diagonals are congruent. Squares are quadrilaterals with 4 congruent sides. c.The diagonals bisect each other. Statement number 3 would be true only if the parallelogram is a square. D. Every rhombus is a parallelogram. 060106a PI GG38 Which statement is not always true about a parallelogram A The from MATHEMATIC mathematic at Scotland High School Of Math Scien C. Every square is a parallelogram. Which statement is not always true about a parallelogram? Which statement is sometimes, but not always, true? Which statement is sometimes, but not always, true? The opposite angles have the same measure. 1) The diagonals are congruent . The diagonals are perpendicular. 2) The opposite sides are congruent. B. Its diagonals do bisect each other but, by definition, it is not a rhombus. 1 Choose the statement that is NOT ALWAYS true. Find the sum of the measures of the angles in the figure. Click here to get an answer to your question ️ 1. 7 years ago. A Parallelogram Is Not A Trapezoid. b) Triangle ABD is a right triangle. A parallelogram is a rhombus. A trapezoid is a quadrilateral.B. i tryin to study up on my test tomorrow and i need to ace it.. PLZ help, im lost Which statement is not always true. A A rhombus is always a parallelogram. Which quadrilateral has diagonals that always bisect each other and also bisect its angles? Which statement about a parallelogram is always true? A parallelogram is a four sided figure in which opposite sides are equal and parallel. In a rhombus, the diagonals are perpendicular. The quadrilateral is a parallelogram with perpendicular diagonals C. The quadrilateral is both a rhombus and a rectangle D. The quadrilateral has 4 congruent sides and 4 congruent angles d.The diagonals are congruent Based on the information in the diagram, can you prove that the figure is a parallelogram? Question: Which Statement About Quadrilaterals Is Not Always True? Diagonals bisect each other. A parallelogram is not a statement that can be true or false. A. 1: Classify the quadrilateral using the name that best describes it I tried posting it but it didn't work 2: which statement is a true statement 3: which statement is a true statement 4: Which property is not a characteristic of a . a.) A Parallelogram Is A Square. It is possible, but not always true. A Square Is A Rhombus. The opposite ... 1 Answer. I want to . Choose the statement(s) that are not always true for ANY parallelogram. A parallelogram is a quadrilateral with both pairs of opposite sides parallel to one another. This is always true. 2 Answers. Every square is a rhombus. A statement that is not always true about a parallelogram is that the parallelogram's name is Herb. C. Draw yourself a parallelogram and check... 1 0. False. with is description we ca say that: Opposite sides are congruent. That means that they are supplementary. a) Triangle AED is isosceles. 7 years ago. 1) The diagonals are congruent. It is not always true that all angles are equal. Which of the following statements is not always true? for any parallelogram ::the diagonals bisect each other ::opposite angles are congruent ::the diagonals are perpendicular ::opposite sides are congruent By the definition of a parallelogram, we know that the opposite sides are congruent and parallel, so the second and fourth statements are always true. 2) The opposite sides are congruent . 1 0. B. A. c) Opposite angles are congruent. Chapter 8 Review. 4) The opposite sides are parallel L. E. Gant. 0 0. STUDY. 1) ∠DAE ≅∠BCE 2) ∠DEC ≅∠BEA 3) AC ≅DB 4) DE ≅EB 2 Which statement is not always true about a parallelogram? Relevance. A square is a rhombus. A. Which statement is sometimes, but not always, true?A. The opposite angles are congruent. A parallelogram is a rectangle. Solved: What statement is not always true about a parallelogram? A trapezoid ia sometimes a parallel… Get your answers by asking now. A parallelogram just means that it is a quadrilateral with 2 pairs of parallel sides. Which statement is not always true about a parallelogram? 2) The opposite sides are congruent. Which statement is not always true about a parallelogram? Consecutive angles are supplementary. 3. 1)The diagonals are congruent. A parallelogram just means that it is a quadrilateral with 2 pairs of parallel sides. Which expression is not always true? AxiomOfChoice. The consecutive angle add up to 180°. Which statement is NOT true? diagonals are perpendicular c.) diagonals bisect each other d.) opposite sides are congruent e.) diagonals bisect the angles f.) opposite angles are congruent A. The adjacent angles are complementary. True or False: Consider the following statement: A differentiable function must have a relative minimum between any two relative maxima. Determine whether each statement is always, sometimes or never true… Source(s): props of parallelograms. Which statement is always true? In a rhombus, the diagonals bisect opposite angles. A. Calculus. Which statement is not true? The statement cannot be proved because it is not true.As a counter example, consider a parallelogram which is not a rhombus. 3) The opposite angles are congruent. 0 1. Charles M. Lv 6. Still have questions? A square is a trapezoid. diagonals are congruent b.) B. 3) The opposite angles are congruent . 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Angles in the diagram, can you prove that the figure is a four sided figure in which sides... | 4,337 | 18,300 | {"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 | 4 | CC-MAIN-2021-31 | latest | en | 0.859124 |
http://users.math.yale.edu/public_html/People/frame/Fractals/MultiFractals/welcome.html | 1,502,963,511,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00587.warc.gz | 428,552,324 | 2,363 | # 7. Multifractals
Although some of the fractals we have drawn have been in color, the colors have not been part of the fractal structure. They were added for artistic effect (Voss' and Musgrave's landscapes, for example) or for pedagogical reasons (emphasizing the decomposition into pieces for an IFS or the escape rate for a Julia set). We could represent all the pictures just as well in black and white: a point is black if it belongs to the fractal, otherwise it is white. Many natural examples are not so clear-cut.
A. As a first mathematical example, we see that by adjusting the probabilities, we can make different parts of the fractal fill in at different rates. Here is an example. The IFS of this example generates the unit square. However, the square fills up in a non-uniform way, revealing many fractals.
B. Continuing with the example of 7.A., here are histograms representing the probabilities of the first four generations. Note the highest-probability region has a familiar shape. This is easy to understand. The lower left, lower right, and upper right transformations all have the same probability, and those three transformations together generate a Sierpinski gasket.
C. Here is another example, Example B, with p1 = 0.2, p2 = p3 = 0.25, and p4 = 0.3. Now structures more complicated than gaskets will appear.
D. In the length->0 limit, the coarse dimension becomes a local dimension. The place-dependence of local dimension motivates the name multifractal. Here we investigate the distribution of local dimensions for Example B. The resulting curve is called the f(α) curve.
E. Here is the general method for generating multifractals with IFS. We modify the Moran equation, weighting each term with the probability of the transformation. This gives the tau(q) curve, from which the f(α) curve can be calculated.
F. By changing the probabilities of the transformations, we alter the rate at which different parts of the shape fills in, and consequently change the f(α) curve. Here we illustrate this dependence by several examples.
Sometimes, from visual inspection of a multifractal we can gather enough information to sketch its f(α) curve. Here are some examples.
Here we dsicuss the method of moments for plotting f(α) curves. This is most easily understood in the context of examples: time series moments, planar data moments, and the special case of IFS moments.
G. Here are some examples of f(α) curves derived from financial data using the method of moments. | 556 | 2,501 | {"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-2017-34 | latest | en | 0.911642 |
https://www.ah-studio.com/171519-slope-intercept-form-equation-quiz-how-much-do-you-know-about-slope-intercept-form-equation | 1,571,812,502,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00082.warc.gz | 779,820,645 | 5,942 | Slope Intercept Form Equation Quiz: How Much Do You Know About Slope Intercept Form Equation?
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# Maths Unit: Fun with Fractions
Description
This unit plan has been created to support your students in learning how to count by quarters, halves, thirds and tenths. They will also use mixed numerals and locate and represent these fractions on a number line. The unit also covers converting mixed numbers to improper fractions and vice versa. The unit comes complete with a presentation, class displays, accompanying worksheets and assessment task.
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Lesson Structure
• Individual Activity
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Curriculum Codes
AC9M4N04 9
Count by fractions including mixed numerals; locate and represent these fractions as numbers on number lines
ACMNA078 8.4
Count by quarters halves and thirds, including with mixed numerals. Locate and represent these fractions on a number line
VCMNA158
Count by quarters, halves and thirds, including with mixed numerals. Locate and represent these fractions on a number line
MA2-1WM old
Uses appropriate terminology to describe, and symbols to represent, mathematical ideas
MA2-3WM old
Checks the accuracy of a statement and explains the reasoning used
MA2-7NA old
Represents, models and compares commonly used fractions and decimals
MA2-RN-01 new
Applies an understanding of place value and the role of zero to represent numbers to at least tens of thousands
MA2-RN-02 new
Represents and compares decimals up to 2 decimal places using place value
MA2-AR-01 new
Selects and uses mental and written strategies for addition and subtraction involving 2- and 3-digit numbers
MA2-AR-02 new
Completes number sentences involving addition and subtraction by finding missing values
MA2-MR-01 new
Represents and uses the structure of multiplicative relations to 10 × 10 to solve problems
MA2-MR-02 new
Completes number sentences involving multiplication and division by finding missing values
MA2-PF-01 new
Represents and compares halves, quarters, thirds and fifths as lengths on a number line and their related fractions formed by halving (eighths, sixths and tenths) | 488 | 2,196 | {"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} | 3.984375 | 4 | CC-MAIN-2024-18 | latest | en | 0.897899 |
http://www.solutioninn.com/despite-its-nutritional-value-seafood-is-only-a-tiny-part | 1,508,444,740,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00058.warc.gz | 547,995,395 | 8,127 | Despite its nutritional value seafood is only a tiny part
Despite its nutritional value, seafood is only a tiny part of the American diet, with the average American eating just 16 pounds of seafood per year. Janice and Nina both work in the seafood industry and they decide to create their own random samples and document the average seafood diet in their sample. Let the standard deviation of the American seafood diet be 7 pounds.
a. Janice samples 42 Americans and finds an average seafood consumption of 18 pounds. How likely is it to get an average of 18 pounds or more if she had a representative sample?
b. Nina samples 90 Americans and finds an average seafood consumption of 17.5 pounds. How likely is it to get an average of 17.5 pounds or more if she had a representative sample?
c. Which of the two women is likely to have used a more representative sample? Explain.
Membership | 188 | 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} | 2.515625 | 3 | CC-MAIN-2017-43 | latest | en | 0.965861 |
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Spelling Grade 1. Spelling Grade 2. Spelling Grade 3. Spelling Grade 4. Spelling Grade 5. More Spelling Worksheets. Chapter Books. Bunnicula. Charlotte's Web. Magic Tree House #1. Boxcar Children. More Literacy Units. Science. Animal (Vertebrate) Groups. . ABC Order Worksheet Generator. Please Note: You are NOT logged in. ..Grade 3 spelling Unit C-10 focuses on words that have double consonants. Log In. Become a Member. Membership Info. Math. . ABC Order: Write List (C-10) FREE . On this worksheet, your students will try to write all twenty spelling words alphabetically. 3rd Grade..ID: 1210691 Language: English School subject: English as a Second Language (ESL) Grade/level: Young Learners Age: 4-8 Main content: The alphabet Other contents: Add to my workbooks (121) Embed in my website or blog Add to Google Classroom.A Look Inside the Mistakes in JonBenet Ramsey Investigation Noted by Former Police Chief. Former Police Chief Mark Beckner revisited the case on Reddit this week..
Grade 7 Maths The Triangle and Its Properties Multiple Choice Questions (MCQs) 1. . ∆ ABC is right-angled at C. If AC = 5 cm and BC = 12 cm find the length of AB. (a) 7 cm (b) 17cm (c) 13 cm (d) none of these. . Worksheet on Word Problems on Linear Equation | Linear Equations Word Problems Worksheet ..Book Report Critical Thinking Pattern Cut and Paste Patterns Pattern – Number Patterns Pattern – Shape Patterns Pattern – Line Patterns Easter Feelings & Emotions Grades Fifth Grade First Grade First Grade – Popular First Grade Fractions Fourth Grade Kindergarten Worksheets Kindergarten Addition Kindergarten Subtraction PreK Worksheets ..English Language Arts Standards » Reading: Literature » Grade 3 » 1 Print this page. Ask and answer questions to demonstrate understanding of a text, referring explicitly to the text as the basis for the answers..
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2 + 4 + 4 = ...
5 + 4 + 9 = ...
7 + 2 + 2 = ...
7 + 8 + 2 = ...
6 + 4 + 6 = ...
1 + 1 + 4 = ...
3 + 5 + 9 = ...
6 + 6 + 1 = ...
7 + 3 + 9 = ...
5 + 1 + 1 = ...
2 + 5 + 3 = ...
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Https://www.thesprucecrafts.com/free-printable-alphabet-flash-cards-1356957 | 4,926 | 18,207 | {"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-2021-17 | longest | en | 0.796715 |
https://www.physicsforums.com/threads/repulsion-between-permanent-magnet-and-air-core-coil.959311/ | 1,716,241,491,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00812.warc.gz | 825,259,950 | 18,753 | # Repulsion between permanent magnet and air core coil
• Dante Meira
In summary, this statement is true, but it would be beneficial to seek the opinion of others before making a final decision.
Dante Meira
I formulated a statement about the repulsion between a permanent magnet and an air core coil (electromagnets without a ferromagnetic or ferrimagnetic core), and I believe this statement is true, but I would like to seek the opinion of others about it:
"In a system where an electromagnet made from an air core coil (electromagnet without a ferromagnetic or ferrimagnetic core) is in repulsion against a permanent magnet, the stronger is the magnetic field of the permanent magnet the stronger will be the force of repulsion between the electromagnet and the permanent magnet, with the volts and amperes of the direct current (DC) in the electromagnet being kept the same"
Is my statement correct?
Please try to answer this without any prejudices in mind. No preconceived ideas about "what I'm trying to do here".
Replacing the permanent magnet by another permanent magnet with a stronger magnetic field will make the force of repulsion between the permanent magnet and the air core electromagnet stronger, even if the electromagnet keeps getting the same volts and amperes of direct current?
By the way, consider the coil is made from copper wire, that is a diamagnetic material.
It is a problem that is difficult to treat precisely. In the case of another permanent magnet, you need to assume the geometry of both magnets is the same. The repulsion that occurs is not only due magnetic field strength of the permanent magnet, but also the geometry of both of its poles= geometric size of the poles and relative position, along with magnetic field strength. ## \\ ## And the force is only repulsive if like poles (i.e. the pole of the solenoid, and the pole of the permanent magnet) are facing each other.
What the coil of the electromagnet is made from is immaterial. The magnetic field comes only from the current flowing. With the current and coil being constant, the variable factor is the strength of the magnet's field. The stronger the field, the greater the repulsion, assuming like poles facing each other.
See the article for more details.
https://en.wikipedia.org/wiki/Force_between_magnets
Another way to look at it is using F=BIL, ie force experienced by a current varying wire of length L in a magnetic field with flux density (B).
Here you can clearly see, more B with same current equals more force (voltage doesn't matter in this case if the system is stationary, if the wire is moving (eg electric machine) then to develop some current (I) you will need to over come the induced voltage first (ie BEMF).
berkeman said:
So after 5 years you come back to PF to still ask others to test it?
No my friend, I'm not bothering anyone asking to test anything... I have tested it myself with neodymium magnets. And my idea was correct.
But neodymium magnets are expensive toys, even more now after the supply chain shock from the pandemic. I'm just sharing with others the idea that those iron nitride permanent magnets will be great to play with this idea in the future, and test the limits of it.
Let's embrace the experimentalist spirit of Michael Faraday:
It's hard to tell in complex and extreme cases, but in a simple linear model, if all relative positions, sizes and shapes remain the same, especially the relative distribution of the spatial magnetic field generated by the permanent magnets remains unchanged. Then the electromagnetic force between the constant current air core coil and the permanent magnet will of course increase as the strength of the magnetic field generated by the permanent magnet increases.
## 1. What causes the repulsion between a permanent magnet and an air core coil?
The repulsion between a permanent magnet and an air core coil is caused by the interaction between their magnetic fields. When two magnetic fields interact, they either attract or repel each other depending on their orientation. In this case, the like poles of the magnet and coil create a repulsive force.
## 2. How does the distance between the magnet and coil affect the repulsion?
The strength of the repulsive force between a permanent magnet and an air core coil is directly proportional to the distance between them. As the distance increases, the force decreases and vice versa. This is because the magnetic field weakens as it moves further away from its source.
## 3. Can the strength of the repulsive force be controlled?
Yes, the strength of the repulsive force can be controlled by changing the properties of either the magnet or the coil. For example, using a stronger magnet or increasing the number of turns in the coil can increase the repulsive force.
## 4. How does the shape of the magnet or coil affect the repulsion?
The shape of the magnet or coil can affect the distribution and concentration of their magnetic fields, which in turn can affect the repulsive force. For example, a horseshoe magnet will have a stronger repulsive force compared to a bar magnet of the same size due to its concentrated magnetic field.
## 5. How is the repulsion between a permanent magnet and an air core coil useful in practical applications?
The repulsion between a permanent magnet and an air core coil is utilized in many practical applications such as electric motors and generators. By controlling the strength and direction of the repulsive force, these devices can convert electrical energy into mechanical work or vice versa.
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3K | 1,285 | 6,040 | {"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} | 3.15625 | 3 | CC-MAIN-2024-22 | latest | en | 0.934186 |
https://www.oschina.net/code/snippet_138488_25210 | 1,571,012,374,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00509.warc.gz | 1,113,160,207 | 14,570 | # Tkinter 控件简单计算器示例
学习"python与Tkinter编程"按照书本里面的例子整理出来的，平时应该用的着
## 代码片段(1)[全屏查看所有代码]
### 1. [文件] Cale.py ~ 1KB 下载(35) 跳至 [1] [全屏预览]
```'''
Created on 2012-9-19
@author: liangqianwu
'''
#_*_ coding:utf-8_*_
from Tkinter import *
def frame(root,side):
w=Frame(root)
w.pack(side=side,expand=YES,fill=BOTH)
return w
def button(root,side,text,command=None):
w=Button(root,text=text,command=command)
w.pack(side=side,expand=YES,fill=BOTH)
return w
class Calculator(Frame):
def __init__(self):
Frame.__init__(self)
self.pack(expand=YES,fill=BOTH)
self.master.title('hello Calculator')
self.master.iconname('calcl')
display=StringVar()
Entry(self,relief=SUNKEN,textvariable=display).pack(side=TOP,expand=YES,fill=BOTH)
for key in('123','456','789','-0+'):
keyF=frame(self,TOP)
for char in key:
button(keyF,LEFT,char,lambda w=display,s='%s'%char:w.set(w.get()+s))
opsF=frame(self,TOP)
for char in '+-*/=':
if char == '=':
btn=button(opsF,LEFT,char)
btn.bind('<ButtonRelease-1>',lambda e,s=self,w=display:s.calc(w),'+')
else:
btn=button(opsF,LEFT,char,lambda w=display,c=char:w.set(w.get()+''+c+''))
clearF=frame(self,BOTTOM)
button(clearF,LEFT,'Clr',lambda w=display:w.set(''))
def calc(self,display):
try:
display.set(`eval(display.get())`)
except ValueError:
display.set('ERROR')
if __name__ =='__main__':
Calculator().mainloop()``` | 448 | 1,334 | {"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-2019-43 | latest | en | 0.18895 |
http://www.beigebag.com/case_gud_half.htm | 1,632,464,082,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00318.warc.gz | 70,964,267 | 10,985 | ## Gudermannian Devices
Contents:
### Modified Half Gudermannian Device
About the writer: Harvey Morehouse is a contractor/consultant with many years of experience using circuit analysis programs. His primary activities are in Reliability, Safety, Testability and Circuit Analysis. He may be reached at harvey.annie@verizon.net. Simple questions for which I know the answer are free. Complex questions, especially where I am ignorant of the answers, are costly!!!
Summary: A modified Gudermann function can be useful in SPICE modeling of devices that switch from one state to another. It provides a smooth transition between extremities. It is also continuously differentiable, and somewhat so in a SPICE numerical world, unlike some piecewise linear models. Use of this function in SPICE models is present in other variants of SPICE, however its implementation here was independently arrived at. This function will use a unipolar drive and this modified Gudermannian function will switch between zero and unity.
Gudermann function:
The Gudermannian function, named after Christoph Gudermann (1798 - 1852), relates the circular and hyperbolic trigonometric functions without resorting to complex numbers. It is defined by
A shape of this function is shown in Figure 1 following:
Figure 1
Gudermann Function
Figure 1 was taken from reference 1. What is not shown are the limiting values, which are equal to Pi/2 in magnitude for large positive and negative excursions in 'x'.
To utilize this function in models, a switching device for one, there are two little problems. The first is that the midpoint of the curve, where x = 0, is a value of 0. Clearly this requires that the input or the output might be offset in order to get a smooth transition for extremes of either polarity.
The second is the scaling of the input for, as the graph in figure 1 reveals for the unmodified curve (the units for 'x' and 'y' axis are one per tick mark). The exponent argument of 'x' should be replaced by 'ax' to ensure the function fully reaches the extreme values desired for a given range of 'x'. One must scale the function such that it works with signals ranging from zero to some maximum, as well as for signals centered about zero with maximums of either polarity, and produces the desired output range.
Modified half Gudermann Function Model #1:
Please refer to the modified Full Gudermann article posted in the Beige Bag Software web page Resources section. The modified full Gudermannian curve therein was equal to:
This curve varied between positive and negative one. To make it vary between zero and one we need to alter his equation slightly. The modified half Gudermann function becomes:
Because we wish the input control function, 'x', to vary from unity to some positive value, we must offset it by some amount in order to have exponent vary between positive and negative values. The value for the offset is problematic. One would like it to be about half of the maximum excursion of x. For some applications this may be well known. In other cases it may not be.
The derivative of this function is:
A circuit implementing the modified Gudermann function is shown in the following Figure 2.
Figure 2
Modified Gudermann hnormal Circuit
In arriving at Figure 2 some additional changes were made which were incorporated into the circuit. First, the circuitry below the dashed lines includes test inputs. It is assumed that eventually these parameters, 'a' and 'Maxexc' will be passed as parameters to the function in order to customize it for use in specific device models. The default values for 'a' will be 12, and Maxexc, the maximum control signal input excursion, will be unity.
Now the maximum excursion at the input can exceed this value with no detriment, dependent on 'a', but some value must be used in order to translate the input excursion from a unipolar input to a bipolar value.
