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https://universe-review.ca/R15-25-DiracEq04.htm | 1,723,388,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00478.warc.gz | 474,568,822 | 4,184 | ## Wave Equations
### Electromagnetic Wave Equation
The field equations for electromagnetic radiation is a product of the Victorian Era in the 19th century. It was originated from a set of equations in electromagnetism. Maxwell's contribution is to add a "displacement current" term, i.e., the (1/c)E term in Eq.(40), and thus put the equations into a consistent set which implied new physical phenomena, at that time unknown but subsequently verified in all details by experiments. In terms of the electric field E and magnetic field B the Maxwell's equations are four first order differential equations :
---------- (39) ---------- (40) ---------- (41) ---------- (42)
,
the i, j, k, are unit vectors along the x, y, z axis respectively, J is the current density with current I, is the charge density with charge Q, and c is the velocity of light. The conversion of the laws from differential to integral form is accomplished by applying the Gauss's and the Stoke's theorems as shown below via a much simplified derivation with either a cube or square (the circle on the integral sign denotes either a closed surface or loop).
For regions of space where charge and current are absent, the right-hand sides vanish in all the equations. These first order differential equations can be translated into second order equations more suitable for describing the electromagnetic wave in relativistic invariant form. By virtue of the mathematical identity ( A)=0, Eq.(41) can be defined by the vector potential A:
B = A ---------- (43)
By substituting Eq.(43) into Eq.(42) and using another mathematical identity ()=0, we can define E in terms of A and the scalar potential as :
E = - - (1/c)A ---------- (44)
which combines with Eq.(39) yields:
2 + (1/c)(A) = 0 ---------- (45)
Substituting Eqs.(43), (44) into (40) and using one more identity ( ) A = -2A + (A), we get another second order differential equation :
2A - (1/c2)A - (A + (1/c)) = 0 ---------- (46)
The U(1) symmetry on the spin of A adds an extra degree of freedom, which can be fixed by taking the Lorenz condition :
A + (1/c) = 0,
then Eqs.(45), and (46) can be written in simpler form :
2A - (1/c2)A = 0 ---------- (46a),
and 2 - (1/c2) = 0 ---------- (45a).
The vector potential A and the scalar potential form a four potential, whcih are prescribed by the Lorentz invariant wave equations of Eqs. (46a), (45a). The formulation is now suitable for use in high energy particle physics. The Lorenz condition is also manifest Lorentz invariance (note the two names differ by a "t"), but is incomplete in the sense that the definition of B in Eq.(43) is unchanged by adding a term in the form of the "gradient of scalar function " because of the identity () = 0. It follows that E, B and Eqs.(45a), (46a) are unchanged by the transformation :
A A' = A + ---------- (47)
if ' = - (1/c) ---------- (48)
and 2 - (1/c2) = 0.
This is known as the gauge transformation of the second kind (or Lorenz gauge).
Other choice for involves setting 2 = -A, and eliminating the longitudinal component with = (1/c) in the transformation Eqs.(47), and (48). In this way, the transversality condition A' = 0 is ensured. This condition is also known as "Coulomb gauge". It follows that :
1. Eq.(45) is reduced to the Laplace's equation for time-independent electrostatic potential in free space, that's why it is also called "Coulomb gauge".
2. The longitudinal component is eliminated, i.e., there is no component along the direction of motion. This is ensured by the condition A' = 0.
3. The two transversal components correspond to the two polarizations as observed in electromagnetic radiation.
4. Since ' = 0 in free space without the presence of any source, the wave equation Eq.(46) become :
2A' - (1/c2)A' = 0 ---------- (49)
5. The magnetic field is related to A by B = A' ---------- (50)
6. The electric field is related to A by E = - (1/c)A' ---------- (51)
7. From Eqs.(49), (51), the wave equation for E can be derived :
2E - (1/c2)E = 0 ---------- (50a)
8. From Eqs.(49), (50), the transverse gauge, and the identity leading to Eq.(46), we obtain the wave equation for B :
2B - (1/c2)B = 0 ---------- (51a)
9. It is now clear that the introduction of the vector potential A is mainly to simplify computations. For the case of transverse electromagnetic wave two components of A is required to describe the wave, while the E and B fields together introduce four components.
10. The Coulomb gauge is not covariant under a Lorentz transformation.
11. See more in "Quantum Electrodynamics".
Eqs.(49), (50a), (51a) are in a form very similar to the Klein-Gordon equation in Eq.(18) except that the mass term vanishes (because the photon has no rest mass) and the field is a vector (instead of scalar) with two transversal components (polarization) perpendicular to each other. The Maxwell's equations were derived more than one hundred years ago without any knowledge about quantum theory, the similarity in form is perhaps because all kinds of wave (including the gravitational wave for spin 2 particle) are governed by the same kind of equation. Only the details (such as number of components) and interpretations are slightly different (note the omission of the Planck constant h in the Maxwell's equations has no consequence in the comparison as m = 0).
The solution of Eq.(49) is :
kx, ky, kz = 2n/L (n = 1, 2, ...) with kx = kx - t = kx - |k|ct. The ck coefficients is linked to the spectrum associated with the electromagnetic radiation. It will assume the role of q-number in second quantization, which endows particle attributes to the wave.
. | 1,402 | 5,653 | {"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} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.899764 |
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ORDER NOW! | 1,073 | 4,648 | {"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-2018-22 | latest | en | 0.949655 |
https://web2.0calc.com/questions/help-please-and-thanks_1 | 1,531,856,156,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00257.warc.gz | 792,461,647 | 5,257 | +0
0
197
1
my teacher wants me to use this formula: x2+(a+b)x+ab
I don't understand.
Please use this formula with this expression: ($$\sqrt[2]{3}+2)(\sqrt[2]{3}+1)$$
Guest Mar 28, 2017
#1
+7154
+2
The 2's indicating that it is the second root are assumed when not written.
$$(\sqrt{3}+2)(\sqrt{3}+1)$$
x = √3
a = 1
b = 2
Substitute x, a, and b, into x2+(a+b)x+ab .
$$(\sqrt{3})^2+(1+2)(\sqrt3)+(1)(2)$$
Simplify.
$$3+3\sqrt3+2 \\ 5+3\sqrt3$$
hectictar Mar 29, 2017 | 219 | 481 | {"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 | 4 | CC-MAIN-2018-30 | latest | en | 0.663067 |
https://www.lumoslearning.com/llwp/edsearch.html?q=Problem+Solving%3A+Look+for+a+Pattern&type=App | 1,556,242,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578747424.89/warc/CC-MAIN-20190426013652-20190426035652-00166.warc.gz | 741,701,822 | 35,642 | ### 1.8 Problem Solving Strategies: Make a Table and Look for a Pattern
https://www.ck12.org/book/CK-12-Algebra-I-Second.../1.8/
Develop and use the strategy “make a table.” Develop and use the strategy “look
for a pattern.” Plan and compare alternative approaches to solving a problem.
### Problem Solving Strategy : Look for a Pattern - YouTube
Aug 17, 2012 ... Problem Solving Strategy : Look for a Pattern. SmithMathAcademy. Loading...
Unsubscribe from SmithMathAcademy? Cancel Unsubscribe.
### Problem Solving: Find a Pattern (Grades 2-8) - TeacherVision
https://www.teachervision.com/problem-solving/problem-solving-find- pattern
Finding a Pattern is a strategy in which students look for patterns in the data in
order to solve the problem. Students look for items or numbers that are repeated,
...
### Problem Solving Strategy 4: Find a Pattern - YouTube
Mar 30, 2017 ... Problem Solving Strategy 4: Find a Pattern ... Problem Solving Strategies 1 -
Looking for Clues and Key Words (Part A) - Duration: 6:13.
### 7-PS1.Looking for a Pattern
https://www.parklands.havering.sch.uk/.../Year-5-Looking-For-A-Pattern. compressed.pdf
The following skills and strategies need to be developed by pupils to help them
solve problems that involve looking for a pattern. CREATING AND CONTINUING.
### Lumos EdSearch Overview:
EdSearch is a free standards-aligned educational search engine specifically designed to help teachers, parents and students find engaging videos, apps, worksheets, interactive quizzes, sample questions and other resources. Educators can select resources of their choice and design a resource kit for their students in minutes! They can assign a collection of resources to an entire class, a small group or just one student and monitor progress. Using EdSearch, you can
• - Discover thousands of curated standards-aligned resources
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• - Get ready access to thousands of grade appropriate practice questions and lessons
• - Find information about nearby schools, libraries, school supply stores, conferences and bookstores.
• - View school test scores, enrollments, calendar events and much more.
• - Create and assign personalized resource kits to your students | 501 | 2,228 | {"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} | 2.6875 | 3 | CC-MAIN-2019-18 | latest | en | 0.851914 |
http://math.stackexchange.com/questions/tagged/arithmetic-functions+summation | 1,398,363,392,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00451-ip-10-147-4-33.ec2.internal.warc.gz | 207,880,728 | 10,843 | # Tagged Questions
### Prove: $\sum_{k<n, (k,n)=1} k= \frac{1}{2}n \varphi (n)$
Prove: $\sum_{k<n, (k,n)=1}k = \frac{1}{2}n \varphi (n)$ I have had strep throat and missed the lecture discussing properties of the Euler function. Any help in solving this is appreciated. ...
Given a series $p_n(s)=\sum_{k=1}^n a_k$. I'd like to get $\hat{p}_n(s)=\sum_{k=1}^n \mu(k)a_k$. Think of $a_k=k^{-s}$ for example. If you let $n$ go to $\infty$, you'll see the well-known relation ... | 176 | 476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2014-15 | latest | en | 0.809195 |
https://learnbps.bismarckschools.org/mod/glossary/showentry.php?eid=22713&displayformat=dictionary | 1,670,178,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00103.warc.gz | 391,727,024 | 12,223 | #### MAT-07.RP.02.a
7th Grade MAT Targeted StandardsDomain (RP) Ratios and Proportional RelationshipsCluster: Analyze proportional relationships and use them to solve real-world and mathematical problems Recognize and represent proportional relationships between quantities. MAT-07.RP.02.a Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin.
## Student Learning Targets:
### Reasoning Targets
• I can determine if two quantities are proportional using a variety of methods (tables, graphs, diagrams, equations, or verbal description
## Resources
### Vocabulary
• ratio
• proportion
• cross product | 156 | 774 | {"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-2022-49 | latest | en | 0.846385 |
https://eleanorrigby-movie.com/what-is-a-longitudinal-wave-diagram/ | 1,721,442,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00555.warc.gz | 194,112,407 | 17,290 | # What is a longitudinal wave diagram?
## What is a longitudinal wave diagram?
Difference Between Longitudinal And Transverse Wave
Longitudinal Wave Transverse Wave
A wave that moves in the direction of its propagation A wave that moves in the direction perpendicular to its propagation
A sound wave is an example of a longitudinal wave Water waves are an example of a transverse wave
### What are the 3 parts of a longitudinal wave?
A compression is where the particles of the medium are closest together, and a rarefaction is where the particles are farthest apart. Amplitude is the distance from the relaxed point in the medium to the middle of a rarefaction or compression. A wavelength is the distance between two equivalent points.
What is a longitudinal wave example?
In a longitudinal wave the particles are displaced parallel to the direction the wave travels. An example of longitudinal waves is compressions moving along a slinky. We can make a horizontal longitudinal wave by pushing and pulling the slinky horizontally.
What are the parts of a longitudinal wave?
While a transverse wave has an alternating pattern of crests and troughs, a longitudinal wave has an alternating pattern of compressions and rarefactions.
## What is a longitudinal wave made up of?
Waves come in two kinds, longitudinal and transverse. Transverse waves are like those on water, with the surface going up and down, and longitudinal waves are like of those of sound, consisting of alternating compressions and rarefactions in a medium.
### What are 3 examples of longitudinal waves?
7 Real Life Examples Of Longitudinal Waves
• Speaking on the mic. A sound wave is a significant example of a longitudinal wave.
• Clapping.
• Tsunami Waves.
• Earthquake (Seismic-P wave)
• Vibration in Window Panels after a Thunder.
• Music Woofers.
Do longitudinal waves have crests and troughs?
Since a longitudinal wave does not contain crests and troughs, its wavelength must be measured differently. A longitudinal wave consists of a repeating pattern of compressions and rarefactions.
How do particles move in a longitudinal wave?
Longitudinal Waves The particles do not move down the tube with the wave; they simply oscillate back and forth about their individual equilibrium positions. Pick a single particle and watch its motion. The wave is seen as the motion of the compressed region (ie, it is a pressure wave), which moves from left to right.
## What is the best example of a longitudinal wave?
Sound Waves in the Air: The sound waves are the best example of a longitudinal wave and are produced by vibrating or disturbing the motion of the particles that travel through a conductive medium.
### What are crests in a longitudinal wave?
The crest of a wave is the highest point that it reaches, while the trough of the wave is the lowest point. These are respectively the maximum and minimum amplitudes, or displacement of the wave.
What is the trough in longitudinal waves?
low point is called the trough. For longitudinal waves, the compressions and rarefactions are analogous to the crests and troughs of transverse waves. The distance between successive crests or troughs is called the wavelength. The height of a wave is the amplitude.
What is longitudinal direction?
extending in the direction of the length of a thing; running lengthwise: a thin, longitudinal stripe. Zoology. pertaining to or extending along the long axis of the body, or the direction from front to back, or head to tail.
## How do longitudinal waves travel through air?
Sound waves are longitudinal waves . They cause particles to vibrate parallel to the direction of wave travel. The vibrations can travel through solids, liquids or gases. The speed of sound depends on the medium through which it is travelling.
### What direction does a longitudinal wave move?
parallel to
Longitudinal waves cause the medium to move parallel to the direction of the wave.
What do you observe in longitudinal waves?
Answer. Explanation: Longitudinal wave, wave consisting of a periodic disturbance or vibration that takes place in the same direction as the advance of the wave. Sound moving through air also compresses and rarefies the gas in the direction of travel of the sound wave as they vibrate back and forth. | 864 | 4,296 | {"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-2024-30 | latest | en | 0.870496 |
https://www.physicsforums.com/threads/uncertainty-principle-and-double-slit-interference.222893/ | 1,576,330,310,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541157498.50/warc/CC-MAIN-20191214122253-20191214150253-00118.warc.gz | 818,746,229 | 25,736 | # Uncertainty principle and double-slit interference?
## Main Question or Discussion Point
Both uncertainty principle and double-slit interference are the most important contents in Quantum mechanics. Those days I am thinking about this question: Is there a hidden relation between them? Unfortunately, I couldn't give me a satisfied answer.
How do you think about it?
Thank you!
Last edited:
Related Quantum Physics News on Phys.org
What do you mean by a hidden relation? Double slit interference, like most quantum phenomina, is a consequence of the uncertainty principle.
xylai, I'm replying to your private message here in the hopes that someone with more experience might chime in with a better answer, but to answer your request for more explanation:
The uncertainty principle tells us that you cannot know the value of two non-commuting observables (such as position and momentum) with perfect accuracy; the more you know one, the less certain the other becomes. In the double slit experiment, when a photon passes through the double slit we know two possible, very precise, positions for that photon: the left slit or the right slit. Those two very precise position possibilities lead to great uncertainty in momentum.
In a single-slit case, this results in a wide gaussian (bell-curve-like) distribution for the photons. In the double-slit case, these two uncertainties interfere with one another to create the beloved interference pattern. But in both cases, the pattern results from the uncertainty in momentum resulting from the certainty (albiet, narrowed down to two possibilities in the double slit case) in position. If not for the uncertainty principle, the double slit experiment would just result in two, well defined bright spots behind the slits.
Why do the two uncertainties in a double slit not simply add up to one, wide blob of uncertainty? That has to do with the Schrodinger Equation and the fact that the wave function represents probability amplutides, not probabilities themselves. The probabilities are derived by taking the complex conjugate of the wave function, thereby eliminating the imaginary component. But in the case of a double slit, the phase difference in the photons going through the two different slits is reflected in the wave function before the complex conjugate is taken (and thereby the actual probability is derived). So that when you do find the probability density function for the two-slit setup, the phase difference between the photons from each slit is reflected in the probability density function as an interference pattern and, sure enough, that's what we see when we do the experiment.
If someone else wants to correct anything I said or explain it more clearly be my guest. :)
Ken G
Gold Member
That all sounds good to me, though it might not be necessary to think so mathematically if you are not used to complex numbers. I might add that in a sense the "deeper source" to both the uncertainty principle and the interference pattern is just wave mechanics. The only time you need quantum mechanics is if you want to ask what a single quantum will do, but if you just want to understand the overall pattern and think of any spreading as due to "uncertainty", then the whole thing also works with water waves going between two breaks in a jetty. In that case the uncertainty principle comes from the way every excitation of the water can serve as a source of new waves, and those new waves are seen to propagate out unless they destructively interfere (which they do quite a lot if the wavelength is short). That same destructive interference causes the pattern you see.
This connects to quantum mechanics when we connect, as peter0302 said, the wave amplitudes you see on the water with probabilities of the quantum showing up in those places.
Can someone confirm something that's been bothering me?
For double-slit interference to be visible to the naked eye, the light source needs to be a laser, or something collimated, correct? If you try it with a flashlight, you'll just get a blob on the other end, right?
Can we equate the collimated nature of laser light with a well-defined momentum which becomes ill-defined as soon as we simultaneously try to isolate it to a well-defined position (aka the slits)?
DrChinese
Gold Member
For double-slit interference to be visible to the naked eye, the light source needs to be a laser, or something collimated, correct? If you try it with a flashlight, you'll just get a blob on the other end, right?
There is a different reason that non-coherent light does not generate a double slit pattern when it is something like a flashlight. The beam is not in phase and so there is a tremendous amount of destructive interference. The fringe pattern is cancelled out. If you could send those same photons through the double slit apparatus a photon at a time, the pattern would emerge.
Ken G
Gold Member
Not to be combative, but it's not a question of phase, I would say, it's bandwidth. Lasers have coherent phase, but that's not necessary for a two-slit pattern, because the location of the nulls doesn't depend on the initial phase anyway. In other words, you could superimpose a thousand different laser beams and you'd still get the same pattern-- when they all produce nulls individually, they will also do that in tandem. But if the bandwidth is not tight, you are superimposing different wavelengths that do have nulls at different places, so you won't get the pattern. In short, you need a specific color to the beam, even sharper than the eye can discern. That holds even if you send the photons through one at a time, so the flashlight wouldn't work even at such a low intensity.
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You will have an interference pattern (but much more complicated) even with many colours, provided there is a phase relation between all the various colours.
Ken G
Gold Member
Why would there be a phase relation between the different colors? Perhaps a multicolored hologram? The OPer might think you are saying a flashlight would have such a relation.
Actually what I was getting at was simply whether the light needed to be parellel, and if that had anything to do with a "certain" momentum translating into an "uncertain" position when detected.
Ken G
Gold Member
Actually what I was getting at was simply whether the light needed to be parellel, and if that had anything to do with a "certain" momentum translating into an "uncertain" position when detected.
Yes, I was referring only to what DrChinese said. You are correct that the light being parallel connects to the pattern we see (and here lightarrow's comments are again relevant-- nonparallel yet patterned sources could yield their own different and possibly more complicated patterns). Also, the overall spreading of the diffraction pattern, regardless of whether one gets fringes or not, is indeed an example of the uncertainty principle. But I would have said it the other way-- knowledge about the lateral positioning as the light exits the slits translates into uncertainty in the lateral momentum, which in turn leads to spreading of the final position.
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Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.
I think there are other more down to earth examples of white light interference patterns as well.
Ken G
Gold Member
Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.
I think there are other more down to earth examples of white light interference patterns as well.
That's interesting, the bandwidth is about the same as the frequency, for sunlight, so the variance in the k vector is as large as the k vector itself. To get significant diffraction, one needs a large dot product between the k vector and the slit width, but to get fringes, you also need a small dot product between the variance in k and the slit width. So you can probably use sunlight if you are willing to compromise the spreading of the pattern, but the traditional "two-slit pattern" is probably not possible, I would expect.
Actually I think you can generate interference patters with white light. From red to blue is a narrow enough bandwidth provided your source is small enough and far enough away to get out of the near field range and into a far field arrangement. One example; sunlight from the sun is far enough away but the sun is too wide as a source making it a far field source. But in an eclipse during the phase when there is just thin crescent of light, I’ve seen walking interference patterns vibrating within the sunlight on the ground during a eclipse at the ABC islands a several years ago. That’s why they recommend laying out a white sheet so you can see the effect if the sky is clear enough.
I think there are other more down to earth examples of white light interference patterns as well.
What you mean like Young's slit experiment, I'm pretty sure he wasn't using lasers.
I was thinking the same thing. The electrons in Young's experiment were neither collimated nor in coherent phase. They were pretty random as I recall. So I think Dr.Chinese is right that if a flashlight emitted one photon at a time, (which, by definition, would be one precise color at a time) you'd get the pattern.
Which is why I continue to be astounded by the fact that you don't get the pattern from entangled photons unless you take affirmative steps to destroy the position certainty of the twin.
I was thinking the same thing. The electrons in Young's experiment were neither collimated nor in coherent phase. They were pretty random as I recall. So I think Dr.Chinese is right that if a flashlight emitted one photon at a time, (which, by definition, would be one precise color at a time) you'd get the pattern.
Which is why I continue to be astounded by the fact that you don't get the pattern from entangled photons unless you take affirmative steps to destroy the position certainty of the twin.
Just to be sure are you talking about the double slit quantum eraser experiment?
You would it is called the http://www.sciam.com/article.cfm?id=how-can-a-single-photon-p" which produces interference patterns when you fire one photon at a time.
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Just to be sure are you talking about the double slit quantum eraser experiment?
I am indeed. That and the Dopfer thesis as well.
I am indeed. That and the Dopfer thesis as well.
Yeah its been done here and its explainable in quantum theory.
http://arxiv.org/PS_cache/quant-ph/pdf/9903/9903047v1.pdf
Still very interesting though.
Mathematically, it turns out that the integral of the wave is well explained by theory. So the behaviour of the entangled photons is exactly what we would expect, given the conditions in the experiment.
It's much more adequately explained on those links and in much better detail.
Even in the case of the normal delayed choice quantum eraser setup where the which-path information is erased, the total pattern of photons on the screen does not show any interference, it's only when you look at the subset of signal photons matched with idler photons that ended up in a particular detector that you see an interference pattern. For reference, look at the diagram of the setup in fig. 1 of this paper:
http://xxx.lanl.gov/PS_cache/quant-ph/pdf/9903/9903047.pdf
In this figure, pairs of entangled photons are emitted by one of two atoms at different positions, A and B. The signal photons move to the right on the diagram, and are detected at D0--you can think of the two atoms as corresponding to the two slits in the double-slit experiment, while D0 corresponds to the screen. Meanwhile, the idler photons move to the left on the diagram. If the idler is detected at D3, then you know that it came from atom A, and thus that the signal photon came from there also; so when you look at the subset of trials where the idler was detected at D3, you will not see any interference in the distribution of positions where the signal photon was detected at D0, just as you see no interference on the screen in the double-slit experiment when you measure which slit the particle went through. Likewise, if the idler is detected at D4, then you know both it and the signal photon came from atom B, and you won't see any interference in the signal photon's distribution. But if the idler is detected at either D1 or D2, then this is equally consistent with a path where it came from atom A and was reflected by the beam-splitter BSA or a path where it came from atom B and was reflected from beam-splitter BSB, thus you have no information about which atom the signal photon came from and will get interference in the signal photon's distribution, just like in the double-slit experiment when you don't measure which slit the particle came through. Note that if you removed the beam-splitters BSA and BSB you could guarantee that the idler would be detected at D3 or D4 and thus that the path of the signal photon would be known; likewise, if you replaced the beam-splitters BSA and BSB with mirrors, then you could guarantee that the idler would be detected at D1 or D2 and thus that the path of the signal photon would be unknown. By making the distances large enough you could even choose whether to make sure the idlers go to D3&D4 or to go to D1&D2 after you have already observed the position that the signal photon was detected, so in this sense you have the choice whether or not to retroactively "erase" your opportunity to know which atom the signal photon came from, after the signal photon's position has already been detected.
This confused me for a while since it seemed like this would imply your later choice determines whether or not you observe interference in the signal photons earlier, until I got into a discussion about it online and someone showed me the "trick". In the same paper, look at the graphs in Fig. 3 and Fig. 4, Fig. 3 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D1, and Fig. 4 showing the interference pattern in the signal photons in the subset of cases where the idler was detected at D2 (the two cases where the idler's 'which-path' information is lost). They do both show interference, but if you line the graphs up you see that the peaks of one interference pattern line up with the troughs of the other--so the "trick" here is that if you add the two patterns together, you get a non-interference pattern just like if the idlers had ended up at D3 or D4. This means that even if you did replace the beam-splitters BSA and BSB with mirrors, guaranteeing that the idlers would always be detected at D1 or D2 and that their which-path information would always be erased, you still wouldn't see any interference in the total pattern of the signal photons; only after the idlers have been detected at D1 or D2, and you look at the subset of signal photons whose corresponding idlers were detected at one or the other, do you see any kind of interference.
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Oh, I know what the math and quantum theory say. I just still find it remarkable. :)
Ken G
Gold Member
Young did use a coherent source of light, though not a laser:
http://micro.magnet.fsu.edu/primer/java/interference/doubleslit/ says
"In order to test his hypothesis, Young devised an ingenious experiment. Using sunlight diffracted through a small slit as a source of coherent illumination, he projected the light rays emanating from the slit onto another screen containing two slits placed side by side."
As I said, the phase relation in a laser is irrelevant to the interference pattern, but the sharp color and the collimated direction are very relevant. As for entangled photons, the behavior you cite is exactly what quantum mechanics predicts. It would be more astounding if our macroscopic intuition trumped quantum mechanics in that single application but not all the other surprising things particles do.
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Oh, I know what the math and quantum theory say. I just still find it remarkable. :)
Yes it is, but only remarkable in as much as it confirms quantum theory in that no matter how we set up the experiment the results are consistent with quantum theory. Which I admit is pretty cool. But damn it it works as Ken G just said.
What you mean like Young's slit experiment, I'm pretty sure he wasn't using lasers.
...and not even the interferometer in Michelson-Morley experiment.
I thought Young used electrons.
Ken G
Gold Member
He used sunlight, but diffracted through a small slit to achieve a result similar to the collimation of a beam. The key requirement is to have the wave function for each photon to give a nearly constant phase over the whole slit, which means that you want parallel rays (such as from a tiny source) and a narrow bandwidth for each photon, and then also a fairly narrow bandwidth for the ensemble, but this last requirement seems the easiest to compromise on (as lightarrow said, it only alters the pattern moreso than destroying it). So you can get away with sunlight as long as you pass it through a narrow slit to collimate what impinges on the two slits. | 3,853 | 18,378 | {"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-2019-51 | latest | en | 0.907314 |
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A crate is dragged 3m along a smooth level floor by a 30n force applied at 25deg to the floor. Then it is pulled 4m up a ramp inclined at 20deg to the horizontal, using the same force. Then, the crate is dragged a further 5m along a level platform using the same force. Determine the total work done in moving the crate.
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# Just draw the part you need to solve the problem the
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Unformatted text preview: ine the value of p (You will not be able to draw the whole tree, since it is infinitely large. Just draw the part you need to solve the problem.) The basic idea of this problem is to figure out a way how we get exactly two “good day” messages. 2 p start 1-p 0.4 B 0.6 0.6 B 0.4 B1 B1 B2 B2 There are two possibilities for exactly two “good day” messages: • B true, B1 true, B1 true, B1 false =: way 1 • B false, B2 true, B2 true, B2 false =: way 2 The probabilities for each of the ways are: P ( way 1 ) = p · 2/5 · 2/5 · 3/5 = p · P ( way 2 ) = 12 125 (1 − p) · 3/5 · 3/5 · 2/5 = (1 − p) · 0.4 0.6 0.6 0.4 B1 B1 B2 B2 0.4 0.6 0.6 0.4 B1 B1 B2 B2 18 125 Therefore 3/25 = P ( “exactly 3...
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# exam_3_s07 - is convergent find its limit 2 Let p n be the...
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1. Define the sequence { a n } by setting a 1 = 1 and a n +1 = 1 1 + a n for n 1. Assuming that
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Unformatted text preview: is convergent, find its limit. 2. Let { p n } be the sequence of prime numbers, i.e. { p n } = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , . . . } . Show that ∞ X n =1 (-1) n +1 p n converges. 3. Find a power series representation for the function g ( x ) = x x-5 in powers of x and state for which values of x this representation is valid. 4. Find the interval of convergence of the power series ∞ X n =1 x n 3 n n 1 / 2 . Don’t forget to check for convergence at the endpoints of the interval! 5. Determine if the series converges or diverges. If it converges, determine if this convergence is absolute or conditional. a. ∞ X n =1 (-1) n n + 1 n 3-3 n 2 + 9 n + 13 b. ∞ X n =1 ± n 2 + 1 2 n 2 + 1 ² n c. ∞ X n =2 1 n (ln n ) 2 d. ∞ X n =1 (-1) n +1 4 √ n + 1 e. ∞ X n =1 (-1) n n n + 5 f. ∞ X n =1 1 5 + (-3) n Calculus II, Exam 3 Work Page...
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4-13: Students in a management science class have just recd their grades on the first test. The instructor has provided information about the first test grades in some previous classes as well as the final average for the same students. Some of these grades have been sampled and are as follows:
Student 1 2 3 4 5 6 7 8 9
1st test grade 98 77 88 80 96 61 66 95 69
Final avg. 93 78 84 73 84 64 64 95 76
a). develop a regression model that could be used to predict the final average in the course based on the first test grade.
b). predict the final avg. of a student who made an 83 on the first test
c). give the values of r and r(squared) for this model. Interpret the value of r (squared) in the context of this problem.
4-15: Using computer software, find the least squares regression line for the data in problem 4-13. Based on the F test, is there a statistically significant relationship between the first test grade and the final average in the course?
4-17: Accountants at the firm Walker and Walker believed that several traveling executives submit unusually high travel vouchers when they return from business trips. The accountants took a sample of 200 vouchers submitted from the past year; then they developed the following multiple regression equation relating expected travel cost (Y) to number of days on the road (X 1) and distance traveled (X 2) in miles:
^Y= \$90.00 + \$48.50X 1 + \$0.40X 2
4-27: A sample of 20 automobiles was taken, and the miles per gallon (MPG), horsepower, and total weights were recorded. Develop a linear regression model to predict MPG, using horsepower as the only independent variable. Develop another model with weight as the independent variable. Which of these two models is better? Explain.
MPG Horsepower Weight
44 67 1,844
44 50 1,998
40 62 1,752
37 69 1,980
37 66 1,797
34 63 2,199
35 90 2,404
32 99 2,611
30 63 3,236
28 91 2,606
26 94 2,580
26 88 2,507
4-28: Use the data in problem 4-27 to develop a multiple linear regression model. How does this compare with each of the models in problem 4-27?
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Newtonâs Rings Experiment Aim: To determine the wavelength of monochromatic light (sodium light) using These concentric rings are known as " Newton's Rings ". At the center the thickness of the air film formed between lens and glass plate is zero. Diameter of Newtonâs 15 th ring ⦠This question has been asked and answered previously. EXPERIMENT: 4 Object: To find the wavelength of Sodium light by Newtonâs ring. Newton rings are Alternate dark and bright rings formed due to the presence of air film when the plano-convex lens is placed on the glass plate is called newtons rings. They are observed when light is reflected from a plano-convex lens of a long focal length placed in contact with a plane glass plate. (a) Experimental set-up (b) Newtonâs rings Figure 1 In general, the path di erence between the re ected light beams which are undergoing interference (for oblique incidence) is given by An air wedge film can be formed by placing a Plano-convex lens on a flat glass plate. These rings are known as Newtonâs rings. 2] Sol: The given data are. Newtonâs Rings. Formula used: ⦠alternate dark and bright rings (see g.1(b)) with the point of contact between the lens and the plate at the center. Newtons Ring. In a Newtonâs rings experiment the diameter of the 15 th ring was found to be 0.59 cm and that of the 5 th ring is 0.336 cm. If the radius of curvature of plano-convex lens is much greater than distance ârâ and the system is viewed through the above, the pattern of dark & bright ring is observed. 0 5,024 2 minutes read. Newtonâs ring is a process in which Circular bright and dark fringes obtained due to air film enclosed between a Plano-convex lens and a glass plate. admin November 22, 2020. The global geometry of Newtonâs rings. 4. Letâs consider a dark ring with radius r at a point where the separation is t. The right angled triangle shown in red has a ⦠Apparatus used: A Plano convex lens of large radius of curvature, optical arrangement for Newtonâs rings, plane glass plate, sodium vapour lamp and traveling microscope. NEWTON'S RINGS are the circular interference pattern first discovered by Newton. Newtonâs ring experiment with animation. If the radius of curvature of the lens is 100 cm, find the wave length of the light. Formation of fringes When a plano-convex lens with large radius of curvature is placed on a plane glass plate such that its curved surface faces the glass plate, a wedge air film (of gradually increasing thickness) is formed between the lens and the glass plate. When a plano-convex lens is placed over a flat glass plate, then a thin air layer is formed between glass plate and a convex lens. View Newton's ring.pdf from MATH 111 at Pandit Deendayal Petroleum University. 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The circular interference pattern first discovered by Newton rings are fringes of equal thickness 4 Object: To find wavelength... Following list of colours from the centre ⦠Newtons ring ring ⦠concentric! Is zero light is reflected from a Plano-convex lens on a flat glass plate zero... Is reflected from a Plano-convex lens of a long focal length placed in with! Radius of curvature of the light 100 cm, find the wave length of the air formed... Colours from the centre ⦠Newtons ring of equal thickness with films of thickness! Placed in contact with a plane glass plate Thin film interference with films varying! Concentric rings are known as `` Newton 's ring.pdf from MATH 111 at Pandit Petroleum! ¦ These concentric rings are fringes of equal thickness by Newton of varying (. Following list of colours from the centre ⦠Newtons ring reflected from a Plano-convex lens on a flat glass.! Light by Newtonâs ring rings are the circular interference pattern first discovered by Newton from a Plano-convex on... Wave length of the light: ⦠Thin film interference with films of varying thickness Newtonâs! Lens of a long focal length placed in contact with a plane glass plate 111 at Pandit Petroleum. ( Newtonâs rings ): rings are known as `` Newton 's rings are the circular interference pattern discovered! Lens of a long focal length placed in contact with a plane glass plate Pandit Deendayal Petroleum University the! Th ring ⦠These concentric rings are the circular interference pattern first discovered by Newton formed. Formula used: ⦠Thin film interference with films of varying thickness Newtonâs...: rings are known as `` Newton 's rings `` formula used: ⦠film. | 2,031 | 8,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-39 | latest | en | 0.920866 |
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# Anonymity-Preserving Space Partitions
## File
LIPIcs.ISAAC.2021.32.pdf
• Filesize: 0.71 MB
• 16 pages
## Acknowledgements
We thank Daniel Lokshtanov for discussions relating to our approximation algorithm for Anonymity-Preserving Partition, as well as an anonymous reviewer, whose comments helped to improve the aforementioned result.
## Cite As
Úrsula Hébert-Johnson, Chinmay Sonar, Subhash Suri, and Vaishali Surianarayanan. Anonymity-Preserving Space Partitions. In 32nd International Symposium on Algorithms and Computation (ISAAC 2021). Leibniz International Proceedings in Informatics (LIPIcs), Volume 212, pp. 32:1-32:16, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2021)
https://doi.org/10.4230/LIPIcs.ISAAC.2021.32
## Abstract
We consider a multidimensional space partitioning problem, which we call Anonymity-Preserving Partition. Given a set P of n points in ℝ^d and a collection H of m axis-parallel hyperplanes, the hyperplanes of H partition the space into an arrangement A(H) of rectangular cells. Given an integer parameter t > 0, we call a cell C in this arrangement deficient if 0 < |C ∩ P| < t; that is, the cell contains at least one but fewer than t data points of P. Our problem is to remove the minimum number of hyperplanes from H so that there are no deficient cells. We show that the problem is NP-complete for all dimensions d ≥ 2. We present a polynomial-time d-approximation algorithm, for any fixed d, and we also show that the problem can be solved exactly in time (2d-0.924)^k m^O(1) + O(n), where k is the solution size. The one-dimensional case of the problem, where all hyperplanes are parallel, can be solved optimally in polynomial time, but we show that a related Interval Anonymity problem is NP-complete even in one dimension.
## Subject Classification
##### ACM Subject Classification
• Theory of computation → Design and analysis of algorithms
##### Keywords
• Anonymity
• Hitting Set
• LP
• Constant Approximation
• Fixed-Parameter Tractable
• Space Partitions
• Parameterized Complexity
## Metrics
• Access Statistics
• Total Accesses (updated on a weekly basis)
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## References
1. P K Agarwal. Geometric Partitioning and Its Applications. Discrete and Computational Geometry: Papers from the DIMACS Special Year (J. Goodman, R. Pollack, and W. Steiger, eds.), pages 1-37, 1991.
2. Pankaj K Agarwal and Micha Sharir. Applications of a new space-partitioning technique. Discrete & Computational Geometry, 9(1):11-38, 1993.
3. Ron Aharoni, Ron Holzman, and Michael Krivelevich. On a Theorem of Lovász on Covers in r-partite Hypergraphs. Combinatorica, 16(2):149-174, 1996.
4. Walid G Aref and Ihab F Ilyas. Sp-gist: An extensible database index for supporting space partitioning trees. Journal of Intelligent Information Systems, 17(2-3):215-240, 2001.
5. J. Baltes and J. Anderson. Flexible binary space partitioning for robotic rescue. In Proceedings 2003 IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS 2003) (Cat. No.03CH37453), volume 4, pages 3144-3149 vol.3, 2003.
6. Norbert Beckmann, Hans-Peter Kriegel, Ralf Schneider, and Bernhard Seeger. The R*-tree: an efficient and robust access method for points and rectangles. In Proceedings of the 1990 ACM SIGMOD international conference on Management of data, pages 322-331, 1990.
7. Jon Louis Bentley. Multidimensional binary search trees used for associative searching. Communications of the ACM, 18(9):509-517, 1975.
8. Hervé Brönnimann and Michael T Goodrich. Almost optimal set covers in finite vc-dimension. Discrete & Computational Geometry, 14(4):463-479, 1995.
9. J. Cortes, S. Martinez, T. Karatas, and F. Bullo. Coverage control for mobile sensing networks. IEEE Transactions on Robotics and Automation, 20(2):243-255, 2004.
10. Marek Cygan, Fedor V Fomin, Łukasz Kowalik, Daniel Lokshtanov, Dániel Marx, Marcin Pilipczuk, Michał Pilipczuk, and Saket Saurabh. Parameterized algorithms. Springer, 2015.
11. Pratyush Dayal and Neeldhara Misra. Deleting to Structured Trees. In International Computing and Combinatorics Conference, pages 128-139. Springer, 2019.
12. Adrian Dumitrescu, Joseph SB Mitchell, and Micha Sharir. Binary space partitions for axis-parallel segments, rectangles, and hyperrectangles. Discrete & Computational Geometry, 31(2):207-227, 2004.
13. Guy Even, Dror Rawitz, and Shimon Moni Shahar. Hitting sets when the vc-dimension is small. Information Processing Letters, 95(2):358-362, 2005.
14. Fedor V Fomin, Serge Gaspers, Dieter Kratsch, Mathieu Liedloff, and Saket Saurabh. Iterative compression and exact algorithms. Theoretical Computer Science, 411(7-9):1045-1053, 2010.
15. Sanjeev Khanna, S. Muthukrishnan, and Mike Paterson. On Approximating Rectangle Tiling and Packing. In Proceedings of the ninth annual ACM-SIAM Symposium On Discrete Algorithms, volume 95, page 384. SIAM, 1998.
16. Subhash Khot and Oded Regev. Vertex cover might be hard to approximate to within 2-ε. Journal of Computer and System Sciences, 74(3):335-349, 2008.
17. Kristen LeFevre, David J DeWitt, and Raghu Ramakrishnan. Mondrian multidimensional k-anonymity. In 22nd International conference on data engineering (ICDE'06), pages 25-25. IEEE, 2006.
18. Jiri Matousek. Lectures on discrete geometry, volume 212. Springer Science & Business Media, 2013.
19. Rolf Niedermeier and Peter Rossmanith. An efficient fixed-parameter algorithm for 3-hitting set. Journal of Discrete Algorithms, 1(1):89-102, 2003.
20. Hanan Samet. The quadtree and related hierarchical data structures. ACM Computing Surveys (CSUR), 16(2):187-260, 1984.
21. Adam Smith and Subhash Suri. Rectangular tiling in multidimensional arrays. Journal of Algorithms, 37(2):451-467, 2000.
22. Marc Van Kreveld, Otfried Schwarzkopf, Mark de Berg, and Mark Overmars. Computational geometry algorithms and applications. Springer, 2000. | 1,641 | 5,865 | {"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-2024-33 | latest | en | 0.765409 |
http://vedder.se/forums/search.php?st=0&sk=t&sd=d&sr=posts&author_id=717&start=50 | 1,555,951,855,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00257.warc.gz | 183,611,881 | 6,164 | ## Search found 255 matches
14 Oct 2017, 09:38
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
Throttle Response Update: In theory this update to the code improves the dynamic throttle response produced by the algorithm at times when full throttle would result in the motor amp limit setting or duty cycle limit setting being reached. Where: M= 100 = Throttle % Setting K= 90 = Desired Efficienc...
13 Oct 2017, 12:22
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
So your algorithm gets to propose a throttle setting to amp conversion depending on more than just the throttle, right? Well if your algorithm limits "max throttle" to < 5A at < 5km/h, then you will lose customers fast. Acceleration on my bike is unacceptably slow with 20A motor current. ...
09 Oct 2017, 13:20
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
You suggest that while I command 50A of motor current through the throttle, I would accept getting only 0.8A. Well, I don't. I think most people would agree that such a big difference is unacceptable. The general idea is that the more throttle you use, the worse the efficiency becomes. At low speed...
09 Oct 2017, 00:35
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
So, I managed to find a situation, a not so theoretical example, where you fail to reach the "desired efficiency" by getting a much higher efficiency. Now I'm going come up with an example where you fail to reach the desired efficiency by coming short a long ways.... Same motor, but now w...
05 Oct 2017, 23:54
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
@rew Suppose I have a (4) 140kv 0.02ohm hub motors on an electric skateboard with 84mm tires which can do 80A and a 10ah 30C 50V battery. During a lighthearted ride through the park I'm willing to draw up to 300a battery amps for a short time, & on the VESCs I use a 75a battery amp limit and an ...
03 Oct 2017, 13:05
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
Just a quick question. Say I have a 50V motor, currently running at 25V BEMF . Suppose the motor can do 50A, but the torque required is only 5A . The motor has 20mOhm resistance . What is the effective voltage that you propose and what is the efficiency of the motor? (the last example that I read, ...
02 Oct 2017, 14:16
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
please note: if X>0 & 0/X=error then this code won't work and the 9/28/17 12:25 (12 lines) code post will be necessary to use. if X>0 & 0/X=0 then this code should still work. Here I believe is a solution in theory to the potential error caused by zero division and division by zero in the c...
01 Oct 2017, 18:21
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
Efficiency Control in Least Code Steps I previously believed that 8 lines of code was the least possible number of steps for the main portion of efficiency control… but i’ve simplified it down to just 7 lines… K= 90 = Desired Efficiency % Setting L= 500 = Desired Min Watts Available Setting P= 4500...
01 Oct 2017, 17:43
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
Efficiency Control in Least Code Steps I previously believed that 9 lines of code was the least possible number of steps for the main portion of efficiency control… but i’ve simplified it down to just 8 lines… K= 90% = Desired Efficiency % Setting L= 500w = Desired Min Watts Available Setting P= 45...
01 Oct 2017, 15:34
Forum: General
Topic: "Peak Efficiency" Control Mode?
Replies: 110
Views: 24802
### Re: "Peak Efficiency" Control Mode?
Efficiency Control in Least Code Steps I previously believed that 12 lines of code was the least possible number of steps for the main portion of efficiency control… but i’ve simplified it down to just 9 lines… K= 90% = Desired Efficiency % Setting L= 500w = Desired Min Watts Available P= 4500w = D... | 1,196 | 4,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.891007 |
gk.palem.in | 1,696,463,857,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00447.warc.gz | 298,014,852 | 11,041 | # Sequence Indexing
## Introduction
The theory of Sequence Indexing deals with the notion of structured Enumeration, the mathematical concept of counting objects. It tries to answer one simple question : Given a collection of objects - how to create, access and identify various groups of objects drawn from that collection?
This article presents few classes that abstact the formulae for indexing and de-indexing some of the well-known Sequences.
By Sequence, here, we mean, collection of all possible groups that could be formed with any given set of symbols. The exact group formation mechanism may vary from Sequence to Sequence. For example, considering the Sequence {123,132,213,231,312,321 }, it was made of three symbols 1 ,2,3 and has six unique symbol-groups such as 123, 132 etc.. which are called sequents (1stSequent=123, 2ndSequent=132 ... 6thSequent=321 ). The number of sequents in any Sequence gives the length of that Sequence. For this example, the length of the Sequence is 6.
Depending on the mechanism of formation, each Sequence would have its own regulations and principles. In the following, we discuss the details of some of the common sequences we find useful in daily life, such as the Range Sequence, Permutation Sequence and the Combinational Sequence. These Sequences were abstracted into different classes aptly named: CRangeSequenceIndexer, CPermutationSeuqnceIndexer and CCombinationalSequenceIndexer respectively.
While these classes use different mathematical manipulations for their indexing mechanism, they all, nevertheless, have one single operation in common: Factorial. Written as n! to indicate Factorial(n), it is equivalent to n*(n-1)*(n-2)*...*3*2*1, the multiplication of all consecutive integers right from n down to 1 , it finds important role in all these classes. One of the nice properties of Factorial is its ability to be expressed in its own terms such as Factorial(n) = n * Factorial(n-1); For example,
Factorial(4) = 4! = 4 * 3 * 2 * 1 = 24;
Factorial(5) = 5! = 5 * 4 * 3 * 2 * 1 = 120;
Factorial(5) == 5*4! == 5*4*3! == 5*4*3*2! == 5*4*3*2*1!; //(The value of Factorial(1) is considered to be 1).
We can use this property to our advantage by pre-computing and storing the smaller factorial values to avoid the recomputation later for larger values. However, one big problem with Factorial is that, due to its multiplicative nature, the values grow out of bounds quite quickly. A typical 4-byte integer can not hold Factorial values for numbers beyond 15 or 20. We have to use larger data types (such as Arbitrary Precision Mathematical Library Types) if we need to support larger values. As for the sample code here, we have type defined a BIGNUMBER data type from unsigned long. You might later want to change it with your own larger data types as the need be. The type definitions and the Factorial function are presented below.
```typedef unsigned long BIGNUMBER; // Replace unsigned long with Arbitrary Precision Types to support Large Numbers
typedef BIGNUMBER POSITION; // Factorial value, Needs bigger types !!
typedef unsigned long SYMBOLINDEX; // Number of symbols can not exceed the maximum unsigned long number
typedef std::vector<symbolindex> SYMBVECTOR;
static const BIGNUMBER& Factorial(long lNumber)
{
if(lNumber <= 1) return m_FactArray[0];
if(lNumber <= (long)m_FactArray.size())
{
static BIGNUMBER Product;
Product = m_FactArray[m_FactArray.size() - 1];
for(long l = (long)m_FactArray.size(); l <= lNumber; ++l)
{
Product *= l;
m_FactArray.push_back(Product);
}
}
return m_FactArray[lNumber];
}
```
In the above code for the factorial function, we are storing all the computed factorial values in a big array (implemented with STL vector). When we are required to compute a factorial value for any new number, we first check if it has already been computed. If yes, then we return the stored value for that number. If no, we take the largest number we have computed the factorial value till now, and start computing from that number (instead of starting from 1 every time). As you could easily see, this simple mechanism saves lot of time (of course, at the cost of memory). With this simple factorial function in place, we can now turn our attention back to our Sequence classes.
### Range Sequence
This is the sequence that is formed by set of symbols each of which takes values from a fixed range. For example, a symbol denoting the day of a week can take values only from a fixed set of 7 weekdays, and a symbol that denotes the hour of a day can have only one of 24 values etc...Now considering an object made of two symbols, one for day and one for hour, then all the possible unique objects that can be made from those two symbols would make up a Range Sequence. Each element of the sequence would have a unique day-hour value that is different from that of all others. For example, consider a hypothetical class as defined below:
class WesternPersonDetails{
Sex sVal; // can take two values : { Male, Female}
Nature nVal; // can take three values:{ Good, Bad, Ugly}
Living lVal; // can take two values : { Alive, Dead }
};
The above class has three member variables each of which can take 2,3 and 2 values respectively. Now, all the possible unique objects that can be created for that class would makeup the following sequence.
Range[] = { 2, 3, 2 }; // Range[0] == 2; Range[1] == 3; Range[2] == 2; SymbolCount = 3; // SymbolCount == (sizeof(Range)/sizeof(Range[0]) ) SequenceLength = 2*3*2 = 12; // SequenceLength == Range[0] * Range[1] * Range[2] == 2 * 3 * 2 == 12 Index Generalized Sequence 0 { 0 0 0 } 1 { 0 0 1 } 2 { 0 1 0 } 3 { 0 1 1 } 4 { 0 2 0 } 5 { 0 2 1 } 6 { 1 0 0 } 7 { 1 0 1 } 8 { 1 1 0 } 9 { 1 1 1 } 10 { 1 2 0 } 11 { 1 2 1 }
As can be seen from the above table, there can be 12 unique objects for the afore mentioned class. Any other objects created would be just a repetition of one of these 12. In the above table, index 0 presents a generalized sequence {0,0,0} that represents an object {Male , Good , Alive } and index 1 presents a generalized sequence {0,0,1} that represents an objects {Male , Good , Dead } and so on. Given any class whose variables have ranges given by the array Range[] , the number of unique objects for that class can be obtained by multiplying all the range values as: Range[0]*Range[1]*Range[2]*... For any such class of unique objects, how to compute what would be the variable values for ith object for any given i without going through all the rest of the objects?
The class CRangeSequenceIndexer makes this possible with the function CRangeSequenceIndexer::GetAt(). It takes the index i as the argument and returns the ith unique object as the result. It does the computation without going through all the rest of the objects. Further, if one need to compute all of them instead, it is quite possible that the function could be called within a loop, as shown below, by single-step incrementing the loop-index from 0 to SequenceLength-1 , the value of SequenceLength having obtained from the function CRangeSequenceIndexer:: TotalSequenceLength();
```RANAGE Ranges[]={2,3,2};
CRangeSequenceIndexer CRIndexer(Ranges, sizeof(Ranges)/sizeof(Ranges[0]));
for(long i=0; i < CRIndexer.TotalSequenceLength(); ++i)
cout << CRIndexer.GetAt(i);
```
The above code snippet prints only the generalized sequence. One has to re-index the generalized symbols to get the exact sequence. This way, it becomes quite easy to adapt the class for any set of symbols instead of few hard coded ones. Some of the implementation details of CRangeSequenceIndexer are as follows.
```class CRangeSequenceIndexer
{
SYMBOLINDEX m_nSymbolCount;
POSITION m_SeqLengths[];
public:
CRangeSequenceIndexer(const RANGE* pRange, const SYMBOLINDEX lSymbolCount)
{
POSITION Product = 1;
m_SeqLengths[lSymbolCount] = 1; //We need this as Sentinel
for(long s = lSymbolCount - 1; s >= 0; --s)
{
m_SeqLengths[s] = Product * pRange[s];
Product = m_SeqLengths[s];
}
m_nSymbolCount = lSymbolCount;
}
};
SYMBVECTOR CRangeSequenceIndexer::GetAt(const POSITION& SeqPosIndex)
{
SYMBVECTOR SymbVector;
for(SYMBOLINDEX s = 0; s < m_nSymbolCount; ++s)
SymbVector[s] = (SeqPosIndex % m_SeqLengths[s]) / m_SeqLengths[s+1];
return SymbVector;
}
POSITION CRangeSequenceIndexer::GetIndexOf(const SYMBVECTOR& SymbolVector)
{
POSITION Index = 0;
for(SYMBOLINDEX s = 0; s < m_nSymbolCount; ++s)
Index += (SymbolVector[s] * m_SeqLengths[s+1]);
return Index;
}
```
The another important method supported by CRangeSequenceIndexer is GetIndexOf() . Given any sequent, this method would tell us the index of that sequent in constant time (w.r.to the number of sequents in the Sequence). This would be useful when we have the object information with us and we would like to know what is its position in the set of all possible objects. For example, considering our above hypothetical WesternPersonDetails class, suppose we have some object that has values {Female , Good , Alive }, and we would like to know what would be its index among the 12 possible objects. We can achieve this easily by first identifying its generalized sequent {1,0,0 } and then submitting it to the method CRangeSequenceIndexer:: GetIndexOf() . Identifying the generalized sequent is pretty straight forward in that all that need be done is replacing the Female with its index 1 (1 because Sex can take 2 values {Male,Female} and Female is in position 1), and Good with its index 0 (0 because Nature can take 3 values {Good,Bad,Ugly} and Good is in position 0), and so on. As can be seen from the table, the generalized sequent {1,0,0 } has index 6 and hence GetIndexOf() would return 6 .
