url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://py.checkio.org/mission/inscribe-a-contour/lang/en/share/db11f178a4973255f9d4aceddd2ae3e7/ | 1,726,716,962,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00205.warc.gz | 437,521,325 | 19,572 | # Inscribe a Contour
Over the past almost 15 years, 3D printing has become very popular, not only amongst the big industrial companies which can afford to buy a super expensive 3D manufacturing complex (e.i. SLM or SLS machines), but also among ordinary people, engineers and enthusiasts. It started in 2005 with the RepRap project within the University of Bath. The project's main objective was to develop an affordable production system that would enable the production of cheap everyday household goods regardless of the household's location. The project was successful and gave birth to the printer called "Darwin" at the beginning of 2008. It was a fully open project under the GNU GPL and by the end of 2008th ~100 copies of "Darwin" have been produced in various countries. Since then, there has been an exponential increase in the number of printers manufactured, both DIY and commercial. Most of the FDM printers you may find today in libraries, universities, workshops, engineering companies, friend's houses are likely to be direct descendants of that very first desktop printer. I do not possess any more up to date information, but in 2013 it's been sold ~72.000 of desktop printers, in 2014 - almost 140.000 (according to the Wohlers Report ).
Leaving aside the 3D design process, which is the first step on the way to something to be printed, we go straight to preparing a 3D model for printing. There's a variety of software applications designed for that purpose called slicers . They slice a 3D model from the bottom to the top with a constant (not always) height and get a sequence of the planar contours, each of which is represented by a sequence of the dots.
These sequences of dots then, obvious, are used for a printing program along with the calculated velocities, accelerations, etc. In order to estimate if a model fits the printer building table (video) and then can be printed, our slicing software should find the optimal orientation of the model on the table, so the model will require the least possible space.
So, in this mission you are given a list of dots, each represents a projection of a 3D model vertex onto a horizontal plane. Your task is to find the smallest rectangle (by its area) into which all the given dots (and thus the projected contour) can be inscribed.
Input: A list of tuples, each tuple contains coordinates of a projected dot (x, y). All given coordinates are integers. Although in the example illustrations the dots are connected, forming sane contours, they can be connected in any arbitrary consequence and that won't affect the result.
Output: The area of the smallest rectangle (with ±0.001 precision) which inscribes the given contour.
Example:
```inscribe([(1, 1), (1, 2), (0, 2), (3, 5), (3, 4), (4, 4)]) == 6.0 #example #1
inscribe([(6, 5), (10, 7), (2, 8)]) == 20.0 #example #2
inscribe([(2, 3), (3, 8), (8, 7), (9, 2), (3, 2), (4, 4), (6, 6), (7, 3), (5, 3)]) == 41.538 #example #3
```
Precondition:
• n ∈ [3; 30] - dots number
• i = range(n)
• x i ∈ [0; 300] - x coordinate of the i-th dot
• y i ∈ [0; 300] - y coordinate of the i-th dot
• there won't be two (or more) similar dots
• there won't be a case with all the dots on the same line
28
Settings
Code:
Other:
Invalid hot key. Each hot key should be unique and valid
Hot keys:
• to Run Code: to Check Solution: to Stop:
CheckiO Extensions
CheckiO Extensions allow you to use local files to solve missions. More info in a blog post.
In order to install CheckiO client you'll need installed Python (version at least 3.8)
Install CheckiO Client first:
`pip3 install checkio_client`
`checkio --domain=py config --key=`
Sync solutions into your local folder
`checkio sync`
(in beta testing) Launch local server so your browser can use it and sync solution between local file end extension on the fly. (doesn't work for safari)
`checkio serv -d`
Alternatevly, you can install Chrome extension or FF addon
`checkio install-plugin`
`checkio install-plugin --ff`
`checkio install-plugin --chromium`
Read more here about other functionality that the checkio client provides. Feel free to submit an issue in case of any difficulties.
Pair Programming (Beta-version)
Welcome to Pair Programming! Engage in real-time collaboration on coding projects by starting a session and sharing the provided unique URL with friends or colleagues. This feature is perfect for joint project development, debugging, or learning new skills together. Simply click 'Start Session' to begin your collaborative coding journey!
Waiting for Pair Programming to start...
You are trying to join a pair programming session that has not started yet.
Please wait for the session creator to join.
Waiting for Pair Programming to reconnect...
It looks like the creator of the pair programming session closed the editor window.
It might happen accidentally, so that you can wait for reconnection. | 1,176 | 4,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.955641 |
https://scienceblogs.com/principles/2007/05/10/many-worlds-many-treats | 1,569,078,219,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00146.warc.gz | 653,377,329 | 22,563 | Many Worlds, Many Treats
I'm sitting at the computer typing, when the dog bumps up against my legs. I look down, and she's sniffing the floor around my feet intently.
"What are you doing down there?"
"I'm looking for steak!" she says, wagging her tail hopefully.
"I'm pretty certain that there's no steak down there," I say. "I've never eaten steak at the computer, and I've certainly never dropped any on the floor."
"You did in some universe," she says, still sniffing.
I sigh. "I'm going to move the quantum physics books to a higher shelf, so you can't reach them."
"It won't matter. I've got Wikipedia."
"All right, what ridiculous theory has your silly little doggy brain come up with?"
"Well, it's possible that you would eat steak at the computer, yes?"
"I do eat steak, yes, and I sometimes eat at the computer, so sure."
"And if you were to eat steak at the computer, you'd probably drop some on the floor."
"Dude, I've seen you eat." Yes, the dog calls me "dude." There may be obedience classes in her future.
"All right, we'll allow the possibility."
"Therefore, it's possible that you dropped steak on the floor. And according to Everett's Many Worlds Interpretation of quantum mechanics, that means that you did drop steak on the floor. Which means I just need to find it."
"Well, technically, what the Many Worlds interpreation says is that there's some branch of the unitarily evolving wavefunction of the universe in which I dropped steak on the floor."
"Right, so I just need to find the unitary whatsis."
"The thing is, though, we can only perceive one branch of the wavefunction."
"Maybe you can only perceive one branch. I have a very good nose. I can sniff into extra dimensions."
"That would explain some of your mystery barking fits, but extra dimensions are a completely different thing. In this case, once there has been sufficient decoherence between the branches of the wavefunction that there's no possibility of interference between the different parts, they're effectively separate and inaccessible universes."
"What do you mean, decoherence?"
"Well, say I did have a piece of steak here-- stop wagging your tail, it's a hypothetical-- quantum mechanics says that if I dropped it on the floor, then picked it back up, there could be an interference between the wavefunction describing the bit of steak that fell and the wavefunction describing the bit of steak that didn't fall. Because, of course, there's only a probability that I'd drop it, so you need both bits."
"What would that mean?"
"Well, the steak would probably produce some sort of interference pattern. I'm not really sure what that would look like. The point is, though, it doesn't really matter. The steak is constantly interacting with its environment-- the air, the desk, the floor--"
"The dog!"
"Whatever. Those interactions are essentially random, and unmeasured. These interactions lead to shifts in the wavefunctions of the different bits of steak, and those shifts make it so the wavefunctions don't interfere cleanly any more. That process is called 'decoherence,' and it happens very fast."
"How fast?" she asks, looking hopeful.
"It depends on the exact situation, but as a rough guess, I'd say about the same time as the lifetime of a bunny made of cheese. 10-30 seconds or less."
"Oh." She deflates a little. "That's fast." She still hasn't caught a bunny made of cheese.
"Yeah. And once that decoherence has happened, the different branches of the wavefunction can't really interact with each other any more. Which means, essentially, that the different branches become separate universes that are completely inaccessible to one another. Things that happen in these other 'universes' have absolutely no effect on what happens in our universe."
"Why do we only see one branch of the whatchamacallit?"
"Ah, now that's the big question. Nobody knows. A lot of people think this means that quantum mechanics is fundamentally incomplete, and there's a whole community of scientists doing research into the fundamentals of quantum theory, and the various interpretations. Matt Leifer has a whole blog talking about this stuff."
"We don't like him. He said mean things about me."
"It wasn't so much mean, as dismissive. But that's not the point. The point is, there's no way you're going to find steak under my desk, so please get out of there."
"Oh. OK." She mopes out from under the desk, head down and tail drooping.
"Hey, look on the bright side," I say. "In the universe where a version of me dropped a piece of steak on the floor, there's also a version of you."
"Yeah. And you're a mighty hunter, so you probbaly got to the steak before I could pick it up."
"Yeah?" Her tail starts wagging.
"Yeah. So, in the universe where I dropped steak, you got to eat steak."
"Oooh!" The tail wags furiously. "I like steak!"
"I know you do." I save what I was working on. "Tell you what, how about we go for a walk?"
"Ooooh! Good plan!" and she's off, clattering down the stairs for the back door and the leash.
She's really a very silly dog.
Tags
More like this
The Law of Burger Attraction
I'm sitting at the dining room table eating lunch, when I get the feeling of being watched. I look around, and see the dog across the room, curled up on her pillows staring at me. She's quietly chanting to herself "I get stuff. I get stuff. I get stuff." "You're not trying that hypnosis thing again…
Many Worlds, Many Treats: The Movie
Another dramatic reading of a chapter from How to Teach Physics to Your Dog, just because. This is Chapter 4, which is based on the original Many Worlds, Many Treats post that kick-started the whole thing: I'm sitting at the computer typing, when Emmy bumps up against my legs. I look down, and she…
Thursday Royalty Blogging 042111
I'm sitting on the couch reading when Emmy trots in looking excited. "Hey, dude, I've got a great idea!" she says cheerily. "You know how your puppy is away for the week?" "Yeah." "Well, I think this is the perfect opportunity to class up your blog a little. I mean, you always post pictures of her…
Bunnies Made of Cheese: The Book
I'm checking a last few things and putting papers into an envelope when the dog wakes up from her nap. "Hey," she says, stretching, "What're you doing?" "I'm getting ready to mail this," I say. "What is it?" "Several copies of a book contract that I just signed." "It's a book about me, right?" she…
Dogs know some very serious tricks.
I suppose you've changed that old Einstein quip about explaining things to your grandmother into explaining them to your dog. Your dog seems to have fantastic physics and speaking abilities, much greater than any other dog I've met.
I've been trying to teach Spike to program, but so far he's more interested in going for runs.
By Ron Avitzur (not verified) on 10 May 2007 #permalink
My dogs go around measuring "no-steak" in the hopes that they're forcing "yes-steak" somewhere else in an entangled system.
By rat-terrier (not verified) on 10 May 2007 #permalink
"We don't like him. He said mean things about me."
Well, that was before I knew she was an expert on quantum foundations, or at least had about as good a grip on them as the majority of physicists :)
To clarify, I have nothing against dogs in general, just the concept of writing posts about them when they are not either discussing physics or being captioned in an amusing way. In any case, they don't allow pets in my apartment building, so my dog related posts would be rather boring.
Wow, that was the most entertaining excuse for posting a crotch shot on one's blog that I think I've ever read....
she barked at a Schrodinger's Cat
You shouldn't have open that box.
By Torbjörn Lars… (not verified) on 10 May 2007 #permalink
I dunno, Kate...I guess your link might be somewhat funnier, but I'd rate the two conversations roughly equivalent in entertainment value as Chad's gets points for novelty.
Oh, and Kate's right..this was an entertaining dialogue, but the one she linked to was downright heelarious, on a par with dogs in elk when it comes to side-splitting doggie stories.
what a GORGEOUS dog!
F-ing hilarious, and I learned something about decoherance as well :)
omg. well, the sweet potato story might get funny + simplicity points, but yours definitely gets funny + originality + educational points. i was a philosophy major and i learned something, for pete's sake. yay you!
I got a question for the pup...
If these are truly physical universes that are created anytime something might happen, where does the matter come from? Doesn't the Conservation of Matter still apply?
ok, considering I was on my way to bed and considering I have had three glasses of wine, I was very amused. Enough to actually follow the text and the gist and not get brain farts or stare blankly into the ether nor did I lose track. I have to say, I loved this piece (coming from my interest in ...wait now...brain fart..have to scroll up...right: quantum physics/mechanics..) and having my own menagerie of cats who like to knead private parts of the body and lay on the warm bits, I did not see anything weird about the crotch shot. Keep up the great writing. Laurie
I guess sweet potato particles don't move as quite quickly as bunnies made of cheese.
You realize that this series of conversations would make for a terrific book on the current state of physics.
:-) nice
I remember one of Lee Smolin's book has quite an amusing passage about gender confused quantum pets (forgot which book. might have been 3 Roads to QG). best,
B.
Dude,
The mystery barking fits are a psychological event not physical.
Dogs are pack animals. They are instinctively built for constant company, not to mention constant activity. Life among humans denies them this for the greater part of each day, causing them stress/distress. Barking is one method of relieving this stress. It is such a common symptom that we have come to expect it of dogs.
Absolutely wonderful! If text-books were written in this style, I'd have paid a lot more attention in school. I agree, conversations with the dog about physics would make a great book in general. :-)
Absolutely wonderful! If text-books were written in this style, I'd have paid a lot more attention in school. I agree, conversations with the dog about physics would make a great book in general. :-)
I live in a universe where I always eat at my computer desk, I always have steak on Friday, I usually drop a bit of steak on the floor every Friday, and I don't own a dog but the cat won't eat anything but catfood... So it goes...
I live in a universe where I always eat at my computer desk, I always have steak on Friday, I usually drop a bit of steak on the floor every Friday, and I don't own a dog but the cat won't eat anything but catfood... So it goes...
Please Don't tell the dog about the universe where you all of a sudden decide to cook her for dinner.
I can't believe Matt thinks what you do is less informative than his boorish, self-indulgent ramblings. I know for a fact that you speak to a much great audience, because you use a relative medium.
Real men have websites! :D
Picked up by Digg.
I can't believe Matt thinks what you do is less informative than his boorish, self-indulgent ramblings. I know for a fact that you speak to a much great audience, because you use a relative medium.
Real men have websites! :D
I reside in a time warp with two rescued dogs of mix breeding who apparently not only know physics but in their own way, they've been trying to teach me...they lose patience with me as I'm too dumb to comprehend their teachings and the subject matter, and I'm sure they use this computer when I'm not around... so much for higher human education..dogs 1, human 0.
My Dog: {Howling]
Me: Why are you howling at the Moon?
Dog: Thanks for asking. Wag, wag! [howls some more]
Me: No, really, why are you howling at the Moon?
Dog: That's what you call it. Well. LOOK at it. It's big! It's bright!
Dog: But it has NO SMELL! No smell at all! That's weird. That's an anomaly. That's outside the paradigm. Might be dangerous. Right? Anyway, this is our territory! [Howls some more]
Me: Okay, I get it now. Let me know when you want to go back inside. The Moon: LOOK at it. It's big! It's bright!
Dog: Glad you finally understand. [wag, wag]. Glad you're in my Pack.
How do I get to that universe wherein I am the master and he is the dog. In this one, it seems the other way round.
Does that dog give you pleasure once in awhile in that position?
Love this. Make me think of my brother in-law who is a true life quantum professor. But he's also a vegetarian, and israeli... and like most tremendous science minds out there, doesn't have much sense of humor. All that said, I laughed out loud reading this.
By 300redbulls (not verified) on 11 May 2007 #permalink
Kudos to the very clever Dr. George Hockney, quantum computing expert, who figured out my dialogue, and emailed me as below.
===========
Aha. That explains why this appeared today. I was wondering about that.
-- George
> Is the moon there when nobody looks?
> (Submitted on 10 May 2007)
>
> Abstract: In 1981, David Mermin described a cleverly simplified version of Bell's theorem. It pointed out in a straightforward way that interpreting entanglement from a local realist point of view can be problematic. We propose here an extended version of Mermin's device that can actually be given a simple local realist interpretation, and we argue that we still have no scientific reason to believe that the moon could possibly not be there when nobody looks.
>
> 11 pages, 8 Figures, 1 Table
> Subjects:
> Quantum Physics (quant-ph)
> Cite as: arXiv:0705.1477v1 [quant-ph]
===========
Hey Chad! Didn't know you had a blog. I found it from the 2nd page under "Top stories in 24 hours." I enjoyed the post, I'm taking Quantum Chemistry right now, so I have a light grasp on what you are talking about. I'll be working with the chemistry and physics departments this summer for research so I'll see you around.
You stole my dog, Ginger. And you've ruined all the work I've done in teachng him classical Neutonian physics. Next thing you know she's going to want to know about spooky action at a distance.
Thanks for nothing.
On the serious side, very nice article. Think I'll send the link to the wife. She likes dogs, maybe she'll grab the science.
I don't want to be an alarmist, but if that were my dog I'd have her checked for cataracts.
By Brook Monroe (not verified) on 13 May 2007 #permalink
Expand this. Weave it into a kids' story for 6-10 year olds. Second book on chemistry from the perspective of physics for the same age group. Then another on math. All can feature your dog. You won't need tenure to have a career and you'll do more good for a lot bigger audience than being a professor. You're on to something. It's highly needed.
By hikingprem (not verified) on 14 May 2007 #permalink
Greetings,
Nice Posted! Its so very informative and knowledgeable for your visitors or readers.
Thank You for sharing.. Keep up the good work..
More Power,
Mara Brown
By Mara Brown (not verified) on 17 May 2007 #permalink
Very interesting Chad and entertaining! Dogs do know this stuff, they just choose not to apply it! 8^) There are some interesting implications here: Those things at absolute zero either exist across all dimensions or none. One might also suppose that black holes do too, as, according to the latest thinking, no information is lost at the event horizon.
Don
Wow - a dog that mis/undertands the quantum world. That's amazing!
I like reading about physics, although there's a lot in quantum physics I distrust. This is an informative, and entertaining blog. Well done! The fun aspect makes the learning easier.
Although, thinking about it some more, most if not all pet owners talk to their pets, and there's no doubt the pets try to communicate back with us. Maybe one day when they can talk, or relay their thoughts to as in another manner, they may actually be able to un-entangle all this quantum stuff for us. | 3,734 | 16,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | longest | en | 0.979824 |
https://en.m.wikipedia.org/wiki/Pseudopotential | 1,618,784,064,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00057.warc.gz | 332,879,989 | 21,980 | # Pseudopotential
In physics, a pseudopotential or effective potential is used as an approximation for the simplified description of complex systems. Applications include atomic physics and neutron scattering. The pseudopotential approximation was first introduced by Hans Hellmann in 1934.[1]
Comparison of a wavefunction in the Coulomb potential of the nucleus (blue) to the one in the pseudopotential (red). The real and the pseudo wavefunction and potentials match above a certain cutoff radius ${\displaystyle r_{c}}$.
## Atomic physics
The pseudopotential is an attempt to replace the complicated effects of the motion of the core (i.e. non-valence) electrons of an atom and its nucleus with an effective potential, or pseudopotential, so that the Schrödinger equation contains a modified effective potential term instead of the Coulombic potential term for core electrons normally found in the Schrödinger equation.
The pseudopotential is an effective potential constructed to replace the atomic all-electron potential (full-potential) such that core states are eliminated and the valence electrons are described by pseudo-wavefunctions with significantly fewer nodes. This allows the pseudo-wavefunctions to be described with far fewer Fourier modes, thus making plane-wave basis sets practical to use. In this approach usually only the chemically active valence electrons are dealt with explicitly, while the core electrons are 'frozen', being considered together with the nuclei as rigid non-polarizable ion cores. It is possible to self-consistently update the pseudopotential with the chemical environment that it is embedded in, having the effect of relaxing the frozen core approximation, although this is rarely done. In codes using local basis functions, like Gaussian, often effective core potentials are used that only freeze the core electrons.
First-principles pseudopotentials are derived from an atomic reference state, requiring that the pseudo- and all-electron valence eigenstates have the same energies and amplitude (and thus density) outside a chosen core cut-off radius ${\displaystyle r_{c}}$ .
Pseudopotentials with larger cut-off radius are said to be softer, that is more rapidly convergent, but at the same time less transferable, that is less accurate to reproduce realistic features in different environments.
Motivation:
1. Reduction of basis set size
2. Reduction of number of electrons
3. Inclusion of relativistic and other effects
Approximations:
1. One-electron picture.
2. The small-core approximation assumes that there is no significant overlap between core and valence wave-function. Nonlinear core corrections[2] or "semicore" electron inclusion[3] deal with situations where overlap is non-negligible.
Early applications of pseudopotentials to atoms and solids based on attempts to fit atomic spectra achieved only limited success. Solid-state pseudopotentials achieved their present popularity largely because of the successful fits by Walter Harrison to the nearly free electron Fermi surface of aluminum (1958) and by James C. Phillips to the covalent energy gaps of silicon and germanium (1958). Phillips and coworkers (notably Marvin L. Cohen and coworkers) later extended this work to many other semiconductors, in what they called "semiempirical pseudopotentials".[4]
### Norm-conserving pseudopotential
Norm-conserving and ultrasoft are the two most common forms of pseudopotential used in modern plane-wave electronic structure codes. They allow a basis-set with a significantly lower cut-off (the frequency of the highest Fourier mode) to be used to describe the electron wavefunctions and so allow proper numerical convergence with reasonable computing resources. An alternative would be to augment the basis set around nuclei with atomic-like functions, as is done in LAPW. Norm-conserving pseudopotential was first proposed by Hamann, Schlüter, and Chiang (HSC) in 1979.[5] The original HSC norm-conserving pseudopotential takes the following form:
${\displaystyle {\hat {V}}_{\textit {ps}}(r)=\sum _{l}\sum _{m}|Y_{lm}\rangle V_{lm}(r)\langle Y_{lm}|}$
where ${\displaystyle |Y_{lm}\rangle }$ projects a one-particle wavefunction, such as one Kohn-Sham orbital, to the angular momentum labeled by ${\displaystyle \{l,m\}}$ . ${\displaystyle V_{lm}(r)}$ is the pseudopotential that acts on the projected component. Different angular momentum states then feel different potentials, thus the HSC norm-conserving pseudopotential is non-local, in contrast to local pseudopotential which acts on all one-particle wave-functions in the same way.
Norm-conserving pseudopotentials are constructed to enforce two conditions.
1. Inside the cut-off radius ${\displaystyle r_{c}}$ , the norm of each pseudo-wavefunction be identical to its corresponding all-electron wavefunction:[6]
${\displaystyle \int _{r ,
where ${\displaystyle \phi _{\mathbf {R} ,i}}$ and ${\displaystyle {\tilde {\phi }}_{\mathbf {R} ,i}}$ are the all-electron and pseudo reference states for the pseudopotential on atom ${\displaystyle \mathbf {R} }$ .
2. All-electron and pseudo wavefunctions are identical outside cut-off radius ${\displaystyle r_{c}}$ .
Pseudopotential representing the effective core charge.
### Ultrasoft pseudopotentials
Ultrasoft pseudopotentials relax the norm-conserving constraint to reduce the necessary basis-set size further at the expense of introducing a generalized eigenvalue problem.[7] With a non-zero difference in norms we can now define:
${\displaystyle q_{\mathbf {R} ,ij}=\langle \phi _{\mathbf {R} ,i}|\phi _{\mathbf {R} ,j}\rangle -\langle {\tilde {\phi }}_{\mathbf {R} ,i}|{\tilde {\phi }}_{\mathbf {R} ,j}\rangle }$ ,
and so a normalised eigenstate of the pseudo Hamiltonian now obeys the generalized equation
${\displaystyle {\hat {H}}|\Psi _{i}\rangle =\epsilon _{i}{\hat {S}}|\Psi _{i}\rangle }$ ,
where the operator ${\displaystyle {\hat {S}}}$ is defined as
${\displaystyle {\hat {S}}=1+\sum _{\mathbf {R} ,i,j}|p_{\mathbf {R} ,i}\rangle q_{\mathbf {R} ,ij}\langle p_{\mathbf {R} ,j}|}$ ,
where ${\displaystyle p_{\mathbf {R} ,i}}$ are projectors that form a dual basis with the pseudo reference states inside the cut-off radius, and are zero outside:
${\displaystyle \langle p_{\mathbf {R} ,i}|{\tilde {\phi }}_{\mathbf {R} ,j}\rangle _{r .
A related technique[8] is the projector augmented wave (PAW) method.
## Fermi pseudopotential
Enrico Fermi introduced a pseudopotential, ${\displaystyle V}$ , to describe the scattering of a free neutron by a nucleus.[9] The scattering is assumed to be s-wave scattering, and therefore spherically symmetric. Therefore, the potential is given as a function of radius, ${\displaystyle r}$ :
${\displaystyle V(r)={\frac {4\pi \hbar ^{2}}{m}}b\,\delta (r)}$ ,
where ${\displaystyle \hbar }$ is the Planck constant divided by ${\displaystyle 2\pi }$ , ${\displaystyle m}$ is the mass, ${\displaystyle \delta (r)}$ is the Dirac delta function, ${\displaystyle b}$ is the bound coherent neutron scattering length, and ${\displaystyle r=0}$ the center of mass of the nucleus.[10] The Fourier transform of this ${\displaystyle \delta }$ -function leads to the constant neutron form factor.
## Phillips pseudopotential
James Charles Phillips developed a simplified pseudopotential while at Bell Labs useful for describing silicon and germanium.
## References
1. ^ Schwerdtfeger, P. (August 2011), "The Pseudopotential Approximation in Electronic Structure Theory", ChemPhysChem, 12 (17): 3143–3155, doi:10.1002/cphc.201100387, PMID 21809427
2. ^ Louie, Steven G.; Froyen, Sverre; Cohen, Marvin L. (August 1982), "Nonlinear ionic pseudopotentials in spin-density-functional calculations", Physical Review B, 26 (4): 1738–1742, Bibcode:1982PhRvB..26.1738L, doi:10.1103/PhysRevB.26.1738
3. ^ Reis, Carlos L.; Pacheco, J. M.; Martins, José Luís (October 2003), "First-principles norm-conserving pseudopotential with explicit incorporation of semicore states", Physical Review B, American Physical Society, 68 (15), p. 155111, Bibcode:2003PhRvB..68o5111R, doi:10.1103/PhysRevB.68.155111
4. ^ M. L. Cohen, J. R. Chelikowsky, "Electronic Structure and Optical Spectra of Semiconductors", (Springer Verlag, Berlin 1988)
5. ^ Hamann, D. R.; Schlüter, M.; Chiang, C. (1979-11-12). "Norm-Conserving Pseudopotentials". Physical Review Letters. 43 (20): 1494–1497. Bibcode:1979PhRvL..43.1494H. doi:10.1103/PhysRevLett.43.1494.
6. ^ Bachelet, G. B.; Hamann, D. R.; Schlüter, M. (October 1982), "Pseudopotentials that work: From H to Pu", Physical Review B, American Physical Society, 26 (8), pp. 4199–4228, Bibcode:1982PhRvB..26.4199B, doi:10.1103/PhysRevB.26.4199
7. ^ Vanderbilt, David (April 1990), "Soft self-consistent pseudopotentials in a generalized eigenvalue formalism", Physical Review B, American Physical Society, 41 (11), pp. 7892–7895, Bibcode:1990PhRvB..41.7892V, doi:10.1103/PhysRevB.41.7892
8. ^ Kresse, G.; Joubert, D. (1999). "From ultrasoft pseudopotentials to the projector augmented-wave method". Physical Review B. 59 (3): 1758–1775. Bibcode:1999PhRvB..59.1758K. doi:10.1103/PhysRevB.59.1758.
9. ^ E. Fermi (July 1936), "Motion of neutrons in hydrogenous substances", Ricerca Scientifica, 7: 13–52
10. ^ Squires, Introduction to the Theory of Thermal Neutron Scattering, Dover Publications (1996) ISBN 0-486-69447-X
## Pseudopotential libraries
• Pseudopotential Library : A community website for pseudopotentials/effective core potentials developed for high accuracy correlated many-body methods such as quantum Monte Carlo and quantum chemistry
• NNIN Virtual Vault for Pseudopotentials : This webpage maintained by the NNIN/C provides a searchable database of pseudopotentials for density functional codes as well as links to pseudopotential generators, converters, and other online databases.
• Vanderbilt Ultra-Soft Pseudopotential Site : Website of David Vanderbilt with links to codes that implement ultrasoft pseudopotentials and libraries of generated pseudopotentials.
• GBRV pseudopotential site : This site hosts the GBRV pseudopotential library
• PseudoDojo : This site collates tested pseudo potentials sorted by type, accuracy, and efficiency, shows information on convergence of various tested properties and provides download options.