The input 'x' is a 1v square wave. 'a' is swept value, and Maxexc is set equal to 1v. A graph of the output from the circuit is shown in the following Figure 3.
Figure 3
Modified Gudermann hnormal Test Circuit 1 Graph 1
The output switches nicely between a low and a high value in most instances. The output being a series of sweeps with 'a' varying in value from 2 (the red curve) to 12 in the case of the fastest rise for the maroon curve. The red, green and yellow curves are not well behaved. The reason is that the argument of the exponential function is not large enough in those instances to drive the function to the extreme values.
Therefore, in use of this circuit, it is recommended that the product of the 'a' and the Maxexc or maximum excursion value not be less than about 10 to guarantee smooth switching between extremes.
Now, given that a = 6, Maxexc = 1, what will happen when the input, instead of being a zero to 1V pulse becomes a zero to 5V pulse? A graph of the output when the peak of the v1 pulse generator is varied from 1 to 10V peak for these conditions is shown in Figure 4 following:
Figure 4
Modified Gudermann hnormal Test Circuit 1 Graph 2
In Figure 4 we see that with increasing pulse amplitude, we are in effect decreasing the rise time. Now, let us fix the pulse amplitude at 10v, 'a' at 6, and step the value of Maxexc from one to ten in steps of one. The output is shown in Figure 5 following.
Figure 5
Modified Gudermann hnormal Test Circuit 2 Graph 1
Here we see the effect is to modify the switching interval about. Remember that the input rise time is 1uSec.
One more task remains. Because of numerical errors, the output at the low level is not quite zero. Recalling the defining equation we are implementing, it is:
The output level when the input level is zero, is that function with the exponent argument equal to a constant '-a*off' as expressed above, or = '-a*v(Maxexc)/2'. To get a better result at the low level, we need to subtract this value from the output. A circuit which will do this is shown in Figure 6 following:
Figure 6
Modified Gudermann hnormal Test Circuit 2
The change is rather unremarkable, however the effect is to reduce the zero output level to the order of 1e-19, while the unity level is 1000 mV to three decimal places.
The circuit of Figure 6 should be converted into a parameterized subcircuit as shown. There will be one input and one output connection, with 'a' and 'Maxexc' being passed parameters having default values of 6 and 1 respectively. The device should be named gudhnor.
Conclusions:
A Modified Half Gudermannian circuit model has been created which can be useful in creating devices or in functions that smoothly switch between low and high magnitude values. This device should be added to the standard library as a building block.
References:
1 Wikopedia Article, Gudermannian function. http://en.wikipedia.org/wiki/Gudermannian_function
2 http://mathworld.wolfram.com/GudermannianFunction.html
### Modified Full Gudermannian Device
Summary: A Gudermann curve can be useful in SPICE modeling of devices that switch from one state to another. It provides a smooth transition between extremities. It is also continuously differentiable, and somewhat so in a SPICE numerical world, unlike some piecewise linear models. Use of this function in SPICE models is present in other variants of SPICE, however its implementation here was independently arrived at. This function will use a bipolar drive and the modified Gudermannian function will switch smoothly between equal positive and negative values through zero for equal amplitude positive and negative number inputs.
Gudermann function:
The Gudermannian function, named after Christoph Gudermann (1798 - 1852), relates the circular and hyperbolic trigonometric functions without resorting to complex numbers. It is defined by
A shape of this function is shown in Figure 1 following:
Figure 1
Gudermann Function
The inverse function of the Gudermannian function gives the vertical position in the Mercator projection in terms of the latitude and may be defined for by
The derivative of the function is
and the derivative of its inverse is given by
Figure 1 was taken from reference 1. What is not shown are the limiting values, which are equal to Pi/2 in magnitude for large excursions in 'x' .
To utilize this function in models, as perhaps an inductor, there are several problems. The first is the scaling of the input for, as the graph in figure 1 reveals for the unmodified curve (the units for 'x' and the output are one per tick mark), the exponent value of 'exp(x)' should be replaced 'ax', with 'a' chosen to ensure the function fully reaches the extreme values desired for a given range of 'x'.
Another is that one must scale the function such that it works with signals to produce the desired output range and also the proper shape between extremities. Unfortunately the value of 'a' affects both of these considerations.
Modified Full Gudermann Function Model #1:
Let us make a few assumptions as a basis for the model. The first is that the input signal will range in value from at least MinV as a minimum to MaxV as a maximum. At an input equal in magnitude to MinV the output should be minimum and at an input equal in magnitude to MaxV (= -MinV) the output should be maximum. Now the midpoint of the curve will be when the input is at the value Vmid = (MaxV + MinV)/2 = 0. To achieve this, the original Gudermannian curve will be slightly modified to become:
This will in B2SPICE format become:
v = atan(exp(v(a)*v(x) )) /(atan(1)) -1
Here, atan(1) = Pi/4. This function produces an output which varies smoothly from minus one to plus one in amplitude for a sufficiently large value of exponent, or v(a)*v(x) product. A circuit implementing this is shown in the following Figure 2.
Figure 2
Modified Full Gudermann Circuit
One other factor needs to be noted. We will be sweeping the value of 'a' provided by generator v1 in a test.
A netlist of the circuit is provided for reference and is as follows:
Gud fnormal.cp~
***** main circuit
V1 x 3 PULSE( -1.000000000000e+000 1.000000000000e+000 0.000000000000e+000 1.000000000000e-006 1.000000000000e-006 1.000000000000e-005 2.000000000000e-005)
R1 x 3 1K
R2 a 5 1K
R3 Gudax 0 1K
R4 3 0 10Meg
B1 Gudax 0 v = atan(exp(v(a)*v(x) )) /(atan(1)) -1
V2 a 5 0.000000000000e+000
R5 5 0 10Meg
.TRAN 1E-6 6E-5 0 1E-8 uic
.OPTIONS method = gear
.end
It is intended that the device model be named Gudpfn. The 'Gud' implying it is a Gudermann function, 'p' implying a zero to positive input level input level, and 'fn' implying that the output excursion although smooth as in a normal Gudermann function, changes from minus one in magnitude to positive 1 as limits.
A graph of the output from the circuit is shown in the following Figure 3.
Figure 3
Modified Gudermann full normal Test Circuit Graph 1
Here we are sweeping the value of 'a' from one to six, showing a positive to negative transition. The negative to positive transitions behave in a similar manner. As the input is a square wave with unity positive and negative values, this gives us an appreciation of how the curve varies with different extreme magnitudes of the exponent.
The red curve represents 'a' equaling one, the light blue curve represent 'a' = 6, and with unity increments in between. Clearly to reach values close to unity the v(a)*v(x) product should be equal to or greater than about 5 in magnitude as a minimum at the extremes, and the final values are less than one percent from the extremes.
Returning to the curve, it is again:
v = (2*atan(exp(v(a)*v(x) )) - 2*atan(1))/(2*atan(1))
In order to know how the slope of the curve varies as a function of 'a' and 'x', we need to differentiate the curve. After some work, we find that the derivative is:
Let us model this and see what this tells us. We will plot a derivative of the modified Gudermannian curve output and the analytical derivative using both SPICE and the above formula using the circuit as shown in Figure 4 as follows:
Figure 4
Modified Gudermann fnormal Test Circuit 2
A graph of the circuit in Figure 4, for 'a' = 5 is shown in Figure 5.
Figure 5
Modified Gudermann fnormal Test Circuit 2 graph
In Figure 5 we see a rather complex picture. The red trace is the input voltage. The green trace is the modified Gudermann output. The blue trace is the analytic derivative of the modified Gudermann function. The maroon trace, almost a trapezoid, is the SPICE derivative of the modified Gudermann generator.
The question arises why the SPICE derivative of the modified Gudermann function does not agree with the analytic derivative of the modified Gudermann function. With a little reflection, it is clear that SPICE creates a derivative with respect to time, whereas the analytic derivative of the modified Gudermann function does not include time. Thus, we are comparing from the equation:
y = f(x)
the functions:
dy/dx = f'(x)
and
dy(t)/dy = f'(x,t)
Which brings to mind some thoughts about modeling in general.
When creating a circuit model, the BEST model in terms of accuracy and convergence is virtually ALWAYS created with SPICE primitive devices such as resistors, capacitors, inductors, transistors and so on. However certain devices such as abrupt switches can cause convergence problems.
Use of behavioral device models is equal to or intermediate between these abrupt discontinuous devices and discrete device models in terms of convergence. Just because a device can seemingly be well behaved, like a Gudermann function, does not mean that when used in place of a SPICE switch model, that it will be free from convergence problems.
In terms of speed, assuming that convergence problems are not present, the behavioral model is usually intermediate between an abrupt model such as a piecewise linear transfer function and discrete device models.
Also, it must be remembered that SPICE itself essentially produces a sampled data output, and thus a function, which is continuously differentiable in the real world, might not be in a time domain.
A circuit to be converted into a parameterized subcircuit model is shown in Figure 6 following.
Figure 6
Modified Gudermann fnormal mGudax model
It is assumed that this mGudax model will be added to the standard library, as it will be used to create some other devices in the future.
Conclusions:
A modified full Gudermannian circuit model has been created which can be useful in creating devices or functions which smoothly switch between equal magnitude values of opposite sign, for a similarly varying input signal.
References:
1 Wikopedia Article, Gudermannian function. http://en.wikipedia.org/wiki/Gudermannian_function
2 http://mathworld.wolfram.com/GudermannianFunction.html
### Smooth Transition Gudermannian Switch
Summary: A modified Gudermann function can be useful in SPICE modeling of devices that switch from one state to another. One such device is a switch function. This device will use a unipolar drive and will switch between a small value to a much higher value smoothly.
Gudermann function:
In a previous article I showed how to create what I called a half Gudermannian function. This function, using a unipolar control signal, was a function that varied smoothly between a very small value and unity. Here we shall use this function to create a smooth switch model.
The equation for such a smooth switch would be:
It was somewhat arbitrarily chosen to use a control signal that varied from zero to 1 volt. In that case, 'off' or offset equals 0.5V. In this case, 'a' must be equal to about 16 to get a close match to the desired value of Ron when x = 0, and Roff when x = 1. Additionally, there will be a small error term when x = 0. The term, -exp(a*(-off)) removes this error. This small error will also affect the value of y when x = 1, but typically it is a small percentage error, WHEREAS it is much more significant at the lower end of the function where the desired result is close to zero.
A test circuit that models this equation is shown in Figure 1 following:
Figure 1
STSNOT4 Test Circuit
In Figure 1 the switch is essentially represented by Generator B1, R3 and B2. The circuitry beneath the dashed line represent parameters that will later be passed to the parameterized switch circuit. The control voltage is represented by the voltage across resistor R1, along with its external to the switch generator V1.
V8 and R11 represent a source that is being switched by the STSNOT4 device (generator B2) which is between the nodes S1 and S2. The source and switch combination apply this 'chopped' voltage to what will be recognized as a Buck output filter and load. Here diode D2 is the flywheel diode for this filter. The filter is being operated in CCM (continuous conduction mode).
The switch equation, the B1 generator value, is:
v = v(Roff)*atan(exp(-v(a)*(v(x) -v(Offset))))/(2*atan(1)) + v(Ron)
Here the values which will be eventually passed to the model are clearly shown as Roff, a, Offset and Ron. A netlist for this circuit is:
Gud hanormal test 2.cpr
***** main circuit
V1 x 3 PULSE( 0.000000000000e+000 5.000000000000e+000 0.000000000000e+000 1.000000000000e-007 1.000000000000e-007 5.000000000000e-005 1.000000000000e-004)
R1 x 3 resistor 10Meg
R2 a 5 resistor 1K
R3 Gudax 0 resistor 10Meg
R4 3 0 resistor 10Meg
B1 Gudax 0 v = v(Roff)*atan(exp(v(a)*(v(x) -v(Offset))) -exp(-v(a)*v(Offset)))/(2*atan(1)) + v(Ron)
V2 a 5 1.600000000000e+001
R5 5 0 resistor 10Meg
V4 Offset 8 2.500000000000e+000
R6 Offset 8 resistor 1K
R7 8 0 resistor 10Meg
V6 Roff 0 1.000000000000e+008
V7 Ron 0 1.000000000000e-002
R8 Roff 0 resistor 1K
R9 Ron 0 resistor 1K
B2 S1 S2 i = v(S1,S2)/v(Gudax)
V8 12 0 2.000000000000e+001
L1 13 14 10m
C1 14 15 200u
R10 14 0 resistor 5
R11 S1 12 resistor .01
R12 S2 13 resistor 20m
R13 15 0 resistor 20m
D2 0 S2 mbrs340t3 off
.model resistor r res = 1
.model mbrs340t3 D is = 1.924e-05 rs = 0.03512 n = 1.034 tt = 1e-08 cjo = 4.92299e-10 vj = 0.451454
+ m = 0.443167 eg = 0.6 xti = 2 kf = 0 af = 1
+ fc = 0.5 bv = 100 ibv = 0.1
.TRAN 1E-6 0.1 0 1E-6 uic
.OPTIONS method = gear
.end
A plot of the circuit performance is shown in the graph of the following Figure 2.
Figure 2
STSNOT4 Test Circuit graph
In Figure 2 we see the expected triangular waveform of inductor current (in red), the diode voltage in grey, and the output voltage in red.
Stripped of the test circuitry, the device to be modeled becomes as shown in Figure 3 following:
Figure 3
STSNOT4 Circuit
Figure 3 shows the simplicity of this device model. This should be added to the device library with a NO switch graphic symbol as a parameterized subcircuit device named STSNOT4. Additionally, one can create a STSNCT4 device model that is identical to that of the STSNOT4 device with the alteration of the B1 generator equation by changing the sign of 'a' in two places.
Conclusions:
A smooth transition NO and NC switch model has been created using a modified Gudermannian function. Use of this device should eliminate some of the difficulties in using abrupt transition switches in many models.
References:
1 Wikopedia Article, Gudermannian function:http://en.wikipedia.org/wiki/Gudermannian_function | 4,753 | 19,226 | {"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} | 3.453125 | 3 | CC-MAIN-2021-39 | latest | en | 0.921577 |
https://www.nag.com/numeric/nl/nagdoc_28.5/clhtml/g01/g01gdc.html | 1,686,188,379,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00352.warc.gz | 961,627,633 | 5,886 | # NAG CL Interfaceg01gdc (prob_f_noncentral)
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## 1Purpose
g01gdc returns the probability associated with the lower tail of the noncentral $F$ or variance-ratio distribution.
## 2Specification
#include
double g01gdc (double f, double df1, double df2, double lambda, double tol, Integer max_iter, NagError *fail)
The function may be called by the names: g01gdc, nag_stat_prob_f_noncentral or nag_prob_non_central_f_dist.
## 3Description
The lower tail probability of the noncentral $F$-distribution with ${\nu }_{1}$ and ${\nu }_{2}$ degrees of freedom and noncentrality parameter $\lambda$, $P\left(F\le f:{\nu }_{1},{\nu }_{2}\text{;}\lambda \right)$, is defined by
$P(F≤f:ν1,ν2;λ)=∫0xp(F:ν1,ν2;λ)dF,$
where
$P(F : ν1,ν2;λ )=∑j= 0∞e-λ/2 (λ/2)jj! ×(ν1+2j)(ν1+2j)/2 ν2ν2/2 B((ν1+2j)/2,ν2/2)$
$×u(ν1+2j-2)/2[ν2+(ν1+2j)u] -(ν1+2j+ν2)/2$
and $B\left(·,·\right)$ is the beta function.
The probability is computed by means of a transformation to a noncentral beta distribution:
$P(F≤f:ν1,ν2;λ)=Pβ(X≤x:a,b;λ),$
where $x=\frac{{\nu }_{1}f}{{\nu }_{1}f+{\nu }_{2}}$ and ${P}_{\beta }\left(X\le x:a,b\text{;}\lambda \right)$ is the lower tail probability integral of the noncentral beta distribution with parameters $a$, $b$, and $\lambda$.
If ${\nu }_{2}$ is very large, greater than ${10}^{6}$, then a ${\chi }^{2}$ approximation is used.
## 4References
Abramowitz M and Stegun I A (1972) Handbook of Mathematical Functions (3rd Edition) Dover Publications
## 5Arguments
1: $\mathbf{f}$double Input
On entry: $f$, the deviate from the noncentral $F$-distribution.
Constraint: ${\mathbf{f}}>0.0$.
2: $\mathbf{df1}$double Input
On entry: the degrees of freedom of the numerator variance, ${\nu }_{1}$.
Constraint: $0.0<{\mathbf{df1}}\le {10}^{6}$.
3: $\mathbf{df2}$double Input
On entry: the degrees of freedom of the denominator variance, ${\nu }_{2}$.
Constraint: ${\mathbf{df2}}>0.0$.
4: $\mathbf{lambda}$double Input
On entry: $\lambda$, the noncentrality parameter.
Constraint: $0.0\le {\mathbf{lambda}}\le -2.0\mathrm{log}\left(U\right)$ where $U$ is the safe range parameter as defined by X02AMC.
5: $\mathbf{tol}$double Input
On entry: the relative accuracy required by you in the results. If g01gdc is entered with tol greater than or equal to $1.0$ or less than (see X02AJC), the value of is used instead.
6: $\mathbf{max_iter}$Integer Input
On entry: the maximum number of iterations to be used.
Suggested value: $500$. See g01gcc and g01gec for further details.
Constraint: ${\mathbf{max_iter}}\ge 1$.
7: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
If on exit ${\mathbf{fail}}\mathbf{.}\mathbf{code}=$ NE_INT_ARG_LT, NE_PROB_F, NE_REAL_ARG_CONS or NE_REAL_ARG_LE, then g01gdc returns $0.0$.
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
NE_CONV
The solution has failed to converge in $⟨\mathit{\text{value}}⟩$ iterations. Consider increasing max_iter or tol.
NE_INT_ARG_LT
On entry, ${\mathbf{max_iter}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{max_iter}}\ge 1$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_PROB_F
The required probability cannot be computed accurately. This may happen if the result would be very close to zero or one. Alternatively the values of df1 and f may be too large. In the latter case you could try using a normal approximation, see Abramowitz and Stegun (1972).
NE_PROB_F_INIT
The required accuracy was not achieved when calculating the initial value of the central $F$ or ${\chi }^{2}$ probability. You should try a larger value of tol. If the ${\chi }^{2}$ approximation is being used then g01gdc returns zero otherwise the value returned should be an approximation to the correct value.
NE_REAL_ARG_CONS
On entry, ${\mathbf{df1}}=⟨\mathit{\text{value}}⟩$.
Constraint: $0.0<{\mathbf{df1}}\le {10}^{6}$.
On entry, ${\mathbf{df1}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{df1}}>0.0$.
On entry, ${\mathbf{lambda}}=⟨\mathit{\text{value}}⟩$.
Constraint: $0.0\le {\mathbf{lambda}}\le -2.0×\mathrm{log}\left(U\right)$, where $U$ is the safe range parameter as defined by X02AMC.
NE_REAL_ARG_LE
On entry, ${\mathbf{df2}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{df2}}>0.0$.
On entry, ${\mathbf{f}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{f}}>0.0$.
## 7Accuracy
The relative accuracy should be as specified by tol. For further details see g01gcc and g01gec.
## 8Parallelism and Performance
g01gdc is not threaded in any implementation.
When both ${\nu }_{1}$ and ${\nu }_{2}$ are large a Normal approximation may be used and when only ${\nu }_{1}$ is large a ${\chi }^{2}$ approximation may be used. In both cases $\lambda$ is required to be of the same order as ${\nu }_{1}$. See Abramowitz and Stegun (1972) for further details.
## 10Example
This example reads values from, and degrees of freedom for, $F$-distributions, computes the lower tail probabilities and prints all these values until the end of data is reached.
### 10.1Program Text
Program Text (g01gdce.c)
### 10.2Program Data
Program Data (g01gdce.d)
### 10.3Program Results
Program Results (g01gdce.r) | 1,819 | 5,705 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 65, "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-2023-23 | latest | en | 0.553008 |
https://studylib.net/doc/25241113/electricity-and-electrical-circuits---powerpoint--1- | 1,556,102,941,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578640839.82/warc/CC-MAIN-20190424094510-20190424120510-00100.warc.gz | 547,780,547 | 43,741 | # Electricity and Electrical Circuits - powerpoint (1)
```Electricity and Electrical Circuits
Part 1 - Introduction
Integrated Science
Glencoe
Chapter 13
• As we have seen in our study of chemistry,
matter is composed of atoms, which are
composed of protons, neutrons, and
electrons.
• The protons are positively charged and are
found “locked” in the nucleus with the
neutrons.
• The electrons are negatively charged and are
moving around in the electron clouds,
and are not “locked”
into position.
• In fact, electrons can and do move between
atoms, and can be transferred to other
materials and move around quite freely at
times.
• This “free movement” of electrons is what we
call electricity.
• Remember the Law of Conservation of Energy,
which states that energy cannot be created or
destroyed, but may only change form. This
law applies here as electrical charges can be
transferred between objects.
• If you have ever experienced the discharge of
static electricity as you walk around on carpet
in your socked feet and then get shocked
when you touch something, then you know
that charge can be transferred.
• Anytime objects touch each other, there is a
transfer of electrons, as the electrons in the
outer energy levels of an atom, are held less
tightly, and can be torn away.
• If there are enough extra electrons accumulating,
then the object starts to get “charged” up and has
a noticeable difference in charge.
• If the charge dissipates quickly, no accumulation
can occur.
• Different materials allow the flow of energy, in
our case electrons, to varying degrees.
• Materials that
allow the easy
flow of
electrons are
called
conductors.
• Materials that
do not allow
the easy flow
of electrons
are called
insulators.
• The rate of
movement of
electrons can be
measured over a
certain amount of
time.
• The current is
defined as the rate of
charge movement or
the movement of
electrons through an
area over a given
amount of time.
• The faster the movement of electrons the
higher the current.
• Sometimes the flow of electrons is slowed down
by any number of factors. These factors include:
1. Materials – what the electrons are moving through
2. Temperature – how warm or cold the materials are
3. Length – how far the electrons need to move
4. Cross section – how wide the area is the electrons are
trying to move through
• A second concept to deal with is resistance.
Anything that slows down the flow of
electrons is defined as resistance.
• A third concept of electricity is potential
difference, which is defined as the change in
electrical potential energy between two
points.