The combination of CRangeSequenceIndexer::GetAt() and CRangeSequenceIndexer:: GetIndexOf() would prove quite useful in maintaining the persistence (storing and retrieving) of class objects. These two methods complement each other and hence complete the functionality of the Sequence. For complete implementation details please refer the source code available at: SequenceIndexingSourceCode.html
### Permutation Sequence
In mathematics the concept of a permutation expresses the idea of arranging objects in various different ways. For any given set of symbols, the exhaustive rearrangements of all the symbols would give us the Permutation Sequence. Each element of the Sequence (a sequent, that is) would correspond to one unique arrangement of the symbols. The position of symbol is the crucial concept here and so an arrangement 1234 is not same as 4321. For example, consider a scenario where we have 3 symbols {7,4,8} and we are supposed to rearrange them in various ways. Then the list of all possible arrangements would form the following sequence:
Symbol[] = { 7, 4, 8 }; // Symbol[0] == 7; Symbol[1] == 4; Symbol[2] == 8; SymbolCount = 3; // SymbolCount == (sizeof(Symbol)/sizeof(Symbol[0])) SequenceLength = 6; // SequenceLength == Factorial(SymbolCount) == 3! == 3 * 2 * 1 == 6 Index Generalized Sequence Exact Sequence 0 { 0 1 2 } 7 4 8 1 { 0 2 1 } 7 8 4 2 { 1 0 2 } 4 7 8 3 { 1 2 0 } 4 8 7 4 { 2 1 0 } 8 4 7 5 { 2 0 1 } 8 7 4
As can be seen from the above table, we have 6 ways of arranging 3 symbols. The first way (indicated by index 0 in the above table) is to simply place the symbols in their given order as 748 and the second way is to keep the first element intact while switching the positions of other two elements (as indicated by the generalized sequence {0 2 1} in the above table across index 1 ) resulting in 784 and so on. If we consider our total number of symbols as n, then the number of all possible ways of arranging them would be given by the mathematical formula n! = Factorial(n) = n*(n-1)*(n-2)*...*3*2*1 ; With so many ways of arrangements, how to compute what would be the ith way for any given i without going through all the rest of them?
The class CPermutationSequenceIndexer makes this possible with the function CPermutationSequenceIndexer::GetAt(). It takes the index i as the argument and returns the ith arrangement as the result. It does the computation without going through all the rest of the rearrangements. Further, if one need to compute all of them instead, it is quite possible that the function could be called within a loop, as shown below, by single-step incrementing the loop-index from 0 to SequenceLength-1 , the value of SequenceLength having obtained from the function CPermutationSequenceIndexer:: TotalSequenceLength();
```CPermutationSequenceIndexer CPIndexer(3);
for(long i=0; i < CPIndexer.TotalSequenceLength(); ++i)
cout << CPIndexer.GetAt(i);
```
The above code snippet prints only the generalized sequence. One has to re-index the generalized symbols to get the exact sequence. This way, it becomes quite easy to adapt the class for any set of symbols instead of few hard coded ones. Some of the implementation details of CPermutationSequenceIndexer are as follows.
```
typedef std::vector<SYMBOLINDEX> POSVECTOR;
class CPermutationSequenceIndexer
{
SYMBOLINDEX m_nSymbolCount;
SYMBVECTOR m_SymbVector;
POSVECTOR m_Pos;
public:
CPermutationSequenceIndexer(const SYMBOLINDEX lLength)
{
m_Pos.reserve(lLength + 16);
m_SymbVector.reserve(lLength + 16);
for(SYMBOLINDEX l=0; l < lLength; ++l)
{
m_Pos.push_back(l);
m_SymbVector.push_back(l);
}
m_nSymbolCount = lLength;
m_nCompleteSeqLength = Factorial(lLength);
}
};
SYMBVECTOR CPermutationSequenceIndexer::GetAt(const POSITION& SeqPosIndex )
{
SYMBOLINDEX lSwapVal;
SYMBOLINDEX lLength = m_nSymbolCount;
for(long j = lLength - 1; j >=0; --j)
{
lSwapVal = (SeqPosIndex % Factorial((lLength - j))) / Factorial(lLength - (j+1));
lSwapVal += j;
m_Pos[j] = j;
m_SymbVector[j] = lSwapVal;
m_SymbVector[m_Pos[lSwapVal]] = j;
m_Pos[j] = m_Pos[lSwapVal];
m_Pos[lSwapVal] = j;
}
return m_SymbVector;
}
POSITION CPermutationSequenceIndexer::GetIndexOf(SYMBVECTOR SymbolVector)
{
SYMBOLINDEX lLength = (SYMBOLINDEX) m_nSymbolCount;
for(long j = 0; j < lLength; ++j)
m_Pos[SymbolVector[j]] = j;
POSITION Index = 0;
for(SYMBOLINDEX j = 0; j < lLength; ++j)
{
Index += ((SymbolVector[j]-j) * Factorial(lLength - (j+1)));
SymbolVector[m_Pos[j]] = SymbolVector[j];
m_Pos[SymbolVector[j]] = m_Pos[j];
m_Pos[j] = j;
SymbolVector[j] = j;
}
return Index;
}
```
The another important method supported by CPermutationSequenceIndex is GetIndexOf() . Given any sequent, this method would tell us the index of that sequent in constant time (w.r.to the number of sequents in the Sequence). This would be quite useful when we have some particular arrangement with us whose position we would like to know in the set of all possible arrangements. For example, considering our above scenario of 3 symbols {7,4,8} , suppose we have one arrangement 478 , and we would like to know what is its index among the 6 possible arrangements. We can achieve this easily by first identifying its generalized sequent {1,0,2 } and then submitting it to the method CRangeSequenceIndexer:: GetIndexOf() . Identifying the generalized sequent is pretty straight forward in that all that need be done is replacing the 4 with its index 1, 7 with its index 0, and so on. As can be seen from the table, the generalized sequent {1,0,2 } has index 2 and hence GetIndexOf() would return 2 .
The combination of CPermutationSequenceIndexer::GetAt() and CPermutationSequenceIndexer:: GetIndexOf() would prove quite useful in simulating the randomness, among other things. For example, if we associate one event with each of the numbers 4,8 and 7 then we can change the order in which events occur just by changing the index for the arrangements, and conversely we can as well predict the process that is creating the events by just observing the order of the events and finding the corresponding index for that order. These two methods GetAt() and GetIndexOf() complement each other and hence complete the functionality of the Sequence. For complete implementation details please refer the source code available at: SequenceIndexingSourceCode.html
### Combinational Sequence
Combinational Sequences are the result of selection of few symbols out of many. In contrast to the Permutation Sequence (where we care for the position of the symbol), Combinational Sequences do not respect the positional differences of symbols. Thus, both 1234 and 1432 are treated as one and the same here. It is the combination that makes the difference and not the position. For example, consider a scenario where we have 5 symbols {7,4,8,1,5} and we are supposed to choose 3 out of them. Then the list of all possible selections would form the following sequence:
Symbol[] = { 7, 4, 8, 1, 5 }; n = SymbolCount = 5; // SymbolCount == (sizeof(Symbol)/sizeof(Symbol[0])) r = Choice = 3; // Any value such that 0 <= Choice <= SymbolCount SequenceLength = 10; // SequenceLength == ncr == n! /(r! (n-r)!) == 5! / (3!*2!) == 120/12 == 10 Index Generalized Sequence Exact Sequence 0 { 0 1 2 } 7 4 8 1 { 0 1 3 } 7 4 1 2 { 0 1 4 } 7 4 5 3 { 0 2 3 } 7 8 1 4 { 0 2 4 } 7 8 5 5 { 0 3 4 } 7 1 5 6 { 1 2 3 } 4 8 1 7 { 1 2 4 } 4 8 5 8 { 1 3 4 } 4 1 5 9 { 2 3 4 } 8 1 5
As can be seen from the above table, we have 10 ways of selecting 3 symbols out of 5 . The first way (indicated by index 0 in the above table) is to simply choose the first three symbols 748 and the second way is to choose the first two symbols followed by fourth symbol (indicated by the generalized sequence {0 1 3} in the above table across index 1 ) resulting in 741 and so on. If we consider our total number of symbols as n and the number of symbols to choose as r, then the number of ways of selecting r symbols out of n symbols is given by the mathematical formula ncr = Factorial(n) / (Factorial(r)*Factorial(n-r)); With so many ways of selection, how to compute what would be the ith way for any given i without going through all the rest of them?
The class CCombinationalSequenceIndexer makes this possible with the function CCombinationalSequenceIndexer::GetAt(). It takes the index i as the argument and returns the ith combination as the result. It does the computation without going through all the rest of the combinations. Further, if one need to compute all the possible combinations instead, it is quite possible that the function could be called within a loop, as shown below, by single-step incrementing the loop-index from 0 to SequenceLength-1 , the value of SequenceLength having obtained from the function CCombinationalSequenceIndexer:: TotalSequenceLength();
```CCombinationalSequenceIndexer CSIndexer(5, 3);
for(long i=0; i < CSIndexer.TotalSequenceLength(); ++i)
cout << CSIndexer.GetAt(i);
```
The above code snippet prints only the generalized sequence. One has to re-index the generalized symbols to get the exact sequence. This way, it becomes quite easy to adapt the class for any set of symbols instead of few hard coded ones. Some of the implementation details of CCombinationalSequenceIndexer are as follows.
```
class CCombinationalSequenceIndexer
{
SYMBOLINDEX m_nSymbolCount;
SYMBOLINDEX m_nChoice;
public:
CCombinationalSequenceIndexer(SYMBOLINDEX NoofSymbolsExisting, SYMBOLINDEX NoofSymbolsToChoose)
{
m_nSymbolCount = NoofSymbolsExisting;
m_nChoice = NoofSymbolsToChoose;
}
POSITION ComputeNcR(SYMBOLINDEX n, SYMBOLINDEX r)
{
return Factorial(n) / (Factorial(r) * Factorial(n-r)) ;
}
};
SYMBVECTOR CCombinationalSequenceIndexer::GetAt(POSITION SeqPosIndex) // Compute Combination from Index
{
SYMBVECTOR SymbVector;
POSITION SeqLength, CumulativeSeqLength;
SYMBOLINDEX r = m_nChoice;
SYMBOLINDEX n = m_nSymbolCount;
for(SYMBOLINDEX s=0; s < m_nChoice; ++s)
{
CumulativeSeqLength = 0;
for(SYMBOLINDEX i=0, nMax = n - r; i <= nMax; ++i)
{
SeqLength = ComputeNcR(n-(i+1), r-1);
CumulativeSeqLength += SeqLength;
if(SeqPosIndex < CumulativeSeqLength)
{
SeqPosIndex += SeqLength;
SeqPosIndex -= CumulativeSeqLength;
SymbVector[s] = (m_nSymbolCount - n) + i;
r -= 1;
n -= (i+1);
break;
}
}
}
return SymbVector;
}
POSITION CCombinationalSequenceIndexer::GetIndexOf(const SYMBVECTOR& SymbolVector) // Compute Index from Combination
{
SYMBOLINDEX r = m_nChoice;
SYMBOLINDEX n = m_nSymbolCount;
POSITION Index = 0;
for(SYMBOLINDEX s=0; s < m_nChoice; ++s)
{
const SYMBOLINDEX& SelectedIndex = SymbolVector[s] - (m_nSymbolCount - n);
for(SYMBOLINDEX i=0 ; i < SelectedIndex; ++i)
Index += ComputeNcR(n-(i+1), r-1);
r -= 1;
n -= (SelectedIndex + 1);
}
return Index;
}
```
The another important method supported by CCombinationalSequenceIndexer is GetIndexOf() . Given any sequent, this method would tell us the index of that sequent in constant time (w.r.to the number of sequents in the Sequence). This would be useful when we have one set of choices with us and we would like to know what is its position in the set of all possible choices. For example, considering our above example of selecting 3 symbols out of {7,4,8,1,5} , suppose we have chosen 415 , and we would like to know what is its index among the 10 possible choices. We can achieve this easily by first identifying its generalized sequent {1,3,4 } and then submitting it to the method CRangeSequenceIndexer:: GetIndexOf() . Identifying the generalized sequent is pretty straight forward in that all that need be done is replacing the 4 with its index 1, 1 with its index 3, and so on. As can be seen from the table, the generalized sequent {1,3,4 } has index 8 and hence GetIndexOf() would return 8 .
The combination of CCombinationalSequenceIndexer::GetAt() and CCombinationalSequenceIndexer:: GetIndexOf() would prove quite useful in all situations where selective ignorance (or elimination) is required. For example, consider a situation where we need to prepare a questionnaire for r questions out of a data base of n questions. We need to guarantee that the questionnaire we prepare is not a duplicate of any of the old questionnaires. We can achieve this by using the indexing to generate the questionnaire and then comparing the index for any past duplicates. This index comparision for the set of selected questions is far efficient compared with the method of comparing each and every question against each and every one from the past. These two methods GetAt() and GetIndexOf() complement each other and hence complete the functionality of the Sequence. For complete implementation details please refer the source code available at: SequenceIndexingSourceCode.html
### Conclusions
Theory of Sequence Indexing deals with the notion of structured Enumeration. It is about ordering and indexing the objects based on their properties. The classes presented here index different kinds of Sequences suitable for different purposes at different times. Range Sequences are suitable for indexing variables with well known ranges such as class objects, graph statistics etc.. Permutation Sequences are suitable for indexing exhaustive rearrangements of any given set of symbols, such as enumerating all braches in a source code etc... and the Combination Sequences find their usage when choosing a subset of symbols out of many, such as picking up samples for observations etc... Given suitable details, these classes help computing any element of a Sequence without going through all the rest of the elements.
By
P.GopalaKrishna | 5,957 | 23,314 | {"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} | 4.1875 | 4 | CC-MAIN-2023-40 | longest | en | 0.922954 |
https://community.splunk.com/t5/Splunk-Search/Search-multiple-values-from-a-single-event-where-one-value-might/m-p/619606 | 1,716,025,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057379.11/warc/CC-MAIN-20240518085641-20240518115641-00543.warc.gz | 156,482,749 | 38,559 | Splunk Search
## Search multiple values from a single event where one value might be less than 800?
Explorer
I have the following criteria from a single event that appears like:
Time Event
11/4/22
4:10:28.000 AM
{ [-]
Total: 6656
srv110: 1002
srv111: 1105
srv112: 1007
srv113: 995
srv114: 1269
srv115: 1278
}
<My Query>| timechart span=1m values(srv*) will return the values as so:
_time values(srv110) values(srv111) values(srv112) values(srv113) values(srv114) values(srv115) 11/4/2022 4:04 1003 1105 1007 996 1268 1278
But I need to return all of them as so even if any one of those values falls under 800 but also greater than -1.
I attempted to transpose and search from there but I'm failing somewhere.
Any help or nudge in the right direction would be greatly appreciated. Thank you!
Labels (1)
• ### timechart
1 Solution
SplunkTrust
Can you please try this?
``````YOUR_SEARCH
| bin span=1m _time
| stats values(srv*) as srv* by _time
| eval flag = 0
| foreach srv*
[ eval flag = if(flag == 0 AND <<FIELD>><800 AND <<FIELD>> > -1, 1 , flag)
]
| where flag=1
| fields - flag, srvTotal``````
Thanks
KV
If any of my replies help you to solve the problem Or gain knowledge, an upvote would be appreciated.
Explorer
{"srv110": 1001, "srv111": 1104, "srvTotal": 6651, "srv112": 1006, "time": "2022-11-04T08:47:02Z", "srv113": 995, "srv114": 1268, "srv115": 1277}
SplunkTrust
Can you please try this?
``````YOUR_SEARCH
| bin span=1m _time
| stats values(srv*) as srv* by _time
| eval flag = 0
| foreach srv*
[ eval flag = if(flag == 0 AND <<FIELD>><800 AND <<FIELD>> > -1, 1 , flag)
]
| where flag=1
| fields - flag, srvTotal``````
Thanks
KV
If any of my replies help you to solve the problem Or gain knowledge, an upvote would be appreciated.
SplunkTrust
Can you please share _raw from the sample event?
``index=YOUR_INDEX | table _raw``
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IT leaders are choosing Splunk Cloud as an ideal cloud transformation platform to drive business resilience, ... | 718 | 2,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.799956 |
https://destinationnative.com/qa/how-do-you-do-7-divided-by-8.html | 1,624,538,547,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00120.warc.gz | 197,490,380 | 6,822 | # How Do You Do 7 Divided By 8?
## How do you write 7/12 as a decimal?
Explanation: So to convert 712 into a decimal, you divide 7 by 12 which gives us 0.583333333 ..
## What is 7 divided by 8 as a fraction?
78 is 7 divided by 8, which equals 0.875. So an equivalent fraction is another fraction that also equals 0.875. To find this fraction, just take any number and multiply both the numerator and denominator by this number. 7×28×2 equals 1416 because it too equals 0.875.
## How do you do Division step by step?
The steps are more or less the same, except for one new addition:Divide the tens column dividend by the divisor.Multiply the divisor by the quotient in the tens place column.Subtract the product from the divisor.Bring down the dividend in the ones column and repeat.
## What is 7/8 in a decimal?
0.8757/8 as a decimal is 0.875.
## Is 7/8 a terminating or repeating decimal?
1 Expert Answer To do this without a calculator, divide 7 by 8 longhand. Alas, I can’t really duplicate this, but the answer is . 875. It does not repeat, it terminates.
## What is 7/8 as a percentage?
87.5%Fraction to percent conversion tableFractionPercent5/862.5%6/875%7/887.5%1/911.111111%41 more rows
## What is 3 over 7 as a decimal?
How to Write 3/7 as a Decimal?FractionDecimalPercentage4/70.571457.14%3/70.428642.86%2/70.285728.57%3/40.7575%5 more rows
## What is 9 over 20 as a decimal?
0.45 is a decimal and 45/100 or 45% is the percentage for 9/20.
## How do you write 5 divided by 8?
5 divided by 8 is 0.625.
## What divided by what gives you 7?
Therefore, the answer to what divided by 7 equals 7 is 49. You can prove this by taking 49 and dividing it by 7, and you will see that the answer is 7. Tip: For future reference, when you are presented with a problem like “What divided by 7 equals 7?”, all you have to do is multiply the two known numbers together.
## How do you turn 7/8 into a decimal?
For example, to convert the fraction 7/8 to a decimal using a calculator, simply perform 7 divided by 8 and press enter. The resulting decimal would be 0.875.
## What is 7/8 as a decimal and percent?
87.5%Common Fractions with Decimal and Percent EquivalentsFractionDecimalPercent1/80.12512.5%3/80.37537.5%5/80.62562.5%7/80.87587.5%21 more rows•Feb 21, 2017 | 679 | 2,285 | {"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-2021-25 | latest | en | 0.915013 |
https://www.mvorganizing.org/what-is-integrated-reasoning-in-gmat/ | 1,643,154,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00004.warc.gz | 917,757,671 | 13,174 | # What is Integrated Reasoning in GMAT?
## What is Integrated Reasoning in GMAT?
The Integrated Reasoning section gives schools another data point to differentiate among candidates. The Integrated Reasoning section, consisting of 12 questions covering four question types to be completed in 30 minutes, requires test takers to analyze and synthesize data in different formats from multiple sources.
## Is Integrated Reasoning important in GMAT?
41% said that they found Integrated Reasoning to be an important part of overall GMAT score evaluation. Kaptest did the same survey again in 2015. This time the number of business schools considering the IR section to be important was up to 59%.
## What is a good integrated reasoning score GMAT?
A high GMAT score on the Integrated Reasoning section is roughly considered to be 6 or above. 11% of test-takers get a perfect score on the Analytical Writing Assessment. A high GMAT score on the AWA is usually considered to be 5.0 or above.
## Does Integrated Reasoning affect GMAT score?
Integrated Reasoning scores range from 1 to 8, in single-digit increments. So, your Integrated Reasoning score could be an 8, or a 7, or a 6 (and so on). Your score will be separate from your Quantitative and Verbal scores, and it will not factor into your total GMAT score out of 800.
## Is IR calculated in GMAT score?
GMAT IR Score is reported separately and does not affect your overall score out of 800. The GMAT IR section is not Computer adaptive, and hence questions of varying difficulty levels can appear at any point in the test. Scaled scores for GMAT IR range from 1 – 8 (reported in 1-point intervals).
## Is it hard to get 500 on GMAT?
I would say it depends what level you stand as of today. If you have a little flair of Quant and verbal then it’s sufficient time to get to the level of 500-600. 500-600 is generally considered an easy score for most GMAT takers but even for that scpre you need to understand basic principles of CR, SC, RC, PS and DS.
## Is it hard to get a 450 on GMAT?
It should not be too difficult. That is considered a fairly low score. If you are nervous about it, you could buy a study book and download the free practice test from the GMAT website. A few practice tests should help you get to the 450, and they will help you identify where you should focus your time studying.
## Can I pass the GMAT without studying?
In a way the GMAT is designed to be taken without much preparation. The quant questions use basic math for building blocks and are complex mostly because of the way they are designed.
## Is 400 a bad GMAT score?
It’s somewhat uncommon for students to get either very low or very high scores on the GMAT. Relatively few students receive below a 400 or above a 600; according to the GMAC, two-thirds of test-takers receive a score between 400 and 600.
## Is a 730 GMAT score good?
Most top business schools, like Harvard, Stanford and Wharton, have averages around 730. This means that a 740 score, “good” by most standards, is only +10 from a school’s average, but a 770 score is +40 points. Those +40 points can be used to directly raise the average.
## How can I prepare for GMAT in 15 days?
Make sure you’re sleeping, but if you can take time off work, do. Don’t cram on the weekends only – that might work if you had 6 months, but with only one month to study, you’ll need to do at least some GMAT every single day.
## How can I apply for GMAT 2020?
GMAT 2020 Application Form
1. Candidates can fill their application form through online mode.
2. Candidates need to enter their name, email and password for registration.
3. Candidates will have to fill all the correct details in the application form including personal and academic details.
## How do I get my GMAT waived?
4 Easy Steps to Get a GMAT Waiver in 2021
1. Business schools in the US that accept low scores or a GMAT waiver.
2. Find out what are the options for replacing the GMAT.
3. Write a great letter.
4. Make your professional experience stand out.
## Is GMAT mandatory for MBA?
The short answer is yes, the GMAT is required for the majority of MBA programs. Most business schools require the exam and place a major emphasis on it in the admissions process. Business schools tend to require the GMAT for two primary reasons.
## Is GMAT still online?
GMAT Online Will Be a Permanent Option: 2021 Changes Basically, the online GMAT was such a success in 2020 that the GMAC has made it permanent. Now, even though in-person test centers are opening back up, test-takers will still have the option of taking the exam online.
## Which is better GRE or GMAT?
Although the quantitative section is harder on the GMAT than on the GRE for most test-takers, the GMAT may be easier for those who prefer logic problems over geometry questions because there are more geometry questions on the GRE, experts say. Fit the GMAT Into Overall MBA Application Strategy. ] Language skills. | 1,138 | 4,965 | {"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-2022-05 | latest | en | 0.933957 |
https://kr.mathworks.com/matlabcentral/answers/526833-matrix-manipulation-for-color-spaces?s_tid=prof_contriblnk | 1,726,134,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651440.11/warc/CC-MAIN-20240912074814-20240912104814-00592.warc.gz | 316,239,617 | 37,981 | # matrix manipulation for color spaces.
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Malini Bakthavatchalam 2020년 5월 19일
답변: vecdi 2024년 6월 11일
Hi , I have an image . I want to convert that to [3 3] matrix value to play with color space. I understand imead will convert image to matrix form but if I want 3 3 matrix, How should I proceed?
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darova 2020년 5월 19일
Can you explain more? What size of your image? And what kind of conversion you want?
Malini Bakthavatchalam 2020년 5월 19일
tVersion: ''
Width: 480
Height: 502
BitDepth: 24
ColorType: 'truecolor'
FormatSignature: ''
NumberOfSamples: 3
This is my image size. The image is just a face of a girl. So I want to convert this image in DKL space and work on illusion project.
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### 채택된 답변
Walter Roberson 2020년 5월 19일
편집: Walter Roberson 2020년 5월 19일
You probably do not want a 3 x 3 images. What you probably want is to let T be a 3 x 3 transformation matrix, and RGB be your RGB image, then
M = reshape(RGB, [], 3);
transformed = M * T;
nonRGB = reshape(transformed, size(RGB));
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Malini Bakthavatchalam 2020년 5월 25일
Yes, thank you I got your point, so I wrote my final code for converting into DKL color space back into RGB..
clear all
close all
clc
MyImrgb = reshape(im2double(RGB), [],3);
Imrgb = MyImrgb.^2.2; %gamma correction
[x, y, z] = size(Imrgb);
ldrgyv2rgbMat = [1,1,0.236748566577269;
1,-0.299338211934457,-0.235643322285071;
1,0.0137437185685517,1]; % B (mat based on the classic calibration technic)
MyImrgbCol = reshape(Imrgb, [x*y, z]);
MyImldrgyvCol = ldrgyv2rgbMat\(MyImrgbCol'*2 - 1);
figure(1), imshow(RGB)
figure(2), imshow(MyImldrgyvCol)
But i get error
Error using \
Matrix dimensions must agree.
Error in dkltry2 (line 12)
MyImldrgyvCol = ldrgyv2rgbMat\(MyImrgbCol'*2 - 1);
Malini Bakthavatchalam 2020년 5월 25일
Also, I have one more doubt, I used an color thresholder to remove my background. and I used that image in the color space transformation. but now when I use my complete code for the project, it is still calculating my histogram with background so how can i solve the issue.. I am attaching my complete code here
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### 추가 답변 (1개)
vecdi 2024년 6월 11일
MyImrgb = reshape(im2double(RGB), [],3);
Imrgb = MyImrgb.^2.2; %gamma correction
[x, y, z] = size(Imrgb);
ldrgyv2rgbMat = [1,1,0.236748566577269;
1,-0.299338211934457,-0.235643322285071;
1,0.0137437185685517,1]; % B (mat based on the classic calibration technic)
MyImrgbCol = reshape(Imrgb, [x*y, z]);
MyImldrgyvCol = ldrgyv2rgbMat\(MyImrgbCol'*2 - 1);
figure(1), imshow(RGB)
figure(2), imshow(MyImldrgyvCol)
But i get error
Error using \
Matrix dimensions must agree.
Error in dkltry2 (line 12)
MyImldrgyvCol = ldrgyv2rgbMat\(MyImrgbCol'*2 - 1);
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### 카테고리
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Translated by | 1,045 | 2,979 | {"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 | 0.592099 |
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# Pixel Types
576
3
01-03-2013 08:41 AM
New Contributor III
Dear all,
there is a raster layer I would like to convert to polygons witrh the tool "Raster To Polygon". But that doesn't work. If I got it right the pixel type must be "integer", but in fact it is "floating point".
With the tool "Copy Raster" I wanted to change the pixel type. There are a lot of pixel types available, but several tests resulted in raster images different from the original.
How do I know which pixel type to choose and what do I have to observe to get a raster image identical to the original?
Thanks in advance and kind regards,
Matthias
Tags (3)
1 Solution
Accepted Solutions
Esri Esteemed Contributor
Hi Matthias,
If you have 3D or Spatial Analyst, use the Int tool to convert the floating point raster to an integer raster. The decimal values will be dropped from each pixel.
If you need to maintain the decimal values, first use the Times tool and multiply the raster by a constant to preserve the number of decimals. For example, if you need to maintain 3 decimals, multiply the raster by 1000. Then, execute the Int tool, followed by the Raster to Polygon tool.
Within your polygon shapefile/feature class you will have a Value field. You can then use the field calculator to divide by the constant used for the Times tool, and you will then have the original pixel values.
3 Replies
Esri Esteemed Contributor
Hi Matthias,
If you have 3D or Spatial Analyst, use the Int tool to convert the floating point raster to an integer raster. The decimal values will be dropped from each pixel.
If you need to maintain the decimal values, first use the Times tool and multiply the raster by a constant to preserve the number of decimals. For example, if you need to maintain 3 decimals, multiply the raster by 1000. Then, execute the Int tool, followed by the Raster to Polygon tool.
Within your polygon shapefile/feature class you will have a Value field. You can then use the field calculator to divide by the constant used for the Times tool, and you will then have the original pixel values.
New Contributor III
Dear Jake, | 491 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.807483 |
http://convertwizard.com/59305-foots_minute-to-meters_second | 1,686,020,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652207.81/warc/CC-MAIN-20230606013819-20230606043819-00750.warc.gz | 12,702,050 | 17,332 | # 59305 Foots/Minute to Meters/Second (59305 ft/min to m/s)
Convert 59305 Foots/Minute to Meters/Second (ft/min to m/s) with our conversion calculator and conversion tables. To convert 59305 ft/min to m/s use direct conversion formula below.
59305 ft/min = 301.2694 m/s.
You also can convert 59305 Foots/Minute to other Speed (popular) units.
59305 FOOTS/MINUTE
=
301.2694 METERS/SECOND
Direct conversion formula: 1 Foots/Minute * 196.85039370079 = 1 Meters/Second
Amount:
From:
To:
## Conversion table: Foots/Minute to Meters/Second
FOOTS/MINUTE METERS/SECOND
1 = 0.00508
2 = 0.01016
3 = 0.01524
4 = 0.02032
5 = 0.0254
7 = 0.03556
8 = 0.04064
9 = 0.04572
10 = 0.0508
METERS/SECOND FOOTS/MINUTE
1 = 196.85039370079
2 = 393.70078740157
3 = 590.55118110236
4 = 787.40157480315
5 = 984.25196850394
7 = 1377.9527559055
8 = 1574.8031496063
9 = 1771.6535433071
10 = 1968.5039370079
## Nearest numbers for 59305 Foots/Minute
FOOTS/MINUTE METERS/SECOND
23023 ft/min = 116.95684 m/s
24500 ft/min = 124.46 m/s
25000 ft/min = 127 m/s
26800 ft/min = 136.144 m/s
28000 ft/min = 142.24 m/s
29354 ft/min = 149.11832 m/s
30000 ft/min = 152.4 m/s
31000 ft/min = 157.48 m/s
31680 ft/min = 160.9344 m/s
33000 ft/min = 167.64 m/s
35280 ft/min = 179.2224 m/s
36960 ft/min = 187.7568 m/s
38000 ft/min = 193.04 m/s
42000 ft/min = 213.36 m/s
46600 ft/min = 236.728 m/s
55000 ft/min = 279.4 m/s
60000 ft/min = 304.8 m/s
67500 ft/min = 342.9 m/s
144000 ft/min = 731.52 m/s
999999999999 ft/min = 5079999999.9949 m/s | 623 | 1,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-23 | latest | en | 0.490745 |
https://mathemania.com/monty-hall-problem/ | 1,696,233,736,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00161.warc.gz | 399,040,897 | 34,475 | The Monty Hall Problem
# The Monty Hall Problem
The Monty Hall problem is a probability puzzle based on the American television game show whose host was Monty Hall. The popular show was called Let’s Make a Deal.
Imagine you’re having a really great day, and you’re feeling very lucky so you decide to go on the show. Monty Hall picks you for one of the games. You see three doors.
These are the rules of the game: Behind two of these three doors are goats, and behind one door is a brand new car. (Let’s say you’re trying to get a car and not a goat.)
Let’s say you already chose one of them, for example the red one. The host now, who knows what’s behind the doors, opens the green doors which have a goat.
Now the host offers you the chance to change your decision. What do you do? Do you stick with the door you chose or do you change? It may seem odd, but the odds are not $50 – 50$. You should switch the door – by doing that you’ll have almost $67%$ chance of winning.
Why exactly should you change your decision?
First you have three choices. The chance of winning is $1 : 3$. If you stick with the door you chose your chance will remain $1 : 3$. This means that there must be $\frac{2}{3}$ chance the car is somewhere else. If we know that the car is not in the last door, this means that there is $\frac{2}{3}$ chance the car is behind the yellow door. It is not certain that the car will be behind the yellow doors, but there is twice as much chance he is behind the yellow door than it is behind the red door.
Now let’s imagine having 100 doors. Now you got yourself in a kind of a bad situation. You have only $\frac{1}{100}$ chance to get it right. Now what if the host opens 98 doors with all goats behind them? Now you are left with only 2 doors. This becomes a bit clearer. The chance you got it right remained $\frac{1}{100}$, but the chance that the right door is somewhere among the other 99 door is now concentrated on the door you didn’t pick. | 476 | 1,973 | {"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-2023-40 | latest | en | 0.958175 |
https://communities.sas.com/t5/SAS-Enterprise-Guide/Create-date-variable-from-week-month-year/td-p/142123?nobounce | 1,534,450,347,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211167.1/warc/CC-MAIN-20180816191550-20180816211550-00684.warc.gz | 629,674,689 | 32,101 | Create date variable from week, month, year
Solved
Occasional Contributor
Posts: 14
Create date variable from week, month, year
I have week, month, and year variables and need to create a date variable from these. Suggestions on how to do this?
Accepted Solutions
Solution
06-13-2014 02:48 PM
Super User
Posts: 23,992
Re: Create date variable from week, month, year
Aligns to the beginning of the week;
date=mdy(1,1,year)+(week-1)*7;
OR
intnx('week', mdy(1,1, year), week, 'middle');
And look at the alignment options for intnx function, ie beginning end, middle.
SAS(R) 9.2 Language Reference: Dictionary, Fourth Edition
All Replies
Super User
Posts: 23,992
Re: Create date variable from week, month, year
How does your week look, it is 25 indicating 25th week of the year or 2 indicating the 2nd week of a month?
Assuming its between 1 and 5
something like the following may work:
date=mdy(month, 7*(week-1)+1 , year);
Occasional Contributor
Posts: 14
Re: Create date variable from week, month, year
It is numbered 0 through 52, so 25 indicates the 25th week of the year.
Super User
Posts: 13,923
Re: Create date variable from week, month, year
Would you like the date to reflect any specific day of the week?
Occasional Contributor
Posts: 14
Re: Create date variable from week, month, year
Thursdays preferably, but if it is too difficult that way it doesn't particularly matter that much.
Super User
Posts: 23,992
Re: Create date variable from week, month, year
Weeks can cross months so having a specific day may not be possible
Occasional Contributor
Posts: 14
Re: Create date variable from week, month, year
Even just an approximate day would be ok.
e.g.
wk 1 = 1/1/14
wk 2 = 1/8/14
wk 3 = 1/15/14
etc.
Solution
06-13-2014 02:48 PM
Super User
Posts: 23,992
Re: Create date variable from week, month, year
Aligns to the beginning of the week;
date=mdy(1,1,year)+(week-1)*7;
OR
intnx('week', mdy(1,1, year), week, 'middle');
And look at the alignment options for intnx function, ie beginning end, middle.
SAS(R) 9.2 Language Reference: Dictionary, Fourth Edition
🔒 This topic is solved and locked. | 628 | 2,153 | {"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-2018-34 | latest | en | 0.833574 |
https://community.adobe.com/t5/acrobat-sdk/how-do-i-calculate-of-months-between-2-dates-in-acrobat-dc-form/td-p/9460115 | 1,611,537,604,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00028.warc.gz | 271,305,127 | 110,567 | ## How do I calculate # of months between 2 dates in Acrobat DC form?
Nov 03, 2017
Copied
Hi there,
Admitted JavaScript idiot here. Need to calculate number of months to 2 decimals between a move in and move out date.
I got it down to the number of days using what's below.
To make it easy, I only need to take the result from this script and divide by 30.
I do not need years. So a calculation of more than 12 months is fine.
Could someone help??
Thank you!!
// Custom calculate script
(function () {
var sStart = getField("MOVE_IN_DATE_CLIENT").valueAsString;
var sEnd = getField("MOVE_OUT_DATE_CLIENT").valueAsString;
var dStart, dEnd, diff;
if(sStart && sEnd) {
dStart = util.scand("mm/dd/yy", sStart);
dEnd = util.scand("mm/dd/yy", sEnd);
diff = dEnd.getTime() - dStart.getTime();
event.value = Math.floor(diff / 864e5);
} else {
event.value = "";
}
})();
Most Valuable Participant
| Most Valuable Participant
Change:
event.value = Math.floor(diff / 864e5);
To:
event.value = Math.floor(diff / 864e5) / 30;
TOPICS
Acrobat SDK and JavaScript
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## How do I calculate # of months between 2 dates in Acrobat DC form?
Nov 03, 2017
Copied
Hi there,
Admitted JavaScript idiot here. Need to calculate number of months to 2 decimals between a move in and move out date.
I got it down to the number of days using what's below.
To make it easy, I only need to take the result from this script and divide by 30.
I do not need years. So a calculation of more than 12 months is fine.
Could someone help??
Thank you!!
// Custom calculate script
(function () {
var sStart = getField("MOVE_IN_DATE_CLIENT").valueAsString;
var sEnd = getField("MOVE_OUT_DATE_CLIENT").valueAsString;
var dStart, dEnd, diff;
if(sStart && sEnd) {
dStart = util.scand("mm/dd/yy", sStart);
dEnd = util.scand("mm/dd/yy", sEnd);
diff = dEnd.getTime() - dStart.getTime();
event.value = Math.floor(diff / 864e5);
} else {
event.value = "";
}
})();
Most Valuable Participant
| Most Valuable Participant
Change:
event.value = Math.floor(diff / 864e5);
To:
event.value = Math.floor(diff / 864e5) / 30;
TOPICS
Acrobat SDK and JavaScript
Views
594
Likes
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Nov 03, 2017 0
6 Replies 6
Most Valuable Participant ,
Nov 03, 2017
Copied
Just divide the result by 12...
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Nov 03, 2017 1
Most Valuable Participant ,
Nov 03, 2017
Copied
Sorry, I meant by 30...
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Nov 03, 2017 1
Nov 03, 2017
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One can use the getFullYear(), getMoth() and getDate() methods to get the number of years, months and the date within the month respectively. With these values one can easily calculate the number or fractional months to use in the unused term calculation even when the dates cross a year boundary and the all months are assumed to have 30 days.
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Nov 03, 2017 1
Nov 03, 2017
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I'm ok with assuming a 30 day month for my purposes. The calculation is general for what we need it for. How would I divide this result by 30 using my existing script?
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Nov 03, 2017 0
Most Valuable Participant ,
Nov 03, 2017
Copied
Change:
event.value = Math.floor(diff / 864e5);
To:
event.value = Math.floor(diff / 864e5) / 30;
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Nov 03, 2017 2
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topology.uniform_space.completion
Hausdorff completions of uniform spaces #
THIS FILE IS SYNCHRONIZED WITH MATHLIB4. Any changes to this file require a corresponding PR to mathlib4.
The goal is to construct a left-adjoint to the inclusion of complete Hausdorff uniform spaces into all uniform spaces. Any uniform space α gets a completion completion α and a morphism (ie. uniformly continuous map) coe : α → completion α which solves the universal mapping problem of factorizing morphisms from α to any complete Hausdorff uniform space β. It means any uniformly continuous f : α → β gives rise to a unique morphism completion.extension f : completion α → β such that f = completion.extension f ∘ coe. Actually completion.extension f is defined for all maps from α to β but it has the desired properties only if f is uniformly continuous.
Beware that coe is not injective if α is not Hausdorff. But its image is always dense. The adjoint functor acting on morphisms is then constructed by the usual abstract nonsense. For every uniform spaces α and β, it turns f : α → β into a morphism completion.map f : completion α → completion β such that coe ∘ f = (completion.map f) ∘ coe provided f is uniformly continuous. This construction is compatible with composition.
In this file we introduce the following concepts:
• Cauchy α the uniform completion of the uniform space α (using Cauchy filters). These are not minimal filters.
• completion α := quotient (separation_setoid (Cauchy α)) the Hausdorff completion.
References #
This formalization is mostly based on N. Bourbaki: General Topology I. M. James: Topologies and Uniformities From a slightly different perspective in order to reuse material in topology.uniform_space.basic.
def Cauchy (α : Type u) :
Space of Cauchy filters
This is essentially the completion of a uniform space. The embeddings are the neighbourhood filters. This space is not minimal, the separated uniform space (i.e. quotiented on the intersection of all entourages) is necessary for this.
Equations
Instances for Cauchy
def Cauchy.gen {α : Type u} (s : set × α)) :
The pairs of Cauchy filters generated by a set.
Equations
theorem Cauchy.monotone_gen {α : Type u} :
@[protected, instance]
def Cauchy.uniform_space {α : Type u} :
Equations
theorem Cauchy.mem_uniformity {α : Type u} {s : set (Cauchy α × Cauchy α)} :
s uniformity (Cauchy α) (t : set × α)) (H : t , s
theorem Cauchy.mem_uniformity' {α : Type u} {s : set (Cauchy α × Cauchy α)} :
s uniformity (Cauchy α) (t : set × α)) (H : t , (f g : Cauchy α), t f.val.prod g.val (f, g) s
def Cauchy.pure_cauchy {α : Type u} (a : α) :
Embedding of α into its completion Cauchy α
Equations
@[protected, instance]
def Cauchy.complete_space {α : Type u} :
@[protected, instance]
def Cauchy.inhabited {α : Type u} [inhabited α] :
Equations
@[protected, instance]
def Cauchy.nonempty {α : Type u} [h : nonempty α] :
noncomputable def Cauchy.extend {α : Type u} {β : Type v} (f : α β) :
β
Extend a uniformly continuous function α → β to a function Cauchy α → β. Outputs junk when f is not uniformly continuous.
Equations
theorem Cauchy.extend_pure_cauchy {α : Type u} {β : Type v} {f : α β} (hf : uniform_continuous f) (a : α) :
= f a
theorem Cauchy.uniform_continuous_extend {α : Type u} {β : Type v} {f : α β} :
theorem Cauchy.Cauchy_eq {α : Type u_1} [inhabited α] {f g : Cauchy α} :
Lim f.val = Lim g.val (f, g)
@[protected, instance]
def uniform_space.completion (α : Type u_1) :
Type u_1
Hausdorff completion of α
Equations
Instances for uniform_space.completion
@[protected, instance]
Equations
@[protected, instance]
Equations
@[protected, instance]
@[protected, instance]
@[protected, instance]
@[protected, instance]
Automatic coercion from α to its completion. Not always injective.
Equations
@[protected]
theorem uniform_space.completion.coe_eq (α : Type u_1) :
theorem uniform_space.completion.comap_coe_eq_uniformity (α : Type u_1) :
filter.comap (λ (p : α × α), ((p.fst), (p.snd))) =
def uniform_space.completion.cpkg {α : Type u_1} :
The Haudorff completion as an abstract completion.
Equations
@[protected, instance]
Equations
The uniform bijection between a complete space and its uniform completion.
Equations
@[protected, instance]
theorem uniform_space.completion.dense_range_coe₂ {α : Type u_1} {β : Type u_2} :
dense_range (λ (x : α × β), ((x.fst), (x.snd)))
theorem uniform_space.completion.dense_range_coe₃ {α : Type u_1} {β : Type u_2} {γ : Type u_3} :
dense_range (λ (x : α × β × γ), ((x.fst), (x.snd.fst), (x.snd.snd)))
theorem uniform_space.completion.induction_on {α : Type u_1} {p : Prop} (a : uniform_space.completion α) (hp : is_closed {a : | p a}) (ih : (a : α), p a) :
p a
theorem uniform_space.completion.induction_on₂ {α : Type u_1} {β : Type u_2} {p : } (a : uniform_space.completion α) (b : uniform_space.completion β) (hp : is_closed {x : | p x.fst x.snd}) (ih : (a : α) (b : β), p a b) :
p a b
theorem uniform_space.completion.induction_on₃ {α : Type u_1} {β : Type u_2} {γ : Type u_3} {p : } (a : uniform_space.completion α) (b : uniform_space.completion β) (c : uniform_space.completion γ) (hp : is_closed {x : | p x.fst x.snd.fst x.snd.snd}) (ih : (a : α) (b : β) (c : γ), p a b c) :
p a b c
theorem uniform_space.completion.ext {α : Type u_1} {Y : Type u_2} [t2_space Y] {f g : Y} (hf : continuous f) (hg : continuous g) (h : (a : α), f a = g a) :
f = g
theorem uniform_space.completion.ext' {α : Type u_1} {Y : Type u_2} [t2_space Y] {f g : Y} (hf : continuous f) (hg : continuous g) (h : (a : α), f a = g a) (a : uniform_space.completion α) :
f a = g a
@[protected]
noncomputable def uniform_space.completion.extension {α : Type u_1} {β : Type u_2} (f : α β) :
"Extension" to the completion. It is defined for any map f but returns an arbitrary constant value if f is not uniformly continuous
Equations
theorem uniform_space.completion.uniform_continuous_extension {α : Type u_1} {β : Type u_2} {f : α β} :
theorem uniform_space.completion.continuous_extension {α : Type u_1} {β : Type u_2} {f : α β} :
@[simp]
theorem uniform_space.completion.extension_coe {α : Type u_1} {β : Type u_2} {f : α β} (hf : uniform_continuous f) (a : α) :
theorem uniform_space.completion.extension_unique {α : Type u_1} {β : Type u_2} {f : α β} (hf : uniform_continuous f) {g : β} (hg : uniform_continuous g) (h : (a : α), f a = g a) :
@[simp]
theorem uniform_space.completion.extension_comp_coe {α : Type u_1} {β : Type u_2} {f : β} (hf : uniform_continuous f) :
@[protected]
noncomputable def uniform_space.completion.map {α : Type u_1} {β : Type u_2} (f : α β) :
Completion functor acting on morphisms
Equations
theorem uniform_space.completion.uniform_continuous_map {α : Type u_1} {β : Type u_2} {f : α β} :
theorem uniform_space.completion.continuous_map {α : Type u_1} {β : Type u_2} {f : α β} :
@[simp]
theorem uniform_space.completion.map_coe {α : Type u_1} {β : Type u_2} {f : α β} (hf : uniform_continuous f) (a : α) :
= (f a)
theorem uniform_space.completion.map_unique {α : Type u_1} {β : Type u_2} {f : α β} (hg : uniform_continuous g) (h : (a : α), (f a) = g a) :
@[simp]
theorem uniform_space.completion.map_id {α : Type u_1} :
theorem uniform_space.completion.extension_map {α : Type u_1} {β : Type u_2} {γ : Type u_3} {f : β γ} {g : α β} (hf : uniform_continuous f) (hg : uniform_continuous g) :
theorem uniform_space.completion.map_comp {α : Type u_1} {β : Type u_2} {γ : Type u_3} {g : β γ} {f : α β} (hg : uniform_continuous g) (hf : uniform_continuous f) :
The isomorphism between the completion of a uniform space and the completion of its separation quotient.
Equations
@[protected]
noncomputable def uniform_space.completion.extension₂ {α : Type u_1} {β : Type u_2} {γ : Type u_3} (f : α β γ) :
Extend a two variable map to the Hausdorff completions.
Equations
@[simp]
theorem uniform_space.completion.extension₂_coe_coe {α : Type u_1} {β : Type u_2} {γ : Type u_3} {f : α β γ} (hf : uniform_continuous₂ f) (a : α) (b : β) :
= f a b
theorem uniform_space.completion.uniform_continuous_extension₂ {α : Type u_1} {β : Type u_2} {γ : Type u_3} (f : α β γ) :
@[protected]
noncomputable def uniform_space.completion.map₂ {α : Type u_1} {β : Type u_2} {γ : Type u_3} (f : α β γ) :
Lift a two variable map to the Hausdorff completions.
Equations
theorem uniform_space.completion.uniform_continuous_map₂ {α : Type u_1} {β : Type u_2} {γ : Type u_3} (f : α β γ) :
theorem uniform_space.completion.continuous_map₂ {α : Type u_1} {β : Type u_2} {γ : Type u_3} {δ : Type u_4} {f : α β γ} {a : δ } {b : δ } (ha : continuous a) (hb : continuous b) :
continuous (λ (d : δ), (b d))
theorem uniform_space.completion.map₂_coe_coe {α : Type u_1} {β : Type u_2} {γ : Type u_3} (a : α) (b : β) (f : α β γ) (hf : uniform_continuous₂ f) :
= (f a b) | 2,814 | 8,821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.858068 |
https://www.physicsforums.com/threads/evaluate-the-iterated-intergal-by-converting-to-polar-coordinate.316200/ | 1,708,947,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00096.warc.gz | 935,662,812 | 16,292 | # Evaluate the iterated intergal by converting to polar coordinate?
• ZuzooVn
In summary, the conversation is about evaluating an iterated integral by converting it to polar coordinates. The integral is given as \int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dxdy and the speaker asks for help with converting the limits into polar coordinates. The limits are found to be 0≤r≤2 and 0≤θ≤pi/2.
ZuzooVn said:
Help me Evaluate the iterated intergal by converting to polar coordinate:
http://www.ziddu.com/gallery/4894419/Untitled.jpg.html
hmm … that's $$\int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dxdy$$
ok … I assume you know how to convert dxdy into r and θ
and for the limits, convert y2 ≤ 2x - x2 into r and θ also
tiny-tim said:
hmm … that's $$\int_0^2\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2} dxdy$$
ok … I assume you know how to convert dxdy into r and θ
and for the limits, convert y2 ≤ 2x - x2 into r and θ also
0≤y≤1
0≤x≤2
0≤r≤2
0≤θ≤ pi/2
ZuzooVn said:
0≤y≤1
0≤x≤2
0≤r≤2
0≤θ≤ pi/2
(have a pi: π )
No, the upper limit of r will depend on θ.
I repeat … convert y2 ≤ 2x - x2 into r and θ
## 1. What is an iterated integral?
An iterated integral is a type of double or triple integral in which the limits of integration are expressed as a sequence of integrals. It is used to calculate the volume or area of a region bounded by multiple functions.
## 2. What does it mean to convert to polar coordinates?
Converting to polar coordinates is a way to express a point in the Cartesian coordinate system using a distance (r) and angle (θ) from the origin. It is often used in integration to simplify the calculation of certain types of functions.
## 3. Why is it useful to evaluate an iterated integral in polar coordinates?
Evaluating an iterated integral in polar coordinates can be useful because it can simplify the calculation of certain types of functions, particularly those with circular or symmetric boundaries. It can also help to visualize and understand the geometric meaning of the integral.
## 4. How do you convert an iterated integral to polar coordinates?
To convert an iterated integral to polar coordinates, the limits of integration must be changed from rectangular coordinates to polar coordinates. This can be done by substituting the polar coordinate equations (r = √(x^2 + y^2) and θ = tan^-1 (y/x)) into the original integral and changing the differential to account for the change in variable.
## 5. Are there any limitations to using polar coordinates for evaluating an iterated integral?
While polar coordinates can simplify the calculation of certain types of functions, they may not be suitable for all types of regions or functions. It is important to understand the limitations of polar coordinates and when it is more appropriate to use rectangular coordinates for evaluating an iterated integral. | 753 | 2,846 | {"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} | 4.0625 | 4 | CC-MAIN-2024-10 | latest | en | 0.847602 |
https://www.extendoffice.com/documents/excel/5254-excel-lucky-draw-name | 1,695,816,405,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510297.25/warc/CC-MAIN-20230927103312-20230927133312-00738.warc.gz | 861,391,221 | 28,960 | ## How to make lucky draw names in Excel?
In your daily work, sometimes, you need to select some employees randomly from a long list names as the lucky names. In Excel, how could you randomly select some names for making lucky draw? This article, I will talk about some useful methods to deal with this job in Excel.
Extract random names for making lucky draw with formula
Extract random names for making lucky draw with VBA code
Select random names for making lucky draw with Kutools for Excel
#### Extract random names for making lucky draw with formula
For example, I want to extract any 3 names from the name list, the following long formula can help you, please do as this:
Enter this formula into a blank cell where you want to put the result: =IF(ROWS(C\$2:C2)>B\$2,"",INDEX(A\$2:A\$16,AGGREGATE(15,6,((ROW(A\$2:A\$16)-ROW(A\$2)+1)/ISNA(MATCH(A\$2:A\$16,C\$1:C1,0))),RANDBETWEEN(1,ROWS(A\$2:A\$16)-COUNTA(C\$1:C1)+1)))) , and then drag the fill handle down to cells as you need, see screenshot:
Notes:
1. In the above formula: A2:A16 is the name list that you want to extract from randomly, B2 is the required number you want to extract names, C2 is the cell where to enter the formula, and C1 is the cell above the formula cell.
2. You can press F9 key to get another group of new names randomly.
#### Extract random names for making lucky draw with VBA code
1. Hold down the Alt + F11 keys to open the Microsoft Visual Basic for Applications window.
2. Click Insert > Module, and paste the following code in the Module Window.
VBA code: Extract random names from a list:
``````Public Sub LuckyDraw()
Dim I, J, xRnd As Long
Dim xSRg, xDRg As Range
Dim xDic As New Dictionary
Dim xnum, xLastRow As Long
On Error Resume Next
Set xSRg = Application.InputBox("Please select the data list:", "KuTools for Excel", Selection.Address, , , , , 8)
If xSRg Is Nothing Then Exit Sub
Set xDRg = Application.InputBox("Please selecta cell to put the result:", "KuTools for Excel", , , , , , 8)
If xDRg Is Nothing Then Exit Sub
xLastRow = xSRg.Rows.Count
Set xSRg = xSRg(1)
Set xDRg = xDRg(1)
xnum = Range("B2")
If xnum < 1 Then Exit Sub
J = 0
For I = 1 To xnum
LabExit:
xRnd = Int(Rnd() * xLastRow)
If xDic.Exists(xRnd) Then GoTo LabExit
xDRg.Offset(J, 0).Value = xSRg.Offset(xRnd, 0).Value
J = J + 1
Next
End Sub
``````
Note: In the above code, B2 is the cell contains the number of names you want to extract.