• SSSP : Standard Solid State Pseudopotentials | 2,685 | 10,367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 28, "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-2021-17 | latest | en | 0.911146 |
https://www.mpgh.net/forum/2-general/860927-fags.html | 1,513,193,177,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948530668.28/warc/CC-MAIN-20171213182224-20171213202224-00182.warc.gz | 804,406,645 | 24,657 | 1. Fags.
In case there are some smart people left on MPGH,I have a question that'll make you work for a little while.
How many '2's appear from 0-332?
It should take you less than 12 minutes.
Google it and you're a faggot.
2. 69,420,666
...
3. Originally Posted by Aborted
69,420,666
...
Close
4. Your mom take less than 12 minutes.
5. Originally Posted by -Lame
Your mom take less than 12 minutes.
6. 0-332?
2 or 48, depending on how you ask the question. :P 48 if you meant all whole numbers ranging from 0 to 332.
- Edit -
Me being a Cocksucker and all, I dang fucked up. Reposting below.
7. Only one unless they teach you how to count differently where you come from.
8. The numbers I came up with:
2 (1)
12 (1)
22 (1)
32 (1)
42 (1)
52 (1)
62 (1)
72 (1)
82 (1)
92 (1)
102 (1)
112 (1)
122 (2)
132 (1)
142 (1)
152 (1)
162 (1)
172 (1)
182 (1)
192 (1)
202 (2)
212 (2)
222 (3)
232 (2)
242 (2)
252 (2)
262 (2)
272 (2)
282 (2)
292 (2)
302 (1)
312 (1)
322 (2)
332 (1)
Well I fucked up and I'm too tired to fix it. I posted a fix below...
So basically you need to add in the values:
20-29
120-129
200-299
320-329
After that you calculate the number of "2's" found in the numbers. Keep in mind some numbers will contain 3 2's such as 222 and a good amount will have 2 "2's" such as 322 or 122 or 22
Feel free to calculate it yourself, I'm pretty sure I set it all up for you
Oh yeah.. Let's assume its all WHOLE NUMBERS. So decimals won't count.
9. Originally Posted by Predator
The numbers I came up with:
2 (1)
12 (1)
22 (1)
32 (1)
42 (1)
52 (1)
62 (1)
72 (1)
82 (1)
92 (1)
102 (1)
112 (1)
122 (2)
132 (1)
142 (1)
152 (1)
162 (1)
172 (1)
182 (1)
192 (1)
202 (2)
212 (2)
222 (3)
232 (2)
242 (2)
252 (2)
262 (2)
272 (2)
282 (2)
292 (2)
302 (1)
312 (1)
322 (2)
332 (1)
Every digit that contains "2" was added together, thus adding up to: 47
would't all the 200's be like 201 202 203 204 205 206 etc etc... all count as well?
or am i just a dumb ass and confused af?
10. Originally Posted by Mr.Seyeko
would't all the 200's be like 201 202 203 204 205 206 etc etc... all count as well?
or am i just a dumb ass and confused af?
Yeah I literally was just about to go to sleep when I realized that...
So basically you need to add in the values:
20-29
120-129
200-299
320-329
11. Originally Posted by Predator
The numbers I came up with:
2 (1)
12 (1)
22 (1)
32 (1)
42 (1)
52 (1)
62 (1)
72 (1)
82 (1)
92 (1)
102 (1)
112 (1)
122 (2)
132 (1)
142 (1)
152 (1)
162 (1)
172 (1)
182 (1)
192 (1)
202 (2)
212 (2)
222 (3)
232 (2)
242 (2)
252 (2)
262 (2)
272 (2)
282 (2)
292 (2)
302 (1)
312 (1)
322 (2)
332 (1)
Well I fucked up and I'm too tired to fix it. I posted a fix below...
So basically you need to add in the values:
20-29
120-129
200-299
320-329
After that you calculate the number of "2's" found in the numbers. Keep in mind some numbers will contain 3 2's such as 222 and a good amount will have 2 "2's" such as 322 or 122 or 22
Feel free to calculate it yourself, I'm pretty sure I set it all up for you
and 20-29
12. Originally Posted by RedSpot
and 20-29
13. New Updated Answer: Updated my answer, its the first post on page 2
Did it in a hurry so I can get some damn sleep...
2 (1)
12 (1)
(11)
32 (1)
42 (1)
52 (1)
62 (1)
72 (1)
82 (1)
92 (1)
102 (1)
112 (1)
(11)
132 (1)
142 (1)
152 (1)
162 (1)
172 (1)
182 (1)
192 (1)
200 - 210 (11)
(11)
(12)
11
66
302 (1)
312 (1)
11
332 (1)
Pretty much followed what I said on my above post. If I got it wrong, then it must be my work. But who cares... its 12:00 and its time for me to sleep
14. Originally Posted by Predator
I refreshed then baam i see your post, apology master.
15. 2, 12, 22, 32, 42, 52, 62, 72, 82, 92...
Total: 11
200 to 299, there are 100 '2's plus the various '2's above
Total: 122
300 to 332, there is 5 '2's
Total: 127
Page 1 of 3 123 Last | 1,361 | 3,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-51 | latest | en | 0.840787 |
https://www.expertsmind.com/library/determine-the-anualvolume-of-solid-waste-produced-by-5529942.aspx | 1,719,108,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00890.warc.gz | 656,543,160 | 15,053 | ### Determine the anualvolume of solid waste produced by
Assignment Help Civil Engineering
##### Reference no: EM13529942
Solid waste management
the Bailey stone work employs six people. assuming that thedensity of uncompacted waste is 480 kg/m^3 determine the anualvolume of solid waste produced by the stone works assuminga wastegeneration rate of 1kg/cap *d
#### Questions Cloud
Calculate the overflow rate detention time and weir loading : A fine sand has a hydraulic condutivity of 3.5 x 10^-5 m/s a hydraulic gradient of 0.00141, and a porosity of 45 percent. Detrmine the Darcy velocity and the averge linear velocity Find the average mechanical power output of the heart : A patients heart pumps 4.2L of blood per minute into the aorta, which has a diameter of 2.2cm. find the average mechanical power output of the heart An unknown immiscible liquid seeps into the bottom of an : An unknown immiscible liquid seeps into the bottom of an openoil tank. Some measurements indicate that the depth of theunkown liquid is 1.5m and the depth of the oil ( specific weight =8.5 Find the magnitude of the hanging blocks acceleration : Two blocks are connected by a string. The smooth inclined surface makes an angle of 35 degree with the horizontal, Find the magnitude of the hanging block''s acceleration Determine the anualvolume of solid waste produced by : The Bailey stone work employs six people. assuming that thedensity of uncompacted waste is 480 kg/m^3 determine the anualvolume of solid waste produced by the stone works assuminga wastegeneration rate of 1kg/cap *d Obtain the potential difference between the two points : A uniformly charged infinite sheet with a surface charge density +? lies flat in the x-y plane. What is the potential difference DV between the two points Give the phase and the missing property of : Give the phase and the missing property of P, T, v and x. Put the values in a table withcolumns P, T, v, and x. What is the maximum amount of soluble : If the solubility product of magnesium hydroxide is 11.25 answerthe following question. If water at 25-degree C is bufferedsuch that the pH remains fixed at 7, what is the maximum amount ofsoluble Mg2+, in grams, in 1L of water? How fast are is he going at the bottom : The skier starts at a snow covered slope which is inclined 20 degrees above the horizontal. The skier has a mass of 100kg. How fast are is he going at the bottom
### Write a Review
#### Find the centroid for the cone of ice cream
A single-scoop ice cream cone is a composite body made from a single scoop of ice cream placed into a cone. Assume that the scoop of ice cream is a sphere with radius = 3.36cm that is placed into a 9.90cm tall cone.
#### Calculate the q-w and the final temperature of paths
One mole of an ideal monoatomic gas initially at 300 K is expanded from an initial pressure of 10 atm to a final pressure of 1 atm. The molar heat capacity for constant volume for the gas is Cv = 3/2 R .
#### Determine the flow rate when the head extends across channel
A 0.75 m high rectangular sharp-crested weir extends across a 2.2m wide rectangular channel. When the head is 32 cm, determine the flow rate
#### Determine humidity ratio-dew point and mass of dry air
An air-water vapor mixture at 0.1 MPa, 35 C and 70% relative humidity occupies a volume of 100 m^3 Determine the humidity ratio, the dew point, the mass of dry air in the volume and the mass of water vapor in the volume.
#### Explain what is the difference between axial and hoop stress
what is the difference between normal stress and allowable shear stress. What is the difference between axial and hoop stress and how do you calculate tensile force per mm.
#### Determining the compressive strengths
Determine the compressive strengths of the following: A) A 6-in. diameter concrete cylinder that failed at ta test load of 135,000 B) A-3-in. diameter specimen that failed at a test load of 30,000
#### Determine the angle theta which the 6-lb
uniform rod makes with the vertical during the roll, and (b) the speed of the roller after it moves 4 ft., starting from rest. The roller rolls without slipping, and the rod hangs vertically (theta = 0) when the system is at rest.
#### What is the concentration of mlvss in the activated sludge
What is the concentration of MLVSS in the activated sludge (mg L-1)
#### Determine the m and o costs for alternative s
Alternative R has a first cost of \$100,000, annual M&O costs of \$50,000, and a \$20,000 salvage value after 5 years. Alternative S has a first cost of \$175,000 and a \$40,000 salvage value after 5 years
#### Determine the components of the reaction at point c
The object ABC consists of two slender rods welded together at point B. Rod AB has a mass of 1 kg and bar BC has a mass of 2 kg. Knowing the magnitude of the angular velocity of ABC is 10 rad/s when thetha is zero.
#### Calculate how long will it take to completely charge tank
Describe the equation for charging a water tank. Use: H= 2m, A_s = 1m^2, density water = 1000kg/m^3, K = 2000kg hr^-1 Mass flow in = K t. Calculate how long will it take to charge completely the tank
#### Write possible goal-scope and functional unit for the lca
You are preparing a life-cycle assessment of different transportation options for getting from your house to work (10 miles each way). The options include carpooling with three or more people, one person in a car, bicycling, or taking the bus.
#### Assured A++ Grade
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd | 1,353 | 5,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.864743 |
https://math.stackexchange.com/questions/3104969/fiber-bundles-over-spheres | 1,716,755,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00124.warc.gz | 333,579,466 | 36,281 | # Fiber Bundles over spheres
Let $$F$$ be any topological space. In many books, for example in http://pi.math.cornell.edu/~hatcher/VBKT/VBpage.html, it is said that a continuous map (characteristic map) (where $$Homeo(F)$$ has the compact-open topology) $$\phi \colon S^{n-1} \to Homeo(F)$$ define a fiber bundle $$\xi_{\phi}$$ over $$S^n$$ by gluing two trivial bundles with fiber $$F$$ over two hemispheres $$D_+$$ and $$D_-$$, whose total space is $$E = ((D_+ \times F) \coprod (D_- \times F)) / R$$ where the equivalence relation is for $$x \in D_+ \cap D_- = S^{n-1}$$ and $$y \in F$$ $$(x,y) \in (D_+ \times F) \sim (x,\phi(x)(y)) \in D_- \times F).$$ In fact, Hatcher and others write that for vector bundles but here it doesn't matter.
I agree with them if you enlarge both hemispheres so that their intersection is an open equatorial zone. But if you use $$D_+$$ and $$D_-$$, I don't understand why $$\xi_{\phi}$$ is a fiber bundle.
The gluing theorem works for an open covering of the base, but here it is a closed covering, and their intersection iso to $$S^{n-1}$$ is also closed.
I don't see how the local triviality of this $$\xi_{\phi}$$ is obtained for $$x \in S^{n-1}$$. Strictly speaking, for me it is not a fiber bundle...
I doubt that is true for arbitrary spaces $$F$$, but it is true for locally compact $$F$$. This covers most relevant cases.
In the sequel let $$F$$ be locally compact. The exponential law tells us that continuous maps $$\phi : S^{n-1} \to Homeo(F)$$ can be identified with continuous maps $$\phi' : S^{n-1} \times F \to F$$ such that each $$\phi'(x,-) : F \to F$$ is a homeomorphism. Note that these maps can be identified with bundle isomorphims $$\psi : S^{n-1} \times F \to S^{n-1} \times F$$ : To $$\phi'$$ associate $$\psi(x,y) = (x,\phi'(y))$$ and to $$\psi$$ associate $$\psi_F = p_F \circ \psi$$ with projection $$p_F : S^{n-1} \times F \to F$$.
Also observe that in many cases one does not work with $$Homeo(F)$$, but with a subset $$S \subset Homeo(F)$$. For example, if $$F = \mathbb{R}^m$$, then one usually takes $$S = GL(\mathbb{R}^m)$$.
Gluing two trivial bundles over $$D_\pm$$ along $$\psi$$ is obvious and yields a space $$E_\psi$$ with projection $$\pi :E_\psi \to S^n$$. We want to show that $$(E_\psi,\pi)$$ is a fiber bundle. It remains to show local triviality around an arbitrary point $$x \in S^{n-1}$$. Let $$U = S^n \setminus \{ northpole, southpole \}$$. There is a canonical retraction $$r : U \to S^{n-1}$$. Define $$\mu : U \times F \to \pi^{-1}(U)$$ by $$\mu(x,y) = \begin{cases} [x,y] & x \in D_+ \cap U \\ [x,\psi_F(r(x),y)] & x \in D_- \cap U \end{cases}$$ This is a well-defined continuous map since on $$S^{n-1} = D_+ \cap D_-$$ we have $$r(x) = x$$ and $$[x,y] = [x,\psi_F(x,y)]$$. An inverse $$\lambda : \pi^{-1}(U) \to U \times F$$ for $$\mu$$ is given by $$\lambda([x,y]) = \begin{cases} (x,y) & x \in D_+ \cap U \\ (x,\psi^{-1}_F(r(x),y)) & x \in D_- \cap U \end{cases}$$ Note that it is induced by a continuous map $$\lambda' : (D_+ \cap U) \times F \coprod (D_- \cap U) \times F \to U \times F$$, i.e. is itself a continuous map.
• OK, but why the restriction to F locally compact ? You need it here to get the exponential law a bijection, but why do you really need it here ? it seems you could do the same using directly $\phi$ instead of your $\Psi_F$ Feb 8, 2019 at 15:59
• We have to know that $\phi' : S^{n-1} \times F \to F$ is continuous if $\phi$ is continuous. This requires $F$ locally compact. Feb 8, 2019 at 16:05
• But in fact, if you ask that $\phi : S^{n-1} \to Homeo(F)$ be such that the map $S^{n-1} \times F \to F$, $(x,y) \mapsto \phi(x)(y)$ is continuous, your demo will still work. It is true that the supplementary condition is met if $F$ is locally compact", but it may still be satisfied in some larger context. Feb 8, 2019 at 16:22
• I doubt that this is true for non-locally compact $F$. Perhaps it is worth to ask another question concerning this point? Feb 8, 2019 at 17:23
• As I said: Perhaps it is worth to ask a question whether the condition of local compactness is necessary. I do not know the answer. Feb 8, 2019 at 23:23 | 1,353 | 4,152 | {"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": 51, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.871916 |
https://myhomeworkgeeks.com/soc-113-sociology-hypothesis-testing-questions/ | 1,642,455,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00172.warc.gz | 477,089,988 | 13,159 | # SOC 113 Sociology Hypothesis Testing Questions
SOC 113 Sociology Hypothesis Testing Questions.
### Question Description
1. Hypothesis testing: how to form hypotheses (null and alternative); what is the meaning of reject the null or fail to reject the null; how to compare the p-value to the significant level (suchlike alpha = 0.05), and what a smaller p-value means.
Haven’t Found The Relevant Content? Hire a Subject Expert to Help You With
SOC 113 Sociology Hypothesis Testing Questions
2. How to interpret the one-sample t-test results: what are Ho and Ha; the standard for determining statistical significance, i.e., t statistic and p-value; what are the steps for the one-sample t test; what a normal distribution looks like.
3. How to interpret the one-way ANOVA results: what are Ho and Ha; the standard for determining statistical significance, i.e., F statistic and p-value; what an F distribution looks like.
4. How to interpret the simple linear regression results: what are Ho and Ha; the standard for determining statistical significance, i.e., t statistic and p-value of the slope; what is the slope and what it means; what is the R-square (not R, it is R-square!) and what it means; what are independent variables and dependent variable, and what their relationships are; how would you plot the relationship between a dependent variable and an independent variable; from a given independent variable, how would you predict the value of a dependent variable.
5. How to interpret the multiple regression results: how to interpret the slope of an independent variable (i.e., the impact of this independent variable, holding other independent variables constance).
SOC 113 Sociology Hypothesis Testing Questions
Calculate the price of your paper
Pages (550 words)
Approximate price: -
Why Hire a Professional Essay Writer from MyHomeworkGeeks
Quality Urgent Essays
Domyhomeworkk online platform provides the best essay writing service ever received elsewhere, thanks to our professional essay writers. Experience gained in the 10+ years of being in the assignment writing industry is also an added advantage.
A lot is done to maintain us as the best essay writing service provider. Besides owning a degree from a recognized university, a writer must pass the rigorous tests we take them through before they are considered eligible to offer urgent essay help on our website.
Affordable Urgent Assignment Help
Who said a reliable website should not be cheap also? While it is human nature to associate prices with quality, that does not always work. At domyhomeworkk.com, we guarantee you cheap and urgent essay help without compromising the quality.
24/7 Available Urgent Essay Writing Service
Whether it is a ‘write my essay for me cheap’ or ‘edit my essay’ order you want to request, always feel free to reach us at any time of the day. Our experts work in shifts to ensure that you access urgent essay writing services at any time of the day.
Plagiarism Free Essays Online
The punitive measures that come with plagiarized content are so harsh, and we understand that. We consequently ensure to write each client’s papers from scratch no matter how urgent the essay is. To acknowledge borrowed content, our quick writer references and cites the work.
Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.
Try it now!
## Calculate the price of your order
We'll send you the first draft for approval by at
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Our Services
So much stress and so little time? Take care of yourself: let us help you with your tasks. We offer all kinds of writing services.
## Custom Essay Writing Services
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 866 | 4,176 | {"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} | 2.515625 | 3 | CC-MAIN-2022-05 | latest | en | 0.871815 |
www.rainboweducation.ie | 1,719,197,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00525.warc.gz | 836,727,099 | 31,758 | # Box of Numicon Shapes 1-10
8.45
SKU 978-0-19-848731-9 Categories
## Description
Transform how well you teach number in the Infant Classes, and beyond, by using a simple box of Numicon 1 to 10 Shapes.
The 10 individual Numicon shapes are KEY to the programme, each representing a value from 1 to 10. The shapes are used to build number experience and competency to 10, onwards through the teen numbers, onwards to 100, tens & units, multiplication, division, fractions etc.
Let’s just focus on INFANTS.
Here are just a few of the mathematical ideas, concepts and activities that can be engaged in:
NUMICON STARTER ACTIVITIES
Here are just a few of the mathematical ideas, concepts and activities that can be engaged in using the shapes 1 to 10.
1. Identify each shape by assigning value to it i.e. 1 to 10 The value of the shape can be determined by counting the holes. (Alternatively there are bags of 80 pegs available for €15.80 (4 colours x 20 of each) and children can stick pegs into the shape holes e.g. stick 5 pegs into the 5 holes in the RED shape. By doing this they are assigning the value of 5 to the 5 shape which in time they will conserve.
2. Trace the shape, colour the shape in and write the numeral beside it .
3. Ask the child to hold up 3, the child will hold up the yellow. Show me 5, the child holds up the red etc.
4. You hold up a red and yellow – ask the child to hold up the shape that is equivalent value i.e. the green 8.
5. The black bag activity : put all 10 shapes into a black FEELY BAG, any bag that can’t be seen into, hand the bag to a child and ask him /her to reach in, find a shape / number, identify it by touch alone and tell you what that number is, pull it out and check if correct. Great whole class activity. Children take it in turns to put in their hand and using only the sense of touch, to take a shape into their fist, tell you what the number is and then extract their hand to see if he/she is correct. (At first they will feel the number of holes to determine the value of the shape. Later they will tell the number by its overall shape e.g. they will quickly come to know the shape of 5, the size of 10 as being the largest etc. They will even begin to tell you if it is an even or an odd number that is in their hand by whether there is a bit sticking up or not i.e. odd or even, odd doesn’t have a matching partner. There are so many angles to this activity.
Follow on: this black feely bag has 3 shapes in it. The total of the shapes is 8. What shapes might be in it? What a lovely open ended task with so many different but equally correct answers. Great discussion question. Repeat for other combinations e.g. in this bag are 4 shapes that total to 10. What are the 4 shapes?
6. Ordering and naming shapes: ask the children to put the numbers in order from 1 to 10 i.e. put down the orange 1, followed by the light blue 2, the yellow 3, the 4 etc. They can see that each is one more than the previous.
Follow on activity: in pairs the children to put the numbers in order from 1 to 10 i.e. put down the orange 1, followed by the light blue 2, the yellow 3, the 4 etc. They can see that each is one more than the previous. One child then turns their back. The second child removes a shape e.g. the red 5 from the sequence and closes the gap by pushing all the remaining shapes together. The first child turns around and tries to work out which shape or number is missing from the sequence.
7. Cut out numbers 1 to 10 and ask the children to match each shape to its written numeral.
8. Addition: if I add 5 and 3, how many does that make? The child combines the yellow with the red and gets 8, can confirm that he / she is correct by overlaying with the green 8 shape.
9. Story of Number: e.g. what are all the ways we can combine numbers 1 to 9 to make 10? The child takes the blue 10 shape and experiments with overlaying it with combinations of smaller numbers e.g. pink 7 and yellow 3 cover the blue 10, so 7 + 3 = 10. Find all the combinations. This is real maths! Repeat for other numbers e.g. what are all the ways we can make 8?
10. Make number bonds e.g. put the blue 6 and yellow 3 on the purple 9. Make observations and record e.g. 6 + 3 = 9, 3 + 6 = 9, 9 – 3 = 6, 9 – 6 = 3:
11. Here’s a really engaging activity that will drive ral mathematical problem solving and plenty of great maths conversation: take a little black bag, into it drop 3 shapes and then pose the following question: In this bag I have three number shapes and their total is 8. What three shapes could be in the bag?
12. Can you draw some odd Numicon Shapes? Can you draw some even Numicon Shapes? Write the numbers beside your shapes.
And on and on goes the number work possibilities.
The Box of 10 Shapes come in a clear plastic box.
FREE CD with Resource Templates and Activities included when you purchase 10+ boxes + Ideas pamphlet to get you started on your NUMICON Journey.
## Reviews
There are no reviews yet. | 1,262 | 5,042 | {"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} | 4.34375 | 4 | CC-MAIN-2024-26 | longest | en | 0.906527 |
https://math.stackexchange.com/questions/1508753/prove-for-the-inequality-between-arithmetic-and-quadratic-means-fracxy2-leq | 1,566,076,042,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027312025.20/warc/CC-MAIN-20190817203056-20190817225056-00308.warc.gz | 563,679,040 | 31,253 | # Prove for the inequality between arithmetic and quadratic means $\frac{x+y}2\leq \sqrt{\frac{x^2+y^2}2}$
Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality $$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$ holds.
After my calculations I'll get:
$$-x^2+2xy-y^2$$ which can't be $\leq 0$.
Completing the square gives:
$-x^2+2xy-y^2=-(x+y)^2\leq 0$
We can square both sides of your inequality, to get
$$\frac{x^2+2xy+y^2}{4} \leq \frac{x^2+y^2}{2}$$
$$\frac{x^2+2xy+y^2}{4} - \frac{x^2+y^2}{4} \leq \frac{x^2+y^2}{2}-\frac{x^2+y^2}{4}$$
$$\frac{xy}{2} \leq \frac{x^2+y^2}{4}$$
$$2xy \leq x^2+y^2$$
$$0 \leq (x-y)^2$$
This inequality is true, since $x-y$ is a real number and all squares of real numbers are non-negative. We have shown that your inequality is equivalent to one that we know is always true. Therefore yours is always true.
• It is not true that $a^2\geq a$ for all real $a$, but it doesn't matter. We can still justify squaring each side of the inequality. If the LHS is negative then the inequality obviously holds. Otherwise, each side is non-negative, and $f (x)=x^2$ is an increasing function for non-negative $x$. – Dylan Nov 2 '15 at 7:50
• @Dylan Thanks for the correction - not sure why I wrote that, seems very silly in hindsight. – Zubin Mukerjee Nov 2 '15 at 9:41 | 478 | 1,344 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-35 | latest | en | 0.876922 |
http://docplayer.net/21112826-Chapter-one-introduction-to-programming.html | 1,537,970,087,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267164925.98/warc/CC-MAIN-20180926121205-20180926141605-00162.warc.gz | 65,232,525 | 25,203 | # Chapter One Introduction to Programming
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 Chapter One Introduction to Programming 1-1 Algorithm and Flowchart Algorithm is a step-by-step procedure for calculation. More precisely, algorithm is an effective method expressed as a finite list of well-defined instructions for calculating a function. Starting from an initial state, the instructions describe a computation that, when executed, will proceed through a finite number of well-defined successive states, eventually producing "output" and terminating at a final ending state. Flowchart is a graphical representation of an algorithm. These flowcharts play a vital role in the programming of a problem and are quite helpful in understanding the logic of complicated and lengthy problems. Once the flowchart is drawn, it becomes easy to write the program in any high level language. Often we see how flowcharts are helpful in explaining the program to others. Hence, it is correct to say that a flowchart is a must for the better documentation of a complex program. Flowcharts are usually drawn using some standard symbols; however, Start or end of the program Computational steps or processing function of a program Input or output operation Decision making and branching Connector or joining of two parts of program 1
2 The following are some guidelines in flowcharting: a. In drawing a proper flowchart, all necessary requirements should be listed out in logical order. b. The flowchart should be clear, neat and easy to follow. c. The usual direction of the flow of a procedure or system is from left to right or top to bottom. d. Only one flow line should come out from a process symbol. or e. Only one flow line should enter a decision symbol, but two or three flow lines, one for each possible answer, should leave the decision symbol. f. Only one flow line is used in conjunction with terminal symbol. h. If the flowchart becomes complex, it is better to use connector symbols to reduce the number of flow lines. Avoid the intersection of flow lines if you want to make it more effective and better way of communication. i. Ensure that the flowchart has a logical start and finish. j. It is useful to test the validity of the flowchart by passing through it with a simple test data. 2
3 Ex 1-1: Write an algorithm and draw the flowchart for finding the sum of two numbers Algorithm: Input: two numbers x and y Output: the sum of x and y Steps: 1. input x 2. input y 3. sum = x + y 4. output sum START Input x Input y Sum = x + y Output Sum END Q1-1: Write an algorithm and draw the flowchart for finding the area of a rectangle. 1-2 C++ Programming Language Programming is a core activity in the process of performing tasks or solving problems with the aid of a computer. [problem or task specification] - COMPUTER - [solution or completed task] Unfortunately things are not (yet) that simple. In particular, the "specification" cannot be given to the computer using natural language. Moreover, it cannot (yet) just be a description of the problem or task, but has to contain information about how the problem is to be solved or the task is to be executed. Hence we need programming languages. There are many different programming languages, but all of them can be classified into: "High-level" programming languages: These are languages whose syntax is relatively close to natural language. Low-level languages: These include many technical references to the 0's and 1's of the computer. 3
4 C++ is one of the high-level languages; it was developed at Bell Laboratories in the early 1980's, and is based on the C language. C++ is intended as an incremental improvement of C. Most of C is a subset of C++, so that most C programs can be compiled using a C++ compiler. 1-3 Introduction to Programming Ex 1-2: Printing line of text. Line 1: // A first program in C+ begins with //, indicating that the remainder of the line is a comment which describes the purpose of the program. A comment beginning with // is called a single-line comment because it terminates at the end of the current line. Note: You also may use C s style in which a comment (possibly containing many lines) begins with /* and ends with */ /* A first program in C++ Date 2 Oct */ Line: #include <iostream.h> notifies the preprocessor to include in the program the contents of the input/output stream header <iostream.h>. This header must be included for any program that outputs data to the screen or inputs data from the keyboard using C++ s stream input/output. 4
5 The main function: int main ( ) is a part of every C++ program. The parentheses after main indicate that main is a program building block called a function. The left brace, {, must begin the body of every function. A corresponding right brace, }, must end each function s body. An Output Statement cout << Welcome to C++!\n"; instructs the computer to perform an action, to print the string of characters contained between the double quotation marks. The entire line, including cout, the << operator, the string "Welcome to C++!\n" and the semicolon ( ; ), is called a statement. Note: Every C++ statement must end with a semicolon, omitting the semicolon at the end of a C++ statement is a syntax error. Executing the last statement will print the string ( Welcome to C++!) on the screen. You noticed that \n has not been printed, it called a escape sequence. The escape sequence \n means newline. It causes the cursor (i.e., the current screen-position indicator) to move to the beginning of the next line on the screen. Some common escape sequences are listed below: The return Statement return 0; // indicate that program ended successfully is one of several means we ll use to exit a function. The right brace, }, indicates the end of function main. 5
6 * The string Welcome to C++! can be printed by a different way: Ex 1-3: Adding two integer numbers. This program uses the input stream object cin to obtain two integers typed by a user at the keyboard, computes the sum of these values and outputs the result using cout. 6
7 Variable Declarations The identifiers integer1, integer2 and sum are the names of variables. A variable is a location in the computer s memory where a value can be stored for use by a program. These declarations specify that the variables integer1, integer2 and sum are data of type int, meaning that these variables will hold integer values. Note: 1. We could have declared all three variables in one declaration by using a comma-separated list as follows: 2. Declarations of variables can be placed almost anywhere in a program, but they must appear before their corresponding variables are used in the program, for example: Displaying the Result The statement above displays the character string Sum is followed by the numerical value of variable sum followed by endl, the name endl is an abbreviation for end line. Note: Calculations can also be performed in output statements: 7
8 1-4 Arithmetic Operators Integer division yields an integer quotient; for example, the expression 7 / 4 evaluates to 1. C++ provides the modulus operator, %, that yields the remainder after integer division. The modulus operator can be used only with integer operands; for example, 7 % 4 yields 3. Parentheses are used in C++ expressions in the same manner as in algebraic expressions. For example, to multiply a times the quantity b + c we write: a * ( b + c ) C++ applies the operators in arithmetic expressions in a precise order determined by these rules of operator precedence, which are generally the same as those in algebra: 8
9 Ex 1-4: Determine arithmetic mean (average) of five terms. Note: If the parentheses are omitted, we obtain a + b + c + d + e / 5, which evaluates as: Ex 1-5: Convert the following algebraic expression to C++ expression and indicate the order in which C++ applies the operators. Q1-2: Convert the second-degree polynomial y = ax 2 + bx + c to C++ expression and indicate the order in which C++ applies the operators. Ex 1-6: Write a C++ program to calculate the sum and average of three integer numbers. 9
10 Note: The variables sum and average are data of type float, meaning that these variables will hold floating point values. Ex 1-7: Write a C++ program to calculate the circumference of a circle. Note: Constant variables can be declared either: const int S=179; const float PI=3.14; or #define S 179 #define PI
### C++ Programming: From Problem Analysis to Program Design, Fifth Edition. Chapter 2: Basic Elements of C++
C++ Programming: From Problem Analysis to Program Design, Fifth Edition Chapter 2: Basic Elements of C++ Objectives In this chapter, you will: Become familiar with the basic components of a C++ program,
### Introduction to C++ Programming
cpphtp5_02.fm Page 36 Wednesday, December 15, 2004 11:11 AM 2 Introduction to C++ Programming OBJECTIVES In this chapter you will learn: To write simple computer programs in C++. To write simple input
### Chapter 4 Operators & Arithmetic
Chapter 4 Operators & Arithmetic 4-1 Arithmetic in C++ Arithmetic expressions can be made up of constants, variables, operators and parentheses. The arithmetic operators in C++ are as follows + (addition)
### Name: Class: Date: 9. The compiler ignores all comments they are there strictly for the convenience of anyone reading the program.