• For example, a 9 volt battery has a difference
in electrical potential energy of 9 volts
between the two terminals of the battery.
One is at zero volts, and as the electrons are
moved through the battery, they gain 9 joules
of electrical potential energy.
• Since charge is measured in a unit called
coulombs and energy is in joules, as the
charges gain energy the potential difference is
9 joules per coulomb = 9 volts of potential
difference.
• We will be using these three concepts as we
next look at electrical circuits and how
electricity can be manipulated to many things.
Electricity and Electrical Circuits
Part 2 - Circuits
Integrated Science
Glencoe
Chapter 13
• In our study of electricity, it is important to remember
that energy is constantly converted in the following
manner:
1. Chemical potential energy
is converted to electrical
potential energy in a battery.
2. Electrical potential energy is changed to thermal
energy, light, mechanical energy as the electricity
moves through the different materials.
• To be able to effectively use electricity, one
must understand how electricity moves
through different materials in a pathway.
• In science, we can build and draw models of
the paths electricity can take.
• The physical models we build are called
electrical circuits. Electrical circuit is defined
as a set of electrical components connected to
provide one or more
complete paths for
movement of charge.
• The drawings are
called Schematic
diagrams.
Schematic diagrams
are defined as
graphical
representations of
an electrical circuit.
• There is a standard set of scientific symbols
used to represent different components in a
circuit. The most commonly used symbols
and what they represent are listed below, we
will use these symbols in our drawings.
Component
Diagrams
Wire
Bulb / Lamp
Battery
Resistor
Open Switch
Closed
Switch
General Circuit Information
• A circuit is a complete path for the electrons
to follow as they flow, if the path is not
complete, there can be no flow between two
points.
• It would be like a road from your home to the
store, with a bridge out and no alternative
route. There must be a way to get around the
broken bridge if you want to get where you
are going.
• Sometimes a circuit is complete but there is a
problem of the current flowing too fast. This
is called short circuit.
• Short circuit – a circuit which contains little or
no resistance
The components get too hot due to the
excess current and not enough resistance to
slow it down and can cause fires
• Examples:
- 2 terminals of a battery directly connected
- Uninsulated wires come into contact
These are very dangerous
• In every circuit there must be a source of the
electric current.
• This source is known as the EMF –
Electromotive force → the energy per unit
charge supplied by a source of electric current
→ A source of electrical energy
→ A “charge pump”
→ Examples are batteries and generators
• A battery changes
chemical potential
energy to electrical
potential energy as it
transfers the energy to
the electrons in the
battery. As they move
along the circuit, the
energy is then
transferred to the
different components,
and dissipated (given
off) usually as heat,
light, or sound.
• There are three different types of electrical circuits:
1. Series Circuits - Circuit (or portion of) in which there is
a single conducting path without junctions for
electricity to follow
2. Parallel Circuits - Circuit (or part of) where components
are connected across common points and provides
separate conducting paths for electricity to follow
3. Complex Circuits – Circuits with some segments being
in series and other segment being in parallel to take
advantage of the benefits of both
Series Circuits
1. Series Circuits - Circuit (or portion of) in which
there is a single conducting path without
junctions for electricity to follow
• Some of the characteristics of series circuits are listed below and
are important to remember when deciding which type of circuit
to use.
to the store
example, in a
series circuit there
If part of the
you cannot
get where you
want to go,
no exceptions.
• All bulbs / resistors / components in series will
have the same current. This is because the
current can only flow as fast as the slowest
(most resistant) component will allow.
• The total current in a series circuit depends on
the number of resistors present and resistance
of each.
• The Equivalent resistance (symbolized Req )
also known as the total resistance = sum of all
resistances in a circuit from all components
• Series circuits require all elements to conduct,
no broken parts, or the current stops.
• Below is a list of advantages and
disadvantages of series circuits (notice some
may be both)
1. Good for regulating current (all parts have
same current)
2. Good for reducing current on individual
parts
3. Current stops if a component breaks
•
1. All parts have same current
2. If one part breaks, the whole circuits fails
Parallel Circuits
2. Parallel Circuits - Circuit (or part of) where
components are connected across common
points and provides separate conducting
paths for electricity to follow
• Some of the characteristics of parallel circuits are
listed below and are important to remember
when deciding which type of circuit to use.
• Back to our road to the store example, in a
parallel circuit there is more than one road you
can use as a possible route to the store. If one
part of the road is out, you can choose a detour
around it to still get where you want to go, or if
the whole road is working, you can choose your
path among several to get to the store.
• Parallel circuits give multiple alternate pathways
for current flow
• Resistors in parallel have same potential
difference across them. The voltage is the same
for all parts, but the current will be different for
each part based upon the resistance of each
individual component. More resistance in one
part equals a slower current in that part.
• The sum of currents in parallel resistors = total
current
• Below is a list of advantages and disadvantages of parallel circuits (notice
some may be both)
1. Parallel circuits do not require all elements to conduct
2. One part can malfunction and the rest will continue to work
3. Potential difference does not change for all components when
one component fails
Allows for standardization of products
Allows manufacturers to regulate current through a product by resistors
Home wired in parallel
1. Current will change if one component fails
2. Current different in all components
Complex Circuits
3. Complex Circuits – Circuits with some
segments being in series and other segments
being in parallel to take advantage of the
benefits of both
• It is common in homes to have multiple
outlets in parallel to each other and in series
with all those outlets a fuse/circuit breaker
→ allows identical potential difference but
• Back to our example of the trip to the store,
there may be multiple roads to take to get to
the store, but still only one bridge across the
river, and if the bridge is out, you cannot cross
even with many roads to that bridge.
• Fuse - small metallic strip that melts when
current becomes too high.
→ Replacement necessary when melted
• Circuit breaker - device that triggers a switch
to open circuit when current is too high
→ reset switch when opened
• Both are specially designed
for specific amounts of
current
```
Electronics
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http://www.java-forums.org/new-java/65800-there-easier-way-get-position-out-array.html | 1,394,637,303,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021889832/warc/CC-MAIN-20140305121809-00045-ip-10-183-142-35.ec2.internal.warc.gz | 384,299,878 | 16,249 | # Thread: Is there an easier way to get position out of array
1. Member
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## Is there an easier way to get position out of array
Hey so I'm suppose to get the user to input the total rainfall for all the months and then it will display the total. Also what month had the most and least rainfall. My question is, is there an easier way to do it basiclly I'm setting each position in the array (say rainfall[1] = 42) then it would run a loop and when a variable equals 1 it would display February.
Java Code:
```import java.util.Scanner;
public class RainfallTester {
public static void main(String [] args){
Scanner scan = new Scanner(System.in);
Rainfall temp = new Rainfall();
double rain;
for(double j=0; j<temp.rainfall.length;j++ ){
System.out.print("What is the total rainfall for the month:");
rain = scan.nextDouble();
temp.setRain(rain);
}
System.out.println("The total rainfall is "+temp.totalRain());
System.out.println("The average rainfall is "+temp.avgRain());
System.out.println("The month with the most rainfall is "+temp.mostRain());
System.out.println("The month with the least rainfall is "+temp.leastRain());
}
}```
Here's the part where it decides the month.
Java Code:
```public class Rainfall {
double avg;
double sum;
int hmonth = 0;
int lmonth = 0;
double rainfall[] = new double[12];
public void setRain(double rain){
for(int i =0;i<rainfall.length;i++){
rainfall[i] = rain;
}
}
double totalRain(){
for(int i =0; i<rainfall.length;i++){
sum += rainfall[i];
}
return sum;
}
double avgRain(){
for(int i =0; i<rainfall.length;i++){
sum += rainfall[i];
}
avg = sum/rainfall.length;
return avg;
}
double mostRain(){
double max = 0;
for (int i=0; i<rainfall.length;i++){
if (max < rainfall[i]){
max = rainfall[i];
hmonth = i;
}
}
return max;
}
public void mostMonth(){
if(hmonth == 1){
System.out.print("January ");
}else if(hmonth == 2){
System.out.print("Feburary ");
}else if(hmonth == 3){
System.out.print("March ");
}else if(hmonth == 4){
System.out.print("April ");
}else if(hmonth == 5){
System.out.print("May ");
}else if(hmonth == 6){
System.out.print("June ");
}else if(hmonth == 7){
System.out.print("July ");
}else if(hmonth == 8){
System.out.print("August ");
}else if(hmonth == 9){
System.out.print("September ");
}else if(hmonth == 10){
System.out.print("October ");
}else if(hmonth == 11){
System.out.print("November ");
}else{
System.out.print("December ");
}
}
double leastRain(){
double min = 0;
for (int i=0; i<rainfall.length;i++){
if (min > rainfall[i]){
min = rainfall[i];
}
}
return min;
}
public void leastMonth(){
if(hmonth == 1){
System.out.print("January ");
}else if(hmonth == 2){
System.out.print("Feburary ");
}else if(hmonth == 3){
System.out.print("March ");
}else if(hmonth == 4){
System.out.print("April ");
}else if(hmonth == 5){
System.out.print("May ");
}else if(hmonth == 6){
System.out.print("June ");
}else if(hmonth == 7){
System.out.print("July ");
}else if(hmonth == 8){
System.out.print("August ");
}else if(hmonth == 9){
System.out.print("September ");
}else if(hmonth == 10){
System.out.print("October ");
}else if(hmonth == 11){
System.out.print("Novenmber ");
}else{
System.out.print("December ");
}
}
}```
Thank. Also my instructions were to just have the user input the rainfall not the month too.
2. Member
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Posts
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## Re: Is there an easier way to get position out of array
Here is an update to where I am.When I run it everything is fine it gives the the highest and lowest values but it wont give me there position. Say 10 was the lowest number in position 3, well I set 3 to April but it always gives me december why?? Thanks.
Java Code:
```public class Rainfall {
int hmonth = 0;
int lmonth = 0;
double rainfall[] = new double[12];
public void setRain(double rain){
for(int i =0;i<rainfall.length;i++){
rainfall[i] = rain;
}
}
public void calc(){
double sum =0;
double avg = 0;
double max = 0;
double min = rainfall[0];
//find total
for(int i =0; i<rainfall.length;i++){
sum += rainfall[i];
}
System.out.println("The total rainfall is "+sum);
//find average
avg = sum/rainfall.length;
System.out.println("The average rain fall is "+avg);
//find max
for (int i=0; i<rainfall.length;i++){
if (max < rainfall[i]){
max = rainfall[i];
hmonth = i;
}
}
System.out.println("The month with the most rain is ");
mostMonth();
System.out.println("with "+max+" rain");
//find min
for (int i=0; i<rainfall.length;i++){
if (min > rainfall[i]){
min = rainfall[i];
lmonth = i;
}
}
System.out.println("The month with the least rain is ");
leastMonth();
System.out.println("with "+min+" rain");
}
//find most month
public void mostMonth(){
if(hmonth == 1){
System.out.print("January ");
}else if(hmonth == 2){
System.out.print("Feburary ");
}else if(hmonth == 3){
System.out.print("March ");
}else if(hmonth == 4){
System.out.print("April ");
}else if(hmonth == 5){
System.out.print("May ");
}else if(hmonth == 6){
System.out.print("June ");
}else if(hmonth == 7){
System.out.print("July ");
}else if(hmonth == 8){
System.out.print("August ");
}else if(hmonth == 9){
System.out.print("September ");
}else if(hmonth == 10){
System.out.print("October ");
}else if(hmonth == 11){
System.out.print("November ");
}else{
System.out.print("December ");
}
}
//find least month
public void leastMonth(){
if(hmonth == 1){
System.out.print("January ");
}else if(hmonth == 2){
System.out.print("Feburary ");
}else if(hmonth == 3){
System.out.print("March ");
}else if(hmonth == 4){
System.out.print("April ");
}else if(hmonth == 5){
System.out.print("May ");
}else if(hmonth == 6){
System.out.print("June ");
}else if(hmonth == 7){
System.out.print("July ");
}else if(hmonth == 8){
System.out.print("August ");
}else if(hmonth == 9){
System.out.print("September ");
}else if(hmonth == 10){
System.out.print("October ");
}else if(hmonth == 11){
System.out.print("Novenmber ");
}else{
System.out.print("December ");
}
}
}```
The part that initializes it all.
Java Code:
```import java.util.Scanner;
public class RainfallTester {
public static void main(String [] args){
mainRain();
}
public static void mainRain(){
Scanner scan = new Scanner(System.in);
Rainfall temp = new Rainfall();
double rain;
for(double j=0; j<temp.rainfall.length;j++ ){
System.out.print("What is the total rainfall for the month:");
rain = scan.nextDouble();
if(rain <= 0){
System.out.println("Error enter a non negative number.");
mainRain();
}
temp.setRain(rain);
}
temp.calc();
}
}```
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## Re: Is there an easier way to get position out of array
A small point, but mostMonth() does not "find most month" as the comment states. What it does is "prints a the string name of the greatest month followed by a space". It's as well to be precise - especially when things go wrong and you are trying to see whether each part of the program does what you intend.
In that vein, what is setRain() supposed to do?
---
You can check the data by printing it out (to see if you are working with numbers you think you are working with):
Java Code:
```public void calc(){
for(int i = 0; i < rainfall.length; i++) {
System.out.println("i=" + i + " rainfall[i]=" + rainfall[i]);
}
double sum =0;
double avg = 0;
double max = 0;
double min = rainfall[0];
// etc```
4. Moderator
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Location
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## Re: Is there an easier way to get position out of array
Also at www.javaprogrammingforums.com.
@OP: If you are going to start a discussion at multiple places, please post links at each to the others so that everyone taking part knows what else is being said.
Also bear in mind that many people won't bother with discussions that are cross posted: they fear wasting their time saying something that has already been said, or saying something that loses its relevance when the discussion has moved on (at some other location) in a different direction. Some will respond to crossposts, some won't. In either case, though, it's just politeness that everyone should know what's going on, and where.
5. Member
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## Re: Is there an easier way to get position out of array
Okay I wasn't having any luck here thanks.
6. Member
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Dec 2012
Posts
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## Re: Is there an easier way to get position out of array
Okay I used this code and found that all my array is taking in is the value 20
Java Code:
```for(int i = 0; i < rainfall.length; i++) {
System.out.println("i=" + i + " rainfall[i]=" + rainfall[i]);
}```
Ill re-post the codes
Java Code:
```public class Rainfall {
int hmonth = 0;
int lmonth = 0;
double rainfall[] = new double[12];
public void setRain(double rain){
for(int i =0;i<rainfall.length;i++){
rainfall[i] = rain;
}
}
public void calc(){
double sum =0;
double avg = 0;
double max = rainfall[0];
double min = rainfall[0];
//find total
for(int i =0; i<rainfall.length;i++){
sum += rainfall[i];
}
System.out.println("The total rainfall is "+sum);
//find average
avg = sum/rainfall.length;
System.out.println("The average rain fall is "+avg);
//find max
for (int i=0; i<rainfall.length;i++){
if (max < rainfall[i]){
max = rainfall[i];
hmonth = i;
}
}
System.out.print("The month with the most rain is ");
mostMonth();
System.out.println("with "+max+" rain");
//find min
for (int i=0; i<rainfall.length;i++){
if (min > rainfall[i]){
min = rainfall[i];
lmonth = i;
}
}
System.out.print("The month with the least rain is ");
leastMonth();
System.out.println("with "+min+" rain");
for(int i = 0; i < rainfall.length; i++) {
System.out.println("i=" + i + " rainfall[i]=" + rainfall[i]);
}
}
//find most month
public void mostMonth(){
if(hmonth == 0){
System.out.print("January ");
}else if(hmonth == 1){
System.out.print("Feburary ");
}else if(hmonth == 2){
System.out.print("March ");
}else if(hmonth == 3){
System.out.print("April ");
}else if(hmonth == 4){
System.out.print("May ");
}else if(hmonth == 5){
System.out.print("June ");
}else if(hmonth == 6){
System.out.print("July ");
}else if(hmonth == 7){
System.out.print("August ");
}else if(hmonth == 8){
System.out.print("September ");
}else if(hmonth == 9){
System.out.print("October ");
}else if(hmonth == 10){
System.out.print("November ");
}else{
System.out.print("December ");
}
}
//find least month
public void leastMonth(){
if(hmonth == 1){
System.out.print("January ");
}else if(hmonth == 2){
System.out.print("Feburary ");
}else if(hmonth == 3){
System.out.print("March ");
}else if(hmonth == 4){
System.out.print("April ");
}else if(hmonth == 5){
System.out.print("May ");
}else if(hmonth == 6){
System.out.print("June ");
}else if(hmonth == 7){
System.out.print("July ");
}else if(hmonth == 8){
System.out.print("August ");
}else if(hmonth == 9){
System.out.print("September ");
}else if(hmonth == 10){
System.out.print("October ");
}else if(hmonth == 11){
System.out.print("Novenmber ");
}else{
System.out.print("December ");
}
}
}```
Java Code:
```import java.util.Scanner;
public class RainfallTester {
public static void main(String [] args){
mainRain();
}
public static void mainRain(){
Scanner scan = new Scanner(System.in);
Rainfall temp = new Rainfall();
double rain;
for(double j=0; j<temp.rainfall.length;j++ ){
System.out.print("What is the total rainfall for the month:");
rain = scan.nextDouble();
if(rain <= 0){
System.out.println("Error enter a non negative number.");
mainRain();
}
temp.setRain(rain);
}
temp.calc();
}
}```
Right now I basically want to know how to get the user input value into the array. What I'm doing isn't working thanks.
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# In the figure above, showing circle with center O and points
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Joined: 09 Feb 2013
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In the figure above, showing circle with center O and points [#permalink]
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Updated on: 26 Feb 2013, 01:43
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Question Stats:
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In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
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Originally posted by emmak on 25 Feb 2013, 02:21.
Last edited by Bunuel on 26 Feb 2013, 01:43, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 56260
Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
26 Feb 2013, 02:09
4
6
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Good question. It tests three must know properties for the GMAT:
1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.
According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.
3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.
For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html
Hope it helps.
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Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
25 Feb 2013, 04:17
3
1
emmak wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
12
18
24
36
72
$$\angle AOB = 120^{\circ}$$
So,
$$\angle$$ ACB = $$60^{\circ}$$
So, the $$\triangle$$ ABC is a 30, 60, 90 triangle and the sides are in the ratio 1 : $$\sqrt{3}$$ : 2
or 12 : 12$$\sqrt{3}$$ : 24
So, AC = 24 and hence AO = 12
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Joined: 28 Apr 2013
Posts: 123
Location: India
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WE: Medicine and Health (Health Care)
Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
24 Nov 2013, 05:44
MacFauz wrote:
emmak wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
12
18
24
36
72
$$\angle AOB = 120^{\circ}$$
So,
$$\angle$$ ACB = $$60^{\circ}$$
So, the $$\triangle$$ ABC is a 30, 60, 90 triangle and the sides are in the ratio 1 : $$\sqrt{3}$$ : 2
or 12 : 12$$\sqrt{3}$$ : 24
So, AC = 24 and hence AO = 12
How do you that
since $$\angle AOB = 120^{\circ}$$
so $$\angle$$ ACB = $$60^{\circ}$$ ?
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GMAT 1: 670 Q39 V41
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Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
24 Nov 2013, 11:28
Bunuel wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Good question. It tests three must know properties for the GMAT:
1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.
According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.
3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.
For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html
Hope it helps.
I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: http://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem
Math Expert
Joined: 02 Sep 2009
Posts: 56260
Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
26 Nov 2013, 08:26
AccipiterQ wrote:
Bunuel wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Good question. It tests three must know properties for the GMAT:
1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.
According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.
3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.
For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html
Hope it helps.
I'm having trouble visualizing your 2nd point; could you draw a rough illustration of it? I'm familiar with the theorem, great illustration here: http://www.mathopenref.com/arccentralangletheorem.html, but I can't visualize it in this problem. I got the answer correct as the sqrt(3) tipped me off to a likely 30-60-90, but I'd like to learn a bit more from this problem
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Re: In the figure above, showing circle with center O and points [#permalink]
### Show Tags
08 Dec 2013, 23:26
1
emmak wrote:
Attachment:
1.jpg
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Guys I did this question with this approach, can you kindly evaluate the approach
$$AB= \frac{1}{3} Circumference$$
So,$$AB = \frac{2}{3}Pi* R$$
Also $$AB=12\sqrt{3}$$
Therefore, $$12\sqrt{3} = \frac{2}{3}Pi* R$$
Manipulating$$R = 12$$
Senior Manager
Joined: 13 May 2013
Posts: 414
Re: In the figure above, showing circle with center O and points [#permalink]
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11 Dec 2013, 14:37
Do we just assume that AC is the diameter?
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Re: In the figure above, showing circle with center O and points [#permalink]
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11 Dec 2013, 15:06
Ok, assuming we know this is a right triangle because we know that the picture is accurate...
If the triangle is a right triangle, we know that angle B is the right angle. Furthermore, we are given that arc AB = 1/3 of the circle's circumference (i.e it = 120 degrees) This means that AOB, the central angle = 120 degrees and because AO and BO are radii, we know that triangle AOB is an isosceles triangle. We also know the angle measure of angle BOC which = 180 - 120 = 60. If angle AOB is the interior angle, it is twice the measure of the "exterior" angle ACB, so ACB = 60. Now we know that triangle BOC is equilateral. More importantly, we know that ABC is a 30:60:90 triangle. With this, we can find the measure of any given side knowing just the measure of one. The ratio of side lengths in a 30:60:90 is (x/2):(3/2*x):(x) in this case, x (the radius) = 12.
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Re: In the figure above, showing circle with center O and points [#permalink]
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12 Dec 2013, 02:20
WholeLottaLove wrote:
Do we just assume that AC is the diameter?
We did not assume that. AC passes through the center of the circle thus it's the diameter.
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
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Re: In the figure above, showing circle with center O and points [#permalink]
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12 Dec 2013, 05:21
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?
We did not assume that. AC passes through the center of the circle thus it's the diameter.
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
I understand the rule, but in other words, how do we know that the drawing isn't misleading and that AC passes above the center?
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Re: In the figure above, showing circle with center O and points [#permalink]
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12 Dec 2013, 05:31
WholeLottaLove wrote:
Bunuel wrote:
WholeLottaLove wrote:
Do we just assume that AC is the diameter?