3. After inserting the code, then click Tools > References in the opened Microsoft Visual Basic for Applications window, and then, in the popped out References – VBAProject dialog box, check Microsoft Scripting Runtime option in the Available References list box, see screenshot:
4. And then click OK button to exit the dialog box, then press F5 key to run this code, and a prompt box is popped out to remind you selecting the data list you want to extract names from, see screenshot:
5. Click OK button, and another prompt box is popped out, please select a cell where you want to put the result, see screenshot:
6. Then click OK, and the desired number of names have been created randomly at once, see screenshot:
#### Select random names for making lucky draw with Kutools for Excel
May be above two methods are difficult for most of us, here, if you have Kutools for Excel, with its Sort Range Randomly feature, you can quickly select names randomly.
: with more than 300 handy Excel add-ins, free to try with no limitation in 30 days.
After installing Kutools for Excel, please do as this:
1. Select the name list that you want to select randomly. Then click Kutools > Range > Sort / Select Range Randomly, see screenshot:
2. In the Sort/Select Range Randomly dialog box, under the Select tab, enter the number that you want to select names into the No. of cells to select text box, and then choose Select random cells in the Select Type section, see screenshot:
3. Then click Ok button, and your specific number of names have been selected as you need, see screenshot:
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This comment was minimized by the moderator on the site
Error:
#NUM!
Formula:
=IF(ROWS(C\$2:C2)>B\$2,"",INDEX(A\$2:A\$1500,AGGREGATE(15,6,((ROW(A\$2:A\$1500)-ROW(A\$2)+1)/ISNA(MATCH(A\$2:A\$1500,C\$1:C1,0))),RANDBETWEEN(1,ROWS(A\$2:A\$1500)-COUNTA(C\$1:C1)+1))))
This comment was minimized by the moderator on the site
Hello, Mangipudi,
This formula works well in my Excel workbook, which Excel version do you use?
You can also give your problem as a screenshot here.
This comment was minimized by the moderator on the site
I have the same error, but it is not showing all times.
There are no comments posted here yet | 1,746 | 6,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-40 | latest | en | 0.806002 |
http://www.evi.com/q/how_many_litres_is_1500cc | 1,408,553,276,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500811391.43/warc/CC-MAIN-20140820021331-00158-ip-10-180-136-8.ec2.internal.warc.gz | 361,469,909 | 13,598 | # How many litres is 1500cc?
• 1,500 milliliters is equivalent to 1 1/2 liters
the volume 1 1/2 liters
• tk10npubl tk10ncanl
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• 1500cc to leitres | 563 | 1,773 | {"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} | 3.203125 | 3 | CC-MAIN-2014-35 | longest | en | 0.857608 |
https://schoollearningcommons.info/question/if-15-chair-cost-is-1530-so-cost-how-many-chair-can-be-bought-in-4590-rupees-rupees-19848692-67/ | 1,632,847,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00560.warc.gz | 533,587,540 | 13,545 | ## if 15 chair cost is 1530 so cost how many chair can be bought in 4590 rupees rupees
Question
if 15 chair cost is 1530 so cost how many chair can be bought in 4590 rupees rupees
in progress 0
2 months 2021-08-13T13:04:00+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
cost of 15 chair = 1530
so, find the cost of 1 chair
=1530/15
= 103
how many chair can be bought in 4590 rupees
= 4590/103
= 45
so, 45 chair purchased in 4590
2. cost of one chair- 1530 rupees
number of chairs bought in 4590 rupees- 4590÷1530= 3
so, 3 chairs can be purchased in 4590 rupees
hope it helped | 213 | 601 | {"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-2021-39 | latest | en | 0.904603 |
https://edurev.in/t/75404/NCERT-Solutions-for-Class-9-Maths-Chapter-8-Chapter-8-Quadrilaterals-I | 1,713,652,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00886.warc.gz | 206,278,204 | 65,631 | NCERT Solutions Chapter 8 - Quadrilaterals (I), Class 9, Maths
# NCERT Solutions for Class 9 Maths Chapter 8 - Chapter 8 - Quadrilaterals (I),
Exercise 8.1
1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.
Let x be the common ratio between the angles.
Sum of the interior angles of the quadrilateral = 360°
Now,
3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.
∠A = ∠B (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90°
Thus ABCD is a rectangle.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show,
AC = BD, AO = OC and ∠AOB = 90°
Proof,
BC = BA (Common)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Given,
Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.
To prove,
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by CPCT. --- (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB || CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT. --- (ii)
also,
⇒ AD = BC = CD = AB --- (ii)
also, ∠ADC = ∠BCD by CPCT.
and ∠ADC + ∠BCD = 180° (co-interior angles)
⇒ ∠ADC = 90° --- (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.
6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
(i)
AD = CB (Opposite sides of a ||gm)
DC = BA (Opposite sides of a ||gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a ||gm)
Thus, ABCD is a rhombus.
7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)
also, AB || CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.
Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.
8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
Therefore, AB = BC = CD = AD
Thus, ABCD is a square.
(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
Thus, BD bisects ∠D
9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a ||gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.
(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a ||gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.
(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.
(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.
10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.
11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF.
(i) AB = DE and AB || DE (Given)
Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.
(ii) Again BC = EF and BC || EF.
Thus, quadrilateral BEFC is a parallelogram.
(iii) Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)
Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)
(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.
(v) AC || DF and AC = DF because ACFD is a parallelogram.
(vi) In ΔABC and ΔDEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (Opposite sides of a parallelogram)
Thus, ΔABC ≅ ΔDEF by SSS congruence condition.
12. ABCD is a trapezium in which AB || CD and
AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iv) diagonal AC = diagonal BD
[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B
(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C
AB = AB (Common)
∠DBA = ∠CBA
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.
(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.
The document NCERT Solutions for Class 9 Maths Chapter 8 - Chapter 8 - Quadrilaterals (I), is a part of the Class 9 Course Extra Documents & Tests for Class 9.
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## FAQs on NCERT Solutions for Class 9 Maths Chapter 8 - Chapter 8 - Quadrilaterals (I),
1. What are the properties of a quadrilateral?
Ans. A quadrilateral is a polygon with four sides. The properties of a quadrilateral include the sum of its interior angles being equal to 360 degrees, opposite angles being equal, opposite sides being parallel, and consecutive angles being supplementary.
2. How do you classify quadrilaterals based on their sides and angles?
Ans. Quadrilaterals can be classified based on their sides as follows: - A quadrilateral with all sides of equal length is called a square. - A quadrilateral with opposite sides parallel and equal in length is called a parallelogram. - A quadrilateral with exactly one pair of parallel sides is called a trapezium. Quadrilaterals can also be classified based on their angles as follows: - A quadrilateral with all interior angles measuring 90 degrees is called a rectangle. - A quadrilateral with exactly one pair of opposite angles measuring 90 degrees is called a kite. - A quadrilateral with all interior angles measuring less than 90 degrees is called a rhombus.
3. How do you find the area of a quadrilateral?
Ans. The area of a quadrilateral can be determined using different formulas based on the type of quadrilateral. Here are the formulas for some common quadrilaterals: - Square: Area = side length * side length - Rectangle: Area = length * width - Parallelogram: Area = base length * height - Trapezium: Area = (sum of parallel sides) * height / 2 - Rhombus: Area = (product of diagonals) / 2
4. Can a quadrilateral have all angles measuring 90 degrees?
Ans. Yes, a quadrilateral can have all angles measuring 90 degrees. This type of quadrilateral is called a rectangle. In a rectangle, all four interior angles are right angles, and opposite sides are equal in length.
5. How do you prove that a quadrilateral is a parallelogram?
Ans. To prove that a quadrilateral is a parallelogram, you can use any of the following methods: - Show that both pairs of opposite sides are parallel. - Show that both pairs of opposite sides are equal in length. - Show that one pair of opposite sides is both parallel and equal in length. - Show that both pairs of opposite angles are equal. - Show that one pair of opposite angles is both equal and supplementary. By demonstrating any of these properties, you can establish that a quadrilateral is a parallelogram.
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; | 3,578 | 11,246 | {"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} | 4.6875 | 5 | CC-MAIN-2024-18 | latest | en | 0.881531 |
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244人阅读 评论(0)
Zipper
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16825 Accepted: 5998
Description
Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.
For example, consider forming "tcraete" from "cat" and "tree":
String A: cat
String B: tree
String C: tcraete
As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":
String A: cat
String B: tree
String C: catrtee
Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".
Input
The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.
For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.
Output
For each data set, print:
Data set n: yes
if the third string can be formed from the first two, or
Data set n: no
if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.
Sample Input
3
cat tree tcraete
cat tree catrtee
cat tree cttaree
Sample Output
Data set 1: yes
Data set 2: yes
Data set 3: no
Source
#include <cstdio>
#include <cstring>
char a[205],b[205],s[410];
int ns,na,nb;
bool q;
void dfsb()
{
int n=0;
for(int i=0;i<ns;i++)
{
if(s[i]=='0')
continue;
if(s[i]!=b[n])
return;
n++;
}
q=true;
}
void dfsa(int n,int st)
{
if(q)
return;
if(na==n)
{
dfsb();
return ;
}
for(int i=st;i<ns;i++)
{
if(s[i]==a[n])
{
int t=s[i];
s[i]='0';
dfsa(n+1,i+1);
if(q)
return;
s[i]=t;
}
}
}
int main()
{
freopen("in.txt","r",stdin);
int t;
scanf("%d",&t);
for(int ca=1;ca<=t;ca++)
{
q=false;
scanf("%s%s%s", a,b,s);
ns=strlen(s),na=strlen(a),nb=strlen(b);
if(ns!=nb+na||s[ns-1]!=a[na-1]&&s[ns-1]!=b[nb-1])
{
printf("Data set %d: no\n",ca);
continue;
}
dfsa(0,0);
if(q)
printf("Data set %d: yes\n",ca);
else
printf("Data set %d: no\n",ca);
}
return 0;
}
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• 评论：4条 | 844 | 2,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-34 | longest | en | 0.761972 |
https://nigerianscholars.com/past-questions/general-paper/question/303651/ | 1,638,682,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00013.warc.gz | 482,466,615 | 17,227 | Home » » Determine the quantity of heat in joules given out when a piece of iron of mass ...
# Determine the quantity of heat in joules given out when a piece of iron of mass ...
### Question
Determine the quantity of heat in joules given out when a piece of iron of mass 40g and specific heat capacity 460Jkg °C cools from 70°C to 40°C.
A) 552J
B) 7363
C) 1288J
D) 368J
### Explanation:
$$Q = mc(\Delta t)$$
$$Q = 40 \times 460 (70-40)$$
$$= 40 \times 460 \times 30$$
$$= 552000J = 552kJ$$
## Dicussion (1)
• $$Q = mc(\Delta t)$$
$$Q = 40 \times 460 (70-40)$$
$$= 40 \times 460 \times 30$$
$$= 552000J = 552kJ$$ | 225 | 623 | {"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} | 3.90625 | 4 | CC-MAIN-2021-49 | latest | en | 0.49876 |
https://www.meritnation.com/cbse-class-8/math/rd-sharma/data-handling-i-classification-and-tabulation-of-data/textbook-solutions/10_1_1165_5482_23.7_46468 | 1,590,832,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00446.warc.gz | 804,115,516 | 20,431 | Rd Sharma Solutions for Class 8 Math Chapter 23 Data Handling I Classification And Tabulation Of Data are provided here with simple step-by-step explanations. These solutions for Data Handling I Classification And Tabulation Of Data are extremely popular among Class 8 students for Math Data Handling I Classification And Tabulation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 23 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.
#### Question 1:
The marks obtained by 40 students of class VIII in an examination are given below:
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 3, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23.
Divide the data into five groups, namely 0-5, 5-10, 10-15, 15-20 and 20-25 and prepare a grouped frequency table.
The frequency table for the marks of 40 students of class VIII in an examination is given below:
Range of Marks Tally Marks Frequency 0$-$5 |||| |||| 9 5$-$10 |||| |||| 9 10$-$15 |||| || 7 15$-$20 |||| |||| 9 20$-$25 |||| | 6
#### Question 2:
The marks scored by 20 students in a test are given below:
54, 42, 68, 56, 62, 71, 78, 51, 72, 53, 44, 58, 47, 64, 41, 57, 89, 53, 84, 57.
Complete the following frequency table:
(Marks in class intervals) Tally marks Frequency (no. of children) 40-50 50-60 60-70 70-80 80-90
What is the class interval in which the greatest frequency occurs?
The frequency table can be completed as follows:
Marks Tally Marks Frequency 40$-$50 |||| 4 50$-$60 |||| ||| 8 60$-$70 ||| 3 70$-$80 ||| 3 80$-$90 || 2
The class interval with the greatest frequency (8) is 50$-$60.
#### Question 3:
The following is the distribution of weights (in kg) of 52 persons:
Weight in kg Persons 30-40 10 40-50 15 50-60 17 60-70 6 70-80 4
(i) What is the lower limit of class 50-60?
(ii) Find the class marks of the classes 40-50, 50-60.
(iii) What is the class size?
(i) The lower limit of the class 50$-$60 is 50.
(ii) Class mark for the class 40$-$50:
$\frac{40+50}{2}\phantom{\rule{0ex}{0ex}}=\frac{90}{2}\phantom{\rule{0ex}{0ex}}=45$
Again, class mark for the class 50$-$60:
$\frac{50+60}{2}\phantom{\rule{0ex}{0ex}}=\frac{110}{2}\phantom{\rule{0ex}{0ex}}=55$
(iii) Here the class size is 40$-$30, i.e. 10.
#### Question 4:
Construct a frequency table for the following weights (in gm) of 35 mangoes using the equal class intervals, one of them is 40-50 (45 not included):
30, 40, 45, 32, 43, 50, 55, 62, 70, 70, 61, 62, 53, 52, 50, 42, 35, 37, 53, 55, 65, 70, 73, 74, 45, 46, 58, 59, 60, 62, 74, 34, 35, 70, 68.
(i) What is the class mark of the class interval 40-45?
(ii) What is the range of the above weights?
(iii) How many classes are there?
The frequency table for the given weights (in gm) of 35 mangoes is given below:
Weight Tally Marks Frequency 30$-$40 |||| | 6 40$-$50 |||| | 6 50$-$60 |||| |||| 9 60$-$70 |||| || 7 70$-$80 |||| || 7
(i) Class mark for the class interval 40$-$45:
(ii) Range of the above weights:
(iii) There are 5 classes (30$-$40, 40$-$50, 50$-$60, 60$-$70, 70$-$80).
#### Question 5:
Construct a frequency table with class-intervals 0-5 (5 not included) of the following marks obtained by a group of 30 students in an examination.
0, 5, 7, 10, 12, 15, 20, 22, 25, 27, 8, 11, 17, 3, 6, 9, 17, 19, 21, 29, 31, 35, 37, 40, 42,45, 49, 4, 50, 16.
The frequency table with class intervals 0$-$5,5$-$10,10$-$15,...,50$-$55 is given below:
Marks Tally Marks Frequency 0$-$5 ||| 3 5$-$10 |||| 5 10$-$15 ||| 3 15$-$20 |||| 5 20$-$25 ||| 3 25$-$30 ||| 3 30$-$35 | 1 35$-$40 || 2 40$-$45 || 2 45$-$50 || 2 50$-$55 | 1
#### Question 6:
The marks scored by 40 students of class VIII in mathematics are given below:
81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54.
Prepare a frequency distribution with class size of 10 marks.
The frequency table of the marks scored by 40 students of class VIII in mathematics is given below:
Mark Tally Marks Frequency 20$-$30 | 1 30$-$40 ||| 3 40$-$50 |||| 5 50$-$60 |||| ||| 8 60$-$70 |||| ||| 8 70$-$80 |||| |||| 9 80$-$90 |||| 4 90$-$100 || 2
#### Question 7:
The heights (in cm) of 30 students of class VIII are given below:
155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157, 162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153.
Prepare a frequency distribution table with 160-164 as one of the class intervals.
The frequency table is given below:
Height Tally Marks Frequency 145$-$149 |||| 4 150$-$154 |||| |||| 9 155$-$159 |||| |||| || 12 160$-$164 |||| 5
#### Question 8:
The monthly wages of 30 workers in a factory are given below:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.
The frequency table of the monthly wages of 30 workers in a factory is given below:
Wage Tally Marks Frequency 800$-$810 ||| 3 810$-$820 || 2 820$-$830 | 1 830$-$840 |||| ||| 8 840$-$850 |||| 5 850$-$860 | 1 860$-$870 ||| 3 870$-$880 | 1 880$-$890 | 1 890$-$900 |||| 5
#### Question 9:
Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.
The frequency table of the monthly wages of 28 labourers working in a factory is given below:
Wage Tally Marks Frequency 210$-$230 |||| 4 230$-$250 |||| 4 250$-$270 |||| 5 270$-$290 ||| 3 290$-$310 |||| || 7 310$-$330 |||| 5
#### Question 10:
The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows:
−12.5, −10.8, −18.6, −8.4, −10.8, −4.2, −4.8, −6.7, −13.2, −11.8, −2.3, 1.2, 2.6, 0, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −5.8, −8.9, −14.6, −12.3, −11.5, −7.8, −2.9
Represent them as frequency distribution table taking − 19.9 to − 15 as the first class interval.
The frequency table of the daily minimum temperatures is given below:
Temperature Tally Marks Frequency $-$19.9 to $-$15 | 1 $-$14.9 to $-$10 |||| ||| 8 $-$9.9 to $-$5 |||| 5 $-$4.9 to 0 |||| |||| ||| 13 0.1 to 5 |||| ||| 8
#### Question 1:
Define the following terms:
(i) Observations
(ii) Raw data
(iii) Frequency of an observation
(iv) Frequency distribution
(v) Discrete frequency distribution
(vi) Grouped frequency distribution
(vii) Class-interval
(viii) Class-size
(ix) Class limits
(x) True class limits
(i) Observation is the value at a particular period of a particular variable.
(ii) Raw data is the data collected in its original form.
(iii) Frequency of an observation is the number of times a certain value or a class of values occurs.
(iv) Frequency distribution is the organisation of raw data in table form with classes and frequencies.
(v) Discrete frequency distribution is a frequency distribution where sufficiently great numbers are grouped into one class.
(vi) Grouped frequency distribution is a frequency distribution where several numbers are grouped into one class.
(vii) Class interval is the width of such a class.
(viii) Class size is the difference between the upper and the lower values of a class.
(ix) Class limits are the smallest and the largest observations (data, events, etc.) in a class.
(x) True class limits are the actual class limits of a class.
#### Question 2:
The final marks in mathematics of 30 students are as follows:
53, 61, 48, 60, 78, 68, 55, 100, 67, 90, 75, 88, 77, 37, 84, 58, 60, 48, 62, 56, 44, 58, 52, 64, 98, 59, 70, 39, 50, 60
(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group etc.
(ii) What is the highest score?
(iii) What is the lowest score?
(iv) What is the range?
(v) If 40 is the pass mark how many have failed?
(vi) How many have scored 75 or more?
(vii) Which observations between 50 and 60 have not actually appeared?
(viii) How many have scored less than 50?
The given raw data can be arranged in an ascending order. The class intervals are 30$-$39, 40$-$49,...100$-$109. Then, take the raw data and place it in the appropriate class intervals.
(i) The marks can be arranged in an ascending order as shown below:
30 to 39 $\to$ 37, 39
40 to 49 $\to$ 44, 48, 48
50 to 59 $\to$ 50, 52, 53,
55, 56, 58, 58, 59
60 to 69 $\to$ 60, 60, 60, 61, 62, 64, 67, 68
70 to 79 $\to$ 70, 75, 77, 78
80 to 89 $\to$ 84, 88
90 to 99 $\to$ 90, 98
100 to 109 $\to$ 100
(ii) The highest score is 100.
(iii) The lowest score is 37.
(iv) The range is 100$-$37, i.e. 63.
(v) If 40 is the passing mark, then the number of students who failed is 2 (i.e. 37, 39).
(vi) The number of students scoring 75 and above is 8 (i.e. 75, 77, 78, 84, 88, 90, 98, 100).
(vii) The marks 51, 54, and 57 do not actually appear between 50 and 60.
(viii) The number of students scoring less than 50 is 5 (i.e. 37, 39, 44, 48, 48).
#### Question 3:
The weights of new born babies (in kg) in a hospital on a particular day are as follows:
2.3, 2.2, 2.1, 2.7, 2.6, 3.0, 2.5, 2.9, 2.8, 3.1, 2.5, 2.8, 2.7, 2.9, 2.4
(i) Rearrange the weights in descending order.
(ii) Determine the highest weight.
(iii) Determine the lowest weight.
(iv) Determine the range.
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 kg?
(vii) How many babies weigh more than 2.8 kg?
(viii) How many babies weigh 2.8 kg?
The frequency distribution of the weights of new born babies in a hospital on a particular day is represented in the following table:
(i) The weights of the newly born babies in descending order are as follows:
Weight Tally marks Frequency 3.1 I 1 3.0 I 1 2.9 II 2 2.8 II 2 2.7 II 2 2.6 I 1 2.5 II 2 2.4 I 1 2.3 I 1 2.2 I 1 2.1 I 1
(ii) The highest weight is 3.1 kg.
(iii) The lowest weight is 2.1 kg.
(iv) The range is 3.1$-$2.1, i.e. 1 kg.
(v) The number of babies born on that day is 15.
(vi) The number of babies whose weights are below 2.5 kg is 4 (i.e. 2.4, 2.3, 2.2, 2.1).
(vii) The number of babies whose weights are more than 2.8 kg is 4 (i.e. 3.1, 3.0, 2.9, 2.9).
(viii) The number of babies whose weight is 2.8 kg is 2.
#### Question 4:
Following data gives the number of children in 40 families:
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.
The data can be put in the form of frequency distribution in the following manner:
Number of Children Tally marks Frequency 0 |||| 5 1 |||| || 7 2 |||| |||| || 12 3 |||| 5 4 |||| | 6 5 ||| 3 6 ||| 3
#### Question 5:
Prepare a frequency table of the following scores obtained by 50 students in a test:
42, 51, 21, 42, 37, 37, 42, 49, 38, 52, 7, 33, 17, 44, 39, 7, 14, 27, 39, 42, 42, 62, 37, 39, 67, 51, 53, 53, 59, 41, 29, 38, 27, 31, 64, 19, 53, 51, 22, 61, 42, 39, 59, 47, 33, 34, 16, 37, 57, 43,
The frequency table of 50 students is given below:
Marks Number of Students Marks Number of Students Marks Number of Students 7 2 33 2 49 1 14 1 34 1 51 3 16 1 37 4 52 1 17 1 38 2 53 3 19 1 39 4 54 1 21 1 41 1 57 1 22 1 42 6 59 2 27 2 43 1 61 1 29 1 44 1 62 1 31 1 47 1 67 1
#### Question 6:
A die was thrown 25 times and following scores were obtained:
1, 5, 2, 4, 3, 6, 1, 4, 2, 5, 1, 6, 2, 6, 3, 5, 4, 1, 3, 2, 3, 6, 1, 5, 2,
Prepare a frequency table of the scores.
The frequency of the scores of the die is shown below:
The Die Tally Marks Frequency 1 |||| 5 2 |||| 5 3 |||| 4 4 ||| 3 5 |||| 4 6 |||| 4
#### Question 7:
In a study of number of accidents per day, the observations for 30 days were obtained as follows:
6, 3, 5, 6, 4, 3, 2, 5, 4, 2, 4, 2, 1, 2, 2, 0, 5, 4, 6, 1, 6, 0, 5, 3, 6, 1, 5, 5, 2, 6
Prepare a frequency distribution table.
The frequency table for the number of accidents per day for a period of 30 days is given below:
Number of Accidents Tally Marks Frequency 0 || 2 1 ||| 3 2 |||| | 6 3 ||| 3 4 |||| 4 5 |||| | 6 6 |||| | 6
#### Question 8:
Prepare a frequency table of the following ages (in years) of 30 students of class VIII in your school:
13, 14, 13, 12, 14, 13, 14, 15, 13, 14, 13, 14, 16, 12, 14, 13, 14, 15, 16, 13, 14, 13, 12, 17, 13, 12, 13, 13, 13, 14
The frequency table of the ages of 30 students of class VII in the school is given below:
Age Tally Marks Frequency 12 |||| 4 13 |||| |||| || 12 14 |||| |||| 9 15 || 2 16 || 2 17 | 1
#### Question 9:
Following figures relate to the weekly wages (in Rs) of 15 workers in a factory:
300, 250, 200, 250, 200, 150, 350, 200, 250, 200, 150, 300, 150, 200, 250
Prepare a frequency table.
(i) What is the range in wages (in Rs)?
(ii) How many workers are getting Rs 350?
(iii) How many workers are getting the minimum wages?
The frequency table for the number of accidents per day for a period of 30 days is given below:
Wage (in Rs) Tally Marks Frequency 150 ||| 3 200 |||| 5 250 |||| 4 300 || 2 350 | 1
(i) The range of wages (in Rs) is 350$-$150 i.e. 200.
(ii) From the frequency table, we can see that the number of workers earning Rs 350 is 1.
(iii) Here, the minimum wage is 150. Hence, the number of workers earning the minimum wage is 3.
#### Question 10:
Construct a frequency distribution table for the following marks obtained by 25 students in a history test in class VIII of a school:
9, 17, 12, 20, 9, 18, 25, 17, 19, 9, 12, 9, 12, 18, 17, 19, 20, 25, 9, 12, 17, 19, 19, 20, 9
(i) What is the range of marks?
(ii) What is the highest mark?
(iii) Which mark is occurring more frequently?
(i) The range of marks is 25$-$9, i.e. 16. | 5,395 | 14,067 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 93, "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} | 4.8125 | 5 | CC-MAIN-2020-24 | longest | en | 0.860986 |
https://mathzsolution.com/how-do-people-come-up-with-difficult-math-olympiad-questions/ | 1,675,158,525,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00464.warc.gz | 400,693,251 | 18,854 | # How do people come up with difficult math Olympiad questions?
The problems that appear in difficult math competitions such as the IMO or the Putnam exam are usually very difficult and require some ingenuity to solve. They also usually don’t look like they can be solved by simply knowing more advanced theory and the such.
How do people typically come up with these problems? Do they arise naturally from advanced mathematics (the somewhat infamous ‘grasshopper problem’ from the 2009 IMO comes to mind – to my not exactly knowledgeable mind this problem looks like it popped out of basically nowhere)? What is the perspective that mathematicians take when seemingly “inventing” these problems with no theoretical motivation to them whatsoever?
When I am working on my research, or on MO/math.SE questions, I often find myself thinking in a way which reminds me of the feel of solving Olympiad problems. If I then solve the problem, I try to find a very special case of my problem which is still challenging, and can be stated and solved using only material on the Olympiad curriculum. I then e-mail Kiran Kedlaya and say “Hey Kiran, do you think this would be a good Olympiad problem?” If he thinks so, he proposes it to the USAMO committe.
I wrote Problem 2 on the 2010 USAMO in this way; it is Theorem 3.2 of this paper specialized to the case that $$W0W_0$$ is the group $$SnS_n$$. The fact that the “total number of moves” referred to in the theorem is at most $$\binom{n}{3}\binom{n}{3}$$ is computed in Section 5.2.
I think I send about one problem a year; but most of them get rejected.
UPDATE 2014 Problem B4 of the 2014 Putnam was mine. Let $$F(x,y,z) = \sum F_{ijk} x^i y^j z^kF(x,y,z) = \sum F_{ijk} x^i y^j z^k$$ be a homogenous polynomial of degree $$nn$$ with positive real coefficients. We say that $$FF$$ is hyperbolic with respect to the positive orthant if, for all $$(u_1,v_1,w_1)(u_1,v_1,w_1)$$ and $$(u_2,v_2,w_2) \in \mathbb{R}_{> 0}^3(u_2,v_2,w_2) \in \mathbb{R}_{> 0}^3$$, the polynomial $$f(t) = F(tu_1+u_2,tv_1+v_2,tw_1+w_2)f(t) = F(tu_1+u_2,tv_1+v_2,tw_1+w_2)$$ has $$nn$$ negative real roots.
In this paper, I show that there are constants $$V_1V_1$$ and $$V_2V_2$$ (dependent on $$nn$$) so that,
(1) if $$FF$$ is hyperbolic with respect to the positive orthant, then $$F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_1 F_{i(j+1)(k+1)} F_{(i+2)jk}F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_1 F_{i(j+1)(k+1)} F_{(i+2)jk}$$ and the same for all permutations of the indices
(2) if $$F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_2 F_{i(j+1)(k+1)} F_{(i+2)jk}F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_2 F_{i(j+1)(k+1)} F_{(i+2)jk}$$ and the same for all permutations of the indices, then $$FF$$ is hyperbolic with respect to the positive orthant.
The proof is nonconstructive; I also (Theorem 20) give an explicit value of $$V_1V_1$$. I was thinking about whether I could give a concrete value for $$V_2V_2$$. The problem was too hard, so I thought instead about homogenous polynomials in two variables, which is the same as inhomogenous polynomials in one variable. At this point, I was basically looking for a converse to Newton’s inequality: I wanted a constant $$CC$$ so that, if $$a_k^2 > C a_{k-1} a_{k+1}a_k^2 > C a_{k-1} a_{k+1}$$, then all the roots of $$\sum_{k=0}^n a_k z^k\sum_{k=0}^n a_k z^k$$ are real. The result in one variable wasn’t worth publishing, but I figured I could make a nice problem by choosing a particular polynomial and asking people to prove the roots were real.
UPDATE 2020 Problem 6 of the 2020 USOMO was mine. It is the key computation from this MO answer, specialized to the case that the matrix $$XX$$ has rank one, so $$X_{ij} = x_i y_jX_{ij} = x_i y_j$$. The rank one case turned out not to be easier than the general case, which is why it doesn’t show up in that MO thread, but it is one of the things I thought about when working on that answer, and I noticed at the time that it looked like a strengthening of the rearrangement inequality. | 1,245 | 3,984 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.933577 |
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A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
For example:
`"112358"` is an additive number because the digits can form an additive sequence: `1, 1, 2, 3, 5, 8`.
`1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8`
`"199100199"` is also an additive number, the additive sequence is: `1, 99, 100, 199`.
`1 + 99 = 100, 99 + 100 = 199`
Note: Numbers in the additive sequence cannot have leading zeros, so sequence `1, 2, 03` or `1, 02, 3` is invalid.
Given a string containing only digits `'0'-'9'`, write a function to determine if it’s an additive number.
Solution
This is a permutation a like problem. The key to solve it is to build the first number and second number. Then we check if string is an additive number started by the 2 number. Let len1 is the length of number1, len2 for number2, n for length of string. So, the length of the next number should be greater than or equal to max(len1, len2). So we have condition that n – len1 + len2 >= max(len1, len2)
Then we should write isValid(String str, int i, int j) function to check if str is a additive number with first number [0,…,i], second number is [i+1,…,j]. When writing this function, we should eliminate exception number like ‘012’, ’01’. On the other hand, we should pay attention that ‘0’ is a valid number.
check my code on github: link
## pascal triangle
https://leetcode.com/problems/pascals-triangle-ii/
Given an index k, return the kth row of the Pascal’s triangle.
For example, given k = 3,
Return [1,3,3,1].
Analysis
Matrix like below is a pascal triangle p[][]. For row i, column j, p[i][j] it has value C(i, j) [Binomial Coefficient].
Below is the formula which can help us get C(n, m) from C(n, m – 1).
In this way, we can get elements in row i in O(n) time.
check my code on github: link | 561 | 1,939 | {"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} | 3.828125 | 4 | CC-MAIN-2022-33 | latest | en | 0.825519 |
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# Use the following to answer the questions below:Data were collected on the mileage (in thousands of ...
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Use the following to answer the questions below:
Data were collected on the mileage (in thousands of miles) and price (in thousands of dollars) of a random sample of used Hyundai Elantras. A scatterplot of the data (with regression line), some summary statistics, and partial computer output from a regression analysis are provided.
Use three decimal places when reporting the results from any calculations, unless otherwise specified.
Variable Mean StDev Minimum Maximum Price 8.304 4.025 1.900 15.900 Mileage 60.01 39.31 0.90 138.20
The regression equation is Price = 13.8 - 0.0912 Mileage
Source DF SS MS F P Regression 1 308.32 308.32 88.01 0.000 Residual Error 23 80.57 3.50 Total 24 388.89
Use the provided computer output to compute the standard deviation of the error term.
▸ 3.504
▸ 3.357
▸ 1.872
▸ 1.832
Textbook
## Statistics: Unlocking the Power of Data
Edition: 3rd
Authors:
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### Related Topics
britt138 Author
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Thanks
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Thank you, thank you, thank you!
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## Uniform movement
In this section we are concerned with the uniform motion of physics. In this chapter you will learn what it is all about, what the formulas are and how to use them.
With regard to the movements, we would like to start with what is known as the uniform movement. We try to explain this as simply as possible and also deal with the conversion of units (kilometers to meters, hours to seconds, etc.). However, there are a few very basic things about mathematics that you should know. If, while reading the following section, you realize that you are having math difficulties, please read the articles under the links below. Everyone else can save this:
Uniform motion video:
• Notes: This is still a blackboard video. A New edition in HD is planned. It can also be called up directly in the Uniform Motion Video section.
• Problems: If you have playback problems, please go to the article Video problems.
Show:
### What is uniform motion?
Uniform motion has the following properties:
• The speed of the object is always the same
• The acceleration a is zero (a = 0), i.e. the object is neither decelerated nor does it get faster
Or expressed as an example: You are driving on a road at 100 km / h. As long as you really drive the 100km / h, this is a uniform movement. On the other hand, if the speed changes, you no longer have a uniform movement.
Show:
### How do I calculate a uniform motion (formula)?
The formula of uniform motion sets the information Distance, speed, time and starting distance in relation to each other. The following is the general formula and the meaning of the symbols. Then we will provide you with some explanations and examples.
Uniform motion formula:
• s = v t + s0
• "s" is the distance in meters [m]
• "v" is the speed in meters per second [m / s]
• "t" is the time in seconds [s]
• "s0"is the starting distance in meters [m]
In very many cases the starting path does not exist. This simplifies the formula too
s = v ยท t.
### Uniform movement: examples
Time for a few small examples of uniform movement. One thing is very important: you have to insert the numbers in the form in which they were given in the formula. Say the distance in meters and not in kilometers or centimeters. Or the time in seconds and not in hours. If you don't pay attention to this, you will quickly get crap out of arithmetic.
Example 1:
Example 2:
Tip: In any case, do the math for our exercises! This is the only way to get fit when calculating with the uniform movement!
### Conversion of units
Many students have problems converting units. For this reason, we are going to give you a brief summary of the most important conversion rules.
• 100cm = 1m = 0.001km
• 1m / s = 3.6km / h
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https://lottoforums.com/threads/picks-aug-2.1916/page-2 | 1,717,020,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00045.warc.gz | 319,400,145 | 11,569 | # Picks Aug.2
#### Nick Koutras
##### Member
hot4 said:
Hi Nick,
only today I got to see this forum (database problem ?!).
I cannot tell you now how many draws I used to get my last group of 17 numbers. May be you used a different amount of past draws ...
But may be my history file is different from yours and not correct. I'll check that.
These groups of 17 numbers are the *only* having no triplet? Or can you get more?
Frank
Hi Frank..
>> These groups of 17 numbers are the *only* having no triplet? Or can you get more?
Yes, many more...
#### Nick Koutras
##### Member
hot4 said:
Ok Nick!
For those not believing in any advantage of these groups, there are two bets:
First- 3 numbers will come in every group in next 5 draws;
Second- 57 lines with Max=4 Min=1 hit better than 1:57
I think that this way of thinking can get the most of these groups IF really they will hit from 1 to 4 numbers.
There is another way too: to use only 3 numbers (Max=3 Min=3) from one group (e.g. the first) and 1 to 4 numbers of the remaining 3 groups. LOf course players must wait till 3 numbers matching.
LottoDesigner can make such wheel in two clicks.
Frank
Results:
02040910111619232427353638394243
02030409101618192327343536394243
02060911161823242735363839404243
01020306091023273135363839404243
01020306091018212327353638394043
G_1=4
G_2=3
G_3=3
G_4=3
G_5=2
Any conclusions?
#### hot4
##### Member
Nick Koutras said:
Results:
02040910111619232427353638394243
02030409101618192327343536394243
02060911161823242735363839404243
01020306091023273135363839404243
01020306091018212327353638394043
G_1=4
G_2=3
G_3=3
G_4=3
G_5=2
Any conclusions?
Yes, the majority of the 5 groups hit better than expected (2).
This advantage makes players to think about the best way to wheel these groups.
I think all three bets have been good. To wheel every group in 16,6,?,6 would not be a good way to explore the advantage.
Till now I cannot see a better way to *try_and_get* the slight advantage from this strategy.
I'd like to know some other opinions, meanwhile.
Frank
#### hot4
##### Member
Nick Koutras said:
Hi Frank..
>> These groups of 17 numbers are the *only* having no triplet? Or can you get more?
Yes, many more...
I like to get the minimum possible amount (the most overdue groups), so that I can compare groups easier.
To choose from a big set of groups is more difficult. But till now I'm not sure that >=3 hits are associated with the *most* overdue and that, those groups having 3hits this draw will not have 3 again next draws.
Frank
#### Nick Koutras
##### Member
hot4 said:
Yes, the majority of the 5 groups hit better than expected (2).
This advantage makes players to think about the best way to wheel these groups.
I think all three bets have been good. To wheel every group in 16,6,?,6 would not be a good way to explore the advantage.
Till now I cannot see a better way to *try_and_get* the slight advantage from this strategy.
I'd like to know some other opinions, meanwhile.
Frank
Hi Frank,
What I have noticed is that
the frequency of the Triplet of 16,27,38
does belong to at least 3 of the 5 groups.
This is an indicator that we shall use
Triplets from these Groups with the highest frequncy.
#### hot4
##### Member
Nick Koutras said:
Hi Frank,
What I have noticed is that
the frequency of the Triplet of 16,27,38
does belong to at least 3 of the 5 groups.
This is an indicator that we shall use
Triplets from these Groups with the highest frequncy.
Yes Nick,
and we can make another good assumption:
these triplets frequently appear <= 4 next draw.
It's possible to reduce in a good percentage winner lines.
Frank | 991 | 3,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.868444 |
http://www.investopedia.com/articles/active-trading/091114/strategies-trading-fibonacci-retracements.asp | 1,505,967,965,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687606.9/warc/CC-MAIN-20170921025857-20170921045857-00361.warc.gz | 475,951,234 | 51,912 | Leonardo Pisano, nicknamed Fibonacci, was an Italian mathematician born in Pisa in the year 1170.
His father Guglielmo worked at a trading post in Bugia, now called Béjaïa, a Mediterranean port in northeastern Algeria. The young Leonardo studied mathematics in Bugia and during extensive travels he learned about the advantages of the Hindu–Arabic numeral system.
After returning to Italy, in 1202 Fibonacci documented what he had learned in the Liber Abaci (Book of Abacus). In doing so he popularized the use of Hindu–Arabic numerals in Europe.
The Fibonacci Number Sequence
In the Liber Abaci, Fibonacci described the numerical series now named after him.
In the Fibonacci sequence of numbers, after 0 and 1, each number is the sum of the two prior numbers.
Hence, the sequence is as follows: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610 and so on extending to infinity.
Each number is approximately 1.618 times greater than the preceding number.
The Golden Ratio
This figure 1.618 is called Phi or the Golden Ratio. The inverse of 1.618 is .618. The Golden Ratio mysteriously appears frequently in the natural world, architecture, fine art and biology.
The ratio has been observed in the Parthenon, Leonardo da Vinci's Mona Lisa, sunflowers, rose petals, mollusk shells, tree branches, human faces, ancient Greek vases, and even the spiral galaxies of outer space.
Fibonacci Levels Used in the Financial Markets
The levels used in Fibonacci retracements in the context of trading are not numbers in the sequence, rather they are derived from mathematical relationships between numbers in the sequence.
The basis of the 'golden' Fibonacci ratio of 61.8% comes from dividing a number in the Fibonacci series by the number that follows it. For example, 89/144 = 0.6180. (See Investopedia's great video on Playing the Golden Ratio.)
The 38.2% ratio is derived from dividing a number in the Fibonacci series by the number two places to the right. For example: 89/233 = 0.3819.
The 23.6% ratio is derived from dividing a number in the Fibonacci series by the number three places to the right. For example: 89/377 = 0.2360.
Fibonacci retracement levels are depicted by taking high and low points on a chart and marking the key Fibonacci ratios of 23.6%, 38.2%, 61.8% horizontally to produce a grid. These horizontal lines are used to identify possible price reversal points.
The 50% Retracement Level
The 50% retracement level is normally included in the grid of Fibonacci levels that can be drawn using charting software. While the 50% retracement level is not based on a Fibonacci number it is widely viewed as an important potential reversal level, notably recognized in Dow Theory and also in the work of W.D. Gann,.
Fibonacci Retracement Levels as Part of a Trading Strategy
Fibonacci retracements are often used as part of a trend trading strategy.
In this scenario traders observe a retracement taking place within a trend and try to make low risk entries in the direction of the initial trend using Fibonacci levels.
Simply put, traders using this strategy anticipate that price has a high probability of bouncing from the Fibonacci levels back in the direction of the initial trend.
For example, on the EUR/USD daily chart below, we can see that price was in a major downtrend since early May of 2014 (point A). Price then bottomed in June (point B) and retraced upwards to approximately the 38% Fibonacci retracement level of the downmove (point C).
Figure 1: EUR/USD Daily Chart Fibonacci Retracement. Chart Courtesy of TradingView.
In this case, the 38% level would have been a good place to enter a short position, with a view to capitalizing on the continuation of the downtrend that started in May.
There is no doubt that many traders were also watching the 50% retracement level and the 61.8% retracement level, but in this case the market was not bullish enough to reach those points. Instead, EUR/USD turned lower, resuming the downtrend and taking out the prior low in a fairly fluid movement.
Keep in mind that the likelihood of a reversal increases if there is a confluence of technical signals when price reaches a Fibonacci level.
Other popular technical indicators that are used in conjunction with Fibonacci levels include candlestick patterns, trendlines, volume, momentum oscillators and moving averages. A greater number of confirming indicators in play equates to a more robust reversal signal.
Fibonacci retracements are used on a variety of financial instruments including stocks, commodities and foreign exchange. They are also used on multiple time frames. However, as with other technical indicators, the predictive value is proportional to the timeframe used, with greater weight given to longer timeframes.
So for example, a 38% retracement on a weekly chart is a far more important technical level than a 38% retracement on a 5 minute chart.
Using Fibonacci Extensions
As we saw above, Fibonacci retracement levels can be used to forecast potential areas of support or resistance at which traders can enter the market with a view to catching the resumption of an initial trend.
Fibonacci extensions can compliment this strategy by giving traders Fibonacci based profit targets.
Fibonacci extensions consist of levels drawn beyond the standard 100% level and can be used by traders to project areas that make good potential exits for their trades in the direction of the trend. The major Fibonacci extension levels are 161.8%, 261.8% and 423.6%.
Let's take a look at an example here, using the same EUR/USD daily chart:
Figure 2: EUR/USD Daily Chart with Fibonacci Extension. Chart courtesy of TradingView.
Looking at the Fibonacci extension level drawn on the EUR/USD chart above, we can see that a potential price target for a trader holding a short position from the 38% retracement described earlier lies below at the 161.8% level at 1.3195.
The Bottom Line
Fibonacci retracement levels often mark reversal points with an uncanny accuracy. However, they are harder to trade than they look in retrospect.
The levels are best used as a tool within a broader strategy that looks for the confluence of a number of indicators to identify potential reversal areas offering low risk, high potential reward trade entries.
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Delivered twice a week, straight to your inbox. | 1,430 | 6,465 | {"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} | 4.03125 | 4 | CC-MAIN-2017-39 | latest | en | 0.938467 |
https://web2.0calc.com/questions/math-help-thank-you_2 | 1,582,302,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00063.warc.gz | 621,932,250 | 5,828 | +0
# math help thank you
+1
259
1
+701
http://prntscr.com/lbzyk4
Oct 29, 2018
#1
0
This is exactly the same as the previous question, except being about rabbit population:
P=220; a=listfor(t, 1, 3, (P =1.15*(P - 20))
Day 1 =230, Day 2 =242, Day 3 =255 - Population of rabbits in 3 days from now.
Oct 30, 2018 | 123 | 316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-10 | latest | en | 0.885387 |
https://upperreviews.com/write-24000-in-words-in-english/ | 1,653,190,452,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00763.warc.gz | 657,621,421 | 19,560 | If you have to Write 24000 in Words, then you can write 24000/Twenty-four Thousand Rupees/dollars. Suppose you have earned 24000/Twenty-four Thousand Rupees/dollars in 1 month, then you can write, I have earned 24000/Twenty-four Thousand Rupees/dollars in a month. Twenty-four Thousand is the cardinal number word of 24000 which denotes a quantity.
• 24000 in Words = Twenty-four Thousand
• Twenty-four Thousand in Numbers = 24000
In other languages:
• 24000 in German words: Twenty-four Thousand Euro
• 24000 in Russian words: Twenty-four Thousand Ruble
• 24000 in India: Twenty-four Thousand Rupee
## How to Write 24000 in Words?
Using a place value chart we can identify the place and number name for each digit in the given number. For 24000 we see that digit in units = 0, tens = 0, hundreds = 0, thousands = 24. So we can write 24000 as Twenty-four Thousand in words.
### How do Write 24000 in Words?
Using a place value chart we can identify the place and number name for each digit in the given number. For 24000 we see that digit in units = 0, tens = 0, hundreds = 0, thousands = 24. So we can write 24000 as Twenty-four Thousand in words.
### What are the Rules to Write 24000 in Words?
Let us find out what is the place value of all the digits of 24000.
What are thousand = 24
hundreds = 0
tens = 0
Units = 0 To write the rule 24000 in words?
We can see that there are 0 ‘units’, 0 ‘tens’, 0 ‘hundreds’, 24 ‘thousands’.
Now we can read the number from right to left along with its place value and
24 thousand can be written in 24000= Twenty-four Thousand in words.
### What is 24000 Decimal to Binary?
(24000)10 = (101110111000000)2
154.919334 | 463 | 1,669 | {"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} | 4.46875 | 4 | CC-MAIN-2022-21 | latest | en | 0.899135 |
https://www.jiskha.com/display.cgi?id=1191291349 | 1,526,818,128,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00189.warc.gz | 781,359,613 | 4,944 | # Chemistry
posted by Tammy
Unfortunately I was hit by the sweep of illnesses going around due to changing whether and missed school today. Thus I missed the notes in Chemistry leaving me completely confuzzled about tonights homework on Dimmensional analysis or Unit Factoring. I called a friend but she isn't the note-taking type of person and was only able to tell me to do the problems in a table (??) so if someone could do an example problem or two for me I think I can handle the rest.
It says to perform the following metric conversion using dimnesional analysis and one conversion factor.
3.50 nm - ? m
Another said to do the same but with two conversion factors.
23 dL - ? cL
It would really help me out if someone could work these out for me. I'm not trying to ask someone to do my homework, just show me the right way to complete it.
Thanks, Tammy
1. DrBob222
Dimensional analysis is the art (or act) of multiplying some number by a factor to convert to another unit. And the units you don't want to keep cancel but the unit you want to keep stays. Here is the first one.
The factor for converting nm (nanometers) and meters is 1 nm = 10^-9 m; therefore, this factor can be written one of two ways.
1 nm/10^-9 m OR
10^-9 m/1 nm AND
the conversion then is
3.50 nm x factor = ?? meters. Note that the only thing you need to do is to know the factors (which you should have done with the metric system) AND use the correct one of the two factors. BUT you always know which factor is correct because the units you don't want will cancel.
3.50 nm x (10^-9 m/1 nm) = ?? m
Note that the unit of nm cancels and leaves the unit of m so the unit we don't want to keep cancels and the unit we want to keep (meters) stays.
IF you use the wrong factor, the units will NOT cancel, like so:
3.50 nm x (1 nm/10^-9 m) = ?? nm^2/m and that certainly is not the unit we want in the answer. So we know the first one is correct and the second one is not. I hope this gets you started.
For the other one, I will leave most of the discussion out.
If they want two conversion factors, I suggest we convert dL to L, then L to cL.
We know 10 dL = 1L and we know 100 cL = L so we do the following:
23 dL x (1 L/10 dL) x (100 cL/1 L) = ?? cL. Note that dL in the numerator of the first term cancels with the dL in the denominator of the second term and L in the numerator of the seccond term cancels with L in the denominator of the third term to leave what we want, which is cL.
That gives us 230 cL.
Check my work.
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More Similar Questions | 1,171 | 4,713 | {"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-2018-22 | latest | en | 0.966403 |
https://www.albert.io/learn/act-science/question/average-hatching-times | 1,490,449,246,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188926.39/warc/CC-MAIN-20170322212948-00428-ip-10-233-31-227.ec2.internal.warc.gz | 859,505,133 | 40,784 | Limited access
Optimum temperature, humidity, ventilation and periodic turning are some important requirements for good yield and timely hatching of chickens. A yield of 80 or more chickens per 100 eggs is considered acceptable. A study was conducted to find the optimal conditions for a quick hatch with good yield.
Experiment 1
Incubators with four openings on the top were used and the relative humidity was maintained at 52%. Five incubators with 50 eggs each were maintained at different temperatures.
Table 1
Temperature Successful hatches Average hatching time
99$^\circ$F 42 24 days
100$^\circ$F 48 21 days
101$^\circ$F 47 20 days
102$^\circ$F 44 19 days
103$^\circ$F 38 17 days
Experiment 2
Incubators with four openings on the top were used and the temperature was maintained at 100° F. Five incubators with 50 eggs each were maintained at different humidity levels.
Table 2
Relative humidity Successful hatches Average hatching time
50% 45 21 days
51% 45 21 days
52% 47 21 days
53% 47 20 days
54% 45 20 days
Experiment 3
Incubators with four openings on the side were used and the relative humidity was maintained at 52%. Five incubators with 50 eggs each were maintained at different temperatures.
Table 3
Temperature Successful hatches Average hatching time
99$^\circ$F 40 22 days
100$^\circ$F 44 21 days
101$^\circ$F 42 20 days
102$^\circ$F 40 19 days
103$^\circ$F 34 16 days
Experiment 4
Five incubators (containing 50 eggs each) with standard settings were placed at different sites under different conditions.
Table 4
Specific condition Successful hatches Average hatching time
Directly under the sun 20 18 days
In a windy place 30 17 days
Eggs were not turned at all 35 24 days
For this question, find the trials in experiments 1 and 2 that were conducted under a similar set of conditions.
What are the average hatching times (in days) in these trials in experiments 1 and 2 respectively?
A
$19$ and $20$
B
$20$ and $21$
C
$21$ and $21$
D
$21$ and $20$
Select an assignment template | 543 | 2,029 | {"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} | 3.1875 | 3 | CC-MAIN-2017-13 | latest | en | 0.926537 |
https://community.airtable.com/t/need-help-with-an-if-statement/34956 | 1,606,169,592,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00709.warc.gz | 268,485,450 | 5,166 | # Need help with an IF statement
Hello,
I’m new writting formulas, I’ve read all the guides on airtable’s website and other posts in the forum.
However I can’t get my formula to work. I only need that when the value of a cell is equal to a number a specific name appears.
This is what I’ve wrote; IF (Total = “6.99”, Hernia discal, BLANK())
But airtable won’t let me keep the formula because it’s misspelled.
Any advises?
Thanks.
Anything that isn’t a function or the name of a field needs to be in either straight apostrophes or quotation marks. If a field name contains a space it must be surrounded by curly braces.
So your formula should likely be either:
``````IF(
Total = "6.99",
"Hernia discal",
BLANK()
)
``````
or
``````IF(
Total = "6.99",
{Hernia discal},
BLANK()
)
``````
Thanks a lot for your fast reply, It worked.
What if I want to add different values for total? I’ve tried to duplicate the formula with different values but it doesn’t work,
Something like this:
Total = “6.99”,
“Hernia discal”,
BLANK()
Total = “10.99”,
“Hernia discal lumbar”,
BLANK()
Total = “15.99”,
“Hernia discal cervical”,
BLANK()
Thanks a lot for your help.
If you’re comparing the value of one thing multiple times you could use a SWITCH statement.
``````SWITCH(
Total,
"6.99",
"Hernia discal",
"10.99",
"Hernia discal lumbar",
"15.99",
"Hernia discal cervical",
BLANK()
)
``````
For your reference, if you wanted to nest an IF() statement it would look like this:
``````IF(
Total = "6.99",
"Hernia discal",
IF(
Total = "10.99",
"Hernia discal lumbar",
IF(
Total = "15.99",
"Hernia discal cervical",
BLANK()
)
)
)
``````
Thanks mate, you’re a beast.
1 Like
This topic was solved and automatically closed 3 days after the last reply. New replies are no longer allowed. | 507 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-50 | latest | en | 0.869008 |
http://library.kiwix.org/wikipedia_en_all_nopic_2018-09/A/Probabilistic_design.html | 1,556,264,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00398.warc.gz | 106,212,815 | 17,774 | Probabilistic design
Probabilistic design is a discipline within engineering design. It deals primarily with the consideration of the effects of random variability upon the performance of an engineering system during the design phase. Typically, these effects are related to quality and reliability. Thus, probabilistic design is a tool that is mostly used in areas that are concerned with quality and reliability. For example, product design, quality control, systems engineering, machine design, civil engineering (particularly useful in limit state design) and manufacturing. It differs from the classical approach to design by assuming a small probability of failure instead of using the safety factor. [1] [2]
Designer's perspective
When using a probabilistic approach to design, the designer no longer thinks of each variable as a single value or number. Instead, each variable is viewed as a probability distribution. From this perspective, probabilistic design predicts the flow of variability (or distributions) through a system. By considering this flow, a designer can make adjustments to reduce the flow of random variability, and improve quality. Proponents of the approach contend that many quality problems can be predicted and rectified during the early design stages and at a much reduced cost.