Name: Class: Date: Exam #1 - Prep True/False Indicate whether the statement is true or false. 1. Programming is the process of writing a computer program in a language that the computer can respond to
### Adjusted/Modified by Nicole Tobias. Chapter 2: Basic Elements of C++
Adjusted/Modified by Nicole Tobias Chapter 2: Basic Elements of C++ Objectives In this chapter, you will: Become familiar with functions, special symbols, and identifiers in C++ Explore simple data types
### Chapter 2: Problem Solving Using C++
Chapter 2: Problem Solving Using C++ 1 Objectives In this chapter, you will learn about: Modular programs Programming style Data types Arithmetic operations Variables and declaration statements Common
### C Programming Dr. Hasan Demirel
C How to Program, H. M. Deitel and P. J. Deitel, Prentice Hall, 5 th edition (3 rd edition or above is also OK). Introduction to C Programming Dr. Hasan Demirel Programming Languages There are three types
### A First Book of C++ Chapter 2 Data Types, Declarations, and Displays
A First Book of C++ Chapter 2 Data Types, Declarations, and Displays Objectives In this chapter, you will learn about: Data Types Arithmetic Operators Variables and Declarations Common Programming Errors
### [Page 70 (continued)] a. are used to document a program and improve its readability. [Page 71]
[Page 70 (continued)] Exercises 2.7 Discuss the meaning of each of the following objects: a. std::cin b. std::cout 2.8 Fill in the blanks in each of the following: a. are used to document a program and
### PROG0101 Fundamentals of Programming PROG0101 FUNDAMENTALS OF PROGRAMMING. Chapter 3 Algorithms
PROG0101 FUNDAMENTALS OF PROGRAMMING Chapter 3 1 Introduction to A sequence of instructions. A procedure or formula for solving a problem. It was created mathematician, Mohammed ibn-musa al-khwarizmi.
### Sources: On the Web: Slides will be available on:
C programming Introduction The basics of algorithms Structure of a C code, compilation step Constant, variable type, variable scope Expression and operators: assignment, arithmetic operators, comparison,
### Chapter 2: Basic Elements of C++
Chapter 2: Basic Elements of C++ Objectives In this chapter, you will: Become familiar with functions, special symbols, and identifiers in C++ Explore simple data types Discover how a program evaluates
### Punctuation in C. Identifiers and Expressions. Identifiers. Variables. Keywords. Identifier Examples
Identifiers and Expressions CSE 130: Introduction to C Programming Spring 2005 Punctuation in C Statements are terminated with a ; Groups of statements are enclosed by curly braces: { and } Commas separate
### Introduction to C++ Programming
2 Introduction to C++ Programming OBJECTIVES In this chapter you will learn: To write simple computer programs in C++. To write simple input and output statements. To use fundamental types. Basic computer
### Programming Fundamental. Instructor Name: Lecture-2
Programming Fundamental Instructor Name: Lecture-2 Today s Lecture What is Programming? First C++ Program Programming Errors Variables in C++ Primitive Data Types in C++ Operators in C++ Operators Precedence
### Control Structure. Pseudocode: Input number If number is even Then Print even Else print odd. Flowchart: Begin. End. Print Odd. number even?
Control Structure Normally, a program is executed in a sequential manner.however, in many cases, a program has to choose among alternative statements C++ provides constructs that enable the programmer
### Lecture 1 Notes: Introduction
Introduction to C++ January 4, 2011 Massachusetts Institute of Technology 6.096 Lecture 1 Notes: Introduction 1 Compiled Languages and C++ 1.1 Why Use a Language Like C++? At its core, a computer is just
### Simple C++ Programs. Engineering Problem Solving with C++, Etter/Ingber. Dev-C++ Dev-C++ Windows Friendly Exit. The C++ Programming Language
Simple C++ Programs Engineering Problem Solving with C++, Etter/Ingber Chapter 2 Simple C++ Programs Program Structure Constants and Variables C++ Operators Standard Input and Output Basic Functions from
### Which of the following is the correct operator to compare two variables? A. := B. = C. equal D. ==
Group 1 Question 1. What is the correct value to return to the operating system upon the successful completion of a program? A. -1 B. 1 C. 0 D. Programs do not return a value. Question 2. What is the only
### 6.096 Lab 1. Due: 7 January 12:30:00. September 20, 2011
6.096 Lab 1 Due: 7 January 1:30:00 September 0, 011 1 Additional Material 1.1 Constants A constant is an expressions with a fixed value. Kinds of constants: Literals: used to express particular values
### GETTING STARTED WITH C++ C++ BASICS - 1 -
- 1 - GETTING STARTED WITH C++ Programming is a core activity in the process of performing tasks or solving problems with the aid of a computer. An idealised picture is: PROBLEM COMPUTER SOLUTION Unfortunately
### Chapter 2: Elements of Java
Chapter 2: Elements of Java Basic components of a Java program Primitive data types Arithmetic expressions Type casting. The String type (introduction) Basic I/O statements Importing packages. 1 Introduction
### Tutorial-3a: First steps with C++ programming
HTTP://WWW.HEP.LU.SE/COURSES/MNXB01 Introduction to Programming and Computing for Scientists (2015 HT) Tutorial-3a: First steps with C++ programming Programming for Scientists Tutorial 3a 1 / 17 Quick
### Problem Solving With C++ Ninth Edition
CISC 1600/1610 Computer Science I Programming in C++ Professor Daniel Leeds dleeds@fordham.edu JMH 328A Introduction to programming with C++ Learn Fundamental programming concepts Key techniques Basic
### Table 3.1: Arithmetic operators
26 Chapter 3 Arithmetic Expressions Java programs can solve mathematical problems. Often programs that appear nonmathematical on the surface perform a large number of mathematical calculations behind the
1 6 JavaScript: Introduction to Scripting 2 Comment is free, but facts are sacred. C. P. Scott The creditor hath a better memory than the debtor. James Howell When faced with a decision, I always ask,
### Dept. of CSE, IIT KGP
Basic Programming Concepts CS10001: Programming & Data Structures Pallab Dasgupta Professor, Dept. of Computer Sc. & Engg., Indian Institute of Technology Kharagpur Some Terminologies Algorithm / Flowchart
### PE1 Worksheet. 3) What are the three control structures for writing an algorithm in pseudocode?
PE1 Worksheet Problem solving 1) What are the four stages of problem solving by programming? What shall be done in each stage? Stage 1: Stage 2: Stage 3: Stage 4: 2) What is the top-down design strategy
### Comp181 Midterm Examination, Fall Name Surname: Number: 90 min. 18/11/09
Comp181 Midterm Examination, Fall 09 10 Name Surname: Number: 90 min. 18/11/09 Answer ALL off the following questions. 1.What does the term hardware refer to? a. The relative difficulty of programming
### Solution Exercise 1.1
Solution Exercise 1.1 Part I Fill in the blanks with the most appropriate answers. 1. A Program is a sequence of step-by-step instructions that a computer executes in order to solve a problem or to perform
### INTRODUCTION TO FLOWCHARTING
CHAPTER 1 INTRODUCTION TO FLOWCHARTING 1.0 Objectives 1.1 Introduction 1.2 Flowcharts 1.3 Types of Flowcharts 1.3.1 Types of flowchart 1.3.2 System flowcharts 1.4 Flowchart Symbols 1.5 Advantages of Flowcharts
### Dept. of CSE, IIT KGP
Programming in C: Basics CS10001: Programming & Data Structures Pallab Dasgupta Professor, Dept. of Computer Sc. & Engg., Indian Institute of Technology Kharagpur Types of variable We must declare the
### C++ Language Tutorial
cplusplus.com C++ Language Tutorial Written by: Juan Soulié Last revision: June, 2007 Available online at: http://www.cplusplus.com/doc/tutorial/ The online version is constantly revised and may contain
### switch Multiple-Selection Statement
switch Multiple-Selection Statement (This feature is rarely useful, although it s perfect for programming the iterative song The Twelve Days of Christmas!) If no match occurs, the default case is executed,
### Chapter 3 Basic Input & Output
Chapter 3 Basic Input & Output 3-1 Basic Output in C++ cout is a predefined variable in C++ that indicates you are going to output a stream of characters to an output device. cout uses an operator called
### CHAPTER 2 PROBLEM SOLVING
CHAPTER 2 PROBLEM SOLVING This chapter will cover the following topics: Problem Solving Concepts for the Computer Pre-Programming Phase Programming Or Implementation Phase What Problem Can Be Solved By
### C++ Basics. C++ Basics: Names: Identifiers. Names: Identifiers. Display 2.1 A C++ Program (1 of 2) Display 2.1 A C++ Program (2 of 2)
C++ Basics C++ Basics: 2.1 Variables and Assignments Variables and Assignments Input and Output Data Types and Expressions Simple Flow of Control Program Style A C++ variable can hold a number or other
### Bachelors of Computer Application Programming Principle & Algorithm (BCA-S102T)
Unit- I Introduction to c Language: C is a general-purpose computer programming language developed between 1969 and 1973 by Dennis Ritchie at the Bell Telephone Laboratories for use with the Unix operating
### Chapter 4 Intro to Programming in C++
Chapter 4 Intro to Programming in C++ 4-1 Computer Programming Earlier we defined an algorithm to be a step by step process for solving a problem. Computer programming is the process of implementing the
1 2 Introduction to Java Applications 2.2 First Program in Java: Printing a Line of Text 2 Application Executes when you use the java command to launch the Java Virtual Machine (JVM) Sample program Displays
### Python to C/C++ Fall 2011
Python to C/C++ Fall 2011 1. Main Program Python: Program code is indented after colon : def main(): body of program C/C++: Have more setup overhead. C: Both require #include directives to access libraries
### 9 Control Statements. 9.1 Introduction. 9.2 Objectives. 9.3 Statements
9 Control Statements 9.1 Introduction The normal flow of execution in a high level language is sequential, i.e., each statement is executed in the order of its appearance in the program. However, depending
### CS101 Computer programming and utilization
CS101 Computer programming and utilization Dr Deepak B Phatak Subrao Nilekani Chair Professor Department of CSE, Kanwal Rekhi Building IIT Bombay Lecture 3, Basic features of C++ Tuesday, 2 August 2011
### Informatica e Sistemi in Tempo Reale
Informatica e Sistemi in Tempo Reale Introduction to C programming Giuseppe Lipari http://retis.sssup.it/~lipari Scuola Superiore Sant Anna Pisa October 25, 2010 G. Lipari (Scuola Superiore Sant Anna)
### Introduction to Programming
Introduction to Programming SS 2012 Adrian Kacso, Univ. Siegen adriana.dkacsoa@duni-siegena.de Tel.: 0271/740-3966, Office: H-B 8406 Stand: April 25, 2012 Betriebssysteme / verteilte Systeme Introduction
### C programming. Intro to syntax & basic operations
C programming Intro to syntax & basic operations Example 1: simple calculation with I/O Program, line by line Line 1: preprocessor directive; used to incorporate code from existing library not actually
### PART-A Questions. 2. How does an enumerated statement differ from a typedef statement?
1. Distinguish & and && operators. PART-A Questions 2. How does an enumerated statement differ from a typedef statement? 3. What are the various members of a class? 4. Who can access the protected members
### The Fundamentals of C++
The Fundamentals of C++ Basic programming elements and concepts JPC and JWD 2002 McGraw-Hill, Inc. Program Organization Program statement Definition Declaration Action Executable unit Named set of program
### BHARATHIAR UNIVERSITY: COIMBATORE CENTRE FOR COLLABORATION OF INDUSTRY AND INSTITUTIONS(CCII) CERTIFICATE IN ADVANCED PROGRAMMING C++ LANGUAGE
Certificate in Advanced Programming - C++ Language Page 1 of 7 BHARATHIAR UNIVERSITY: COIMBATORE 641046 CENTRE FOR COLLABORATION OF INDUSTRY AND INSTITUTIONS(CCII) CERTIFICATE IN ADVANCED PROGRAMMING C++
### Primitive Data Types Summer 2010 Margaret Reid-Miller
Primitive Data Types 15-110 Summer 2010 Margaret Reid-Miller Data Types Data stored in memory is a string of bits (0 or 1). What does 1000010 mean? 66? 'B'? 9.2E-44? How the computer interprets the string
### ALGORITHMS AND FLOWCHARTS
ALGORITHMS AND FLOWCHARTS A typical programming task can be divided into two phases: Problem solving phase produce an ordered sequence of steps that describe solution of problem this sequence of steps
### BITG 1113: Problem Solving
BITG 1113: Problem Solving 1 Learning Outcomes At the end of this lecture, you should be able to: explain the software development methodology identify basic problem solving techniques describe the steps
### Appendix K Introduction to Microsoft Visual C++ 6.0
Appendix K Introduction to Microsoft Visual C++ 6.0 This appendix serves as a quick reference for performing the following operations using the Microsoft Visual C++ integrated development environment (IDE):
### Part IA Computing Course Tutorial Guide to C++ Programming
Part IA Computing Course Tutorial Guide to C++ Programming Roberto Cipolla Department of Engineering University of Cambridge September 1, 2004 This document provides an introduction to computing and the
### Introduction to C++ Programming
Introduction to C++ Programming C++ is one of the most popular programming languages in use in industry today. According to the TIOBE Programming Community Index, which can be accessed at this url: http://www.tiobe.com/index.php/content/paperinfo/tpci/
### Algorithms and Flowcharts Week 2
Algorithms and Flowcharts Week 2 1 Principles of Programming The program or set of programs in a computer that helps in processing the information is called SOFTWARE. Software is a detailed writing of
### INDEX. C programming Page 1 of 10. 5) Function. 1) Introduction to C Programming
INDEX 1) Introduction to C Programming a. What is C? b. Getting started with C 2) Data Types, Variables, Constants a. Constants, Variables and Keywords b. Types of Variables c. C Keyword d. Types of C
### Introduction to Visual C++.NET Programming. Using.NET Environment
ECE 114-2 Introduction to Visual C++.NET Programming Dr. Z. Aliyazicioglu Cal Poly Pomona Electrical & Computer Engineering Cal Poly Pomona Electrical & Computer Engineering 1 Using.NET Environment Start
### Computers process data under the control of sets of instructions called computer programs.
Introduction Computer is a device capable of performing computations and making logical decisions at speeds millions and even billions of times faster than human beings. Computers process data under the
### JavaScript: Control Statements I
1 7 JavaScript: Control Statements I 7.1 Introduction 2 The techniques you will learn here are applicable to most high-level languages, including JavaScript 1 7.2 Algorithms 3 Any computable problem can
### A Rudimentary Intro to C programming
A Rudimentary Intro to C programming Wayne Goddard School of Computing, Clemson University, 2008 Part 1: Variables and Data-types 1 Getting Started............................... A1 2 Some Basics.................................
### Programming Language: Syntax. Introduction to C Language Overview, variables, Operators, Statements
Programming Language: Syntax Introduction to C Language Overview, variables, Operators, Statements Based on slides McGraw-Hill Additional material 2004/2005 Lewis/Martin Modified by Diana Palsetia Syntax
### Going from Python to C
Going from Python to C Darin Brezeale December 8, 2011 Python is a high-level, interpreted language. C has many of the same types of programming constructs as in Python: arrays, loops, conditionals, functions,
### C Syntax and Semantics
C Syntax and Semantics 1 C Program Structure C Language Elements Preprocessor directives Function Header, and Function body Executable statements Reserved word, Standard identifiers, user defined identifiers
### Today s topics. Java programs. Java Virtual Machine (JVM) Bytecodes. HelloWorld.java javac HelloWorld.class
Today s topics Java programs Parsing Java Programming Notes from Tammy Bailey Reading Great Ideas, Chapter 3 & 4 Java programs are created as text files using a text editor (like emacs) Save to disk with.java
### C programming: exercise sheet L2-STUE (2011-2012)
C programming: exercise sheet L2-STUE (2011-2012) Algorithms and Flowcharts Exercise 1: comparison Write the flowchart and associated algorithm that compare two numbers a and b. Exercise 2: 2 nd order
### MIT Aurangabad FE Computer Engineering
MIT Aurangabad FE Computer Engineering Unit 1: Introduction to C 1. The symbol # is called a. Header file c. include b. Preprocessor d. semicolon 2. The size of integer number is limited to a. -32768 to
### C AND C++ PROGRAMMING
C AND C++ PROGRAMMING Bharathidasan University A Courseware prepared by University Informatics Centre Part I - Programming in C Getting Started This courseware is intended to be an introduction to C programming
### QUIZ-II QUIZ-II. Chapter 5: Control Structures II (Repetition) Objectives. Objectives (cont d.) 20/11/2015. EEE 117 Computer Programming Fall-2015 1
QUIZ-II Write a program that mimics a calculator. The program should take as input two integers and the operation to be performed. It should then output the numbers, the operator, and the result. (For
### Input and Output! Objectives" (and predefined functions)" 1E3! Topic 4! n Learn what a stream is and examine input and output streams!
Input and Output! (and predefined functions)" 1E3! Topic 4! 4 I/O 1 Objectives" n Learn what a stream is and examine input and output streams! n Explore how to read data from the standard input device!
### Overview of a C Program
Overview of a C Program Programming with C CSCI 112, Spring 2015 Patrick Donnelly Montana State University Programming with C (CSCI 112) Spring 2015 2 / 42 C Language Components Preprocessor Directives
### Final Exam Review. CS 1428 Fall Jill Seaman. Final Exam
Final Exam Review CS 1428 Fall 2011 Jill Seaman 1 Final Exam Friday, December 9, 11:00am to 1:30pm Derr 241 (here) Closed book, closed notes, clean desk Comprehensive (covers entire course) 25% of your
### Pseudocode, Flowcharts and Python
3 Pseudocode, Flowcharts and Python In Chapter 2, we learned how to store information in the computer and the rules governing the manipulation of numbers and logical values. Now we will look at how to
### Java Basics: Data Types, Variables, and Loops
Java Basics: Data Types, Variables, and Loops If debugging is the process of removing software bugs, then programming must be the process of putting them in. - Edsger Dijkstra Plan for the Day Variables
### Branching. We ll continue to look at the basics of writing C++ programs including. Overview
Branching Overview We ll continue to look at the basics of writing C++ programs including Boolean Expressions (need to know this before we can using branching and looping!) Branching J.S. Bradbury CSCI
### Lecture 1 Introduction to Java
Programming Languages: Java Lecture 1 Introduction to Java Instructor: Omer Boyaci 1 2 Course Information History of Java Introduction First Program in Java: Printing a Line of Text Modifying Our First
### Outline TOPIC 2 INTRODUCTION TO JAVA AND DR JAVA. What is DrJava? Dr Java
1 Outline TOPIC 2 INTRODUCTION TO JAVA AND DR JAVA Notes adapted from Introduction to Computing and Programming with Java: A Multimedia Approach by M. Guzdial and B. Ericson, and instructor materials prepared
### Lecture Set 2: Starting Java
Lecture Set 2: Starting Java 1. Java Concepts 2. Java Programming Basics 3. User output 4. Variables and types 5. Expressions 6. User input 7. Uninitialized Variables CMSC 131 - Lecture Outlines - set
### 6.1. Introduction to Void Functions (Procedures) L E S S O N S E PURPOSE PROCEDURE
L E S S O N S E 6.1 T Introduction to Void Functions (Procedures) PURPOSE PROCEDURE 1. To introduce the concept of void functions (procedures) 2. To work with void functions (procedures) that have no parameters
### OUTCOMES BASED LEARNING MATRIX
Course: CTIM371 Programming in C++ OUTCOMES BASED LEARNING MATRIX Department: Computer Technology and Information Management Course Description: This is the first course in the C++ programming language.
### The American University in Cairo Computer Science & Engineering Department CSCE 106 Fundamentals of Computer Science
Dr. KHALIL The American University in Cairo Computer Science & Engineering Department CSCE 106 Fundamentals of Computer Science Problem Solving & Assignment (2) Programming in C++ Group 1 Write a C++ program
### 3. BINARY NUMBERS AND ARITHMETIC
3. BINARY NUMBERS AND ARITHMETIC 3.1. Binary Numbers The only place in a computer where you will find the number 9 is on the keyboard. Inside the computer you will only find 0 s and 1 s. All computer memory
### ()A Crash Course in Programming with C++ and the Ubuntu O.S. June 8, / 47
A Crash Course in Programming with C++ and the Ubuntu O.S. Dr. Daniel A. Ray MCS Dept UVa-Wise June 8, 2010 ()A Crash Course in Programming with C++ and the Ubuntu O.S. June 8, 2010 1 / 47 Outline What
### Flowchart Techniques
C H A P T E R 1 Flowchart Techniques 1.1 Programming Aids Programmers use different kinds of tools or aids which help them in developing programs faster and better. Such aids are studied in the following
### Introduction to Python Programming. CSE 110: Introduction to Computer Science
Introduction to Python Programming CSE 110: Introduction to Computer Science Announcements Labs begin on Wednesday Labs will meet in CS 2129, NOT the CS SINC site! Homework 1 is due in class on Friday
### Introduction to Programming
Introduction to Programming SS 2012 Adrian Kacso, Univ. Siegen adriana.dkacsoa@duni-siegena.de Tel.: 0271/740-3966, Office: H-B 8406 Stand: April 12, 2012 Betriebssysteme / verteilte Systeme Introduction
### C++ Programming: From Problem Analysis to Program Design, Fifth Edition. Chapter 3: Input/Output
C++ Programming: From Problem Analysis to Program Design, Fifth Edition Chapter 3: Input/Output Objectives In this chapter, you will: Learn what a stream is and examine input and output streams Explore
### Chapter 8. Arithmetic in C++
Christian Jacob Chapter 8 Arithmetic in C++ 8.1 The C++ Vocabulary 8.2 Variables and Types 8.2.1 Data Objects 8.2.2 Variables 8.2.3 Declaration of Variables 8.3 Elementary C++ Data Types 8.3.1 Integers
### Algorithms, Flowcharts & Program Design. ComPro
Algorithms, Flowcharts & Program Design ComPro Definition Algorithm: o sequence of steps to be performed in order to solve a problem by the computer. Flowchart: o graphical or symbolic representation of
### ADVANCED SCHOOL OF SYSTEMS AND DATA STUDIES (ASSDAS) PROGRAM: CTech in Computer Science
ADVANCED SCHOOL OF SYSTEMS AND DATA STUDIES (ASSDAS) PROGRAM: CTech in Computer Science Program Schedule CTech Computer Science Credits CS101 Computer Science I 3 MATH100 Foundations of Mathematics and
### Programming for MSc Part I
Herbert Martin Dietze University of Buckingham herbert@the-little-red-haired-girl.org July 24, 2001 Abstract The course introduces the C programming language and fundamental software development techniques.
### Question 2. Question 3. 0 out of 1 points. The basic commands that a computer performs are, and performance of arithmetic and logical operations.
The basic commands that a computer performs are, and performance of arithmetic and logical operations. input, file Question 2 input, output, storage output, folder storage, directory Main memory is called.
### Topics. Parts of a Java Program. Topics (2) CS 146. Introduction To Computers And Java Chapter Objectives To understand:
Introduction to Programming and Algorithms Module 2 CS 146 Sam Houston State University Dr. Tim McGuire Introduction To Computers And Java Chapter Objectives To understand: the meaning and placement of
### Embedded Systems. Review of ANSI C Topics. A Review of ANSI C and Considerations for Embedded C Programming. Basic features of C
Embedded Systems A Review of ANSI C and Considerations for Embedded C Programming Dr. Jeff Jackson Lecture 2-1 Review of ANSI C Topics Basic features of C C fundamentals Basic data types Expressions Selection
### Moving from CS 61A Scheme to CS 61B Java
Moving from CS 61A Scheme to CS 61B Java Introduction Java is an object-oriented language. This document describes some of the differences between object-oriented programming in Scheme (which we hope you
### Basic C Syntax. Comp-206 : Introduction to Software Systems Lecture 10. Alexandre Denault Computer Science McGill University Fall 2006
Basic C Syntax Comp-206 : Introduction to Software Systems Lecture 10 Alexandre Denault Computer Science McGill University Fall 2006 Next Week I'm away for the week. I'll still check my mails though. No | 7,814 | 34,388 | {"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-2018-39 | longest | en | 0.888985 |
https://pingpdf.com/pdf-buffer-capacity-buffer-ph-depends-on-the-ratio-of-acid-to-base.html | 1,568,797,623,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573264.27/warc/CC-MAIN-20190918085827-20190918111827-00257.warc.gz | 627,284,625 | 6,840 | # BUFFER CAPACITY - Buffer pH depends on the RATIO of acid to base
180
BUFFER CAPACITY
- A buffer is good only as long as there is a significant concentration of both the acidic and basic species - buffer capacity: how much acid or base can a buffer resist before losing its ability to buffer
- Buffer pH depends on the RATIO of acid to base! HendersonHasselbalch Equation Ratio determines pH; the actual concentrations don't! - So, if you make a buffer with 1.0M HA and 1.0M A- , it will have the same pH as a buffer with 2.0M HA and 2.0M A- .... but the 2M buffer will have a higher BUFFER CAPACITY it will resist more additions of acid or base.