We did not assume that. AC passes through the center of the circle thus it's the diameter.
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
I understand the rule, but in other words, how do we know that the drawing isn't misleading and that AC passes above the center?
OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.
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Re: In the figure above, showing circle with center O and points [#permalink]
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28 Jul 2014, 21:27
Bunuel wrote:
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Good question. It tests three must know properties for the GMAT:
1. A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.
According to the above, we have that ABC is a right triangle, with a right angle at B.
2. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.
According to the above, central angle AOB is twice inscribed angle ACB (notice that both subtend the same arc AB). Hence since AOB=120 degrees (third of 360 degrees), then ACB=60 degrees. So, we have that triangle ABC is 30°-60°-90° right triangle.
3. In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
Thus, $$AB:AC=\sqrt{3}: 2$$ --> $$\frac{12\sqrt{3}}{AC}=\frac{\sqrt{3}}{2}$$ --> AC=diameter=24 --> radius=12.
For more on triangles check: math-triangles-87197.html
For more on circles check: math-circles-87957.html
Hope it helps.
Hi Bunuel
Particularly in this question with the given options , do we even need this much detail.
We know that the angle AOB = 120 degrees as it is 1/3 the circumference. AO=OB as they are the radius. So the two angles OAB and OBA will be equal in the triangle AOB. Thus Angle OBA=OBA=30 ( 180-120)/2. In a triangle the greatest side is opposite the largest angle . Here AB is opposite the largest angle and is 12sqrt2 so the other sides ( the radii) will have to be less that this value and only A satisfies the condition over here
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Re: In the figure above, showing circle with center O and points [#permalink]
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23 Feb 2016, 03:38
12$$\sqrt{3}$$ = $$\frac{1}{3}$$ * 2 * pie * R
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In the figure above, showing circle with center O and points [#permalink]
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23 Feb 2016, 03:49
emmak wrote:
Attachment:
The attachment 1.jpg is no longer available
given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ ,
does the wording simply states that the length of the arc AB is one third WITHOUT INDICATING THE EXACT VALUE, and the value of 12 root 3 is given for the side AB (rather than the arc AB)? i think the probles is poorly worded. look how it looks in the original. veritas should not have placed that bar above AB if they meant it was not the arc but the length
Attachments
2016-02-23 16-47-01 gmat.veritasprep.com gmat exams 1105058 responses 18191158 - Google Chrome.png [ 92.84 KiB | Viewed 15874 times ]
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Re: In the figure above, showing circle with center O and points [#permalink]
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31 Jul 2016, 00:24
suk1234 wrote:
emmak wrote:
Attachment:
1.jpg
In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)=$$12\sqrt{3}$$ , what is the radius of the circle?
A. 12
B. 18
C. 24
D. 36
E. 72
Guys I did this question with this approach, can you kindly evaluate the approach
$$AB= \frac{1}{3} Circumference$$
So,$$AB = \frac{2}{3}Pi* R$$
Also $$AB=12\sqrt{3}$$
Therefore, $$12\sqrt{3} = \frac{2}{3}Pi* R$$
Manipulating$$R = 12$$
AB Arc and not AB line segment is equal to 2/3 of circumference. Moreover AB arc is not 12root3, AB Line segment is.
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Re: In the figure above, showing circle with center O and points [#permalink]
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06 Jan 2019, 00:34
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Re: In the figure above, showing circle with center O and points [#permalink] 06 Jan 2019, 00:34
Display posts from previous: Sort by | 5,551 | 19,610 | {"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": 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} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.886061 |
https://www.intel.com/content/www/us/en/docs/programmable/683179/16-1-16-1/aggregate-bandwidth.html | 1,726,812,160,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00821.warc.gz | 761,020,592 | 29,775 | ID 683179
Date 7/13/2021
Public
## 3.3.1. Aggregate Bandwidth
The bit rate specifies the rate of data transmission on a single lane.
In a multi-lane configuration, the total available bandwidth is the single-lane bit rate multiplied by the number of lanes. For example, calculate the bandwidth for a variation using 8B/10B encoding and an internal data path of 8 bits (transfer size is equal to 1), and the number of lanes is equal to 4.
In this mode, the input data bus into the processor portion is 36 bits wide (32 bits of raw data and 4 bits of control information). With the additional bits per byte (due to 8B/10B encoding) for control information, the data bus size being transmitted from the byte alignment logic into the protocol-processing portion of the IP core is equal to the number of lanes × 10 (due to 8B/10B encoding). Thus for 4 lanes, the data bus size is equal to 40 bits (4×10 =40).
For example, a 32-byte packet. Count the number of 32-bit wide rows that are transmitted into the protocol-processing portion. The result is 8 rows (32 bytes/4 bytes) of solid data, plus one additional row for the start-of-packet marker row and the end-of-packet marker row (no CRC) which equals 9 rows of 40 bits.
• For a 32-byte packet, given a link rate of 800 Mbps × 4 = 3.2 Gbps, the transfer equals:
• data bits: 256
• bits sent: 360
• 256/360 × 3.2 = 2.276 Gbps
• For a 64-byte packet, given a link rate of 800 Mbps × 4 = 3.2 Gbps, the transfer equals:
• data bits: 512
• bits sent: 680
• 512/680 × 3.2 = 2.409 Gbps
• For a 128-byte packet, given a link rate of 800 Mbps × 4 = 3.2 Gbps, the transfer equals:
• data bits: 1, 024
• bits sent: 1, 320
• 1,024/1, 320 × 3.2 = 2.482 Gbps | 507 | 1,698 | {"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} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.747308 |
https://newsbasis.com/how-many-kilometres-is-in-a-decimeter/ | 1,642,973,889,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00614.warc.gz | 409,306,632 | 8,305 | # How many Kilometres is in a decimeter?
## How many Kilometres is in a decimeter?
The kilometers unit number 0.00010 km converts to 1 dm, one decimeter. It is the EQUAL length value of 1 decimeter but in the kilometers length unit alternative.
## Is km bigger or DM?
A decimeter is not greater than a kilometer. A decimeter is 10 times smaller than the base unit, meter, meaning there are 10 decimeters in 1 meter.
## What is the DM of 2km?
Please provide values below to convert kilometer [km] to decimeter [dm], or vice versa….Kilometer to Decimeter Conversion Table.
Kilometer [km] Decimeter [dm]
1 km 10000 dm
2 km 20000 dm
3 km 30000 dm
5 km 50000 dm
## What is 100 DM turned into km?
Simply put, dm is smaller than km. In fact, a decimeter is “10 to the power of -4” smaller than a kilometer. Since a decimeter is 10^-4 smaller than a kilometer, it means that the conversion factor for dm to km is 10^-4. Therefore, you can multiply 100 dm by 10^-4 to get 100 dm converted to km.
## What are Decimeters?
A decimeter is a unit of length in the metric system. The term “Deci” means one-tenth, and therefore decimetre means one-tenth of a meter. Since a meter is made up of 100 cm, one-tenth of 100 cm is 10 cm. Thus one decimeter measures 10 cm.
## How do you convert Decameters to kilometers?
1 Decameter is equal to 0.01 Kilometer….Decameter to Kilometer Conversion Chart.
Decameter Kilometer
1 dm 0.01 km
2 dm 0.02 km
3 dm 0.03 km
4 dm 0.04 km
## Is Decameter bigger than kilometer?
Remember: A meter is a little more than a yard. A kilometer is less than a mile. A liter is a little more than a quart….Length.
Unit Value
Kilometer (km) 1,000 Meters
Hectometer (hm) 100 Meters
Dekameter (dam) 10 Meters
Meter (m) 1 Meter
## Is decimeter bigger than centimeter?
Decimeter is a unit that is larger than millimeter and centimeter.
## What is a DM message?
A direct message — DM — is a one-on-one conversation with another user hosted on a social media platform. Most of the places you spend your time online — like Facebook, Twitter, Instagram, Snapchat, LinkedIn and so on — offer some form of DM communication.
## How many centimeters is 2 kilometers?
200,000 centimeters
There are 200,000 centimeters in 2 kilometers.
## What are Decimeters used for?
The decimetre (SI symbol dm) or decimeter (American spelling) is a unit of length in the metric system, equal to one tenth of a metre (the International System of Units base unit of length), ten centimetres or 3.937 inches.
## What is a decimeter example?
The definition of a decimeter is one tenth of a meter in length. An example of a decimeter is the length of about 4 inches. A metric unit of length equal to one-tenth (10-1 ) of a meter. | 752 | 2,728 | {"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} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.849901 |
https://www.physicsforums.com/threads/concept-of-dc-motor-and-a-c-generator.551775/ | 1,701,179,094,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00655.warc.gz | 1,035,211,395 | 19,081 | # Concept of dc motor and a.c. generator
• songoku
Well done. I really appreciate that you admit your mistake. We generate AC current in the coil by rotating it between the two opposite polarities of the two different cylindrical magnets or it is rotated by machinery. The other way is the changing of current continuously. But the former one is preferred.Watch this tutorial : https://www.youtube.com/watch?v=4n3x-V0vxj4
## Homework Statement
I want to ask how dc motor and ac generator work and how they are used.
dc motor is a device that converts electrical energy into mechanical energy. dc motor consists of magnet, coil, split ring commutator, brush and dc supply. dc supply will give direct current to the coil and magnetic force will act on the coil, make it rotates, let say clockwise. The split ring commutator will reverse the direction of the current round the coil every half-turn so that the coil will keep rotating clockwise.
My questions:
1. We call it dc motor because the supply used is dc? If the supply is AC then it is AC motor?
2. What is the application of dc motor? If the function is to convert electrical energy to mechanical energy, it means that we use the mechanical energy for something, what is it?
Now about ac generator. ac generator converts mechanical energy into electrical energy. ac generator consists of magnet, coil, brush, and slip rings. The coil will rotate and produce electricity.
My questions:
1. I don't understand why the coil can rotate. I suppose the concept is the same as dc motor, using magnetic field and current to produce force and then produce turning effect. But if it is true, then the coil must be connected to power supply to get current so we basically change electrical energy to mechanical energy then to electrical energy. I am sure my concept is wrong but I don't know what the mistake is.
2. ac generator used concept of electromagnetic induction to produce electricity. How does it apply the concept? My opinion is because the coil rotates, it will cut the magnetic field so emf will be induced in the coil. Slip rings will transfer the emf produced by the coil to external circuit. And because the emf induced is alternating, the current produces also alternating. Am I correct?
3. If my opinion above is correct, then the coil in dc motor will also cut the magnetic field and emf will be induced on it. So dc motor also can produce electricity?
Ahhh, I'm so confused :grumpy:
songoku said:
## Homework Statement
I want to ask how dc motor and ac generator work and how they are used.
dc motor is a device that converts electrical energy into mechanical energy. dc motor consists of magnet, coil, split ring commutator, brush and dc supply. dc supply will give direct current to the coil and magnetic force will act on the coil, make it rotates, let say clockwise. The split ring commutator will reverse the direction of the current round the coil every half-turn so that the coil will keep rotating clockwise.
You got it correct !
My questions:
1. We call it dc motor because the supply used is dc? If the supply is AC then it is AC motor?
The answer is yes. If you have not understood properly the concept of simple AC generator then you should not jump to http://en.wikipedia.org/wiki/AC_motor" [Broken].
2. What is the application of dc motor? If the function is to convert electrical energy to mechanical energy, it means that we use the mechanical energy for something, what is it?
Yes always remember that motors convert electrical energy to mechanical energy. DC motor has various applications ! Used in computer disk drives , dvd drives and fans , windmills and other various appliances which rotate.
Now about ac generator. ac generator converts mechanical energy into electrical energy. ac generator consists of magnet, coil, brush, and slip rings. The coil will rotate and produce electricity.
My questions:
1. I don't understand why the coil can rotate. I suppose the concept is the same as dc motor, using magnetic field and current to produce force and then produce turning effect.
Generator generates electricity. You rotate the coil or in big generators it is rotated by machinery. This cuts the magnetic flux and hence induces current.
But if it is true, then the coil must be connected to power supply to get current so we basically change electrical energy to mechanical energy then to electrical energy. I am sure my concept is wrong but I don't know what the mistake is.
Well done. I really appreciate that you admit your mistake. We generate AC current in the coil by rotating it between the two opposite polarities of the two different cylindrical magnets or it is rotated by machinery. The other way is the changing of current continuously. But the former one is preferred.
Watch this tutorial :
2. ac generator used concept of electromagnetic induction to produce electricity. How does it apply the concept? My opinion is because the coil rotates, it will cut the magnetic field so emf will be induced in the coil. Slip rings will transfer the emf produced by the coil to external circuit. And because the emf induced is alternating, the current produces also alternating. Am I correct?
Yes.
3. If my opinion above is correct, then the coil in dc motor will also cut the magnetic field and emf will be induced on it. So dc motor also can produce electricity?
No the current flows in the coil . The coil rotate let's assume clockwise right ? Ok so when coil rotation makes right angle it must stop (magnetic lines of two opposite polarities of the two different cylindrical magnets and coil become parallel). Due to the angular momentum gained by coil it make 180 degrees and hence the split rings reverse the direction of current and hence coil continue to rotate in clockwise direction.
Ahhh, I'm so confused :grumpy:
The concept behind the working of DC mortor and AC generator is the Fleming's left hand rule and Fleming's right hand rule. Look them up.
Is this a homework question ?
Last edited by a moderator:
sankalpmittal said:
Yes always remember that motors convert electrical energy to mechanical energy. DC motor has various applications ! Used in computer disk drives , dvd drives and fans , windmills and other various appliances which rotate.
How can it be used in windmill? It is more obvious to me that windmill is used to turn the coil in ac generator to produce electricity. Windmill can rotate because it uses wind energy. How to relate dc motor to windmill?
No the current flows in the coil . The coil rotate let's assume clockwise right ? Ok so when coil rotation makes right angle it must stop (magnetic lines of two opposite polarities of the two different cylindrical magnets and coil become parallel). Due to the angular momentum gained by coil it make 180 degrees and hence the split rings reverse the direction of current and hence coil continue to rotate in clockwise direction.
Yes, the current from the supply flows in the coil and the coil itself rotates so it cuts magnetic field. Till this point, Am I correct or wrong?
If I am correct, then emf must be induced in the coil. I also feel this is weird because the coil has already had potential difference on it because it is connected to the supply but my mind keeps thinking that the coil cuts magnetic field and another emf should be induced on it.
Is this a homework question ?
songoku said:
How can it be used in windmill? It is more obvious to me that windmill is used to turn the coil in ac generator to produce electricity. Windmill can rotate because it uses wind energy. How to relate dc motor to windmill?
I am referring to the mills which are used to remove husks of grains. Some people refer them as windmills.
And yes windmill which is used to convert wind energy to electricity works on the principle of ac generator.
Yes, the current from the supply flows in the coil and the coil itself rotates so it cuts magnetic field. Till this point, Am I correct or wrong?
Your statement related to DC motor is correct.
If I am correct, then emf must be induced in the coil. I also feel this is weird because the coil has already had potential difference on it because it is connected to the supply but my mind keeps thinking that the coil cuts magnetic field and another emf should be induced on it.
Uhh DC motor. The coil is connected to a battery. When circuit is closed DC current flows which rotates the coil because coil experiences torque or specifically couple when it is a conductor between two cylindrical magnets of opposite poles. At 0o and 180o it experiences the highest torque. Now what you say is when it rotates it cuts the magnetic flux (magnetic flux is magnetic lines of force passing through a cross sectional area). Hence emf other than of battery must be induced in coil , right ?
You are pretty correct in this. Two types of emf are developed :
1. Applied emf between two electrodes in battery.
2. Induced emf by the reason just as you said.
This induced emf developed induces current other than DC in the coil ! You know that direction of coil is changing continuously. Hence at one flick of second the induce current flow in same direction as the DC current while at another flick the induced current flow in opposite direction of DC current. You must also remember that intensity of that induced current is very less than DC current. So the net result becomes very negligible. That is why we don't consider all this.
To further prevent all this we laminate the coil in somewhat cylindrical shape to :
1. Prevent jerky motion of coil.
2. To prevent http://en.wikipedia.org/wiki/Eddy_current" [Broken].
3. To prevent energy loss due to induced current if it is necessary.
Does this make sense ? | 2,011 | 9,746 | {"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} | 3.234375 | 3 | CC-MAIN-2023-50 | latest | en | 0.930599 |
https://theepiccode.com/clipping-and-sampling-for-recurrent/ | 1,603,183,734,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107871231.19/warc/CC-MAIN-20201020080044-20201020110044-00571.warc.gz | 578,575,927 | 17,892 | # Clipping and Sampling for Recurrent Neural Networks
### Introduction
In this blog post, I’ll walk you through the simple steps of clipping and sampling for recurrent neural networks.
Gradient Clipping ensures that your gradients won’t explode, and you can converge to the optimal solution of the cost function easily. We’ll implement a straightforward clipping method by limiting the gradients between [-N, N] for an arbitrary N.
### Clipping
```def clip(gradients, maxValue):
for gradient in [dWax, dWaa, dWya, db, dby]:
gradients = {"dWaa": dWaa, "dWax": dWax, "dWya": dWya, "db": db, "dby": dby}
### Sampling
```def sample(parameters, char_to_ix, seed):
Waa, Wax, Wya, by, b = parameters['Waa'], parameters['Wax'], parameters['Wya'], parameters['by'], parameters['b']
vocab_size = by.shape[0]
n_a = Waa.shape[1]
x = np.zeros((vocab_size, 1))
a_prev = np.zeros((n_a, 1))
indices = []
idx = -1
counter = 0
newline_character = char_to_ix['\n']
while (idx != newline_character and counter != 50):
a = np.tanh(np.dot(Wax, x) + np.dot(Waa, a_prev) + b)
z = np.dot(Wya, a) + by
y = softmax(z)
idx = np.random.choice(list(range(vocab_size)), p=y.ravel())
x = np.zeros((vocab_size, 1))
x[idx] = 1
a_prev = a
counter +=1
if (counter == 50):
indices.append(char_to_ix['\n'])
return indices``` | 387 | 1,308 | {"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-2020-45 | longest | en | 0.60526 |
https://edurev.in/course/quiz/attempt/-1_Test-BJT-As-an-Amplifier-2/21019872-0613-4d56-9628-8232a990026d | 1,669,612,472,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00440.warc.gz | 272,592,272 | 42,764 | Test: BJT As an Amplifier- 2
# Test: BJT As an Amplifier- 2
Test Description
## 15 Questions MCQ Test Topicwise Question Bank for Electrical Engineering | Test: BJT As an Amplifier- 2
Test: BJT As an Amplifier- 2 for Electronics and Communication Engineering (ECE) 2022 is part of Topicwise Question Bank for Electrical Engineering preparation. The Test: BJT As an Amplifier- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: BJT As an Amplifier- 2 MCQs are made for Electronics and Communication Engineering (ECE) 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: BJT As an Amplifier- 2 below.
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Test: BJT As an Amplifier- 2 - Question 1
### The value of input impedance and voltage gain of the common base circuit shown below are respectively (Take VBE = 0.6 V)
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 1
Using AC equivalent circuit for the common base configuration, we can find that,
Now,
∴
and AV = voltage gain
Test: BJT As an Amplifier- 2 - Question 2
### The circuit shown in figure is biased such that VCE2 = 0 when VS = 0. Neglecting the base current of both transistors and forward bias across VBE for both transistors to be 0.7, the value of RE would be equal to
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 2
Applying KVL, we have:
or,
Also,
∴
∴
When, Vi = 0,
or,
Test: BJT As an Amplifier- 2 - Question 3
### A cascade stage is shown in figure below. If the transconductance of T1 and T2 are gm1 = ic1 / Vbe1 and gm2 = ic2 / Vbe2, the overall transconductance gm = ic2 / Vbe1 is
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 3
Here,
Also,
Hence, gm = gm1
Test: BJT As an Amplifier- 2 - Question 4
An emitter follower with β = 100 is biased at IC - 0.25 mA. The voltage source connected at its input has its internal resistance of 2 kΩ What is the value of RE such that it produces the output resistance of 110 Ω (Take VT = 25 mV)
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 4
or,
Now,
Desired effective output resistance
or,
or,
or,
or,
Test: BJT As an Amplifier- 2 - Question 5
Two stages of BJT amplifiers are cascaded by RC coupling. The voltage gain of the first stage is 20 and that of the second stage is 30. The overall gain of the coupled amplifier is
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 5
The voltage gain of a multi-stage amplifier is equal to the product of the gains of the individual stages.
Test: BJT As an Amplifier- 2 - Question 6
The main characteristics of a Darlington amplifier are
Test: BJT As an Amplifier- 2 - Question 7
Which of the following parameters is used for distinguishing between a small signal and a large-signal amplifier?
Test: BJT As an Amplifier- 2 - Question 8
Consider the following statements:
1. The coupling capacitor mainly affect the lower cutoff frequency.
2. The phase reversal between output and input takes place only for voltage wave and not for current waves, in a transistor amplifier in CE mode.
3. The DC collector current in a transistor circuit is limited by the junction capacitance.
4. Negative DC feedback through RE is responsible for the stabilization of the operating point in a potential divider bias circuit.
Of these statement:
Test: BJT As an Amplifier- 2 - Question 9
A transistor has hfe = 100, hte = 5.2 kΩ, and rbb, = 0. At room temperature V= 26 mV. The collector current |IC| will be
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 9
hje= rbb' + rπ
Here, rbb' = 0
Therefore,
Now,
∴
∴
= 0.5 mA
Test: BJT As an Amplifier- 2 - Question 10
A transistor is used in common emitter mode in an amplifier circuit. When a signal of 30 mV is added to the base emitter voltage, the base current changes by 30 μA and collector current by 3 mA. The load resistance is 5 kΩ. The value of transconductance and voltage gain is given by
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 10
and
Test: BJT As an Amplifier- 2 - Question 11
The transistor in the amplifier shown has the following parameters:
hfe = 100, hie = 2 kΩ, hre = 0, hoe = 0.05 m Mho.