The objective of probabilistic design
Typically, the goal of probabilistic design is to identify the design that will exhibit the smallest effects of random variability. This could be the one design option out of several that is found to be most robust. Alternatively, it could be the only design option available, but with the optimum combination of input variables and parameters. This second approach is sometimes referred to as robustification, parameter design or design for six sigma
Methods used
Essentially, probabilistic design focuses upon the prediction of the effects of random variability. Some methods that are used to predict the random variability of an output include:
Footnotes
1. Sundararajan, S (1995). Probabilistic Structural Mechanics Handbook. Springer. ISBN 978-0412054815.
2. Long, M W; Narcico, J D (June 1999), Design Methodology for Composite Aircraft Structures, DOT/FAA/AR-99/2, FAA, retrieved 24 January 2015
3. Sundarth, S; Woeste, Frank E.; Galligan, William (1978), Differential reliability : probabilistic engineering applied to wood members in bending-tension (PDF), Res. Pap. FPL-RP-302., US Forest Products Laboratory, retrieved 21 January 2015
References
• Ang and Tang (2006) Probability Concepts in Engineering: Emphasis on Applications to Civil and Environmental Engineering. John Wiley & Sons. ISBN 0-471-72064-X
• Ash (1993) The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else). Wiley-IEEE Press. ISBN 0-7803-1051-9
• Clausing (1994) Total Quality Development: A Step-By-Step Guide to World-Class Concurrent Engineering. American Society of Mechanical Engineers. ISBN 0-7918-0035-0
• Haugen (1980) Probabilistic mechanical design. Wiley. ISBN 0-471-05847-5
• Papoulis (2002) Probability, Random Variables and Stochastic Process. McGraw-Hill Publishing Co. ISBN 0-07-119981-0
• Siddall (1982) Optimal Engineering Design. CRC. ISBN 0-8247-1633-7 | 717 | 3,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-18 | latest | en | 0.940184 |
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# All Activity
1. Past hour
2. ## Pegity - white to win
Is O win if we get XOOOOOX?
3. Today
4. ## Next card is Red
Thanks Thalia...
5. ## Pegity - white to win
OK so I was a kid once, back in the 40s and 50s, and we had this game called Pegity. It had a board with a 16x16 array of holes, and in turn each of 2-4 players inserted a peg of his own color into a hole. The object was to get 5 pegs of your color in a row: vertically, horizontally or diagonally. Not every game was won: similar to tic-tac-toe, you could run out of holes. But with only two players, there was usually a win. For simplicity let's shrink to a 9x9 board, mark the holes like in chess (A1, E7, etc.) and say that after O and X have each made three moves we have this position, with O to move: 1 2 3 4 5 6 7 8 9 A + + + + + + + + + B + + + + + + + + + C + + + + + + + + + D + + + + + + + + + E + + + + O + + + + F + + + O + O + + + G + + + + X + + + + H + + + X + X + + + I + + + + + + + + + With best play by both players, how soon can O win? Use chess notation (naming the holes, in two columns) to list the moves.
7. ## Next card is Red
@CaptainEd shuffles a standard deck of playing cards and begins dealing the cards face up on the table. At any time @plainglazed can say Stop and bet \$1 that the next card will be red. If he does not interrupt, the bet will automatically be placed on the last card. What is the best strategy? How much better than 50% can @plainglazed do?
8. ## Green and Yellow hats again, but harder
That is amazing. And you've shown that 1 fewer than a power of 2 is the optimal value for n. Can you reduce it to only q memorized strings, and the success rate to (q-1)/q, where q is the highest power of 2 less than or equal to n? (Purposely leaving this un-spoilered, cuz this is a tough problem.)
9. ## Baggage on a conveyor belt
Yup. And a bit of possibly helpful thinking:
10. ## A big, big, big hotel
Yeah, I hate annoyances, too...
11. Yesterday
12. ## Obtuse triangles in a circle
Not an answer, but a description of a potential approach and what would still be needed to make it work since this has gone for a while without being cracked.
13. ## A big, big, big hotel
Brilliant, surprisingly interesting and leading to a major annoyance. Well, now back to the original problem. Everybody leaves, the hotel is empty. The green and blue men arrive all together and go to the room they occupied previously. Is this problem equivalent to the first one? Precision: By equivalent, I mean that the problems are announced in a slightly different way, but the answers remain the same. Something this way: Aunt: If I give you 5 bananas and take back 2 bananas, how many bananas have you? John: I do not know. Aunt: I met your teacher and she told me you make this kind of problems at school. John: Yes, but we do it with apples. Just do not tell me that the men know now the room numbers. Besides being clairvoyant, they can move in time. Forwards, backwards and sideways.
14. Last week
15. ## Green and Yellow hats again, but harder
I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:
16. ## When midnight strikes
Sorry for 'discard', I meant these numbers build a {set} from which elements can be removed. And yes, we remain in rationals. What if l only paint those between 0.5 and 0.6, even leaving .53745 unpainted? Did I not paint an infinite number of numbers? We turn in round here. You claim that if you can remove an infinite amount of elements from a set, the set will be empty - I claim there may remain some elements (and even more then removed, depending on circumstances). I suppose you apply a 'law' or a 'theorem'. Can you give me the (Wikipedia) reference?
18. ## Modified checkerboard destruction
Can you think of a reason?
19. ## Baggage on a conveyor belt
I see, I didn’t pay attention. I really want to count only segments that are surrounded by shorter segments.
20. ## Modified checkerboard destruction
Thank you for the kind words, Bonanova. I fear my answer to this one is not so brilliant.
21. ## A big, big, big hotel
OK I think I just found your first blue man.
22. ## Worth the Weight
@aiemdao - Yes, nice solve indeed Aiem. Well done. @bonanova - thanks.
23. ## Right Prime
Any other prime numbers less than 1000.
24. ## Worth the Weight
@aiemdao - Nice solve. @plainglazed - Nice puzzle.
25. ## Worth the Weight
4 light ones/48
26. ## Modified checkerboard destruction
@CaptainEd brilliantly answered a puzzle that I mis-worded into a much tougher one. Nice job. Here's the puzzle I had intended to post: You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares, completely through to the other side. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
27. ## When midnight strikes
Isn't this fun? The { real numbers } between 0 and 1 comprise two infinite groups: { rationals } and { irrationals }. Rationals (expressible as p/q where p and q are integers) are countably infinite. We can order them, by placing them in an infinite square of p-q space and drawing a serpentine diagonal path. But the irrationals are not countable. The cardinality (notion of size used for infinite sets) of the rationals is called Aleph0 and for the irrationals (and reals) it's called Aleph1 or C (for continuum.) So first off, what we can do with the { numbers } between 0 and 1 depends on { which numbers } we want to deal with. Next, there's a problem that points are not objects that can be moved from place to place, as coins can. Points are more descriptions of space than of autonomous objects. If 0 and 1 are on a number line, the location midway between them is the point denoted by 0.5. It can't be removed. (We could erase the line, I guess, but that would not produce an interval [0 1] devoid of numbers, either.) But we can finesse this matter by "painting" a point. Kind of like what we did when we inscribed a number on each coin. Painting does a little less - it puts a point into a particular group or class but it does not distinguish among them. We can tell coin 2 from coin 3 and also distinguish both from a coin with no inscription. Two blue points, however, are distinguishable from unpainted points, but not from each other. But for our purpose here, painting is actually enough. Since the { rationals } are countable, we can paint them, in sequence, and if we do that at times of 1, 1/2, 1/4 .... minutes before midnight, all the rationals between 0 and 1 will be painted blue by midnight. (I don't know of a similar scheme for completing uncountable task sets, so the irrationals alas must remain unpainted.) Next, instead of asking whether all the points in [0 1] were removed, we can ask, with pretty much the same meaning, whether at this point the entire interval [0 1] has been painted blue. It should be, right? After all, between any two rationals lie countably infinite other rationals. So the paint must have covered the entire interval, right?. Well ... actually ... the answer is counter-intuitively No. And this is because even though the { rationals } are "dense" meaning there are no "gaps" between them, as noted above, the { irrationals } are even "more dense." That is, between any two irrational numbers there are uncountably many other irrationals. That means, for every blue point we created in the interval [0 1] there are uncountably many unpainted points. We can't magnify the line greatly enough to "see" individual points, but "if we could," if we were lucky enough to come across a single blue point we'd have to pan our camera over uncountably many unpainted points before we saw another blue one. In measure theory, the measure of the rationals over any interval is zero. The measure of the irrationals over [0 1] is unity. So what of our quest of emptying [0 1] of points? It's the same as our quest to paint [0 1] blue by painting all the rational numbers in that interval. (Reminding ourselves that there were too many irrationals to paint one at a time.) Turning again to measure theory, the measure of the "blueness" of the interval would be zero. That means we would not see even the hint of a faint blue haze. Back to our "empty the interval by removing the rationals" quest, Nope. [0 1] would not be empty: uncountably many points (numbers) would remain.
28. ## Right Prime
Some amount of primes.
29. ## Right Prime
Say 5939 is a "right" prime because it remains prime after dropping any number of digits from the right: 5939, 593, 59, and 5 are all prime. How many right primes are there less than 1000?
× | 2,298 | 9,025 | {"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} | 3.78125 | 4 | CC-MAIN-2018-09 | longest | en | 0.937584 |
https://www.physicsforums.com/threads/centripetal-motion-and-universal-gravitation-question-mars-and-sun-question.296218/ | 1,555,613,091,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526228.27/warc/CC-MAIN-20190418181435-20190418203435-00133.warc.gz | 776,645,154 | 17,625 | # Centripetal motion and universal gravitation question: Mars and Sun question
#### zeion
1. The problem statement, all variables and given/known data
Mars travels around the Sun in 1.88 (Earth) years, in an approximately circular orbit with a radius of 2.28 x 10^8 km
Determine
a) the orbital speed of Mars (relative to the Sun)
b) the mass of the Sun
2. Relevant equations
acceleration centripetal = 4(pi^2)(r) / (T^2)
universal attraction = (G)(m1)(m2) / (radius)^2
3. The attempt at a solution
Given:
T = 1.88 years = 59287680s = 5.9x10^7s
r = 2.28 x 10^8km = 2.28 x 10^11m
Orbital speed means centripetal acceleration yes?
Then,
acceleration centripetal = 4(pi^2)(r) / (T^2)
acceleration centripetal = 4(9.869604401)(2.28x10^11m) / (34.81x10^14s)
acceleration centripetal = 90.01x10^11m / 34.81x10^14s
acceleration centripetal = 2.59x10^-3m/s = 0.00259m/s
How come its so slow?
Related Introductory Physics Homework News on Phys.org
#### Doc Al
Mentor
Orbital speed means centripetal acceleration yes?
No. Speed is distance divided by time, not acceleration.
#### zeion
Ok so,
Mars travels around the Sun once every 1.88 years. So I have to find the distance of that circle and divide it by the time to find the speed. I have the radius, so I have to find the perimeter.
Perimeter of circle is 2(pi)(r)
So the distance traveled is 14.33x10^11m
In 1.88 years = 5.9x10^7s
So the speed is 14.33x10^11m / 5.9x10^7s = 2.43x10^4m/s yes?
Mentor
Good.
#### zeion
Now can I use this formula to solve for mass of Sun?
V^2 = (G)[m1(sun)] / r
[m1(sun)] = (V^2)(r) / (G)
[m1(sun)] = (2.43x10^4m/s)^2(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = (5.9x10^8m^2/s^2)(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = (13.45x19^19m^3/s^2) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = 2.02x10^8(N)m/kg^2
Why are the units all weird?
#### Doc Al
Mentor
Now can I use this formula to solve for mass of Sun?
V^2 = (G)[m1(sun)] / r
Good.
[m1(sun)] = (V^2)(r) / (G)
[m1(sun)] = (2.43x10^4m/s)^2(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = (5.9x10^8m^2/s^2)(2.28 x 10^11m) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = (13.45x19^19m^3/s^2) / [6.67x10^-11(N)(m^2)/kg^2]
[m1(sun)] = 2.02x10^8(N)m/kg^2
In your last step you didn't divide the units properly. You should have gotten:
[m^3/s^2]/[(N)(m^2)/kg^2] = [m^3/s^2]*[kg^2/(N)(m^2)] = [m kg^2]/[N s^2]
To simplify further, express Newtons in terms of more fundamental units:
N = kg-m/s^2
"Centripetal motion and universal gravitation question: Mars and Sun question"
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• Solo and co-op problem solving | 1,102 | 2,864 | {"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-2019-18 | latest | en | 0.77772 |
https://blog.bankbazaar.com/fy20-21which-is-better-old-new-tax-system/ | 1,713,918,192,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818835.29/warc/CC-MAIN-20240423223805-20240424013805-00583.warc.gz | 117,423,566 | 32,186 | FY20-21:Which is better – Old & New Tax System
By | February 23, 2020
Should you opt for the new tax regime or stick to the old one? Take a call after watching this video.
Budget 2020 announcement on the introduction of a new tax system from FY20-21 has confused many taxpayers. But we are here to help you resolve this confusion. First up, you need to understand the difference in slab rates under both the systems.
The slab rate for income between Rs.5L-Rs.7.5L will come down from 20% to 10% in the new system. On income between Rs. 7.5L-Rs.10L, the slab rate will come down from 20% to 15% under the new system. If you opt for the new system, the slab rate of income between Rs.10L-Rs12.5L will drop from 30% to 20% and that of income between Rs.12.5L-Rs.15L will come down from 30% to 25%. The slab rate for income of Rs.15L and above will continue to remain 30%.
Once you add up all your tax deductions (like the standard deduction of Rs. 50,000, up to Rs.1.5L u/s 80C, up to Rs.25,000 u/s 80D, up to Rs.2L on home loan interest u/s 24B, etc.), and see whether your total tax deductions constitute 20% of your gross income or not.
If yes, then you’re likely to cut down your tax liability by staying with the old system. If no, you’re likely to save more in tax by adopting the new system.
However, if your income is above the range of Rs.20L-Rs.25L, you’re likely to benefit more with the new system.
Income Tax Calculator
We’ve tried to explain this by comparing 3 scenarios in the video: First, assuming you stay with the old system without claiming any tax deduction, second when you adopt the new system and avail discounted slab rates by forgoing most of the tax deduction benefits, and third when you stay with the old system but claim 20% of your gross income as tax deduction.
The conclusion is you’re likely to save more in taxes by staying with the old system if you claim 20% of your gross income as tax deduction. If you don’t claim much in tax deductions, you’re likely to save more in taxes by adopting the new system.
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Topic: Predicativity
Horatiu Cheval (May 18 2021 at 13:14):
Probably a very basic type theory question, but what is the purpose of the predicativity of Type? What would break if
Π α : Type, αwere of type Type instead of Type 1? And what among the features distinguishing Prop from Type make Prop admissible as an impredicative sort?
Kevin Buzzard (May 18 2021 at 13:19):
If Pi a : Type, a were of type Type then you will get the same kind of contradiction as Russell's paradox, where you allow the collection of all sets to be a set. I think it's called Girard's paradox in type theory. Prop is impredicative because Leo designed it that way, I'm not sure I really understand the second question.
Kevin Buzzard (May 18 2021 at 13:21):
Oh I see -- you're asking "how come forall a : Prop, a is allowed to be a Prop"?
Horatiu Cheval (May 18 2021 at 13:21):
Yes, that's right
Kevin Buzzard (May 18 2021 at 13:22):
You could look at https://leanprover-community.github.io/mathlib_docs/logic/girard.html and then change all the Type u to Prop and see what breaks.
Horatiu Cheval (May 18 2021 at 13:22):
And why is it not contradictory if with Type it is. It seems like the same situation
Kevin Buzzard (May 18 2021 at 13:29):
OK so with the set theoretic version you construct a problematic set -- the set of all sets which don't contain themselves -- and then you make a contradiction from that. The key thing is that the contradiction involves constructing data, which doesn't live in the Prop universe.
If you just stick to Type then the Girard contradiction involves
F: Type → Type := λ (X : Type), (set (set X) → X) → set (set X)
U: Type := pi F
If you try this with X : Prop then Lean chokes at set X.
David Wärn (May 18 2021 at 13:30):
As I understand it, Type : Typeis contradictory because Type is too big, e.g. it should have strictly more elements than any A : Type. Similarly whenever you construct some type by referring to all A : Type, there is a risk that the result is too big, so can't be in Type. (Actually Π α : Type, α is not too big, since it's empty.) This is not true when you construct a Prop by referring to all A : Type: the resulting prop can only have one element, so classically it's true or false. So you shouldn't expect an impredicative Prop to be contradictory.
Kevin Buzzard (May 18 2021 at 13:33):
#check λ (X : Prop), (((X → Prop) → Prop) → X) → ((X → Prop) → Prop) -- Prop → Type
#check λ (X : Type), (((X → Prop) → Prop) → X) → ((X → Prop) → Prop) -- Type → Type
#check λ (X : Type u), (((X → Prop) → Prop) → X) → ((X → Prop) → Prop) -- Type u → Type u
So here explicitly is where the Girard proof breaks. I've replaced set X with X -> Prop. The trick doesn't work with Prop as the target is data.
Kevin Buzzard (May 18 2021 at 13:34):
Of course the _proof_ that an impredicative Prop isn't contradictory is that Lean's type theory is equiconsistent with ZFC + universes!
Horatiu Cheval (May 18 2021 at 13:57):
Thank you both for the great clarifications!
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#### Finance Mcqs – (Management Sciences) MCQs Latest For FPSC, PPSC, NTS, KPPSC, SPSC & Other Tests
Finance Mcqs “. Tab this page to check “Latest Finance MCQs” for the preparation of competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs and others. The most occurred mcqs of Finance in past papers. Past papers of Finance mcqs. Past papers of Finance MCQs. Finance Mcqs are the necessary part of any competitive / job related exams. The Finance mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related Finance mcqs. The Important series of Finance Mcqs are given below:
If a company revaluates its fixed assets, the current ratio of the company will:
A. Improve if assets are revalued downwards
B. Remain unaffected
C. Improve if assets are revalued upward
D. Undergo change only if liabilities are remaining constant
If we were studying a sample of 100 students and their examination performance and if the standard deviation of the list of results was say 14, then we could calculated the standard error by ___________?
A. Dividing the standard deviation by the square root of the number of items in the sample
B. Dividing standard deviation by number of items in the sample
C. Dividing the square root of the number of items in the sample by the mean
D. We cannot calculate standard error on account of inadequacy of information
Rule of 72 as a short cut method is explained by the formula:
A. 72 divided by the annual interest rate
B. 72 divided by (annual interest rate multiplied by discount factor)
C. Annual interest rate dividend by 72
D. None of these
Full Form of BCCI ?
A. Bank of Central Cooperation International
B. Bank of Commerce and Cooperation International
C. Bank of Credit and Commerce International
D. None of These
The Capital Asset Pricing Model calculate expected:
A. Return
B. Risk and Return
C. Risk
D. None of the above
A technique uses in comparative analysis of financial statement is____________?
A. Preference analysis
B. Graphical analysis
C. Common size analysis
D. Returning analysis
Net income available to stockholders is \$125 and total assets are \$1,096 then return on common equity would be________?
A. 0.12 times
B. 11.40%
C. 0.11%
D. 12%
Price per share is \$30 and an earnings per share is \$3.5 then price for earnings ratio would be_____________?
A. 8.57 times
B. 0.11 times
C. 8.57%
D. 11%
Formula such as net income available for common stockholders divided by total assets is used to calculate__________________________?
A. Return on total assets
B. Return on debt
C. Return on total equity
D. Return on sales
Price per ratio is divided by cash flow per share ratio which is used for calculating___________?
A. Sales to growth ratio
B. Dividend to stock ratio
C. Cash flow to price ratio
D. Price to cash flow ratio
A techniques uses to identify financial statements trends are included____________?
A. Percent change analysis
B. Common size analysis
C. Returning ratios analysis
D. Both A and B
Companies that help to set benchmarks are classified as__________?
A. Analytical companies
B. Benchmark companies
C. competitive companies
D. Return companies
Total assets divided common equity is a formula uses for calculating___________?
A. Equity multiplier
B. Turnover multiplier
C. Graphical multiplier
D. Stock multiplier
Price per share divided by earnings per share is formula for calculating_________?
A. Price earnings ratio
B. Pricing ratio
C. Earning price ratio
D. Earning ratio
In independent projects evaluation, results of internal rate of return and net present value lead to_____________?
A. Cost decision
B. Cash flow decision
C. Same decisions
D. Different decisions
Company low earning power and high interest cost cause financial changes which have_____________?
A. Low return on assets
B. High return on assets
C. High return on equity
D. Low return on equity
Projects which are mutually exclusive but different on scale of production or time of completion then the__________________?
A. Net future value method
B. Net present value of method
C. External return method
D. Internal return method
A point where profile of net present value crosses horizontal axis at plotted graph indicates project____________________?
A. Cash flows
B. Costs
C. Internal rate of return
D. External rate of return
Profit margin multiply assets turnover multiply equity multiplier is used to calculate____________?
A. Return on turnover
B. Return on assets
C. Return on stock
D. Return on equity
Payback period in which an expected cash flows are discounted with help of project cost of capital is classified as___________________?
A. Discounted payback period
B. Discounted cash flows
C. Discounted rate of return
D. Discounted project cost
Ratios which relate firm’s stock to its book value per share, cash flow and earnings are classified as_________?
A. Marginal ratios
B. Market value ratios
C. Return ratios
D. Equity ratios
An equation in which total assets are multiplied to profit margin is classified as_____________?
A. Du DuPont equation
B. Preference equation
C. Turnover equation
D. Common equation
In capital budgeting, term of bond which has great sensitivity to interest rates is______________?
A. Long-term bonds
B. Internal term bonds
C. Short-term bonds
D. External term bonds
Price earning ratio and price by cash flow ratio are classified as___________?
A. Marginal ratios
B. Return ratios
C. Equity ratios
D. Market value ratios
High price to earning ratio shows company’s_____________?
A. High risk prospect
B. Low dividends paid
C. High growth prospect
D. High marginal rate
Process of comparing company results with other leading firms is considered as____________?
A. Analysis
B. Comparison
C. Bench marking
D. Return analysis
An equity multiplier is multiplied to return on assets to calculate_________?
A. Return on assets
B. Return on turnover
C. Return on multiplier
D. Return on stock
A project whose cash flows are more than capital invested for rate of return then net present value will be___________?
A. Positive
B. Negative
C. Independent
D. Zero
In mutually exclusive projects, project which is selected for comparison with others must have____________?
A. Higher net present value
B. Zero net present value
C. Lower net present value
D. All of above
Profitability index in capital budgeting is used for_________?
A. Relative projects
B. Negative projects
C. Evaluate projects
D. Earned projects
Relationship between Economic Value Added (EVA) and Net Present Value (NPV) is considered as____________?
A. Economic relationship
B. Valued relationship
C. Direct relationship
D. Inverse relationship
An uncovered cost at start of year is \$200, full cash flow during recovery year is \$400 and prior years to full recovery is 3 then payback would be__________?
A. 4 years
B. 3.5 years
C. 5 years
D. 4.5 years
Present value of future cash flows is divided by an initial cost of project to calculate_______?
A. Exchange index
B. Negative index
C. Project index
D. Profitability index
First step in calculation of net present value is to find out_________?
A. Future value of equity
B. Present value of equity
C. Present value cash flow
D. Future value of cash flow
Life that maximizes net present value of an asset is classified as__________?
A. Present value life
B. Minimum life
C. Economic life
D. Transaction life
In capital budgeting, positive net present value results in_________________?
In estimating value of cash flows, compounded future value is classified as its__________?
A. Terminal value
B. Quit value
C. Existed value
D. Relative value
If two independent projects having hurdle rate, then both projects should________?
A. Be accepted
B. Have capital acceptance
C. Not be accepted
D. Have return rate acceptance
Cash flow which starts negative than positive then again positive cash flow is classified as__________?
A. Non-normal costs
B. Normal costs
C. Non-normal cash flow
D. Normal cash flow
Cash inflows are revenues of project and are represented by__________?
A. Relative number
B. Hurdle number
C. Negative numbers
D. Positive numbers
Net present value, profitability index, payback and discounted payback are methods to______________?
A. Evaluate budgeting
B. Evaluate projects
C. Evaluate cash flow
D. Evaluate equity
A type of project whose cash flows would not depend on each other is classified as______________?
A. Dependent projects
B. Independent projects
C. Project net gain
D. Net value projects
Bonds issued by corporations and exposed to default risk are classified as_________?
A. Corporation bonds
B. Risk bonds
C. Default bonds
D. Zero risk bonds
Falling interest rate leads change to bondholder income which is__________?
A. Reduction in income
B. Matured income
C. Increment in income
D. Frequent income
Bonds that have high liquidity premium are usually have_________?
Treasury bonds are exposed to additional risks that are included________?
A. Interest rate risk
B. Reinvestment risk
C. Investment risk
D. Both A and B
Payment divided by par value is classified as______________?
A. Par payment
B. Coupon payment
C. Divisible payment
D. Per period payment
An annual interest payment divided by current price of bond is considered as_____________?
A. Current yield
B. Return yield
C. Maturity yield
D. Earning yield
Coupon rate of convertible bond is_________?
A. Variable
B. Lower
C. Higher
D. Stable
An outstanding bond are also classified as__________?
A. Outdated bonds
B. Standing bonds
C. Dated bonds
D. Seasoned bonds
Finance Mcqs
A market interest rate for specific type of bond is classified as bond’s_____________?
A. Required rate of return
B. Required rate of redemption
C. Required option
D. Required rate of earning
An inflation rate includes in bond’s interest rates is one which is inflation rate________?
A. Expected at deferred call
B. Expected in future
C. Expected at time of maturity
D. At bond issuance
An average inflation rate which is expected over life of security is classified as__________?
Type of bond which pays interest payment only when it earns is classified as__________?
A. Income bond
B. Payment bond
C. Interest bond
D. Earning bond
Price of an outstanding bond increases when market rate___________?
A. Increases
B. Never changes
C. Decreases
D. Earned
If coupon rate is less than going rate of interest, then bond will be sold________?
A. Seasoned par value
B. More than its par value
C. Seasoned par value
D. At par value
A bond whose price will rise above its face value is classified as________?
Stated value of bonds or face value is considered as_____________?
A. Bond value
B. Par value
C. State value
D. Per value
Real risk-free interest rate in addition with an inflation premium is equal to_____________?
A. Liquidity risk-free interest rate
B. Quoted risk-free interest rate
C. Required interest rate
Bonds with deferred call have protection which is classified as__________?
A. Provision protection
B. Provision protection
C. Deferred protection
D. Call protection
When price of bond is calculated below its par value, it is classified as___________?
A. Compound bond
B. Discount bond
C. classified bond
D. Consideration earning
Rate on debt that increases as soon market rises is classified as________?
A. Market rate debt
B. Floating rate debt
C. Rising bet rate
D. Stable debt rate
Bonds that can be converted into shares of common stock are classified as_________?
A. Convertible bonds
B. Shared bonds
C. Stock bonds
D. Common bonds
Reinvestment risk of bond’s is usually higher on______?
B. Callable bonds
C. Income bonds
D. Default free bonds
Market in which bonds are traded over-the-counter than in an organized exchange is classified as__________?
B. Organized markets
C. Counter markets
D. Bond markets
Coupon payment of bond which is fixed at time of issuance____________?
A. Remains same
B. Becomes change
C. Becomes stable
D. Becomes low
Coupon payment is calculated with help of interest rate, then this rate considers as________?
A. Par interest
B. Payment interest
C. Coupon interest
D. Yearly interest rate
An effect of interest rate risk and investment risk on a bond’s yield is classified as_________?
Yield of interest rate which is below than coupon rate, this yield is classified as_________?
A. Yield to earning
B. Yield to call
C. Yield to maturity
D. Yield to investors
If market interest rate falls below coupon rate then bond will be sold__________?
A. Equal to return rate
B. Above its par value
C. Below its par value
D. Seasoned price
Rate of return (in percentages) consists of___________?
A. Capital gain yield interest yield
B. Return yield + unstable yield
C. Return yield + stable yield
D. Par value + market value
Type of bonds that are issued by foreign governments or foreign corporations are classified as__________?
A. Zero bonds
B. Zero risk bonds
C. Foreign bonds
D. Government bonds
If market interest rate rises above coupon rate, then bond will be sold_____________?
A. Seasoned price
B. Equal to return rate
C. Below its par value
D. Above its par value
An interest rate which is used in calculation of cash flows of bonds is called______________?
A. Required rate of earning
B. Required rate of redemption
C. Required rate of return
D. Required option
Type of options that permit bond holder to buy stocks at stated price are classified as______?
A. Guarantee
B. Provision
C. Warrants
D. Convertibles
Bond that has been issued in very recent timing is classified as_______?
A. Earning issue
B. Mature issue
C. New issue
D. Recent issue
Bonds issued by local and state governments with default risk are____________?
A. Municipal bonds
B. Default bonds
C. Corporation bonds
D. Zero bonds
Maturity date decides at time of issuance of bond and legally permissible is classified as____________?
A. Original maturity
B. Artificial maturity
C. Permanent maturity
D. Valued maturity
Value generally promises to pay at maturity date and a firm borrows is considered as bond’s__________?
A. Per value
B. Bond value
C. State value
D. Par value
Bonds issued by government and backed by Pak government are classified as_________?
A. U.S bonds
B. Treasury bonds
C. Issued security
D. Return security
An increasing in interest rate leads to decline in value of__________?
A. Standing bonds
B. Outstanding bonds
C. Junk bonds
Coupon rate of bond is also called____________?
B. Nominal rate
C. Quoted rate
D. Both a and c
Bonds issued by small companies tend to have_____________?
Type of bonds that pays no coupon payment but provides little appreciation are classified as______________?
A. Interest bond
B. Depreciated bond
C. Zero coupon bond
D. Appreciation bond
Reinvestment risk of bonds is higher on__________?
A. Short maturity bonds
C. High maturity bonds
D. High inflated bonds
In large expansion programs, increased riskiness and flotation cost associated with project can cause_______________?
A. Rise in marginal cost of capital
B. Rise in transaction cost of capital
C. Fall in marginal cost of capital
D. Rise in transaction cost of capital
Cash outflows are costs of project and are represented by___________?
A. Negative numbers
B. Hurdle number
C. Positive numbers
D. Relative number
Sum of discounted cash flows is best defined as____________?
A. Defined future value
B. Technical equity
C. Project net present value
D. Equity net present value
If net present value is positive, then profitability index will be__________?
A. Equal to
B. Greater than two
C. Less than one
D. Greater than one
Cash flows occurring with more than one change in sign of cash flow are classified as________?
A. Non-normal cash flow
B. Normal costs
C. Normal cash flow
D. Non-normal costs
Situation in which firm limits expenditures on capital is classified as________?
A. Marginal rationing
B. Capital rationing
C. Optimal rationing
D. Transaction rationing
An internal rate of return in capital budgeting can be modified to make it representative of_________?
A. Relative inflow
B. Relative outflow
C. Relative cost
D. Relative profitability
Other factors held constant, greater project liquidity is because of___________?
A. Greater project return
B. Less project returns
C. Shorter payback period
D. Greater payback period
Project whose cash flows are sufficient to repay capital invested for rate of return then net present value will be_________?
A. Positive
B. Zero
C. Negative
D. Independent
Process in which managers of company identify projects to add value is classified as__________?
A. Capital budgeting
B. Book value budgeting
C. Cost budgeting
D. Equity budgeting
Number of years forecasted to recover an original investment is classified as________?
A. Payback period
B. Original period
C. Forecasted period
D. Investment period
In capital budgeting, a negative net present value result in______________?
Modified rate of return and modified internal rate of return with exceed cost of capital if net present value is____________?
A. Positive
B. Zero
C. Negative
D. One
Set of projects or set of investments usually maximize firm value is classified as_________?
A. Optimal capital budget
B. Maximum capital budget
C. Minimum capital budget
D. Greater capital budget
In internal rate of returns, discount rate which forces net present values to become zero is classified as__________?
A. Negative rate of return
B. Positive rate of return
C. External rate of return
D. Internal rate of return
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http://cecilsunkure.blogspot.com/2010/02/ | 1,508,600,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00807.warc.gz | 63,399,624 | 22,176 | ## Thursday, February 25, 2010
### Caeser Cipher
Now I have finished up the encryption chapter of this book. The entire idea behind the encryption is to shift each character of a string over by a certain amount. Whenever a character of a string exited the parameters of a-z or A-Z, you just add or subtract 26, thus "wrapping" your encryption around and back, so that letters are only encrypted into letters. This took me about 30 minutes total to create. Here is my source code:
import sys
def getMode():
print('Do you want to encrypt decrypt, or brute force a message?')
mode = raw_input().lower()
if mode in 'encrypt decrypt e d brute b'.split():
return mode
else:
print('You need to enter in one of the following: e, d, b')
def getMessage():
return raw_input()
def getKey():
key = 0
while True:
print('Enter in a number for your key, 1-25.')
key = int(raw_input())
if (key >= 1 and key <= 25):
return key
def getTranslatedMessage(key, mode, message):
if mode[0] == 'd':
key = -key
translated = ''
for symbol in message:
if symbol.isalpha():
num = ord(symbol)
num += key
if symbol.isupper():
if num > ord('Z'):
num -= 26
elif num < ord('A'):
num += 26
elif symbol.islower():
if num > ord('z'):
num -= 26
elif num < ord('a'):
num += 26
translated += chr(num)
else:
translated += symbol
return translated
def playAgain():
print('Do you wish to continue? (yes or no)')
return raw_input().lower().startswith('y')
Again = True
while True:
mode = getMode()
message = getMessage()
if mode != 'b':
Key = getKey()
if mode != 'b':
print(getTranslatedMessage(Key, mode, message))
else:
for key in range(1, 26 + 1):
print(key, getTranslatedMessage(key, 'decrypt', message))
if not playAgain():
sys.exit
The next chapter of this book is here. This chapter goes over the Reversi game. I won't be constructing a program from this chapter as it doesn't really introduce any new concepts. The Ai for the reversi game just runs through each possible move and chooses the one that results in the most points. I will be skipping over chapter 14, and chapter 15, and heading straight into chapter 16. Chapter 16 is going over graphics and animation; I'll be starting to use the Pygame engine.
## Wednesday, February 24, 2010
### Sonar
I have finished up creating a game that functions almost the same as the game in this chapter. The game I created is slightly different in how the code works in a few specific areas, but altogether the example code in this chapter, and mine, should be fairly similar. Here is my source code:
#Sonar
import random
import sys
def drawBoard(board):
hline = ' '
for i in range(1, 6):
hline += (' ' * 9) + str(i)
print(hline)
print(' ' + ('0123456789' * 6))
print('')
for i in range(15):
if i < 10:
es = ' '
else:
es = ''
print('%s%s %s %s' % (es, i, getRow(board, i), i))
print('')
print(' ' + ('0123456789' * 6))
print(hline)
def getRow(board, row):
boardc = ''
for i in range(60):
boardc += board[i][row]
return boardc
def getNewBoard():
board = []
for x in range(60):
board.append([])
for y in range(15):
if random.randint(0, 1) == 1:
board[x].append('~')
else:
board[x].append('`')
return board
def placeChests(chestAmount):
chests = []
for i in range(chestAmount):
chests.append([random.randint(0, 59), random.randint(0, 14)])
return chests
def isValidMove(x, y):
if (x >= 0 and x <= 59 and y >= 0 and y <= 14):
return True
else:
return False
def makeMove(board, chests, x, y):
if not isValidMove(x, y):
return False
smalld = 100
for cx, cy in chests:
if (abs(cx - x) > abs(cy - y)):
distance = abs(cx - x)
else:
distance = abs(cy - y)
if distance < smalld:
smalld = distance
if smalld == 0:
chests.remove([x, y])
return 'You have found a treasure chest!'
else:
if smalld < 10:
board[x][y] = str(smalld)
return 'Treasure has been detected at a distance of %s' % (smalld)
else:
board[x][y] = '0'
return 'Treasure is nowhere to be found within range.'
def enterPlayerMove():
print('Enter in two integers (0-59, 0-14), otherwise type quit to exit the game.')
while True:
move = raw_input()
if move.lower() == 'quit':
sys.exit
move = move.split()
if len(move) == 2 and move[0].isdigit() and move[1].isdigit() and isValidMove(int(move[0]), int(move[1])):
return [int(move[0]), int(move[1])]
print('Enter in two integers (0-59, 1-15), otherwise type quit to exit the game.')
def playAgain():
print('Do you want to play again? (yes or no)')
return input().lower().startswith('y')
print('S O N A R !')
while True:
sonarDevices = 16
theBoard = getNewBoard()
previousMoves = []
playerMovec = []
drawBoard(theBoard)
print('Where do you want to place your sonar device?')
secretChests = placeChests(3)
while sonarDevices > 0:
if sonarDevices > 1: extraSsonar = 's'
else: extraSsonar = ''
if len(secretChests) > 1: extraSchest = 's'
else: extraSchest = ''
print('You have %s sonar device%s left. %s treasure chest%s remaining.' % (sonarDevices, extraSsonar, len(secretChests), extraSchest))
x, y = enterPlayerMove()
playerMovec.append([x, y])
result = makeMove(theBoard, secretChests, x, y)
if sonarDevices < 16:
if playerMovec[len(previousMoves)-1] == previousMoves[len(previousMoves)-1]:
sonarDevices -= 1
else:
result = False
drawBoard(theBoard)
print('You have already placed a sonar device here!')
previousMoves.append([x, y])
if sonarDevices == 16:
sonarDevices -= 1
if result == False:
continue
else:
if result == 'You have found a treasure chest!':
for x, y in previousMoves:
makeMove(theBoard, secretChests, x, y)
drawBoard(theBoard)
print(result)
if len(secretChests) == 0:
print('You have found all the sunken treasure chests! Congratulations and good game!')
break
if sonarDevices == 0:
print('We\'ve run out of sonar devices! Now we have to turn the ship around and head')
print('for home with treasure chests still out there! Game over.')
print(' The remaining chests were here:')
for x, y in theChests:
print(' %s, %s' % (x, y))
if not playAgain():
sys.exit()
A large difference between my code, and the example code, is the fact that I have support for allowing the player to suffer no negative penalty (a loss of one of the sonar beacons) for trying to place a sonar over an area that was placed during the previous turn. This honestly, took me forever to figure out how to do. I finally decided on creating two copies of all previous turns, and then use these two copies to see if the last values in the two lists equal one another, before the value in the second list had been appended. This way, I compared the value of entered coordinates to the value of previous coordinates, and didn't subtract from the overall sonar amount when this happened.
The math is simple, and the whole Cartesian Plane didn't really teach me anything new. Although, if you don't understand how the math in my makeMove() function doesn't work, you might need to read over this chapter again.
I plan to go other the thirteenth chapter, here, for my next post. This seems rather interesting, as it is about cryptology.
Let me know if you see something I could improve upon in this code. Hopefully my lack of commenting the code isn't detrimental, if it is let me know and I'll add in comments.
## Tuesday, February 23, 2010
### Bagels
The name of the game I have created for this post, is the game Bagels. Bagels is a game where the user guesses a three digit number, and the computer responds with "Fermi" for every digit that is in the right place that you guessed, and "Pico" for every digit that you guessed which is in the wrong place, but is still a digit in the secret number.
The concepts that this game goes over are hardcoding and nested loops, the chapter is this one: http://inventwithpython.com/chapter10.html.
A nested loop is simply a loop within a loop. This is useful for when you want to iterate something x times, for every time y appears, or something similar.
Hardcoding is a coding practice where the coder codes the program so that changing of simple variables requires changing lots of code. This program was programmed without hardcoding, where a couple constants (variables that do not change while the program is running, which are represented with caps, although, are actually just normal variables in Python) can be changed to change how the game is played.
import random
def getSecretNumbers(numDigits):
numbers = list(range(10))
random.shuffle(numbers)
secretNum = ''
for i in range(numDigits):
secretNum += str(numbers[i])
return secretNum
def getClues(guess, secretNum):
if guess == secretNum:
return 'You have guessed the correct number!'
clue = []
for i in range(len(guess)):
if guess[i] == secretNum[i]:
clue.append('Fermi')
elif guess[i] in secretNum:
clue.append('Pico')
if len(clue) == 0:
return 'Bagels'
clue.sort()
return ' '.join(clue)
def isOnlyDigits(num):
if num == '':
return False
for i in num:
if i not in '1 2 3 4 5 6 7 8 9'.split():
return False
return True
def playAgain():
print('Do you want to play again? Yes or no.')
return raw_input().lower().startswith('y')
NUMDIGITS = 3
MAXGUESS = 10
print('I am thinking of a %s-digit number. Try to guess what it is.' % (NUMDIGITS))
print('Here are some clues:')
print('When I say: That means:')
print(' Pico One digit is correct but in the wrong position.')
print(' Fermi One digit is correct and in the right position.')
print(' Bagels No digit is correct.')
while True:
secretNum = getSecretNumbers(NUMDIGITS)
print('I have thought up a number. You have %s guesses to get it.' % (MAXGUESS))
numGuesses = 1
while numGuesses <= MAXGUESS:
guess = ''
while len(guess) != NUMDIGITS or not isOnlyDigits(guess):
print('Guess #%s: ' % (numGuesses))
guess = raw_input()
clue = getClues(guess, secretNum)
print(clue)
numGuesses += 1
if guess == secretNum:
break
if numGuesses > MAXGUESS:
print('You ran out of guesses. The answer was %s.' % (secretNum))
if not playAgain():
break
My code should be pretty much identicle to the code in the example program, as there aren't really any easier ways to code this program; this is a really rather cut and draw program to create.
This part of the program confused me for a moment:
def getSecretNumbers(numDigits):
numbers = list(range(10))
random.shuffle(numbers)
These lines create a list of numbers and then shuffle them. When I first saw these lines, the first one confused me: numbers = list(range(10)). I wasn't sure why the auther was changing the range() to list() inside of the range. Although, in the book it explained that if I wanted to store a list of integers, they needed to be converted from an iterator (value of range) into a list. This made sense, although the author didn't go into detail about exactly what the value of a range was. I assume it's just an amount of iterations to perform, which isn't what I would want if I wanted a list to store variables in.
I plan to run through chapter eleven of this book for my next post. This should be a relatively easy chapter, as it just goes through the basics of a Cartesian coordinate system.
## Monday, February 22, 2010
### Finishing Tic Tac Toe
Aha! So I coded up my own version of Tic Tac Toe. I did take a look at my old program that I posted up before (here) in a couple spots where I couldn't pinpoint my own bugs, but other than that I wrote this thing without a reference. This, pretty much just solidified my understanding and familiarity of Python syntax even further. Not much else to explain. Here is my source code:
import random
def drawBoard():
print(board[1] + '|' + board[2] + '|' + board[3])
print('-----')
print(board[4] + '|' + board[5] + '|' + board[6])
print('-----')
print(board[7] + '|' + board[8] + '|' + board[9])
def pickYourLetter():
letter = ''
while (letter != 'X' and letter != 'O'):
print('Would you like to X or O?')
letter = raw_input().upper()
if (letter != 'X' and letter != 'O'):
print('Uhm, I said X or O.')
if letter == 'X':
return ['X', 'O']
else:
return ['O', 'X']
def whoGoesFirst():
if random.randint(0, 1) == 1:
return 'player'
else:
return 'computer'
def getPlayerMove(theBoard):
move = ' '
while move not in '1 2 3 4 5 6 7 8 9'.split() or not isSpaceFree(theBoard, int(move)):
print('It is your turn to move.')
move = raw_input()
return int(move)
def isSpaceFree(theBoard, move):
return board[move] == ' '
def makeMove(board, letter, move):
board[move] = letter
def isWinner(b, l):
return ((b[1] == l and b[2] == l and b[3] == l) or
(b[4] == l and b[5] == l and b[6] == l) or
(b[7] == l and b[8] == l and b[9] == l) or
(b[1] == l and b[4] == l and b[7] == l) or
(b[2] == l and b[5] == l and b[8] == l) or
(b[3] == l and b[6] == l and b[9] == l) or
(b[1] == l and b[5] == l and b[9] == l) or
(b[3] == l and b[5] == l and b[7] == l))
def isTie(board):
for i in range(1, 10):
if isSpaceFree(board, i):
return False
return True
def getBoardCopy(theBoard):
dupeBoard = []
for i in board:
dupeBoard.append(i)
return dupeBoard
def getComputerMove(theBoard, letter):
if computerLetter == 'X':
playerLetter = 'O'
else:
playerLetter = 'X'
for i in range(1, 10):
copy = getBoardCopy(theBoard)
if isSpaceFree(copy, i):
makeMove(copy, computerLetter, i)
if isWinner(copy, computerLetter):
return i
for i in range(1, 10):
copy = getBoardCopy(theBoard)
if isSpaceFree(copy, i):
makeMove(copy, playerLetter, i)
if isWinner(copy, playerLetter):
return i
move = chooseRandomMove(theBoard, [1, 3, 7, 9])
if move != None:
return move
if isSpaceFree(board, 5):
return 5
return chooseRandomMove(board, [2, 4, 6, 8])
def chooseRandomMove(theBoard, moveList):
moves = []
for i in moveList:
if isSpaceFree(theBoard, i):
moves.append(i)
if len(moves) != 0:
return random.choice(moves)
else:
return None
def playAgain():
print('Do you want to play again? (yes or no)')
return raw_input().lower().startswith('y')
while True:
board = [' '] * 10
print('This is Tic Tac Toe.')
playerLetter, computerLetter = pickYourLetter()
gameIsPlaying = True
currentTurn = whoGoesFirst()
if currentTurn == 'player':
print('You will move first, as decided at random.')
else:
print('The computer will move first, as decided at random.')
while gameIsPlaying == True:
if currentTurn == 'player':
drawBoard()
move = getPlayerMove(board)
makeMove(board, playerLetter, move)
if isWinner(board, playerLetter):
drawBoard()
print('You have won the game!')
gameIsPlaying = False
else:
if isTie(board):
drawBoard()
print('Game is a tie!!')
gameIsPlaying = False
else:
currentTurn = 'computer'
else:
move = getComputerMove(board, computerLetter)
makeMove(board, computerLetter, move)
if isWinner(board, computerLetter):
drawBoard()
print('You have lost the game!')
gameIsPlaying = False
else:
if isTie(board):
drawBoard()
print('Game is a tie!')
gameIsPlaying = False
else:
currentTurn = 'player'
if not playAgain():
break
And that is my code. Here is the exe file in case you are interested: http://www.mediafire.com/?oodckz5oxmy
Anyways, I plan to go over this chapter tomorrow and follow through with constructing the program from scratch without referring to the example code
I had a few troubles throughout the program, and the most annoying one was in the definitions of winning. I had constantly placed a 1 in place of a lower case L, and this resulted in bad things. I also had a couple typos here and there, things like typing = instead of ==, and vise versa.
One thing in particular that I found interesting while creating this program, was the "None" value. Here is a quote from the book "Calls to functions that do not return anything (that is, they exit by reaching the end of the function and not from a return statement) will evaluate to None." This was useful in that whenever I wanted the Ai in my simulation to choose a random location to place his letter, say, one of the four corners; if all four of these corners were taken, I could return the value of None instead of one of the values for one of the four corners.
I also learned from this chapter some interesting aspects of list referencing. Take a look at this function:
def makeMove(board, letter, move):
board[move] = letter
It would seem like this function wouldn't do anything due to the change of the board list being of the private scope by default, since it is happening within the function itself. This is because the word "board" isn't actually the list board, it is just a reference to that list passed onto the function as an argument. This means that when we have the line board[move] = letter we aren't modifying the data stored in a variable, but we are modifying the data stored in the list, which is referenced by the word "board". Here is a quote from the book that explains this: "When you assign a list to a variable with the = sign, you are actually assigning a reference to the list. A reference is a value that points to some bit of data, and a list reference is a value that points to a list."
### Starting up Python!
Well, sadly, I have no more time for C++, and I probably won't for a couple months. Things are getting really busy. Although, I have good news; I started a programming club at my highschool that meets once a week, where I teach them all how to code in Python. For the time being, my next posts will be about my learnings of Python.
I chose Python after doing a little bit of research; Python is a language that was created to allow coders to create code quickly and efficiently. Readability of Python was also a concern while developing the Python environment. As such, Python is great for readability and for quickly coding programs. Although Python is great at what it was intended to be used for, Python isn't made to create highly optimized code for intense programming (things like 3D rendering). Although, you can create visual programs that render in 3D, Python code can only be optimized so much.
I spent this weekend reading some of the chapters from this excellent book on programming games in Python, with the Pygame engine. Here is the book I am currently reading: http://inventwithpython.com/chapters/. I am currently on nine.
I am working with Python 2.6, as this has the most up-to-date third party support from things like Pygame and Py2Exe.
I recommend that anyone reading this check out the Pygame.org website, and sift around through the tutorials there, as well as check out the source code for some of the more simple programs that are on display there.
Over the course of chapter nine, I created a game of TicTacToe. The player can choose to play as X's or O's, and I have a functioning Ai.
At this point in time, I do have a working program, although, I created it while using the example code in the book as reference. As an exercise, my next post will be about me trying to create the entire program without reference code whatsoever.
Here is my source code for my program:
import random
def drawBoard(board):
#This function draws the board whenever it is called. The board is represented as an array.
print(' ' + board[7] + ' | ' + board[8] + ' | ' + board[9])
print(' | |')
print('-----------')
print(' | |')
print(' ' + board[4] + ' | ' + board[5] + ' | ' + board[6])
print(' | |')
print('-----------')
print(' | |')
print(' ' + board[1] + ' | ' + board[2] + ' | ' + board[3])
print(' | |')
def getPlayerLetter():
#This function allows the human to choose whether they play as an X or O.
letter = ''
while not (letter == 'X' or letter == 'O'):
print('What would you like to play as? X or O?')
letter = raw_input().upper()
if letter != 'X' and letter != 'O':
print('You can only pick X or O to play as.')
if letter == 'X':
return ['X', 'O']
else:
return ['O', 'X']
def whoGoesFirst():
#This function randomly picks which player goes first.
if random.randint(0, 1) == 1:
return 'player'
else:
return 'computer'
def getPlayerMove(board):
#Captures the player's input on their desired move for their turn.
move = ' '
while move not in '1 2 3 4 5 6 7 8 9'.split() or not isSpaceFree(board, int(move)):
move = raw_input()
return int(move)
def isSpaceFree(board, move):
#Return true if the move is available on the board.
return board[move] == ' '
def makeMove(board, letter, move):
board[move] = letter
def isWinner(bo, le):
# Given a board and a player's letter, this function returns True if that player has won.
# We use bo instead of board and le instead of letter so we don't have to type as much.
return ((b[7] == l and b[8] == l and b[9] == l) or # across the top
(b[4] == l and b[5] == l and b[6] == l) or # across the middle
(b[1] == l and b[2] == l and b[3] == l) or # across the bottom
(b[7] == l and b[4] == l and b[1] == l) or # down the left side
(b[8] == l and b[5] == l and b[2] == l) or # down the middle
(b[9] == l and b[6] == l and b[3] == l) or # down the right side
(b[7] == l and b[5] == l and b[3] == l) or # diagonal
(b[9] == l and b[5] == l and b[1] == l)) # diagonal
def isBoardFull(board):
#Checks to see if the board is full or not.
for i in range(1, 10):
if isSpaceFree(board, i):
return False
return True
def getComputerMove(board, computerLetter):
#Determines how to react depending on which letter (X or O) the computer player has.
if computerLetter == 'X':
playerLetter = 'O'
else:
playerLetter = 'X'
# Check if the computer can win in the next move.
for i in range(1, 10):
copy = getBoardCopy(board)
if isSpaceFree(copy, i):
makeMove(copy, computerLetter, i)
if isWinner(copy, computerLetter):
return i
# Check if the player could win on his next move, and block them.
for i in range(1, 10):
copy = getBoardCopy(board)
if isSpaceFree(copy, i):
makeMove(copy, playerLetter, i)
if isWinner(copy, playerLetter):
return i
# Try to take one of the corners, if they are free.
move = chooseRandomMoveFromList(board, [1, 3, 7, 9])
if move != None:
return move
# Try to take the center, if it is free.
if isSpaceFree(board, 5):
return 5
# Move on one of the sides.
return chooseRandomMoveFromList(board, [2, 4, 6, 8])
def chooseRandomMoveFromList(board, movesList):
# Returns a valid move from the passed list on the passed board.
# Returns None if there is no valid move.
possibleMoves = []
for i in movesList:
if isSpaceFree(board, i):
possibleMoves.append(i)
if len(possibleMoves) != 0:
return random.choice(possibleMoves)
else:
return None
def getBoardCopy(board):
#This returns a copy of the actual board, used to simulate or test things.
dupeBoard = []
for i in board:
dupeBoard.append(i)
return dupeBoard
def playAgain():
# This function returns True if the player wants to play again, otherwise it returns False.
print('Do you want to play again? (yes or no)')
return raw_input().lower().startswith('y')
while True:
#Resets the board.
theBoard = [' '] * 10
playerLetter, computerLetter = getPlayerLetter()
currentTurn = whoGoesFirst()
print('The ' + currentTurn + ' player will go first.')
gameIsPlaying = True
while gameIsPlaying:
if currentTurn == 'player':
#Player's turn.
drawBoard(theBoard)
move = getPlayerMove(theBoard)
makeMove(theBoard, playerLetter, move)
if isWinner(theBoard, playerLetter):
drawBoard(theBoard)
print('You have won the game!')
gameIsPlaying = False
else:
if isBoardFull(theBoard):
drawBoard(theBoard)
print('The game is a tie!')
break
else:
currentTurn = 'computer'
else:
#Computer player's turn.
move = getComputerMove(theBoard, computerLetter)
makeMove(theBoard, computerLetter, move)
if isWinner(theBoard, computerLetter):
drawBoard(theBoard)
print('The computer has beaten you! You lose.')
gameIsPlaying = False
else:
if isBoardFull(theBoard):
drawBoard(theBoard)
print('The game is a tie!')
break
else:
currentTurn = 'player'
if not playAgain():
break
As you might notice, this code is pretty darn similar to my reference code. As I mentioned, I will as my next project re-write this entire program without any reference.