181
Buffer calculation: Tris buffer - Tris(hydroxymethyl)-aminomethane
tris base
tris-HCl (conjugate acid of tris base)
Calculate the pH of a buffer made from 50 mL of 0.10M tris and 50 mL of 0.15M tris-HCl. Assume volumes add.
182
Take 100. mL of the previous buffer (0.05 M tris / 0.075 M tris-HCl), and add 5.0 mL of 0.10 M HCl. What is the pH of the mixture? The HCl reacts with the basic component of the buffer, changing it to its conjugate acid (tris-HCl).
We need to find out the NEW concentrations of the species in the buffer system
The volume of the system is now 105 mL due to the added 5.0 mL of HCl! Now, find pH using Henderson-Hasselbalch equation:
The pH of the original buffer was 7.88, so the pH has decreased by 0.07 pH units.
183
Compare this 0.07 unit pH change with adding 5.0 mL of 0.10 M HCl to 100. mL of pure water.
... which is a change of 4.68 pH units from water's original pH of 7.00!
184
INDICATORS -Instead of using a pH meter to monitor acidity, we may choose to use an acid-base INDICATOR.
- Acid-base indicators are weak acids or weak bases which are highly colored. - The color of the undissociated indicator MUST BE DIFFERENT than the color of the dissociated form!
The indicator must be present in very low concentrations so that the indicator's equilibrium DOES NOT CONTROL the pH of the solution!
185
Look at the Henderson-Hasselbalch equation - we want to know how much of the red form and how much of the blue form are present!
When does the color of the indicator change?
IF the pH is > pKa, then the log term above must be both large AND positive!
- What color is the solution?
... so the SOLUTION IS BLUE! - So, the color changes when the pH of the solution is near the pKa of the indicator, BUT we can only DETECT the change when enough of the other form is present.
186
Titration
- also called volumetric analysis. See the end of Ebbing chapter 4 for more details. - frequently used to determine concentration of unknown acids or bases. - typically react a basic sample with a STRONG ACID, or an acidic sample with a STRONG BASE Example: Titrate 20 mL of vinegar (acetic acid) with 0.35 M NaOH. Let's study this titration. What happens to the pH of the solution during the titration? How does an indicator work?
187
Vinegar is typically about 0.88M acetic acid. What would the EQUIVALENCE POINT (the point where we react away all of the acetic acid) be?
But how do we tell the titration is over if we don't already know the concentration of the acid? In the lab, we have used phenolphthalein indicator for vinegar titrations. Phenolphthalein changes from colorless to pink over the range of about pH 9 to pH 10. How does this indicator show where the endpoint is? Let's look at the pH of the solution during the titration- that may show us what's going on!
188
Titration curve for the titration of 20 mL of 0.88 M acetic acid with 0.35 M sodium hydroxide
buffer region: With a moderate amount of NaOH added, we have a solution that contains significant amounts of both acetic acid and its conjugate base (acetate ion). We have a buffer.
189
The equivalence point:
buffer region
EQUIVALENCE POINT
Equivalence point: We're reacting away more and more of the original acetic acid and converting it to acetate ion. At the equivalence point, all of the acetic acid has been converted, and we have only a solution of acetate ion.
190
Let's calculate the pH at the equivalence point.
At the equivalence point, we have 17.6 mmol of ACETATE ION in 70.3 (20+50.3) mL of solution.
Once you figure out the concentration of acetate ion, this is simply the calculation of the pH of a salt solution!
191
PINK
phenolphthalein color change
CLEAR
buffer region
EQUIVALENCE POINT
Near the equivalence point, a very small volume of base added (a drop!) will change the pH from slightly over 6 to near 12. Since phenolphthalein changes colors at about pH 9-10, we can stop the titration within a drop of the equivalence point.
192
Another interesting point: The halfway point
phenolphthalein color change
buffer region
EQUIVALENCE POINT
What's special about it? It's the point where we have added half the required acid to reach the equivalence point
8.8 millimoles is also the amount of acid left, and the added base gets converted to acetate ion!
193
phenolphthalein color change
buffer region
EQUIVALENCE POINT
The total volume is 25.15 mL, and both the acid and base are present at the same concentration. We have a BUFFER. Find the pH of this buffer using the Henderson-Hasselbalch equation.
At the halfway point, the pH = pKa of the acid!
Useful for finding acid ionization constants! | 1,366 | 5,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-39 | latest | en | 0.875949 |
https://math.stackexchange.com/questions/2407210/matrix-calculus-differentiate-powered-quadratic-form | 1,623,553,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487598213.5/warc/CC-MAIN-20210613012009-20210613042009-00244.warc.gz | 367,304,093 | 37,067 | # Matrix Calculus - Differentiate powered quadratic form
I want to differentiate:
$$(w'Aw)^{y}$$
with respect to w (w is a nx1 vector, A is a nxn matrix, A does not depend on w, w' means the transpose of w, y belongs to the set {-1,+1}
I know how to do it when there is no y but since y is present, I have no idea how to handle this. Really appreciate any help
The usual chain rule is going to work here; the answer will be $$\partial_\mu \left( w_\nu A^\nu_\tau w^\tau \right)^n = n \left( A^\mu_\tau w^\tau + w_\nu A^\nu_\mu \right)\left( w_\nu A^\nu_\tau w^\tau \right)^{n-1}$$ | 187 | 585 | {"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} | 2.921875 | 3 | CC-MAIN-2021-25 | latest | en | 0.877659 |
https://www.stefanelli.eng.br/en/category/metrology/page/8/ | 1,579,963,089,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00448.warc.gz | 1,085,616,143 | 12,284 | Top
Commented Solution of the Micrometer in Hundredth Millimeter Exercises – 2
Commented solution of the exercises of reading and interpretation of micrometer in hundredth millimeter - List 2 Solved exercises of reading and interpretation of micrometer in tenth millimeter List Results 2 exercises: micrometer - Exercises of reading and interpretation in millimeter, hundredth resolution - Metrology [pdf-embedder url="/assets/pdf/micrometro-milimetro-centesimo-exercicio-resultado-2.pdf" title="Commented solution...
Commented Solution of Exercise: Reading of Micrometer in Thousandth Millimeter – 2
Commented Solution of List 2 of Exercise: Reading and Interpretation of Micrometer in Thousandth Millimeter [pdf-embedder url="/assets/pdf/micrometro-milimetro-milesimal-exercicio-resultado-2.pdf" title="Commented solution of the exercises of micrometer in thousandth millimeter" description="Commented solution of the List 2 of exercises of reading and interpretation of micrometer in thousandth millimeter"] Micrometer - Exercises...
Exercises: Reading of Vernier Caliper in Decimal Millimeter – 3
list 3 of exercises of reading and interpretation of vernier caliper in millimeter, resolution decimal - 0.1 mm [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-milimetro-exercicios-decimal-3.pdf" title="Exercises of vernier caliper in millimeter decimal" description="List of printable exercises to help the performance of the professor in regular classes, e-learning and to practice/ self-assessment...
Commented Solution: Reading Exercises of Vernier Caliper in Millimeter Decimal – 3
Solution commented list 3 of reading exercises and interpretation of universal caliper with vernier Decimal millimeter - 0.1 mm [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-milimetro-resultado-exercicio-decimal-3.pdf" title="Caliper with vernier in millimeter and decimal resolution" description="Results and comments of the list of reading and interpretation exercises of universal caliper with vernier in...
Exercises Reading of Micrometer in Thousandth Inch – 2
Exercises: Reading and Interpretation of Micrometer in Thousandth Inch - List 2 [pdf-embedder url="/assets/pdf/micrometro-polegada-milesimal-exercicio-2.pdf" title="Exercises of reading and interpretation of micrometer in thousandth inch" description="List 2 of printable exercises to help the performance of the professor in class, e-learning and for practicing / self assessment of...
Exercises: Reading and Interpretation of Micrometer in Tenths of Inch – 2
Exercises of reading and interpretation of micrometer in tenths of inch - list 2 Micrometer in tenths of thousandths of inch. [pdf-embedder url="/assets/pdf/micrometro-polegada-decimo-milesimo-exercicio-2.pdf" title="Exercises of reading and interpretation of micrometer in tenths of inch - list 2" description="List 2 of printable exercises to help the performance of...
Commented Results of Exercises Reading of Micrometer in Thousandth Inch – 2
Commented Results of Reading and Interpretation of Micrometer in Thousandth Inch - List 2 of Exercises [pdf-embedder url="/assets/pdf/micrometro-polegada-milesimo-exercicio-resultado-2.pdf" title="Results of the reading and interpretation exercises of micrometer in thousandth inch" description="Commented results of the printable list 2 of exercises to help the performance of the professor...
Commented Exercises Results – Reading Micrometer in Tenths of Thousandths of Inch – 2
Micrometer in tenths of thousandths of inch. Commented results of the list 2 Solutions of list 2 of exercises of reading and interpretation of micrometer in tenths of thousandth of inch [pdf-embedder url="/assets/pdf/micrometro-polegada-decimo-milesimo-exercicio-resultado-2.pdf" title="Commented results of exercises of micrometer in tenths of thousandths of inch" description="Commented results of...
Exercises of Vernier Caliper in Thousandth Inch 0.001 – List 3
Exercises of reading and interpretation of vernier caliper in thousandth inch - 0.001 in - with 25 divisions - list 3 [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-polegada-milesimal-exercicios-3.pdf" title="Exercises of vernier caliper in thousandth inch 0.001 in" description="Exercises of reading and interpretation of vernier caliper in thousandth inch - 0.001 in - with...
Commented Solution: Reading of Vernier Caliper in Thousandth Inch Exercises – List 3
Commented solution of list 3 of exercises of reading of vernier caliper with vernier scale in thousandth inch Commented results of the exercises list Caliper - Exercises of reading and interpretation in thousandth inch - 3 [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-polegada-milesimal-resultado-exercicio-3.pdf" title="Results of reading exercises of vernier caliper with...
Exercises: Reading of Vernier Caliper in Hundredth Millimeter 0.02 mm – 3
list 3 of exercises of reading and interpretation of vernier caliper in hundredth millimeter, resolution 0.02 mm [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-milimetro-exercicios-02-3.pdf" title="Exercises of reading and interpretation of vernier caliper in millimeter with vernier scale of centesimal 0.02 mm" description="List of printable exercises to help the performance of the professor...
Commented Solution: Reading Exercises of Vernier Caliper in Millimeter 0.02 mm – 3
Commented solution of the list 3 of reading exercises of vernier caliper in millimeter resolution 0.02 mm [pdf-embedder url="/assets/pdf/paquimetro-nonio-vernier-milimetro-resultado-exercicio-02-3.pdf" title="Vernier caliper, graduated ruler in millimeter and resolutions 0.02 mm" description="Commented solution of the list 3 of reading exercises of vernier caliper in millimeter 0.02 mm"] Commented results... | 1,384 | 5,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-05 | latest | en | 0.713091 |
https://www.kopykitab.com/blog/rs-aggarwal-solutions-class-8-maths-chapter-5-playing-with-numbers/ | 1,675,102,752,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499826.71/warc/CC-MAIN-20230130165437-20230130195437-00657.warc.gz | 877,963,056 | 31,038 | # RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Numbers (Updated For 2021-22)
RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Numbers: If you think Class 8 Maths can be quite tricky, we recommend you to study the RS Aggarwal Solutions Class 8 Maths. You can prepare for your Maths exam or finish your assignments with the RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Number. All the solutions are well-explained, credible and accurate.
You can download the Free PDF of RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Number by using the link given in the blog. To know more, read the whole blog.
## Download RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Numbers PDF
RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Numbers
## RS Aggarwal Solutions Class 8 Maths Chapter 5 Playing With Numbers – Overview
In RS Aggarwal Solutions Class 8 Maths Chapter 5, you will cover the following topics:
• Interchanging the digits of a number.
• The generalized form of numbers.
• Tests of divisibility by numbers.
• Cryptarithms.
The last unit of Chapter 5 is based on addition and multiplication of cryptarithms, (cryptarithms are puzzles, operated on various mathematical operations and numbers, like in which letters take the place of digits and one has to find out which letter represents which digit).
You must learn about Applications of cryptarithms as they are very important in higher grades. You can cover all this in the RS Aggarwal Solutions Class 8 Maths.
This is the complete blog on the RS Aggarwal Solutions Class 8 Maths for Chapter 5 Playing With Numbers. To know more about the CBSE Class 8 Maths exam, ask in the comments. | 388 | 1,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-06 | latest | en | 0.906781 |
https://physics.stackexchange.com/questions/605880/why-does-a-1d-hardcore-bosonic-chain-have-different-ground-state-energy-in-boson | 1,713,116,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.19/warc/CC-MAIN-20240414161724-20240414191724-00019.warc.gz | 431,577,897 | 40,973 | # Why does a 1D hardcore bosonic chain have different ground state energy in bosonic and spin representations?
Consider a simple periodic 1D chain with four sites with periodic boundary condition. The Hamiltonian reads
$$H = t c_1^\dagger c_2 + c_2^\dagger c_3 + t c_3^\dagger c_4 + c_4^\dagger c_1 + h.c.$$
where $$t$$ is the hopping strength. In terms of matrix, it is simply
$$H = \left[\begin{array}{cccc} 0 & t & 0 & 1 \\ t & 0 & 1 & 0 \\ 0 & 1 & 0 & t \\ 1 & 0 & t & 0 \end{array}\right]$$
with eigenvalues $$[-1-t,-1+t,1-t,1+t]$$. For $$|t|<1$$, the ground state will be a two-body state, which fills the lowest two eigenlevels such that the ground state energy is $$E=-2$$.
Now let's assume the operators $$c_i$$ are hardcore-bosonic, such that we can make the following Jordan-Wigner type of transformation
$$c^\dagger_j = (X_j + i Y_j)/2\\ c_j = (X_j - i Y_j)/2$$
where $$X_j,Y_j$$ are the Pauli operators at site $$j$$. As one can check, the operators satisfy $$[c_i^\dagger,c_j]=0$$ if $$i\neq j$$ and $$\left\{c_i^\dagger,c_j\right\}=1$$ if $$i=j$$. In this representation, we shall make the following identification
$$c_1^\dagger c_2 = \frac{X_1+iY_1}{2}\otimes\frac{X_2-iY_2}{2}\otimes I_3\otimes I_4$$
similarly for other terms, such that the Hamiltonian reads
\begin{align*} H =\ &t \frac{X_1+iY_1}{2}\otimes\frac{X_2-iY_2}{2}\otimes I_3\otimes I_4\\ &+ I_1\otimes\frac{X_2+iY_2}{2}\otimes \frac{X_3-iY_3}{2}\otimes I_4 \quad \\ &+ t I_1\otimes I_2\frac{X_3+iY_3}{2}\otimes\frac{X_4-iY_4}{2} \\ &+ \frac{X_1-iY_1}{2}\otimes I_2\otimes I_3\otimes\frac{X_4+iY_4}{2} \\ &+ h.c. \end{align*}
which is a $$16\times 16$$ matrix that contains all the possible many-body state. One could also diagonalize it to obtain the ground state energy, and for $$|t|<1$$, I found $$E=-2\sqrt{1+t^2}$$, which is different from the previous result.
One could repeat the calculation for chains of length $$2L$$, and I found that the two methods wil give the same ground state energy if $$L$$ is odd, while there seems to be discrepancy if $$L$$ is even. This is quite strange to me, and I could not find any mistake. Any help is greatly appreciated!
• Hardcore bosons have a different spectrum than soft bosons Jan 7, 2021 at 14:23
• @JahanClaes Sure, and I explicitly use the commutator for the hard boson here. But this does not explain why the spectrum changes when I change from hard boson representation to spin representation, when L is even.
– fagd
Jan 7, 2021 at 18:25
• Fermions? Bosons? Hardcore bosons? (And which of those where?) The mapping from spins/hardcore bosons to fermions can have antiperiodic boundary conditions depending on the parity. Jan 7, 2021 at 22:19
• @NorbertSchuch The system is hardcore boson, and this is the basis for the first Hamiltonian. Then I try to transform it to spins, which is the basis for the second Hamiltonian. There is no fermion involved here. This is supposed to be trivial, and I thought the boundary condition does not play a role here. Am I making a mistake here?
– fagd
Jan 8, 2021 at 0:10
• You say (correctly) that $H$ is a $16\times16$ matrix, but the matrix you wrote explicitly is $4\times4$. Clearly they are not the same matrix.
– fqq
Jan 8, 2021 at 1:48
## 1 Answer
In your second approach, you correctly solve the hardcore boson problem (in essence, hardcore bosons are exactly two-level spin systems).
In your first approach, on the other hand, you diagonalize the single-particle problem. As an approach to solve the many-particle problem, where you subsequently fill all the negative energy modes, this only works for non-interacting particles, either bosons or fermions. It will, however, not work for hardcore bosons (which are bosons with an infinite on-site repulsion).
However, you are also not solving free bosons in your first approach: In that case, you would have to put an infinite number of bosons in each mode with negative energy, leading to an energy $$-\infty$$. (Differently speaking, for free bosons, the single-mode energies of the Hamiltonian must all be positive).
So what you are doing is solving the free fermion problem, where you fill each negative energy mode with one fermion.
So how does this compare to the hard-core boson problem (and why do you get the same result for odd $$L$$)?
When you do the Jordan-Wigner transformation from fermions to spins ( = hard-core bosons), you will get pretty much the same Hamiltonian (probably with a overall minus sign, but your free fermion energies are symmetric around zero, so this is no big business). However, the term across the boundary will be different: It will be either periodic or antiperiodic, depending on the total fermion parity of the ground state. If I am not mistaken, it should be periodic for an odd ground state parity. This is what is going on for odd $$L$$: Half of the modes are filled in the ground state, so there is an odd number of fermions, and the fermionic problem does indeed map to the hardcore bosons, and thus has the same energy.
• Thanks for the reply, it at least gives some explanation for the problem. So essentially there is no single-body method to solve the many-body hardcore boson problem, because it is intrinsically interacting problem? I thought after we take into account the onsite repulsion, the hardcore problem can be treated as noninteracting.
– fagd
Jan 8, 2021 at 18:32
• @fagd Nope, it cannot. That's why these problems are hard. (Not "super-hard", because they don't have a sign problem and you can do Monte Carlo, as opposed to interacting fermions, but still hard.) And what do you mean by "it at least gives some explanation for the problem" -- what is missing? Jan 8, 2021 at 20:06
• Yes, I do have another question. Based on the logic above, if I have even L and periodic boundary condition, then the spin Hamiltonian can be mapped to neither hardcore boson nor fermion. How about if I have even L and anti-periodic boundary condition? Can it be mapped to hard-core boson problem?
– fagd
Jan 8, 2021 at 20:09
• You have explained it very well, and really appreciate that. But it is counter-intuitive, as I thought the boundary condition is irrelevant because the transformation for the boson case is local. Now I realize it is actually not the case. And in fact, I have tested that if the boundary condition is open, then regardless of whether L is even or odd, the spectrum is the same.
– fagd
Jan 8, 2021 at 20:13
• @fagd Pretty much. Well, almost: In one dimension, there is no difference between fermions and hard-core bosons, except for the boundary conditions. So if you get your boundary conditions right, you can solve your hard-core bosons by mapping them to free fermions. But this approach only works in one dimension. Jan 8, 2021 at 21:17 | 1,904 | 6,767 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-18 | latest | en | 0.759166 |
https://nzmaths.co.nz/connected-level-2 | 1,702,111,704,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00733.warc.gz | 471,548,056 | 12,573 | # Connected Level 2
## Connected Number and Algebra, Level 2
### Amazing Algorithms
Connected Level 2, 2018: Step by Step
ContextAlgorithms
MathematicsAlgebra: Using step by step instructions to solve number problems.
Cross CurriculumTechnology
### Busy Bees
Connected Level 2, 2012: The Buzz of Bees, p.9
ContextHoney bees
MathematicsNumber: Students calculate honey-producing rates of, and distances travelled by, honey bees.
Cross CurriculumScience: Living World
### Living with a Volcano
Connected 1, 2011: Rūaumoko Rages, p.28
ContextVolcanoes
MathematicsNumber: The number of passengers grounded at Heathrow Airport, as a result of a volcanic eruption in Iceland, is calculated from data.
Cross CurriculumScience: Physical World
### Finding Friday
Connected 1, 2007, p.26
ContextA classroom is kept tidy by sorting and counting classroom materials.
MathematicsNumber: Stage 5 number strategies are applied to practical classroom tasks.
Cross CurriculumKey competencies
### Multiplication Monday
Connected 1, 2004, p.2
ContextStudents look for number patterns about them in the school grounds and the classroom.
MathematicsNumber: A connection is developed between skip counting and multiplication.
Cross CurriculumN/A
### A Bird’s breakfast
Connected 1, 2004, p.8
ContextFictional birds are catching worms and sharing worm segments fairly.
MathematicsNumber: Division problems posed and these are supported by precise drawings that show the numbers in each problem. Connections are made between fractions, division and multiplication.
Cross CurriculumScience: Living World
### The Monster Birthday Party
Connected 1, 2003, p.2
ContextA monster mother is providing party food for young monsters and ensuring it is fairly shared.
MathematicsNumber: Simple division problems are posed within the context of sharing different kinds of party food.
Cross CurriculumN/A
### Out of Step
Connected 1, 2002, p.26
ContextThis is a fictional story about three animals that escape from a zoo and are being pursued.
MathematicsAlgebra: Problems posed in the story require the reader to continue different sequential patterns and explore relationships between them.
Cross CurriculumEnglish
### Baby Maths
Connected 1, 2001, p.12
ContextA short poem tells about a baby’s body.
MathematicsNumber/Geometry: The numbers in the poem highlight the body’s symmetry.
Cross CurriculumEnglish
### Snails
Connected 1, 2001, p.14
ContextA boy earns money by removing unwanted snails from his father’s garden.
MathematicsNumber: Place value is explored through collections of ten snails, each of which is worth ten cents.
Cross CurriculumScience: Living World
### The Case of the 13-year-old Grandmother
Connected 1, 2000, p.28
ContextA teacher has a leap year birthday.
MathematicsNumber/Measurement: The multiplicative relationship between the number of leap years and the number of actual years is represented and explored with a double number line.
Cross CurriculumSocial Sciences
### Numbers, Numbers Everywhere
Connected 1, 2000, p.30
ContextNumbers are found in many real life situations.
MathematicsNumber: Readers are encouraged to recognize where we find numbers in our daily lives and how they help us.
Cross CurriculumSocial Sciences
### Eggs in the Nest
Connected 1, 1999, p.8
ContextA game for two players
MathematicsNumber: Single digit subtraction strategies are investigated.
Cross CurriculumN/A
### My Amazing Plant
Connected 1, 1999, p.9
ContextA plant is growing quickly in a glasshouse.
MathematicsNumber: Doubling and halving are investigated.
Cross CurriculumScience: Living World
### Racehorse Rhyme
Connected 1, 1999, p.29
ContextA rhyme, a riddle and a joke.
MathematicsNumber: This is a play on single digit number words.
Cross CurriculumEnglish
### Patterns
Connected 1, 1999, p.30
ContextPhotographs of patterns made with physical objects.
MathematicsAlgebra: Pattern attributes of size, shape and colour are identified and recorded on a chart.
Cross CurriculumN/A
### A Number Grid Game
Connected 1, 1998, p.26
ContextA Number Grid Game
MathematicsNumber: This game involves the consecutive addition of 21 single digit numbers.
Cross CurriculumN/A
## Connected Geometry and Measurement, Level 2
### Squawkzilla
Connected Level 2, 2020: Digging Deeper
ContextFossils
MathematicsMeasurement: Sizes of bones from different birds are compared.
Cross CurriculumScience: Living World
### Designer Homes
Connected Level 2, 2014
ContextDesigning a children’s playground.
MathematicsGeometry: Students make a 3D model of an original playground design that is informed by survey data.
Cross CurriculumSocial Sciences and The Arts
### Designer Homes
Connected Level 2, 2012: The Buzz of Bees, p.28
ContextHoney bees
MathematicsGeometry: The features of regular 2D shapes are explored, squares are counted to calculate their area, and the tessellations of hexagons and circles are compared.
Cross CurriculumScience: Living World
### Bumble Bees
Connected Level 2, 2012: The Buzz of Bees, p.8
ContextBees
MathematicsMeasurement: The body measurements of bumble bees and honey bees are compared.
Cross CurriculumScience: Living World
### Investigating insulation
Connected 1, 2010: Staying Warm and Keeping Cool, p.23
ContextA science experiment is undertaken to find the best material to trap heat.
MathematicsMeasurement: Students read and record temperatures at regular intervals.
Cross CurriculumScience: Physical/Material World
### Heat Thieves
Connected 1, 2010: Staying Warm and Keeping Cool, p.23
ContextExploring conditions that effect our body temperature.
MathematicsMeasurement: Students read a wind-chill table, relating wind speed and air temperature.
Cross CurriculumScience: Physical/Material World
### Monsters of the Deep
Connected 1, 2009, p.2
ContextSquid
MathematicsMeasurement: The body measurements of different kinds of squid are presented for discussion.
Cross CurriculumScience: Living World
### Te Papa’s Colossal Squid
Connected 1, 2009, p.10
ContextThe capture, preservation and examination of a colossal squid.
MathematicsMeasurement: A range of measurements (length, capacity, temperature, weight) of a colossal squid is presented for discussion.
Cross CurriculumScience: Living World
### Bird Land
Connected 1 and 2, 2008, p.8
ContextFossils reveal the nature of some NZ creatures from the past.
MathematicsMeasurement: A limited number of time and measurement data are given and these can form the basis of discussion.
Cross CurriculumScience: Living World
### Nailing it down
Connected 1, 2006, p.20
ContextA father and son are building a wooden paling fence.
MathematicsMeasurement: A fence length is measured using metres and centimetres, and skip counting and multiplication are applied to solve a problem.
Cross CurriculumTechnology
### Centre Stage
Connected 1, 2005, p.20
ContextA school idol competition is being performed on a circular rotating stage.
MathematicsGeometry: The full rotation of a circular disk is completed by 60 small turns. Quarter, half and three-quarter turns are explored with reference to this measure.
Cross CurriculumThe Arts
### Finding the Shapes in Buildings
Connected 1, 2002, p.30
ContextGeometric shapes are identified in photographs of buildings from different cultures.
MathematicsGeometry: Two-dimensional and three-dimensional shapes are defined.
Cross CurriculumThe Arts and Technology
### Baby Maths
Connected 1, 2001, p.12
ContextA short poem tells about a baby’s body.
MathematicsNumber/Geometry: The numbers in the poem highlight the body’s symmetry.
Cross CurriculumEnglish
### The Giant Who Needed a Bed
Connected 1, 2000, p.22
ContextThe fictional story tells about a giant who accepts hospitality from the people of a village.
MathematicsMeasurement: It is inferred that body measurements can be calculated from the length of a footprint.
Cross CurriculumEnglish
### The Case of the 13-year-old Grandmother
Connected 1, 2000, p.28
ContextA teacher has a leap year birthday.
MathematicsNumber/Measurement: The multiplicative relationship between the number of leap years and the number of actual years is represented and explored with a double number line.
Cross CurriculumSocial Sciences
### Make a spinner
Connected 1, 1999, p.6
ContextInstructions are given to make a paper spinner.
MathematicsMeasurement: A circle, its diameter and radius are informally investigated in this practical task.
Cross CurriculumScience: Physical World
### What do you see?
Connected 1, 1998, p.12
ContextOptical illusions
MathematicsGeometry: The length of lines on two trapeziums is compared.
Cross CurriculumThe Arts
### Clocks with a Difference
Connected 1, 1998, p.14
ContextTime is shown in different ways on several unusual clocks.
MathematicsMeasurement: Digital and analogue time is explored in an unusual way.
Cross CurriculumTechnology
### How Do You Measure a Dinosaur?
Connected 1, 1998, p.20
ContextA practical investigation into the actual size of Tyrannosaurus rex.
MathematicsMeasurement: A picture of a dinosaur is enlarged onto a grid using metre measures.
Cross CurriculumScience: Living World
### Sun Facts
Connected 1, 1998, p.6
ContextSun facts
MathematicsMeasurement: Mathematical facts about the sun are presented for discussion.
Cross CurriculumScience: Planet Earth and Beyond
## Connected Statistics, Level 2
### City of Bugs
Connected Level 2, 2020: Digging Deeper
ContextBugs
MathematicsStatistics: The number and type of invertebrates in different areas are counted.