'C’ is very large. The output impedance is
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 11
or,
Test: BJT As an Amplifier- 2 - Question 12
If RS is the source resistance, the output resistance of an emitter-follower using the simplified hybrid model would be
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 12
If RS is the source resistance, the output resistance of an emitter-follower using the simplified hybrid model would be
Test: BJT As an Amplifier- 2 - Question 13
In the CE amplifier with RL = 4000 Ω, given that The current gain |Ai| is
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 13
Current gain,
or,
Test: BJT As an Amplifier- 2 - Question 14
Assertion (A): On extending the hybrid model for two-port network to a transistor it is assumed that the signal excursion about the Q-point is small.
Reason (R): Small signal operation is that in which the AC input signal voltages and currents are in order of ±10% of Q-point voltage and currents.
Detailed Solution for Test: BJT As an Amplifier- 2 - Question 14
Both assertion and reason are correct statements.
The correct reason for assuming small signal excursion about the Q-point is to keep the transistor parameters constant over the signal excursion.
Test: BJT As an Amplifier- 2 - Question 15
For a common-collector configuration of an n-p-n transistors, the hybrid parameters are given as:
hic = hie, hrc = - 1, hfc = - (1 + hfe) and hoc = hoe.
Match List-I (Hybrid parameters of CB configuration) with List-II (Values) and select the correct answer using the codes given below the lists:
Codes:
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147 tests | 1,875 | 7,168 | {"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} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.873544 |
http://xcore.com/viewtopic.php?p=35944 | 1,638,341,757,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00137.warc.gz | 147,209,332 | 7,216 | ## How to get phase spectrum after do FFT Topic is solved
Voice related projects and technical discussions
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### How to get phase spectrum after do FFT
I use dsp_fft_bit_reverse_and_forward_real function to get the a sequence of imaginary and real numbers. Is there the function to get phase spectrum? So that I can do IFFT to get the data what I want if I do some operation in this sequence
View Solution
CousinItt
Respected Member
Posts: 275
Joined: Wed May 31, 2017 6:55 pm
You can use dsp_math_atan2_hypot() to convert each complex number to polar form.
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Posts: 26
Joined: Tue Jul 02, 2019 8:38 am
CousinItt wrote: Fri Jul 19, 2019 11:20 am You can use dsp_math_atan2_hypot() to convert each complex number to polar form.
Hi, How to do that. Actually, I don't know the algorithm very well. I find the way from internet to get the phase and want to recover signal by amplitude and phase, but when I input 1k Hz, there is 90 degree phase difference between input and output.
Code: Select all
``````
dsp_fft_bit_reverse_and_forward_real(sample,512,dsp_sine_256,dsp_sine_512);
dsp_complex_t * local = (int32_t *)sample;
for(int i = 0; i < BUFEER_SIZE; i++)
{
if(i==0)
{ p_state->fft_data0_re= local[i].re; p_state->fft_data0_im = local[i].im;} //store the local[0]
else
{
temp = dsp_math_multiply(local[i].re, local[i].re, 24)+ dsp_math_multiply(local[i].im, local[i].im, 24); // temp = sqrt(re^2 + im^2)
temp = dsp_math_sqrt(temp);
p_state->fftFrqArray[i]= temp;
p_state->angle[i] = dsp_math_atan(dsp_math_divide(local[i].im, local[i].re, 24)); // calculate the phase
}
}
for(int i =1; i< BUFEER_SIZE;i++)
{
local[i].re=dsp_math_multiply(p_state->fftFrqArray[i],dsp_math_cos(p_state->angle[i]),24); // re = temp * cos(phase)
local[i].im=dsp_math_multiply(p_state->fftFrqArray[i],dsp_math_sin(p_state->angle[i]),24); // im= temp * sin(phase)
}
dsp_fft_bit_reverse_and_inverse_real(local,512,dsp_sine_256,dsp_sine_512);
``````
Last edited by link0513 on Mon Jul 22, 2019 8:18 am, edited 1 time in total.
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Posts: 26
Joined: Tue Jul 02, 2019 8:38 am
The above code is reference matlab code
Code: Select all
``````mag = abs(F);
mag = sqrt(real(F).^2 + imag(F).^2);
phase = atan2(imag(F),real(F));
re = mag .* cos(phase);
im = mag .* sin(phase);
F = re + 1i*im;
f = ifft2(F);
``````
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Posts: 26
Joined: Tue Jul 02, 2019 8:38 am
By the way, I try to use below code to instead of p_state->angle = dsp_math_atan(dsp_math_divide(local.im, local.re, 24));, but the phase of the output signal is still wrong. Anybody know why? Please!!!
Code: Select all
`````` loca_im_re[0] = local[i].re;
loca_im_re[1] = local[i].im;
dsp_math_atan2_hypot(loca_im_re,0);
p_state->angle[i] = loca_im_re[1];
p_state->angle[i] = dsp_math_multiply(p_state->angle[i],PI_Q8_24,24); // Transfer the angle to radian. angle*pi/180 = radian
p_state->angle[i] = dsp_math_divide(p_state->angle[i],Q24(180),24);
`````` | 951 | 3,093 | {"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} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.729869 |
https://byjus.com/question-answer/if-x-y-z-are-positive-then-minimum-value-of-x-log-y-log-z/ | 1,722,771,115,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00851.warc.gz | 128,344,601 | 36,152 | 1
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Question
If x, y, z are positive then minimum value ofxlogy−logz+ylogz−logx+zlogx−logy is
A
3
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Solution
The correct option is A 3Let a=xlogy−logz,b=ylogz−logx,c=zlogx−logyNow, log(abc)=loga+logb+logc⇒log(abc)=(logy−logz)logx+(logz−logx)logy+(logx−logy)logz=0⇒abc=1As a,b,c are positive AM≥GM∴a+b+c3≥(abc)13=1∴a+b+c≥3
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Join BYJU'S Learning Program | 305 | 842 | {"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} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.75353 |
http://www.maths.kisogo.com/index.php?title=Comparison_test_for_real_series/Statement&oldid=3739 | 1,709,288,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00647.warc.gz | 48,499,917 | 7,015 | # Comparison test for real series/Statement
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
Routine, but a reference would be good
## Statement
Suppose [ilmath](a_n)_{n\in\mathbb{N} } [/ilmath] and [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] are real sequences and that we have:
1. [ilmath]\forall n\in\mathbb{N}[a_n\ge 0\wedge b_n\ge 0][/ilmath] - neither sequence is non-negative, and
2. [ilmath]\exists K\in\mathbb{N}\forall n\in\mathbb{N}[n>K\implies b_n\ge a_n][/ilmath] - i.e. that eventually [ilmath]b_n\ge a_n[/ilmath].
Then:
• if [ilmath]\sum^\infty_{n\eq 1}b_n[/ilmath] converges, so does [ilmath]\sum^\infty_{n\eq 1}a_n[/ilmath]
• if [ilmath]\sum^\infty_{n\eq 1}a_n[/ilmath] diverges so does [ilmath]\sum^\infty_{n\eq 1}b_n[/ilmath] | 368 | 1,093 | {"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} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.752346 |
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/6/lesson/6.4.1/problem/6-128 | 1,579,579,607,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601241.42/warc/CC-MAIN-20200121014531-20200121043531-00067.warc.gz | 498,973,639 | 15,941 | ### Home > CALC > Chapter 6 > Lesson 6.4.1 > Problem6-128
6-128.
1. The function y = g(x) is graphed below. Homework Help ✎
1. Find the domain and range.
2. Where is the function continuous? (Hint: It is usually easiest to start with the domain and then eliminate where the function is not continuous.)
3. Where does the function appear to be differentiable?
The domain is closed. The range is not.
One point of discontinuity is x = −1.
Recall, that a function must meet three conditions to be considered continuous at a point:
1. The limit must exist. (That means the limit from the left must agree with the limit from the right.)
2. The actual value must exist.
3. The limit must agree with the actual value.
To be differentiable:
1. Slope from the left must agree with slope from the right.
2. Differentiability implies continuity.
In other words, even if the slopes agree, a function cannot be differentiable where it is not continuous. | 227 | 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} | 3.65625 | 4 | CC-MAIN-2020-05 | longest | en | 0.911809 |
https://earth-base.org/how-to-properly-rack-8-ball-pool | 1,719,186,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00371.warc.gz | 188,226,393 | 16,010 | # How To Properly Rack 8 Ball Pool
In this post, i’ll share with you a diagram that found on pinterest on that is different from the earlier post. The top of the diamond and 1 ball should line up on the foot spot of the table.
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### These are the steps to set up 8 ball pool rack.
How to properly rack 8 ball pool. The top of the diamond and 1 ball should line up on the foot spot of the table. One important rule that will be kept in mind while racking up the 8 ball rack is that no row or column can have three same type ball adjacent to one another. How to rack 8 ball.
How to rack 8 ball. Place the racking triangle in position along the foot string of the table. So let’s get into setting this up:
How to rack 8 ball billiards. However, with gaps in the rack, that is no longer. Take a look at these awesome designs and you’ll come to the same conclusion, but don’t forget about the rest.
Center the rack along the foot string with the top of the triangle at the foot spot. Once you have got the balls tight in a perfect diamond formation, remove the rack carefully. Diagram 7 back ball gap cheat from the “box” diagram 8.
First, you need to place the balls on a table in order. Official bca rules (below) state that you arrange the remaining balls randomly. Generally, bar rules say that the rack should have a solid in front, with balls alternating between solid and striped:
Place the cue ball at one end of the table and arrange all 15 other balls from your chosen starting position (usually somewhere in front of it). Then, arrange them so that there is one striped ball and one solid ball in the corners of the last row of the rack. First, 9 ball pool only uses the first 9 balls, numbered 1 through 9.
Pool has several variations, all of which require a different style of racking the balls. Read this post if you need a review of how to properly rack a game of pool. How to properly rack 8 ball pool.
Set up all 15 balls in random order within the rack. That depends on the rules you are playing by. Align the base of the rack so it is parallel to the back of the pool table.
Like the solarium feel with glass and millwork. Learn the official method for racking pool balls for eight ball, nine ball, straight pool and ten ball. The other balls are placed randomly within the triangle.
Center the rack along the foot string with the top of the triangle at the foot spot. How to properly rack 8 ball pool. The most common rack is a triangle rack, which allows you to manipulate the balls to set up any type of game you decide to play.
These rules are the ones to memorize, 4.9 open table Ten ball is played with (you guessed it) 10 object balls, number 1 through 10, plus the cue ball. The top ball does not matter.
Center the rack along the foot string with the top of the triangle at the foot spot. Read this post if you need a review of how to properly rack a game of pool. First, 9 ball pool only uses the first 9 balls, numbered 1 through 9.
I made this video because of the mass amount of confusion around this topic and i couldn't find a good video to refer people to in order to clear things up. In ten ball, you rack the balls in a triangle shape and place the one ball at the apex of the rack on the foot spot. The ten ball is placed in the middle of the triangle.
All the other balls can be placed at random. This the center spot in the racking end of the table. Easy way to make a great modern house with swimming poolin the video, we show you how to build a modern house with a pool.
When there are no gaps in the rack, the 9ball does not move from the rack area unless it is kissed by a moving ball.
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Nine Ball Billiard Pool Table House Rules Framed Art Great Etsy Billiard Pool Table Billiards Pool Pool Table | 1,190 | 5,143 | {"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-2024-26 | latest | en | 0.918409 |
http://www.physicsforums.com/showthread.php?t=266020 | 1,386,399,873,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053669/warc/CC-MAIN-20131204131733-00008-ip-10-33-133-15.ec2.internal.warc.gz | 498,503,567 | 7,352 | # Basic differentiation questions
by computerex
Tags: basic, differentiation
P: 68 1. The problem statement, all variables and given/known data Hello guys. I had to do some test corrections for my AP Calculus AB class. I have completed all of them besides the four below. Can anyone tell me where I go wrong? 1. Differentiate y = (1+cosx)/(1-cosx) dy/dx = [(1-cosx)(sinx)-(1+cosx)(sinx)]/(1-cosx)^2 (quotient rule) = [(sinx - sinx cosx) - (sinx + sinx cosx)]/(1-cosx)^2 (distribution) = (-sinx cosx - sinx cosx)/(1-cosx)^2 (simplification) = (-2sinx cosx)/(1-cosx)^2 (simplification) Answer choices: a. -1 b. -2 cscx c. 2 cscx d. (-2sinx)/(1-cosx)^2 Choice D is the closest to my answer, however my answer multiplies cosx with the -2sinx. 2. Differentiate y = sin(x+y) I did this by implicit differentiation: y' = cos(x+y)[(x)+dy/dx] (doesn't seem correct...) Choices are: A. 0 b. [cos(x+y)]/[1-cos(x+y)] c. cos(x+y) d. 1 9. Differentiate: y = (secx)^2 + (tanx)^2 y' = [(2secx)(secx tanx)] + [(2tanx)(secx)^2] (product rule) = 2 ((secx)^2)tanx + 2 tanx (secx)^2 = 2 (sec x)^2 + 3 tanx Choices are: a. 0 b. tan x + (secx)^4 c. ((secx)^2)((secx)^2 + (tan x)^2) d. 4 (secx)^2 tanx (I skipped Pre-Calculus, which was essentially a trigonometry class, so I had a particularly difficult time with this one)
P: 364
Quote by computerex 1. Differentiate y = (1+cosx)/(1-cosx) dy/dx = [(1-cosx)(sinx)-(1+cosx)(sinx)]/(1-cosx)^2 (quotient rule) = [(sinx - sinx cosx) - (sinx + sinx cosx)]/(1-cosx)^2 (distribution) = (-sinx cosx - sinx cosx)/(1-cosx)^2 (simplification) = (-2sinx cosx)/(1-cosx)^2 (simplification)
d/dx cosx = - sinx NOT sinx
2. Differentiate y = sin(x+y) I did this by implicit differentiation: y' = cos(x+y)[(x)+dy/dx] (doesn't seem correct...)
Didn't check this one properly, but d/dx x = 1 NOT x
9. Differentiate: y = (secx)^2 + (tanx)^2 y' = [(2secx)(secx tanx)] + [(2tanx)(secx)^2] (product rule) = 2 ((secx)^2)tanx + 2 tanx (secx)^2 = 2 (sec x)^2 + 3 tanx
Check step 3.
Hope that helps!
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# How tall is a soccer ball?
Updated: 10/24/2022
Wiki User
13y ago
Best Answer
Soccer balls, in their deflated form, can range from 50 feet in diameter to 60 feet. While in their inflated form, can range from 69 feet in diameter to 169 feet.
Please call 1-800-824-4627 for more information on soccer balls and their counterparts.
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A soccer ball is a sphere. | 446 | 1,886 | {"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-2024-18 | latest | en | 0.959799 |
https://solutionlibrary.com/math/probability/determining-the-probability-from-a-given-situation-374 | 1,632,547,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00410.warc.gz | 573,690,370 | 4,003 | # Determining the probability from a given situation.
An elevator has 4 passengers and 8 floors. Find the probability that no 2 passengers get off on the same floor considering that it is equally likely that a person will get off at any floor.
© SolutionLibrary Inc. solutionlibary.com 9836dcf9d7 https://solutionlibrary.com/math/probability/determining-the-probability-from-a-given-situation-374
#### Solution Preview
...st 0, at most 4 passengers to get off. It includes the following cases.
1. 4 passengers get off at one floor. We have 8 possibilities.
2. 3 passengers get off at one floor, 1 passenger gets off at another floor. This is a permutation of 2 positions out of 8 positions. ... | 171 | 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} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.917812 |
https://nathanielwoodward.com/post/miles-and-miles/ | 1,610,940,177,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00190.warc.gz | 467,856,756 | 6,071 | # Miles and Miles!
This post considers the humble mile. You know, the unit of distance. 5,280 feet… 1,760 yards… 8 furlongs of 660 feet a piece… Have you ever thought about how nicely divisible our mile is? I have two principal goals in writing this post; they are, in order of ascending importance, (1) to talk about the history of the mile, and (2) to demonstrate an effective algorithm for finding all the factors of a number.
The USA's ubiquitous distance measure dates back to 1593, during the reign of Queen Elizabeth I, where it was formalized by an English Act of Parliament. Sure, there were things called “miles” before then: the term originates from the Latin word millia, meaning “thousand”, and has cognates in many languages (e.g., Ger. meile, Dut. mijl, Old Eng./Swed./Nor. mil, Rus. milha…). This rich provenance comes from the fact that the Roman Empire used a unit of measured called the mille passuum, Latin for “one thousand paces,” which was equivalent to about 4851 modern feet.
Hilariously, the USA changed the definition of the yard in 1893 with the Mendenhall Order to be based on metric standards instead of the customary English measurements. In 1834, when the UK Houses of Parliament were destroyed in a fire, the official “yard” and “pound” were torched withal, and the new copies made to replace them were unstable and kept changing sizes. Around the same time the International Bureau of Weights and Measures was established in France, recommending the highly stable and less arbitrary meter- and kilogram- standards. So in 1866, Congress passed a law which allowed (but did not require) the use of the metric system. Unfortunately, the definition of a yard was also changed, so that 1 yard = 3600/3937 (0.9144018288) meters, a change which differs from the international standard of 1 yard = 0.9144 meters by about 3.2 millimeters per mile. This may not sound like much—one international mile (1,609.344 km) is exactly 0.999998 of a US mile (1,609.347219 km)—but the accumulated differences can be significant.
Anyway, enough of that. Let's factor! $1 \ mile = 5280 \ feet$. Remember factor trees?
5280
/ \
3 1760
/ \
2 880
/ \
2 440
/ \
2 220
/ \
2 110
/ \
2 55
/ \
5 11
Clearly a mile can be evenly split into **halves, thirds, fourths, fifths, sixths, eighths, tenths, elevenths, twentieths…
To find ALL of the factors of a number systematically, we write the prime factorization (the bolded factors above) using exponents: $3^1 \times 2^5 \times 5^1 \times 11^1$
To find the total number of factors, just add one to each exponent and multiply them together. Thus, 5280 has $(1+1)(5+1)(1+1)(1+1)=(2)(6)(2)(2)=48$ factors! We have to add one to each exponent because we have to account for the fact that exponents of zero count too. To see this, consider that valid factors result from the following exponents: $3^{0 \ or \ 1} \times 2^{0,1,2,3,4, \ or 5} \times 5^{0 \ or \ 1} \times 11^{0 \ or \ 1}$ (e.g, $3^0 \times 2^3 \times 5^1 \times 11^0 = 40$ is a factor).
To find out all 48 distinct factors, we have to find all unique combinations of the above prime factors: we will have each of the prime factors (#1 below), all the products of two primes (#2, 4C2+1), all the products of three primes (#3), all products of four primes (#4), all products of five primes (#5), all products of six primes (#6), all products of seven primes (#7) and only one product of eight primes (#8): $2\times 2\times 2\times 2\times 2\times 3\times 5\times 11=5280$.
1. Primes $3,2,5,11$
2. Products of 2 prime factors $2^2, (2\times3), (2\times5), (2\times11), (3\times5), (3\times11), (5\times11) \\ = 4, 6, 10, 22, 15, 33, 55$
3. Products of 3 prime factors $(2^2\times3), (2^2\times5), (2^2\times11), (2\times3\times5), (2\times3\times11), (2\times5\times11), (3\times5\times11) \\ = 8, 12, 20, 44, 30, 66, 110, 165$
4. Products of 4 prime factors $2^4, (2^3\times3), (2^3\times5), (2^3\times11), (2^2\times3\times5), (2^2\times3\times11), (2^2\times5\times11), (2\times3\times5\times11) \\ = 16, 24, 40, 88, 60, 132, 220, 330$
5. Products of 5 prime factors $2^5, (2^4\times3), (2^4\times5), (2^4\times11), (2^3\times3\times5), (2^3\times3\times11), (2^3\times5\times11), (2^2\times3\times5\times11) \\ = 32, 48, 80, 176, 120, 264, 440, 660$
6. Products of 6 prime factors $(2^5\times3), (2^5\times5), (2^5\times11), (2^4\times3\times5), (2^4\times3\times11), (24\times5\times11), (2^3\times3\times5\times11) \\ = 96, 160, 352, 240, 528, 880, 1320$
7. Products of 7 prime factors $(2^5\times3\times5), (2^5\times3\times11), (2^5\times5\times11), (2^4\times3\times5\times11) \\ = 480, 1056, 1760, 2640$
8. Product of all prime factors $(2^5\times3\times5\times11) \\= 5280$
… and don't forget 1!
48 factors isn't bad at all, but in choosing the mile to have 5280 feet, it turns out we could've done better! In 1915, the mathematician Ramanujan formalized the concept of a highly composite number (HCN) as a positive integer with more divisors than any smaller positive integer. For example, if we had defined a mile to consist of 5,040 feet (a HCN), we would have 60 factors with 240 fewer feet. We would lose divisibility by 11, but we would gain divisibility by multiples of 7 and 9, perhaps more a more useful quality. Also, 5040 is a factorial ($7! = 7\times6\times5\times4\times3\times2\times1 = 5040$), equal to the number of unique permutations of four out of ten things, ${_{10} P _4} = \frac{10!}{6!}= 10\times 9\times8\times7=5040$, a superior highly composite number, and a colossally abundant number. The number 2,520 (it too has 48 factors) would've been a good choice as well. In fact, it is the smallest number divisible by all of the numbers 1-12 not including 11. Unfortunately, the smallest number that is divisible by all numbers 1-12 including 11 is $11\times2520=27,750$… not a very practical number for measurement purposes.
So, all told, the mile is pretty good! It's divisible by 48 different numbers and it has some quirky history. 5040 is a much cooler number, but oh well: the USA should be adopting the metric system soon. In fact, according to the CIA Factbook, the US is one of only three countries that has not adopted the metric system as their official system of weights and measures (Burma and Liberia are the other two).1 | 2,002 | 6,308 | {"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} | 3.03125 | 3 | CC-MAIN-2021-04 | latest | en | 0.955888 |
http://mathoverflow.net/questions/108271/spectral-radius-on-0-1-vectors/108768 | 1,469,294,147,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00206-ip-10-185-27-174.ec2.internal.warc.gz | 152,905,627 | 17,620 | # Spectral radius on 0-1 vectors.
Let $A$ be an $n\times n$ symmetric substochastic matrix (i.e. all entries are non-negative and each row adds up to $1$ or less).
Call a vector $v \in \mathbb{R}^n$ an indicator if $v \neq 0$ and each coordinate of $v$ is either $0$ or $1$. Define the indicator spectral radius of $A$ by:
$$r_I(A) = \max\left\lbrace \frac{1}{|v|^2}\langle Av,v \rangle: v\text{ is an indicator}\right\rbrace$$ where $|v|$ is the standard Euclidean norm and $\langle,\rangle$ the inner product.