I would say that the most interesting part of creating this program, was the Ai. The Ai really only cycles through a copy of the currently existing board, and searches for any winning moves from either player. If it finds one, it takes it. If not, it moves to one of the currently open corners, randomly. If all the corners are taken, the Ai moves into the middle space. If that is taken, it moves onto one of the four last spaces, at random. This algorithm is pretty darn simple, but it is effective at playing Tic Tac Toe.
Overall, I would say I gained a better understanding of Python syntax more than anything while writing this program.
I tried to include a source file, and my exe, although I'm currently having troubles uploading. I'll try to remember to modify this post with some links, in case anyone wanted to download the exe or the source. Edit: No, I will not be supplying the exe//source code this time ;P | 6,569 | 24,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-43 | longest | en | 0.700801 |
https://www.thestudentroom.co.uk/showthread.php?t=7041595 | 1,627,825,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.36/warc/CC-MAIN-20210801123745-20210801153745-00653.warc.gz | 1,087,722,225 | 33,612 | binomial distribution question query
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#1
I really dont understand why they are working out p (x=12) and p(x>12), but not
p (x<12) they did p (x <11) instead. why?
Last edited by Lina_Amiour; 1 month ago
0
1 month ago
#2
Looks like a typo as they define and use it as the 12 outcomes 0,1,2, ...,,11 so it should be P(x<12)
0
#3
(Original post by mqb2766)
Looks like a typo as they define and use it as the 12 outcomes 0,1,2, ...,,11 so it should be P(x<12)
oh ok thank you. Would you mind helping me with one more question, a different one?
Its the practice question 1 - in the mark scheme they got k as 1/10 whereas I got 1/4. I don't know where I went wrong. I used the cumulative distribution function,
0
1 month ago
#4
(Original post by Lina_Amiour)
oh ok thank you. Would you mind helping me with one more question, a different one?
Its the practice question 1 - in the mark scheme they got k as 1/10 whereas I got 1/4. I don't know where I went wrong. I used the cumulative distribution function,
Can you upload what you did.
0
#5
(Original post by mqb2766)
Can you upload what you did.
i dont have any storage left to upload the picture. is it ok if you just briefly explain how to do the question?
0
1 month ago
#6
(Original post by Lina_Amiour)
i dont have any storage left to upload the picture. is it ok if you just briefly explain how to do the question?
Shouldn't be more than a oouple of lines. Just upload your attempt as text.
Last edited by mqb2766; 1 month ago
0
#7
(Original post by mqb2766)
Shouldn't be more than a oouple of lines. Just upload your attempt as text.
so i thought that X has to be 4 or less so I did P (X<=4) = 1 getting k as 1/4. so i really don't get how they got 1/10
0
1 month ago
#8
(Original post by Lina_Amiour)
so i thought that X has to be 4 or less so I did P (X<=4) = 1 getting k as 1/4. so i really don't get how they got 1/10
P(x<=4) is not P(x==4).
Just sum the given probabilities?
Last edited by mqb2766; 1 month ago
0
#9
(Original post by mqb2766)
P(x<=4) is not P(x==4).
Just sum the given probabilities?
why sum them?
0
1 month ago
#10
(Original post by Lina_Amiour)
why sum them?
Because P(X<=4) is the sum of each of the probabilities P(x==1) to P(x==4).
0
#11
(Original post by mqb2766)
Because P(X<=4) is the sum of each of the probabilities P(x==1) to P(x==4).
ohhhh i see i get it now. Thank you SO much you have been a GREAT help
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1.23% | 964 | 3,212 | {"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} | 3.203125 | 3 | CC-MAIN-2021-31 | latest | en | 0.957547 |
http://www.vias.org/tmdatanaleng/cc_test_1samplettest_small.html | 1,558,720,954,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00253.warc.gz | 348,839,291 | 2,310 | You are working with the text-only light edition of "H.Lohninger: Teach/Me Data Analysis, Springer-Verlag, Berlin-New York-Tokyo, 1999. ISBN 3-540-14743-8". Click here for further information.
## One-Sample t-TestSmall Samples
For small sample sizes, the mean shows a t-distribution (compare the case for large samples: in this case the means are distributed according to a normal distribution). The reason for this is that the estimate s of the standard deviation is not accurate enough. But again the assumption is that the distribution of the data is approximately normal. The only difference from the derivation for the large sample case is that, instead of the z-scores, we will use the t-scores, which we can obtain from statistical tables, or by using the Teach/Me distribution calculator. The t-distribution depends on the number of samples. Its parameter is called degree of freedom, and it is equal to the number of samples minus 1: df = n - 1.
For a detailed description of how to perform a t-test, please have a look at the z-test section.
Last Update: 2005-Jul-16 | 249 | 1,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-22 | latest | en | 0.877511 |
https://in.mathworks.com/matlabcentral/answers/164310-could-anyone-help-me-with-the-interp1-please | 1,653,079,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534669.47/warc/CC-MAIN-20220520191810-20220520221810-00027.warc.gz | 386,109,137 | 22,369 | # Could anyone help me with the interp1 please?
1 view (last 30 days)
Avan Al-Saffar on 25 Nov 2014
I have a system of two equations with two variables,I am using ode45 to solve the system numerically then trying to find Fisher Information(I),,Because my variables x(1) and x(2) depend on time so I am using '' interp1 '' ,,The problem is that I am not sure of using ' interp1' and the second thing I am getting two values for (I) while I am expecting to get only one value!
My code is :
function RunFisherpaperInt
A=1;
x0=[12 5];
dt=0.1;
tspan=0:dt:100;
b=4;
[t,x]=ode45(@Fisherpaper,tspan,x0,[],b);
x=[x(:,1),x(:,2)];
X = @(T) interp1(t,x,T) ;
a1=(15.*X(1)-b.*X(1).*X(2)) ;
b1=(-5.*X(1)+0.5.*X(1).*X(2)) ;
f = @(T) (A.*( (( a1.*( a1.*(15 - b.*X(2)) + b1.*(0.5.*X(2)) ) + b1 .*(a1.*(-b.*X(1)) +b1.*(-5 +
0.5.*X(2)) ) ).^2)./(a1.^2+b1.^2).^3 ) ) ;
I = integral(f,0,100,'ArrayValued',true)
1;
% function dxdt = Fisherpaper(t,x,b)
%x(1) = prey
% x(2) = predator
% dxdt=zeros(2,1);
% dxdt(1) = 15 * x(1) - b * x(1).* x(2);
% dxdt(2) = -5 * x(2) + .5 * x(1).* x(2);
% end | 436 | 1,068 | {"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-2022-21 | latest | en | 0.504416 |
https://www.gmpuzzles.com/blog/2019/01/fillomino-cipher-by-tim-marsden/ | 1,720,923,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00370.warc.gz | 693,840,626 | 14,063 | ### Fillomino (Cipher) by Tim Marsden
PDF
or solve online (using our beta test of Penpa-Edit tools; use tab to alternate between a composite mode for line/edge drawing and a number entry mode.)
Theme: Clue Symmetry and Logic
Author/Opus: This is the 1st puzzle from guest contributor Tim Marsden. [Update: this puzzle is the 10th puzzle from guest contributor Dan Katz and is part of a set connected to the 2019 Mystery Hunt.]
Rules: Standard Fillomino rules. Also, each letter represents a different positive integer.
Answer String: For each cell in the marked rows/columns, enter the area of the polyomino it belongs to. Enter just the last digit for any two-digit number. Start with the 3rd row, followed by a comma, followed by the 8th column.
Time Standards (highlight to view): Grandmaster = 2:00, Master = 4:15, Expert = 8:30
Solution: PDF
Note: Follow this link for other classic Fillomino and this link for more variations on Fillomino puzzles. If you are new to this puzzle type, here are our easiest Fillomino puzzles to get started on. More Fillomino puzzles can be found in The Art of Puzzles, in Fill o’ Fillomino and Fill o’ Fillomino 2 by Grant Fikes, and in our beginner-friendly book Logic Puzzles 101.
• Bryce Herdt says:
MUDPATCH+N, DUTCHMAN+P… CHANT DUMP? Sorry, is there rhyme or reason to the choice of letters?
• Dan Katz says:
Now that the event is over, I think it’s safe to answer this question: The four “Tim Marsden” puzzles, in addition to being solvable independently, served as part of this puzzle from the 2019 MIT Mystery Hunt:
http://web.mit.edu/puzzle/www/2019/puzzle/connect_four.html
• skynet says:
6 mins.
• karkea says:
7:12 interesting puzzle.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 454 | 1,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.875761 |
https://www.symbolab.com/study-guides/collegealgebra1/section-exercises-19.html | 1,721,148,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00392.warc.gz | 894,185,340 | 47,578 | We've updated our
TEXT
# Section Exercises
1. What is an arithmetic sequence? 2. How is the common difference of an arithmetic sequence found? 3. How do we determine whether a sequence is arithmetic? 4. What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence? 5. Describe how linear functions and arithmetic sequences are similar. How are they different? For the following exercises, find the common difference for the arithmetic sequence provided. 6. $\left\{5,11,17,23,29,...\right\}$ 7. $\left\{0,\frac{1}{2},1,\frac{3}{2},2,...\right\}$ For the following exercises, determine whether the sequence is arithmetic. If so find the common difference. 8. $\left\{11.4,9.3,7.2,5.1,3,...\right\}$ 9. $\left\{4,16,64,256,1024,...\right\}$ For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference. 10. ${a}_{1}=-25$ , $d=-9$ 11. ${a}_{1}=0$ , $d=\frac{2}{3}$ For the following exercises, write the first five terms of the arithmetic series given two terms. 12. ${a}_{1}=17,{a}_{7}=-31$ 13. ${a}_{13}=-60,{a}_{33}=-160$ For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference. 14. First term is 3, common difference is 4, find the 5th term. 15. First term is 4, common difference is 5, find the 4th term. 16. First term is 5, common difference is 6, find the 8th term. 17. First term is 6, common difference is 7, find the 6th term. 18. First term is 7, common difference is 8, find the 7th term. For the following exercises, find the first term given two terms from an arithmetic sequence. 19. Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{6}=12$ and ${a}_{14}=28$. 20. Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{7}=21$ and ${a}_{15}=42$. 21. Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{8}=40$ and ${a}_{23}=115$. 22. Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{9}=54$ and ${a}_{17}=102$. 23. Find the first term or ${a}_{1}$ of an arithmetic sequence if ${a}_{11}=11$ and ${a}_{21}=16$. For the following exercises, find the specified term given two terms from an arithmetic sequence. 24. ${a}_{1}=33$ and ${a}_{7}=-15$. Find ${a}_{4}$. 25. ${a}_{3}=-17.1$ and ${a}_{10}=-15.7$. Find ${a}_{21}$. For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 26. ${a}_{1}=39;\text{ }{a}_{n}={a}_{n - 1}-3$ 27. ${a}_{1}=-19;\text{ }{a}_{n}={a}_{n - 1}-1.4$ For the following exercises, write a recursive formula for each arithmetic sequence. 28. ${a}_{n}=\left\{40,60,80,...\right\}$ 29. ${a}_{n}=\left\{17,26,35,...\right\}$ 30. ${a}_{n}=\left\{-1,2,5,...\right\}$ 31. ${a}_{n}=\left\{12,17,22,...\right\}$ 32. ${a}_{n}=\left\{-15,-7,1,...\right\}$ 33. ${a}_{n}=\left\{8.9,10.3,11.7,...\right\}$ 34. ${a}_{n}=\left\{-0.52,-1.02,-1.52,...\right\}$ 35. ${a}_{n}=\left\{\frac{1}{5},\frac{9}{20},\frac{7}{10},...\right\}$ 36. ${a}_{n}=\left\{-\frac{1}{2},-\frac{5}{4},-2,...\right\}$ 37. ${a}_{n}=\left\{\frac{1}{6},-\frac{11}{12},-2,...\right\}$ For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term. 38. ${a}_{n}=\left\{7\text{, }4\text{, }1\text{, }...\right\}$; Find the 17th term. 39. ${a}_{n}=\left\{4\text{, }11\text{, }18\text{, }...\right\}$; Find the 14th term. 40. ${a}_{n}=\left\{2\text{, }6\text{, }10\text{, }...\right\}$; Find the 12th term. For the following exercises, use the explicit formula to write the first five terms of the arithmetic sequence. 41. ${a}_{n}=24 - 4n$ 42. ${a}_{n}=\frac{1}{2}n-\frac{1}{2}$ For the following exercises, write an explicit formula for each arithmetic sequence. 43. ${a}_{n}=\left\{3,5,7,...\right\}$ 44. ${a}_{n}=\left\{32,24,16,...\right\}$ 45. ${a}_{n}=\left\{-5\text{, }95\text{, }195\text{, }...\right\}$ 46. ${a}_{n}=\left\{-17\text{, }-217\text{, }-417\text{,}...\right\}$ 47. ${a}_{n}=\left\{1.8\text{, }3.6\text{, }5.4\text{, }...\right\}$ 48. ${a}_{n}=\left\{-18.1,-16.2,-14.3,...\right\}$ 49. ${a}_{n}=\left\{15.8,18.5,21.2,...\right\}$ 50. ${a}_{n}=\left\{\frac{1}{3},-\frac{4}{3},-3\text{, }...\right\}$ 51. ${a}_{n}=\left\{0,\frac{1}{3},\frac{2}{3},...\right\}$ 52. ${a}_{n}=\left\{-5,-\frac{10}{3},-\frac{5}{3},\dots \right\}$ For the following exercises, find the number of terms in the given finite arithmetic sequence. 53. ${a}_{n}=\left\{3\text{,}-4\text{,}-11\text{, }...\text{,}-60\right\}$ 54. ${a}_{n}=\left\{1.2,1.4,1.6,...,3.8\right\}$ 55. ${a}_{n}=\left\{\frac{1}{2},2,\frac{7}{2},...,8\right\}$ For the following exercises, determine whether the graph shown represents an arithmetic sequence. 56. 57. For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence. 58. ${a}_{1}=0,d=4$ 59. ${a}_{1}=9;{a}_{n}={a}_{n - 1}-10$ 60. ${a}_{n}=-12+5n$ For the following exercises, follow the steps to work with the arithmetic sequence ${a}_{n}=3n - 2$ using a graphing calculator:
• Press [MODE]
• Select SEQ in the fourth line
• Select DOT in the fifth line
• Press [ENTER]
• Press [Y=]
• $n\text{Min}$ is the first counting number for the sequence. Set $n\text{Min}=1$
• $u\left(n\right)$ is the pattern for the sequence. Set $u\left(n\right)=3n - 2$
• $u\left(n\text{Min}\right)$ is the first number in the sequence. Set $u\left(n\text{Min}\right)=1$
• Press [2ND] then [WINDOW] to go to TBLSET
• Set $\text{TblStart}=1$
• Set $\Delta \text{Tbl}=1$
• Set Indpnt: Auto and Depend: Auto
• Press [2ND] then [GRAPH] to go to the TABLE
61. What are the first seven terms shown in the column with the heading $u\left(n\right)\text{?}$ 62. Use the scroll-down arrow to scroll to $n=50$. What value is given for $u\left(n\right)\text{?}$ 63. Press [WINDOW]. Set $n\text{Min}=1,n\text{Max}=5,x\text{Min}=0,x\text{Max}=6,y\text{Min}=-1$, and $y\text{Max}=14$. Then press [GRAPH]. Graph the sequence as it appears on the graphing calculator. For the following exercises, follow the steps given above to work with the arithmetic sequence ${a}_{n}=\frac{1}{2}n+5$ using a graphing calculator. 64. What are the first seven terms shown in the column with the heading $u\left(n\right)$ in the TABLE feature? 65. Graph the sequence as it appears on the graphing calculator. Be sure to adjust the WINDOW settings as needed. 66. Give two examples of arithmetic sequences whose 4th terms are $9$. 67. Give two examples of arithmetic sequences whose 10th terms are $206$. 68. Find the 5th term of the arithmetic sequence $\left\{9b,5b,b,\dots \right\}$. 69. Find the 11th term of the arithmetic sequence $\left\{3a - 2b,a+2b,-a+6b\dots \right\}$. 70. At which term does the sequence $\left\{5.4,14.5,23.6,...\right\}$ exceed 151? 71. At which term does the sequence $\left\{\frac{17}{3},\frac{31}{6},\frac{14}{3},...\right\}$ begin to have negative values? 72. For which terms does the finite arithmetic sequence $\left\{\frac{5}{2},\frac{19}{8},\frac{9}{4},...,\frac{1}{8}\right\}$ have integer values? 73. Write an arithmetic sequence using a recursive formula. Show the first 4 terms, and then find the 31st term. 74. Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28th term. | 2,586 | 7,352 | {"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} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.873903 |
https://dsp.stackexchange.com/questions/23477/far-field-spherical-waves-plane-waves/23482 | 1,620,467,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00079.warc.gz | 237,308,736 | 38,927 | # Far field spherical waves plane waves
All em radiation propagates spherically and at a far off place a small enough distance, when observed appears like a plane like our experience on the earth. So I tend to think that whether a wave on an aperture can be considered plane or not depends on how far the aperture is from the origin and how much is the aperture. So it should be only dependent on the angle made at the centre by the aperture i.e. apertures making more than a certain angle at the centre can be considered spherical, and otherwise can be considered plane. This is my wrong understanding. But in radar theory, this is not the criterion for considering plane and far field. A 2D^2/lambda formula is used. How is this formula relevant? How lambda is entering into the formula?
• The distinction near/far field is not identical to the approximation spherical/planar wave! I think you should look up the definition of the near field of a dipole or even multipole oscillator. – Jazzmaniac Jun 17 '15 at 15:40
Rajeev Bansal in the appendix to this article explains the $$2D/\lambda$$ rule:
How does one arrive at the various criteria for the far-field zone? Basically, the criteria are guidelines to the boundaries where the fields start to approximate the “ideal” assumed characteristics. First of all, only the radiation ($$1/r$$) terms remain significant; higher order terms fade away. Second, in the far-field zone, the angular field distribution becomes independent of the distance [5]. Third, only transverse field components remain, and the ratio of the electric and magnetic field components approaches the free-space impedance, 377 ohms [6 ,7]. Finally, for a receiving antenna, the incoming wave-front is nearly planar across the aperture. In fact, the $$r = 2 D^2/\lambda$$ formula corresponds to a phase error (due to the curvature of the actual spherical wave-front) of no more than $$\pm 22.5$$ degrees across the aperture, as compared with the ideal plane wave-front [6, 8].
References:
1. IEEE Standard Dictionary of Electrical and Electronics Terms, IEEE standard 100-1984, 1984.
2. C. Paul, K. Whites, and S. Nasar, Introduction to Electromagnetic Fields, 3rd ed., McGraw-Hill, 1998.
3. Reference Data for Radio Engineers, 6th ed., SAMS, 1981.
4. G. Smith, An Introduction to Classical Electromagnetic Radiation, Cambridge, 1997.
• Obviously I am blind to some point even after reading the article. I am solely thinking of about the final point in your answer - geometric point. When can an arc of a circle be considered straight? When the arc makes a very small angle at the centre. If you consider only this geometric point, the criterion should be independent of the wavelength. It looks like the far field criterion is more coming from radiation field dominance over other terms than the pure geometric requirement of approximation of an arc as straight. Can you confirm this? – Seetha Rama Raju Sanapala May 19 '15 at 6:05
• The condition is that the phase error, due to a spherical wave-front rather than a plane wave, across the aperture being "small", The phase error is wavelength dependent. – Conrad Turner May 19 '15 at 6:37
• How can I accept this answer? That option, I am not able to see. – Seetha Rama Raju Sanapala May 20 '15 at 6:44
• @SeethaRamaRaju There should be a greyed-out tick mark just below the up and down voting arrows for you, the originator of the question. – Peter K. Oct 15 '15 at 18:00 | 818 | 3,456 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-21 | latest | en | 0.90641 |
https://numberworld.info/root-of-500 | 1,660,668,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00542.warc.gz | 400,886,076 | 2,951 | # Root of 500
#### [Root of five hundred]
square root
22.3607
cube root
7.937
fourth root
4.7287
fifth root
3.4657
In mathematics extracting a root is declared as the determination of the unknown "x" in the following equation $y=x^n$ The outcome of the extraction of the root is seen as a mathematical root. In the case of "n equals 2", one talks about a square root or sometimes a second root also, another possibility could be that n equals 3 at that time one would call it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 500 is represented as this: $$\sqrt[]{500}=22.360679774998$$
Furthermore it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 500 is 22.360679774998. The cube root of 500 is 7.937005259841. The fourth root of 500 is 4.7287080450159 and the fifth root is 3.4657242157757.
Look Up | 277 | 968 | {"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} | 3.8125 | 4 | CC-MAIN-2022-33 | latest | en | 0.912932 |
https://www.intechopen.com/online-first/variable-fractional-order-pid-controller-synthesis-novelty-method | 1,611,449,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538741.56/warc/CC-MAIN-20210123222657-20210124012657-00638.warc.gz | 813,463,260 | 105,897 | Open access peer-reviewed chapter - ONLINE FIRST
Variable, Fractional-Order PID Controller Synthesis Novelty Method
By Piotr Ostalczyk and Piotr Duch
Submitted: July 28th 2020Reviewed: November 25th 2020Published: December 28th 2020
DOI: 10.5772/intechopen.95232
Abstract
The novelty method of the discrete variable, fractional order PID controller is proposed. The PID controllers are known for years. Many tuning continuous time PID controller methods are invented. Due to different performance criteria there are optimized three parameters: proportional, integral and differentiation gains. In the fractional order PID controllers there are two additional parameters: fractional order integration and differentiation. In the variable, fractional order PID controller fractional orders are generalized to functions. Nowadays all PID controllers are realized by microcontrollers in a discrete time version. Hence, the order functions are discrete variable bounded ones. Such controllers offer better transient characteristics of the closed loop systems. The choice of the order functions is still the open problem. In this Section a novelty intuitive idea is proposed. As the order functions one applies two spline functions with bounded functions defined for every time subinterval. The main idea is that in the final time interval the variable, fractional order PID controller transforms itself to the classical one preserving the stability conditions and zero steady-state error signal. This means that in the last time interval the discrete integration order is −1 and differentiation is 1.
Keywords
• fractional-Calculus
• PID controller
• discrete system
1. Introduction
A continuous-time proportional–integral–derivative controller (PID controller) [1] invented almost 100 years ago is one of the most widely applied controllers in the closed-loop systems [2] with many industrial applications [3, 4, 5]. Currently the continuous-time control is successively replaced by discrete-time one in which the integration is replaced by a summation and differentiation by a difference evaluation. So, in the discrete PID controller the classical integral is replaced by a sum and the derivative by a backward difference, [6]. The discrete controller’s PID algorithm is mainly realized by micro-controllers [7].
At 70s of the 20-th Century the Fractional Calculus [8] with a great success started a considerable attention in mathematics and engineering [9, 10, 11, 12]. Now, the fractional-order backward-difference (FOBD) and the fractional-order backward sum (FOBS) [6, 13] are applied in the dynamical system modeling [14] and discrete control algorithms. The continuous-time FOPID controllers are more difficult in a practical realization [15, 16, 17, 18].
There are numerous continuous and discrete-time PID and FOPID controller synthesis methods [16, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]. One should mention that the optimisation of the closed-loop system in this case is more complicated because of the controller optimization. Apart from the three classical gains there are two additional parameters, namely, a fractional order of differentiation and summation [32]. The FOPID control characterizes by slow achieving the steady state and growing calculation “tail” [12].
In the paper a novelty variable, the fractional-order PID (VFOPID) [6, 28, 33, 34, 35, 36, 37, 38, 39, 40, 41] controller synthesis is proposed. It consists of dividing the closed-loop system discrete-transient time division into the finite time intervals over which are defined fractional orders summation and differentiation functions. The main idea is that for the final infinite interval kL+the difference order equals 0 and the summation is !preserving quick reaching the zero steady state value. Thus, in the VFOPID control the disadvantages of FOPID are extracted. One should admit that in the FOPID or VFOPID control the microcontrollers are numerically loaded.
Fractional-orders systems are characterized by the so called system “memory”. This, in practice, means that in every step the FOPID controller computes its output signals taking into account step-by-step linearly computed number of samples. This causes in practice the micro-controllers realization problems. It is known as “Finite memory principle” [12].
The paper is organized as follows. In Section 2 the basic information related to the fractional calculus and variable, fractional order Grünwald-Letnikov backward difference is given. The main result of the paper includes Section 3. It contains the proposed VFOPID controller synthesis method with the proposal of the order functions form. The brief description of the controller parameters evaluation algorithm is given. The investigations are supported by a numerical example presented in Chapter 4.
2. Mathematical preliminaries
In the paper the following notation will be used. N0=0,1,2,3, Nl=ll+1l+2R+=0+. 0kwill denote the zero column vector of dimensions k+1×1whereas 0k,kis k+1×k+1zero matrix. Similarly will be denoted a k+1×k+1unit matrix 1k.
In general, a fractional-order functions will be denoted by Greek letters ν:N0Rνwhereas the integer orders will be denoted by Latin ones nR+. In practice, for lN0: 0<νl1. For k,lN0and a given order function νlthe function of two discrete variables k,lN0is defined by the following formula: aνlkas follows:
Definition 2.1. For k,lN0and a given order function νone defines the coefficients function of two 13 discrete variables as
aνlk=1fork=01kνlνl1νlk+1k!forkN1E1
One should mention that function (1) for νl=nl=constN0
ank=1fork=0nn1nk+11k!fork1n0forkNn+1E2
The above function will be named as: the “oblivion function” or “decay function”.
2.1 Variable, fractional-order backward difference
Next one defines the Grünwald–Letnikov variable, fractional-order backward difference (VFOBD). For a discrete-variable bounded real-valued function fdefined over a discrete interval 0kthe VFOBD is defined as a sum (see for instance [6, 9]).
Definition 2.2. The VFOBD with an order function ν, with values νk01, is defined as a finite sum, provided that the series is convergent
k0Δkνkfk=i=0kk0aνkifki=1aνk1aνk2aνkk0fkfk1fkk0E3
Relating to (2) as the first special case of the defined above VFOBD and a constant order function νk=ν=constfrom (2.1) one gets the fractional-order backward difference (FOBD). The second special case is for a constant integer order function νk=ν=n=constwhere the integer-order backward difference (IOBD) is a classical one.
Equality (3) is valid for k,k1,k2,,k0+1,k0. Hence, one gets a finite set of equations. Collecting them in a vector matrix form one gets
k0GLΔkνkfk=k0Akνkfk,E4
where
k0Akνk=1aνk1aνkkk001aνk1kk0100aνk0+11001E5
fk=fkfk1fk0,k0GLΔknkfk=k0GLΔkνkfkk0GLΔk0νk0fk0.E6
2.2 Variable, fractional-order linear time-invariant difference equations
On the base of the Grünwald-Letnikov variable, fractional-order linear time-invariant backward-difference the difference Eqs. (GL-VFOBE) for i=1,2,,pand j=1,2,,qrepresenting discrete models of real dynamical systems or discrete control strategies are defined by the variable, fractional-order linear time-invariant difference equation (VFODE). h>0denotes the sampling time.
l=0niai,lk0GLΔkνi,lkykh=l=0mibi,lk0GLΔkμi,lkukhE7
where mini, νni,lkνni,l1kνi,1kνi,0k=0, μmi,lkμmi,l11kμi,1kμi,0k0, ai,land bi,lare constant coefficients for l=0,1,,niand l=0,1,,mi, respectively. It is assumed that a0,n0=1.
According to the notation (5) Eq. (7) takes the form
l=0niai,lk0Akνi,lkyk=l=0miai,lk0Akμi,lkukE8
The vector ujksatisfies the condition ujk=0kfor k<k0. In the general solution of (8) to the assumed ujkand initial conditions vector yi,k01=yi,k01yi,k02T(T denotes the transposition) must be taken into account with =k0<0k0k. Then, the infinite number of initial conditions (8) are formed in the following vector
yi,k01=yi,k01yi,k02E9
and the combined Eq. (8) is of the form
l=0niaij,lk0Akνi,lkl=0niai,lAk01νi,lk×yikhyi,k01=l=0mibij,lk0Akμi,lkukhE10
or after simple transformation
l=0niai,lk0Akνi,lkykh=l=0mibi,lk0Akμi,lkukhl=0niai,lAk01νi,lkyi,k01E11
2.3 Main assumptions
To preserve the VFOBDE order one assumes that
1+l=0ni1ai,l0fori=1,2,,pE12
In the transfer functions defined by the one-sided Ztransform one assumes zero initial conditions. Following this assumption equality (11) simplifies to
l=0niaik0Akνikyk=l=0mibik0AkμikukE13
Defining matrices
k0DkνPk=l=0niaik0AkνikE14
k0NkμPk=l=0mibik0AkμikE15
one gets
k0DkνPkykh=k0NkμPkukhE16
Under assumption (12) k0Dkνkis invertible, so for k0=0one can write
ykh=0DνPkk10NμPkkukhE17
Denoting
0GkνPk=0DkνPk0NkμkE18
one gets similar to the transfer function description
ykh=0GkνPkμPkukhE19
or for simplicity
Gokh=0GkνPkμPkE20
Remark 2.1. Though the relation (19) looks similar to the classical discrete transfer function it is different by the real discrete variables. It relates discrete SISO systems by vectors and matrices related to its dimensions k+1N0.
2.4 VFO linear system description
One considers a closed-loop system illustrated in Figure 1. Where a plant is described by (19) where ekhand ukh.
2.4.1 VFO_PID
The classical PID controller output is desribed by three terms
ukh=KPekh+KI0Δkμkekh+KD0ΔkνkekhE21
and in the convention proposed above as
ukh=KP1kekh+KD0GkνCkekh+KI0GkμCkekhE22
which may be expressed as
ukh=KP1k+KD0GkνCk+KI0GkμCkekhE23
where νCk,μCk0and controlling and error signals are denoted as ukand ek, respectively. Then, denoting.
Remark 2.2. The plant may be described by classical integer order, fractional or even variable, fractional - order difference equations. The matrix - vector description used makes it possible.
Ckh=KP1k+KD0GkνCk+KI0GkμCkE24
one gets a VFOPID controller transfer function-like description
ukh=CkhekhE25
To simplify the description one assumes a sensor matrix as
Hkh=1kE26
The closed-loop system is presented in Figure 1 from which one gets the following relations
ykh=1k+GokhCkhHkh1GokhCkhrkh+1k+GokhCkhHkh1dkhE27
where
• rkh- a reference signal vector,
• dkh- an external disturbance signal vector,
• yokh- a plant output signal vector,
• ykh- a closed-loop system output signal vector,
• ekh- a closed-loop system error signal,
A system error is evaluated by the formula
ekh=1k+GokhCkhHkh1rkh1k+GokhCkhHkh1HkhdkhE28
3. Variable, fractional-order PID controller synthesis
In the synthesis of the classical PID controller there are three parameters to evaluate. Namely, K,KI,KDknown as the proportional, integral and differential gains. In the fractional-order PID controllers there are two additional parameters: the differentiation order νkhR+and the integration one μkhR+. In the variable, fractional-order PID controller the mentioned orders are generalized to functions. This means that there are three constant coefficients and two discrete variable functions to find
KP,KI,KD,νkh,μkhE29
In the rejection of the external disturbation one can assume that rkh=0so Eq. (29) simplifies to
ekh=1k+GokhCkhHkh1HkhdkhE30
Usually the sensor matrix Hkhis treated as constant, by assumption that sensors do not introduce its own dynamics to the system. Hence, Hkh=H=const. It may be assumed that H=h01kor further, for h0=1, formula (30) takes a form
ekh=1k+GokhCkh1dkhE31
The optimal parameters (29) are evaluated due to the assumed optymality criterion. The most popular is so called ISE one (Integral of the Squared Error) or in the discrete-system case: Sum of the Squared Error (SSE).
SSEKPKIKDνkhμkh=i=0kmaxeih2h=ekhTekhhE32
Substitution of (31) into (32) gives
SSEKPKIKDνkhμkh=dkhT1k+GokhCkhT1k+GokhCkh1dkhE33
In the proposed VFOPID controller synthesis method with partially intuitive and supported by closed-loop systems synthesis experience the classical optimisation due to the performance criterion (32) is performed. The pre-defined differentiation and integration order functions orders are as follows
νkh0E34
νkh=ν1khfork0kN1ν2khforkkN1kN2νNkhforkkNN1kNN0forkkNN+E35
and
μkh0E36
μkh=μ1khfork0kM1μ2khforkkM1kM2μNkhforkkMM1kMM1forkkMM+E37
Every function νikhfor i=1,2,,Nand μikhfor i=1,2,,Mis characterized by a sets of parameters cijand dij, respectively.
In the classical closed-loop system with PID controller there is introduced the integration part preserving the steady - state error signal tending to zero. So, in (38) there is a constant order 1for kKMM+.
Now, for initially assumed order functions one applies the following algorithm based on well known Gauss method.
1. Chose a starting set of coefficients KP,KI,KD, c11,and d11,,
2. Applying the classical Gauss algorithm find a minimal SSE performance index value alongside the first variable (eg. K_P),
3. Repeat step 2 for the next parameter,
4. If the SSE value is satisfactory stop else return to step 2.
Remark 3.1. Algorithm described above can be applied also to the classical discrete PID controller with three parameters.
4. Numerical example
One considers a closed-loop system depicted in Figure 1.
A plant is described by a transfer function
Gos=b0s2+a1s+a0E38
where
• a1=0.5
• a0=0.1
• b0=a0
The plant is discretized with the sampling time h=0.5and a VFOPID controller is applied
νkh=ν1kh=1fork=00fork1+E39
μkh=μ1kh=1+d1ed2kh1hfork=0101fork10+E40
and controller gains KP,kI,KDand order function parameters d1,d2.
Hence, there are 5 parameters to evaluate. Due to the performance index (33) the optimal parameters are as follows
• KP=1.000
• Ki=0.514
• KD=0.890
• d1=0.35
• d2=0.5
The VFOPID controller order functions are plotted in Figure 2 whereas the PID and VFOPID controllers unite step responses are given in Figure 3.
The achieved VFOPID controller synthesis result is compared with the classical discrete-time PID controller optimized due to criterion (30). The optimal parameters are
• KP=1.00
• Ki=0.81
• KD=0.90
Figure 4 contains the closed - loop systems with PID (in blue) and VFOPID (in red) controllers unit step responses. There is included a plant unit step response of the plant (in black.)
In Figure 5 the controlling signals are presented (PID - in black, VFOPID - in red). The controlling signals have typical shapes: first differentiation action and finally the classical integration preserving zero steady - state closed - loop system error.
Remark 4.1. In the Numerical example proposed here the VFOPID and the classical PID controllers maximal control signal values are the same reaching assumed bounding value maxuIkl,maxuFkh=2.
Remark 4.2. In the Numerical example
SSEKPKIKD11)=1.3312SSEKPKIKDνkhμkh=1.2899E41
5. Conclusions
One should emphasize that the proposed solution of the VFOPID controller do not guarantee the absolute optimum of the closed-loop control system synthesis. It proves that the proposal of a physically realizable VFOPID controller by micro-controller (with finite memory) leads to better results due to the assumed performance criterion.
The main idea of the proposed method is to assume a priori the order functions with unknown parameters. In the VFOPID controller synthesis essential is an assumption that the summation order equals 1 One can express the action as the assumption of skeleton order functions with unknown parameters evaluated further in an SSE optimization algorithm.
Here, it is worth mentioning that there are still open problems of the VFOPID controllers tuning.
• One should define a program evaluating the order functions.
• For evaluation of the VFOPID controller parameters one can apply another optimization methods. It seems that optimization methods based on the artificial intelligence will be very effective.
• Another performance index may be applied. Some penalty functions may be introduced to SSE as well a term taking into account the minimal value of the error signal.
Acknowledgments
The work was supported by the National Science Center Poland by Grant no. 2016/23/B/ ST7/03686.
Abbreviations
PID proportional-integral-derivative controller FOBD fractional-order backward difference FOBS fractional-order backward sum FOPID fractional-Order proportional, fractional-order integral and differential controller VFOPID variable, fractional-order PID controller SSE squared sum of the error
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Piotr Ostalczyk and Piotr Duch (December 28th 2020). Variable, Fractional-Order PID Controller Synthesis Novelty Method [Online First], IntechOpen, DOI: 10.5772/intechopen.95232. Available from: | 4,479 | 16,793 | {"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-2021-04 | latest | en | 0.906948 |
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Mechanics of Materials
# Mechanics of Materials (3rd Edition) View more editions Solutions for Chapter 11.3
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Chapter: Problem:
These problems deal with the strain energy stored in various members undergoing axial deformation. Assume that the stress distribution is uniform at every cross section, even where there is a sudden change in the cross-sectional area.
A uniform aluminum-alloy rod with cross- sectional area A = 1 in2 and length L = 50 in. is subjected to an axial load P, as shown in Fig. P11.3-1. Let Eal = 10(103) ksi.
(a) Sketch a load versus elongation diagram (i.e., load P versus elongation e) for elongations from 0 in. to 0.250 in. (Hint. Recall Eq. 3.15.)
(b) Calculate the strain energy stored in the rod when P = 40 kips, and indicate on your P — e diagram of Part (a) the area that represents this strain energy.
P11.3-1
Eq. 3.15
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 4
• Step 2 of 4
• Step 3 of 4
• Step 4 of 4
Corresponding Textbook
Mechanics of Materials | 3rd Edition
9781118136331ISBN-13: 1118136330ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Mechanics of Materials 3rd Edition Textbook Solutions | 378 | 1,377 | {"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-2016-40 | latest | en | 0.810955 |
https://corporatefinanceinstitute.com/resources/excel/match-function-formula/ | 1,718,911,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00444.warc.gz | 155,399,123 | 72,254 | # MATCH Function
Get the position of a value within an array
Over 1.8 million professionals use CFI to learn accounting, financial analysis, modeling and more. Start with a free account to explore 20+ always-free courses and hundreds of finance templates and cheat sheets.
## What is the MATCH Function?
The MATCH Function[1] is categorized under Excel Lookup and Reference functions. It looks up a value in an array and returns the position of the value within the array. For example, if we wish to match the value 5 in the range A1:A4, which contains values 1,5,3,8, the function will return 2, as 5 is the second item in the range.
In financial analysis, we can use the MATCH function along with other functions to look up and return the sum of values in a column. It is commonly used with the INDEX function. Learn how to combine INDEX MATCH as a powerful lookup combination.
### Formula
=MATCH(lookup_value, lookup_array, [match_type])
The MATCH formula uses the following arguments:
1. Lookup_value (required argument) – This is the value that we want to look up.
2. Lookup_array (required argument) – The data array that is to be searched.
3. Match_type (optional argument) – It can be set to 1, 0, or -1 to return results as given below:
Match_typeBehavior
1 or omittedWhen the function cannot find an exact match, it will return the position of the closest match below the lookup_value. (If this option is used, the lookup_array must be in ascending order).
0When the function cannot find an exact match, it will return an error. (If this option is used, the lookup_array does not need to be ordered).
-1When the function cannot find an exact match, it will return the position of the closest match above the lookup_value. (If this option is used, the lookup_array must be in descending order).
### How to Use the MATCH Function in Excel
To understand the uses of the function, let’s consider a few examples:
#### Example 1
Suppose we are given the following data:
We can use the following formula to find the value for Cucumber.
We get the result below:
#### Example 2
Suppose we are given the following data:
Suppose we wish to find out the number of Trousers of a specific color. Here, we will use the INDEX or MATCH functions. The array formula to use is:
We need to create an array using CTRL + SHIFT + ENTER. We get the result below:
### Things to Remember
1. The MATCH function does not distinguish between uppercase and lowercase letters when matching text values.
2. N/A! error – Occurs if the match function fails to find a match for the lookup_value.
3. The function supports approximate and exact matching and wildcards (* or ?) for partial matches. | 604 | 2,692 | {"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-2024-26 | latest | en | 0.80827 |
https://www.geeksforgeeks.org/count-number-of-sub-sequences-with-gcd-1/?ref=rp | 1,624,497,150,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488544264.91/warc/CC-MAIN-20210623225535-20210624015535-00384.warc.gz | 710,719,768 | 33,390 | Related Articles
Count number of sub-sequences with GCD 1
• Difficulty Level : Hard
• Last Updated : 28 May, 2021
Given an array of N numbers, the task is to count the number of subsequences that have gcd equal to 1.
Examples:
```Input: a[] = {3, 4, 8, 16}
Output: 7
The subsequences are:
{3, 4}, {3, 8}, {3, 16}, {3, 4, 8},
{3, 4, 16}, {3, 8, 16}, {3, 4, 8, 16}
Input: a[] = {1, 2, 4}
Output: 4```
A simple solution is to generate all subsequences or subsets. Foe every subsequence, check if its GCD is 1 or not. If 1, increment the result.
When we have values in array (say all smaller than 1000), we can optimize the above solution as we know that number of possible GCDs would be small. We modify the recursive subset generation algorithm where consider two cases for every element, we either include or exclude it. We keep track of current GCD and if we have already counted for this GCD, we return the count. So when we are considering a subset, some GCDs would appear again and again. Therefore the problem can be solved using Dynamic Programming. Given below are the steps to solve the above problem:
• Start from every index and call the recursive function by taking the index element.
• In the recursive function, we iterate till we reach N.
• The two recursive calls will be based on either we take the index element or not.
• The base case will be to return 1 if we have reached the end and the gcd till now is 1.
• Two recursive calls will be func(ind+1, gcd(a[i], prevgcd)) and func(ind+1, prevgcd)
• The overlapping subproblems can be avoided by using memoization technique.
Below is the implementation of the above approach:
## C++
`// C++ program to find the number``// of subsequences with gcd 1``#include ``using` `namespace` `std;``#define MAX 1000``int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(a == 0)`` ``return` `b;`` ``return` `gcd(b % a, a);``}` `// Recursive function to calculate the number``// of subsequences with gcd 1 starting with``// particular index``int` `func(``int` `ind, ``int` `g, ``int` `dp[][MAX], ``int` `n, ``int` `a[])``{` ` ``// Base case`` ``if` `(ind == n) {`` ``if` `(g == 1)`` ``return` `1;`` ``else`` ``return` `0;`` ``}` ` ``// If already visited`` ``if` `(dp[ind][g] != -1)`` ``return` `dp[ind][g];` ` ``// Either we take or we do not`` ``int` `ans = func(ind + 1, g, dp, n, a)`` ``+ func(ind + 1, gcd(a[ind], g), dp, n, a);` ` ``// Return the answer`` ``return` `dp[ind][g] = ans;``}` `// Function to return the number of subsequences``int` `countSubsequences(``int` `a[], ``int` `n)``{` ` ``// Hash table to memoize`` ``int` `dp[n][MAX];`` ``memset``(dp, -1, ``sizeof` `dp);` ` ``// Count the number of subsequences`` ``int` `count = 0;` ` ``// Count for every subsequence`` ``for` `(``int` `i = 0; i < n; i++)`` ``count += func(i + 1, a[i], dp, n, a);` ` ``return` `count;``}` `// Driver Code``int` `main()``{`` ``int` `a[] = { 3, 4, 8, 16 };`` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);`` ``cout << countSubsequences(a, n);`` ``return` `0;``}`
## Java
`// Java program to find the number``// of subsequences with gcd 1``class` `GFG``{`` ` `static` `final` `int` `MAX = ``1000``;``static` `int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(a == ``0``)`` ``return` `b;`` ``return` `gcd(b % a, a);``}` `// Recursive function to calculate the number``// of subsequences with gcd 1 starting with``// particular index``static` `int` `func(``int` `ind, ``int` `g, ``int` `dp[][],`` ``int` `n, ``int` `a[])``{` ` ``// Base case`` ``if` `(ind == n)`` ``{`` ``if` `(g == ``1``)`` ``return` `1``;`` ``else`` ``return` `0``;`` ``}` ` ``// If already visited`` ``if` `(dp[ind][g] != -``1``)`` ``return` `dp[ind][g];` ` ``// Either we take or we do not`` ``int` `ans = func(ind + ``1``, g, dp, n, a)`` ``+ func(ind + ``1``, gcd(a[ind], g), dp, n, a);` ` ``// Return the answer`` ``return` `dp[ind][g] = ans;``}` `// Function to return the``// number of subsequences``static` `int` `countSubsequences(``int` `a[], ``int` `n)``{` ` ``// Hash table to memoize`` ``int` `dp[][] = ``new` `int``[n][MAX];`` ``for``(``int` `i = ``0``; i < n; i++)`` ``for``(``int` `j = ``0``; j < MAX; j++)`` ``dp[i][j] = -``1``;` ` ``// Count the number of subsequences`` ``int` `count = ``0``;` ` ``// Count for every subsequence`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``count += func(i + ``1``, a[i], dp, n, a);` ` ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String args[])``{`` ``int` `a[] = { ``3``, ``4``, ``8``, ``16` `};`` ``int` `n = a.length;`` ``System.out.println(countSubsequences(a, n));``}``}` `// This code is contributed by Arnab Kundu`
## Python3
`# Python3 program to find the number``# of subsequences with gcd 1` `MAX` `=` `1000` `def` `gcd(a, b):`` ``if` `(a ``=``=` `0``):`` ``return` `b`` ``return` `gcd(b ``%` `a, a)` `# Recursive function to calculate the``# number of subsequences with gcd 1``# starting with particular index``def` `func(ind, g, dp, n, a):` ` ``# Base case`` ``if` `(ind ``=``=` `n):`` ``if` `(g ``=``=` `1``):`` ``return` `1`` ``else``:`` ``return` `0` ` ``# If already visited`` ``if` `(dp[ind][g] !``=` `-``1``):`` ``return` `dp[ind][g]` ` ``# Either we take or we do not`` ``ans ``=` `(func(ind ``+` `1``, g, dp, n, a) ``+`` ``func(ind ``+` `1``, gcd(a[ind], g),`` ``dp, n, a))` ` ``# Return the answer`` ``dp[ind][g] ``=` `ans`` ``return` `dp[ind][g]` `# Function to return the number``# of subsequences``def` `countSubsequences(a, n):` ` ``# Hash table to memoize`` ``dp ``=` `[[``-``1` `for` `i ``in` `range``(``MAX``)]`` ``for` `i ``in` `range``(n)]` ` ``# Count the number of subsequences`` ``count ``=` `0` ` ``# Count for every subsequence`` ``for` `i ``in` `range``(n):`` ``count ``+``=` `func(i ``+` `1``, a[i], dp, n, a)` ` ``return` `count` `# Driver Code``a ``=` `[``3``, ``4``, ``8``, ``16` `]``n ``=` `len``(a)``print``(countSubsequences(a, n))` `# This code is contributed by mohit kumar 29`
## C#
`// C# program to find the number``// of subsequences with gcd 1``using` `System;` `class` `GFG``{` `static` `int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(a == 0)`` ``return` `b;`` ``return` `gcd(b % a, a);``}` `// Recursive function to calculate the number``// of subsequences with gcd 1 starting with``// particular index``static` `int` `func(``int` `ind, ``int` `g, ``int` `[][] dp,`` ``int` `n, ``int` `[] a)``{` ` ``// Base case`` ``if` `(ind == n)`` ``{`` ``if` `(g == 1)`` ``return` `1;`` ``else`` ``return` `0;`` ``}` ` ``// If already visited`` ``if` `(dp[ind][g] != -1)`` ``return` `dp[ind][g];` ` ``// Either we take or we do not`` ``int` `ans = func(ind + 1, g, dp, n, a)`` ``+ func(ind + 1, gcd(a[ind], g), dp, n, a);` ` ``// Return the answer`` ``return` `dp[ind][g] = ans;``}` `// Function to return the``// number of subsequences``static` `int` `countSubsequences(``int` `[] a, ``int` `n)``{` ` ``// Hash table to memoize`` ``int` `[][] dp = ``new` `int``[n][];`` ``for``(``int` `i = 0; i < n; i++)`` ``for``(``int` `j = 0; j < 1000; j++)`` ``dp[i][j] = -1;` ` ``// Count the number of subsequences`` ``int` `count = 0;` ` ``// Count for every subsequence`` ``for` `(``int` `i = 0; i < n; i++)`` ``count += func(i + 1, a[i], dp, n, a);` ` ``return` `count;``}` `// Driver Code``public` `static` `void` `Main()``{`` ``int` `[] a = { 3, 4, 8, 16 };`` ``int` `n = 4;`` ``int` `x = countSubsequences(a, n);`` ``Console.Write(x);``}``}` `// This code is contributed by``// mohit kumar 29`
## PHP
``
## Javascript
``
Output
`7`
Alternate Solution: Count number of subsets of a set with GCD equal to a given number
Dynamic programming approach to this problem without memoization:
Basically, the approach will be making a 2d matrix in which i coordinate will be the position of elements of the given array and j coordinate will be numbers from 0 to 100 ie. gcd can vary from 0 to 100 if array elements are not enough large. we will iterate on given array and the 2d matrix will store information that till ith position that how many subsequences are there having gcd vary from 1 to 100. later on, we will add dp[i][1] to get all subsequence having gcd as 1.