Cross CurriculumScience: Living World
Technology
### Sea Science
Connected Level 2, 2019: Wild Discoveries
ContextMarine debris
MathematicsStatistics: Data on the types of marine debris found at different locations.
Cross CurriculumScience: Nature of science
### The War on Weeds
Connected Level 2, 2018: Step by Step
ContextMarine pollution
MathematicsStatistics: Using a database of weeds.
Cross CurriculumScience: Living World
Technology
### Why do our muscles get tired?
Connected Level 2, 2015
ContextInvestigating muscle fatigue.
MathematicsStatistics: Designing an investigation, collecting and analyzing (finger) fitness data.
Cross CurriculumHealth and PE
### Operation Duck Pond
Connected Level 2, 2015
ContextBreeding habitats for ducks.
MathematicsStatistics: Describes the process of collecting scientific data and introduces outliers.
Cross CurriculumScience
### The Takeaway Table
Connected Level 2, 2013: I Spy..., p.20
ContextBirds within a school environment.
MathematicsStatistical Investigations: Students undertake an investigation and collect data to inform them of the range of bird types in their school environment. This enables students to develop an appropriate bird table and feeding plan.
Cross CurriculumScience: Living World
### Look out for Monarchs
Connected Level 2, 2013: I Spy..., p.28
ContextMonarch Butterflies.
MathematicsStatistical Investigations: Students are encouraged to gather data about tagged and numbered monarch butterflies anywhere in NZ. By becoming ‘citizen scientists’, they are helping the Monarch Butterfly New Zealand Trust.
Cross CurriculumScience: Living World
### What’s for lunch?
Connected 1, 2006, p.26
ContextDiscussion of the food pyramid begins an investigation into the nature of the lunches of the students in one classroom.
MathematicsStatistical Investigations: Food-type data are collected using tally charts, displayed in a bar graph, and discussed.
Cross CurriculumHealth and PE
### Bat Maths
Connected 1, 2002, p.24
ContextInformation is presented about the number of bat calls recorded in two locations: above a road and through a forest.
MathematicsStatistical Literacy: The reader is prompted to consider the conclusions drawn about bat call data that are presented in a table and on a bar graph.
Cross CurriculumScience: Living World
### How the Children of Room 3 Keep Their Clothes On.
Connected 1, 1999, p.25
ContextButton, zips, velcro, buckles, laces and elastic help to keep our clothes on.
MathematicsStatistical Investigations: Data are collected using a tally chart, and are presented on a bar graph.
Cross CurriculumSocial Sciences
### How Much Do They Eat?
Connected 1, 1998, p.32
ContextA comparison is made of the amount eaten by a cow, a sheep, a dog, a cat and a hen.
MathematicsStatistical Literacy: The reader interprets a bar graph showing the kilogram amounts of food eaten by selected animals, and answers questions posed.
Cross CurriculumScience: Living World | 2,791 | 12,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | latest | en | 0.851629 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-10-section-10-4-adding-subtracting-and-multiplying-radical-expressions-exercise-set-page-710/31 | 1,531,725,003,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00432.warc.gz | 902,769,951 | 14,338 | ## Algebra: A Combined Approach (4th Edition)
$6\sqrt[3] {11}-4\sqrt {11}$
$6\sqrt[3] {11}+8\sqrt {11}-12\sqrt {11}$ Combine like radicals. $6\sqrt[3] {11}+(8-12)\sqrt {11}$ $6\sqrt[3] {11}-4\sqrt {11}$ | 91 | 203 | {"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} | 3.796875 | 4 | CC-MAIN-2018-30 | latest | en | 0.655994 |
https://www.cfd-online.com/Forums/fluent/40947-cavitation-venturi.html | 1,513,603,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948616132.89/warc/CC-MAIN-20171218122309-20171218144309-00740.warc.gz | 725,686,515 | 15,400 | # Cavitation in venturi
Register Blogs Members List Search Today's Posts Mark Forums Read
May 12, 2006, 05:08 Cavitation in venturi #1 pranav Guest Posts: n/a Hi I am working on venturi cavitation by using fluent. Does anybody know how to get the temperature at collapse phase of Cavitation? Thanks
May 13, 2006, 12:51 Re: Cavitation in venturi #2 CFDtoy Guest Posts: n/a Hi Pranav: In reality, the temperature increase incase of venturi cav is that not high. Thts why people decouple energy eqn, If you want to solve for temp. Include, Energy Eqn and make vapor phase compressible. You will get what you want CFDtoy !
May 13, 2006, 15:59 Re: Cavitation in venturi #3 pranav Guest Posts: n/a Hi CFDtoy, Thank you for your response. I am interested on vapour phase temperature & pressure not for mixtures. In reality, the temperature & pressure rise is more in vapour phase while collapsing stage of cavitation bubbles. Please share with me, If you have any other ideas. Pranav
May 14, 2006, 11:45 Re: Cavitation in venturi #4 Donald Guest Posts: n/a It is not possible to do this in fluent without a correct equation of state. You need gas E.O.S. more adanced than ideal gas. D.
May 15, 2006, 01:19 Re: Cavitation in venturi #5 mateus Guest Posts: n/a Hi! I think that what you try to do is wery interesting but also very hard. There was a paper of cfd simulation including thermal effects at the CAV2003 conference (just write cav2003 in google - you´ll find a link to the papers there). But there they used a different approach of modelling - the barotropic cavitation model. You´ll probably also not get far with vof and bubble dynamics (reyleigh plesset equation) because you´d need a very dense mesh for that - cell size 1/10 of the bubble size...that´s approximatelly 10^-5m (or even less in the beginning and end stage of bubble evolution). I hope I helped a bit... If you have any questions about cavitation modeling i am heppy to help, just write me an email - i did a lot of work in this area (but none considering thermal effects. By mateus
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules
Similar Threads Thread Thread Starter Forum Replies Last Post xhliu1 NUMECA 2 June 27, 2005 17:16 xhliu1 Siemens 11 March 16, 2005 08:19 Liu, L. CFX 2 November 29, 2000 14:50 Liu, L. Siemens 2 November 1, 2000 22:51 Liu, L. Main CFD Forum 7 November 1, 2000 22:26
All times are GMT -4. The time now is 09:17. | 735 | 2,678 | {"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-2017-51 | latest | en | 0.895076 |
https://maksimdan.gitbook.io/interview-practice-problems/coding_practice_questions/stacks_and_queues/stack-of-plates | 1,708,530,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00306.warc.gz | 404,616,462 | 169,097 | # Stack of Plates
3.3 Stack of Plates: Imagine a (literal) stack of plates. If the stack gets too high, it might topple. Therefore, in real life, we would likely start a new stack when the previous stack exceeds some threshold. Implement a data structure SetOfStacks that mimics this. SetOfStacks should be composed of several stacks and should create a new stack once the previous one exceeds capacity. SetOfStacks. push () and SetOfStacks. pop () should behave identically to a single stack (that is, pop ( ) should return the same values as it would if there were just a single stack).
FOLLOW UP Implement a function popAt (int index) which performs a pop operation on a specific sub-stack.
template <class T>
class Stack {
public:
Stack(int capacity) {
cap_s = capacity;
iter_s = 0;
data = new T[capacity];
}
~Stack() {
delete[]data;
}
bool push_s(T thing) {
if (full()) {
return false;
}
else {
data[iter_s] = thing;
iter_s++;
return true;
}
}
bool pop_s() {
if (empty_s()) {
return false;
}
else {
iter_s--;
return true;
}
}
T top_s() {
if (empty_s()) {
return INT_MAX;
}
return data[iter_s - 1];
}
bool empty_s() {
return iter_s == 0;
}
private:
// array
T *data;
int cap_s;
int iter_s;
bool full() {
return iter_s == cap_s;
}
};
template <class T>
class setOfStacks {
public:
setOfStacks(const int capacity) {
Stack<T> *s = new Stack<T>(capacity);
stacks.push_back(s);
cap = capacity;
iter = 0;
}
void push(T thing) {
// current stack is full
if (stacks.at(iter)->push_s(thing) == false) {
Stack<T> *s = new Stack<T>(cap);
stacks.push_back(s);
iter++;
stacks.at(iter)->push_s(thing);
}
}
void pop() {
// everything is empty
if (empty()) {
cout << "stack is empty" << endl;
return;
}
// current stack is empty
else if (subEmpty()) {
// delete current stack, and move on the previous
delete stacks.at(iter);
iter--;
stacks.at(iter)->pop_s();
}
else {
stacks.at(iter)->pop_s();
}
}
T top() {
// everything is empty
if (empty()) {
cout << "stack is empty" << endl;
return INT_MAX;
}
// current stack is empty
else if (subEmpty()) {
// pop or top can be preformed at any point so
// the share the same three lines
delete stacks.at(iter);
stacks.resize(stacks.size() - 1);
iter--;
stacks.at(iter)->top_s();
}
// current stack is empty
else {
return stacks.at(iter)->top_s();
}
}
bool empty() {
return (stacks.size() == 1 && stacks.at(iter)->empty_s());
}
private:
bool subEmpty() {
return (stacks.at(iter)->empty_s());
}
int cap;
int iter;
vector<Stack<T>*> stacks;
};
int main() {
cout << "// testing stack" << endl;
Stack<int> s(50);
for (int i = 0; i <= 50; i++) {
s.push_s(i);
}
// overflow
cout << s.push_s(51);
pause();
while (!s.empty_s()) {
cout << s.top_s() << " ";
s.pop_s();
}
// underflow
cout << s.pop_s();
pause();
cout << "// testing set of Stacks" << endl;
setOfStacks<int> ss(5);
for (int i = 0; i <= 100; i++) {
ss.push(i);
cout << ss.top() << " ";
//pause();
}
// next stack
ss.push(101);
//pause();
while (!ss.empty()) {
cout << ss.top() << " ";
ss.pop();
}
pause();
// completely empty
ss.pop();
pause();
} | 865 | 3,030 | {"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-2024-10 | latest | en | 0.629278 |
https://us.metamath.org/ileuni/resin4p.html | 1,717,077,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00229.warc.gz | 501,960,450 | 8,679 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > resin4p GIF version
Theorem resin4p 11431
Description: Separate out the first four terms of the infinite series expansion of the sine of a real number. (Contributed by Paul Chapman, 19-Jan-2008.) (Revised by Mario Carneiro, 30-Apr-2014.)
Hypothesis
Ref Expression
efi4p.1 𝐹 = (𝑛 ∈ ℕ0 ↦ (((i · 𝐴)↑𝑛) / (!‘𝑛)))
Assertion
Ref Expression
resin4p (𝐴 ∈ ℝ → (sin‘𝐴) = ((𝐴 − ((𝐴↑3) / 6)) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
Distinct variable groups: 𝐴,𝑘,𝑛 𝑘,𝐹
Allowed substitution hint: 𝐹(𝑛)
Proof of Theorem resin4p
StepHypRef Expression
1 resinval 11428 . 2 (𝐴 ∈ ℝ → (sin‘𝐴) = (ℑ‘(exp‘(i · 𝐴))))
2 recn 7760 . . . . 5 (𝐴 ∈ ℝ → 𝐴 ∈ ℂ)
3 efi4p.1 . . . . . 6 𝐹 = (𝑛 ∈ ℕ0 ↦ (((i · 𝐴)↑𝑛) / (!‘𝑛)))
43efi4p 11430 . . . . 5 (𝐴 ∈ ℂ → (exp‘(i · 𝐴)) = (((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6)))) + Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘)))
52, 4syl 14 . . . 4 (𝐴 ∈ ℝ → (exp‘(i · 𝐴)) = (((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6)))) + Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘)))
65fveq2d 5425 . . 3 (𝐴 ∈ ℝ → (ℑ‘(exp‘(i · 𝐴))) = (ℑ‘(((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6)))) + Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
7 1re 7772 . . . . . . 7 1 ∈ ℝ
8 resqcl 10367 . . . . . . . 8 (𝐴 ∈ ℝ → (𝐴↑2) ∈ ℝ)
98rehalfcld 8973 . . . . . . 7 (𝐴 ∈ ℝ → ((𝐴↑2) / 2) ∈ ℝ)
10 resubcl 8033 . . . . . . 7 ((1 ∈ ℝ ∧ ((𝐴↑2) / 2) ∈ ℝ) → (1 − ((𝐴↑2) / 2)) ∈ ℝ)
117, 9, 10sylancr 410 . . . . . 6 (𝐴 ∈ ℝ → (1 − ((𝐴↑2) / 2)) ∈ ℝ)
1211recnd 7801 . . . . 5 (𝐴 ∈ ℝ → (1 − ((𝐴↑2) / 2)) ∈ ℂ)
13 ax-icn 7722 . . . . . 6 i ∈ ℂ
14 3nn0 9002 . . . . . . . . . 10 3 ∈ ℕ0
15 reexpcl 10317 . . . . . . . . . 10 ((𝐴 ∈ ℝ ∧ 3 ∈ ℕ0) → (𝐴↑3) ∈ ℝ)
1614, 15mpan2 421 . . . . . . . . 9 (𝐴 ∈ ℝ → (𝐴↑3) ∈ ℝ)
17 6re 8808 . . . . . . . . . 10 6 ∈ ℝ
18 6pos 8828 . . . . . . . . . . 11 0 < 6
1917, 18gt0ap0ii 8397 . . . . . . . . . 10 6 # 0
20 redivclap 8498 . . . . . . . . . 10 (((𝐴↑3) ∈ ℝ ∧ 6 ∈ ℝ ∧ 6 # 0) → ((𝐴↑3) / 6) ∈ ℝ)
2117, 19, 20mp3an23 1307 . . . . . . . . 9 ((𝐴↑3) ∈ ℝ → ((𝐴↑3) / 6) ∈ ℝ)
2216, 21syl 14 . . . . . . . 8 (𝐴 ∈ ℝ → ((𝐴↑3) / 6) ∈ ℝ)
23 resubcl 8033 . . . . . . . 8 ((𝐴 ∈ ℝ ∧ ((𝐴↑3) / 6) ∈ ℝ) → (𝐴 − ((𝐴↑3) / 6)) ∈ ℝ)
2422, 23mpdan 417 . . . . . . 7 (𝐴 ∈ ℝ → (𝐴 − ((𝐴↑3) / 6)) ∈ ℝ)
2524recnd 7801 . . . . . 6 (𝐴 ∈ ℝ → (𝐴 − ((𝐴↑3) / 6)) ∈ ℂ)
26 mulcl 7754 . . . . . 6 ((i ∈ ℂ ∧ (𝐴 − ((𝐴↑3) / 6)) ∈ ℂ) → (i · (𝐴 − ((𝐴↑3) / 6))) ∈ ℂ)
2713, 25, 26sylancr 410 . . . . 5 (𝐴 ∈ ℝ → (i · (𝐴 − ((𝐴↑3) / 6))) ∈ ℂ)
2812, 27addcld 7792 . . . 4 (𝐴 ∈ ℝ → ((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6)))) ∈ ℂ)
29 mulcl 7754 . . . . . 6 ((i ∈ ℂ ∧ 𝐴 ∈ ℂ) → (i · 𝐴) ∈ ℂ)
3013, 2, 29sylancr 410 . . . . 5 (𝐴 ∈ ℝ → (i · 𝐴) ∈ ℂ)
31 4nn0 9003 . . . . 5 4 ∈ ℕ0
323eftlcl 11401 . . . . 5 (((i · 𝐴) ∈ ℂ ∧ 4 ∈ ℕ0) → Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘) ∈ ℂ)
3330, 31, 32sylancl 409 . . . 4 (𝐴 ∈ ℝ → Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘) ∈ ℂ)
3428, 33imaddd 10739 . . 3 (𝐴 ∈ ℝ → (ℑ‘(((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6)))) + Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))) = ((ℑ‘((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6))))) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
3511, 24crimd 10756 . . . 4 (𝐴 ∈ ℝ → (ℑ‘((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6))))) = (𝐴 − ((𝐴↑3) / 6)))
3635oveq1d 5789 . . 3 (𝐴 ∈ ℝ → ((ℑ‘((1 − ((𝐴↑2) / 2)) + (i · (𝐴 − ((𝐴↑3) / 6))))) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))) = ((𝐴 − ((𝐴↑3) / 6)) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
376, 34, 363eqtrd 2176 . 2 (𝐴 ∈ ℝ → (ℑ‘(exp‘(i · 𝐴))) = ((𝐴 − ((𝐴↑3) / 6)) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
381, 37eqtrd 2172 1 (𝐴 ∈ ℝ → (sin‘𝐴) = ((𝐴 − ((𝐴↑3) / 6)) + (ℑ‘Σ𝑘 ∈ (ℤ‘4)(𝐹𝑘))))
Colors of variables: wff set class Syntax hints: → wi 4 = wceq 1331 ∈ wcel 1480 class class class wbr 3929 ↦ cmpt 3989 ‘cfv 5123 (class class class)co 5774 ℂcc 7625 ℝcr 7626 0cc0 7627 1c1 7628 ici 7629 + caddc 7630 · cmul 7632 − cmin 7940 # cap 8350 / cdiv 8439 2c2 8778 3c3 8779 4c4 8780 6c6 8782 ℕ0cn0 8984 ℤ≥cuz 9333 ↑cexp 10299 !cfa 10478 ℑcim 10620 Σcsu 11129 expce 11355 sincsin 11357 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 603 ax-in2 604 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-13 1491 ax-14 1492 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2121 ax-coll 4043 ax-sep 4046 ax-nul 4054 ax-pow 4098 ax-pr 4131 ax-un 4355 ax-setind 4452 ax-iinf 4502 ax-cnex 7718 ax-resscn 7719 ax-1cn 7720 ax-1re 7721 ax-icn 7722 ax-addcl 7723 ax-addrcl 7724 ax-mulcl 7725 ax-mulrcl 7726 ax-addcom 7727 ax-mulcom 7728 ax-addass 7729 ax-mulass 7730 ax-distr 7731 ax-i2m1 7732 ax-0lt1 7733 ax-1rid 7734 ax-0id 7735 ax-rnegex 7736 ax-precex 7737 ax-cnre 7738 ax-pre-ltirr 7739 ax-pre-ltwlin 7740 ax-pre-lttrn 7741 ax-pre-apti 7742 ax-pre-ltadd 7743 ax-pre-mulgt0 7744 ax-pre-mulext 7745 ax-arch 7746 ax-caucvg 7747 This theorem depends on definitions: df-bi 116 df-dc 820 df-3or 963 df-3an 964 df-tru 1334 df-fal 1337 df-nf 1437 df-sb 1736 df-eu 2002 df-mo 2003 df-clab 2126 df-cleq 2132 df-clel 2135 df-nfc 2270 df-ne 2309 df-nel 2404 df-ral 2421 df-rex 2422 df-reu 2423 df-rmo 2424 df-rab 2425 df-v 2688 df-sbc 2910 df-csb 3004 df-dif 3073 df-un 3075 df-in 3077 df-ss 3084 df-nul 3364 df-if 3475 df-pw 3512 df-sn 3533 df-pr 3534 df-op 3536 df-uni 3737 df-int 3772 df-iun 3815 df-br 3930 df-opab 3990 df-mpt 3991 df-tr 4027 df-id 4215 df-po 4218 df-iso 4219 df-iord 4288 df-on 4290 df-ilim 4291 df-suc 4293 df-iom 4505 df-xp 4545 df-rel 4546 df-cnv 4547 df-co 4548 df-dm 4549 df-rn 4550 df-res 4551 df-ima 4552 df-iota 5088 df-fun 5125 df-fn 5126 df-f 5127 df-f1 5128 df-fo 5129 df-f1o 5130 df-fv 5131 df-isom 5132 df-riota 5730 df-ov 5777 df-oprab 5778 df-mpo 5779 df-1st 6038 df-2nd 6039 df-recs 6202 df-irdg 6267 df-frec 6288 df-1o 6313 df-oadd 6317 df-er 6429 df-en 6635 df-dom 6636 df-fin 6637 df-pnf 7809 df-mnf 7810 df-xr 7811 df-ltxr 7812 df-le 7813 df-sub 7942 df-neg 7943 df-reap 8344 df-ap 8351 df-div 8440 df-inn 8728 df-2 8786 df-3 8787 df-4 8788 df-5 8789 df-6 8790 df-n0 8985 df-z 9062 df-uz 9334 df-q 9419 df-rp 9449 df-ico 9684 df-fz 9798 df-fzo 9927 df-seqfrec 10226 df-exp 10300 df-fac 10479 df-ihash 10529 df-cj 10621 df-re 10622 df-im 10623 df-rsqrt 10777 df-abs 10778 df-clim 11055 df-sumdc 11130 df-ef 11361 df-sin 11363 This theorem is referenced by: sin01bnd 11470
Copyright terms: Public domain W3C validator | 3,962 | 6,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.125079 |
https://allanimalsguide.com/what-are-base-rates-2/ | 1,720,898,091,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00499.warc.gz | 82,557,201 | 20,482 | # What Are Base Rates
What Are Base Rates?
Base rates are the percentage of people in a population who have a certain characteristic. In business base rates are often used to make decisions about marketing sales and product development. For example a company might use base rates to decide which products to sell in which markets.
Base rates can be useful but they have limitations. First base rates can be difficult to calculate. For example it can be hard to know how many people in a population have a certain characteristic. Second base rates can be misleading. For example a small base rate might be misleading if the population is very large.
Third base rates can change over time. For example the base rate of people with a certain characteristic might increase as the population ages.
Fourth base rates can be affected by other factors. For example the base rate of people with a certain characteristic might be different in different countries.
Finally base rates can be used in different ways. For example a company might use base rates to decide which products to sell in which markets. Or a company might use base rates to decide which customers to target with a new product.
Base rates can be a useful tool but it’s important to understand their limitations.
## What is the definition of a base rate?
A base rate is the rate that applies before any modifications or additions.
## What is the base rate for electric service in Texas?
The base rate for electric service in Texas is 10.
5 cents per kilowatt-hour.
## What is the base rate for natural gas service in Texas?
The base rate for natural gas service in Texas is \$3.
50 per month plus \$0.
032 per therm.
## What is the base rate for water service in Texas?
The base rate for water service in Texas is \$26.
00 for the first 2000 gallons of water used.
## What is the base rate for sewer service in Texas?
The base rate for sewer service in Texas is \$26.
00 per month.
## What is the base rate for trash service in Texas?
The base rate for trash service in Texas is \$8.
25 per month.
## What is the base rate for phone service in Texas?
The base rate for phone service in Texas is \$9.
99 per month.
## What is the base rate for internet service in Texas?
The base rate for internet service in Texas is \$19.
99 per month.
## What is the base rate for cable TV service in Texas?
The base rate for cable TV service in Texas is \$29.
99 per month.
## What is the base rate for electricity in California?
The base rate for electricity in California is 13.
3 cents per kilowatt-hour.
## What is the base rate for natural gas in California?
The base rate for natural gas in California is \$5.
50 per month plus \$0.
030 per therm.
## What is the base rate for water in California?
The base rate for water in California is \$35.
00 for the first 2000 gallons of water used.
## What is the base rate for sewer in California?
The base rate for sewer in California is \$35.
00 per month.
## What is the base rate for trash in California?
The base rate for trash in California is \$9.
25 per month.
## What is the base rate for phone in California?
The base rate for phone in California is \$10.
99 per month. | 688 | 3,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.937871 |
https://www.briangwilliams.us/climates-weather/the-coriolis-effect-and-the-oceans.html | 1,638,001,887,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358153.33/warc/CC-MAIN-20211127073536-20211127103536-00190.warc.gz | 768,665,502 | 10,340 | # The Coriolis Effect And The Oceans
So far we have considered ocean temperature and salinity, two of the factors controlling currents in the sea, which in turn govern coastal climates. A third factor is the 'Coriolis effect', the 'apparent deflection of moving objects, due to the observer being on a rotating Earth', named after Gaspard de Coriolis (1792-1843).
Unfortunately, it is not easy to understand immediately, being different from commonsense observation, and so various explanations are offered in what follows.
As a preliminary, think of the Earth rotating once each day, and an observer looking at the Sun. There is a paradox, since we know that it is the Earth (and observer) which actually turn, yet the observer sees the Sun as moving. Reality differs from appearance. Likewise, if you sit on a roundabout, it is the rest of the world which seems to spin. In brief, the 'rotation of the observer is perceived as the rotation of the observed'.
In the same way, a rotating observer, looking at an object which is really moving in a straight line, sees the object as turning. This is illustrated in Figure 11.10. Drawing along a straight ruler creates a curved line on a rotating disc. Laying the ruler in any direction over a disc turning clockwise, and moving the pencil in either direction along the ruler, always causes the trace to bend to the left, whereas turning the disc counter-clockwise instead always causes sidling to the right. These different directions of rotation correspond to conditions in the two hemispheres of the globe, where an observer at the South Pole sees the Earth turning clockwise (i.e. the left hand advances towards the Sun), whereas an observer's right hand moves forward at the North Pole (Figure 11.11). In between, at the equator, there is no turning round by the observer, i.e. neither the left hand moves to take the place of the right nor vice versa.
The Coriolis effect can also be understood in terms of a projectile from a cannon at A fixed facing a target B (Figure 11.12), all on a disc representing the Earth. The cannon and the target turn a little while the shell is in the air, and during that time the target moves from B to B', so that when the shell lands at B it is behind the target. As a result, the actually straight-line trajectory AB (as seen from space) seems to curve (A'B), deflected to the left when observed where the Earth rotates clockwise. Figure 11.12 and Note 11.D show that this is true whichever way the cannon is pointing, whether zonally (i.e. east-west) or meridionally (i.e. north-south).
Theory shows that the effect depends solely on latitude and the object's velocity (Note 11 .D). The outcome is that any straight-line motion (as observed from a fixed point in space, such as the Sun) appears circular to a person on Earth,
Figure 11.10 Demonstration of the Coriolis effect. A line drawn along the straight ruler registers a curve on the rotating disc, i.e. straight-line motion appears curved when viewed from a rotating platform.
as though the moving object is continually pushed to one side by a force. This hypothetical force is called the Coriolis force and is most important in understanding oceanic and atmospheric motions. It is summarised in Ferrel's Law, that 'all motion suffers a bias towards the left in the southern hemisphere (and right in the northern)'.
The Coriolis force is negligible near the equator and on the scale of water going down a plug-hole, for instance (Note 11.D). It is significant only on a scale of many kilometres, affecting global winds and large-scale currents in the oceans, particularly at high latitudes.
### Upwelling
One important consequence of the Coriolis effect is the way in which winds over the ocean move the surface water. Friction at the surface drags the uppermost water along with the wind, but simultaneously the Coriolis effect operates, deflecting the moving water to the left (in the southern hemisphere). As a result, that top layer of the ocean slowly moves at an angle approaching 45 degrees to the wind. The top
Figure 11.11 Explanation of the opposite directions of rotation in the two hemispheres of the Earth. The left hand advancing in the southern hemisphere means a clockwise rotation there. At the equator, there is no rotation about an axis perpendicular to the surface, and therefore no Coriolis effect. (A radian is an angle of 57.3°.)
Figure 11.11 Explanation of the opposite directions of rotation in the two hemispheres of the Earth. The left hand advancing in the southern hemisphere means a clockwise rotation there. At the equator, there is no rotation about an axis perpendicular to the surface, and therefore no Coriolis effect. (A radian is an angle of 57.3°.)
layer in turn drags the layer beneath, which again is affected by the Coriolis force, so that the second layer moves at a greater angle to the wind. Similarly for lower layers, each moving more slowly and more at an angle than the layer above. With each layer's movement represented by an arrow of appropriate direction and a length proportional to the speed, we have the arrangement shown in Figure 11.13. The tips of the arrows trace an 'Ekman spiral', named after the Swedish oceanographer Vagn Ekman (1874-1954). He proposed this spiral in 1902 to explain Fridtjof Nansen's observation that icebergs move at an angle of around 30° to the right of the wind in the northern hemisphere. The outcome of a complete spiral is an average movement, called Ekman transport, which for the whole spiral amounts to motion at right angles to the wind, and the top layer of the ocean (i.e. the Ekman layer) is driven towards that direction. This is a surprising result, that the Coriolis effect causes the ocean to move perpendicular to the wind—towards the left in the southern hemisphere. But a complete spiral develops only in deep water. Often a shallow thermocline, or the sea-bed in shallow waters, limits downwards transfer of momentum, and then surface ocean currents are more closely aligned with the wind.