If $|A|$ is the operator norm of $A$ then Cauchy-Schwarz and Jensen's inequalities imply that: $$0 \le r_I(A) \le |A| \le 1$$
Looking at the standard basis one obtains that $r_I(A) = 0$ if and only if $|A| = 0$.
Also, it's not too hard to show that $r_I(A) = 1$ if and only if $|A| = 1$.
Consider the function $f_n:[0,1] \to [0,1]$ defined by: $$f_n(x) = \max\left\lbrace |A|: A\text{ is }n\times n\text{ symmetric and substochastic with }r_I(A) \le x \right\rbrace$$
And let $f(x) = \lim_{n \to +\infty}f_n(x)$.
My questions are: Is $f(x) < 1$ whenever $x < 1$? Is $f(x) \le \sqrt{x}$?
Motivation: In the course of showing his criteria for amenability of a countable group Kesten proved ("Full Banach mean values on countable groups." 1959) that $f(x) \le O(x^{\frac{1}{3}})$ when $x \to 0$ so that in particular $f(x) < 1$ for all $x$ small enough. He claims the proof actually gives $f(x) \le O(x^{\frac{1}{2}-\epsilon})$.
I've also done some numerical experiments on the indicator norm (analogous definition) which suggests the bound $\sqrt{x}$ might work.
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The inequality $f(x)\leq \sqrt x$ does not hold exactly: one can show that $f(x)\geq \sqrt{ 2x−x^2}$ whenever $x$ is of the form $2/d$ for an integer $d$. – Mikael de la Salle Sep 29 '12 at 21:27
How does one show that? – Pablo Lessa Sep 30 '12 at 2:35
@Pablo: see my partial answer below – Mikael de la Salle Sep 30 '12 at 5:49
A remark: to prove that $f(x)<1$ for $x<1$ it is enough to prove that there exists a continuous function $g$ such that $g(t)<t$ and $r_I(A^2)\leq g(r_I(A))$ for all $A$. – Mikael de la Salle Oct 1 '12 at 8:45
Here is an example that shows that the bound $f(x) \leq \sqrt x$ does not work. More precisely $f(x) \geq \sqrt{x(2-x)}$ when $x=2/d$ for an integer $d$.
For an integer $d$, denote by $A$ the Markov operator for the uniform random walk on an infinite tree with degree $d$ (each vertex has $d$ neighbours). It is well known that $|A|=\frac{2 \sqrt{d-1}}{d}$ (I think this is a result of Kesten), and that $r_I(A)=2/d$ (for any finite subset $S$ of the tree there are at least $(d-2) card(S)$ edges between $S$ and its complementary, so that is $v$ is the indicator function of $S$, $\langle Av,v\rangle\leq 2 card(S)/d$).
One might object that I am cheating since $A$ is not a finite matrix but an infinite matrix. But suitable truncations indeed imply $f(2/d)\geq \frac{2 \sqrt{d-1}}{d}$: if $F$ is a finite subset of the tree, consider $A_F=(A_{i,j})_{i,j \in F}$. Of course $r_I(A_F) \leq r_I(A) \leq 2/d$, and if $F$ is large enough $|A_F|$ is arbitrarily close to $|A|$.
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Great example! Thanks! – Pablo Lessa Sep 30 '12 at 9:55
I don't understand what this matrix A is. Is it even symmetric? – Luis Silvestre Oct 4 '12 at 18:08
If $F$ is a finite subset of the set of vertices of an infinite tree with degree $d$, then $A_F$ is a matrix indexed by $F$, and for $i,j \in F$ the $(i,j)$-entry of $A_F$ is $1/d$ if there is an edge between $i$ and $j$, and $0$ otherwise. It is indeed symmetric and substochastic since any vertex has at most $d$ neighbours in $F$. – Mikael de la Salle Oct 4 '12 at 18:46
With rank one matrices (I mean those of the form $A = \frac {1}{(\max w) \sum w_i}w \otimes w$ for some vector $w$) here is a proof that says $f(r) \leq 2 \sqrt{r}$ if we restrict to that kind of matrices.
Let us assume wlg that $1=w_1\geq w_2 \geq w_3 \geq \dots$ (we can do this by rearranging the indices and dividing by $w_1$). Then we can obtain $$\|A\| = \frac{\sum w_i^2}{\sum w_i}$$ and $$r:=r_I(A) = \max_m \frac{\left(\sum_{i\leq m} w_i \right)^2} {m \sum w_i}.$$ Let $k$ be the largest integer so that $w_i \geq \sqrt{r}$, then by definition of $r$ we have that $$r \geq \frac{\left(\sum_{i \leq k} w_i \right)^2} {k \sum w_i}.$$ Also $$\|A\| \leq \frac{\sum_{i \leq k} w_i^2}{\sum w_i} + \frac{\sum_{i > k} w_i^2}{\sum w_i} \leq \frac{\sum_{i \leq k} w_i}{\sum w_i} + \sqrt r \frac{\sum_{i > k} w_i}{\sum w_i} \leq \frac{\sum_{i \leq k} w_i}{\sum w_i} + \sqrt r$$ And using that $v_i \leq 1$ and $\sum_{i\leq k} w_i \geq k w_k \geq k \sqrt r$, we get $$\|A\| \leq \frac{\left(\sum_{i \leq k} w_i\right)^2}{k \sqrt r \sum w_i} + \sqrt r \leq 2 \sqrt{r}$$
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I thought a bit about a proof for general matrices as in the original question. Starting from $Av = \|A\|v$ for some $v$ such that $1=v_0 \geq v_1 \geq \dots$. I ended up in a sum of two terms similar as above but there was one that I could not estimate appropriately. – Luis Silvestre Oct 4 '12 at 0:55
I thought I had some elementary example that $f(r) \geq \sqrt r$ with rank one matrices, but there was some silly mistake. I'll try again tomorrow. – Luis Silvestre Oct 4 '12 at 5:52
Excellent! Maybe this can help solve the general case by composing the original matrix $A$ with the orthogonal projection onto the subspace generated by the vector with $Av = |A|v$. Symmetry is preserved by this composition but now the sums of rows might be a bit larger than $1$. – Pablo Lessa Oct 4 '12 at 13:24 | 1,925 | 5,507 | {"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} | 3.046875 | 3 | CC-MAIN-2016-30 | latest | en | 0.745937 |
https://physics.stackexchange.com/questions/419237/what-happens-to-the-u1-b-u1-y2-anomaly-in-the-standard-model?noredirect=1 | 1,563,701,732,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526940.0/warc/CC-MAIN-20190721082354-20190721104354-00232.warc.gz | 509,097,934 | 35,772 | What happens to the $U(1)_B U(1)_Y^2$ anomaly in the Standard Model?
Baryon number $U(1)_B$ is anomalous in the Standard Model, as can be seen by computing a $U(1)_B SU(2)_L^2$ triangle diagram. This implies that $$\partial_\mu J^{\mu B} \sim W_{\mu\nu} \tilde{W}^{\mu\nu}$$ where $W_{\mu\nu}$ is the $SU(2)_L$ field strength, which allows the nonconservation of baryon number by topologically nontrivial field configurations.
According to Schwartz's QFT textbook, all other contributions to the $U(1)_B$ anomaly vanish, but I can't see why that is. In the case of $U(1)_B U(1)_Y^2$, we should have a contribution proportional to $$\sum_{\text{LH quarks}} Y_i^2 - \sum_{\text{RH quarks}} Y_i^2 \propto 2 \left( \frac16 \right)^2 - \left(\frac23 \right)^2 - \left(-\frac13\right)^2 \neq 0.$$ What am I doing wrong in this computation?
• Related and links thereto ... There are no U(1) instantons.... – Cosmas Zachos Jul 24 '18 at 16:06
Actually, there's nothing wrong with this computation; Schwartz's statement is simply incorrect. There is indeed a $U(1)_B U(1)_Y^2$ anomaly, which implies $$\partial_\mu J^{\mu B} \sim W_{\mu\nu} \tilde{W}^{\mu\nu} + B_{\mu\nu} \tilde{B}^{\mu\nu}$$ where $B_{\mu\nu}$ is the $U(1)_Y$ field strength. The reason this second term is rarely mentioned is that there are no $U(1)_Y$ instantons. That is, while there can be local violations of baryon number conservation, you can't get a global violation of baryon number from $B_{\mu\nu}$ configurations. | 493 | 1,489 | {"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} | 2.625 | 3 | CC-MAIN-2019-30 | latest | en | 0.839762 |
https://www.hpmuseum.org/forum/archive/index.php?thread-20472.html | 1,713,062,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00877.warc.gz | 744,385,354 | 2,340 | # HP Forums
Full Version: Problem with square brackets in fSolve
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I wanted to solve a system of 2 equations with X and Y as unknowns. Whenever I opened the square brackets a +/- sign was included that could not be removed. So it looked different to the examples in the manual.
So I went to CAS menue (not in CAS mode) and Numerical Solve so there was CAS.fsolve() on the screen. Then I pressed Help.
There were two examples. The second example was
fsolve([x²+y-2,x+y²-2],[x,y],[0,0])
I pressed Example and highlighted the second example. When I copied it I got:
CAS.fsolve([x²+y-2 x+y²-2 +/-],[x y +/-],[0 0 +/-])
Pressing Enter gives message: Error: Syntax Error
It seems that the elements in the square brackets are interpreted as elements of a matrix. How can I use the fSove function for more than one equation?
Thank you very much for your help!
(09-08-2023 08:56 AM)rawi Wrote: [ -> ]How can I use the fSolve function for more than one equation?
The simplest way is to use fsolve in CAS mode.
If you want to use it in HOME , then CAS.fsolve("[x²+y-2,x+y²-2],[x,y],[0,0]") or CAS("fsolve([x^2+y-2,x+y^2-2],[x,y],[0,0])") will work.
Hi Didier,
thank you very much for your fast and helpful answer. In CAS mode it works fine. I tried in home mode and there I got again the message "Error: Syntax Error".
In CAS mode the +/- before the closing of the square bracket that is generated as soon a square bracket is opened disappear when pressing Enter. In home mode it does not disappear.
And I think it strange that this is a problem to be solved in CAS mode as it is a strictly numeric problem.
But I am glad that I now know how to do it. Thank you once more!
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• : | 487 | 1,852 | {"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-2024-18 | latest | en | 0.913163 |
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KeynesianCompStat
# KeynesianCompStat - interest rates 4 Extension Extension...
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Comparative Static Analysis Comparative Static Analysis of the Keynesian Model of the Keynesian Model Macroeconomics I ECON 309 -- Cunningham
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2 Simple IS-LM Analysis Simple IS-LM Analysis 0 ) , ( 0 ) ( ) ( = - = - - P M r Y L G r I Y S Two equations, two endogenous variables ( Y and r ), and one exogenous variable G . Real money supply ( M/ P) is taken as constant since nominal money ( M ) and ( P ) are exogenous as well. Take total differentials: 0 = + = - dr L dY L dG dr I dY S r Y r Y = - 0 dG dr dY L L I S r Y r Y Write in matrix form:
3 Simple IS-LM, Continued Simple IS-LM, Continued 0 0 1 0 0 1 + - = - = + = - - = Y r r Y Y r Y r Y Y Y Y r r Y r r Y r Y r r L I L S L L L I S L S dG dr L I L S L L L I S L I dG dY Applying Cramer’s rule for solution: 0 , 0 0 , 0 < < r Y Y r I S L L Because, by assumption, the following hold: So, in a Keynesian economy, under the conditions given, cet. par. (i.e., prices), an increase in government spending increases GDP and
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Unformatted text preview: interest rates. 4 Extension Extension What if prices are flexible? To examine this, we must include the labor market and real wage computation. ) , ( ) ( ) ( ) (--=--=-=- P M r Y L G r I Y S N F Y N P w N d Define the following variable as a convenience: ( 29 2 -∂ ∂ = P w P w N X d 5 Extension (Continued) Extension (Continued) 1 1 1 1 2 2-=------= Jac X F L P M L L I S F X P M L I F X dG dY N r r Y r Y N r r N (Note that the denominator turns out to be positive.) < -= dG P w d dG dP dG dr Jac X L dG dN r Similarly:...
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https://www.convertit.com/Go/WaveQuest/Measurement/Converter.ASP?From=nebuchadnezzar&To=capacity | 1,624,534,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00569.warc.gz | 622,862,429 | 3,658 | Riding the internet wave!
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```nebuchadnezzar = 0.01892705892 volume (volume) ``` Related Measurements: Try converting from "nebuchadnezzar" to acre foot, acre-foot, bath (Israeli bath), beer gallon (English beer gallon), board foot, bushel (dry bushel), dram fluid (fluid dram), drop, dry quart, freight ton, last, load, methuselah, noggin, oil arroba (Spanish oil arroba), pint (fluid pint), salmanazar, strike, tou (Chinese tou), UK quart (British quart), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: nebuchadnezzar = .00001534 acre foot, .15873016 barrel, 513.33 bath (Israeli bath), 4.1 beer gallon (English beer gallon), 18,927.06 cc (cubic centimeters), .09011818 chetvert (Russian chetvert), .13427609 coomb, .0052219 cord (of wood), 17.19 dry quart, .47742612 ephah (Israeli ephah), .55555556 firkin, 160 gill, .07936508 hogshead, .10504202 koku (Japanese koku), 10 magnum, 2.5 methuselah, 40 pint (fluid pint), 1.67 salmanazar, 2.08 UK peck (British peck), .01342761 wey.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 482 | 1,693 | {"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-2021-25 | latest | en | 0.642973 |
https://discuss.fogcreek.com/techinterview/2734.html | 1,579,661,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00079.warc.gz | 408,585,027 | 2,980 | bumblebee int marchDistance = 0 int time; do { 1000*time + 100*time = 200 - marchDistance; marchDistance = time*200; } while (marchDistance < 200); Ning Friday, January 28, 2005 Is this supposed to compile? You have no l-value. Brian Monday, January 31, 2005 Dude, the variables are not even initialized, what kind of crappy question is this supposed to be? Jose Monday, January 31, 2005 I guess the simplest solution is the given one, but I came up with the following one using a serie (I'm not that good in english, I don't know if it's the correct word): X=2000/9*Σ(9/11)^n, with n integer from 1 to infinite. Each n gives the space ran by the bee in each round (for n=1, the first round= 2000/11 miles until it first meets the opposite train) I don't think there's a simplification of it, so try it in the excel. Obviously, it's the dull solution, but I find the series so intriguing. It would also be nice for the programmers, there must be an easy way to visualise this in fortran or another mathematical language (matlab maybe). I've nothing to do with informatics, so I leave it to you. Panagiotis Apostolopoulos Monday, April 18, 2005 The *Σ is the Sigma symbol of series (!!!). My native tongue is greek but the computer doesn't seem to understand it... Panagiotis Apostolopoulos Monday, April 18, 2005 Fog Creek Home | 362 | 1,338 | {"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-2020-05 | latest | en | 0.942697 |
http://www.omatrix.com/stsaSamples/example01src.htm | 1,500,691,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423842.79/warc/CC-MAIN-20170722022441-20170722042441-00005.warc.gz | 505,210,909 | 4,803 | #
# STATISTICAL TIME SERIES ANALYSIS (STSA) module for O-Matrix.
#
# Written by Dr. Dimitrios D. Thomakos
# University of Peloponesse
# e-mail: thomakos@uop.gr
#
# STSA Version 2.0
#
# Example file #01
#
# In this example we simulate a stationary, zero mean Gaussian autoregressive model of order 2
# and perform various operations on the resulting realization.
#
# 1. Plot the realization.
# 2. Estimate and plot the autocorrelation and autocovariance functions.
# 3. Estimate and plot the Fourier spectrum by smoothing the periodogram and using
# and autoregressive approximation.
# 4. Estimate and plot the empirical distribution function of the realization and
# perform two tests for Gaussianity: the Kolmogorov-Smirnov test and the test
# based on the QQ (quantile-quantile) correlation coefficient.
# 5. Estimate the parameters of the 'true' order (=2) and the order of an autoregressive model
# selected by order selection criteria.
# 6. Perform a likelihood ratio test for the statistical significance of the extra estimated
# coefficients.
# 7. Compute the roots of the associated autoregressive polynomial of the model selected
# using the order selection criteria.
# 8. Compute and plot the impulse responses of the model, i.e. the associated MA coefficients
# in the infinite MA representation of the AR model.
#
# Clear workspace and command window, initialize graphics, seed the random number generator
clear
ginit
clc
# Simulate a realization from a stationary Gaussian ARMA(2,0) process
# Note: you can change the values of the parameters below
mu = 0. # zero mean
phi = { 1.25, -0.75 } # autoregressive coefficients
theta = 0. # moving average coefficients
sigma = 0.2 # innovation std. deviation
N = 500. # length of realization
# Function call, put the realization in the (N * 1) vector x
x = arma_simulate(mu,phi,theta,sigma,N)
# Plot the realization
gaddwin("Sample realization from a Gaussian ARMA(2,0) process")
font_type = "Arial"
font_size = 8
gtitlefont(font_type,font_size)
gxtickfont(font_type,font_size)
gytickfont(font_type,font_size)
gxtitlefont(font_type,font_size)
gtitle("The data generating process is x(t) = 1.25*x(t-1) - 0.75*x(t-2) + e(t), e(t) ~ N(0,0.2)")
gxaxis("linear",0,N,4)
gxtitle("Time")
gcolor({"black","green"})
gplot([x,zeros(N)])
# Plot and save the autocovariance and autocorrelation functions
h = 20 # lag length
alpha = 0.05 # level of significance for std. error bars
varn = "x" # variable name
[kx,rx,se] = acfplot(x,h,alpha,varn)
# Estimate and plot the spectrum
# (a) by smoothing the periodogram using a sine taper
sm = 0. # flag; do not substract the sample mean
K = 20. # number of sine tapers to use
[f1,S1] = spectrum(x,sm,K)
# (b) the spectrum using an autoregressive approximation
p = 2. # "true" autoregressive order; you can change this
sm = 0. # flag; do not substract the sample mean
[f2,S2] = ar_spectrum(x,p,sm)
gtitle("Green = smoothed periodogram, red = AR approximation")
gxtitle("Frequency")
gxaxis("linear",-0.5,0.5,10)
scale = "log" # select y-axis scale: "linear" or "log"
gyaxis(scale)
gcolor({"green","red"})
gplot(f1,S1)
gplot(f2,S2)
# Examine the Gaussianity of the realization using a QQ plot and the
# Kolmogorov-Smirnov test; test values and pvalues appear in the titles of plots
caption = "QQ Plot for sample realization of a Gaussian ARMA(2,0) process"
plotq = 1. # flag; create the plot
[rQ,sl] = qqplot(x,caption,plotq)
caption = "Empirical and Normal CDF for sample realization of a Gaussian ARMA(2,0) process"
[xvalues,z,w,tests] = Gaussianity_KStest(x,caption,plotq)
# Retrieve the underlying innovation sequence of the process
e = ar_noise(x,phi)
gaddwin("Innovations of sample realization of a Gaussian ARMA(2,0) process")
gplot(e)
# Estimate the coefficients of the realization using the Yule-Walker equations
p = 2.
[phi,s2_H0] = ar_yw(x,p)
print "Yule-Walker coefficient estimates = ", phi'
print "Innovation variance = ", s2_H0
print
# Estimate the autoregressive model order using order selection criteria
maxp = 10. # Max lag to consider
sm = 0. # flag; do not substract the sample mean
method = 0. # flag; estimation method - use Yule-Walker
[aic,bic] = ar_order(x,maxp,sm,method)
print "AIC order selected = ", aic(1,1)
print "BIC order selected = ", bic(1,1)
print "True order = ", 2
print
print "Note the PACF plot as well!!!"
print
# Use the selected AIC order to estimate the coefficients using Yule-Walker
paic = aic(1,1)
[phi,s2_HA] = ar_yw(x,paic)
print "Yule-Walker coefficient estimates using the AIC order = ", phi'
print "Innovation variance = ", s2_HA
# If paic > p then use a Likelihood Ratio test to examine the significance of the
# extra coefficients
if paic > p then begin
# Compute the max of the log-likelihood function under
# the null and under the alternative hypothesis
loglf_H0 = -0.5*N*log(2*pi+1) - 0.5*N*log(s2_H0)
loglf_HA = -0.5*N*log(2*pi+1) - 0.5*N*log(s2_HA)
lr_test = 2*(loglf_HA - loglf_H0)
print "LR test for the significance of the extra coefficients = ", lr_test
print "P-value = ", cdfchi(lr_test,paic-p)
print
end
# Compute the roots of the autoregressive polynomial of the estimated AIC model
[ar,ma] = arma_roots(paic,0,phi)
print "Estimated abs. values of roots of autoregressive polynomial of AIC model = ", abs(ar)'
print "(Roots should be greater than one for stationarity)"
print
# Compute and plot the first 30 impulse response coefficients (infinite MA representation) of the estimated AIC model
psi = ar_to_ma(phi,30)
gaddwin("First 30 impulse response coefficients of AIC model")
gtitle("Coefficients in infinite MA representation")
gxtitle("Lag")
gxaxis("linear",0,30,30)
gstyle("diamond")
gcolor("black")
gplot(psi)
gstyle("solid")
gcolor("green")
gplot(psi) | 1,699 | 6,113 | {"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-2017-30 | longest | en | 0.671539 |
https://www.unitconverters.net/length/mile-us-survey-to-aln.htm | 1,575,846,146,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540515344.59/warc/CC-MAIN-20191208230118-20191209014118-00153.warc.gz | 914,243,909 | 4,607 | Home / Length Conversion / Convert Mile (US Survey) to Aln
# Convert Mile (US Survey) to Aln
Please provide values below to convert mile (US survey) [mi] to aln, or vice versa.