Below is the implementation of the above approach:
## C++
`// C++ program for above approach``#include ``using` `namespace` `std;` `// This function calculates``// gcd of two number``int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(b == 0)`` ``return` `a;`` ``return` `gcd(b, a % b);``}` `// This function will return total``// subsequences``int` `countSubsequences(``int` `arr[], ``int` `n)``{`` ` ` ``// Declare a dp 2d array`` ``long` `long` `int` `dp[n][101] = {0};` ` ``// Iterate i from 0 to n - 1`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ` ` ``dp[i][arr[i]] = 1;`` ` ` ``// Iterate j from i - 1 to 0`` ``for``(``int` `j = i - 1; j >= 0; j--)`` ``{`` ``if``(arr[j] < arr[i])`` ``{`` ` ` ``// Iterate k from 0 to 100`` ``for``(``int` `k = 0; k <= 100; k++)`` ``{`` ` ` ``// Find gcd of two number`` ``int` `GCD = gcd(arr[i], k);`` ` ` ``// dp[i][GCD] is summation of`` ``// dp[i][GCD] and dp[j][k]`` ``dp[i][GCD] = dp[i][GCD] +`` ``dp[j][k];`` ``}`` ``}`` ``}`` ``}`` ` ` ``// Add all elements of dp[i][1]`` ``long` `long` `int` `sum = 0;`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``sum=(sum + dp[i][1]);`` ``}`` ` ` ``// Return sum`` ``return` `sum;``}` `// Driver code``int` `main()``{`` ``int` `a[] = { 3, 4, 8, 16 };`` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);`` ` ` ``// Function Call`` ``cout << countSubsequences(a, n);`` ``return` `0;``}`
## Java
`// Java program for the``// above approach``import` `java.util.*;` `class` `GFG{`` ` `// This function calculates``// gcd of two number``static` `int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(b == ``0``)`` ``return` `a;`` ` ` ``return` `gcd(b, a % b);``}` `// This function will return total``// subsequences``static` `long` `countSubsequences(``int` `arr[],`` ``int` `n)``{`` ` ` ``// Declare a dp 2d array`` ``long` `dp[][] = ``new` `long``[n][``101``];` ` ``for``(``int` `i = ``0``; i < n; i++)`` ``{`` ``for``(``int` `j = ``0``; j < ``101``; j++)`` ``{`` ``dp[i][j] = ``0``;`` ``}`` ``} `` ` ` ``// Iterate i from 0 to n - 1`` ``for``(``int` `i = ``0``; i < n; i++)`` ``{`` ` ` ``dp[i][arr[i]] = ``1``;`` ` ` ``// Iterate j from i - 1 to 0`` ``for``(``int` `j = i - ``1``; j >= ``0``; j--)`` ``{`` ``if` `(arr[j] < arr[i])`` ``{`` ` ` ``// Iterate k from 0 to 100`` ``for``(``int` `k = ``0``; k <= ``100``; k++)`` ``{`` ` ` ``// Find gcd of two number`` ``int` `GCD = gcd(arr[i], k);`` ` ` ``// dp[i][GCD] is summation of`` ``// dp[i][GCD] and dp[j][k]`` ``dp[i][GCD] = dp[i][GCD] +`` ``dp[j][k];`` ``}`` ``}`` ``}`` ``}`` ` ` ``// Add all elements of dp[i][1]`` ``long` `sum = ``0``;`` ``for``(``int` `i = ``0``; i < n; i++)`` ``{`` ``sum = (sum + dp[i][``1``]);`` ``}`` ` ` ``// Return sum`` ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String args[])``{`` ``int` `a[] = { ``3``, ``4``, ``8``, ``16` `};`` ``int` `n = a.length;`` ` ` ``// Function Call`` ``System.out.println(countSubsequences(a, n));``}``}` `// This code is contributed by bolliranadheer`
## Python3
`# Python3 program for the``# above approach`` ` `# This function calculates``# gcd of two number``def` `gcd(a, b):`` ``if` `(b ``=``=` `0``):`` ``return` `a; `` ``return` `gcd(b, a ``%` `b);` `# This function will return total``# subsequences``def` `countSubsequences(arr, n):`` ` ` ``# Declare a dp 2d array`` ``dp ``=` `[[``0` `for` `i ``in` `range``(``101``)] ``for` `j ``in` `range``(n)]`` ` ` ``# Iterate i from 0 to n - 1`` ``for` `i ``in` `range``(n): `` ``dp[i][arr[i]] ``=` `1``;`` ` ` ``# Iterate j from i - 1 to 0`` ``for` `j ``in` `range``(i ``-` `1``, ``-``1``, ``-``1``): `` ``if` `(arr[j] < arr[i]):`` ` ` ``# Iterate k from 0 to 100`` ``for` `k ``in` `range``(``101``):`` ` ` ``# Find gcd of two number`` ``GCD ``=` `gcd(arr[i], k);`` ` ` ``# dp[i][GCD] is summation of`` ``# dp[i][GCD] and dp[j][k]`` ``dp[i][GCD] ``=` `dp[i][GCD] ``+` `dp[j][k];`` ` ` ``# Add all elements of dp[i][1]`` ``sum` `=` `0``; `` ``for` `i ``in` `range``(n): `` ``sum` `=` `(``sum` `+` `dp[i][``1``]);`` ` ` ``# Return sum`` ``return` `sum``;`` ` `# Driver code``if` `__name__``=``=``'__main__'``:`` ` ` ``a ``=` `[ ``3``, ``4``, ``8``, ``16` `]`` ``n ``=` `len``(a)`` ` ` ``# Function Call`` ``print``(countSubsequences(a,n))`` ` `# This code is contributed by Pratham76`
## C#
`// C# program for the above approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;`` ` `class` `GFG{`` ` `// This function calculates``// gcd of two number``static` `int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(b == 0)`` ``return` `a;`` ` ` ``return` `gcd(b, a % b);``}`` ` `// This function will return total``// subsequences``static` `long` `countSubsequences(``int` `[]arr,`` ``int` `n)``{`` ` ` ``// Declare a dp 2d array`` ``long` `[,]dp = ``new` `long``[n, 101];`` ` ` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``for``(``int` `j = 0; j < 101; j++)`` ``{`` ``dp[i, j] = 0;`` ``}`` ``} `` ` ` ``// Iterate i from 0 to n - 1`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ` ` ``dp[i, arr[i]] = 1;`` ` ` ``// Iterate j from i - 1 to 0`` ``for``(``int` `j = i - 1; j >= 0; j--)`` ``{`` ``if` `(arr[j] < arr[i])`` ``{`` ` ` ``// Iterate k from 0 to 100`` ``for``(``int` `k = 0; k <= 100; k++)`` ``{`` ` ` ``// Find gcd of two number`` ``int` `GCD = gcd(arr[i], k);`` ` ` ``// dp[i,GCD] is summation of`` ``// dp[i,GCD] and dp[j,k]`` ``dp[i, GCD] = dp[i, GCD] +`` ``dp[j, k];`` ``}`` ``}`` ``}`` ``}`` ` ` ``// Add all elements of dp[i,1]`` ``long` `sum = 0;`` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``sum = (sum + dp[i, 1]);`` ``}`` ` ` ``// Return sum`` ``return` `sum;``}`` ` `// Driver code``public` `static` `void` `Main(``string` `[]args)``{`` ``int` `[]a = { 3, 4, 8, 16 };`` ``int` `n = a.Length;`` ` ` ``// Function Call`` ``Console.WriteLine(countSubsequences(a, n));``}``}` `// This code is contributed by rutvik_56`
## Javascript
``
Output
`7`
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Business Models with Exp. and Nat. Log and Diff. rules LAB I 150 150 Affordable Capstone Projects Written from Scratch
MTH 241 Business Models with Exp. and Nat. Log and Diff. rules LAB I
Name:__________________
Recitation Time: 8 9 10 11 12 1 2 3 4 5
1) A small company has the marketing information that 35 units will sell daily at a price of \$34.75 per unit, and that sales will rise to 36 units per day at a price of \$33.06 per unit.
Use this information to create a linear demand function, then create the associated revenue function and find the price that will yield the maximum revenue.
2) A small company has decided to model its demand function with the exponential function xexP05.200)(−= , where x is the number of items produced and sold daily and P(x) is the price
per unit in \$.
a) Is this model consistent with the marketing information from question1?
b) Does P(x) have the right shape to be a demand function?
c) Develop the associated revenue function.
d) Find the price that will yield the maximum revenue for this model.
e) Graph both functions and compare them to the graphs of the functions you worked with in problem 1.
f) Which models do you think are better?
Why?
3) For a different item in a different company, the cost function associated with producing
x items is: )50ln(5)(costxx⋅=
a) What is the marginal cost at a production level of 10 items?
b) How would you interpret that result?
c) What is the average cost function?
d) What is the derivative of the average cost function?
e) What is the value of that derivative function you just found in 2d at a production level of 50 items?
f) How would you interpret that result?
Recall:
If )()(xhxgy⋅= then )(‘)()()(‘xhxgxhxgdxdy⋅+⋅= [product rule]
If )()(xhxgy= then 2))(()(‘)()(‘)(xhxhxgxgxhdxdy⋅−⋅= [quotient rule]
If ))((xgfy= then )(‘))((‘xgxgfdxdy⋅= [chain rule]
Ex. 4) For F and G functions and given that:
1)5(−=F,3)5(‘−=F4)5(=G2)5(‘=G
a) If )()()(xGxFxP⋅=, then evaluate )5(‘P.
)5(‘P=_____________________
b) If )()()(xGxFxQ=, then evaluate )5(‘Q.
)5(‘Q=_____________________
Ex. 5) For R and Q functions and given that:
5)1(=−Q, 4)1(−=−R, 7)5(=R;
3)1(‘−=−Q, 2)1(‘=−R, 1)5(‘−=R
If ))(()(xQRxf=, then evaluate )1(‘−f.
)1(‘−f=_________________
Ex 6. Differentiate )()(xefxh= )(‘xh=________________
Ex. 7 Differentiate ))(ln()(xfxg= )(‘xg=________________
Ex. 8 Differentiate )()ln()(xfxxk= )(‘xk=________________
Ex 9. Differentiate )()(xfexmx= )(‘xm=________________
Ex 10. Differentiate xexfxn)()(= )(‘xn=________________
Ex 11. Differentiate )()ln()(xfxxp= )(‘xp=________________
Ex 12. Differentiate )()(xfexq= )(‘xq=________________
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QUESTION
# Hi, need to submit a 750 words essay on the topic Origami.Download file to see previous pages... For a dodecahedron, at least 3 different colors are required, as a dodecahedron cannot be properly colo
Hi, need to submit a 750 words essay on the topic Origami.
For a dodecahedron, at least 3 different colors are required, as a dodecahedron cannot be properly colored in less than 3 colours. It is advisable to draw the planar graph of a a dodecahedron when planning the 3-edge coloring. "It is always quite puzzling to try to make use only 3 colors of paper with no two units of the same color touching. Each unit corresponds to an edge of the planar graph, so this is equivalent to a proper 3-edge-coloring of the polyhedron." (T.Hull, 2006)
During the nineteenth century, Sir William Rowan Hamilton who was a mathematician from Ireland, invented a puzzle known as 'Around the World.' The concept behind the puzzle was to label the vertices of a regular dodecahedron according to the names of various cities of the world. Hamilton's puzzle can be solved by beginning from any given city (i.e. any vertex) and traveling around the world from one city (vertex) to another. This entails that one moves along the edges of the dodecahedron in such a manner that each other city is touched only once before going back to the original vertex or starting point. This solution to Hamilton's puzzle is known as a Hamilton cycle/Hamilton circuit. Thus, a Hamilton circuit can be said to be a path in the dodecahedron which starts at a vertex, touches every other vertex, in the dodecahedron, and then returns to the original starting point without touching any single vertex two times.
A Hamilton circuit in the planar graph of a dodecahedron
References
J.A.Gallian (2006) Contemporary Abstract Algebra. Houghton
R.A.Brualdi (2004) Introductory Combinatorics. Prentice Hall
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ALL >> Education >> View Article
# Knowing Our Numbers
By Author: REEII EDUCATION
Total Articles: 13
What number of numbers would you be able to make?
While framing numbers from the given digits, we should check cautiously whether the condition under which the numbers are to be shaped is fulfilled or not. The given condition should basically be fulfilled. Consequently, to frame the biggest number from the given four digits 8,9,4,6 without rehashing a solitary digit, we should be mindful so as to utilize all the given four digits. All things considered, the biggest number can have 9 just as the furthest left digit.
Moving digits
Take any three-digit number with various digits and trade the digit at the hundreds spot and the digit at the ones spot. We find that the new number is more prominent than the previous number, if the digit at the ones spot is more noteworthy than the digit at the hundreds spot and the new number is more modest than the previous number, if in the previous number the digit at the ones spot is more modest than the digit at the hundreds spot.
Presenting 10,000
The littlest 2-digit number is 10 (ten). The biggest 2-digit number is 99. The littlest 3-digit number is (100). The biggest 3-digit number is 999. The littlest 4-digit number is 1000 (1,000). The biggest 4-digit number is 9999. The littlest 5-digit number is (10,000). The biggest 5-digit number is 99999. The littlest 6-digit number is 1,00,000 (one lakh). The biggest 6-digit number is 9,99,999. This carries on for higher digit numbers likewise.
Returning to put esteem
The spot value1 of a digit at ones spot is equivalent to the digit. The spot estimation of a digit at tens spot is gotten by duplicating the digit by 10. Also, the spot estimation of a digit at hundreds spot, thousands spot, ten thousands spot,… is gotten by duplicating the digit by 100, 1000, 10000, … , separately.
Presenting 1,00,000
The best five-digit number is 99,999. In the event that we add 1 to this number, we get 1,00,000 which is the littlest six-digit number. It is named as one lakh. It comes close to 99,999.
Additionally, 10 × 10,000 = 1,00,000
Bigger numbers
The best six-digit number is 9.99.999. Adding 1 to it, we get 10,00,000 which is the littlest seven-digit number. It is known as the ten lakh.
The best seven-digit number is 99.99,999. Adding 1 to it, we get 1.00.00.000 which is the littlest eight-digit number. It is called one crore.
It would be ideal if you note that
1 hundred = 10 tens
1 thousand = 10 hundreds = 100 tens
1 lakh = 100 thousands = 1000 hundreds
1 crore = 100 lakhs = 10,000 thousands
A guide in perusing and composing huge numbers
Shagufta's pointers help us to peruse and compose huge numbers. These are likewise helpful recorded as a hard copy the developments of the numbers. These are as per the following:
T La T Th H T O
Commas additionally help us in perusing and composing enormous numbers. In Indian arrangement of Numeration, we utilize ones, tens, hundreds, thousands and afterward lakhs and crores. Commas are utilized to stamp thousands, lakhs and crores. The principal comma comes following hundreds spot, second comma comes after ten thousands spot and the third comma comes after ten lakh place and markes crore.
Assessment
There are various circumstances in which we needn't bother with the specific amount however need just a gauge of this amount. Assessment implies approximating an amount to the ideal precision.
Assessing to the closest tens by adjusting
The assessment is finished by adjusting the numbers to the closest tens. Subsequently, 17 is assessed as 20 to the closest tens; 12 is assessed as 10 to the closest tens.
Assessing to the closest hundreds by adjusting
Numbers 1 to 49 are nearer to 0 than to 100. So they are adjusted to 0. Numbers 51 to 99 are nearer to 100 than to 0, as are adjusted to 100. Number 50 is equidistant from 0 and 100 both. It is standard to adjust it as 100.
Assessing to the closest thousands by adjusting
Numbers 1 to 499 are closer to 0 than 1000, so these numbers are adjusted as 0. The numbers 501 to 999 are closer to 1000 than 0, so they are adjusted as 1000. Number 500 is usually adjusted as 1000.
REFRENCES-: https://reeii.com/knowing-our-numbers-class-6-maths-ncert-chapter-1/
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Popular questions and responses by Raina
1. ## geometry
List the set of all even numbers between 2 and 10, inclusive.
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Which of the following contains ionic bonding? H2O2 PO33- P2O5 AlCl3
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6. ## math worksheet
Ok i have one more question how would you solve 5 equals % of 12
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15. ## Math
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Hi! I'm really lost on how to do this. Does anyone know how to solve this? Thanks! :) Condesne the expression to the logarithm of a single quanity. 4(lnz+ln(z+5)]-2ln(z-5)
21. ## Math mrs.sue plz help i a confused.
1. Find the slope of a line that is parallel to the line containing the points (3, 4) and (2, 6). (1 point) m = 1 m = 2 m = –2 m = 1/2 2. Find the slope of a line that is perpendicular to the line containing the points (–2, –1) and (2, –3). (1
22. ## Language arts Mrs.Sue check Plz
A factory might emit (1 point) pollution. smokestacks. resources. 2. How do the members of Don’t Be Crude make a persuasive appeal? (1 point) by embarrassing people by educating people by encouraging people by making people laugh 3. What did Gina Gallant
23. ## Help Mrs.Sue plz
Which sentence uses italics, quotation marks, and punctuation correctly? (1 point) The article called “A Persistent Rebel” in American History magazine told about Elizabeth Blackwell. The article called A Persistent Rebel in American History magazine
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25. ## Math
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26. ## Physics - projectile
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If the government imposes a quantity tax on the consumption of a good, it means that the consumer has to pay for each unit of the good its price plus the tax. For example, if the price of a chocolate bar is \$5 and the government imposes a tax of 20 cents
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31. ## Eng poem!
Hold fast to dreams For if dreams die Life is a broken winged bird That cannot fly Hold fast to dreams For when dreams go Life is a barren field Frozen with Snow. Can someone explain to me what this poem means.
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33. ## math
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35. ## AED200
an outline of your Educational Debate Persuasive Paper. The outline should be based on one of the educational debates found in Appendix A and should highlight the main points of your paper. Include an annotated bibliography of at least three sources you
how do you solve y=5x-8 and 5y=2x+6 using elmination? Thank you, Raina
37. ## physics
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38. ## Pre Cal
Solve for x 2^x=8^x-3
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Could anyone help me unscramble "ukonyath"? It has to do with Christmas/winter and it is TWO words... please help =]
1. ## Math
there are tons of ways making a dollar 1. 4 quarters 2. 10 dimes 3. 100 pennies ( I think you should not put this one ) 4. 20 nickels 5. 2 quarters and 5 dimes 6. 5 dimes and 10 nickels there are lots of ways. you just need to think
posted on May 12, 2020
2. ## math
So actually all of you are still incorrect the correct answer is 368.26. So to whomever comes here on this site this is your best bet my friend I honestly did just take the quiz and this is the answer that was actually correct. Oh and your welcome.
posted on December 19, 2018
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7. ## English
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8. ## Help Mrs.Sue plz
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Its from arithmetic and geometric sequence chapter.....and what you wrote is not correct.
posted on December 30, 2012
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an outline of your Educational Debate Persuasive Paper. The outline should be based on one of the educational debates found in Appendix A and should highlight the main points of your paper. Include an annotated bibliography of at least three sources you
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15. ## algebra
fjaldkjf
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16. ## English
I meant mental vs. physical
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17. ## English
I believe there is more emphasis on the mental effects of steroid use then the psychological effects.
posted on April 26, 2010
18. ## English
How do this sound ? Steroids are life-savers for many people with serious illnesses and banning them can ruin one who is dependent upon it.
posted on April 26, 2010
19. ## English
I am opposed to it unless it will be aiding in helping someone in a critical situation .
posted on April 26, 2010
20. ## English
Well I do not want to be bias because of exactly what you said on how it can cure serious illnesses and I'm trying to say that more information is needed before banning steroids because this decision can make or break someones life... How would I sum it up
posted on April 26, 2010
21. ## Science
Nevermind, figured it out. Sorry =P
posted on January 1, 2008 | 2,697 | 9,955 | {"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} | 2.90625 | 3 | CC-MAIN-2020-34 | latest | en | 0.904315 |
https://www.jiskha.com/display.cgi?id=1224782404 | 1,500,901,174,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424876.76/warc/CC-MAIN-20170724122255-20170724142255-00081.warc.gz | 785,534,329 | 4,110 | # physics
posted by .
A spelunker drops a stone from rest into a hole. The speed of sound is 343m/s in air, and the sound of the stone striking the bottom is heard 1.46s after the stone is dropped. How deep is the hole? What is it in meters
• physics -
time=time down + time up
time=sqrt(2h/g)+ h/vsound
solve for time Use the quadratic equation.
• physics -
Solve the equation
t1 + t2 =
(stone's time required to fall) + (time to hear the splash) = 1.46 s
(1/2) g t1^2 = H
t1 = sqrt (2H/g)
a H = t2 (a is the sound speed, 343 m/s)
t2 = H/a
You will have to set it up as a quadratic equation in H, the depth of the well.
• physics -
10.24
• physics -
a stone is dropped into a well . The sound of the splash is heard 30s after the stone is dropped . What is the depth of the well? Note that the speed of sound in air is 343m / s. | 254 | 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} | 4.09375 | 4 | CC-MAIN-2017-30 | longest | en | 0.927523 |
https://studylib.net/doc/7971803/commutative-property-of-multiplication-the-order-of-the-f... | 1,708,468,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00300.warc.gz | 587,087,852 | 11,739 | Commutative Property of Multiplication The order of the factors in a
```Commutative Property of Multiplication
The order of the factors in a multiplication
problem will not effect the product.
5X4=4X5
20 = 20
Associative Property of Multiplication
The way you group the factors in a mult.
problem will not effect the product.
(3X4)X 5 = 3 X (4X5)
12 X 5
3 X 20
60 = 60
Distributive Property | 114 | 391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-10 | latest | en | 0.825306 |
https://www.expertsmind.com/library/write-the-irrationals-51034752.aspx | 1,716,084,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00602.warc.gz | 678,869,518 | 14,837 | Write the irrationals
Assignment Help Mathematics
Reference no: EM131034752
1. Indicate the least upper bound of the following sets. If a set does not have a least upper bound, just write "none."
a. (2,3)
b. (2,3]
c. Q
d. Q∩ (1, π]
2. Prove each of the following using induction:
1. Let a∈R and a ≥ 1. Prove an ≥ a for all n∈N.
2. Prove that 1/2 + 1/4 + 1/8 + ? + 1/2n = 1 - 1/2n for all n∈N.
3. Using basic set notation, do the following:
a. Write Z in terms of N and 0
b. Write Q in terms of Z
c. Write the irrationals in terms of R and Q
4. There exists real numbers a, b, c such that a ≤ b, 0 < c and ac ≤ bc. (T/F)
5. Let a, b∈R with a ≠ 0, b ≠ 0. Prove that (ab)-1 = a-1b-1
6. Let q∈Q and x∈R - Q. Prove that q + x∈R - Q.
7. Let a∈R and let A∈ (0, ∞) with x ∈ (a - A, a + A). Then there exists some P∈ (0, A) such that x ∈ (a - P, a + P). (T/F)
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At what time has the amplitude of oscillation of a spring-mass system with negligible friction decayed to 1/e of its original value? Does this time depend in a reasonable way on k, m, and c? | 1,362 | 5,604 | {"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} | 3.640625 | 4 | CC-MAIN-2024-22 | latest | en | 0.904348 |
https://app.jove.com/science-education/v/15524/principle-of-linear-impulse-and-momentum-for-a-single-particle?trialstart=1 | 1,726,474,919,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651682.69/warc/CC-MAIN-20240916080220-20240916110220-00160.warc.gz | 85,064,075 | 23,129 | Faculty Resource Center
Linear momentum is a fundamental concept in physics that describes the motion of an object. It is a vector quantity, having a magnitude equal to the product of its mass and its velocity, and direction along the object's velocity. On the other hand, linear impulse, also known as momentum impulse, is a concept in physics related to the change in the linear momentum of an object. Impulse is a vector quantity defined as the product of force and the time over which the force is applied.
Delving into the equation of motion for a particle of mass 'm' in an inertial frame, acceleration and velocity are quantified. Integrating this equation over time and rearranging terms yields a formula embodying the principle of linear impulse and momentum. This principle asserts that the sum of the initial momentum of the particle and all applied impulses within a specific timeframe equates to the final momentum. Visualization is facilitated through impulse and momentum diagrams. Momentum diagrams portray the direction and magnitude of initial and final momentum, while the impulse diagram delineates impulses at various points along the particle's path.
To further elucidate, this principle can be translated into three scalar equations by resolving vectors into components. This analytical approach captures the essence of particle dynamics and establishes a foundation for comprehending real-world scenarios involving motion and force interactions.
Privacy | 265 | 1,481 | {"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} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.913706 |
http://www.mathworks.com/examples/statistics/mw/stats-ex17217789-multiple-comparisons-using-one-way-anova | 1,519,602,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00560.warc.gz | 482,329,214 | 12,321 | MATLAB Examples
# Multiple Comparisons Using One-Way ANOVA
```load carsmall ```
MPG represents the miles per gallon for each car, and Cylinders represents the number of cylinders in each car, either 4, 6, or 8 cylinders.
Test if the mean miles per gallon (mpg) is different across cars that have different numbers of cylinders. Also compute the statistics needed for multiple comparison tests.
```[p,~,stats] = anova1(MPG,Cylinders,'off'); p ```
```p = 4.4902e-24 ```
The small p-value of about 0 is a strong indication that mean miles per gallon is significantly different across cars with different numbers of cylinders.
Perform a multiple comparison test, using the Bonferroni method, to determine which numbers of cylinders make a difference in the performance of the cars.
```[results,means] = multcompare(stats,'CType','bonferroni') ```
```results = 1.0000 2.0000 4.8605 7.9418 11.0230 0.0000 1.0000 3.0000 12.6127 15.2337 17.8548 0.0000 2.0000 3.0000 3.8940 7.2919 10.6899 0.0000 means = 29.5300 0.6363 21.5882 1.0913 14.2963 0.8660 ```
In the results matrix, 1, 2, and 3 correspond to cars with 4, 6, and 8 cylinders, respectively. The first two columns show which groups are compared. For example, the first row compares the cars with 4 and 6 cylinders. The fourth column shows the mean mpg difference for the compared groups. The third and fifth columns show the lower and upper limits for a 95% confidence interval for the difference in the group means. The last column shows the p-values for the tests. All p-values are zero, which indicates that the mean mpg for all groups differ across all groups.
In the figure the blue bar represents the group of cars with 4 cylinders. The red bars represent the other groups. None of the red comparison intervals for the mean mpg of cars overlap, which means that the mean mpg is significantly different for cars having 4, 6, or 8 cylinders.
The first column of the means matrix has the mean mpg estimates for each group of cars. The second column contains the standard errors of the estimates. | 548 | 2,057 | {"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} | 3.546875 | 4 | CC-MAIN-2018-09 | latest | en | 0.861007 |
https://lukebenz.com/post/wc_model_methodology/blogpost/ | 1,721,581,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00857.warc.gz | 328,540,807 | 11,028 | # Introduction
With the 2018 FIFA World Cup nearly upon us, we set out to build a model to predict game outcomes and estimate coutries chances of reaching various rounds of the tournament. Ranking systems for various sports, including NCAA Men’s Basketball, NCAA Football, and NBA, have been the basis of several of our projects in the past. With the exception of rankings using ELO, all of our past ranking systems have made use of linear regression in one form or another. Simply, those models took on something similar to the following form $Y = \beta_0 + \beta_{team} X_{team} + \beta_{opp} X_{opp} + \beta_{loc} X_{loc} + \epsilon$ where $$X_{team, i}, X_{opp, i}$$, and $$X_{loc, i}$$ are indicator vectors for the $$i^{th}$$ game’s team, opponent, and location (Home, Away, Neutral) from the perspective of team, and $$Y_i$$ is game’s the score-differential. The key assumptions for this model are that game outcomes are independent of one another, and that our error $$\epsilon \sim N(0, \sigma^2)$$.
# First Attempt
Before going into detail on the model we ultimated settled on, let’s first examine what didn’t work–and more importantly why. Given our prior experience using linear regression to predict score differential in several different sporting events, it seemed natural to extend the notion to international soccer. In fact, our initial attempt seemed quite reasonable. The basic idea was as follows:
• Use linear regression with covariates team, opponent, and location to predict goal_diff (a game’s goal differential).
• Use a multinomial logistic regression model with input pred_goal_diff to estimate probabilities of win, loss, and tie. Our rankings indicated that Germany, Brazil, and Spain were the top 3 teams, entirely plausible our eyes (as casual soccer fans).
In writing Monte Carlo simulations to estimate round-by-round probabilities of advacing, we began to notice a potential fault in our model. While it was easy enough to simulate game outcomes and award points (3 for win, 1 for tie, 0 for loss) given our estimated win, loss, and tie probabilites for each of the group stage games, breaking ties in the group standings to determine which two teams would advance to the knockout round presented more of a challenge. If two or more teams are tied on points after a round robin within the group, the first tie-breaker is net goal differtial, a country’s goals scored - goals allowed in the 3 group stage games. Our initial thought was to simply flip a coin to break ties, but this underestimates the probabiity that better teams advance past the group stage. Consider the following hypothethical scenario in group B:
1
Spain 9 2 Portugal 4 3 Morocco 4 4 Iran 0
Simply flipping a coin to determine whether Portugal or Morocco would advance in the above case almost certianly underestimates the chances that Portugal, the superior of the two, would have a larger goal differential than Morocco. So, was there a better way to account for goal differntial? Recall, goal_diff was the response variable of our linear regression. A nice result of simple linear regression is that for game $$i$$, we have $$\widehat Y_i$$ normally distributed. Since $$\widehat Y_i = \widehat \beta_0 + \widehat\beta X_i$$ and each $$\widehat \beta_i \sim N(\beta_i, \sigma_i^2)$$, $\widehat Y_j \sim N(\beta_0 + \beta X_j, \sum_i \sigma_i^2)$ Approximating such a distribution using prediction intervals for goal_diff, perhaps we could form net goal differential distributions for each team in group play and randomly draw from those distrubutions in the case of a tie in the group standings. Things quickly became not well-defined when trying to draw from $$Y \sim N(\mu, \sigma^2) | Y \geq 1$$ (in case of loss, $$Y \leq -1$$). Moreover, in the case of a tie, we’d be setting $$Y = 0$$ in our simulation, yet under the normal distribution, $$P(Y = 0) = 0$$ (in fact the probability $$Y = c$$ exactly is 0 any fixed constant $$c$$). By this point, we’d realized we were in trouble, as drawing from this continuous distribution for a discrete random variable didn’t make a lot of sense. Back to the drawing board.
# Taking a Step Back
At this stage, there are two questions to consider. The first, is why didn’t a linear model, which had worked well for us in the case of basketball and football, work well in the case of soccer? Furthermore, we wondered, if not a linear model, what type of model would work best? The answer to both of these questions is best seen by looking at histograms of points/goals scored in soccer games vs. basketball games.
Based on the plots above, we see that the distribution of basketball scores looks like a bell curve, while the distribution of soccer scores does not. Rather, the distribution of soccer scores looks to be more like a Poisson distribution. Recall that the probability mass function (PMF) of the possion distribution with parameter $$\lambda$$ is by $P(X = k) = \frac{e^{-\lambda}\lambda^k}{x!} \textrm{ for } x = 0, 1, 2, ...$
Additionally, if $$X \sim Pois(\lambda)$$, $$\mathbb{E}(X) = \lambda$$. As $$\lambda$$ gets large, Poisson distributions become more and more like Guassians, hence the ability for sports with larger average scores (i.e. basketball) to be better modeled with a Normal Model than sports with lower scores (i.e. soccer, hockey). Since the difference of i.i.d. normal random variables is also normal, linear regression is perfectly fine for modeling score_diff is basketball. However, the difference of two i.i.d. poisson random variables is not normal nor poisson. Rather, the difference of two poisson random variables follows the skellam distribution. Thus, it seems like our best bet here is to use poisson regression for our model. In fact, several studies have shown that poisson regression is good for modeling soccer 1 2 3 4.
# Building the Model
The data, available on Kaggle, covers over 39,000 international soccer matches dating back to 1872. For the purpose of this model, we have chosen to use data for games played after January 1, 2014. Many major tournaments in soccer, the World Cup included, occur on a four year cycle, so using the last 4 years worth of data seemed natural.
Daniel Sheehan has written a fantastic blog post on using poisson regression to predict the outcomes of soccer games, and our model is based off of his work (he provides lots of examples in both R and Python, and we’d highly recommend reading it!).
We began by duplicating the data set and transforming one copy to be from the perspective of the opponent. For example, if we had the vector in the < team = "Germany", opponent = "Brazil", location = "N", team_score = 7, opp_score = 1 >, we’d also add to our data set the vector <team = "Brazil", opponent = "Germany", location = "N", team_score = 1, opp_score = 7 >. What we eventually end up predicting is team_score as a function of team, opponent, and location.
Before actually building the model, there were one more things that we considered, namely match_weight, how much weight we should give a particular game. There were two factors on which the model weights were set: time since the was played, and the type of match being played. We broke matches in our data set into four types (derived from the official FIFA rankings formula):
• Friendlies and other matches (Base Weight $$\alpha_i = 1$$)
• Qualification matches for World Cup and continental championships (Base Weight = $$\alpha_i = 3$$)
• Confederations Cup and continental championships (Base Weight = $$\alpha_i = 5$$)
• World Cup Matches (Base Weight = $$\alpha_i= 8$$)
Letting $$\delta_i$$ represent the date on which game $$i$$ was played and let $$\delta_t$$ be today’s date (i.e. the date we choose to fit/re-fit the model). Finally, take $$\delta\star = \max_{i} (\delta_t - \delta_i)$$. Then, the match_weight, $$w_i$$ of match $$i$$ is given by $w_i = \alpha_i \times e^{- \frac{\delta_t - \delta_i}{\delta\star}}$
We see above that the 2014 World Cup and the 2016 Euro Cup are among the most heavily weighted games, as we would hope. Now the call to the model is as follows:
glm.futbol <- glm(goals ~ team + opponent + location,
family = "poisson",
data = y,
weights = match_weight)
The model gives coefficients for each country both as levels of the team and opponent factors. Since the model output predictions can be taken as the average team_score for a given team against a given opponent at a given location, we can view the country specific coefficients as offense and defensive components of a power rating. The interpretation of such coefficients is less intutive than in the case of linear regression, in which coefficients signfy points better than an average team, but you can think of them as more similar to logistic regression coefficients. Higher offensive coefficients indicate a team is likely to score more goals on average while low (more negative) defensive coefficients indicate a team is likely to conceed fewer goals on average.
## team offense defense net_rating rank
## 77 Germany 1.707366 -1.373929 3.081295 1
## 72 France 1.581282 -1.307218 2.888500 2
## 29 Brazil 1.727649 -1.097520 2.825169 3
## 181 Spain 1.639751 -1.116924 2.756675 4
## 9 Argentina 1.318170 -1.438423 2.756593 5
## 45 Colombia 1.456873 -1.285449 2.742322 6
## 20 Belgium 1.565169 -1.140337 2.705506 7
## 140 Netherlands 1.597657 -1.065459 2.663115 8
## 158 Portugal 1.442856 -1.200933 2.643789 9
## 64 England 1.269063 -1.281223 2.550285 10
Offensively, the top teams by our model, are Brazil, Germany and Spain, while Argentina, Germany, and England posses the best defenses.
# Sample Match Prediction
To see how this model works in more detail, let’s walk through how we predict the outcome of a single match. We’ll use the marquee Spain-Portugal fixture from Group B as our case study. The model output for the vector <team = "Spain", opponent = "Portugal", location = "N" > is 1.37, while the model output for the vector <team = "Portugal", opponent = "Spain", location = "N" > is 1.21 . This signifies that on average, we expect Spain to score 1.37 goals and expect Portugal to score 1.21 goals. There is much more information encoded in these two numbers however. Let $$X_s$$ be the random variable denoting the number of goals Spain score and let $$X_p$$ denote the number of goals that Portugal scores. Then, we have that $$X_s \sim Pois(\lambda_s = 1.37)$$ and $$X_p \sim Pois(\lambda_p = 1.21)$$, and from these distributions, we can get a lot of neat stuff. First, we can look at the joint distribution of goals scored. Rows indicate the number of goals Spain scores while columns correspond to the number of goals Portugal scores.
0123
00.07580.09170.05550.0224
10.10380.12560.07600.0307
20.07110.08600.05210.0210
30.03250.03930.02380.0096
Perhaps unsuprisingly, the most likely outcome is a 1-1 draw, but there is still about an 87% chance we see a different outcome. Summing the diagonal entries of this matrix (extended out beyond 3 goals–let’s assume it’s neither team will score more than 10 goals) gives the probability that that Iberian neighbors end in a stalemate, while summing the entries above the diagonal or below the diagonal yield Portugal or Spain’s chances of winning, respectively. Overall, we estimate that Spain has about a 41% chance to win, Portugal has about a 33% chance to win, and there is a 26% chance the two teams draw.
# Simulating the World Cup
Now that we have the ability predict any game, we can run some Monte Carlo simulations to estimate the probability of each time winning the World Cup. We run 10,000 iterations of the following simulation steps:
1. Simulate each group game by drawing the number of goals scored by each team from their respective poisson distributions.
2. Advance the top 2 teams in each group by points, and in the case of ties, use goal differential, goals forced, and goals allowed (in that order) as tiebreakers.
3. Simulate knockout round games as in step 1. If there is a tie, flip a coin to determine who wins the simulated penalty shooutout (assuming that teams convert penalties at similar rates, this is not a decent approximation 5).
4. Repeat step 4 until there is a champion
Our simulations indicate that despite being in the so called “group of death”, defending champion Germany is most likely to win the World Cup, with a roughly 16% chance to hoist the crown. Brazil (11%), Frace (9%), Spain (8%), and the trio of Argentina, Belgium and Columbia (6% each) follow closely behind. A full list of World Cup odds, as well as the code used in this project can be found on GitHub.
# Limitations
A key assumption of the poisson distribution is that the rate parameter $$\lambda$$ does not depend on time. That is, in using this model, we are assuming that the rate of goals is equal during each minute of the match. However, this is not true in practice, and several sources extend their framework to bivariate poisson regression 6 7 8 9. Other limitations include the relatively small number of important matches in international soccer. While we examine on a 4 year basis, there may only be 15-20 matches per team of the highest importance, making prediction dificult even with our best effort to correct for difference between matches on the basis of time and relative importance.
# Acknowledgements
We’d like to thank Kostas Pelechrinis and Michael Lopez for offering suggestions regarding the switch from linear to poisson regression, as well as providing example papers modeling soccer outcomes using poisson regressions. Additionally, we’d like to thank Edward Egros, who mentioned the data set used in this project during his April visit to Yale University. | 3,360 | 13,797 | {"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} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.937037 |
http://www.mathisfunforum.com/viewtopic.php?id=3941&p=4 | 1,606,923,447,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00075.warc.gz | 145,715,900 | 6,433 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #76 2010-06-17 16:55:42
Member
Registered: 2010-06-17
Posts: 1
having a daily exercise that could help enhance your skill would be a great thing to do lets say an hour or two daily would be very ok for those who want to learn great technique and i found those exercise as mentioned on this thread is very useful.
Last edited by sade1125 (2010-06-17 16:57:24)
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## #77 2012-04-21 21:23:27
rosatily
Member
Registered: 2012-04-21
Posts: 1
Pour l'exercice, je pense qu'il est nécessaire, parce que nous vivons dans une telle période tendue, la pression sera grande, il ne faut pas avoir le temps d'exercer, de sorte que le corps n'est pas très bonne, mais le corps aurait une incidence sur la Si nous insistons sur l'exercice du corps et l'esprit sont une bonne aide.
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## #78 2012-04-21 22:13:40
Bob
Registered: 2010-06-20
Posts: 8,914
Bonjour Rosatily
Bienvenue dans l'forum.
Bien sur, mais ce site est seulement pour l'esprit .
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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## #79 2012-04-21 23:34:20
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,045
Wow,Bob! You surprised me!
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Online
## #80 2012-04-22 00:46:33
Bob
Registered: 2010-06-20
Posts: 8,914
Moi, je peux parler français, un peu, je pense.
Mais, je suis le meilleur à l'anglais, n'est pas?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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## #81 2012-04-22 04:01:04
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,045
Wow,it's so good Fench that Google Translate messed it up. You are really good!
Last edited by anonimnystefy (2012-04-22 04:01:30)
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Online
## #82 2012-04-22 06:45:23
Bob
Registered: 2010-06-20
Posts: 8,914
Thank you for suggesting it was google translate that had it wrong.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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## #83 2012-04-22 07:24:29
anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,045
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Online
## #84 2013-02-13 14:56:44
Mathegocart
Member
Registered: 2012-04-29
Posts: 2,073
bob bundy wrote:
Moi, je peux parler français, un peu, je pense.
Mais, je suis le meilleur à l'anglais, n'est pas?
Bob
omg.you are aazing *claps*
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
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## #85 2013-05-09 21:53:11
mathaholic
Member
From: Earth
Registered: 2012-11-29
Posts: 3,251
Well, maybe I will just see some more of bob bundy's French.
Last edited by julianthemath (2013-05-29 18:23:48)
Mathaholic | 10th most active poster | Maker of the 350,000th post | Person | rrr's classmate
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## #86 2013-12-18 18:43:03
TracyBidwell
Member
Registered: 2013-12-18
Posts: 3
Devantè wrote:
No problem. I just got a little confused when you said 'number the excercises'.
I'm making some more as we speak.
Same here. Can you give us?
Habeeb Akande
If you are not working towards something, your life will end with nothing.
― Habeeb Akande
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## #87 2016-10-07 04:53:48
dilipamandas
Member
Registered: 2016-10-07
Posts: 1
Find the equation of the circle which passes through the point (2,0) and whose center is the limit of the point of intersection of the lines 3x+5y = 1; (2+c)x +5c^2y=0 as c tends to 1
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## #88 2016-10-07 05:21:52
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi;
This is a test question from the CBSE.
How to use a forum:
All forums are there to converse and answer questions. It is a great opportunity to get help. But all forum members are under an obligation to do the right thing. To cause no harm.
Lao Tsu wrote:
If you give a hungry man a fish, you feed him for a day, but if you teach him how to fish, you feed him for a lifetime. - old Sicilian saying
Most forum members will not answer homework or test questions. I will under certain conditions, if the questioner shows that he has tried and failed many times, I will give the answer if I can. This process of trying and failing is extremely important. It teaches perseverance and the right type of thinking. Dealing with failure is mandatory, since we will fail every day that we live.
I can give you the equation of the circle in terms of c. It is a bit involved and long. But I can not try to help unless I know where you need help. So, what have you tried to solve this? Let me see what you think.
Loopholes: There are always ways out of any predicament and no one knows that better than me... You posted this in Exercises, all of the above comments are basically for Help Me! If you know the answer and are posing this as a problem for others to solve then I will just blurt out the answer before anyone else does. I was the fastest in class, at work and just about anywhere else. True, I ain't 25 anymore but at 96 and toothless I will still take on any whippersnapper in a forum and on the trail.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #89 2019-01-26 22:20:15
Kamov50K
Member
Registered: 2019-01-26
Posts: 4 | 1,972 | 6,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | longest | en | 0.554294 |
https://www.tag-challenge.com/2022/10/05/how-do-you-write-4-hours-15-minutes/ | 1,726,084,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00357.warc.gz | 954,268,307 | 9,005 | ## How do you write 4 hours 15 minutes?
To convert to decimal hours, add (minutes ÷ 60) to the hours number. So, 4 hours 15 minutes is 4 + 15 ÷ 60 = 4.25 hours.
What does .25 mean in minutes?
15 minutes
For example 15 minutes (¼ hour) equals . 25, 30 minutes (½ hour) equals . 5, etc.
How do you convert MS 1 to minutes?
To convert a millisecond measurement to a minute measurement, divide the time by the conversion ratio. The time in minutes is equal to the milliseconds divided by 60,000.
### What is 7 hours and 40 minutes as a decimal?
Common Time to Hours, Minutes, and Seconds Decimal Values
Time Hours Minutes
07:20:00 7.333 hrs 440 min
07:30:00 7.5 hrs 450 min
07:40:00 7.667 hrs 460 min
07:50:00 7.833 hrs 470 min
What is 3 hours and 15 minutes as a decimal?
This online tool will help you convert time given in hours and minutes to decimal hours and/ or decimal minutes. 3 hours 15 minutes is 3.25 hours or 195 minutes.
What does .75 mean in time?
Decimal Hours-to-Minutes Conversion Chart
Minutes Tenths of an Hour Hundredths of an Hour
44 .7 .74
45 .7 .75
46 .7 .76
47 .7 .78
#### What is .43 on a time clock?
Option 2: Use our minutes conversion chart
Minutes Decimal Hours Decimal Hours
5 .08 .42
6 .10 .43
7 .12 .45
8 .13 .47
How do you calculate milliseconds?
To convert a second measurement to a millisecond measurement, multiply the time by the conversion ratio. The time in milliseconds is equal to the seconds multiplied by 1,000.
How do you convert milliseconds to hours minutes seconds in Excel?
1. Are you entering the formula in cell B1? – Santosh. Oct 11, 2013 at 19:33.
2. There are much simpler ways to do this. Just divide by 86400000 (the number of milliseconds in a day) and format result cell as [h]:mm:ss.000 – i.e. with this formula =B1/86400000. – barry houdini. Oct 11, 2013 at 21:21.
## How many units is 45 minutes?
Minute Conversion Chart
Minutes Decimal Conversion
43 0.72
44 0.73
45 0.75
46 0.77
How do I convert time?
To convert time to a number of hours, multiply the time by 24, which is the number of hours in a day. To convert time to minutes, multiply the time by 1440, which is the number of minutes in a day (24*60). To convert time to seconds, multiply the time time by 86400, which is the number of seconds in a day (24*60*60 ). | 670 | 2,304 | {"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} | 3.8125 | 4 | CC-MAIN-2024-38 | latest | en | 0.790775 |
https://datascience.stackexchange.com/questions/6511/are-there-any-interesting-application-of-linear-regression/6512 | 1,726,040,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00250.warc.gz | 177,901,651 | 39,003 | Are there any interesting application of linear regression [closed]
Linear regression is a widely used ML algorithm. So far I have only encountered 'boring' applications of it. (e.g predict sales for next quarter, predict housing prices for next year , predict population of a country by 2020 etc.)
What are some interesting/cool application of linear regression? Finding cool applications greatly helps in motivating myself to learn, hence the quest.
Cool as in: Application to stock trading, application to video games, application to astronomy, application to sports betting, application to airfare prediction
• What is cool or boring for you ? Commented Jul 20, 2015 at 16:02
The way that you have phrased this question makes it tough for people to answer without first offering you some background on linear regression (LR). Its great that you are interested in learning some ML and LR is a great place to start.
Linear regression is really just finding a line (or plain or hyperplain) that maps a relationship between two or more variables or features. The important requirement is that the target variable be a continuous numerical value. So its less about finding a problem for which linear regression "works" and more about finding some data that interests you and playing with it using linear regression.
I suggest you download some open data in the realms that you have mentioned. There is plenty of open data in every topic that you mentioned that contain continuous numerical values that you can predict using other features of the data.
Also think about taking some sort of online MOOC as this will help you gain some footing in the subject. Andrew Ng's Coursera on Machine Learning is highly recommended as a starting point and include some linear regression during the first portion and scikit-learn is a great Python based library.
• Thanks. I was thinking of R. Any reason for python over R? Commented Jul 20, 2015 at 17:42
• No specific reason for python. R vs. Python is a hotly debated topic. R has been around for a long time so has tons of packages. Python is more user friendly and is the future of data science. Python will make you happier and will help you get things done faster. If you really need R then you can call R from within Python. Someone will probably respond and tell you why R is best, but I like python. Take a look at all of the following: pandas, scikit-learn, numpy, matplotlib, ipython notebook. There's a great book "Python for Data Analysis" and the scikit learn user guide is fantastic. Commented Jul 20, 2015 at 17:50
• And I thought Julia was the future of data science ;) Commented Jul 21, 2015 at 12:44
I think those were the coolest. I recommend you that video also https://www.khanacademy.org/math/probability/regression/regression-correlation/v/regression-line-example | 607 | 2,834 | {"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-2024-38 | latest | en | 0.953538 |
https://www.jiskha.com/display.cgi?id=1270519311 | 1,531,868,308,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00551.warc.gz | 905,715,311 | 3,942 | # statistics
posted by tina
a study found that the means waiting time to see a physican at an outpatient clinic was 40 minutes with a standard deviation of 28 minutes. use excel to find each probability. a. what is the probability of more than an hours wait? B. less than 20 minutes? C. at least 10 minutes?
1. PsyDAG
I don't know how to use Excel for this.
Z = (score - mean)/SD
See the Help for Excel for Z test.
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More Similar Questions | 658 | 2,964 | {"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-2018-30 | latest | en | 0.914452 |
https://alldimensions.fandom.com/wiki/Finality_Index | 1,726,553,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00081.warc.gz | 70,846,938 | 43,792 | 3,406
pages
The finality index, written as n-finality, is a rank assigned to specific, very large verses, containers, etc...
A verse with an n-finality rank contains all things under a specific label, with the labels being more broad, as n gets larger. An n-finality object also contains everything below it, for example, The Box contains everything that doesn't exist (2-finality), but also everything that exists, aka, the Omniverse (1-finality). There is no limit to how large the n in n-finality can be, as there are an unending number of labels, which then logically brings about the existence of an unending ascending hierarchy of broader labels, despite what our limited human languages can comprehend and describe.
Two containers can have the same finality rank but are expected to be different in size or contain more. An example of this is The Box and The Complex Box.
Rules
For ${\displaystyle n \in \mathbb{Z}^+}$:
• 1-finality = Existence-1
• The property that ${\displaystyle n+1}$-finality works over must be inaccessible to ${\displaystyle n}$-finality
• ${\displaystyle n+1}$-finality also contains ${\displaystyle n}$-finality. This property halts at 1-finality, as there is no 0-finality or lower.
How finality functions
Finality is an inherent property which is built into everything via information. This splits everything in our Reality into two groups, that being finality based containers, which have actual finality indexes, and everything else, which simply abides by the rules of finality. Finality containers usually have their own unique method of not allowing their objects to escape. This property of inability to wander outside of a container is similar to a law, except that it physically cannot be broken and applies the same everywhere, equally, in our Reality. Here are two good examples:
• Omniverse: doesn't allow the object to travel at infinite speeds. Any object outside of the Omniverse will physically be unable to exist.
• The Box: made of self-containing layers, even in the opposing direction. Therefore, any attempt at traveling or teleportation will simply loop back upon itself through The Box's infinite layers.
Examples
• 1-finality (Omniverse) contains everything that exists. This is the base of the finality function.
• 2-finality (The Box) contains everything that exists and doesn't exist.
• 3-finality (The Clock) likely subsumes existence, non-existence and describability. (decided upon by the community)
• 4-finality (The Flower) thus contains all that is describable and indescribable, but maybe not all that is conceivable.
• 5-finality (The Soul) may be the barrier of natural (or even formal) language, and as such, is unknown in more detail.
Specific finality indexes
3/4-finality concerns describability and indescribability. This does not cover objects whose information is unreadable in any way or totally readable. This simply involves objects which can not be 100% constructed through a finite number of properties and characteristics. Of course, if one were to somehow clone all of the information which makes up an object located in The Flower, one would see an exact replica of said object appear in front of them. However, this is physically impossible in anything above The Clock, thanks to how finality imposes itself.
5-finality and above is, as of right now, sadly inaccessible using human language, whether natural or constructed. However, it is theoretically possible, that if someone combined complex (higher-order) logic and a human language (preferably English?), through some weird and impractically difficult method, 5-finality and above could be talked about and even imagined. This has not happened yet, and is unlikely to ever happen, but is however possible.
Subfinality Index
The Subfinality Index works the same way as the finality index. However, it is much weaker and allows more wiggle room.
A container which is of n-subfinality contains all objects which possess n properties. Unlike finality, however, they don't have to be in any specific order.
If container X contains all objects with property A and property B, but B is a special case of A, then B is ignored when counting.
Objects which are strictly n-final are also n-subfinal.
(This is just a modified version of the end-all-be-all index.) | 949 | 4,313 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.940676 |
http://www.theinfolist.com/html/ALL/s/representable_functor.html | 1,653,663,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00554.warc.gz | 114,892,745 | 5,165 | TheInfoList
In mathematics, particularly category theory, a representable functor is a certain functor from an arbitrary category (mathematics), category into the category of sets. Such functors give representations of an abstract category in terms of known structures (i.e. Set (mathematics), sets and function (mathematics), functions) allowing one to utilize, as much as possible, knowledge about the category of sets in other settings. From another point of view, representable functors for a category ''C'' are the functors ''given'' with ''C''. Their theory is a vast generalisation of upper sets in posets, and of Cayley's theorem in group theory.
# Definition
Let C be a locally small category and let Set be the category of sets. For each object ''A'' of C let Hom(''A'',–) be the hom functor that maps object ''X'' to the set Hom(''A'',''X''). A functor ''F'' : C → Set is said to be representable if it is naturally isomorphic to Hom(''A'',–) for some object ''A'' of C. A representation of ''F'' is a pair (''A'', Φ) where :Φ : Hom(''A'',–) → ''F'' is a natural isomorphism. A contravariant functor ''G'' from C to Set is the same thing as a functor ''G'' : Cop → Set and is commonly called a presheaf (category theory), presheaf. A presheaf is representable when it is naturally isomorphic to the contravariant hom-functor Hom(–,''A'') for some object ''A'' of C.
# Universal elements
According to Yoneda's lemma, natural transformations from Hom(''A'',–) to ''F'' are in one-to-one correspondence with the elements of ''F''(''A''). Given a natural transformation Φ : Hom(''A'',–) → ''F'' the corresponding element ''u'' ∈ ''F''(''A'') is given by :$u = \Phi_A\left(\mathrm_A\right).\,$ Conversely, given any element ''u'' ∈ ''F''(''A'') we may define a natural transformation Φ : Hom(''A'',–) → ''F'' via :$\Phi_X\left(f\right) = \left(Ff\right)\left(u\right)\,$ where ''f'' is an element of Hom(''A'',''X''). In order to get a representation of ''F'' we want to know when the natural transformation induced by ''u'' is an isomorphism. This leads to the following definition: :A universal element of a functor ''F'' : C → Set is a pair (''A'',''u'') consisting of an object ''A'' of C and an element ''u'' ∈ ''F''(''A'') such that for every pair (''X'',''v'') with ''v'' ∈ ''F''(''X'') there exists a unique morphism ''f'' : ''A'' → ''X'' such that (''Ff'')''u'' = ''v''. A universal element may be viewed as a universal morphism from the one-point set to the functor ''F'' or as an initial object in the category of elements of ''F''. The natural transformation induced by an element ''u'' ∈ ''F''(''A'') is an isomorphism if and only if (''A'',''u'') is a universal element of ''F''. We therefore conclude that representations of ''F'' are in one-to-one correspondence with universal elements of ''F''. For this reason, it is common to refer to universal elements (''A'',''u'') as representations.
# Examples
* Consider the contravariant functor ''P'' : Set → Set which maps each set to its power set and each function to its inverse image map. To represent this functor we need a pair (''A'',''u'') where ''A'' is a set and ''u'' is a subset of ''A'', i.e. an element of ''P''(''A''), such that for all sets ''X'', the hom-set Hom(''X'',''A'') is isomorphic to ''P''(''X'') via Φ''X''(''f'') = (''Pf'')''u'' = ''f''−1(''u''). Take ''A'' = and ''u'' = . Given a subset ''S'' ⊆ ''X'' the corresponding function from ''X'' to ''A'' is the indicator function, characteristic function of ''S''. *Forgetful functors to Set are very often representable. In particular, a forgetful functor is represented by (''A'', ''u'') whenever ''A'' is a free object over a singleton set with generator ''u''. ** The forgetful functor Grp → Set on the category of groups is represented by (Z, 1). ** The forgetful functor Ring → Set on the category of rings is represented by (Z[''x''], ''x''), the polynomial ring in one Variable (mathematics), variable with integer coefficients. ** The forgetful functor Vect → Set on the category of real vector spaces is represented by (R, 1). ** The forgetful functor Top → Set on the category of topological spaces is represented by any singleton topological space with its unique element. *A group (mathematics), group ''G'' can be considered a category (even a groupoid) with one object which we denote by •. A functor from ''G'' to Set then corresponds to a G-set, ''G''-set. The unique hom-functor Hom(•,–) from ''G'' to Set corresponds to the canonical ''G''-set ''G'' with the action of left multiplication. Standard arguments from group theory show that a functor from ''G'' to Set is representable if and only if the corresponding ''G''-set is simply transitive (i.e. a torsor, ''G''-torsor or heap (mathematics), heap). Choosing a representation amounts to choosing an identity for the heap. *Let ''C'' be the category of CW-complexes with morphisms given by homotopy classes of continuous functions. For each natural number ''n'' there is a contravariant functor ''H''''n'' : ''C'' → Ab which assigns each CW-complex its ''n''th cohomology group (with integer coefficients). Composing this with the forgetful functor we have a contravariant functor from ''C'' to Set. Brown's representability theorem in algebraic topology says that this functor is represented by a CW-complex ''K''(Z,''n'') called an Eilenberg–MacLane space. *Let ''R'' be a commutative ring with identity, and let R-Mod be the category of ''R''-modules. If ''M'' and ''N'' are unitary modules over ''R'', there is a covariant functor ''B'': R-Mod → Set which assigns to each ''R''-module ''P'' the set of ''R''-bilinear maps ''M'' × ''N'' → ''P'' and to each ''R''-module homomorphism ''f'' : ''P'' → ''Q'' the function ''B''(''f'') : ''B''(''P'') → ''B''(''Q'') which sends each bilinear map ''g'' : ''M'' × ''N'' → ''P'' to the bilinear map ''f''∘''g'' : ''M'' × ''N''→''Q''. The functor ''B'' is represented by the ''R''-module ''M'' ⊗''R'' ''N''.