The result of Ekman transport is upwelling of deeper, cold water to the surface near some coasts, as illustrated in Figure 11.14. The upwelling happens wherever there is either a polewards wind parallel to an east coast, or an equatorwards wind parallel to a west coast, as in north Chile or Namibia. The wind creates Ekman transport of the warm surface water away from the land, and cold deep water rises to takes its place at a rate of a metre per day or so. This is one explanation for the low
Figure 11.12 Demonstrations of the apparent deflections of actually straight-line motions (relative to the Sun, for instance) which are either (a) radial (corresponding to movement along a line of longitude), or (b) circumferential (along a line of latitude). In both cases, the rotation is shown as clockwise (as in the southern hemisphere—see Figure 11.11), and in both cases the apparent deflection is to the left.
Figure 11.12 Demonstrations of the apparent deflections of actually straight-line motions (relative to the Sun, for instance) which are either (a) radial (corresponding to movement along a line of longitude), or (b) circumferential (along a line of latitude). In both cases, the rotation is shown as clockwise (as in the southern hemisphere—see Figure 11.11), and in both cases the apparent deflection is to the left.
Figure 11.13 The Ekman spiral in the southern hemisphere. The wind is towards the top left (parallel to the right-hand long side of the base), and Ekman transport is towards the bottom left, parallel to the nearest short side of the base.
temperature of waters within 20 km of the Peruvian coast, and for the climate of Lima at 12°S (Note 11.E, Table 11.3). The upwelling off Lima results in an SST which is 3 K less than that 2,400 km nearer the South Pole at Antafagasta in Chile. Similarly, pleasantly cool conditions near Rio de Janeiro are induced by the upwelling caused by occasional north winds, even though they come from the equator. Coastal upwelling off south-west
upwelling
Figure 11.14 Wind-induced ocean currents and upwelling at a coast in the southern hemisphere.
upwelling
Figure 11.14 Wind-induced ocean currents and upwelling at a coast in the southern hemisphere.
Africa, south of 15°S, leads to SSTs near the shore which are as much as 5 K lower than at 320 km out to sea, particularly in summer and at 26°S. This leads to low temperatures in the atmospheric PBL, under a strong inversion (Figure 7.10).
Upwelling also occurs along the equator in the eastern Pacific, because winds from the east deflect the water southward (to the left) just south of the equator, and northward (to the right) just north of the equator, where the Coriolis force though weak acts in the opposite direction. This creates a furrow in the equatorial surface water, which in turn causes upwelling there, and consequently surface temperatures of only about 19°C off the Galapagos Islands, on the equator in the Pacific ocean. Decline of the easterly winds prior to an El Niño (Note 11.C) leads to a rapid end to upwelling and
Table 11.3 Comparison of the average climates of Darwin and Lima, both at 1 2°S
Darwin Lima
Table 11.3 Comparison of the average climates of Darwin and Lima, both at 1 2°S
Darwin Lima
July January July January Daily mean temp. (°C) 25 28 15 21 Sunshine (hours/day) 9.8 6.1 3.4 5.1 Raindavs/month <1 19 1 <1 Rainfall (mm/mo) 1 391 2 1.2 Wind direction at 3 p.m. E NW S S Wind speed (m/s) 2.5 2.6 2.5 3.5
consequently a dramatic rise of sea-surface temperature.
There are plenty of fish where there is upwelling, partly because the deep water which is brought up contains the deposited nutritious debris of previous generations of fish. Another reason is that cold waters contain more oxygen, e.g. water at 0°C can hold twice as much as water at 25°C. Ninety per cent of the world's fish are caught in the 15 per cent of the oceans where upwelling takes place.
+2 0
### Responses
Can oceans transfer heat by producing the coriolis effect?
9 months ago
• leigh
Does corlolis effect surface water?
2 years ago
• tristan
Which weather feature is Coriolis Force negligible?
2 years ago
• juho
How does the coriolis effect affect the oceans?
2 years ago
• Ulla-Maj
Which direction does s csnnon shell due to coriolus?
2 years ago | 2,413 | 10,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-49 | latest | en | 0.924772 |
http://www.stata.com/statalist/archive/2009-09/msg00032.html | 1,496,099,177,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00355.warc.gz | 812,918,666 | 6,582 | [Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
# Re: st: Survival analysis
From moleps islon To statalist@hsphsun2.harvard.edu Subject Re: st: Survival analysis Date Wed, 2 Sep 2009 11:32:44 +0200
```Thanks go out to both Kieran and Marten for providing immense help.
In doing a multivariate survival analysis with several time-varying
covariates (found using estat phtest,det and graphs after "regular"
stcox) i fitted the model
(1) xi:stcox a b c d e i.g, tvc(b d e i.g)
after first having done univariate modeling with
(2) stcox a, tvc(a) etc
finding significant interaction with time for b d e & i.g (hence the tvc var).
-How do I know from this that (1) fulfills the PH assumption and that
(_t) is the "correct" form of interaction with time (instead of ln _t
etc)?
-When the time interaction is significant in the univariate model, but
not in the multivariate model-how do I determine the reason for this ?
Appreciate your time and efforts,
M
On Wed, Sep 2, 2009 at 12:08 AM, Kieran McCaul<Kieran.McCaul@uwa.edu.au> wrote:
> ...
>
> There isn't anything counterintuitive here: look at the K-M graph.
>
> The failures in icp=0 occur in greater number then those in icp=1. They also occur more quickly. In icp=1 the failures take longer to accrue. So overall, icp looks good, but if you start conditioning on some initial period of survival, icp is going to start to look "bad". That's because the flat region in the survival curve for those on icp=0 occurs earlier than it does for those on icp=1. So if you condition on about 2 days of survival, you are in the flat region of icdp=0 (essential no more failures occurring after this), but there are still failure occurring in icp=1. So icp=1 starts to look "bad".
>
> It isn't: icp is doing two things. First, it's reducing the risk of failure overall and second, it's delaying failure in those who do ultimately fail.
>
> If this were a disease, I would say that without icp people most people survive, but those who don't succumb quickly. It's like cholera: most people survive, but those who die, die quickly. So, people have an ability to fight off the disease. With icp, more people survive and those who eventually fail take longer to fail. So, if this were cholera, icp would be like a treatment that tended to reduce the severity of the cholera symptoms and increased the ability of people to fight off the disease.
>
>
> ______________________________________________
> Kieran McCaul MPH PhD
> WA Centre for Health & Ageing (M573)
> University of Western Australia
> Level 6, Ainslie House
> 48 Murray St
> Perth 6000
> Phone: (08) 9224-2701
> Fax: (08) 9224 8009
> email: Kieran.McCaul@uwa.edu.au
> http://myprofile.cos.com/mccaul
> http://www.researcherid.com/rid/B-8751-2008
> ______________________________________________
> If you live to be one hundred, you've got it made.
> Very few people die past that age - George Burns
>
>
> -----Original Message-----
> From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of moleps islon
> Sent: Wednesday, 2 September 2009 2:55 AM
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: Survival analysis
>
> Dear Marten and other listers,
> Somehow that looks contraintuitive looking at the graphs (and from
> what i understand I cannot post graphs or links here). But if you look
> at the following output you`ll see that from patients surviving >.2
> ,.5,1 and 4 days the logrank test points in the direction of a
> beneficial effect first, but detrimental effect afterwards. The PH
> assumtion is not fulfilled initially, but later. Isn't this suggestive
> of a breakpoint somewhere around 1 day ??
>
>
>
> Regards,
> M
>
>
> patients surviving >.2 days
>
> failure _d: dod
> analysis time _t: cox
>
> Iteration 0: log likelihood = -2393.672
> Iteration 1: log likelihood = -2374.6741
> Iteration 2: log likelihood = -2374.4875
> Iteration 3: log likelihood = -2374.4874
> Refining estimates:
> Iteration 0: log likelihood = -2374.4874
>
> Cox regression -- Breslow method for ties
>
> No. of subjects = 971 Number of obs = 971
> No. of failures = 357
> Time at risk = 248731.5
> LR chi2(1) = 38.37
> Log likelihood = -2374.4874 Prob > chi2 = 0.0000
>
> ------------------------------------------------------------------------------
> _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> icp | .4842698 .0598328 -5.87 0.000 .3801186 .616958
> ------------------------------------------------------------------------------
>
> Test of proportional-hazards assumption
>
> Time: Time
> ----------------------------------------------------------------
> | chi2 df Prob>chi2
> ------------+---------------------------------------------------
> global test | 25.20 1 0.0000
> ----------------------------------------------------------------
>
> failure _d: dod
> analysis time _t: cox
>
>
> Log-rank test for equality of survivor functions
>
> | Events Events
> icp | observed expected
> ------+-------------------------
> 0 | 270 214.69
> 1 | 87 142.31
> ------+-------------------------
> Total | 357 357.00
>
> chi2(1) = 38.71
> Pr>chi2 = 0.0000
> patients surviving >.5 days
>
> failure _d: dod
> analysis time _t: cox
>
> Iteration 0: log likelihood = -1506.3678
> Iteration 1: log likelihood = -1505.0847
> Iteration 2: log likelihood = -1505.0842
> Refining estimates:
> Iteration 0: log likelihood = -1505.0842
>
> Cox regression -- Breslow method for ties
>
> No. of subjects = 842 Number of obs = 842
> No. of failures = 228
> Time at risk = 248667
> LR chi2(1) = 2.57
> Log likelihood = -1505.0842 Prob > chi2 = 0.1091
>
> ------------------------------------------------------------------------------
> _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> icp | .8043308 .1102187 -1.59 0.112 .614884 1.052146
> ------------------------------------------------------------------------------
>
> Test of proportional-hazards assumption
>
> Time: Time
> ----------------------------------------------------------------
> | chi2 df Prob>chi2
> ------------+---------------------------------------------------
> global test | 10.23 1 0.0014
> ----------------------------------------------------------------
>
> failure _d: dod
> analysis time _t: cox
>
>
> Log-rank test for equality of survivor functions
>
> | Events Events
> icp | observed expected
> ------+-------------------------
> 0 | 143 131.12
> 1 | 85 96.88
> ------+-------------------------
> Total | 228 228.00
>
> chi2(1) = 2.64
> Pr>chi2 = 0.1044
> patients surviving >1 days
>
> failure _d: dod
> analysis time _t: cox
>
> Iteration 0: log likelihood = -994.44857
> Iteration 1: log likelihood = -992.59906
> Iteration 2: log likelihood = -992.59879
> Refining estimates:
> Iteration 0: log likelihood = -992.59879
>
> Cox regression -- Breslow method for ties
>
> No. of subjects = 766 Number of obs = 766
> No. of failures = 152
> Time at risk = 248591
> LR chi2(1) = 3.70
> Log likelihood = -992.59879 Prob > chi2 = 0.0544
>
> ------------------------------------------------------------------------------
> _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> icp | 1.366676 .2218148 1.92 0.054 .9942913 1.878528
> ------------------------------------------------------------------------------
>
> Test of proportional-hazards assumption
>
> Time: Time
> ----------------------------------------------------------------
> | chi2 df Prob>chi2
> ------------+---------------------------------------------------
> global test | 2.66 1 0.1032
> ----------------------------------------------------------------
>
> failure _d: dod
> analysis time _t: cox
>
>
> Log-rank test for equality of survivor functions
>
> | Events Events
> icp | observed expected
> ------+-------------------------
> 0 | 74 85.81
> 1 | 78 66.19
> ------+-------------------------
> Total | 152 152.00
>
> chi2(1) = 3.79
> Pr>chi2 = 0.0516
> patients surviving >2 days
>
> failure _d: dod
> analysis time _t: cox
>
> Iteration 0: log likelihood = -788.57192
> Iteration 1: log likelihood = -783.00942
> Iteration 2: log likelihood = -783.00902
> Refining estimates:
> Iteration 0: log likelihood = -783.00902
>
> Cox regression -- Breslow method for ties
>
> No. of subjects = 735 Number of obs = 735
> No. of failures = 121
> Time at risk = 248529
> LR chi2(1) = 11.13
> Log likelihood = -783.00902 Prob > chi2 = 0.0009
>
> ------------------------------------------------------------------------------
> _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> icp | 1.840018 .3397673 3.30 0.001 1.28128 2.64241
> ------------------------------------------------------------------------------
>
> Test of proportional-hazards assumption
>
> Time: Time
> ----------------------------------------------------------------
> | chi2 df Prob>chi2
> ------------+---------------------------------------------------
> global test | 0.65 1 0.4187
> ----------------------------------------------------------------
>
> failure _d: dod
> analysis time _t: cox
>
>
> Log-rank test for equality of survivor functions
>
> | Events Events
> icp | observed expected
> ------+-------------------------
> 0 | 50 68.29
> 1 | 71 52.71
> ------+-------------------------
> Total | 121 121.00
>
> chi2(1) = 11.33
> Pr>chi2 = 0.0008
> patients surviving >4 days
>
> failure _d: dod
> analysis time _t: cox
>
> Iteration 0: log likelihood = -669.88384
> Iteration 1: log likelihood = -661.82737
> Iteration 2: log likelihood = -661.82737
> Refining estimates:
> Iteration 0: log likelihood = -661.82737
>
> Cox regression -- Breslow method for ties
>
> No. of subjects = 717 Number of obs = 717
> No. of failures = 103
> Time at risk = 248467
> LR chi2(1) = 16.11
> Log likelihood = -661.82737 Prob > chi2 = 0.0001
>
> ------------------------------------------------------------------------------
> _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]
> -------------+----------------------------------------------------------------
> icp | 2.229562 .4553754 3.93 0.000 1.494055 3.327152
> ------------------------------------------------------------------------------
>
> Test of proportional-hazards assumption
>
> Time: Time
> ----------------------------------------------------------------
> | chi2 df Prob>chi2
> ------------+---------------------------------------------------
> global test | 0.09 1 0.7633
> ----------------------------------------------------------------
>
> failure _d: dod
> analysis time _t: cox
>
>
> Log-rank test for equality of survivor functions
>
> | Events Events
> icp | observed expected
> ------+-------------------------
> 0 | 38 58.28
> 1 | 65 44.72
> ------+-------------------------
> Total | 103 103.00
>
> chi2(1) = 16.37
> Pr>chi2 = 0.0001
>
> .
> end of do-file
>
> .
>
>
> On Mon, Aug 31, 2009 at 11:52 AM, Maarten buis<maartenbuis@yahoo.co.uk> wrote:
>> -----------------------------------------
>> Maarten L. Buis
>> Institut fuer Soziologie
>> Universitaet Tuebingen
>> Wilhelmstrasse 36
>> 72074 Tuebingen
>> Germany
>>
>> http://www.maartenbuis.nl
>> -----------------------------------------
>>
>>
>> --- moleps islon wrote:
>>> > This is the ouput I´m getting using your approach:
>>> >
>>> > n=896, failures=292
>>> >
>>> > stcox var,tvc(var) texp((_t>1)_t)
>>> >
>>> > rh
>>> >
>>> > var HR 0.64, p=0.005, CI 0.47-0.87
>>> >
>>> > t
>>> > var HR 1.01,p=0.001,CI 1.01-1.03
>>> >
>>> > So as far as I understand this the interpretation is
>>> > that the -var- is protective within the first 24hrs,
>>> > but detrimental afterwards ??
>>
>> --- On Mon, 31/8/09, Maarten buis wrote:
>>> No, the coefficient in the t equation is an interaction
>>> effect. So from t =0 to t=1 the hazard ratio increased
>>> with 1%. So at t=0 the hazard ratio for var is
>>> 0.64/1.01=0.62. In other words, in the first 24hrs var
>>> was even more protective than afterwards (but only very
>>> little, so I doubt whether that has any practical
>>> relevance).
>>
>> Sorry, I did not see that you turned around the inquality
>> sign (from < to >). So, in your case you assume that the
>> PH assumption holds in the first 24hrs, and that
>> afterwards the log hazard ratio changes linearly with time.
>> So, from t=0 to t=1 the hazard ratio of var is .64, and
>> after t=1 the hazard ratio increases by 1% every day. At
>> t=2 the hazard ratio of var is 1.01*.64=.646, at t=3
>> 1.01^2*.64=.653, at t=4 1.01^3*.64=.659, etc.
>>
>> To get the interpretation I gave in my previous post you
>> have to replace
>> stcox var,tvc(var) texp((_t>1)_t)
>>
>> with
>> stcox var,tvc(var) texp((_t<1)_t)
>>
>> Hope this helps,
>> Maarten
>>
>>
>>
>>
>>
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
© Copyright 1996–2017 StataCorp LLC | Terms of use | Privacy | Contact us | What's new | Site index | 4,552 | 16,016 | {"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-2017-22 | latest | en | 0.921944 |
https://www.coursehero.com/file/6474336/Samp-Exam2/ | 1,500,558,328,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423183.57/warc/CC-MAIN-20170720121902-20170720141902-00657.warc.gz | 748,663,719 | 76,649 | Samp_Exam2
# Samp_Exam2 - SAMPLE EXAM 2(This sample exam has a lot of...
This preview shows pages 1–2. Sign up to view the full content.
SAMPLE EXAM 2 (This sample exam has a lot of choices to give you practice – however the actual exam will have limited choice) ECON 80A Professor: Paroma Sanyal Microeconomic Theory Sachar 126 Brandeis University (781) 736-2268 Srping 2009 [email protected] NAME : Total 100 points Time : 2hrs Section 1: Multiple Choice Questions (2 points each) (10 Points) Please answer all questions in this section. Write your name and ID number clearly on Page 1 and on your blue books. Circle your answers (in pen) on the question paper (pg. 1 & 2) and put hand these two pages along with your blue book. NO CALCULATORS ALLOWED. 1.If the variable cost curve is a straight line then A) the marginal cost curve will be U-shaped. C) the marginal cost curve will be horizontal. B) the marginal cost curve may be U-shaped. D)none of the above. 2. Output for a simple production process is given by Q = 2KL, where K denotes capital, and L denotes labor. The price of capital is \$25 per unit and capital is fixed at 8 units in the short run. The price of labor is \$5 per unit. What is the total cost of producing 80 units of output? A) \$525 B) \$200 C) \$233 D) \$185 E) none of the above 3. A production function for which proportional changes in all inputs leads to a more-than-proportional change in output is said to exhibit A) diminishing returns. C) constant returns to scale. B) decreasing returns to scale. D) increasing returns to scale. 4. In a short-run production function before diminishing returns set in, A) both MP L and AP L will have positive slopes and MP L will lie above AP L . B) both MP L and AP L will have positive slopes and AP L will lie above MP L . C) both MP
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
### Page1 / 4
Samp_Exam2 - SAMPLE EXAM 2(This sample exam has a lot of...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 559 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-30 | latest | en | 0.862594 |
https://lujackhyundai.com/service-and-maintenance/your-question-how-much-does-damage-reduce-car-value.html | 1,652,855,789,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00132.warc.gz | 445,142,445 | 19,224 | # Your question: How much does damage reduce car value?
Contents
Following a car collision, your vehicle will depreciate 10% to 25% more than the average rate. Factors that can influence this depreciation percentage include the car’s age and its condition after the accident.
## How much is my car worth after an accident?
The formula broken down into steps:
1. Find out just how much your car was worth prior to the accident. You can check NADA or Kelley Blue Book to find your car’s value prior to the accident. …
2. Calculate the 10% cap that is immediately placed on your car’s value. …
3. Multiply the number you got from step 2 by the Damage Modifier.
## How do I calculate the diminished value of my car after an accident?
Example of a diminished value calculation
1. Step One: Check your car’s value. \$20,000.
2. Step Two: Calculate the base loss of value. \$20,000 x 10% = \$2,000.
3. Step Three: Apply a damage multiplier. \$2,000 x 0.75 = \$1,500.
4. Step Four: Apply a mileage multiplier. \$1,500 x 0.40 = \$600.
## How much does an accident reduce trade in value?
Putting it simply, any similar model to your car that wasn’t in a wreck is going to have much higher resale value. Dealers many times will cut about 10-30% off a trade, so if your car is valued at \$20,000 undamaged then an \$18,000 offer would be pretty much par for the course when trading it in post-accident.
## How much is my car worth if it’s totaled?
To get an idea of what your totaled car is worth, find the Kelley Blue Book value for it in fair condition. Figure out what the 20 to 40 percent fair condition value is. Depending on the amount of damage done to your vehicle, it’s likely going to be closer to the 20 percent range, according to CarBrain.
## Can you fight your car being totaled?
If you think your totaled car is valuable enough to justify a repair, you can contest your insurance company’s decision to declare it a total loss, but be prepared to provide evidence that the car is worth the effort.
## Is a diminished value claim worth it?
By filing a diminished value claim, you might be able to recoup some of the car’s depreciated value. If you’re successful, the insurance company pays you the difference between the car’s value before and after the accident.
Step 3: Apply a damage multiplier.
Multiplier Damage level
0.00 No structural damage
## How much will I get for diminished value?
As a general rule, you should expect to recover 10% to 25% of the fair market value of your vehicle. That means if your vehicle has a fair market value of \$30,000, your diminished value after an accident could be as high as \$7,500.
## Can you get diminished value from your own insurance?
In California, damages for diminished value can only be recovered against the party who negligently damages your car; diminished value is not recoverable against your own Collision Coverage except in very limited circumstances.
THIS IS IMPORTANT: How do I remove a windshield and install it?
## Do all accidents show up on Carfax?
Yes. If an accident has been reported to CARFAX it will be included in the CARFAX Vehicle History Report. … However, we do not have all accidents as many have never been reported, or may only have been reported to a source to which CARFAX does not have access.
## How much does a minor accident affect resale value?
A car with an accident on the vehicle history report or still evident on the vehicle simply doesn’t command the same resale price. At any stage, the car depreciation rate is about 10 to 25 percent more than the normal rate.
## Can you sell a car that has been in an accident?
Selling a car that has been in an accident without disclosing the damage puts you at risk for legal action from the buyer. If a person buys a car and later discovers that it’s been in an accident, you could end up with charges pressed against you. | 880 | 3,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.923417 |
https://math.stackexchange.com/questions/3676630/if-there-are-more-than-frac3n-12-edges-then-there-exist-veritces-u-v | 1,718,240,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00701.warc.gz | 375,643,270 | 37,435 | # If there are more than $\frac{3(n-1)}{2}$ edges then there exist veritces $u,v$ with $3$ vertex-disjoint $(u,v)$-paths
If a graph $$G$$ with $$n \geq 4$$ vertices has more than $$\frac{3(n-1)}{2}$$ edges then there exist $$u,v \in V(G)$$ with $$3$$ vertex-disjoint $$(u,v)$$-paths.
I tried induction but didn't work. Can someone provide me with a hint on how to start?
• Sure got the right type of induction? That one you consider a case for $n=k$ when assuming cases all for $n<k$ already true, so you can delete any vertex you want from the graph, but not add. I'm sure there have to be induction proof if the statement is true. Commented May 15, 2020 at 18:02
• @AlexeyBurdin yes, I tried deleting a vertex s.t. the remaing edges would still satisfy the inequality of the induction hypothesis, but I couldn't prove the existence of such vertex. Commented May 15, 2020 at 18:39
• I'm mostly just spitballing, but if your condition on the size of the graph implied the existence of an $S_3$ minor (the complement of the net), you'd be done. I seem to recall seeing something similar to this problem in one of Bollobás' books (either extremal graph theory or modern graph theory), and that's the way it was handled. Sorry I'm not of much more help. Commented May 19, 2020 at 0:49
• The answer by Hagen von Eitzen is correct. Commented May 20, 2020 at 7:22
Note that the restriction $$n\ge4$$ is irrelevant because for $$1\le n\le 3$$, we have $${n\choose 3}\le\frac {3(n-1)}2$$.
Assume $$G$$ contains a triangle, i.e., vertices $$u,v,w$$ with edges $$uv,vw,wu$$. Let $$G'$$ be $$G$$ with these three edges removed. Assume there is a path in $$G'$$ betwen two of the vertices $$u,v,w$$ and wlog $$uz_1\ldots z_mv$$ is the shortest such path. Then none of the $$z_i$$ equals $$w$$ and so on $$G$$ we have $$uz_1\ldots z_mv$$, $$uwv$$ and $$uv$$ and are done. Hence $$G'$$ splits into three disjoint graphs $$G_u,G_v,G_w$$ with $$n_u,n_v,n_w$$ vertices and $$e_u,e_v,e_w$$ edges. We have \begin{align}e_u+e_v+e_w&=e-3\\&>\frac{3(n-3)}2\\&=\frac{3(n_u-1)}2+\frac{3(n_v-1)}2+\frac{3(n_w-1)}2\end{align} so that at least one of the $$G_x$$ has $$e_x>\frac{3(n_x-1)}2$$ and we are done. We may therefore assume henceforth that $$G$$ contains no triangle.
Let $$uv$$ be any edge. Then $$u,v$$ have no neighbours in common. Hence we obtain a simple graph $$G'$$ with $$n'=n-1$$ vertices and $$e'=e-1>\frac{3(n-1)}2-1>\frac{3(n'-1)}2$$ edges by merging $$u$$ and $$v$$ (into $$u$$). By induction hypothesis, $$G'$$ has the desired property. Let $$xz_1\ldots z_my$$, $$xz_1'\ldots z_{m'}'y$$, $$xz_1''\ldots z''_{m''}y$$ be three vertex disjoint paths in $$G'$$. If any of the edges involved is from $$u$$ to a neighbour of $$v$$, we can insert $$v$$ between these points in the path and obtain three paths in $$G$$. If we have to add $$v$$ this way at most once, we are done. Hence suppose we have to add $$v$$ at least twice. Then $$u$$ occurs twice in the pats and must be an endpoint, wlog. $$u=x$$. If we need to insert $$v$$ all three times, we have $$vz_1\ldots z_m y$$, $$vz_1'\ldots z'_{m'} y$$, $$vz_1''\ldots z''_{m''} y$$, and are done. Hence assume we insert $$v$$ only twice, say in the $$z'$$ and the $$z''$$ path. Then we have $$vz_1'\ldots z'_{m'} y$$, $$vz_1''\ldots z''_{m''} y$$, $$vuz_1\ldots z_m y$$, and are done.
• 13 minutes since the answer, still can't read "Hence $G′$ splits into three disjoint graphs" -- how? and "If any of the edges involved is from $u$ to a neighbour of $v$, we can insert $v$ between these points in the path and obtain three paths in $G$", ) Commented May 15, 2020 at 20:08
• @AlexeyBurdin The answer is correct. The graph $G'$ splits into connected components. The claim about the insertion of $v$ is correct too, because $v$ is adjacent both to its neighbor and to $u$. Commented May 20, 2020 at 7:20 | 1,252 | 3,849 | {"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": 62, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-26 | latest | en | 0.911004 |
https://myassignmenthelp.com/free-samples/budget-assignment | 1,560,661,278,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997731.69/warc/CC-MAIN-20190616042701-20190616064701-00240.warc.gz | 546,393,978 | 42,602 | Budget Assignment Add in library
240 Download4 Pages 779 Words
Question:
A consulting firm produces a service that requires the use of labor and materials. Each unit of service requires a standard labor time of 30 minutes (0.5 hours). The average pay rate for a labor hour is £20. The consulting firm considers all materials that are required for the service as variable overheads (OH), the cost of which is directly associated with the labor hours worked. It has been estimated that variable OH rate is £10 per service hour.