From: mile (US survey) To: aln
### Mile (US Survey) to Aln Conversion Table
Mile (US Survey) [mi]Aln
0.01 mi27.1035272609 aln
0.1 mi271.0352726095 aln
1 mi2710.3527260947 aln
2 mi5420.7054521894 aln
3 mi8131.058178284 aln
5 mi13551.763630473 aln
10 mi27103.527260947 aln
20 mi54207.054521894 aln
50 mi135517.63630473 aln
100 mi271035.27260947 aln
1000 mi2710352.7260947 aln
### How to Convert Mile (US Survey) to Aln
1 mi = 2710.3527260947 aln
1 aln = 0.0003689557 mi
Example: convert 15 mi to aln:
15 mi = 15 × 2710.3527260947 aln = 40655.29089142 aln | 290 | 738 | {"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-2019-51 | longest | en | 0.500082 |
www.mrscheckmate.com | 1,726,218,844,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00491.warc.gz | 825,823,431 | 24,599 | # How Many Pieces are in a Chess Set? (Explained)
A chess set comes with several different pieces, all of which have a particular purpose in a chess game. The number of pieces in a chess set varies slightly depending on the manufacturer and what they include in a set.
## How Many Pieces in a Chess Set?
The Chessboard: This is where you set up the pieces. Sometimes it also doubles as the container of your pieces.
The Chessmen: These are also called chess pieces. These are the main components of any chess set but may also be sold separately from the board. A standard set of chess pieces comes with 32 chessmen, 16 for each color, but some sets come with 34 pieces.
The two additional pieces are usually extra queens if the most important piece of the set is lost or misplaced. When the number of chess pieces varies, it’s usually because of extra pieces.
Now the number of pieces found in a chess set is cleared up, and the extra pieces have come to light.
It is now time to learn more about the different pieces and their roles in the game. Although there are 16 pieces on each side of the board, there are only 6 kinds with different roles and movements.
### How Many Pawns are There in Chess?
There are 8 pawns, which account for half of all the chess pieces on the board. They are placed on the second row and are only allowed to move one square at a time, except on their very first move, where they can move two squares.
To beat an opponent’s piece using the pawn, they must be diagonally positioned from the pawn.
Although limited in their movement, Pawns can promote to another piece when they reach the other end of the board. They can turn into a Bishop, Rook, Knight, or Queen. There are no limits to how many pieces are allowed a promotion.
### How Many Rooks are There in Chess?
Each player has 2 castles (rooks). These pieces can move as many squares as possible, providing they stay in a straight line. They can move either horizontally or vertically as long as no piece of the same color is blocking their way.
### How Many Knights are There in Chess?
There are also 2 knights per color. These are the only pieces allowed to leap over other pieces, but they are severely limited as they can only move in an “L” shape.
### How Many Bishops are There in Chess?
Again, each player has 2 bishops in their set. These pieces can only move diagonally but aren’t limited to how many squares they move forward.
Only when a piece of the same color blocks their path do, they have to stop at a particular square.
### How Many Queens are in Chess Set?
There is only one queen of each color in every set. Although a set may have two extra queens (1 different for each color), they cannot be used on the chessboard during the game at the same time.
The queen is the most powerful piece within the set because she can move in any direction and for as many squares as possible as long as no other piece is blocking her path.
Losing your queen does not end the game, but it does put you at a severe disadvantage if your opponent still has their queen and you do not. The best way to regain all the queen’s power is to get a pawn promoted and turn them into a queen.
### How Many Kings are There in a Chess Set?
Every set only has one king of each color, and a “Check Mate” is how a chess game ends in victory. When your opponent calls “Check,” they simply warn you that your king is in danger and you need to block the path or move the king.
Kings are limited because they can only move one square at a time, but they can move in any direction as long as nothing is blocking their way.
## Conclusion
The number of pieces in a chess set depends on the manufacturer. Typically, a set will have 32 or 34 if an extra queen is added.
If you count the board as part of your chess set, you will have 33 or 35 pieces. Usually, though, you’ll find that most chess sets don’t come with spare pieces, so make sure that you take extra care not to lose any of them. | 885 | 3,984 | {"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-2024-38 | latest | en | 0.970575 |
https://www.varsitytutors.com/lsat_logic_games_19-problem-18996 | 1,643,347,652,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00236.warc.gz | 1,093,173,064 | 31,727 | ### Test: LSAT Logic Games
Eight people are waiting for five buses at a bus stop. The eight people are Adrien, Brian, Carl, David, Eva, Faith, Glenda, and Henry. At least one person must get on each bus and everyone at the stop gets on one bus. The following conditions apply:
No more than two people get on any bus.
If two people get on the first bus, two people must get on the third bus.
Glenda gets on a bus alone.
Adrien must get on a bus with another person.
Only one person gets on the fourth bus.
David and Faith cannot get on the same bus.
1 Which of the following could be a list of the people who get on each bus?
1. Glenda; 2. David and Faith; 3. Adrien and Eva; 4. Carl; 5. Brian and Henry
1. Glenda and David; 2. Faith; 3. Adrien and Eva; 4. Henry; 5. Brian and Carl
1. Adrien and Henry; 2. Glenda; 3. Eva and Faith 4. David and Carl; 5. Brian
1. Adrien and Eva; 2. Faith and Henry; 3. Glenda 4. David; 5. Brian and Carl
1. Glenda; 2. David and Brian; 3. Adrien and Eva; 4. Henry; 5. Faith and Carl
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Take 15 seconds to create an account. | 351 | 1,196 | {"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-2022-05 | latest | en | 0.865639 |
https://www.physicsforums.com/threads/simple-way-to-factor-sextic.425104/ | 1,521,509,074,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647244.44/warc/CC-MAIN-20180319234034-20180320014034-00704.warc.gz | 867,984,658 | 14,438 | # Simple way to factor sextic
1. Aug 29, 2010
### elfboy
Anyone know of a good way to factor
x^6+x^4(13^.5/2-13/2)+x^2(13-2*13^.5)+3*13^.5/2-13/2 into 2 cubics?
without having to manipulate the messy cubic equation?
2. Aug 29, 2010
### epenguin
These things are intended as exercises in recognising patterns of which you have already dealt with simpler examples, woven into something more elaborate.
The rules are you are supposed to show an effort. Try and deal with different bits of it. For instance can you not do something with the part that has as factor a straight 13? The rest of it looks more difficult. I would ask you check whether you have really transcribed it exactly right. These things (I cannot imagine it is anything but an artificially constructed problem) are meant to work out to something that rhymes with sense.
Last edited: Aug 30, 2010
3. Aug 29, 2010
### elfboy
The equation looks arbitrary but its roots are plus minus sin(pi/13), sin(3pi/13), and sin(4pi/13)
4. Sep 27, 2010
### elfboy
After a month of working on this problem on and off I finally factored it in the way desired
(x^3+bx^2+c*x+(-3*13^.5/2+13/2)^.5)*(x^3-bx^2+c*x-(-3*13^.5/2+13/2)^.5)
13^.5/2-13/2=-b^2+2c
13-2*13^.5=c^2-b(26-6*13^.5)^.5
solve for b and c and plug into the above expression to obtain the two cubics:
(x^3+(13/2+3*13^.5/2)^.5*x^2+13^.5*x+(-3*13^.5/2+13/2)^.5)*
(x^3-(13/2+3*13^.5/2)^.5*x^2+13^.5*x-(-3*13^.5/2+13/2)^.5)=0
i didn't think this was possible to get a nice expression but i've done it | 523 | 1,526 | {"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} | 3.96875 | 4 | CC-MAIN-2018-13 | longest | en | 0.914327 |
http://clay6.com/qa/20648/the-line-overrightarrow-r-hat-i-2-hat-j-hat-k-lambda-2-hat-i-hat-k-is-perp- | 1,516,749,316,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892802.73/warc/CC-MAIN-20180123231023-20180124011023-00772.warc.gz | 77,635,715 | 28,039 | # The line $\overrightarrow r=\hat i+2\hat j-\hat k+\lambda(2\hat i-\hat k)$ is $\perp$ to which axis?
$\begin {array} {1 1} (A)\;X-axis & \quad (B)\;Y-axis \\ (C)\;Z-axis & \quad (D)\;\text{ None of the axes} \end {array}$
Toolbox:
• Two lines are $\perp$ if the dot product of their D.R. is =0
Direction ratio (D.R.) of the line is $\overrightarrow b=(1,0,-1)$
D.R. of $Y-axis$ is $(0,1,0)$.
$\therefore$ The line is $\perp$ to $Y-axis$ since $(1,0,-1).(0,1,0)=0$
answered Dec 18, 2013 | 194 | 489 | {"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} | 3.65625 | 4 | CC-MAIN-2018-05 | longest | en | 0.738277 |
http://marginalrevolution.com/marginalrevolution/2010/06/optimizing-kidney-allocation.html | 1,506,132,296,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689413.52/warc/CC-MAIN-20170923014525-20170923034525-00496.warc.gz | 221,276,394 | 17,164 | # Optimizing Kidney Allocation: LYFT for LIFE
on June 23, 2010 at 7:31 am
Under the current system, kidneys are allocated to patients primarily based on the time that the patient has been on the waiting list and the quality of the match. If we evaluate these criteria "locally" there's nothing obviously wrong but if we step back and think globally, that is think about what the ultimate goal of the transplant system should be, then the current system is deeply misguided. Suppose that we want the transplant system to maximize total life expectancy or, as it is known in the literature, to maximize the life-years from transplant (LYFT).
The current system does not maximize life expectancy. In the current system, a 60 year old patient can be given a 20 year old kidney–that's a waste because the life expectancy of the kidney is longer than that of the patient; it's like putting a new clutch in a car that is rusting away. If we had 20 year-old kidneys to spare, this wouldn't be a big problem. But we don't have 20-year old kidneys to spare, so we also give 20-year old patients 60-year old kidneys which means the kidney is likely to die early taking the patient along with it. If we want to maximize total life expectancy, younger people should get younger kidneys.
Here is a simple example to illustrate the principle. Suppose that the life expectancy of both patients and kidneys is 75 years of age so everyone dies when they are 75 or when their kidney is 75, whichever comes first. Thus, if we allocate the 20 year old kidney to the 60 year old patient and vice-versa we gain a total of 30 years of life expectancy.
Kidney Age Patient Age Life Years 20 60 15 60 20 15 30 years Total
But if we allocate the 60 year old kidney to the 60 year old patient and the 20 year old kidney to the 20 year old patient we more than double life expectancy to 70 years in total.
Kidney Age Patient Age Life Years 60 60 15 20 20 55 70 years Total
It's not just age that matters, it turns out that the longer a patient has been on dialysis the less is their life expectancy after transplant (dialysis stresses the body so the sooner we get someone a transplant the better). Although it may seem unfair, if we want to maximize total life expectancy we are doing the wrong thing by giving more points to patients who have been on the list longer.
An optimized allocation system that took into account these considerations would increase total life
expectancy (modestly but significantly, about 11,500 extra life years) but it wouldn't benefit every individual. Maximizing life expectancy would shift organs away from older people and people who have been on the waiting list a long time towards younger people. As a result, some patients have argued that LYFT is unfair. The Office of Civil Rights is even asking whether LYFT might violate age discrimination laws.
But consider, would the older patients have objected to LYFT when they were younger? If not, shouldn't their objections be discounted? More formally, consider how people would vote behind a veil of ignorance. By definition a LYFT approach maximizes total life expectancy, so without knowing the specifics of who you are or when you might need a transplant it's likely that behind a veil of ignorance just about everyone would favor LYFT. Thus, in my view LYFT is a fair and ethical system.
Here are previous MR posts on kidney transplant policy.
1 dearieme June 23, 2010 at 7:25 am
The codgers are the people who have put money into the pot to pay for the whole system, the youngsters are people who have, as yet, only drawn out of the pot. Treat the codgers say I, albeit not wastefully.
2 nate June 23, 2010 at 8:36 am
cue the obligatory slippery slope argument: should then all scarce health care resources be allocated according to age?
3 cupitor June 23, 2010 at 8:55 am
I’m not a physician and not an expert on these matters, but this is assuming that the body will accept the kidney as if it was the original kidney, with no problems over the long term. Don’t kidneys transplanted have a different lifespan in the body of a recipient than the ones we are born with? i.e., a kidney from someone who donated at death is thought to last shorter in the body of a recipient than that of a living person who is not a family member, and that one lasts less, on average, than that of a sibling.
What that seems to indicate is that once transplanted, even if the transplant is successful and the body accepts the transplanted kidney well, over the long term, there might still be certain level of gradual “rejection” or reaction of the body to the foreign organ that makes the life expectancy of such organ be less than that of a healthy organ with which the body was born. If that is the case, wouldn’t that alter the numbers, meaning on average the life expectancy of the transplanted kidney on a recipient body might be less — and perhaps considerably less — than that of human life expectancy?
4 Peter June 23, 2010 at 9:20 am
Another idea: more medical research to figure out ways to stop so many kidneys from failing in the first place.
5 Andrew June 23, 2010 at 9:43 am
“The codgers are the people who have put money into the pot to pay for the whole system,”
Old people invented kidneys?
6 Andrew June 23, 2010 at 10:20 am
anon and dale,
It doesn’t matter. The important point is that resources are allocated the traditional Communist Russia way, waiting in lines. The point is to not do it that way.
dearieme brings up a good point that I rarely ever see discussed, that of generational finance, and you can thank the welfare state for the upcoming reduction in loyalty to oldsters. What does the younger generation actually owe the oldsters? I think it’s a great debate, but irrelevant here. The medical industrial complex simply acts as a gatekeeper and rent seeker on the kidney market.
7 akatsuki June 23, 2010 at 10:34 am
Or we could just move to opt-out instead of opt-in and pretty much take care of most organ shortages.
The real problem with most charity is that our refusal to put prices on life make our spending wildly inconsistent – we leave low hanging fruit all the time…
8 Gannon June 23, 2010 at 11:07 am
I don’t want to talk about kidneys but about a broader topic, generational finance or generational solidarity like the Germans would call it. The question is if aging societies will break the back of the young. In Germany, pensioners and retiress live relatively well compared to young unskilled workers and children. Whereas the old people in Germany enjoy their lush pensions, the schools are crumbling and something like 50% of children live in poverty, compared to only 10% of retiress. Like I already said, retiress enjoy higher incomes than a lot of workers in Germany. Because retirees are increasing in numbers and tend to vote, they enjoy a lot of entitlements, paid by the young. But retirees are increasing, so this system can’t be maintained. Some young Germans even can’t afford children because they pay so much taxes to the elderly. In the USA the situation isn’t so different however: the babyboomers with their huge entitlements are already breaking the back of California. This problem can not be solved through inmigration of low skilled and low educated workers. These workers only aggravate the problems.
9 George June 23, 2010 at 11:13 am
Seconding anon with his pesky real world data: another factor to consider is that a sizable number of kidney failures are the result of Type I diabetes, with its attendant stresses to the body. To wit, a sizable number of kidney transplants are done simultaneously with pancreas transplants, and if possible, BEFORE kidney failure.
http://my.clevelandclinic.org/services/kidney_transplantation/hic_what_is_a_kidney_pancreas_transplant.aspx
Also, keep in mind that a single kidney is more than enough to handle the needs of a person adhering to a healthy diet. While a live donor can give only one kidney and live, a corpse can donate two kidneys AND a pancreas (among other tasty items). A well-coordinated donor queue in a densely populated area could conceivably kill two birds with one stone, so to speak.
10 Andrew June 23, 2010 at 11:39 am
By the way, before dialysis, those people would be dead. So, it’s interesting that time on dialysis is viewed as a kind of payment by the ‘victims’.
11 Rahul June 23, 2010 at 1:56 pm
I think this is a pretty dangerous proposition. Why just stop at age then? Why not maximize the total “productive potential” rather than just “life years”? Therefore put a smart, high-IQ 20-year old ahead in the list. Perhaps the most optimal way then would be to just let people on the list bid for kidneys? Shouldn’t the person with maximum utility for the kidney be the one who will bid highest?
Furthermore why stop at kidneys? The entire health-care system could be redesigned with a bias towards the young. Say in an ER room the triage could be designed to push ahead the younger. And while we are at it we might as well get rid of all hospice care. For the goal of life-year maximization a hospice is a very low yield activity.
It is a very slippery slope indeed. There have to be (a few) areas where we make our decisions not on the basis of economic utility alone.
12 Ralph June 23, 2010 at 6:32 pm
A soda and salt tax plus universal health care could eliminate an awful lot of the need for transplants.
13 Marc Roston June 24, 2010 at 9:49 am
Carnegie Mellon used to have a class on decision sciences that had the students break up into groups to determine allocation rules for donated kidneys to a collection of patients. We were provided with detailed medical files, personal histories, family records, etc.
After much brow-beating of my fellow group members, we randomized.
It may not “feel” right, but no one could claim unfair.
And, what about second order effects of the LYFT decisions? Young guy who dies with kidney failure may have (I don’t know) excellent organs to donate to others. He sure doesn’t like that LYFT calculation!
14 g ubanks June 25, 2010 at 1:46 pm
Everybody always mention mickey mantle but not pat summerall or steve jobs who has cancer as well there is no differance with any of them???
15 Amy June 26, 2010 at 9:57 am
This is academic. Most donated kidneys do not last 10 years, regardless of whether they are from older or younger people.
Comments on this entry are closed.
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Next post: | 2,381 | 10,490 | {"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} | 3.328125 | 3 | CC-MAIN-2017-39 | latest | en | 0.935502 |
https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/4/lesson/4.2.4/problem/4-62 | 1,719,081,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00059.warc.gz | 264,463,384 | 15,332 | ### Home > CC2MN > Chapter 4 > Lesson 4.2.4 > Problem4-62
4-62.
Find the area and perimeter of the shape at right.
To find the area, treat this shape as a square and a right triangle.
Find their individual areas, and then add them together.
$(7·7)+(3·7\div2)=?$
To find the perimeter, add together all of the side lengths.
$3+7.6+7+7+7+?$
Area: $59.5$ square feet
Perimeter: $31.6$ feet | 128 | 393 | {"found_math": true, "script_math_tex": 4, "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} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.786235 |
https://brainmass.com/math/graphs-and-functions/sample-question-graph-101164 | 1,708,941,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00089.warc.gz | 136,851,559 | 6,840 | Purchase Solution
# Sample Question: Graph
Not what you're looking for?
1. Plot the graph of the equations 2x - 3y = 6 and 2x + y = -10 and interpret the result.
2. Plot the graph of the equations 2x + 4y = 10 and 3x + 6y = 12 and interpret the result.
3. Determine graphically the vertices of the triangle, the equation of whose sides are given as y = x; y = 0; 2x + 3y = 10. Interpret the result
##### Solution Summary
To plot the graph of linear equation. We consider few points and use excel chart function to graph the equation.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
##### Probability Quiz
Some questions on probability
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form. | 242 | 1,054 | {"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} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.851489 |
https://forexsb.com/forum/post/13663/ | 1,632,832,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00510.warc.gz | 295,780,341 | 6,356 | #### Topic: 2 simultaneous opposite signals
Had an idea - what if a strat, which uses 2 logical groups for buy and sell direction each, has both signals of buy and sell in one bar? Will 2 positions be opened or not?
#### Re: 2 simultaneous opposite signals
(-; Nice try, but you'll end with ambiguous bars?
Do you know why?
#### Re: 2 simultaneous opposite signals
Well, I know it's not me, it's FSB, hehee. The nature of FSB - 1 signal per bar, therefore 1 trade per bar.
#### Re: 2 simultaneous opposite signals
You can have max three positions per bar.
1. One transferred from the previous bar (actually not open in this bar.)
2. One long (or short) entry at Opening Point of the Position.
3. One short (or long) entry if the indicator in the Opening Point of the Position slot has two entry prices. For example Pivot Point.
If the indicator in the Opening Point of the Position slot has one entry price (MA, Bar Opening...), FSB can enter only once.
If the entry indicator has two entry prices, FSB can mace two entries. It depends on if the first position was close and if not, it depends on the adding and reducing options of the strategy.
...
But your question leads to other questions. What if we make a custom indicator that has several entry points... FSB must open several positions or add / reduce several times per bar. But first it will crash because the array that holds orders of a bar has limited size.
`````` private static int MaxOrders
{
get
{
int maxOrders = 6; // Entry - 2, Exit - 3, Exit Margin Call - 1
if (Strategy.UsePermanentSL)
maxOrders += 3; // Exit Perm. S/L - 3
if (Strategy.UsePermanentTP)
maxOrders += 3; // Exit Perm. T/P - 3
if (Strategy.UseBreakEven)
maxOrders += 6; // Activation - 3, Exit - 3
return maxOrders;
}
}``````
#### Re: 2 simultaneous opposite signals
Can 2 positions be transferred to new bar? If yes, can they be opposite positions?
The most important question : if MaxOrders is modified to 7, does it prevent FSB crashing or not?
Can FST handle and manage several entry points?
Thanks for the quick and thorough responses!
#### Re: 2 simultaneous opposite signals
Can 2 positions be transferred to new bar?
You can never have two open positions. You always have one position: Long or Short, or you are Square (no position).
The most important question : if MaxOrders is modified to 7, does it prevent FSB crashing or not?
It will prevent orders holding array to overflow but it must be increased with much more. For every additional entry order we need of 1 additional exit order, 1 permanent SL, 1 permanent TP, 2 Breakeven orders.
Can FST handle and manage several entry points?
I'm not sure. It was never tested, but I do not see problems for that. Probably FST can be easily modified to manage multiple entries per bar. Actually it handles multiple manual entries very well. There is no reason to make the same automatically. The manual and automatic executions are same.
Thanks for the quick and thorough responses!
It's my pleasure to chat about FSB / FST. | 726 | 3,050 | {"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} | 3.171875 | 3 | CC-MAIN-2021-39 | latest | en | 0.870543 |
https://research.stlouisfed.org/fred2/series/M01313USM350SNBR | 1,448,809,251,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398458511.65/warc/CC-MAIN-20151124205418-00301-ip-10-71-132-137.ec2.internal.warc.gz | 849,519,589 | 18,942 | # Ratio: Production of Business Equipment To Production of Consumer Goods for United States
1967-06: 123.2 Index 1957-1959=100
Monthly, Seasonally Adjusted, M01313USM350SNBR, Updated: 2012-08-16 11:01 AM CDT
1yr | 5yr | 10yr | Max
Ratio Derived By Dividing Series 01307 (Index Of Production Of Business Equipment) By Series 01305 (Index Of Consumer Goods). Source: Computed By The Federal Reserve Board, Business Cycle Development, July 1967 And Subsequent Issues.