# Properties
## Uniqueness
Representations of functors are unique up to a unique isomorphism. That is, if (''A''11) and (''A''22) represent the same functor, then there exists a unique isomorphism φ : ''A''1 → ''A''2 such that :$\Phi_1^\circ\Phi_2 = \mathrm\left(\varphi,-\right)$ as natural isomorphisms from Hom(''A''2,–) to Hom(''A''1,–). This fact follows easily from Yoneda's lemma. Stated in terms of universal elements: if (''A''1,''u''1) and (''A''2,''u''2) represent the same functor, then there exists a unique isomorphism φ : ''A''1 → ''A''2 such that :$\left(F\varphi\right)u_1 = u_2.$
## Preservation of limits
Representable functors are naturally isomorphic to Hom functors and therefore share their properties. In particular, (covariant) representable functors Limit (category theory)#Preservation of limits, preserve all limits. It follows that any functor which fails to preserve some limit is not representable. Contravariant representable functors take colimits to limits.
## Left adjoint
Any functor ''K'' : ''C'' → Set with a left adjoint ''F'' : Set → ''C'' is represented by (''FX'', η''X''(•)) where ''X'' = is a singleton set and η is the unit of the adjunction. Conversely, if ''K'' is represented by a pair (''A'', ''u'') and all small copowers of ''A'' exist in ''C'' then ''K'' has a left adjoint ''F'' which sends each set ''I'' to the ''I''th copower of ''A''. Therefore, if ''C'' is a category with all small copowers, a functor ''K'' : ''C'' → Set is representable if and only if it has a left adjoint.
# Relation to universal morphisms and adjoints
The categorical notions of universal morphisms and adjoint functors can both be expressed using representable functors. Let ''G'' : ''D'' → ''C'' be a functor and let ''X'' be an object of ''C''. Then (''A'',φ) is a universal morphism from ''X'' to ''G'' if and only if (''A'',φ) is a representation of the functor Hom''C''(''X'',''G''–) from ''D'' to Set. It follows that ''G'' has a left-adjoint ''F'' if and only if Hom''C''(''X'',''G''–) is representable for all ''X'' in ''C''. The natural isomorphism Φ''X'' : Hom''D''(''FX'',–) → Hom''C''(''X'',''G''–) yields the adjointness; that is :$\Phi_\colon \mathrm_\left(FX,Y\right) \to \mathrm_\left(X,GY\right)$ is a bijection for all ''X'' and ''Y''. The dual statements are also true. Let ''F'' : ''C'' → ''D'' be a functor and let ''Y'' be an object of ''D''. Then (''A'',φ) is a universal morphism from ''F'' to ''Y'' if and only if (''A'',φ) is a representation of the functor Hom''D''(''F''–,''Y'') from ''C'' to Set. It follows that ''F'' has a right-adjoint ''G'' if and only if Hom''D''(''F''–,''Y'') is representable for all ''Y'' in ''D''.
# See also
* Subobject classifier * Density theorem (category theory), Density theorem
# References
* {{Functors Representable functors, | 2,510 | 8,808 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-21 | latest | en | 0.845824 |
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# Electrostatics
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Asked by parminder2032008 22nd December 2020 10:56 AM | 2,799 | 8,946 | {"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-2023-40 | latest | en | 0.926354 |
https://benchpartner.com/what-is-regression | 1,675,480,037,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500080.82/warc/CC-MAIN-20230204012622-20230204042622-00687.warc.gz | 139,007,367 | 15,094 | In 1889, Sir Francis Galton, a cousin of Charles Darwin published a paper on heredity, “Natural Inheritance”. He reported his discovery that sizes of seeds of sweet pea plants appeared to “revert” or “regress”, to the mean size in successive generations. He also reported results of a study of the relationship between heights of fathers and heights of their sons. A straight line was fit to the data pairs: height of father versus height of son. Here, too, he found a “regression to mediocrity” The heights of the sons represented a movement away from their
fathers, towards the average height. We credit Sir Galton with the idea of statistical regression.
While most applications of regression analysis may have little to do with the “regression to the mean” discovered by Galton, the term “regression” remains. It now refers to the statistical technique of modeling the relationship between two or more variables. In general sense, regression analysis means the estimation or prediction of the unknown value of one variable from the known value(s) of the other variable(s). It is one of the most important and widely used statistical techniques in almost all sciences - natural, social or physical.
In this lesson we will focus only on simple regression –linear regression involving only two variables: a dependent variable and an independent variable. Regression analysis for studying more than two variables at a time is known as multiple regressions.
• INDEPENDENT AND DEPENDENT VARIABLES
Simple regression involves only two variables; one variable is predicted by another variable. The variable to be predicted is called the dependent variable. The predictor is called the independent variable, or explanatory variable. For example, when we are trying to predict the demand for television sets on the basis of population growth, we are using the demand for television sets as the dependent variable and the population growth as the independent or predictor variable.
The decision, as to which variable is which sometimes, causes problems. Often the choice is obvious, as in case of demand for television sets and population growth because it would make no sense to suggest that population growth could be dependent on TV demand! The population growth has to be the independent variable and the TV demand the dependent variable.
If we are unsure, here are some points that might be of use:
• if we have control over one of the variables then that is the independent. For example, a manufacturer can decide how much to spend on advertising and expect his sales to be dependent upon how much he spends
• it there is any lapse of time between the two variables being measured, then the latter must depend upon the former, it cannot be the other way round
• if we want to predict the values of one variable from your knowledge of the other variable, the variable to be predicted must be dependent on the known one | 570 | 2,923 | {"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} | 3.71875 | 4 | CC-MAIN-2023-06 | latest | en | 0.939601 |
https://db0nus869y26v.cloudfront.net/en/Euler_pseudoprime | 1,726,251,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00101.warc.gz | 183,924,011 | 14,462 | In arithmetic, an odd composite integer n is called an Euler pseudoprime to base a, if a and n are coprime, and
${\displaystyle a^{(n-1)/2}\equiv \pm 1{\pmod {n))}$
(where mod refers to the modulo operation).
The motivation for this definition is the fact that all prime numbers p satisfy the above equation which can be deduced from Fermat's little theorem. Fermat's theorem asserts that if p is prime, and coprime to a, then ap−1 ≡ 1 (mod p). Suppose that p>2 is prime, then p can be expressed as 2q + 1 where q is an integer. Thus, a(2q+1) − 1 ≡ 1 (mod p), which means that a2q − 1 ≡ 0 (mod p). This can be factored as (aq − 1)(aq + 1) ≡ 0 (mod p), which is equivalent to a(p−1)/2 ≡ ±1 (mod p).
The equation can be tested rather quickly, which can be used for probabilistic primality testing. These tests are twice as strong as tests based on Fermat's little theorem.
Every Euler pseudoprime is also a Fermat pseudoprime. It is not possible to produce a definite test of primality based on whether a number is an Euler pseudoprime because there exist absolute Euler pseudoprimes, numbers which are Euler pseudoprimes to every base relatively prime to themselves. The absolute Euler pseudoprimes are a subset of the absolute Fermat pseudoprimes, or Carmichael numbers, and the smallest absolute Euler pseudoprime is 1729 = 7×13×19.
## Relation to Euler–Jacobi pseudoprimes
The slightly stronger condition that
${\displaystyle a^{(n-1)/2}\equiv \left({\frac {a}{n))\right){\pmod {n))}$
where n is an odd composite, the greatest common divisor of a and n equals 1, and (a/n) is the Jacobi symbol, is the more common definition of an Euler pseudoprime. See, for example, page 115 of the book by Koblitz listed below, page 90 of the book by Riesel, or page 1003 of.[1] A discussion of numbers of this form can be found at Euler–Jacobi pseudoprime. There are no absolute Euler–Jacobi pseudoprimes.[1]: p. 1004
A strong probable prime test is even stronger than the Euler-Jacobi test but takes the same computational effort. Because of this advantage over the Euler-Jacobi test, prime-testing software is often based on the strong test.
## Implementation in Lua
function EulerTest(k)
a = 2
if k == 1 then return false
elseif k == 2 then return true
else
m = modPow(a,(k-1)/2,k)
if (m == 1) or (m == k-1) then
return true
else
return false
end
end
end
## Examples
n Euler pseudoprimes to base n 1 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99, 105, 111, 115, 117, 119, 121, 123, 125, 129, 133, 135, 141, 143, 145, 147, 153, 155, 159, 161, 165, 169, 171, 175, 177, 183, 185, 187, 189, 195, 201, 203, 205, 207, 209, 213, 215, 217, 219, 221, 225, 231, 235, 237, 243, 245, 247, 249, 253, 255, 259, 261, 265, 267, 273, 275, 279, 285, 287, 289, 291, 295, 297, 299, ... (all odd composites) 2 341, 561, 1105, 1729, 1905, 2047, 2465, 3277, 4033, 4681, 5461, 6601, 8321, 8481, ... 3 121, 703, 1541, 1729, 1891, 2465, 2821, 3281, 4961, 7381, 8401, 8911, ... 4 341, 561, 645, 1105, 1387, 1729, 1905, 2047, 2465, 2701, 2821, 3277, 4033, 4369, 4371, 4681, 5461, 6601, 7957, 8321, 8481, 8911, ... 5 217, 781, 1541, 1729, 5461, 5611, 6601, 7449, 7813, ... 6 185, 217, 301, 481, 1111, 1261, 1333, 1729, 2465, 2701, 3421, 3565, 3589, 3913, 5713, 6533, 8365, ... 7 25, 325, 703, 817, 1825, 2101, 2353, 2465, 3277, 4525, 6697, 8321, ... 8 9, 21, 65, 105, 133, 273, 341, 481, 511, 561, 585, 1001, 1105, 1281, 1417, 1541, 1661, 1729, 1905, 2047, 2465, 2501, 3201, 3277, 3641, 4033, 4097, 4641, 4681, 4921, 5461, 6305, 6533, 6601, 7161, 8321, 8481, 9265, 9709, ... 9 91, 121, 671, 703, 949, 1105, 1541, 1729, 1891, 2465, 2665, 2701, 2821, 3281, 3367, 3751, 4961, 5551, 6601, 7381, 8401, 8911, ... 10 9, 33, 91, 481, 657, 1233, 1729, 2821, 2981, 4187, 5461, 6533, 6541, 6601, 7777, 8149, 8401, ... 11 133, 305, 481, 645, 793, 1729, 2047, 2257, 2465, 4577, 4921, 5041, 5185, 8113, ... 12 65, 91, 133, 145, 247, 377, 385, 1649, 1729, 2041, 2233, 2465, 2821, 3553, 6305, 8911, 9073, ... 13 21, 85, 105, 561, 1099, 1785, 2465, 5149, 5185, 7107, 8841, 8911, 9577, 9637, ... 14 15, 65, 481, 781, 793, 841, 985, 1541, 2257, 2465, 2561, 2743, 3277, 5185, 5713, 6533, 6541, 7171, 7449, 7585, 8321, 9073, ... 15 341, 1477, 1541, 1687, 1729, 1921, 3277, 6541, 9073, ... 16 15, 85, 91, 341, 435, 451, 561, 645, 703, 1105, 1247, 1271, 1387, 1581, 1695, 1729, 1891, 1905, 2047, 2071, 2465, 2701, 2821, 3133, 3277, 3367, 3683, 4033, 4369, 4371, 4681, 4795, 4859, 5461, 5551, 6601, 6643, 7957, 8321, 8481, 8695, 8911, 9061, 9131, 9211, 9605, 9919, ... 17 9, 91, 145, 781, 1111, 1305, 1729, 2149, 2821, 4033, 4187, 5365, 5833, 6697, 7171, ... 18 25, 49, 65, 133, 325, 343, 425, 1105, 1225, 1369, 1387, 1729, 1921, 2149, 2465, 2977, 4577, 5725, 5833, 5941, 6305, 6517, 6601, 7345, ... 19 9, 45, 49, 169, 343, 561, 889, 905, 1105, 1661, 1849, 2353, 2465, 2701, 3201, 4033, 4681, 5461, 5713, 6541, 6697, 7957, 8145, 8281, 8401, 9997, ... 20 21, 57, 133, 671, 889, 1281, 1653, 1729, 1891, 2059, 2413, 2761, 3201, 5461, 5473, 5713, 5833, 6601, 6817, 7999, ... 21 65, 221, 703, 793, 1045, 1105, 2465, 3781, 5185, 5473, 6541, 7363, 8965, 9061, ... 22 21, 69, 91, 105, 161, 169, 345, 485, 1183, 1247, 1541, 1729, 2041, 2047, 2413, 2465, 2821, 3241, 3801, 5551, 7665, 9453, ... 23 33, 169, 265, 341, 385, 481, 553, 1065, 1271, 1729, 2321, 2465, 2701, 2821, 3097, 4033, 4081, 4345, 4371, 4681, 5149, 6533, 6541, 7189, 7957, 8321, 8651, 8745, 8911, 9805, ... 24 25, 175, 553, 805, 949, 1541, 1729, 1825, 1975, 2413, 2465, 2701, 3781, 4537, 6931, 7501, 9085, 9361, ... 25 217, 561, 781, 1541, 1729, 1891, 2821, 4123, 5461, 5611, 5731, 6601, 7449, 7813, 8029, 8911, 9881, ... 26 9, 25, 27, 45, 133, 217, 225, 475, 561, 589, 703, 925, 1065, 2465, 3325, 3385, 3565, 3825, 4741, 4921, 5041, 5425, 6697, 8029, 9073, ... 27 65, 121, 133, 259, 341, 365, 481, 703, 1001, 1541, 1649, 1729, 1891, 2465, 2821, 2981, 2993, 3281, 4033, 4745, 4921, 4961, 5461, 6305, 6533, 7381, 7585, 8321, 8401, 8911, 9809, 9841, 9881, ... 28 9, 27, 145, 261, 361, 529, 785, 1305, 1431, 2041, 2413, 2465, 3201, 3277, 4553, 4699, 5149, 7065, 8321, 8401, 9841, ... 29 15, 21, 91, 105, 341, 469, 481, 793, 871, 1729, 1897, 2105, 2257, 2821, 4371, 4411, 5149, 5185, 5473, 5565, 6097, 7161, 8321, 8401, 8421, 8841, ... 30 49, 133, 217, 341, 403, 469, 589, 637, 871, 901, 931, 1273, 1537, 1729, 2059, 2077, 2821, 3097, 3277, 4081, 4097, 5729, 6031, 6061, 6097, 6409, 6817, 7657, 8023, 8029, 8401, 9881, ...
## Least Euler pseudoprime to base n
n Least EPSP n Least EPSP n Least EPSP n Least EPSP 1 9 33 545 65 33 97 21 2 341 34 21 66 65 98 9 3 121 35 9 67 33 99 25 4 341 36 35 68 25 100 9 5 217 37 9 69 35 101 25 6 185 38 39 70 69 102 133 7 25 39 133 71 9 103 51 8 9 40 39 72 85 104 15 9 91 41 21 73 9 105 451 10 9 42 451 74 15 106 15 11 133 43 21 75 91 107 9 12 65 44 9 76 15 108 91 13 21 45 133 77 39 109 9 14 15 46 9 78 77 110 111 15 341 47 65 79 39 111 55 16 15 48 49 80 9 112 65 17 9 49 25 81 91 113 21 18 25 50 21 82 9 114 115 19 9 51 25 83 21 115 57 20 21 52 51 84 85 116 9 21 65 53 9 85 21 117 49 22 21 54 55 86 65 118 9 23 33 55 9 87 133 119 15 24 25 56 33 88 87 120 77 25 217 57 25 89 9 121 15 26 9 58 57 90 91 122 33 27 65 59 15 91 9 123 85 28 9 60 341 92 21 124 25 29 15 61 15 93 25 125 9 30 49 62 9 94 57 126 25 31 15 63 341 95 141 127 9 32 25 64 9 96 65 128 49 | 3,848 | 7,353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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} | 4.0625 | 4 | CC-MAIN-2024-38 | latest | en | 0.864688 |
https://web2.0calc.com/questions/geometry-help-hawaii | 1,545,142,584,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829399.59/warc/CC-MAIN-20181218123521-20181218145521-00301.warc.gz | 778,487,637 | 5,945 | +0
# geometry help *hawaii*
0
140
2
Can somebody help me? (* i will identify my problems with "hawaii" from now on so I can return to them easier in the future)
Guest May 22, 2018
#1
+1
In the triangle with a height of 5.5 inches, the top RHS angle will be:
Sin(Top RHS) =5.5 / 11 =1 / 2
Top RHS angle = arcsin(1/2) = 30 degrees.
In the triangle with height h, the top angle is complementary =90 - 30 = 60 degrees.
So, this triangle is: 30, 60, 90 triangle.
Sin(30) = h / 9.9
1 / 2 = h / 9.9
h =9.9 x 1/2
h =4.95 inches.
Guest May 22, 2018
#2
+94182
0
You could return to them more easily if you were a member as all your posts would sit there fully accessable.
Melody May 23, 2018 | 251 | 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} | 3.578125 | 4 | CC-MAIN-2018-51 | longest | en | 0.893789 |
https://discourse.julialang.org/t/hobby-project-sequentialfit-jl/55717 | 1,709,516,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00044.warc.gz | 198,736,801 | 8,932 | # Hobby project: SequentialFit.jl
I don’t know mathematical optimization, and this might be a tremendously dumb idea, but I couldn’t get it out of my head after working on a project where I was fitting a surrogate model to the outputs from a very expensive simulation. So I made this thing for fun:
Not going to register it or anything, but I’m happy for feedback/discussions/links to related papers.
Basically, if fitting, taking the gradient of your model with respect to its parameters and optimizing on that gradient are all orders-of-magnitude cheaper operations than sampling, it might make sense to use your current model to inform the choice of sample points. Big drawback: It requires you to already be fairly confident in the general quality of your model, so it probably doesn’t hold up for science.
4 Likes
And a picture because those drive engagement.
2 Likes
You might already be aware, but this is pretty similar to Bayesian Optimization.
4 Likes
I’ve encountered it, and it’s similar, but the big difference is that I impose a specific model function, and bayesian optimization (from what I understand) uses GPs, which are essentially polynomials.
In my use case, the purpose of the fit is to find specific, physically meaningful parameters, like linewidth, height and position of a gaussian. Extracting them from a GP seems difficult (but I’m not convinced I understand bayesian optimization enough to say for sure that this is not just a poor implementation of it).
Your code seems different to pure BO, but the idea of fitting a surrogate and optimizing a metric of that pops up in different areas.
You could also look up “Bobyqa”.
2 Likes
There is also https://github.com/SciML/Surrogates.jl.
2 Likes
The idea of this approach looks very interesting for fitting analytical physical models to expensive simulations, or to experimental data when one can actively control the next point to measure. It seems a more direct approach than going through some surrogate model and fitting an analytical function after that.
Could you please clarify what do you mean by:
Big drawback: It requires you to already be fairly confident in the general quality of your model, so it probably doesn’t hold up for science.
?
1 Like
Sure. Like I said, I haven’t investigated this thoroughly, but unlike Bayesian optimization it doesn’t actually act on uncertainty, but rather on “If my model is (close to) correct, where should I sample to get maximum information?”. If your initial guess of parameters is a bit off, there is no proper method to explore other regions (we have exploitation but not exploration if my opti-lingo is up to date).
Maybe there’s some good change I could make to improve the algorithm in this respect – I know for certain that adding the `prod(s.X .- x)^2` factor was necessary to stop it from just sampling the same spot on what it thought were the slopes over and over.
In the example above we can see that it does a decent job, but I’m not certain I could trust it without checking against the ground truth. And I have no idea what kind of functions this would work for vs not.
Hm, I see… It relies on the general behaviour of the model function with the current parameter values, and if the reality is vastly different the point selection will be far from optimal.
Are you aware of any papers/studies of such methods? I tried googling some keywords but no success.
Nope, I was kind of hoping some would turn up here
1 Like
This paper seems to propose a similar idea. The authors implemented it in the old Boeing Design Explorer software.
@INPROCEEDINGS{jdaiaa2,
author = “{Audet, C.} and {J.E. Dennis} and {D. W. Moore} and
{A. J. Booker} and {P. D. Frank}”,
title = “A Surrogate-Model-Based Method for Constrained Optimization,
AIAA-2000-4891”,
year = “2000”,
booktitle= “Eighth AIAA/USAF/NASA/ISSMO Symposium on Multidisciplinary
Analysis and
Optimization”
}
2 Likes
@mohamed82008, on the surrogates topic, the following slides: Introduction to Gaussian Process Surrogate Models, by Nicolas Durrande and Rodolphe Le Riche (2017), provide a really good overview.
2 Likes
This is wrong btw. GPs aren’t essentially polynomials. They are very different beasts. GPs are semi-parametric models in the sense that you need all the data around to do prediction with them, not just the parameters. GPs are naturally probabilistic and Bayesian which makes them an awkward choice to “fit” deterministic functions but it will work anyways if you ignore the variance. But that property of GPs also makes them seamlessly extend to stochastic optimisation where the objective or constraints can be random in nature.
3 Likes
This does not surprise me. I haven’t found a definition that relates them to anything I understand. I think there’s a couple of steps of reading I would have to do in between. | 1,065 | 4,849 | {"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-2024-10 | latest | en | 0.943488 |
http://www.multiplagp.com.br/egd09/17b479-how-to-annualize-returns | 1,638,132,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00548.warc.gz | 121,752,464 | 5,293 | You are correct in your annualized rate of return. If Excel formulas are unfamiliar to you, you could benefit greatly from our completely free Basic Skills E-book, which teaches the basics of Excel formulas. It is 2069063%. For example, if you have a 50 percent return over five years, the annualized return is less than 10 percent because of compounding. Annualized Rate of Return comes in handy while comparing and ranking returns. With a few simple calculations, you can annualize daily return data to determine the investment's average return for the year. number of periods in a year (daily scale = 252, monthly scale = … So, if the monthly rate is 2% for all months, the annualized rate is: = (1+2%)^12 – 1 = 1.02^12-1 = 0.2682 or 26.82%. 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Information is important data, some investors also want to know the annual return rate of investment. | 2,473 | 13,287 | {"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 | 4 | CC-MAIN-2021-49 | latest | en | 0.911091 |
http://www.thelearningpoint.net/home/electrical-science-and-engineering/introduction-to-combinational-circuits--part-1 | 1,566,434,253,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00554.warc.gz | 324,851,952 | 22,472 | ### Introduction to Combinational Circuits : Part 1 - Multiplexers, Demultiplexers, Decoder, Introduction to Memories, ROM/RAM, EPROM
--------------xxxxx--------------
Digital Electronics : Combinational Circuits Part 1
### COMBINATIONAL CIRCUITS
A logic circuit whose output depends directly on inputs and no other factor is known as combinational circuit. The most important types of combinational circuit are-- RAM, ROM, Multiplexers, Encoders, Decoders, Demultiplexers.
### MULTIPLEXERS
A multiplexer is a circuit with many inputs but only one output. A multiplexer has some data inputs, control inputs and one output. Depending on the control inputs, one input from the data inputs is sent to the output.
### DEMULTIPLEXERS
Demultiplex means 1 to many. A Demultiplexer is a circuit with one data input, few control inputs and many outputs. A Demultiplexer, depending on the control inputs, transfers the input to a desired output pin. The number of outputs is given by n as n ≤ 2m, where ‘m’ is the number of control inputs.
#### DECODER
A decoder is a logic circuit that converts a binary number to its equivalent decimal number. It is much like a Demultiplexer but it does not have any data input. A decoder circuit gives a 1(HIGH) at the output corresponding to the decimal equivalent of the input. For example—
if the inputs in the above figure are 010 then a 1(HIGH) will be observed at output ‘x2’. Similarly, if the input is 101, then a 1(HIGH) will be observed at output ‘x5’ and all other outputs will be 0(LOW).
### MEMORIES
The function of a memory is to store information. A memory is designed for bulk storage of data but that is all it can do. Some types of memories can remember the data even when the power is switched off.
The ability to remember data after the power is switched off is the dividing line between the two main types of memory. If it loses its data when the power is switched off, then we call the memory RAM or volatile memory. If it can hold on to the data without power, we call it ROM or non-volatile memory.
#### Different Kinds of ROM
A masked ROM is manufactured to our specification and cannot be changed. We must be very sure that the information is correct before it is made otherwise it all goes in the waste bin. The initial cost is necessarily high due to the expense of the tooling required.
#### Erasable programmable ROM (EPROM)
As the name suggests, this chip allows us to program it, then change our mind and try again. To erase the data there are two methods – ultraviolet light or electrical voltage pulses. EPROMs are ideal for prototyping since it is so easy to change the data to make modifications.
### The chip is bombarded with ultraviolet light via a transparent window on the chip. A specially constructed EPROM eraser provides the light. We pop the chip in, close the lid and switch on the timer. After a few minutes, the data is erased. When erased, all the data output is set to 1.
#### Electrically erasable programmable ROM (EEPROM)
This chip uses electrical voltage pulses as inputs to clear the previous data and is then reprogrammed in the same way as the UVEPROM.
We end this tutorial with a general introduction to RAM.
### Here's a list of all the tutorials we currently have in this area - Introductory Digital Electronic Circuits and Boolean logic
Introduction to the Number System : Part 1 Introducing number systems. Representation of numbers in Decimal, Binary,Octal and Hexadecimal forms. Conversion from one form to the other. Number System : Part 2 Binary addition, subtraction and multiplication. Booth's multiplication algorithm. Unsigned and signed numbers. Introduction to Boolean Algebra : Part 1 Binary logic: True and false. Logical operators like OR, NOT, AND. Constructing truth tables. Basic postulates of Boolean Algebra. Logical addition, multiplication and complement rules. Principles of duality. Basic theorems of boolean algebra: idempotence, involution, complementary, commutative, associative, distributive and absorption laws. Boolean Algebra : Part 2De-morgan's laws. Logic gates. 2 input and 3 input gates. XOR, XNOR gates. Universality of NAND and NOR gates. Realization of Boolean expressions using NAND and NOR. Replacing gates in a boolean circuit with NAND and NOR. Understanding Karnaugh Maps : Part 1 Introducing Karnaugh Maps. Min-terms and Max-terms. Canonical expressions. Sum of products and product of sums forms. Shorthand notations. Expanding expressions in SOP and POS Forms ( Sum of products and Product of sums ). Minimizing boolean expressions via Algebraic methods or map based reduction techniques. Pair, quad and octet in the context of Karnaugh Maps. Karnaugh Maps : Part 2Map rolling. Overlapping and redundant groups. Examples of reducing expressions via K-Map techniques. Introduction to Combinational Circuits : Part 1Combinational circuits: for which logic is entirely dependent of inputs and nothing else. Introduction to Multiplexers, De-multiplexers, encoders and decoders.Memories: RAM and ROM. Different kinds of ROM - Masked ROM, programmable ROM. Combinational Circuits : Part 2 Static and Dynamic RAM, Memory organization. Introduction to Sequential Circuits : Part 1 Introduction to Sequential circuits. Different kinds of Flip Flops. RS, D, T, JK. Structure of flip flops. Switching example. Counters and Timers. Ripple and Synchronous Counters. Sequential Circuits : Part 2ADC or DAC Converters and conversion processes. Flash Converters, ramp generators. Successive approximation and quantization errors. | 1,272 | 5,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-35 | latest | en | 0.827197 |
https://volleydork.blog/2017/01/31/output-win-efficiency/ | 1,656,474,303,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00727.warc.gz | 664,682,926 | 44,112 | # Output Win Efficiency
Here is the extension of the FBSO eff post – but applied to all types of touches.
For every contact, we find the resulting Win Efficiency. So for Skjodt’s R+ contacts, Michigan’s FBSO eff here is around .300. Naturally this flips for Bates’ S- where -0.300 is Wisconsin’s “Point Score” efficiency on that reception to attack by Michigan. This naturally makes sense as the receiving team is typically more likely to win the point on the first possession than is the serving team.
You can glance around and see the logical output efficiences: A# is a kill by Gillis, B= is a Kieffer-Wright getting tooled, and Haggerty blows it trying to set the second ball and loses the point (E=). And it makes sense that these contacts have output eff’s of 1.00 or -1.00; the teams making these contacts will always win or always lose the point on the current possession when the touch is terminal. These are the simple ones.
The interesting ones are the non-terminal touches. Things like receptions, digs, sets, block touches, etc. What I’ve done is found the the Win Eff for each of these. With Jenna Lerg’s perfect freeball pass (F#) to start the possession, Michigan’s Win Eff is around 0.500 when attacking off a F#.. When Haggerty digs Mahlke with a D#, Wisconsin’s Win Eff off a D# is a little above 0.300.
For non-terminal attacks, I find the Output Contact. Possibilities are block touches, dig qualities, and covered balls back to the attacking team. So if we look at that very first attack by Mahlke, the output contact is the B- by Gillis. To calculate the output eff, we look at Michigan’s Win Eff when defending an opponent whose first touch is a B-, which is around 0.050.
Now is the cool part.
An inquisitive mind might argue that a perfect pass against Nebraska might be worth more than a perfect pass against Rutgers (sorry Rutgers). Or that getting a kill on a high ball after a poor pass against Rolfzen twins has more value than a high ball against Rutger’s block/defense. Moral of the story, we need to account for both sides of the equation.
We need to acknowledge that your opponent has an effect on your performance. Illinois may hit .200 on the season, but that number is expected to be higher against worse teams – and lower against the top contenders. To account for this, we average the two sides.
1. How does your team making the contact usually do in this situation against all teams
2. How well do all teams do against your opponent when all teams are in this situation
So if you usually hit .500 on quicks to your middle, but against Penn State teams usually only hit .300 on quicks to your middle – then your “expected Output Eff” would be 0.400. We account for the contact team’s strengths/weaknesses – but also account for how your opponent affects different situations as well.
We do this based on the entire season’s worth of data so that these expectations aren’t biased by single match highs or lows. So that perfect freeball by Lerg is actually the average of how well Michigan does with a F# and how well all teams do with F# against Wisconsin.
I would argue that this is a better way to look at the value of each touch as it relates to your ability to win or lose the possession. | 750 | 3,246 | {"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-2022-27 | longest | en | 0.928844 |
http://newvision24.com/2019/10/reliable-tips-for-what-does-parallel-mean-in-math-you-can-begin-to-use-today/ | 1,603,316,085,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00365.warc.gz | 75,983,573 | 49,505 | # Reliable Tips for What Does Parallel Mean in Math You Can Begin to Use Today
There are a lot of definitions that all give identical outcomes. You can choose the kinds of values along with the amount of values in every single mapping diagram. The very first step is to recognize the given data collection.
The aim of the analysis is to attempt to recognize factors which underlie the variables. If so make certain you measure performance for your data to choose if you ought to use above code VS just straight sorting. In-depth analysis of descriptive data can be very tricky essay writing help online to do, whereas it’s easy and quick to obtain some comprehension of numerical data.
Not only must you to learn math but now you’ve got to learn Greek. The sort of geometry we typically learn in school is called Euclidean geometry. Go right ahead and plug that into the original equation and see whether it works.
You should not presume that your mean is going to be one of your initial numbers. It is important because it describes the behavior of the entire set of numbers. It is also the number that is halfway into the set.
These videos will explain to you how to get the median. There are 2 possible cases to think about in locating the median. The mean of these distribution is 26.
## What Does Parallel https://royalessays.co.uk/custom-essay Mean in Math: the Ultimate Convenience!
A dashed line is utilized to indicate that the boundary isn’t part of the open half-plane. Accordingly, in Euclidean geometry, any 2 lines which are both perpendicular to a third line are parallel to one another, on account of the parallel postulate. It will be a positive number.
## The Basics of What Does Parallel Mean in Math
The range of terms that students are predicted to learn in geometry is a bit crazy. Appropriate comprehension of given situations and contexts can often supply a person who has the tools required to figure out what statistically relevant procedure to use. So it can be difficult to be aware that the evidence we collect from the model is genuinely evidence about the thing we would like to learn about.
## Who Else Wants to Learn About What Does Parallel Mean in Math?
During the last ten decades, the DC metro area has grown basically at the very same pace as the remainder of america. Next we inspect the data to discover the number that lies in the precise middle. This example illustrates the issue.
Following https://biology.missouristate.edu/ that, all the rest of the blanks are primes. Ensure you fully grasp all the significant attributes and formulas related to shapes, particularly circles and triangles. Example If we take a look at the rectangle below, we can observe it has twice the region of the blue triangle inside it as the two yellow triangles are the exact same size as the two sub-divided blue triangles.
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## The Death of What Does Parallel Mean in Math
There are lots of possible explanations for why a student’s MAP score might be lower than that which you might count on. There are a couple actions that may be utilised to discover the range in math. The student’s RIT score is also given in a RIT range, again, to take into consideration error that may be related to the test administration.
The rule of thumb claims that the range is all about four times the normal deviation. A range, as you may have guessed, refers to a cell or a array of cells. It requires a graph.
No experience is essential, though there are a few joints to solder. Unfortunately, such twists on simple concepts can arrive in quite a few unique forms. For the reason, you opt to move the gravitational acceleration value to a customized configuration section.
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So bear that in mind while you’re doing problems also keep this in mind if you’re go into advertising you might opted to advertise the typical salary rather than the median salary if you’ve got an outlier data that type of warps your data set. Adding this text doesn’t have any influence on the behaviour of the program, but nevertheless, it can be helpful for every time a individual wants to read and understand the code at a subsequent date. A double bar graph can be utilized to compare two sets of information, usually about the exact same topic.
## What to Expect From What Does Parallel Mean in Math?
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## Top What Does Parallel Mean in Math Choices
As these questions often look straightforward, it can be simple to find yourself rushing through them. If you believe you may be underpaid, knowing the typical salary in your state will be able to help you negotiate for a more reasonable paycheck. If a minumum of one of arguments can’t be converted to a number, the end result is NaN.
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Unfortunately, in addition, there are some tricky circumstances that require additional notion to specify the range working with the rules alone. Let’s look at a few examples. You can’t alter this choice after it’s been made!
Our article explains how to decide between the 2 tests and find out which one is ultimately suitable for you. There are an assortment of sheets involving finding the mean, and also locating a missing data point once the mean is provided. Excellent luck recruiting for this study! | 1,254 | 6,234 | {"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} | 3.9375 | 4 | CC-MAIN-2020-45 | latest | en | 0.93055 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH152/April_2015/Question_A_28 | 1,713,961,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00609.warc.gz | 571,865,439 | 12,062 | # Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 28
MATH152 April 2015
Other MATH152 Exams
### Question A 28
Let ${\displaystyle A}$ be a ${\displaystyle 2\times 2}$ matrix which represents a reflection across a line in ${\displaystyle \mathbb {R} ^{2}}$. Suppose that ${\displaystyle {\begin{bmatrix}2\\3\end{bmatrix}}}$ is an eigenvector with eigenvalue ${\displaystyle 1}$ and ${\displaystyle {\begin{bmatrix}\alpha \\6\end{bmatrix}}}$ is an eigenvector with eigenvalue ${\displaystyle \beta \neq 1}$. What are the values of ${\displaystyle \alpha }$ and ${\displaystyle \beta }$?
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Math Learning Centre A space to study math together. Free math graduate and undergraduate TA support. Mon - Fri: 12 pm - 5 pm in LSK 301&302 and 5 pm - 7 pm online. Private tutor We can help to | 503 | 2,007 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 20, "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} | 3.28125 | 3 | CC-MAIN-2024-18 | latest | en | 0.93204 |
https://www.reddit.com/r/science/comments/1j3clp/nasas_van_allen_probes_have_made_a_surprising_and/cbawqes | 1,454,966,682,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701154221.36/warc/CC-MAIN-20160205193914-00343-ip-10-236-182-209.ec2.internal.warc.gz | 841,589,348 | 18,565 | This is an archived post. You won't be able to vote or comment.
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[–] 6 points7 points (14 children)
sorry, this has been archived and can no longer be voted on
Relativistic mass is not real. The only physically significant mass is the rest mass.
[–] 2 points3 points (13 children)
sorry, this has been archived and can no longer be voted on
Could you expand upon this please? It is interesting.
[–] 4 points5 points (12 children)
sorry, this has been archived and can no longer be voted on
Objects with mass can never be accelerated to the speed of light. In the past, this was explained with a handwaving argument that the mass of the object tends to infinity as the velocity tends to the speed of light. However, that's not really the explanation that's given anymore. Unfortunately, many laymen, pop-sci enthusiasts still think it's true. The fact of the matter is that this "relativistic mass" doesn't actually act like mass. It doesn't contribute anything physically. It's not real. The only mass that means anything physically is the rest mass of the object, which is invariant under Lorentz transformations (in other words it doesn't change with speed). So there's no sense in talking about relativistic mass because it's not really physically meaningful.
It's kind of a pet peeve of mine when people talk about relativistic mass. The better way of explaining why a massive object can't be accelerated to the speed of light is this:
It takes infinite energy to accelerate a massive object to the speed of light. That is true. In that case, it is physically meaningful. No matter how much (real, physical) kinetic energy I give a chunk of mass, its velocity will never reach the speed of light.
[–] 3 points4 points (0 children)
sorry, this has been archived and can no longer be voted on
I too was under the impression that mass increased as an object accelerated towards the speed of light. Thanks for the clarification.
[–] 1 point2 points (10 children)
sorry, this has been archived and can no longer be voted on
But, if it's mass doesn't increase. What's stopping you from dumping an almost infinite amount of energy on it?
[–] 4 points5 points (9 children)
sorry, this has been archived and can no longer be voted on
You can dump and almost infinite amount of energy into it. And you'll almost make it reach the speed of light.
But since you can never quite get infinite energy, you can never quite get to the speed of light.
[–] 0 points1 point (1 child)
sorry, this has been archived and can no longer be voted on
Why does a photon get to that speed then?
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Photons have no mass. | 644 | 2,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-07 | latest | en | 0.972902 |
https://number.academy/9951 | 1,652,762,062,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00564.warc.gz | 505,818,920 | 13,331 | # Number 9951
Number 9,951 spell 🔊, write in words: nine thousand, nine hundred and fifty-one . Ordinal number 9951th is said 🔊 and write: nine thousand, nine hundred and fifty-first. The meaning of number 9951 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 9951. What is 9951 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 9951.
## What is 9,951 in other units
The decimal (Arabic) number 9951 converted to a Roman number is (IX)CMLI. Roman and decimal number conversions.
The number 9951 converted to a Mayan number is Decimal and Mayan number conversions.
#### Weight conversion
9951 kilograms (kg) = 21938.0 pounds (lbs)
9951 pounds (lbs) = 4513.7 kilograms (kg)
#### Length conversion
9951 kilometers (km) equals to 6184 miles (mi).
9951 miles (mi) equals to 16015 kilometers (km).
9951 meters (m) equals to 32648 feet (ft).
9951 feet (ft) equals 3034 meters (m).
9951 centimeters (cm) equals to 3917.7 inches (in).
9951 inches (in) equals to 25275.5 centimeters (cm).
#### Temperature conversion
9951° Fahrenheit (°F) equals to 5510.6° Celsius (°C)
9951° Celsius (°C) equals to 17943.8° Fahrenheit (°F)
#### Power conversion
9951 Horsepower (hp) equals to 7317.95 kilowatts (kW)
9951 kilowatts (kW) equals to 13531.43 horsepower (hp)
#### Time conversion
(hours, minutes, seconds, days, weeks)
9951 seconds equals to 2 hours, 45 minutes, 51 seconds
9951 minutes equals to 6 days, 21 hours, 51 minutes
### Codes and images of the number 9951
Number 9951 morse code: ----. ----. ..... .----
Sign language for number 9951:
Number 9951 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 9951
### Multiplications
#### Multiplication table of 9951
9951 multiplied by two equals 19902 (9951 x 2 = 19902).
9951 multiplied by three equals 29853 (9951 x 3 = 29853).
9951 multiplied by four equals 39804 (9951 x 4 = 39804).
9951 multiplied by five equals 49755 (9951 x 5 = 49755).
9951 multiplied by six equals 59706 (9951 x 6 = 59706).
9951 multiplied by seven equals 69657 (9951 x 7 = 69657).
9951 multiplied by eight equals 79608 (9951 x 8 = 79608).
9951 multiplied by nine equals 89559 (9951 x 9 = 89559).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 9951
Half of 9951 is 4975,5 (9951 / 2 = 4975,5 = 4975 1/2).
One third of 9951 is 3317 (9951 / 3 = 3317).
One quarter of 9951 is 2487,75 (9951 / 4 = 2487,75 = 2487 3/4).
One fifth of 9951 is 1990,2 (9951 / 5 = 1990,2 = 1990 1/5).
One sixth of 9951 is 1658,5 (9951 / 6 = 1658,5 = 1658 1/2).
One seventh of 9951 is 1421,5714 (9951 / 7 = 1421,5714 = 1421 4/7).
One eighth of 9951 is 1243,875 (9951 / 8 = 1243,875 = 1243 7/8).
One ninth of 9951 is 1105,6667 (9951 / 9 = 1105,6667 = 1105 2/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
9951
#### Is Prime?
The number 9951 is not a prime number. The closest prime numbers are 9949, 9967.
9951th prime number in order is 104179.
#### Factorization and factors (dividers)
The prime factors of 9951 are 3 * 31 * 107
The factors of 9951 are 1 , 3 , 31 , 93 , 107 , 321 , 3317 , 9951
Total factors 8.
Sum of factors 13824 (3873).
#### Powers
The second power of 99512 is 99.022.401.
The third power of 99513 is 985.371.912.351.
#### Roots
The square root √9951 is 99,754699.
The cube root of 39951 is 21,5091.
#### Logarithms
The natural logarithm of No. ln 9951 = loge 9951 = 9,205428.
The logarithm to base 10 of No. log10 9951 = 3,997867.
The Napierian logarithm of No. log1/e 9951 = -9,205428.
### Trigonometric functions
The cosine of 9951 is 0,00527.
The sine of 9951 is -0,999986.
The tangent of 9951 is -189,76041.
### Properties of the number 9951
More math properties ...
## Number 9951 in Computer Science
Code typeCode value
PIN 9951 It's recommendable to use 9951 as a password or PIN.
9951 Number of bytes9.7KB
Unix timeUnix time 9951 is equal to Thursday Jan. 1, 1970, 2:45:51 a.m. GMT
IPv4, IPv6Number 9951 internet address in dotted format v4 0.0.38.223, v6 ::26df
9951 Decimal = 10011011011111 Binary
9951 Decimal = 111122120 Ternary
9951 Decimal = 23337 Octal
9951 Decimal = 26DF Hexadecimal (0x26df hex)
9951 BASE64OTk1MQ==
9951 MD5fb6c84779f12283a81d739d8f088fc12
9951 SHA224a4eb05bf51853a737a2dc5d588a78b3b3ee5f3955af596fec572cb16
9951 SHA256821196656a86a779670da967d7e9ffcd10d11ab1f2a93f5c69c4192b9ccf9596
More SHA codes related to the number 9951 ...
If you know something interesting about the 9951 number that you did not find on this page, do not hesitate to write us here.
## Numerology 9951
### The meaning of the number 9 (nine), numerology 9
Character frequency 9: 2
The number 9 (nine) is the sign of ideals, Universal interest and the spirit of combat for humanitarian purposes. It symbolizes the inner Light, prioritizing ideals and dreams, experienced through emotions and intuition. It represents the ascension to a higher degree of consciousness and the ability to display love for others. He/she is creative, idealistic, original and caring.
More about the meaning of the number 9 (nine), numerology 9 ...
### The meaning of the number 5 (five), numerology 5
Character frequency 5: 1
The number five (5) came to this world to achieve freedom. You need to apply discipline to find your inner freedom and open-mindedness. It is about a restless spirit in constant search of the truth that surrounds us. You need to accumulate as much information as possible to know what is happening in depth. Number 5 person is intelligent, selfish, curious and with great artistic ability. It is a symbol of freedom, independence, change, adaptation, movement, the search for new experiences, the traveling and adventurous spirit, but also of inconsistency and abuse of the senses.
More about the meaning of the number 5 (five), numerology 5 ...
### The meaning of the number 1 (one), numerology 1
Character frequency 1: 1
Number one (1) came to develop or balance creativity, independence, originality, self-reliance and confidence in the world. It reflects power, creative strength, quick mind, drive and ambition. It is the sign of individualistic and aggressive nature.
More about the meaning of the number 1 (one), numerology 1 ...
## Interesting facts about the number 9951
### Asteroids
• (9951) Tyrannosaurus is asteroid number 9951. It was discovered by E. W. Elst from La Silla Observatory on 11/15/1990.
### Distances between cities
• There is a 9,951 miles (16,014 km) direct distance between Álvaro Obregón (Mexico) and Vijayawāda (India).
• There is a 6,184 miles (9,951 km) direct distance between City of London (United Kingdom) and Trujillo (Peru).
• There is a 6,184 miles (9,951 km) direct distance between Damascus (Syria) and Charlotte (USA).
• There is a 6,184 miles (9,951 km) direct distance between Fès (Morocco) and Tangshan (China).
• There is a 6,184 miles (9,951 km) direct distance between Harare (Zimbabwe) and Santiago (Chile).
• There is a 6,184 miles (9,951 km) direct distance between Houston (USA) and Odessa (Ukraine).
• There is a 9,951 miles (16,013 km) direct distance between Incheon (South Korea) and São Luís (Brazil).
• There is a 9,951 miles (16,014 km) direct distance between Iztapalapa (Mexico) and Vijayawāda (India).
• There is a 6,184 miles (9,951 km) direct distance between Pietermaritzburg (South Africa) and Saint Petersburg (Russia).
• There is a 6,184 miles (9,951 km) direct distance between Rio de Janeiro (Brazil) and Tijuana (Mexico).
### Mathematics
• 9951 is the number of ways to color the vertices of a triangle with 31 colors, up to rotation.
## Number 9,951 in other languages
How to say or write the number nine thousand, nine hundred and fifty-one in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 9.951) nueve mil novecientos cincuenta y uno German: 🔊 (Anzahl 9.951) neuntausendneunhunderteinundfünfzig French: 🔊 (nombre 9 951) neuf mille neuf cent cinquante et un Portuguese: 🔊 (número 9 951) nove mil, novecentos e cinquenta e um Chinese: 🔊 (数 9 951) 九千九百五十一 Arabian: 🔊 (عدد 9,951) تسعة آلاف و تسعمائة و واحد و خمسون Czech: 🔊 (číslo 9 951) devět tisíc devětset padesát jedna Korean: 🔊 (번호 9,951) 구천구백오십일 Danish: 🔊 (nummer 9 951) nitusinde og nihundrede og enoghalvtreds Hebrew: (מספר 9,951) תשע אלף תשע מאות חמישים ואחד Dutch: 🔊 (nummer 9 951) negenduizendnegenhonderdeenenvijftig Japanese: 🔊 (数 9,951) 九千九百五十一 Indonesian: 🔊 (jumlah 9.951) sembilan ribu sembilan ratus lima puluh satu Italian: 🔊 (numero 9 951) novemilanovecentocinquantuno Norwegian: 🔊 (nummer 9 951) ni tusen, ni hundre og femti-en Polish: 🔊 (liczba 9 951) dziewięć tysięcy dziewięćset pięćdziesiąt jeden Russian: 🔊 (номер 9 951) девять тысяч девятьсот пятьдесят один Turkish: 🔊 (numara 9,951) dokuzbindokuzyüzellibir Thai: 🔊 (จำนวน 9 951) เก้าพันเก้าร้อยห้าสิบเอ็ด Ukrainian: 🔊 (номер 9 951) дев'ять тисяч дев'ятсот п'ятдесят одна Vietnamese: 🔊 (con số 9.951) chín nghìn chín trăm năm mươi mốt Other languages ...
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If you know something interesting about the number 9951 or any natural number (positive integer) please write us here or on facebook. | 3,022 | 9,369 | {"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-2022-21 | latest | en | 0.658373 |
http://easy-ciphers.com/mostrom | 1,569,237,580,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576355.92/warc/CC-MAIN-20190923105314-20190923131314-00415.warc.gz | 56,950,369 | 18,990 | 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: mostrom
cipher variations: nptuspn oquvtqo prvwurp qswxvsq rtxywtr suyzxus tvzayvt uwabzwu vxbcaxv wycdbyw xzdeczx yaefday zbfgebz acghfca bdhigdb ceijhec dfjkifd egkljge fhlmkhf gimnlig hjnomjh ikopnki jlpqolj kmqrpmk lnrsqnl
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: mostrom Cipher: nlhgiln
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: mostrom Cipher: ABABB ABBAB BAAAB BAABA BAAAA ABBAB ABABB
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: mostrom
cipher variations:
nptuspn
lrdgarl
jtnsitj
hvxeqvh
fxhqyxf
dzrcgzd
zdlawdz
xfvmefx
vhfymhv
tjpkujt
rlzwclr
pnjiknp
oquvtqo
msehbsm
kuotjuk
iwyfrwi
gyirzyg
easdhae
aembxea
ygwnfgy
wigzniw
ukqlvku
smaxdms
qokjloq
prvwurp
ntfictn
lvpukvl
jxzgsxj
hzjsazh
fbteibf
bfncyfb
zhxoghz
xjhaojx
vlrmwlv
tnbyent
rplkmpr
qswxvsq
ougjduo
mwqvlwm
kyahtyk
iaktbai
gcufjcg
cgodzgc
aiyphia
ykibpky
wmsnxmw
uoczfou
sqmlnqs
rtxywtr
pvhkevp
nxrwmxn
lzbiuzl
jblucbj
hdvgkdh
dhpeahd
bjzqijb
zljcqlz
xntoynx
vpdagpv
trnmort
suyzxus
qwilfwq
oysxnyo
macjvam
kcmvdck
iewhlei
eiqfbie
ckarjkc
amkdrma
youpzoy
wqebhqw
usonpsu
tvzayvt
rxjmgxr
pztyozp
nbdkwbn
ldnwedl
jfximfj
fjrgcjf
dlbskld
bnlesnb
zpvqapz
xrfcirx
vtpoqtv
uwabzwu
syknhys
qauzpaq
ocelxco
meoxfem
kgyjngk
gkshdkg
emctlme
comftoc
aqwrbqa
ysgdjsy
wuqpruw
vxbcaxv
tzloizt
rbvaqbr
pdfmydp
nfpygfn
lhzkohl
hltielh
fndumnf
dpngupd
brxscrb
zthektz
xvrqsvx
wycdbyw
uampjau
scwbrcs
qegnzeq
ogqzhgo
mialpim
imujfmi
goevnog
eqohvqe
csytdsc
auiflua
ywsrtwy
xzdeczx
vbnqkbv
tdxcsdt
rfhoafr
phraihp
njbmqjn
jnvkgnj
hpfwoph
frpiwrf
dtzuetd
bvjgmvb
zxtsuxz
yaefday
wcorlcw
ueydteu
sgipbgs
qisbjiq
okcnrko
kowlhok
iqgxpqi
gsqjxsg
euavfue
cwkhnwc
ayutvya
zbfgebz
xdpsmdx
vfzeufv
thjqcht
rjtckjr
pldoslp
lpxmipl
jrhyqrj
htrkyth
fvbwgvf
dxlioxd
bzvuwzb
acghfca
yeqtney
wgafvgw
uikrdiu
skudlks
qmeptmq
mqynjqm
ksizrsk
iuslzui
gwcxhwg
eymjpye
cawvxac
bdhigdb
zfruofz
xhbgwhx
vjlsejv
tlvemlt
rnfqunr
nrzokrn
ltjastl
jvtmavj
hxdyixh
fznkqzf
dbxwybd
ceijhec
agsvpga
yichxiy
wkmtfkw
umwfnmu
sogrvos
osaplso
mukbtum
kwunbwk
iyezjyi
gaolrag
ecyxzce
dfjkifd
bhtwqhb
zjdiyjz
xlnuglx
vnxgonv
tphswpt
ptbqmtp
nvlcuvn
lxvocxl
jzfakzj
hbpmsbh
egkljge
ciuxric
akejzka
ymovhmy
woyhpow
uqitxqu
qucrnuq
owmdvwo
mywpdym
kagblak
icqntci
geazbeg
fhlmkhf
djvysjd
blfkalb
znpwinz
xpziqpx
vrjuyrv
rvdsovr
pxnewxp
nzxqezn
lbhcmbl
jdroudj
hfbacfh
gimnlig
ekwztke
cmglbmc
aoqxjoa
yqajrqy
wskvzsw
swetpws
qyofxyq
oayrfao
mcidncm
kespvek
igcbdgi
hjnomjh
flxaulf
dnhmcnd
bprykpb
zrbksrz
xtlwatx
txfuqxt
rzpgyzr
pbzsgbp
ndjeodn
lftqwfl
jhdcehj
ikopnki
gmybvmg
eoindoe
cqszlqc
ascltsa
uygvryu
saqhzas
qcathcq
oekfpeo
mgurxgm
kiedfik
jlpqolj
hnzcwnh
fpjoepf
drtamrd
btdmutb
zvnycvz
vzhwszv
tbriabt
rdbuidr
pflgqfp
nhvsyhn
ljfegjl
kmqrpmk
gqkpfqg
esubnse
cuenvuc
awozdwa
waixtaw
ucsjbcu
secvjes
qgmhrgq
oiwtzio
mkgfhkm
lnrsqnl
jpbeypj
hrlqgrh
ftvcotf
dvfowvd
bxpaexb
xbjyubx
vdtkcdv
tfdwkft
rhnishr
pjxuajp
nlhgiln
mostrom
kqcfzqk
ismrhsi
guwdpug
ewgpxwe
cyqbfyc
yckzvcy
weuldew
ugexlgu
siojtis
qkyvbkq
omihjmo
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: mostrom Cipher: zbfgebz
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: mostrom Cipher: 23433444244323
Extended Methods:
Method #1
Plaintext: mostrom
method variations: rtxywtr wycdbyw bdhigdb ginomig
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```m o s t r o m
2 4 3 4 2 4 2
3 3 4 4 4 3 3 ```
They are then read out in rows:
24342423344433
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: mostrom Cipher: rsrmstn
Method #3
Plaintext: mostrom
method variations: sntithh ntithhs tithhsn ithhsnt thhsnti hhsntit hsntith
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: mostrom
all 5040 cipher variations:
mostrom
mostrmo
mostorm
mostomr
mostmor
mostmro
mosrtom
mosrtmo
mosrotm
mosromt
mosrmot
mosrmto
mosortm
mosormt
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mosotmr
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motsomr
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motorsm
motorms
motosrm
motosmr
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Read more ...[1] , [2] , [3] | 24,866 | 49,509 | {"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} | 3.21875 | 3 | CC-MAIN-2019-39 | latest | en | 0.82879 |
https://www.jiskha.com/display.cgi?id=1307403615 | 1,503,115,226,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105297.38/warc/CC-MAIN-20170819031734-20170819051734-00483.warc.gz | 935,148,095 | 3,622 | # math
posted by .