The budgeted and actual costs, revenue and units for the month November are given in the table below:
Original Budget Actual Units of Service 1,500 1,600 Sales Revenue £120,000 £124,400 Labor hours 750 860 Labor cost £15,000 £20,210 Variable OH costs £7,500 £8,170 Fixed Cost £68,000 £68,000 Total Cost £90,500 £96,380 Operating Profit £29,500 £28,020
1. Calculate the flexed budget and the key variances between budgeted and actual results.
2. Reconcile the original budget and present the relationship between the budgeted and the actual profit for the month November
3. Discuss the calculated variances, and provide suggestions for better cost management.
Answers:
1. Flexible Budget:-
Particulars Budgeted data for 1,500 units(I) Flexible budget for 1,600 units(II) Actual data for 1,600 units(III) Sales revenue(Note 1) 120,000 128,000 124,400 Units 1,500 1,600 1,600 Selling price per unit(Note 2) 80 80 77.75 Labour hours 750 800 860 Labour hours per unit(Note 3) 0.5 0.5 0.5375 Labour cost(Note 4) 15,000 16,000 20,210 Labour cost per hour(Note 5) 20 20 23.5 Variable overhead cost(Note 6) 7,500 8,000 8,170 Variable rate per hour(Note 7) 10 10 9.5 Fixed cost 68,000 68,000 68,000 Profit(Note 8) 29,500 36,000 28,020
Note: 1-Sales revenue
For (I)-1,500*80
For (II)-1,600*80
For (III)-1,600*77.75
Note: 2-Selling price per unit
For (I)-120,000/1,500
For (II)-128,000/1,600
For (III)-124,400/1,600
Note: 3-Labour hours per unit
For (I)-750/1,500
For (III)-800/1,600
For (III)-860/1,600
Note: 4-Labour cost
For (I)-750*20
For (II)-800*20
For (III)-860*23.5
Note: 5-Labour cost per hour
For (I)-15,000/750
For (II)-16,000/800
For (III)-20,210/860
Note: 6-Variable overhead cost
For (I)-750*10
For (II)-800*10
For (III)-860*9.5
Note: 7-Variable rate per hour
For (I)-7,500/750
For (II)-8,000/800
For (III)-8,170/860
Note: 8-Profit
For (I)-120,000-15,000-7,500-68,000=29,500
For (II)-128,000-16,000-8,000-68,000=36,000
For (III)-124,400-20,210-8,170-68,000=28,020
Key variances between budgeted and actual:-
1. Sales price variance-128,000-124,400=3,600(adverse)
2. Direct labour total variance-16,000-20,210=4,210(adverse)
• Variable overhead total variance-8,000-8,170=170(adverse)
2. According to the flexible budget, the profit for 1,600 units should have been 36,000. But in actual, the profit is 28,020. The difference has arisen because of variances in sales revenue as well as labour cost as well as variable cost. The difference in budgeted profit can be calculated as under:
There is difference in profit is of 7,980. There is as shortfall in profit. The shortfall is caused because of adverse variances. All the three variances calculated are adverse. Due to which the profit is adverse. The calculation is shown below:
7,980=3,600+4,210+170.
3. The variances calculated are sales price variance, direct labour total variance and variable overhead total variance. All the three variances are adverse. Variable expenses and labour expenses are incurred in excess of what should have been actually incurred. The selling price is less compared to the budgeted selling price. Units are same. Therefore, the sales price variance is adverse. Now to achieve the targeted sales revenue, company should try to sell units in an area which can provide the targeted selling price per unit. To increase the selling price is a tough decision to make as it can affect the sales units. Therefore, improving sales value variance is a difficult task.
Another two adverse variances are direct labour total variance and variable overhead total variance. The cost incurred is more than what should have been actually spent as per budgeted data for standard production. Labour cost per hour is 20 while actual labour cost per hour 23.5. To make the variance positive, the labourers should be given incentives and motivation to produce more units in one hour so that per unit labour cost can be decreased.
The most relevant variance to the business is variable overhead cost variance. Because, it depends upon the variable cost incurred by the company. Variable expenses incurred can be reduced to some extent by the organisation. Labour and sales variance can’t be controlled as compared to variable overhead. Variable overhead can be controlled by the company. Company should analyse day to day expenses carefully so that it can cut off expenses somewhere. Maintenance expenses, supplies, material expenses should be controlled to cut off variable expenses.
References:-
ANON, N.D., “variance analysis”, Accessed on 4th February 2015, <https://accounting-simplified.com/management/variance-analysis/>
ANON, N.D., “variance analysis”, Accessed on 4th February 2015, <https://classes.bus.oregonstate.edu/spring-07/ba422/Management%20Accounting%20Chapter%205.htm>
OR
Most Downloaded Sample of Management
277 Download1 Pages 48 Words
Toulin Method Of Argumentation
You are required to write a researched argument essay that convinces persuades the reader of your position / stance. This is an academic, researched and referenced do...
Tags: Australia Arlington Management Management University of New South Wales Management
202 Download9 Pages 2,237 Words
Consumer Behavior Assignment
Executive Summary The purpose of this report is to elaborate the factors which are considered by individuals before selecting an occupation. Choosing an occupati...
Tags: Australia Arlington Management Management University of New South Wales Management
367 Download13 Pages 3,112 Words
Internet Marketing Plan For River Island
Introduction With the increase enhancement in the field of technology, it has been considered essential by the businesses to implement such technology in their b...
Tags: Australia Arlington Management Management University of New South Wales Management
328 Download9 Pages 2,203 Words
Strategic Role Of HR In Mergers & Acquisitions
Executive Summary In a merger & acquisition, role of an HR has emerged as a very critical function. At each stage of merger and acquisition process, HR plays a s...
Tags: Australia Arlington Management Management University of New South Wales Management
357 Download7 Pages 1,521 Words
Relationship Between Knowledge Management, Organization Learning And HRM
Introduction In this competitive business environment where every business organization is trying to attract the customers of each other, it becomes essential for ...
Tags: Australia Arlington Management Management University of New South Wales Management
Next | 1,798 | 7,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-26 | latest | en | 0.771659 |
https://math.meta.stackexchange.com/questions/37568/why-would-someone-close-this-question | 1,721,055,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00114.warc.gz | 334,965,533 | 25,010 | # Why would someone close this question?
I have a question about my Mathematics Stack Exchange post: Can all the sides of a right triangle be hypotenuse numbers?
I don’t understand why this question was deleted, how, or by who. Two people had thoughtfully responded and I wanted to reply and reframe the question. All I could see was a generic “does not meet guidelines.”
I’m bringing the question elsewhere, but am curious why it was buried here.
• 1. It's not deleted - it's currently closed. Closure happened because 5 community members voted to close. Closure can be undone by a vote of community members, too, but you need to convince people to do that, probably by editing the post. 2. Did you see the link to "provide additional context" in the banner? If so, you can click that link to see a bit more about what we're expecting here. If not, please edit a screenshot of what you see in to your post here. Commented Jul 1 at 3:26
• [assuming you're young] Hi Roz! First of all, your question is great! If all 3 sides were the same, all three angles would be the same angle -- and they wouldn't be 90 degrees! BUT there's another, more direct proof. Can you figure it out? (Hint: when can sides A + B be less than the hypotenuse? Ever? Never?) If you peek in the rabbit hole far enough, you'll see that it IS possible, if the triangles extend into both space and time. I hope you someday dig that deep so you can feel how exciting dimensionality is in this universe we were given. Commented Jul 1 at 11:37
• @Miss, according to math.meta.stackexchange.com/users/790478/roz Roz has been a member for over four years. If Roz is a kid now, Roz must have been in diapers on first joining. Anyway, before Roz' question got closed, other users did post examples of right triangles where all sides are hypotenuse numbers, and they did it without needing to posit triangles extending into both space and time ... whatever the heck that's supposed to mean. It is a good question, and I welcome your support of Roz – but not your name-calling, and not your pseudo-mathematics. Commented Jul 1 at 13:04
• For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. Commented Jul 9 at 4:27 | 582 | 2,494 | {"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} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.973882 |
http://badhessian.org/2015/05/concor-in-r/ | 1,585,772,337,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00141.warc.gz | 19,486,804 | 11,681 | In network analysis, blockmodels provide a simplified representation of a more complex relational structure. The basic idea is to assign each actor to a position and then depict the relationship between positions. In settings where relational dynamics are sufficiently routinized, the relationship between positions neatly summarizes the relationship between sets of actors. How do we go about assigning actors to positions? Early work on this problem focused in particular on the concept of structural equivalence. Formally speaking, a pair of actors is said to be structurally equivalent if they are tied to the same set of alters. Note that by this definition, a pair of actors can be structurally equivalent without being tied to one another. This idea is central to debates over the role of cohesion versus equivalence.
In practice, actors are almost never exactly structural equivalent to one another. To get around this problem, we first measure the degree of structural equivalence between each pair of actors and then use these measures to look for groups of actors who are roughly comparable to one another. Structural equivalence can be measured in a number of different ways, with correlation and Euclidean distance emerging as popular options. Similarly, there are a number of methods for identifying groups of structurally equivalent actors. The `equiv.clust` routine included in the `sna` package in R, for example, relies on hierarchical cluster analysis (HCA). While the designation of positions is less cut and dry, one can use multidimensional scaling (MDS) in a similar manner. MDS and HCA can also be used in combination, with the former serving as a form of pre-processing. Either way, once clusters of structurally equivalent actors have been identified, we can construct a reduced graph depicting the relationship between the resulting groups.
Yet the most prominent examples of blockmodeling built not on HCA or MDS, but on an algorithm known as CONCOR. The algorithm takes it name from the simple trick on which it is based, namely the CONvergence of iterated CORrelations. We are all familiar with the idea of using correlation to measure the similarity between columns of a data matrix. As it turns out, you can also use correlation to measure the degree of similarity between the columns of the resulting correlation matrix. In other words, you can use correlation to measure the similarity of similarities. If you repeat this procedure over and over, you eventually end up with a matrix whose entries take on one of two values: 1 or -1. The final matrix can then be permuted to produce blocks of 1s and -1s, with each block representing a group of structurally equivalent actors. Dividing the original data accordingly, each of these groups can be further partitioned to produce a more fine-grained solution.
Insofar as CONCOR uses correlation as a both a measure of structural equivalence as well as a means of identifying groups of structurally equivalent actors, it is easy to forget that blockmodeling with CONCOR entails the same basic steps as blockmodeling with HCA. The logic behind the two procedures is identical. Indeed, Breiger, Boorman, and Arabie (1975) explicitly describe CONCOR as a hierarchical clustering algorithm. Note, however, that when it comes to measuring structural equivalence, CONCOR relies exclusively on the use of correlation, whereas HCA can be made to work with most common measures of (dis)similarity.
Since CONCOR wasn’t available as part of the `sna` or `igraph` libraries, I decided to put together my own CONCOR routine. It could probably still use a little work in terms of things like error checking, but there is enough there to replicate the wiring room example included in the piece by Breiger et al. Check it out! The program and sample data are available on my GitHub page. If you have `devtools` installed, you can download everything directly using R. At the moment, the `concor_hca` command is only set up to handle one-mode data, though this can be easily fixed. In an earlier version of the code, I included a second function for calculating tie densities, but I think it makes more sense to use `concor_hca` to generate a membership vector which can then be passed to the `blockmodel` command included as part of the `sna` library.
```#REPLICATE BREIGER ET AL. (1975)
#INSTALL CONCOR
devtools::install_github("aslez/concoR")
#LIBRARIES
library(concoR)
library(sna)
data(bank_wiring)
bank_wiring
#CHECK INITIAL CORRELATIONS (TABLE III)
m0 <- cor(do.call(rbind, bank_wiring))
round(m0, 2)
#IDENTIFY BLOCKS USING A 4-BLOCK MODEL (TABLE IV)
blks <- concor_hca(bank_wiring, p = 2)
blks
#CHECK FIT USING SNA (TABLE V)
#code below fails unless glabels are specified
blk_mod <- blockmodel(bank_wiring, blks\$block,
glabels = names(bank_wiring),
plabels = rownames(bank_wiring[[1]]))
blk_mod
plot(blk_mod)
```
The results are shown below. If you click on the image, you should be able to see all the labels.
• Tim
I’ve been hitting an error with your code while using it to blockmodel trade data from the Correlates of War project. For some years of data, I am able to use this smoothly, but in other years I receive the following error message: “Error in while (any(abs(mi) <= cutoff) & iter <= max.iter) { : missing value where TRUE/FALSE needed". Do you have any insight into why I'm seeing this, and how I could resolve this problem? I'm happy to provide my base code and data file if necessary. Many thanks!
If it’s working in some years and not others, the problem is probably with the data. The error indicates that the any(mi) <= cutoff statement isn't turning out a logical statement, which points to the abs(mi) command, since cutoff should be an unchanging scalar. mi is a correlation matrix. My guess would be that there is something in the data that is causing every entry in the correlation matrix to turn into an NA. Just take the trade matrix for one of the bad years and dump it into a cor command and see what happens. If you get all NAs, you'll have to figure out what's causing it and then clean up the data accordingly.
Two other things to note…
First, if you are using this for real work, keep in mind that the program is only set up to produce an even number of blocks, which may have implications for the results. You can get an uneven number by, say, partitioning the data into two groups, and then rerunning the code using only one of the two groups. You would need to do this my hand, which would pose some problems if you wanted to automate the procedure to work across time.
Second, looking at the code, I see a couple of lines that could be cleaned up. I may have to post a new version. I'll update the post here if I do. There is, however, no reason to think the current version is producing bad results.
• Tim
You’re absolutely right about the data issues- thanks for the quick response. There’s something producing NA’s in my correlation matrix, so I’m looking into this while attempting other methods of block modeling. Thanks!
• Leah Reisman
Hello! Thanks so much for this. Have you by any chance updated concor_hca to handle 2-mode data, or could you point me in the right direction regarding how to do this? I’m a bit of a beginner to social network analysis and would really appreciate any help you can provide. | 1,604 | 7,343 | {"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-2020-16 | longest | en | 0.95158 |
http://roseindia.net/java/java-biginteger/java-biginteger-max-value.shtml | 1,501,209,679,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436321.71/warc/CC-MAIN-20170728022449-20170728042449-00122.warc.gz | 271,666,542 | 16,446 | ## Online Hibernate Training just in Rs. 1000
One week online training in Hibernate ORM framework. Learn Hibernate Framework and enhance your skills. Best Online Hibernate training at 90% discount. Join Now!! Offer is valid Only for limited Period!!!
# Java BigInteger max value
We can use the max() method of the BigInteger class to get the maximum value of the two BigInteger values.
Tutorials
# Java BigInteger max value
We can use the max() method of the BigInteger class to get the maximum value of the two BigInteger values. It returns the BigInteger value whose value is greater of this and the passed variable. If in case these are equal, any of them may be returned.
Syntax of max() method of BigInteger class is as follows:
public BigInteger max(BigInteger val)
In our java example code we have created three BigInteger variables and we have then compared them with the use of max() method.
BigInteger maxResult = bigInteger1.max(bigInteger2);
Variable maxResult returns the BigInteger value of either of the bigIntger1 or of bigInteger2 which is maximum.Here is the full example code of the example code as follows:
``` import java.math.BigInteger; public class BigIntegerMaxValue { public static void main(String[] args) { BigInteger bigInteger1 = new BigInteger ("123456789"); BigInteger bigInteger2 = new BigInteger ("112334"); BigInteger bigInteger3 = new BigInteger ("987654321"); BigInteger maxResult = bigInteger1.max(bigInteger2); System.out.println("Max big integer in"+bigInteger1+ " and "+bigInteger2+" is ==>" + maxResult); maxResult = bigInteger1.max(bigInteger3); System.out.println("Max big integer in"+bigInteger1+ " and "+bigInteger3+" is ==>" + maxResult); } }```
Output of the example is as follows:
C:\biginteger>javac BigIntegerMaxValue.java C:\biginteger>java BigIntegerMaxValue Max big integer in123456789 and 112334 is ==>123456789 Max big integer in123456789 and 987654321 is ==>987654321
# Java BigInteger max value
Posted on: November 4, 2008 If you enjoyed this post then why not add us on Google+? Add us to your Circles
Related Tutorials
Discuss: Java BigInteger max value
Your Name (*) :
Subject (*): | 551 | 2,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.408498 |
https://andrewcharlesjones.github.io/journal/quasilikelihoods.html | 1,719,174,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00777.warc.gz | 82,904,720 | 5,674 | # Quasi-likelihoods
As their name suggests, “quasi-likelihoods” are quantities that aren’t formally likelihood functions, but can be used as replacements for formal likelihoods in more general settings.
A major reason that quasi-likelihoods are useful is that they don’t require the statistician to specify a full probability distribution for the data – instead, only the first and second moments must be specified. This allows the modeler to have more flexibility without having to impose unnecessary (and incorrect) assumptions.
Most results for quasi-likelihoods are in the context of the exponential family, which we review here first.
## Exponential family
Recall the form of the exponential family of distributions:
$f(y; \theta) = \exp\left\{ a(y) b(\theta) + c(\theta) + d(y) \right\}.$
for some functions $a$, $b$, $c$, and $d$. In canonical form, we have $a(y) = y$ and $b(\theta) = \theta$. The form of the likelihood then simplifies to
$f(y; \theta) = \exp\left\{ \frac{y \theta - b(\theta)}{a(\phi)} + c(y, \phi) \right\}$
where we call $\phi$ a “dispersion parameter”.
## Likelihood for GLMs
Generalized linear models (GLMs) extend the classical linear model to allow for various types of response distributions. For a response vector $\mathbf{Y}$, a matrix of covariates $\mathbf{X}$, and a vector of coefficients $\boldsymbol{\beta}$, a GLM has the model structure of
$\mathbb{E}[\mathbf{Y} | \mathbf{X}] = \mu(\mathbf{X}) = g^{-1}(\mathbf{X} \boldsymbol{\beta})$
where $g(\cdot)$ is a link function. Equivalently, we have $g(\mu(\mathbf{X})) = \mathbf{X}\boldsymbol{\beta}$, which we typically define to be the parameter $\theta$: $\theta = g(\mu(\mathbf{X})) = g(\mu)$. We also define $\eta$ to be the linear predictor: $\eta = \mathbf{X} \boldsymbol{\beta}$.
Additionally, $\mathbf{Y} | \mathbf{X}$ is typically assumed to come from an exponential family distribution. For a sample of size $n$, the likelihood is then
$L(\theta) = \prod\limits_{i=1}^n \exp\left\{ \frac{y_i \theta_i - b(\theta_i)}{a(\phi)} + c(y_i, \phi) \right\}.$
The corresponding log-likelihood is
$\ell(\theta) = \sum\limits_{i=1}^n \left[\frac{y_i \theta_i - b(\theta_i)}{a(\phi)} + c(y_i, \phi)\right].$
To perform maximum likelihood estimation, we take the derivative w.r.t. the coefficients $\beta_1, \dots, \beta_p$ and set these expressions to zero. This can be accomplished with the chain rule:
$\frac{\partial \ell_i}{\partial \beta_j} = \frac{\partial \ell_i}{\partial \theta_i} \cdot \frac{\partial \theta_i}{\partial \mu_i} \cdot \frac{\partial \mu_i}{\partial \eta_i} \cdot \frac{\partial \eta_i}{\partial \beta_j}.$
Breaking these partial derivatives down, we have
\begin{align} \frac{\partial \ell_i}{\partial \theta_i} &= \frac{y_i - b’(\theta_i)}{a(\phi)} = \frac{y_i - \mu_i}{a(\phi)} \\ \frac{\partial \theta_i}{\partial \mu_i} &= g’(\mu_i) = (b’‘(y_i))^{-1} = \frac{a(\phi)}{\text{Var}(y_i)} = \frac{1}{v(y_i)} \\ \frac{\partial \eta_i}{\partial \beta_j} &= X_{ij} \\ \end{align}
The expression for $\frac{\partial \ell_i}{\partial \theta_i}$ follows because $b’(\theta)$ is equal to the mean $\mu$ in the exponential family (see here for an explanation). We won’t specify $\frac{\partial \mu_i}{\partial \eta_j}$ further, just to allow for an arbitrary relationship between $\mu$ and $\eta$.
Putting these together, we have
\begin{align} \frac{\partial \ell_i}{\partial \beta_j} &= \frac{y_i - \mu_i}{a(\phi)} \cdot \frac{1}{v(y_i)} \cdot \frac{\partial \mu_i}{\partial \eta_j} \cdot X_{ij} \\ &= \frac{(y_i - \mu_i) X_{ij}}{a(\phi) v(\mu_i)} \cdot \frac{\partial \mu_i}{\partial \eta_j} \end{align}
Setting these equations to zero gives us the score equations needed to perform MLE:
$\sum\limits_{i=1}^n \frac{(y_i - \mu_i) X_{ij}}{a(\phi) v(\mu_i)} \cdot \frac{\partial \mu_i}{\partial \eta_j} = 0, \;\;\; j = 1, \dots, p$
Notice that these equations depend on the specified probability distribution through $\mu_i$ and the variance function $v(\mu)$. Specifically, once a distribution is specified, this variance function is automatically specified as well.
We now turn to quasi-likelihoods, which allow us to avoid specifying an entire distribution. Instead, we directly specify the mean-variance relationship.
## Quasi-likelihoods
A quasi-likelihood is defined as
$\int_y^\mu \frac{y - t}{a(\phi) V(t)} dt$
where $V$ is a function that relates the variance to the mean, and $a(\phi)$ is a (unknown) dispersion parameter. Differentiating, we have
$\frac{\partial Q}{\partial \mu} = \frac{y - \mu}{a(\phi) V(\mu)}$
We can see already that this expression is equivalent to the typical likelihood case, $\frac{\partial \ell}{\partial \mu}$.
To get the full “quasi-score equations”, we have to perform one more step of the chain rule, by multiplying by $\frac{\partial \mu}{\partial \beta}$:
$\sum\limits_{i=1}^n \frac{(y_i - \mu_i)}{a(\phi) v(\mu_i)} \cdot \frac{\partial \mu_i}{\partial \beta_j} = 0, \;\;\; j = 1, \dots, p$
Notice that the coefficient estimates $\hat{\beta}$ are the same for the full GLM case and for quasi-likelihoods. However, they will have different variance estimates.
## Quasi-likelihoods corresponding to well-known likelihoods
To build intuition, it’s useful to note that many well-known distributions correspond to a particular choice of the mean-variance relationship in the world of quasi-likelihood world. The converse is not true: there are many quasi-likelihood specifications that do not map onto any known distributions.
Table 9.1 from Peter McCullagh’s book Generalized Linear Models shows several such examples:
The column $Q(\mu; y)$ here is calculated by computing the integral
$\int_y^\mu \frac{y - t}{a(\phi) V(t)} dt$
for the specified function $V(\mu)$, and it is assumed $a(\phi) = 1$ for simplicity.
For example, in the Gaussian case we have $V(\mu) = 1$, which implies
\begin{align} \int_y^\mu \frac{y - t}{a(\phi) V(t)} dt &= \int_y^\mu \frac{y - t}{1 \cdot 1} dt \\ &= \int_y^\mu (y - t) dt \\ &= -\frac12 (y - t)^2 \big\rvert_y^\mu \\ &= -\frac12 (y - \mu)^2 \end{align}
In the Poisson case, we have $V(\mu) = \mu$, which implies
\begin{align} \int_y^\mu \frac{y - t}{a(\phi) V(t)} dt &= \int_y^\mu \frac{y - t}{1 \cdot t} dt \\ &= \int_y^\mu \left(\frac{y}{t} - 1\right) dt \\ &= y \log \mu - \mu \\ \end{align}
However, notice that the table also includes variance function with no corresponding closed-form likelihood. An interesting case of this is when $V(\mu) = \mu^\zeta$, where $\zeta \neq 0, 1, 2$. This power law mean-variance relationship occurs often in ecology through Taylor’s law, and more generally in statistics it appears in the family of distributions known as the Tweedie distribution.
## References
• McCullagh, Peter. Generalized linear models. Routledge, 2018.
• Agresti, Alan. Foundations of linear and generalized linear models. John Wiley & Sons, 2015.
• Islam, M. Ataharul, and Rafiqul I. Chowdhury. Analysis of repeated measures data. Singapore: Springer, 2017. | 2,082 | 7,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.818066 |
http://mathhelpforum.com/calculus/53712-help-understand-answer.html | 1,480,912,313,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00501-ip-10-31-129-80.ec2.internal.warc.gz | 175,481,720 | 11,757 | solve ln(x+2)+ln(x-1)=1
it can be written as (x+2)(x-1)=e apparently
What rule is this?
and the answer is x = (sqrt[9+4e]-1)/2
2. note the sum rule for logs ...
$\log(a) + \log(b) = \log(a \cdot b)$
$\ln(x+1) + \ln(x-1) = 1$
$\ln[(x+1)(x-1)] = 1$
change the log equation to an exponential ...
$e^1 = (x+1)(x-1)$
3. o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?
and how did he get that answer? i put it into the quadratic formula but am not getting that same one.
o ok. how do u change it to an exponential? just put e as the base and it gets rid of ln?
and how did he get that answer? i put it into the quadratic formula but am not getting that same one.
You should know that $\log_A B = C$ is equivalent to $A^C = B$.
5. $(x+2)(x-1) = e$
$x^2 + x - 2 - e = 0$
$x^2 + x - (2+e) = 0$
$a = 1$, $b = 1$, $c = -(2+e)$
$x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot -(2+e)}}{2}$
$x = \frac{-1 \pm \sqrt{9+4e}}{2}$
since $x > 1$ ...
$x = \frac{-1 + \sqrt{9+4e}}{2}$
6. ooooooo very true. thx a lot
7. wait, but how do u know x>1?
8. $\ln(x+2) + \ln(x-1)$ ... what is the domain?
9. o ok, i see. I can never find the domain without a calculator.
10. I can never find the domain without a calculator.
learn the concept of domain of a function ... the "machine" will put you at a great disadvantage.
11. yea true, i should. its just hard when you get stuff involving e and log and stuff like that | 518 | 1,454 | {"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": 17, "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-2016-50 | longest | en | 0.87553 |
http://forums.devshed.com/python-programming-11/array-manipulation-229074.html | 1,387,612,093,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345775355/warc/CC-MAIN-20131218054935-00065-ip-10-33-133-15.ec2.internal.warc.gz | 75,910,421 | 20,646 | Forums: » Register « | Free Tools | User CP | Games | Calendar | Members | FAQs | Sitemap | Support |
User Name: Password: Remember me
New Free Tools on Dev Shed!
We're Excited to announce that Dev Shed now has 70 free tools on the site. To learn more, click here!
Dev Shed Forums Sponsor:
#1
February 20th, 2005, 04:26 PM
fupduck
Registered User
Join Date: Feb 2005
Posts: 4
Time spent in forums: 1 h 16 m 38 sec
Reputation Power: 0
Array manipulation
I have a 2D array similar to the following:
A = [[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]
[13 14 15 16]]
I need to swap quadrants 1 and 3 and then 2 and 4 to get some like:
A = [[ 11 12 9 10]
[ 15 16 13 14]
[ 3 4 1 2]
[ 7 8 5 6]]
Any help is greatly appreciated.
f
#2
February 20th, 2005, 07:37 PM
sfb
Contributing User
Join Date: Nov 2003
Posts: 624
Time spent in forums: 3 Days 6 h 10 m 21 sec
Reputation Power: 34
This should do it:
Code:
```>>> a = [[1,2,3,4],
... [5,6,7,8],
... [9,10,11,12],
... [13,14,15,16]]
>>>
>>> r1, r2, r3, r4 = a
>>> a = [r3[2:]+r3[:2], r4[2:]+r4[:2], r1[2:]+r1[:2], r2[2:]+r2[:2]]
>>>
>>> a
[[11, 12, 9, 10], [15, 16, 13, 14], [3, 4, 1, 2], [7, 8, 5, 6]]
>>> ```
But it's a bit ugly. It takes each row out, then builds a new list by taking the parts of each row needed.
The new first row being: r3[2:]+r3[:2]
(the third row after the second character followed by the third row up to the second character).
#3
February 20th, 2005, 07:42 PM
netytan
Hello World :)
Join Date: Mar 2003
Location: Hull, UK
Posts: 2,537
Time spent in forums: 1 Week 2 Days 18 h 17 m 47 sec
Reputation Power: 68
The simplest way I can think of. You can also do it using several temp variables to hold each element and put them into the list in their new position. Although I think this is a little more elegant .
Code:
```>>> target = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
>>> target
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
>>>
>>> for each in range(2): target.append(target.pop(0))
...
>>> target
[[9, 10, 11, 12], [13, 14, 15, 16], [1, 2, 3, 4], [5, 6, 7, 8]]
>>>
>>> for eachList in target:
... for index in range(2):
... eachList.append(eachList.pop(0))
...
>>> target
[[11, 12, 9, 10], [15, 16, 13, 14], [3, 4, 1, 2], [7, 8, 5, 6]]
>>> ```
Enjoy,
Mark.
__________________
programming language development: www.netytan.com Hula
#4
February 20th, 2005, 08:01 PM
sfb
Contributing User
Join Date: Nov 2003
Posts: 624
Time spent in forums: 3 Days 6 h 10 m 21 sec
Reputation Power: 34
Code:
```>>> rows = [[1,2,3,4], [5,6,7,8], [9,10,11,12], [13,14,15,16]]
>>>
>>> for row in rows:
... row[:] = [row[2], row[3], row[0], row[1]]
...
>>> rows = [rows[2], rows[3], rows[0], rows[1]]
>>> rows
[[11, 12, 9, 10], [15, 16, 13, 14], [3, 4, 1, 2], [7, 8, 5, 6]]```
#5
February 20th, 2005, 08:29 PM
fupduck
Registered User
Join Date: Feb 2005
Posts: 4
Time spent in forums: 1 h 16 m 38 sec
Reputation Power: 0
Thanks for the help, but I left out an important point. I need to eventually use the code on any 2D matrix. I have Matlab code that does this but haven't been able to convert it Python.