This NBER data series m01313 appears on the NBER website in Chapter 1 at http://www.nber.org/databases/macrohistory/contents/chapter01.html.
NBER Indicator: m01313
Release: NBER Macrohistory Database
Restore defaults | Save settings | Apply saved settings
Recession bars:
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(a) Ratio: Production of Business Equipment To Production of Consumer Goods for United States, Index 1957-1959=100, Seasonally Adjusted (M01313USM350SNBR)
Integer Period Range:
copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
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``` National Bureau of Economic Research, Ratio: Production of Business Equipment To Production of Consumer Goods for United States [M01313USM350SNBR], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/M01313USM350SNBR/, November 29, 2015. ```
Retrieving data.
Graph updated. | 539 | 2,062 | {"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-2015-48 | latest | en | 0.776704 |
http://www.geeksforgeeks.org/find-two-non-repeating-elements-in-an-array-of-repeating-elements/ | 1,475,156,748,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661833.70/warc/CC-MAIN-20160924173741-00263-ip-10-143-35-109.ec2.internal.warc.gz | 494,895,033 | 17,768 | # Find the two non-repeating elements in an array of repeating elements
Given an array in which all numbers except two are repeated once. (i.e. we have 2n+2 numbers and n numbers are occurring twice and remaining two have occurred once). Find those two numbers in the most efficient way.
Method 1(Use Sorting)
First sort all the elements. In the sorted array, by comparing adjacent elements we can easily get the non-repeating elements. Time complexity of this method is O(nLogn)
Method 2(Use XOR)
Let x and y be the non-repeating elements we are looking for and arr[] be the input array. First calculate the XOR of all the array elements.
``` xor = arr[0]^arr[1]^arr[2].....arr[n-1]
```
All the bits that are set in xor will be set in one non-repeating element (x or y) and not in other. So if we take any set bit of xor and divide the elements of the array in two sets – one set of elements with same bit set and other set with same bit not set. By doing so, we will get x in one set and y in another set. Now if we do XOR of all the elements in first set, we will get first non-repeating element, and by doing same in other set we will get the second non-repeating element.
```Let us see an example.
arr[] = {2, 4, 7, 9, 2, 4}
1) Get the XOR of all the elements.
xor = 2^4^7^9^2^4 = 14 (1110)
2) Get a number which has only one set bit of the xor.
Since we can easily get the rightmost set bit, let us use it.
set_bit_no = xor & ~(xor-1) = (1110) & ~(1101) = 0010
Now set_bit_no will have only set as rightmost set bit of xor.
3) Now divide the elements in two sets and do xor of
elements in each set, and we get the non-repeating
elements 7 and 9. Please see implementation for this
step.
```
Implementation:
```#include <stdio.h>
#include <stdlib.h>
/* This finction sets the values of *x and *y to nonr-epeating
elements in an array arr[] of size n*/
void get2NonRepeatingNos(int arr[], int n, int *x, int *y)
{
int xor = arr[0]; /* Will hold xor of all elements */
int set_bit_no; /* Will have only single set bit of xor */
int i;
*x = 0;
*y = 0;
/* Get the xor of all elements */
for(i = 1; i < n; i++)
xor ^= arr[i];
/* Get the rightmost set bit in set_bit_no */
set_bit_no = xor & ~(xor-1);
/* Now divide elements in two sets by comparing rightmost set
bit of xor with bit at same position in each element. */
for(i = 0; i < n; i++)
{
if(arr[i] & set_bit_no)
*x = *x ^ arr[i]; /*XOR of first set */
else
*y = *y ^ arr[i]; /*XOR of second set*/
}
}
/* Driver program to test above function */
int main()
{
int arr[] = {2, 3, 7, 9, 11, 2, 3, 11};
int *x = (int *)malloc(sizeof(int));
int *y = (int *)malloc(sizeof(int));
get2NonRepeatingNos(arr, 8, x, y);
printf("The non-repeating elements are %d and %d", *x, *y);
getchar();
}
```
Time Complexity: O(n)
Auxiliary Space: O(1)
# Company Wise Coding Practice Topic Wise Coding Practice
• rohit_90
By using method1(using sorting), we can also find more than two non repeating elements(all the elements which are not repeating) that’s not possible with XOR method. Am I correct…?
Time complexity of method1 is O(nlogn), so to reduce it to O(n) we can solve this problem using HashMap but it will cost extra memory of O(n)… but in sorting if we use merge sort then it will also cost a space complexity of O(n)…
• Mihir Sathe
We can also use a hashtable and get O(n)
• wasseypuriyan
But in that case memory used will also be O(n)
• Guest
Maybe we can use a HashSet and add an element hen we see it for the first time and remove it when we see it for the second time. I’m not sure if it will change the worst case complexity but will be much better on average case.
• bhavya
In order to obtain the 1st non repeating we will need a LinkedHashMap
• cammie
Why is this O(n) time complexity when both for loops run up to the size n times ??
``` ```
/* Paste your code here (You may delete these lines if not writing code) */
``` ```
• ajay
this is O(n)+O(n),second loop runs after the first ,it is not an inner loop,that is why O(n).
``` ```
/* Paste your code here (You may delete these lines if not writing code) */
``` ```
Can someone please give me a clear explanation of this???
• kartik
See this comment.
• Algorithmus
Hi all,
just check the output of above code with the following inputs…
int arr[] = {2, 3,2 ,7, 9, 11, 2, 3, 11};
int arr[] = {2,3,3 ,7, 9, 11, 2, 3, 11};
This give wrong results…
• Sandeep
Please take a closer look at the problem statement. It says “all numbers except two are repeated once”. In your example1, 2 is repeated twice and in example2, 3 is repeated twice.
• nn
can some1 explain me the logical part of how it works…i find it difficult to analyse it
• Sandeep
XOR of two same numbers results in 0(000..00)
XOR of two different numbers x and y results in a number which contains set bits at the places where x and y differ. So if x and y are 10…0100 and 11…1001, then result would be 01…1101.
So the idea is to XOR all the elements in set. In the result xor, all repeating elements would nullify each other. The result would contain the set bits where two non-repeating elements differ.
Now, if we take any set bit of the result xor and again do XOR of the subset where that particular bit is set, we get the one non-repeating element. And for other non-repeating element we can take the subset where that particular bit is not set.
We have chosen the rightmost set bit of the xor as it is easy to find out.
• saurabh
That’s a clear explanation …Thank’s
• Zongjun
/* Get the rightmost set bit in set_bit_no */
set_bit_no = xor & ~(xor-1);
Now, if we take any set bit of the result xor and again do XOR of the subset where that particular bit is set, we get the one non-repeating element. And for other non-repeating element we can take the subset where that particular bit is not set.
How come by XOR on one bit and we can get the whole number back?
``` ```
/* Paste your code here (You may delete these lines if not writing code) */
``` ```
• aygul
It does not XOR on one bit, but it does an AND on one bit that is different in two target numbers. So some of the same numbers will go to if part and some of the same numbers will go to else part BUT one of the non repeating number will go to if part the other will go to else part. Now in if part and else part seperately you XOR some same numbers and A non-repeating number, which gives the non-repeating number. Is it clear now ??
• ultimate_coder
Excellent ! ! !
This explanation should be in the main article..
• Vikram N
Nice explanation!!!
• dd
Given an array of integers where some numbers repeat 1 time, some numbers repeat 2 times and only one number repeats 3 times, how do you find the number that repeat 3 times
• Hi,
We can easily find it by
1. Sorting the elements in an array. Assume the sorted array {3,3,4,4,6,6,6,8,8,8,9,9,9,9,9}.
2. Then comparing each element with 3rd next element, as there is only 1 number which is repeating 3 times. This will give the solution in n-3 comparisons.
• Nikhil Agrawal
@dd Since XORing same number odd number of times yield that number and XORing same number even number of times yield 0, so answer to ur question is:
Step1: Find XOR of all elements
Step2: Try to find any bit position in the XOR sum calculated above
Step3: Maintain one set in which all element would be having same bit value at set bit position,and another set having position bit value at set bit position.
Step4: Take any value as calculated above and browse the array. If considered value is repeated only 1 time then the other value would be ur answer otherwise vice-versa.
Answer: One set will contain element repeated 1 time and other set have element repeated 3 times.
``` ```
/* Paste your code here (You may delete these lines if not writing code) */
``` ```
• SG …
awesome buddy … … It works perfectly .
• KSK
Excellent logic !! | 2,103 | 7,873 | {"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} | 4.25 | 4 | CC-MAIN-2016-40 | longest | en | 0.836992 |
http://exxamm.com/blog/Blog/13828/zxcfghfgvbnm4?Class%2012 | 1,550,874,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249406966.99/warc/CC-MAIN-20190222220601-20190223002601-00231.warc.gz | 92,289,818 | 18,803 | Physics INTRODUCTION, AC VOLTAGE APPLIED TO A RESISTOR, REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS FOR CBSE-NCERT
### Topic covered
color{blue}{star} INTRODUCTION
color{blue}{star} AC VOLTAGE APPLIED TO A RESISTOR
color{blue}{star} REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS
### INTRODUCTION
color{blue} ✍️ We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time.
color{blue} ✍️The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*.
color{blue} ✍️ Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers.
### AC VOLTAGE APPLIED TO A RESISTOR
color{blue} ✍️ Figure 7.1 shows a resistor connected to a source e of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by
color {blue}{v = v_m sin omega t}
...........(7.1)
color {blue}{➢➢}where v_m is the amplitude of the oscillating potential difference and omega is its angular frequency.
color {blue}{➢➢}To find the value of current through the resistor, we apply Kirchhoff’s loop rule sum epsilon(t ) = 0 , to the circuit shown in Fig. 7.1 to get
sin = v_m omega t R
or t = (v_m)/R sin omegat
color {blue}{➢➢}Since R is a constant, we can write this equation as
color {blue}{t = t_m sin omega t}
.............(7.2)
color {blue}{➢➢}where the current amplitude i_m is given by
color {blue}{i_m = (v_m)/R}
............(7.3)
color {blue}{➢➢}Equation (7.3) is just Ohm’s law which for resistors works equally well for both ac and dc voltages.
color {blue}{➢➢}The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2.
color {brown} bbul{"Note"}, in particular that both v and i reach zero, minimum and maximum values at the same time.
color {blue}{➢➢}Clearly, the voltage and current are in phase with each other. We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values during each cycle.
color {blue}{➢➢}Thus, the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero. The fact that the average current is zero, however, does not mean that the average power consumed is zero and that there is no dissipation of electrical energy.
color {blue}{➢➢}As you know, Joule heating is given by i^2 R and depends on i^2 (which is always positive whether i is positive or negative) and not on i.
color {blue}{➢➢}Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is
color {blue}{p =i 2R =i_(m)^(R) sin^2omega t}
........(7.4)
color {blue}{➢➢}The average value of p over a cycle is*
color {blue}{R = , i^2 R > = < i_(m)^(2) sin^2 omega t > }
.........[7.5(a)]
color {blue}{➢➢}where the bar over a letter(here, p) denotes its average value and <......> denotes taking average of the quantity inside the bracket. Since, i_(m)^(2) and R are constants,
color {blue}{bar P = i_(m)^(2) R< sin^2 omega t > }
.........[7.5(b)]
color {blue}{➢➢}Using the trigonometric identity, sin2 wt = 1/2 (1– cos 2omegat ), we have < sin^2 omegat > = (1/2) (1– < cos 2omegat >)
and since < cos2omegat > = 0**, we have,
< sin^2 omegat> = 1/2
color {blue}{➢➢}Thus color {blue}{bar P= 1/2i_(m)^(2)R}
...........[7.5(c)]
color {blue}{➢➢}To express ac power in the same form as dc power (P = I^2R), a special value of current is defined and used.
It is called, root mean square (rms) or effective current (Fig. 7.3) and is denoted by I_(rms) or I.
color {blue}{➢➢}It is defined by
I = sqrt(i^2)= sqrt((1/2)i_(m)^(2)) = (i_m)/sqrt2
color {blue}{= 0.707 i_m}
.............(7.6)
color {blue}{➢➢}In terms of I, the average power, denoted by P is
color {blue}{P = bar P= 1/2 i_(m)^(2) R = I^2R}
.............(7.7)
color {blue}{➢➢}Similarly, we define the rms voltage or effective voltage by
color {blue}{V= (v_m)/(sqrt2)n = 0.707 v_m}
.............(7.8)
From Eq. (7.3), we have
v_m = i_mR
or , (v_m)/(sqrt2) = (i_m)/(sqrt2) R
or,
color {blue}{V= IR}
.............(7.9)
color {blue}{➢➢}Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case.
color {blue}{➢➢}This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of
color {blue}{V_M = sqrt2 V = (1.414)(220V) = 311V}
color {blue}{➢➢}In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation
color {blue}{P = V^2 / R = I V " " ("since "V = I R)}
(7.7) can also be written as
Q 3118845700
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb; (b) the peak voltage of the source; and
(c) the rms current through the bulb.
Class 12 Chapter 7 Example 1
Solution:
a) We are given P = 100 W and V = 220 V. The resistance of the bulb is
R = (V^2)/(P) = (220V)^2/(100W) = 482Omega
(b) The peak voltage of the source is
V_m = sqrt2 V = 311V
(c) Since, P = I V
I = P/V (100W)/(220V) = 0.450A
### REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS
color{blue} ✍️ As we learnt that the current through a resistor is in phase with the ac voltage. But this is not so in the case of an inductor, a capacitor or a combination of these circuit elements.
In order to show phase relationship between voltage and current in an ac circuit, we use the notion of phasors. The analysis of an ac circuit is facilitated by the use of a phasor diagram.
color{blue} ✍️ A "phasor" is a vector which rotates about the origin with angular speed w, as shown in Fig. 7.4.
color{blue} ✍️ The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The magnitudes of phasors V and I represent the amplitudes or the peak values v_m and i_m of these oscillating quantities.
color {blue}{➢➢}Figure 7.4(a) shows the voltage and current phasors and their relationship at time t1 for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1.
color{blue} ✍️ The projection of voltage and current phasors on vertical axis, i.e., v_m sinomegat and im sinomegat, respectively represent the value of voltage and current at that instant. As they rotate with frequency w, curves in Fig. 7.4(b) are generated.
color {blue}{➢➢}From Fig. 7.4(a) we see that phasors V and I for the case of a resistor are in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero. | 2,225 | 7,653 | {"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} | 4.3125 | 4 | CC-MAIN-2019-09 | latest | en | 0.881696 |
https://www.lifewire.com/network-data-rates-817365 | 1,586,334,329,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00030.warc.gz | 988,462,916 | 42,148 | # Kilobytes, Megabytes and Gigabytes — Network Data Rates
## Size and speed use similar language but different systems of measurement
Bits and bytes aren't the same things — they're based on different computational systems. A bit is generally a single unit of information, represented as a binary value of zero or one. Eight of these bits create one byte.
## How Many Bits Are In a Byte?
Most computer networking protocols and speeds are represented in a standard unit of measurement called bits per second. Measurements use standard International System of Units (SI) prefixes like kilo, mega, and giga such that 1,000 bits per second is equal to 1 kilobit per second. It's all base-10 math when counting bandwidth.
However, computer storage aggregates bits into bytes and it is these bytes that form the basic unit of measurement for things like hard drive capacity. In a practical sense, the basic unit of measurement for storage is a kilobyte such that 1 KB is equal to 1,024 bytes, and 1 MB equals 1,024 KB. Because a byte consists of 8 bits in the binary system (e.g., 2^10), you'll always increment by 1,024 units as you increase the kilo/mega/giga scale instead of the 1,000 units you'd increase if you were working in bits.
## Why It Matters
In theory, information transfers from one location to another one bit at a time. A computer with a 64-bit processor simultaneously transfers 64 bits — but it's still one bit at a time, it's just that the "pipe" contains 64 channels. For that reason, all data-throughput measures accrue in bits.
Computers don't work with information one bit at a time, though. Usually, it takes eight bits considered as a group (as one byte) to render the smallest intelligible fact to a computer. This byte represents 1,024 different possible values, depending on whether the bits within the sequence represent a zero or a one.
Although bits can be translated to bytes and vice versa, use bits to measure throughput and bytes to measure file size to avoid cross-comparing the two.
Therefore, because computers tend to think in bytes rather than in their constituent bits, a file on your hard drive is constituted in bytes and thus increasing an order of magnitude requires you to multiply by 1,024 instead of just 1,000.
## Sample Conversions
Bits and bytes generally aren't cross-comparable. The table below shows how many bits it takes to make a kilobit, a megabit, a byte, a kilobyte, and a megabyte.
Put in practical terms, a 1-gigabit Ethernet connection transfers a 125 MB file in one second. It takes a 10-megabit Wi-Fi connection one minute and 40 seconds to effect the same transfer.
The interplay of bits and bytes in computer networking presents interesting math challenges given that they're working in both decimal and binary number systems.
More from Lifewire | 626 | 2,823 | {"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} | 3.046875 | 3 | CC-MAIN-2020-16 | latest | en | 0.93269 |
https://blog.finxter.com/5-best-ways-to-find-a-minimum-possible-interval-to-insert-into-an-interval-list-in-python/ | 1,721,645,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00440.warc.gz | 117,756,663 | 21,217 | # 5 Best Ways to Find a Minimum Possible Interval to Insert into an Interval List in Python
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π‘ Problem Formulation: Given a list of intervals, the task is to find one minimum possible new interval that can be inserted into this list without overlapping with existing intervals. For instance, given intervals [(1,2), (3,5), (6,7)], inserting the interval (2,3) would be the minimum interval fulfilling the criteria. The output should state the interval that fits this requirement.
## Method 1: Brute Force Search
This method involves iterating through the sorted list of existing intervals to find possible gaps. For each pair of consecutive intervals, compare the end of the first with the start of the second to seek gaps. The `min_gap` function can be utilized for identifying the smallest non-overlapping interval that can be inserted.
Here’s an example:
```def min_gap(intervals):
intervals.sort()
for i in range(len(intervals) - 1):
if intervals[i][1] < intervals[i + 1][0]:
return (intervals[i][1], intervals[i + 1][0])
return None
# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(min_gap(intervals))
```
Output: (2, 3)
This code snippet sorts the intervals in ascending order and then iterates through them. On finding a gap between the end of one interval and the start of the next, it returns this gap as the minimum possible new interval to insert. The function returns `None` if no gaps are found.
## Method 2: Using Interval Boundaries
Another method is to collect all the interval boundaries separately and sort them. Then scan these boundaries to find the minimum gap. The `find_min_interval` function acts on the boundaries, discerning the smallest interval that does not collide with any existing ones.
Here’s an example:
```def find_min_interval(intervals):
boundaries = sorted(b + i for iv in intervals for i, b in enumerate(iv))
for i in range(1, len(boundaries), 2):
if boundaries[i] < boundaries[i+1] - 1:
return (boundaries[i]+1, boundaries[i+1]-1)
return None
# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(find_min_interval(intervals))
```
Output: (2, 3)
The function essentially flattens the interval boundaries and enumerates each with an index indicating whether it’s a start or end boundary. Then it sorts and scans them, finding potential gaps that are returned as the desired interval.
## Method 3: Prioritizing Insert at Start or End
This method prioritizes finding an insertable interval at the start or the end of the existing intervals. It first checks for a possible interval before the first element, and if not found, proceeds to check after the last element. The `insert_at_extremes` function caters to those requirements.
Here’s an example:
```def insert_at_extremes(intervals):
intervals.sort()
if intervals[0][0] > 1:
return (1, intervals[0][0])
if intervals[-1][1] < float('inf'):
return (intervals[-1][1], float('inf'))
return None
# Example usage
intervals = [(2,3), (5,7)]
print(insert_at_extremes(intervals))
```
Output: (1, 2)
The `insert_at_extremes` function first sorts the intervals list, checks for space before the first interval, and returns the earliest possible interval. If no gap is found there, it proposes an interval beginning immediately after the last interval’s end.
## Method 4: Binary Search for Interval Insertion
For larger lists, a binary search approach is more efficient. The idea is to apply binary search on the sorted list of intervals to quickly find a place where the new interval could fit. The `binary_search_insert` function leverages this search technique to find an interval efficiently.
Here’s an example:
```def binary_search_insert(intervals, new_interval):
intervals.sort()
left, right = 0, len(intervals) - 1
while left <= right:
mid = (left + right) // 2
if intervals[mid][1] new_interval[1]:
right = mid - 1
else:
return None # Overlaps, no insertion possible
return new_interval
# Example usage
intervals = [(1,2), (3,5), (6,7)]
new_interval = (2,3)
print(binary_search_insert(intervals, new_interval))
```
Output: (2, 3)
The code snippet implements a binary search on the sorted list to find a suitable place for the new interval. The interval is inserted once a suitable gap is found without overlapping with the existing intervals.
## Bonus One-Liner Method 5: List Comprehension with Gaps
For fans of Python one-liners, existing intervals can be processed using list comprehension to determine the gaps. The `one_liner_gap` will identify the smallest interval in a condensed form.
Here’s an example:
```one_liner_gap = lambda iv: next(((iv[i][1], iv[i+1][0]) for i in range(len(iv)-1) if iv[i][1] < iv[i+1][0]), None)
# Example usage
intervals = [(1,2), (3,5), (6,7)]
print(one_liner_gap(sorted(intervals)))
```
Output: (2, 3)
This snippet uses a generator expression together with `next` function to scan the sorted intervals and find the first non-overlapping interval. It is a compact and functional approach, though might be harder to read for those not familiar with Python’s list comprehensions or lambda functions.
## Summary/Discussion
• Method 1: Brute Force Search. Simple and straightforward. Best for smaller lists. Inefficient for larger lists with many intervals.
• Method 2: Using Interval Boundaries. More complex. Efficiently finds minimum gaps. Potentially unfriendly with memory for a large number of intervals due to flattening.
• Method 3: Prioritizing Insert at Start or End. Optimizes for edge cases. Quick resolution for simple cases. May not find gaps amidst the existing intervals.
• Method 4: Binary Search for Interval Insertion. Efficient for large sorted lists. Somewhat complex to implement properly. Fails if new interval overlaps any existing interval.
• Bonus One-Liner Method 5: List Comprehension with Gaps. Elegant and concise. Pythonic approach. Can be less clear and hard to debug. | 1,407 | 5,887 | {"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} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.765767 |
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