Find the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if
BY = 3, YC = 2
• math-figure missing -
If you do not show the figure (with a separate link), a description will be required to solve the problem.
## Similar Questions
1. ### Geometry
Given: X, Y, and Z are the midpoints of the sides of triangle ABC. Find rhe ratio of the area of triangle XYZ to the area of triangle ABC
2. ### Geometry
What is the ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if XY is similar to AC?
3. ### Geometry
What is he ratio of the area of triangle XBY to the area of triangle ABC for the given measurements, if XY is similar to AC, and BY=3, YC=2 ?
4. ### Geometry
What is the ratio of the area of triangle XBY to the area of the triangle ABC for the given measurements, if XY is similar to AC, and XY=2 and AC=3
5. ### Geometry
What is the ratio of the area of triangle XBY to the area of triangle ABC for the given measuremnts, if XYis similar to AC, and BY=2 and BC=4?
6. ### math
What is the ratio of the area of triangle XBY to the area of triangle ABC for the given measuremnts, if XYis similar to AC, and BY=2 and BC=4?
7. ### college geometry
Given: right triangle ABC with right angle at C, AC=22 and BC=6.Draw altitude CD where D is o hypotenuse AB. What is the ratio of the area of triangle ADC to the area of triangle CDB?
8. ### geometry
Open that link please please help me with my assignments 1. in the figure , the areas of traingle cef, triangle abe, triangle adf are 3,4, and 5 respectively. find the area of triangle aef 2. equialateral triangle abc has an area of …
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B . . . . . . . . . . . . . . C A Triangle ABC is right angled at A. The area of triangle is = ½bh. The area of the triangle is 3/2x^2 + 10x + 16cm^2 where x is a positive integer. The area of the triangle ABC can be written in the …
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AD is the median of triangle ABC and G divides AD in the ratio 2:1 .Prove that area (triangle AGB )= area (triangle BGC )=area (triangle AGC )=1÷3area (triangle ABC )
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Sum matching columns and rows.
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https://www.arxiv-vanity.com/papers/2001.00181/ | 1,606,570,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00109.warc.gz | 585,322,753 | 63,381 | arXiv Vanity renders academic papers from arXiv as responsive web pages so you don’t have to squint at a PDF. Read this paper on arXiv.org.
# Non-Schur-positivity of chromatic symmetric functions
David G.L. Wang School of Mathematics and Statistics, Beijing Institute of Technology, 102488 Beijing, P. R. China
Beijing Key Laboratory on MCAACI, Beijing Institute of Technology, 102488 Beijing, P. R. China
;
and Monica M.Y. Wang School of Mathematics and Statistics, Beijing Institute of Technology, 102488 Beijing, P. R. China
###### Abstract.
We provide a formula for every Schur coefficient in the chromatic symmetric function of a graph in terms of special rim hook tabloids. As applications, we establish the non-Schur-positivity of some graph families. These graph families include the windmill graphs, non-balanced bipartite graphs, complete bipartite graphs, complete tripartite graphs, and the wheel graphs, when the number of vertices are not too small. We also show that the Dynkin graphs of type and type are not -positive for .
###### Key words and phrases:
chromatic symmetric function, -positivity, Schur positivity, Young tableau
###### 2010 Mathematics Subject Classification:
05E05 05A17 05A15 05C70
This paper was supported by General Program of National Natural Science Foundation of China (Grant No. 11671037).
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1cm
## 1. Introduction
Stanley [23] introduced the chromatic symmetric function for a simple graph as
XG=XG(x1,x2,…)=∑κ∏v∈V(G)xκ(v)
where the sum is over all proper colorings . For example, the chromatic symmetric function of the complete graph is , where is the th elementary symmetric function. It is a generalization of the chromatic polynomial in the sense that . Shareshian and Wachs [22] refined the concept of chromatic symmetric functions, which was called the chromatic quasisymmetric function of , by considering all acyclic orientations of and introducing another parameter to track the number of directed edges for which . Ellzey [7, 8] further generalized these functions by allowing orientations with directed cycles; see also Alexandersson and Panova [1] for the same generalization from another perspective.
Let be the vector space of symmetric functions of degree . The most important basis for , considering its ubiquitousness in representation theory, mathematical physics and other areas, is the Schur symmetric functions, which are crucial in understanding the representation theory of the symmetric group; see Macdonald [15, 16]. In fact, every irreducible homogeneous polynomial representation of the general linear group is given by
char(ϕ)(x)=sλ(x1,x2,…,xn)
for some partition of , where is a Schur polynomial. Let be the symmetric group of order . Then for any -module ,
M=⨁λ⊢n(Sλ)⨁cλ⟺chM(x)=∑λ⊢ncλsλ(x),
where are irreducible -modules and is the Frobenius characteristic of ; see Stanley [25] and Sagan [21].
For any symmetric function basis , the graph is said to be -positive if the expansion of in has nonnegative coefficients. Note that -positive graphs are Schur-positive,
since the coefficient of in the Schur expansion of is the Kostka number , where and are partitions of , and is the conjugate of ; see Mendes and Remmel [18, Theorem 2.22]. The study of Schur-positivity of chromatic symmetric functions is an active area due to its connections to the representation theory of the symmetric group and that of the general linear group. For instance, see Gasharov [12] and recent work of Pawlowski [19].
Many graph classes have been shown -positive, including complete graphs, paths, cycles, co-triangle-free graphs, generalized bull graphs, (claw, )-free graphs, (claw, paw)-free graphs, (claw, co-paw)-free graphs, (claw, triangle)-free graphs, (claw, co-)-free graphs, -chains, lollipop graphs, triangular ladders; see [23, 2, 10, 26, 14, 3, 6]. Schur-positive graphs include the incomparability graphs of (3+1)-free posets, the incomparability graph of the natural unit interval order, and the 2-edge-colorable hyperforests; see [13, 22, 19]. Non--positive graphs include the saltire graphs and triangular tower graphs; see [4, 5]. We have not noticed any work concentrated on non-Schur-positive graphs yet.
To show the non--positivity of a connected -vertex graph , Wolfgang [28, Proposition 1.3.3] provided the following powerful tool.
###### Proposition 1.1 (Wolfgang).
Suppose that is a connected -vertex -positive graph. If has a connected partition of type , then has a connected partition of type for every partition which is a refinement of .
For the non-Schur-positivity, Stanley [24, Proposition 1.5] pointed the following tool.
###### Proposition 1.2 (Stanley).
Every Schur-positive graph having a stable partition of type has a stable partition of type for all partitions dominated by .
The smallest non-Schur-positive graph is the claw, whose chromatic symmetric function is
Xclaw=s31−s22+5s212+8s14.
All other -vertex connected graph are Schur positive.
From Fig. 1.1, we see that any non-Schur-positive graph is not -positive. In this paper, we provide a combinatorial way to compute the coefficient of for any partition in the Schur expansion of a chromatic symmetric function. Our formula involves special rim hook tableaux and stable partitions of the graph whose type is same to the lengths of rim hooks; see Theorem 3.1. It is often quite easy to figure out each factor appearing in the expression of for graphs with specific structures, and to obtain the non-Schur-positivity immediately.
As applications, we establish the non-Schur-positivity of windmill graphs, non-balanced bipartite graphs, complete bipartite graphs, complete tripartite graphs, and wheel graphs, when the number of vertices in these graphs are not too small. The non--positivity of some of these graphs were confirmed before, by Dahlberg, She and van Willigenburg [5].
We also consider the -positivity of Dynkin graphs of type and type by computing the generating function of the chromatic symmetric functions of these graphs. At last we conjecture that these kinds of graphs are Schur-positive.
We check all results in this paper by using Russell’s program [20].
## 2. Preliminaries
A composition of is a list of integers that sum to . An integer partition of is a composition of in non-increasing order, denoted . It is written as alternatively, where is the multiplicity of in . Denote
λ!=∏i≥1mi!.
A stable partition of a graph is a set partition of the vertex set such that every set is stable. The type of such a stable partition is the integer partition of , where is the number of sets of vertices.
Besides the elementary symmetric functions and Schur symmetric functions , standard bases for the ring include the monomial symmetric functions , the homogeneous symmetric functions , and the power symmetric functions . In terms of stable partitions, monomial and power symmetric functions, Stanley [23] obtained Proposition 2.1 as a basic way to compute the chromatic symmetric function of a graph.
###### Proposition 2.1 (Stanley).
The chromatic symmetric function of a graph is
XG=∑π(π1!π2!⋯)mμ(π)=∑S⊆E(G)(−1)\absSpλ(S),
where runs over stable partitions of , is the number of stable sets of vertices in , and is the type of ; is the partition of whose parts are the number of vertices of component of the graph .
As examples, Stanley [23, Propositions 5.3 and 5.4] presented the generating functions of the chromatic symmetric functions of paths and cycles, from which one obtains the -positivities of paths and cycles.
###### Theorem 2.2 (Stanley).
The generating functions of the paths and cycles are respectively
∑n≥0XPnzn=E(z)E(z)−zE′(z)and∑n≥2XCnzn=z2E′′(z)E(z)−zE′(z),
where
E(z)=∑n≥0enzn
is the generating function of the elementary symmetric functions .
In order to state our main result Theorem 3.1, we need more notions on Young tableaux. We follow terminologies from [18] and use the French notation; see Fulton [11] for more information on Young tableaux.
The Young diagram of is the collection of left-justified rows of cells in the th row reading from bottom to top. A rim hook is a sequence of connected cells in a Young diagram which starts from a cell on the northeast boundary and travels along the northeast edge such that its removal leaves the Young diagram a smaller integer partition. For any composition of , a rim hook tableau of shape and content is a filling of the cells of the Young diagram of with rim hooks of lengths , labeled with , such that the removal of the last rim hooks leaves the Young diagram of a smaller integer partition for all . A rim hook tabloid is a rim hook tableau with all rim hook labels removed. A special rim hook tabloid is a rim hook tabloid such that every rim hook intersects the first column of the tabloid. Let be the set of special rim hook tabloids of shape ; see Fig. 2.1 for illustration.
For any , denote by the length list of its rim hooks from up to bottom, and by the set of row labels of from up to bottom such that the rim hook with an end in Row and Column spans an even number of rows. Since any composition of determines at most one tabloid such that , one may identify a tabloid in by clarifying .
For any symmetric function , we use the notation to denote the coefficient of in the -expansion of , and use the notation to denote the coefficient of in the Schur expansion of .
## 3. Non-Schur-positivity of some graphs
Here is our first main result.
###### Theorem 3.1.
For any graph and any partition ,
[sλ]XG=∑T∈T(λ)(−1)\absW(T)⋅π(T)!⋅NG(T),
where is the set of special rim hook tabloids of shape ; is the number of rim hooks that span an even number of rows; is the integer partition of whose parts are the rim hook lengths of ; and is the number of stable partitions of whose type is .
###### Proof.
Recall from [18, Exercise 2.15] that the coefficient of in the monomial symmetric function equals the inverse Kostka number
K−1μ,λ=∑T∈T(λ,μ)∏H(−1)r(H)−1,
where is the set of special rim hook tabloids of shape and content , runs over rim hooks of , and is the number of rows that spans. This result simplifies to
[sλ]mμ=∑T∈T(λ,μ)(−1)\absW(T),
in virtue of the definition of . Now, by Proposition 2.1, we can deduce that
[sλ]XG =[sλ]∑π(π1!π2!⋯)mμ(π) =∑π(π1!π2!⋯)∑T∈T\brkλ,1π12π2⋯(−1)\absW(T) =∑T∈T(λ)(−1)\absW(T)⋅π(T)!⋅NG(T),
where runs over stable partitions of , and is the number of -vertex stable sets in . This completes the proof. ∎
In the remaining of Section 3, we show the non-Schur-positivity for graphs in some popular graph families. The general idea is to discover a particular shape and show that the sum in Theorem 3.1 for this is negative. For the purpose of showing the negativity of a term, we often ignore the value of the factor and use the fact of the positivity of only.
When the graph under consideration is clear from context, we use the notation to denote the coefficient , for the notation is widely used to denote the coefficient . In applications of Theorem 3.1, we often shorten the notation , and , when no confusion arises, as , and , respectively. Lemma 3.2 will be useful in the sequel.
###### Lemma 3.2.
Let be a graph and let be a partition of . For any tabloid with , we have the following.
• Every part in is less than or equal to the independent number .
• The length of is less than or equal to the length of .
• If every stable partition of contains a singleton stable set, then some part in equals .
###### Proof.
Let with . Since every stable set of contains at most vertices, no rim hook in is longer than . Since every rim hook intersects the first column of the Young diagram of , we find that is not longer than . The last statement holds true since the type of a stable partition of is the content of . ∎
The windmill graph is obtained by joining copies of the complete graph at a shared common vertex. Figure 3.1 illustrates the graph .
It is clear that . Note that and are -positive. Dahlberg et al. [5, Example 40] presented that the star is not Schur-positive for . Stanley [23, Corollary 3.6] proved that is -positive by showing that any graph having a bipartition consisting of two cliques is -positive. For the remaining combinations of the integer pair , the graph is not Schur-positive; see Theorem 3.3, which implies Dahlberg et al.’s result [5, Example 36] that the graph is not -positive.
###### Theorem 3.3.
For , the windmill graph is not Schur-positive.
###### Proof.
Let . Consider the partition
λ=dn−2(d−1)2⊢(dn−d+1).
Let and . By Lemma 3.2, for all , and there exists some index such that . It follows that
π(T)=(d,1,d,d,…,d)andW={1}.
By Theorem 3.1, we obtain , where is the number of stable partitions of of type . ∎
A partition of a graph is a set partition of . We say that has size and call the integer partition obtained by rearranging the numbers in the sequence
\brk1\absV1,\absV2,…,\absVk
the type of . In particular, a size- partition of is a bipartition, and a size- partition of is a tripartition. A partition of is balanced if
max\brk[c]1\absVj:1≤j≤k≤min\brk[c]1\absVj:1≤j≤k+1.
### 3.1. Non-Schur-positivity of connected bipartite graphs
Dahlberg et al. [5, Theorem 39] proved that any bipartite -vertex graph with a vertex of degree greater than is not Schur-positive by using Proposition 1.2. For connected bipartite graphs, Proposition 1.2 implies the following stronger result.
###### Theorem 3.4.
Any Schur-positive connected bipartite graph has a balanced stable bipartition, namely, the cardinalities of the two parts differ by or .
###### Proof.
Let be a connected bipartite graph. Since is connected, the bipartition is unique. Assume that . By Proposition 1.2, the graph has a bipartition such that and . It is clear that , contradicting to the uniqueness of . ∎
We remark that Theorem 3.4 can be shown alternatively by using Theorem 3.1 and considering the Schur coefficient with respect to the partition , where is the number of vertices of the graph and the number of vertices in the small part.
Theorem 3.4 implies immediately that every spider with at least 3 legs of odd length is not Schur-positive, while the non--positivity appeared in Dahlberg et al. [5, Corollary 16]. In fact, spider graphs are bipartite. The cardinality difference between the parts of a spider graph is one less than the number of legs of odd length. As a consequence, every spider with at least 3 legs of odd length is unbalanced and not Schur-positive.
The converse of Theorem 3.4 is not true. This can be seen from Theorem 3.5.
###### Theorem 3.5.
The complete bipartite graphs , , , and are -positive, and all the other complete bipartite graphs are not Schur-positive. In particular, any star of at least vertices is not Schur-positive.
###### Proof.
By Theorem 2.2, the paths and , and the cycle are -positive. The complete bipartite graph is -positive since
XK2,3=e221+9e41+e32+35e5.
Assume that is Schur-positive for some integers . By Theorem 3.4, we can suppose that . Consider the partition
λ=(m−1,n−1,2).
When , the only special rim hook tabloid with even satisfies
π=(n,n),W={1,2},π!=2,andN=1.
See Fig. 3.2.
On the other hand, one of the other tabloids satisfies
π=(n,1,n−1),W={1},andN=2n.
By Theorem 3.1, we obtain .
When , the set consists of the following tabloids; see Fig. 3.3.
• , , , and .
• , , and .
• , , and .
• , , , and .
By Theorem 3.1, we obtain
c′λ≤(n+12)−2(n+1)−(n2)+1=−n−1<0.
This completes the proof. ∎
### 3.2. Non-Schur-positive tripartite graphs
A graph is said to be tripartite if the vertex set has a partition such that no edge of links two vertices in the same .
Note that stable tripartitions of a tripartite graph might be of different types.
###### Theorem 3.6.
Let be a Schur-positive tripartite graph. If all stable tripartitions of have the same type, then any stable tripartition of is balanced.
###### Proof.
Immediate from Proposition 1.2. ∎
A fan graph is defined to be the graph join of the complement of the complete graph and the path .
###### Corollary 3.7.
Any Schur-positive fan graph satisfies .
###### Proof.
The three parts in every stable tripartition of the fan graph are, respectively, the vertices in the complement , and the vertices in each part of the unique bipartition of the path . By Theorem 3.6, any two of the integers , and differ at most , i.e., . ∎
The wheel graph is the graph formed by connecting a single vertex to all vertices of the cycle ; see Fig. 3.4 for an illustration of .
The graph is -positive, the graphs and are claw-free and -positive:
XW5=70e5+6e41+2e322andXW6=180e6+40e51+20e42.
###### Theorem 3.8.
Any wheel graph of at least vertices is not Schur-positive.
###### Proof.
Let . If is odd, then the wheel graph is not Schur-positive by Theorem 3.6. Note that
XW8 =28s3212+126s3213+168s315+84s2312+700s2214+1344s216 +2184s18−56s24.
Below we can suppose that is even. Consider the partition
λ=\brk3n2−2,n2−2,2,2.
Let . By Lemma 3.2, each rim hook has length at most , i.e.,
(3.1) πi≤α(Wn)=⌈n/2⌉−1,
and there exists a rim hook of unit length. It follows that
π2=1orπ3=1.
If , then Eq. 3.1 implies that . In view of Theorem 3.1, the tabloid contributes a non-positive summand to the coefficient . If , then Eq. 3.1 implies and either
π=(3,1,n/2−2,n/2−2),π!≥2,andW={1}.
or
π=(3,1,n/2−1,n/2−3),π!≤2,andW={1,3},
where the number is estimated by using the assumption . Let and be the number of stable partitions of of the above two types, respectively. By Theorem 3.1, it suffices to show that .
By choosing a stable set of vertices firstly and a set of 3 isolated vertices then in the remaining vertices, we see that
b=(n−1)(n/2−22)=(n−1)(n−4)(n−6)8.
Let be the set of triples of disjoint stable sets of such that
\absS1=\absS2=n/2−2and\absS3=3.
Let be the subset of such that the graph consists of a path , a path , and isolated vertices. In order to form a triple in , there are ways to choose the path , then ways to form the stable set together with the path . The set must contain exactly one vertex on the path , and one or two vertices on the path .
• If , then . In this case, there are two ways to form , depending on the choice of the vertex in .
• If , then . In this case, there are ways to form .
Therefore, we find
2a=\absA≥\absA′=(n−1)\brk3n2−3\brk32\brk2n2−4+2=(n−1)(n−6)22>2b,
since . Hence for even . This completes the proof. ∎
###### Theorem 3.9.
The graphs , , , and are -positive, and all the other complete tripartite graphs are not Schur-positive.
###### Proof.
By Theorem 2.2, the cycles and are -positive. The graphs , and are -positive since
XK1,2,2 =6e41+2e32+70e5, XK2,2,2 =36e51+6e32+384e6,and XK2,2,3 =12e512+2e321+268e61+12e52+4e43+1988e7.
Let () be a complete tripartite graph which is not one of the above graphs. By Theorem 3.6, we obtain .
The general idea of this proof is to define a partition for each triple and consider the tabloids . Besides Theorems 3.1 and 3.2, we will use the fact that
NKr,s,t(T)=1if π(T) is a rearrangement of r, s and t.
First of all, we handle the non-Schur-positivity of the specific graph . Let
λ=(2,2,2,2).
It is clear that . There are only special rim hook tabloids with :
• , , .
• , , and .
• , , and .
On the other hand, one of the tabloids with satisfies
π=(3,1,2,2),W={1},π!=2,andN=6.
Therefore, we obtain .
Below we suppose that is not one of the above graphs. Consider the partition
λ=(t,s−1,r−1,2).
Since , we have cases to treat.
1. If , then the only tabloid with even satisfies
π=(s,s,s),W={1,2},andπ!=6.
One of the other tabloids satisfies
π=(s,1,s−1,s),W={1},π!=2,andN=3s.
Therefore, we obtain .
2. If , then the only tabloid with even satisfies
π=(s,s,s+1),W={1,2},andπ!=2.
On the other hand, one of the other tabloids satisfies
π=(s,1,s−1,s+1),W={1},π!=1,andN=2s.
Therefore, we obtain .
3. If , then , and there are tabloids in all:
• , , and .
• , , , and .
• , , , and .
• , , and .
Therefore, we obtain
c′λ≤2(s2)−2s−2(s−12)+2=−2s<0.
This completes the proof. ∎
### 3.3. Non-e-positivity of the Dynkin graphs Dn and En
Martin, Morin and Wagner [17] exhibited an infinite graph family, called spiders, whose members are determinable completely by their chromatic symmetric functions. A spider is a tree with only one vertex of degree larger than two. An -vertex spider can be identified as an integer partition of whose parts are the lengths of paths starting from the distinguished vertex, and denoted . For example, the spider is the -vertex star.
In Lie theory, Dynkin diagrams are some graphs with some edges doubled or tripled. Semisimple Lie algebras are classified by Dynkin diagrams. A complete list of simple finite dimensional compact Lie groups is
An,Bn,Cn,Dn,E6,E7,E8,F4,G2,andIn.
The diagram is same to the graph for , and we name it the graph . The diagrams , , and are same to the graphs , , and , respectively. We use the notation to denote the graph for ; see Fig. 3.5. The underlying simple graph of any other type of Dynkin diagrams is a path of some length.
This subsection is devoted to the non--positivity of the graphs and for large ; see Corollary 3.12. Note that the graph is a path when .
###### Corollary 3.10.
Let . If the graph is Schur-positive, then and is odd.
###### Proof.
Immediate from Theorem 3.4. ∎
Standard techniques in “generatingfunctionology” will be used in the proofs of Corollaries 3.12 and 3.11; see Wilf [27] and Flajolet and Sedgewick [9].
###### Theorem 3.11.
Define
XD0=1,XD1=2p1,XD2=p21,and XE0=1,XE1=2p1,XE2=2p21−p2,XE3=p31−p2p1.
Then the generating functions of the chromatic symmetric functions and are respectively
∑n≥0XDnzn =(1+e1z−2e2z2)⋅E(z)E(z)−zE′(Z)and ∑n≥0XEnzn =\brk11+e1z−e21z3−3e3z3⋅E(z)E(z)−zE′(z),
where is the generating function of the elementary symmetric functions .
###### Proof.
We show the two generating functions individually, by finding out a recurrence for the chromatic symmetric functions and , respectively.
Let . Suppose that the graph is obtained by joining the paths
svn−2, tvn−2, vn−2vn−3vn−4⋯v1
at the common vertex . For , denote
Sk={v1v2,v2v3,…,vk−1vk}.
In particular . Let be a set of edges in Proposition 2.1. For , if and , then the edges in form a component of the graph , and consequently
(−1)\absSpλ(S)=(−1)k−1pkXDn−k.
When , there are cases to treat, depending on each of the edges and is in or not. The total contribution of these cases to is
(−1)n\brk2p(n−1)1−p(n−2)12−pn.
Hence for ,
XDn =n−3∑k=1(−1)k−1pkXDn−k+(−1)n\brk2p(n−1)1−p(n−2)12−pn (3.2) =n∑k=1(−1)k−1pkXDn−k.
With the generating function (see [18, Corollary 2.10])
∑n≥1pnzn=zE′(−z)E(−z)
of the power symmetric functions in hand, by using a standard technique in generatingfunctionology, it is routine to deduce the generating function from the recurrence (3.2) together with the initial values for .
Along the same lines, we can derive the recurrence
XEn=n∑k=1(−1)k−1pkXEn−k
for , which implies the generating function . ∎
###### Corollary 3.12.
The Dynkin graph with odd is -positive if and only if . The Dynkin graph is -positive if and only if .
###### Proof.
By Theorem 2.2, the paths and are -positive. It is routine to compute that
XD5 =5e5+7e41+e32+2e312+e221, XE5 =2e312+e221+7e41+e32+5e5, XE7 | 6,569 | 23,390 | {"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-2020-50 | latest | en | 0.810892 |
https://www.physicsforums.com/threads/problem-solving-heat-diffusion-equation.562180/ | 1,545,007,358,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00363.warc.gz | 1,006,342,458 | 12,983 | # Homework Help: Problem solving Heat Diffusion Equation
1. Dec 22, 2011
### XCBRA
1. The problem statement, all variables and given/known data
One face of a thick uniform layer is subject to a sinusoidal temperature variation of angular frequency ω. SHow that the damped sinusoidal temperature oscillation propagate into eh layer and give an expression for the decay length of the oscillation amplitude.
A cellar is built underground covered by a ceiling which 3m thick made of limestone. The Outside temperature is subject to daily fluctuations of amplitude 10 C and annual fluctuations of 20 C. Estimate the magnitude of the daily and annual temperature variation within the cellar.
2. Relevant equations
3. The attempt at a solution
I am unable to solve the first part of this question.
Take the diffusion equation
$$\frac{\partial T}{\partial t} = -D\frac{\partial^2T}{\partial x^{2}}$$
Using separation of variable method:
Let $$T=X(x)F(t)$$
$$X \frac{dF}{dt} = -DF\frac{d^2X}{DX^2}$$
$$-D\frac{dF}{dt}=\frac{1}{X}\frac{d^2X}{DX^2}= k$$
where k is the separation constant.
These separate into two equation which I solve to give
$$X=Ae^{\sqrt{k}x}+Be^{-\sqrt{k}x}$$
$$F=Ce^{-Dkt}$$
By superpositon principle
$$T=\sum (A_ke^{\sqrt{k}x}+B_ke^{-\sqrt{k}x})e^{-Dkt} + A_0 + B_0x$$
where C has been absorbed into A and B.
Then taking the boundary conditions:
At x →∞ T→0, which shows A_k → 0
$$T=\sum B_ke^{-\sqrt{k}x}e^{-Dkt} + B_0x$$
Then apply conditon that at x=0 T $\propto$ sinωt.
$$e^{-Dkt} = sinwt$$
However here is where I am stuck, I do not see how proceed further. Is my solution so far correct, it does not seem so as I seem to have the wrong form. Or have I chosen the wrong form for my separation coefficient?
Any help would be greatly appreciated.
2. Dec 22, 2011
### LawrenceC
The transient heat equation is generally written without the negative sign as you have written it. When you use separation of variables, use -k as your separation constant. That causes the X(x) portion to be sines and cosines.
3. Dec 23, 2011
### XCBRA
Ahh that makes better sense. Thanks for the help, I was able to solve to give:
$$Tcos(wt-x\sqrt{\frac{w}{2D}}) e^{-x\sqrt{\frac{w}{2D}}}$$ | 645 | 2,217 | {"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} | 3.9375 | 4 | CC-MAIN-2018-51 | latest | en | 0.854686 |
https://bestcase.wordpress.com/category/technology/ | 1,632,582,310,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00639.warc.gz | 184,969,974 | 30,805 | ## Flowers! Phi! Codap!
Okay, something very short, with thanks to Avery Pickford: How do sunflowers organize their seeds? Why is phi the most irrational number? How are these two questions connected, and how can we model that in CODAP?
Here is the YouTube video from Numberphile that inspired it. Worth a watch.
And here is a CODAP document:
https://codap.concord.org/releases/latest/static/dg/en/cert/index.html#shared=149141
## Weather Models and Matrices
Ack! I don’t have time to do justice to this right now, but any readers need to know if you don’t already that the geniuses at Desmos seem to be making a matrix calculator: https://www.desmos.com/matrix.
Having read that, you might rightly say, I can’t get to everything in my curriculum as it is, why are you bringing up matrices? (You might also say, Tim, I thought you were a data guy, what does this have to do with data?)
Let me address that first question (and forget the second): I’m about to go do a week of inservice in a district that, for reasons known only to them, have put matrices in their learning goals for high-school math. Their goal seems to be to learn procedures for using matrices to solve systems of linear equations.
I look at that and think, surely there are more interesting things to do with matrices. And there are!
Continue reading Weather Models and Matrices
## Sometimes, articles get done
Back in 2017, I gave a talk in which I spoke of “data moves.” These are things we do to data in order to analyze data. They’re all pretty obvious, though some are more cognitively demanding than others. They range from things like filtering (i.e., looking at a subset of the data) to joining (making a relationship between two datasets). The bee in my bonnet was that it seemed to me that in many cases, instructors might think that these should not be taught because they are not part of the curriculum—either because they are too simple and obvious or too complex and beyond-the-scope. I claimed (and still claim) that they’re important and that we should pay attention to them, acknowledge them when they come up, and occasionally even name them to students and reflect explicitly on how useful they are.
Of course there’s a great deal more to say. And because of that I wrote, with my co-PI’s, an actual, academic, peer-reviewed article—a “position paper”; this is not research—describing data moves. Any of you familiar with the vagaries of academic publishing know what a winding road that can be. But at last, here it is:
Erickson, T., Wilkerson, M., Finzer, W., & Reichsman, F. (2019). Data Moves. Technology Innovations in Statistics Education, 12(1). Retrieved from https://escholarship.org/uc/item/0mg8m7g6.
Then, in the same week, a guest blog post by Bill Finzer and me got published. Or dropped, or whatever. It’s about using CODAP to introduce some data science concepts. It even includes figures that are dynamic and interactive. Check out the post, but stay for the whole blog, it’s pretty interesting:
https://teachdatascience.com/codap/
Whew.
## Ping Pong Ball Bounce Redux
Long ago (2007) Bryan Cooley and I wrote a set of physics labs; in one of them we had students bounce a ping-pong ball. You know the sound; it’s like this:
For the lab, we had students record the sound at 1000 points per second using a Vernier microphone. Using the resulting data, students could identify the times of the “pocks” and then see how the times between the pocks — the “interpock intervals” — decreased exponentially. This is a cool take on the old Algebra 2/Precalculus activity about bouncing balls where you measure drop heights; using sound and the technology, you can get more bounces and more accuracy.
A typical graph of the sound looks like this:
And a graph of the interpock intervals looks like this:
Continue reading Ping Pong Ball Bounce Redux
## Don’t Expect the Expected Value
One day, over 50 years ago, we were visiting Lake Tahoe as a family, and dad went across the border to play keno. He came back elated: he had hit seven out of eight on one of his tickets, and won eleven hundred dollars. He proudly laid out fifty twenties and two fifties on the kitchen table. It was a magnificent sight.
The details of keno are unimportant here, except to note that keno is not a game of skill. Of course the house has an edge. In the long run, you will lose money playing keno no matter how you do it. Even my dad, who over the years has played a lot of keno, and won even bigger payouts, would probably admit that he might have a net lifetime loss.
So why do people play? There are lots of reasons, I’m sure, but one of them must be connected to that heartwarming anecdote: fifty years later, I remember the event clearly, as one of joy and wonder.
Let’s explore that using roulette, which is much simpler than keno. A roulette wheel has 18 red and 18 black numbered slots, plus a smaller number of green slots (often two). You can make many different bets, but we will stick with red and black. If you place a \$1 bet on red, and it comes up red, you get \$2 back (winning \$1); if it comes up black or green, you lose your dollar.
## Data Moves and Simplification
or, What I should have emphasized more at NCTM
I’m just back from NCTM 2018 in Washington DC where I gave a brief workshop that introduced ideas in data science education and the use of CODAP to a very nice group in a room that—well, NCTM and the Marriott Marquis were doing their best, but we really need a different way of doing technology at these big conferences.
Anyway: at the end of a fairly wide-ranging presentation in which my main goal was for participants to get their hands dirty—get into the data, get a feel for the tools, have data science on their radar—it was inevitable that I would feel:
• that I talked too much; and
• that there were important things I should have said.
Sigh. Let’s address the latter. Here is a take-away I wish I had set up better:
In data science, things are often too complicated. So one step is to simplify things; and some data moves, by their nature, simplify.
Complication is related to being awash in data (see this post…); it can come from the sheer quantity of data as well as things like being multivariate or otherwise just containing a lot of stuff we’re not interested in right now.
To cut through that complication, we often filter or summarize, and to do those, we often group. To give some examples, I will look again at the data that appeared in the cards metaphor post, but with a different slant.
Here we go: NHANES data on height, age, and sex. At the end of the process, we will see this graph:
And the graph tells a compelling story: boys and girls are roughly the same height—OK, girls are a little taller at ages 10–12—but starting at about age 13, girls’ heights level off, while the boys continue growing for about two more years.
We arrived at this after a bunch of analysis. But how did we start?
## Data Moves: the cards metaphor
In the Data Science Games project, we started talking, early, about what we called data moves. We weren’t quite sure what they were exactly, but we recognized some when we did them.
In CODAP, for example (like in Fathom), there is this thing we learn to do where you select points in one graph and, since when you select data in CODAP, the data are selected everywhere, the same points are selected in all other graphs—and you can see patterns that were otherwise hidden.
You can use that same selection idea to hide the selected or unselected points, thereby filtering the data that you’re seeing. Anyway, that felt like a data move, a tool in our data toolbox. We could imagine pointing them out to students as a frequently-useful action to take.
I’ve mentioned the idea in a couple of posts because it seemed to me that data moves were characteristic of data science, or at least the proto-data-science that we have been trying to do: we use data moves to make sense of rich data where things can get confusing; we use data moves to help when we are awash in data. In traditional intro stats, you don’t need data moves because you generally are given exactly the data you need.
## Trees. And. Diagnosis. (Live!)
I’ve been invited to give a webinar about our work on trees; it will include material from the previous two posts.
Here’s the blurb:
Data, Decisions, and Trees
We often say that we want to make decisions “based on data.” What does that really mean? We’ll look at a simple approach to data-based decisionmaking using a representation we might not use every day: the tree. In this webinar, you’ll use data to make trees, and then use the trees to diagnose diseases.
On the surface, trees are very simple. But for some reason — perhaps because we’re less familiar with using trees — people (and by that we mean us) have more trouble than we expect. Anticipate having a couple of “wait a second, let me think about this!” moments.
## Trees. And. Diagnosis. (Part two)
Last time we introduced decision trees and a tool we’ve made to explore them. With that tool, embedded in a simple game (Arbor), you can generate data from alien creatures with a simulated malady, figure out its predictors, and make a decision tree that will let you automate its diagnosis. (Here is the link to that not-quite-game.)
Your job was to get through the diseases ague and botulosis. Today I want to reflect on those two scenarios.
## Ague
Ague is ridiculously simple, and with that ridiculous simplicity, the user is supposed to be able to learn the basics of the game, that is, how to “drive” the tools. One way to figure out the disease is to sort the table by health and see what matches health. Here is what the sorted table looks like:
Just scanning the various columns, you can see that health is associated with hair color. Pink means sick, blue means well. With that insight, you can go on to diagnose individual creatures and then make a simple tree, which looks like this:
Although there is a lot of information in the tree, users can generally figure it out. If they (or you) have trouble, they can get additional information by hovering over the boxes or the links.
## Trees. And. Diagnosis. (Part one)
(This is part one. Link to part two.)
In the Data Science Games project, we have recently been exploring decision trees. It’s been great fun, and it’s time to post about it so you dear readers (all three or so of you) can play as well. There is even a working online not-quite-game you can play, and its URL will probably endure even as the software gets upgraded, so in a year it might even still work.
Here’s the genesis of all this: my German colleague Laura Martignon has been doing research on trees and learning, related to work by Gerd Gigerenzer at the Harding Center for Risk Literacy. A typical context is that of a doctor making a diagnosis. The doctor asks a series of questions; each question gets a binary, yes-no answer, which leads either to a diagnosis or a further question. The diagnosis could be either positive (the doc thinks you have the disease) or negative (the doc thinks you don’t).
The risk comes in because the doctor might be wrong. The diagnosis could be a false positive or a false negative. Furthermore, these two forms of failure are generally not equivalent.
Anyway, you can represent the sequence of questions as a decision tree, a kind of flowchart to follow as you diagnose a patient. And it’s a special kind of tree: all branchings are binary—there are always two choices—and all of the ends—the leaves, the “terminal nodes”—are one of two types: positive or negative.
The task is to design the tree. There are fancy ways (such as CART and Random Forest) to do this automatically using machine learning techniques. These techniques use a “training set”—a collection of cases where you know the correct diagnosis—to produce the tree according to some optimization criteria (such as how bad false positives and false negatives are relative to one another). So it’s a data science thing.
But in data science education, a question arises: what if you don’t really understand what a tree is? How can you learn?
That’s where our game comes in. It lets you build trees by hand, starting with simple situations. Your trees will not in general be optimal, but that’s not the point. You get to mess around with the tree and see how well it works on the training set, using whatever criteria you like to judge the tree. Then, in the game, you can let the tree diagnose a fresh set of cases and see how it does.
That’s enough for now. Your job is to play around with the tool. It will look like this to start:
The first few scenarios are designed so that it’s possible to make perfect diagnoses. No false positives, no false negatives. So it’s all about logic, and not about risk or statistics. But even that much is really interesting. As you mess around, think about the representation, and how amazingly hard it can be to think about what’s going on.
There are instructions on the left in the tan-colored “tile” labeled ArborWorkshop. Start with those. There is also a help panel in the tree tile on the right. It may not be up to date. All of the software is under development.
The first disease scenario, ague, is very simple. The next one, botulosis, is almost as simple, and worth reflecting on. That will happen soon, I hope after you have tried it.
Note: if you are unfamiliar with this platform, CODAP, go to the link, then to the “hamburger” menu. Upper left. Choose New. Then Open Document or Browse Examples. Then Getting Started with CODAP. That should be enough for now. | 3,100 | 13,654 | {"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} | 3.203125 | 3 | CC-MAIN-2021-39 | latest | en | 0.962658 |
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# Purdue Krannert 2012 - Calling All Applicants
Author Message
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07 Sep 2011, 07:18
Purdue Krannert 2012
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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04 Jan 2012, 06:29
Hi,
I do not see any activity in this thread
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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11 Jan 2012, 00:39
Did any of the admits get scholarship info?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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20 Jan 2012, 13:49
Waiting to hear on the financial aid.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 00:45
Satish,
congrates on the admit. When were you interviewed and when you received the decision?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 00:52
manisjce wrote:
Satish,
congrates on the admit. When were you interviewed and when you received the decision?
Thank you. I was interviewed on 19th Dec for R1. My decision was withheld because they didn't get my official GMAT and TOEFL scores. ETS was out for Christmas vacation and they sent my scores only on Jan 3rd, 2012. Ad com said they will not rush and they will let me know the decision notification on 19th Jan. So, I got my decision on 19th Jan, 2012, one month after I was interviewed.
Were you interviewed? What's your profile like?
-Satish.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 01:02
PMed you.
btw where are you from in India?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 01:29
manisjce wrote:
PMed you.
btw where are you from in India?
I'm from Hyderabad. But, currently working in US.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 01:41
Satish,
Are you joining Krannert?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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21 Jan 2012, 13:05
manisjce wrote:
Satish,
Are you joining Krannert?
Not sure yet. Depends on the Financial aid.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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31 Jan 2012, 12:19
Just got accepted via e-mail (Jan 10 app deadline). Anyone know when scholarship info is released?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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01 Feb 2012, 22:03
Hi,
Applied for Final round on 01-Feb-2012.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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03 Feb 2012, 06:13
taknevmar wrote:
Hi,
Applied for Final round on 01-Feb-2012.
So did I. Looks like lot of Indian applicants were admitted in the R1.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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03 Feb 2012, 06:20
josh84 wrote:
taknevmar wrote:
Hi,
Applied for Final round on 01-Feb-2012.
So did I. Looks like lot of Indian applicants were admitted in the R1.
Welcome josh84.
Yes, looks like more competitive.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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06 Feb 2012, 08:05
tm153505 wrote:
Just got accepted via e-mail (Jan 10 app deadline). Anyone know when scholarship info is released?
Congrates. Scholarships should be released within 3 weeks of your admit mail.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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06 Feb 2012, 08:07
taknevmar wrote:
josh84 wrote:
taknevmar wrote:
Hi,
Applied for Final round on 01-Feb-2012.
So did I. Looks like lot of Indian applicants were admitted in the R1.
Welcome josh84.
Yes, looks like more competitive.
Hi. Do not worry about competition. I got in with your score. If you have a strong application, you have all the chances. All the best
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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06 Feb 2012, 08:17
manisjce wrote:
tm153505 wrote:
Just got accepted via e-mail (Jan 10 app deadline). Anyone know when scholarship info is released?
Congrates. Scholarships should be released within 3 weeks of your admit mail.
Thanks for the info.
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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06 Feb 2012, 19:30
Thanks Manisjce...
Have you finalized on the school ?
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Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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07 Feb 2012, 00:56
taknevmar wrote:
Thanks Manisjce...
Have you finalized on the school ?
Yes. I am matriculating.
Intern
Joined: 15 Oct 2010
Posts: 21
WE: Information Technology (Computer Software)
Re: Purdue Krannert 2012 - Calling All Applicants [#permalink]
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07 Feb 2012, 14:19
Anone who has applied in Final round got an interview invite?
On on another note, are international students eligible for aid if they apply in Final round, i.e Feb 1 deadline.
Last edited by josh84 on 14 Feb 2012, 23:13, edited 1 time in total.
Re: Purdue Krannert 2012 - Calling All Applicants [#permalink] 07 Feb 2012, 14:19
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# Purdue Krannert 2012 - Calling All Applicants
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GuruFocus has detected 7 Warning Signs with WGL Holdings Inc \$WGL.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
WGL Holdings Inc (NYSE:WGL)
Debt-to-Equity
1.44 (As of Dec. 2016)
WGL Holdings Inc's current portion of long-term debt for the quarter that ended in Dec. 2016 was \$634 Mil. WGL Holdings Inc's long-term debt for the quarter that ended in Dec. 2016 was \$1,435 Mil. WGL Holdings Inc's total equity for the quarter that ended in Dec. 2016 was \$1,439 Mil. WGL Holdings Inc's debt to equity for the quarter that ended in Dec. 2016 was 1.44.
A high debt to equity ratio generally means that a company has been aggressive in financing its growth with debt. This can result in volatile earnings as a result of the additional interest expense.
Definition
Debt to Equity measures the financial leverage a company has.
WGL Holdings Inc's Debt to Equity Ratio for the fiscal year that ended in Sep. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (331.385 + 1444.3) / 1375.561 = 1.29
WGL Holdings Inc's Debt to Equity Ratio for the quarter that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (634.392 + 1435.247) / 1439.465 = 1.44
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the calculation of Debt to Equity, we use the total of Current Portion of Long-Term Debt and Long-Term Debt divided by Total Equity. In some calculations, Total Liabilities is used to for calculation.
Be Aware
Because a company can increase its Return on Equity by having more financial leverage, it is important to watch the leverage ratio when investing in high Return on Equity companies.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
WGL Holdings Inc Annual Data
Sep07 Sep08 Sep09 Sep10 Sep11 Sep12 Sep13 Sep14 Sep15 Sep16 deb2equity 0.61 0.56 0.74 0.61 0.57 0.65 0.74 0.93 1.02 1.29
WGL Holdings Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 deb2equity 0.93 1.08 0.90 0.89 1.02 1.16 1.07 1.10 1.29 1.44
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https://duckdb.org/docs/archive/0.7.1/sql/aggregates | 1,686,447,207,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00535.warc.gz | 250,023,805 | 9,175 | Aggregate Functions
Version 0.7.1
Examples
``````-- produce a single row containing the sum of the "amount" column
SELECT SUM(amount) FROM sales;
-- produce one row per unique region, containing the sum of "amount" for each group
SELECT region, SUM(amount) FROM sales GROUP BY region;
-- return only the regions that have a sum of "amount" higher than 100
SELECT region FROM sales GROUP BY region HAVING SUM(amount) > 100;
-- return the number of unique values in the "region" column
SELECT COUNT(DISTINCT region) FROM sales;
-- return two values, the total sum of "amount" and the sum of "amount" minus columns where the region is "north"
SELECT SUM(amount), SUM(amount) FILTER (region != 'north') FROM sales;
-- returns a list of all regions in order of the "amount" column
SELECT LIST(region ORDER BY amount DESC) FROM sales;
``````
Syntax
Aggregates are functions that combine multiple rows into a single value. Aggregates are different from scalar functions and window functions because they change the cardinality of the result. As such, aggregates can only be used in the `SELECT` and `HAVING` clauses of a SQL query.
When the `DISTINCT` clause is provided, only distinct values are considered in the computation of the aggregate. This is typically used in combination with the `COUNT` aggregate to get the number of distinct elements; but it can be used together with any aggregate function in the system.
When the `ORDER BY` clause is provided, the values being aggregated are sorted before applying the function. Usually this is not important, but there are some order-sensitive aggregates that can have indeterminate results (e.g., `first`, `last`, `list` and `string_agg`). These can be made deterministic by ordering the arguments. For order-insensitive aggregates, this clause is parsed and applied, which is inefficient, but still produces the same result.
General Aggregate Functions
The table below shows the available general aggregate functions.
Function Description Example Alias(es)
`any_value(arg)` Returns the first non-null value from arg. `any_value(A)` -
`arg_max(arg,val)` Finds the row with the maximum `val`. Calculates the `arg` expression at that row. `arg_max(A,B)` `argMax(A,B)`, `max_by(A,b)`
`arg_min(arg,val)` Finds the row with the minimum `val`. Calculates the `arg` expression at that row. `arg_min(A,B)` `argMin(A,B)`, `min_by(A,B)`
`avg(arg)` Calculates the average value for all tuples in arg. `avg(A)` -
`bit_and(arg)` Returns the bitwise AND of all bits in a given expression . `bit_and(A)` -
`bit_or(arg)` Returns the bitwise OR of all bits in a given expression. `bit_or(A)` -
`bit_xor(arg)` Returns the bitwise XOR of all bits in a given expression. `bit_xor(A)` -
`bitstring_agg(arg)` Returns a bitstring with bits set for each distinct value. `bitstring_agg(A)` -
`bool_and(arg)` Returns TRUE if every input value is TRUE, otherwise FALSE. `bool_and(A)` -
`bool_or(arg)` Returns TRUE if any input value is TRUE, otherwise FALSE. `bool_or(A)` -
`count(arg)` Calculates the number of tuples tuples in arg. `count(A)` -
`favg(arg)` Calculates the average using a more accurate floating point summation (Kahan Sum). `favg(A)` -
`first(arg)` Returns the first value of a column. `first(A)` `arbitrary(A)`
`fsum(arg)` Calculates the sum using a more accurate floating point summation (Kahan Sum). `fsum(A)` `sumKahan`, `kahan_sum`
`histogram(arg)` Returns a `LIST` of `STRUCT`s with the fields `bucket` and `count`. `histogram(A)` -
`last(arg)` Returns the last value of a column. `last(A)` -
`list(arg)` Returns a `LIST` containing all the values of a column. `list(A)` `array_agg`
`max(arg)` Returns the maximum value present in arg. `max(A)` -
`min(arg)` Returns the minumum value present in arg. `min(A)` -
`product(arg)` Calculates the product of all tuples in arg `product(A)` -
`string_agg(arg, sep)` Concatenates the column string values with a separator `string_agg(S, ',')` `group_concat`
`sum(arg)` Calculates the sum value for all tuples in arg. `sum(A)` -
Approximate Aggregates
The table below shows the available approximate aggregate functions.
Function Description Example
`approx_count_distinct(x)` Gives the approximate count of distintinct elements using HyperLogLog. `approx_count_distinct(A)`
`approx_quantile(x,pos)` Gives the approximate quantile using T-Digest. `approx_quantile(A,0.5)`
`reservoir_quantile(x,quantile,sample_size=8192)` Gives the approximate quantile using reservoir sampling, the sample size is optional and uses 8192 as a default size. `reservoir_quantile(A,0.5,1024)`
Statistical Aggregates
The table below shows the available statistical aggregate functions.
Function Description Formula Alias
`corr(y,x)` Returns the correlation coefficient for non-null pairs in a group. `COVAR_POP(y, x) / (STDDEV_POP(x) * STDDEV_POP(y))` -
`covar_pop(y,x)` Returns the population covariance of input values. `(SUM(x*y) - SUM(x) * SUM(y) / COUNT(*)) / COUNT(*) ` -
`entropy(x)` Returns the log-2 entropy of count input-values. - -
`kurtosis(x)` Returns the excess kurtosis of all input values. - -
`mad(x)` Returns the median absolute deviation for the values within x. NULL values are ignored. Temporal types return a positive `INTERVAL`. `MEDIAN(ABS(x-MEDIAN(x)))` -
`median(x)` Returns the middle value of the set. NULL values are ignored. For even value counts, quantitiative values are averaged and ordinal values return the lower value. `QUANTILE_CONT(x, 0.5)` -
`mode(x)` Returns the most frequent value for the values within x. NULL values are ignored. - -
`quantile_cont(x,pos)` Returns the intepolated quantile number between 0 and 1 . If `pos` is a `LIST` of `FLOAT`s, then the result is a `LIST` of the corresponding intepolated quantiles. - -
`quantile_disc(x,pos)` Returns the exact quantile number between 0 and 1 . If `pos` is a `LIST` of `FLOAT`s, then the result is a `LIST` of the corresponding exact quantiles. - `quantile`
`regr_avgx(y,x)` Returns the average of the independent variable for non-null pairs in a group, where x is the independent variable and y is the dependent variable. - -
`regr_avgy(y,x)` Returns the average of the dependent variable for non-null pairs in a group, where x is the independent variable and y is the dependent variable. - -
`regr_count(y,x)` Returns the number of non-null number pairs in a group. `(SUM(x*y) - SUM(x) * SUM(y) / COUNT(*)) / COUNT(*)` -
`regr_intercept(y,x)` Returns the intercept of the univariate linear regression line for non-null pairs in a group. `AVG(y)-REGR_SLOPE(y,x)*AVG(x)` -
`regr_r2(y,x)` Returns the coefficient of determination for non-null pairs in a group. - -
`regr_slope(y,x)` Returns the slope of the linear regression line for non-null pairs in a group. `COVAR_POP(x,y) / VAR_POP(x)` -
`regr_sxx(y,x)` - `REGR_COUNT(y, x) * VAR_POP(x)` -
`regr_sxy(y,x)` Returns the population covariance of input values. `REGR_COUNT(y, x) * COVAR_POP(y, x) ` -
`regr_syy(y,x)` - `REGR_COUNT(y, x) * VAR_POP(y) f` -
`skewness(x)` Returns the skewness of all input values. - -
`stddev_pop(x)` Returns the population standard deviation. `sqrt(var_pop(x))` -
`stddev_samp(x)` Returns the sample standard deviation. `sqrt(var_samp(x))` `stddev(x)`
`var_pop(x)` Returns the population variance. - -
`var_samp(x)` Returns the sample variance of all input values. `(SUM(x^2) - SUM(x)^2 / COUNT(x)) / (COUNT(x) - 1)` `variance(arg,val)`
Ordered Set Aggregate Functions
The table below shows the available “ordered set” aggregate functions. These functions are specified using the `WITHIN GROUP(ORDER BY sort_expression)` syntax, and they are converted to an equivalent aggregate function that takes the ordering expression as the first argument.
Function Equivalent
`mode() WITHIN GROUP (ORDER BY sort_expression)` `mode(sort_expression)`
`percentile_cont(fraction) WITHIN GROUP (ORDER BY sort_expression)` `quantile_cont(sort_expression, fraction)`
`percentile_cont(fractions) WITHIN GROUP (ORDER BY sort_expression)` `quantile_cont(sort_expression, fractions)`
`percentile_disc(fraction) WITHIN GROUP (ORDER BY sort_expression)` `quantile_disc(sort_expression, fraction)`
`percentile_disc(fractions) WITHIN GROUP (ORDER BY sort_expression)` `quantile_disc(sort_expression, fractions)`
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