Matlab code:
for jj=1:ny
for ii=1:kx-1
data(ii+kx+1,jj)=dat(ii,jj);
end
for ii=kx:nx
data(ii-kx+1,jj)=dat(ii,jj);
end
end
where,
ny = number columns
kx = nx/2
This swaps quadrant 1 and 3. The code for 2 and 4 is basically the same. Again, thanks for the help.
p.s. The end result is correctly calculate the power spectrum using Fourier transform. Matlab has a function fftshift() but I haven't seen anything within any Python module.
f
#6
February 21st, 2005, 03:34 AM
DevCoach
Contributing User
Join Date: Feb 2004
Location: London, England
Posts: 1,585
Time spent in forums: 2 Weeks 4 Days 2 h 58 m 23 sec
Reputation Power: 1372
Have you looked at the numpy or numarray libraries? Both of these are high performance C extensions for manipulating arrays.
You can extend the python code in the previous posts to handle N*N arrays [1], but any python implementation will be probably be far too slow for a useful FFT algorithm. Both numpy and numarray are orders of magnitude faster.
Dave - The Developers' Coach
[1] e.g. in netytan's post, replace both the range(2) with range(int(N/2)).
Viewing: Dev Shed Forums > Programming Languages > Python Programming > Array manipulation
## Developer Shed Advertisers and Affiliates
Thread Tools Search this Thread Search this Thread: Advanced Search Display Modes Rate This Thread Linear Mode Rate This Thread: 5 : Excellent 4 : Good 3 : Average 2 : Bad 1 : Terrible
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts vB code is On Smilies are On [IMG] code is On HTML code is Off
View Your Warnings | New Posts | Latest News | Latest Threads | Shoutbox Forum Jump Please select one User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home -------------------- Programming Languages PHP Development PHP FAQs and Stickies Perl Programming Perl FAQs and Stickies C Programming C Programming FAQs and Stickies Java Help Java FAQs Python Programming Python Programming FAQs Ruby Programming Ruby Programming FAQs Game Development Game Development FAQs Programming Languages - More ASP Programming ASP Programming FAQs .Net Development .Net Development FAQs Visual Basic Programming Visual Basic Programming FAQs Software Design Software Design FAQs ColdFusion Development ColdFusion Development FAQs Delphi Programming Delphi Programming FAQs Regex Programming Regex Programming FAQs XML Programming XML Programming FAQs Other Programming Languages Other Programming Languages FAQs Web Design HTML Programming HTML Programming FAQs JavaScript Development JavaScript Development FAQs CSS Help CSS Help FAQs Flash Help Flash Help FAQs Photoshop Help Photoshop Help FAQs Web Design Help Web Design Help FAQs Website Critiques Website Critiques FAQs Search Engine Optimization Search Engine Optimization FAQs Mobile Programming Mobile Programming Mobile Programming FAQs iPhone SDK Development iPhone SDK Development FAQs Android Development Android Development FAQs BlackBerry Development BlackBerry Development FAQs Web Site Management Business Help Business Help FAQs Development Software Development Software FAQs Scripts Scripts FAQs Databases Database Management Database Management FAQs DB2 Development DB2 Development FAQs MySQL Help MySQL Help FAQs PostgreSQL Help PostgreSQL Help FAQs Firebird SQL Development Firebird SQL Development FAQs MS SQL Development MS SQL Development FAQs Oracle Development Oracle Development FAQs LDAP Programming LDAP Programming FAQs System Administration Mail Server Help Mail Server Help FAQs Apache Development Apache Development FAQs Security and Cryptography Security and Cryptography FAQs Antivirus Protection Antivirus Protection FAQs DNS DNS FAQs IIS IIS FAQs Networking Help Networking Help FAQs FTP Help FTP Help FAQs Operating Systems BSD Help BSD Help FAQs Linux Help Linux Help FAQs UNIX Help UNIX Help FAQs Windows Help Windows Help FAQs Mac Help Mac Help FAQs Web Hosting Web Hosting Web Hosting FAQs Free Web Hosting Free Web Hosting FAQs Web Hosting Requests Web Hosting Requests FAQs Web Hosting Offers Web Hosting Offers FAQs Computer Hardware Computer Hardware CPUs CPUs FAQs Cooling Cooling FAQs Embedded Programming Embedded Programming FAQs Motherboards Motherboards FAQs Multimedia Hardware Multimedia Hardware FAQs Other Dev Shed Lounge Dev Shed Lounge FAQs Development Articles Development Articles FAQs Beginner Programming Beginner Programming FAQs Hire A Programmer Hire A Programmer FAQs Project Help Wanted Project Help Wanted FAQs Latest News Updated Hourly Technology News Business News Science News Forum Information Forum Rules/Guidelines Forum Rules/Guidelines FAQs Forum Announcements Forum Announcements FAQs Dev Shed Gaming Center Go to the Dev Shed Battle Arena Go to the Dev Shed Arcade Games Go to the Legend of the Green Dragon Suggestions & Feedback Suggestions & Feedback FAQs
Forums: » Register « | Free Tools | User CP | Games | Calendar | Members | FAQs | Sitemap | Support | | 2,481 | 8,900 | {"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-2013-48 | longest | en | 0.873652 |
https://www.coursehero.com/file/6593692/204-812c82-dist-sample-mean-CLT/ | 1,516,377,896,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888041.33/warc/CC-MAIN-20180119144931-20180119164931-00426.warc.gz | 900,969,076 | 26,378 | 204 8.1%2c8.2 dist. sample mean CLT
# 204 8.1%2c8.2 dist. sample mean CLT - 8.1 Mean And Standard...
This preview shows pages 1–2. Sign up to view the full content.
Assume we have a population with the values 2,3,4,5,6. This population has μ = 4 and σ 2 = 2 = = We now take all possible non ordered samples of size two from the population and calculate the mean of each pair. pair 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 5,6 x 2.5 3 3.5 4 3.5 4 4.5 4.5 5 5.5 The statistic x also has a mean, μ, and variance σ 2 . For the above pairs μ= [2.5 + 3 + 2(3.5) + 2(4) + 2(4.5) + 5 + 5.5]/10= 4 σ 2 = .75 Note that μ = μ= 4. In contrast, the variance of is less than the variance of the population (2). The variances can be equated through the following formulas: for infinte populations σ 2 = n 2 σ σ= for finite populations σ 2 = 1 N n N n 2 - - σ = Since our population has N = 5 and n = 2 we obtain: σ 2 = o.75 = . For practical purposes we assume a population is infinite if n<.05N The following is a Histogram of the above distribution of sample means. 2
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 2
204 8.1%2c8.2 dist. sample mean CLT - 8.1 Mean And Standard...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 501 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-05 | latest | en | 0.809848 |
https://ventolaphotography.com/what-is-gravity-for-middle-school/ | 1,675,839,300,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00702.warc.gz | 596,839,829 | 15,108 | ## What is gravity for middle school?
Gravity is the force of attraction between two objects. It is what makes things fall and what keeps us from floating off into space. Gravity is a fundamental force of nature.
How do you do a gravity experiment?
First, tie one end of a piece of string to a paperclip and tie the other end around the stick. Repeat twice more so that the stick has three paper clips attached. Hold the stick up in the air, allowing the paperclips hang freely. Tilt the stick back and forth….For this experiment you’ll use:
1. A stick.
2. Paper clips.
3. String.
What is gravity experiment for kids?
Gravity is defied in this super simple gravity-free water experiment. Children use the effects of air pressure to make it look as though there is zero gravity in an upside-down glass of water. Get your children to fill a glass with water, place a piece of card over it and turn it upside down.
### How do you introduce gravity to students?
Put a globe in the center of the floor and tell students that gravity is pulling them towards it really really hard. One group at a time, ask them to form themselves around the globe so that they are as close to it as possible. Explain that Earth’s gravity is incredibly strong and pulls objects towards its center.
What is gravity lesson?
Everything is attracted to the Earth due to the force of gravity. The word gravity comes from the Latin word “gravis”, which means heavy. Gravity is heavy stuff. The bigger an object, the bigger the gravity force it exerts. The force of gravity is also present on the Moon.
What are the activities that involve gravitational force?
Some examples of the force of gravity include:
• The force that holds the gases in the sun.
• The force that causes a ball you throw in the air to come down again.
• The force that causes a car to coast downhill even when you aren’t stepping on the gas.
• The force that causes a glass you drop to fall to the floor.
#### How is gravity explained?
On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth’s mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects. At Earth’s surface the acceleration of gravity is about 9.8 metres (32 feet) per second per second.
How does gravity work?
Earth’s gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That’s what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.
What is the easiest way to explain gravity?
The answer is gravity: an invisible force that pulls objects toward each other. Earth’s gravity is what keeps you on the ground and what makes things fall. An animation of gravity at work. Albert Einstein described gravity as a curve in space that wraps around an object—such as a star or a planet.
## How do you explain gravity to kindergarten?
“It makes things fall,” children may answer if you ask what gravity does. They might have a little more trouble telling you what this enigmatic force actually is. Scientists don’t fully understand it either, but in simple terms, gravity is an invisible attractive force that causes objects to move toward each other.
What is a simple definition of gravity?
gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. It is by far the weakest known force in nature and thus plays no role in determining the internal properties of everyday matter.
What is one example of gravity that you experience in your daily life?
The force that holds the gases in the sun. The force that causes a ball you throw in the air to come down again. The force that causes a car to coast downhill even when you aren’t stepping on the gas. The force that causes a glass you drop to fall to the floor.
### How do you measure gravity in Middle School?
Create a Gravity Measuring Device Middle school students can perform more advanced experiments. Students can explore the concept of how gravity affects balance and create their own gravity device with a candle, a needle, two glasses and two saucers. Cut off the bottom half-inch of a long candle to expose the wick.
How can students learn more about the power of gravity?
Students can conduct this experiment to learn even more about the power of gravity. This experiment is messy, so make sure you choose your location wisely. Distribute cups and scissors to your students and have them make some holes near the bottom of their cups. Have them consider what will happen if they pour water in the cups.
What is the gravity project?
The Gravity Project was initiated in November 2018 by SIREN with funding from the Robert Wood Johnson Foundation to convene broad stakeholder groups in identifying and harmonizing social risk factor data for interoperable electronic health information exchange.
#### Why gravity experiments for kids?
However, through the use of the following gravity experiments for kids, children will gain a better grasp of gravity’s role in our everyday lives while also having some fun! Most gravity experiments don’t require many materials. | 1,060 | 5,215 | {"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-2023-06 | latest | en | 0.9562 |
http://www.dbai.tuwien.ac.at/marchives/fuzzy-mail98/0340.html | 1,539,816,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511314.51/warc/CC-MAIN-20181017220358-20181018001858-00055.warc.gz | 453,016,821 | 2,104 | # Re: Functions defined by fuzzy rules
Dimitri Lisin (dima@wpi.edu)
Mon, 6 Apr 1998 02:22:44 +0200 (MET DST)
That's the problem. The responses were either "why would you want to do
this in the first place", or "I also want to know how to do this". So far
the only sensible solution I have, is to use equal isosceles triangles for
mambership funtions, which would approximate a function with streight line
segments. I was kind of hoping to do better than that...
Dimay
On 30 Mar 1998, Gal Kaminka wrote:
> Dimitri,
>
> I am also interested in this. I'd appreciate it (and I'm sure
> others will too) if you post a summary of responses back to the
> forum.
>
> Thanks,
>
> Gal
>
> Dimitri Lisin (dima@wpi.edu) wrote:
> : Greetings, All.
>
> : I am rather new to this field, as the case seems to be with most people
> : who post questions here.
>
> : The little that I understand abut fuzzy logic is that a fuzzy inferencing
> : system, i.e. a set of fuzzy rules that relate, say, two variables
> : approximate a function. What I may need to do, is take a regular
> : algebraic function, for example y = x^2, and generate a set of fuzzy rules
> : relating x and y that approximate the function (on some interval). Can
> : anyone point me to any literature on the subject?
>
> : Another question, more out of curiosity than necessity. If a function is
> : defined in terms of a set of fuzzy rules, is it possible to differentiate
> : or integrate it? Any literature discussing this?
>
> : Dima Lisin
> : dima@cs.wpi.edu
> : http://www.wpi.edu/~dima/
>
>
> --
> ----------------------------------------------------------------------------
> Gal A. Kaminka galk@isi.edu/galk@usc.edu http://www.isi.edu/soar/galk/
> Research Assistant, Applied Philosopher USC Information Sciences Institute
> "Death is an engineering problem." -- Bart Kosko, "Fuzzy Thinking"
> "But life is not an engineering task." -- Gal A. Kaminka
>
>
>
>
>
>
> | 520 | 1,930 | {"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-2018-43 | latest | en | 0.882763 |
https://books.google.la/books?id=TwgAAAAAQAAJ&dq=editions:OCLC1012786239&output=html_text&source=gbs_navlinks_s | 1,721,555,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00765.warc.gz | 118,982,916 | 10,834 | # Easy Introduction to Mathematics, Volume 1
Barlett & Newman, 1814
### Contents
ARITHMETIC xvi HISTORICAL INTRODUCTION 1 VULGAR FRACTIONS 4 DECIMALS 10 LOGARITHMICAL ARITHMETIC 13 WHOLE NUMBERS 15 TABLES of MONEY 51 Cloth 63
73 110 175 159 255 239 264 246 LOGARITHMS 254 299 272 Involution 308 HISTORICAL INTRODUCTION 316
Multiplication 67 Goods sold by Tale 71 Proportion Direct and Inverse 73 Subtraction 85 Multiplication 93 Division 103
Binomials 385 REDUCTION OF SIMPLE EQUATIONS 408 Two unknown quantities 425 Promiscuous Examples 432 PROBLEMS 443
### Popular passages
Page xxii - Just so it is in the mind; would you have a man reason well, you must use him to it betimes, exercise his mind in observing the connection of ideas and following them in train. Nothing does this better than mathematics, which therefore I think should be taught all those who have the time and opportunity, not so much to make them mathematicians as to make them reasonable creatures...
Page 48 - LIQUID MEASURE 4 gills (gi.) = 1 pint (pt.) 2 pints = 1 quart (qt...
Page 98 - Then multiply the second and third terms together, and divide the product by the first term: the quotient will be the fourth term, or answer.
Page 42 - AVOIRDUPOIS WEIGHT. 16 drams, dr. make 1 ounce, - - - - oz. 16 ounces - - - 1 pound, - - - - Ib. 28 pounds - - - 1 quarter, - - - qr. 4 quarters - - - 1 hundred weight, - cwt. 20 hundred weight, 1 ton, T.
Page 448 - What number is that, which, being divided by the product of its digits, the quotient is 3 ; and if 18 be added to it, the digits will be inverted ? Ans.
Page 54 - M. 60 minutes, 1 hour, h. 24 hours, 1 day, d. 7 days, . 1 week, w. 4 weeks, 1 month, mo. 13 months, 1 day and 6 hours, 1 Julian year, yr. Thirty days hath September, April, June and November ; February twenty-eight alone, all the rest have thirtyone.
Page 106 - State and reduce the terms as in the Rule of Three Direct. 2. Multiply the first and second terms together, and divide the product by the third ; the quotient will be the answer in the same denomination as the middle term was reduced into.
Page 234 - ... and to the remainder bring down the next period for a dividend. 3. Place the double of the root already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained...
Page 432 - A hare is 50 leaps before a greyhound, and takes 4 leaps to- the greyhound's 3, but 2 of the greyhound's leaps are as much as 3 of the hare's ; how many leaps must the greyhound take to catch the hare ? Ans. 300.
Page 138 - To reduce a whole number to an equivalent fraction, having a given denominator. RULE. Multiply the whole number by the given denominator, and place the product over the said denominator, and it will form the fraction required. | 770 | 2,787 | {"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 | 0.871179 |
http://schoolbag.info/mathematics/ap_calculus/28.html | 1,485,241,526,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284352.26/warc/CC-MAIN-20170116095124-00219-ip-10-171-10-70.ec2.internal.warc.gz | 251,641,601 | 5,761 | IMPLICIT DIFFERENTIATION - Differentiating - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC
## Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 4. Differentiating
IMPLICIT DIFFERENTIATION
At this point in your long and prosperous calculus life, you are no longer intimidated by questions such as “Find dy/dx: y = 2sin x + 3x2.” Without a second thought (hopefully), you would answer dy/dx = 2cos x + 6x. Sometimes, however, the questions are not solved for a variable like the above problem was solved for y. In fact, some things you differentiate aren’t even functions, like the above example is a function of x. In such circumstances, you will employ implicit differentiation. In order to differentiate implicitly, you need to have a basic understanding of what it means to differentiate with respect to a variable (the previous topic in this chapter).
Steps for Success in Implicit Differentiation
ALERT! You must be at least 18 to purchase a book that discusses “explicit” differentiation.
(These steps assume that you are finding dy/dx, as it is your goal 85 percent of the time. If other variables are used, adjust accordingly.)
1. Find the derivative of the entire equation with respect to x.
2. Solve for dy/dx.
3. If a specific solution is required, substitute in the corresponding x and y values.
Example 17: Find the equation of the tangent line to the circle
x2 + y2 = 9 when x = 1
ALERT! Don’t forget to take the derivative of the constant:
Solution: In order to find the equation of a line, you need a point and a slope. It is very simple to find the point. Substitute x = 1 into the formula to find its corresponding y value. This coordinate pair (1, √8) marks the point of tangency, as shown in the figure below.
All that remains is to find the slope of the tangent line, which is given by the derivative, dy/dx. You need to find this derivative implicitly, however. To do so, first find the derivative of everything with respect to x:
To complete the problem, solve for dy/dx:
Hence, the derivative for any point (x,y) on the circle is given by the formula —x/y. Therefore, the slope of the tangent line at (1, √8) is —1/√8. Using point-slope form for a line, the equation of the tangent line is
NOTE. A normal line is perpendicular to the tangent line at the point of tangency.
Example 18: Find the equation of the normal line to sin (x) + exy = 3 when x = π.
Solution: In order to find the slope of the normal line, you need to take the opposite reciprocal of the slope of the tangent line (the derivative), since they are perpendicular. It is far from easy to solve this equation for y, so you should differentiate implicitly. Again, start by finding the point on the normal line by plugging in x = π:
TIP. To keep resulting answers accurate, you shouldn’t round to .350 until the problem is completely over.
Thus, the tangent and normal lines both pass through (π, .3496991526).
Now, find the derivative of the equation with respect to x to get the slope of the tangent line, dy/dx:
Distribute exy and solve for dy/dx:
YIP. You must use the Product Rule to differentiate xy.
Plug in the coordinate (π, .3496991526) for (x,y) to get the slope of the tangent line there:
If this is the slope of the tangent line at (π, .3496991526), then the slope of the normal line is the negative reciprocal of —.0052094021, or 191.9606091 Therefore, the equation of the normal line is
y — .350 = 191.961(x - π)
(You are allowed to round at the very end of the problem on the AP test.)
EXERCISE 8
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
YOU MAY USE A GRAPHING CALCULATOR FOR PROBLEM 4 ONLY.
1. Find dy/dx:
2. What is the slope of the tangent line to ln (xy) + y2 = 2y at the point (e,1)?
3. Find if x2 - y2 = 16.
4. Dennis Franz High School (“Home of the Ferocious Prairie Dogs”) has had a top-notch track team ever since they installed their elliptical track. Its dimensions, major axis length 536 feet and minor axis length 208 feet, are close to an actual track. Below is a diagram of the Prairie Dogs’ track superimposed on a coordinate plane. If the northern boundaries of DF High are linear and tangent to the track at x = ±250 feet, find the equations of the northern property lines.
1. Find the derivative with respect to x.
Now, solve for dy/dx. You’ll need to factor dy/dx out of both terms on the left side of the equation.
2. First, find d/dx. Remember to use the Product Rule for d/dx(xy):
Distribute 1/xy and solve for dy/dx:
At this point, you can solve for dy/dx, but there is no real need to do so in this problem. You just want the derivative at (e, 1), so plug in those values for x and y:
3. means the second derivative, so begin by finding dy/dx, the first derivative:
Use the Quotient Rule to find the second derivative:
TIP. When entering fractions into the calculator, it’s a good idea to surround the numerator and denominator with parentheses.
You already know that so plug it in:
Get common denominators for y and x2/y and simplify:
Now, the original problem states that x2 — y2 = 16, so y2 — x2 = —16. Substitute this, and you’re finished:
4. Begin by creating the equation of the ellipse. Remember, standard form of an ellipse is where a and b are half the lengths of the axes. Therefore, the track has the following equation:
Now find the y that corresponds to both x = 250 and x = —250:
Now, you have points (±250,37.471379). Find dy/dx at these points to get the slope of the tangent lines. (Because the graph of an ellipse is y-symmetric, the slopes at x = 250 and x = —250 will be opposites.)
The property lines have equations y — 37.414 = —1,006(x — 250) and y — 37.414 = 1.006(x + 250).
| 1,458 | 5,817 | {"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} | 4.8125 | 5 | CC-MAIN-2017-04 | longest | en | 0.924422 |
http://cnx.org/content/m11209/latest/ | 1,397,792,704,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00218-ip-10-147-4-33.ec2.internal.warc.gz | 45,984,323 | 9,419 | # OpenStax_CNX
You are here: Home » Content » Squared Differences Simulation
### Recently Viewed
This feature requires Javascript to be enabled.
# Squared Differences Simulation
Module by: David Lane. E-mail the author
Begin by answering the questions, even if you have to guess. The first time you answer the questions you will not be told whether you are correct or not.
## General Instructions
This demonstration allows you to examine the sum of absolute deviations from a given value. The graph to the left shows the numbers 1, 2, 3, 4, and 5 and their deviations from 0. The first number, 1, is represented by a red dot. The deviation from 0 is represented by a red line from the red dot to the black line. The value of the black line is 0.
Similarly, the number 2 is represented by a blue dot and its deviation from 0 is represented by a blue line. The graph with the colored rectangles shows the sum of the absolute deviations. Since the deviations for the numbers 1, 2, 3, 4, and 5 from 0 are the numbers themselves, the sum of the deviations is equal to 1 + 2 + 3 + 4 + 5 = 15 as shown by the height of the colored bar.
In this demonstration, you can move the black bar by clicking on it and dragging it up or down. To see how it works, move it up to 1.0. The deviation of the red point from the black bar is now 0 since they are both 1. The sum of the deviations is now 10.
As you move the bar up and down, the value of the sum of absolute deviations changes. See if you can find the placement of the black bar that produces the smallest value for the sum of the absolute deviations. To check and see if you found the smallest value, click the "OK" button at the bottom of the graph. It will move the bar to the location that produces the smallest sum of absolute deviations.
You can also move the individual points. Click on one of the points and move it up or down and note the effect.
Your goal for this demonstration is to discover a rule for determining what value will give you the smallest sum of absolute deviations.
When you have discovered the rule, go back and answer the questions again.
## Step by Step Instructions
If it is not already there, move the black bar at the bottom of the graph up so that it crosses the Y axis at 1. The bar should go right through the red circle. Notice the numerical indicator of the black bar immediately to its right.
The deviation of the red circle from the bar is 0, so you won't see a red rectangle on the right-hand portion of the graph. The line between the bar and the blue circle is the deviation of the circle from the bar. It has a length of 1. Notice that the height of the blue rectangle is 1.
The green line has a length of 2 and the height of the green rectangle is also 2. The total height of the rectangle is the sum of all the line lengths: 0 + 1 + 2 + 3 + 4 = 10. This height is the sum of the absolute deviations from the bar. It is marked below the rectangle.
Your goal is to find the placement of the bar that gives you the shortest rectangle. This will be the value that minimizes the sum of the rectangles. Move the bar up and down until you think you have found this value. Then, to make sure you are correct, click on the "OK" button at the bottom of the graph. This will move the black bar to the correct location. If nothing changes, you found the correct location on your own.
Now, change the value of the green circle from 3 to somewhere between 2 and 3. You move the circle by clicking on it and dragging it. Notice that the value of the point is shown at the bottom of the graph in green.
Next, find the value that minimizes the sum of absolute deviations for the new data. Once again it is the same value as the green circle.
Now move the blue circle to somewhere between 3 and 4 and again find the value that minimizes the sum of absolute differences. This time it is the value of the blue circle.
How do you know which point it will be? The correct placement of the bar will always be at the value of the circle with the middle value. That is, the circle that has two point higher than it and two points lower than it.
Why is this? If the bar is at the circle with the middle value, then moving the bar will bring the bar closer to two points but farther from three points. So, any movement of the bar from the middle value increases the sum of absolute deviations.
## Summary
The middle number minimizes the sum of the absolute deviations. If you have 7 numbers, say 0, 1, 2, 5, 6, 9, and 12, then the middle number is the fourth highest which is 5. Therefore 5 is the value that minimizes the sum of the absolute deviations. This middle number is called the median.
If you have an even number of numbers such as the four numbers: 2, 4, 6, and 9, then any value between 4 and 6 will give you the same minimum sum of the absolute deviations. By convention, we define the median as the average of the two numbers closest the middle (2 and 4 in this case). So the median of the numbers 2, 4, 6, and 9 is 5.
## Content actions
PDF | EPUB (?)
### What is an EPUB file?
EPUB is an electronic book format that can be read on a variety of mobile devices.
My Favorites (?)
'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.
| A lens I own (?)
#### Definition of a lens
##### Lenses
A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.
##### What is in a lens?
Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.
##### Who can create a lens?
Any individual member, a community, or a respected organization.
##### What are tags?
Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.
| External bookmarks | 1,426 | 6,148 | {"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} | 4.28125 | 4 | CC-MAIN-2014-15 | latest | en | 0.916862 |
https://www.physicsforums.com/threads/finding-tension-between-garfield-and-john-at-the-olympics-parking-lot.661597/ | 1,721,266,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00485.warc.gz | 799,422,616 | 20,617 | # Finding Tension Between Garfield and John at the Olympics Parking Lot
• Lolagoeslala
In summary: Maybe a1 and a2).Sorry, for the first part it is supposed to be mg (typo).I have no idea how far x and y are because the sketch is just rough.a1= acceleration of john and the skateboarda2= acceleration of garfieldIf we are not given the distance how can we have an equation?Sorry, for the first part it is supposed to be mg (typo).I have no idea how far x and y are because the sketch is just rough.a1= acceleration of john and the skateboarda2= acceleration of garfieldIf we are not given the distance how can we have an equation?You don't need to be given the
Lolagoeslala
## Homework Statement
Garfield (a cat) and John (his human) did not go to the Olympics (not feline up to
it). However, they got in trouble. John sits on a frictionless skateboard on a
horizontal parking lot. A light strong rope tied to a solid post passes horizontally
to the skate board, passes around a pulley on the skateboard, returns horizontally
to pass above the post, around a second pulley and vertically down into a yawning
chasm. Garfield hangs on tightly to the bottom end of the rope. Garfield has a
mass of 20 kg (too much lasagna), John and the skateboard have a combined mass
of 60 kg. Calculate the tension in the rope before John hits the post or Garfield
bottoms out.
## The Attempt at a Solution
My FBD diagrams
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/c_zps40e98e7e.png.html
Ok so this is what i came up with to find the tensin between the cat and the by and skateboard
m
Fnet = Fg - T
ma = md - T
M
Fnet = Fg + Fn + T
Fnet = T
Ma = T
Combining
a = (mg/m+M)
To find the tension
Ma = T
M(mg/m+M) = T
T = (M x m/M +m)g
T = ( 60000g x 20000g/60000g + 20000g)(9.8m/s^2)
T = 147000N
so that is the tension between john/skateboard and the cat
however how do i account for the nail and the tension between john/skateboard and the nail ?
Here is a rough sketch of what i think it looks like
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/f_zpsafd5a9e3.png.html
Is there a mechanical advantage involved? How many rope segments are pulling on each mass? How might this affect your FBD?
Are the accelerations of the two objects the same? If not, how are they related?
gneill said:
Is there a mechanical advantage involved? How many rope segments are pulling on each mass? How might this affect your FBD?
Are the accelerations of the two objects the same? If not, how are they related?
Wait are those questions that you are asking me? or? Because all its asking for is the tension before john/skateboard hit the post
Lolagoeslala said:
Wait are those questions that you are asking me? or? Because all its asking for is the tension before john/skateboard hit the post
You need to address the questions in order to solve the problem correctly; their answers affect the equations you write to determine the tension.
Last edited:
gneill said:
You need to address the questions in order to solve the problem correctly; their answers affect the equations you write to determine the tension.
I looked at all the prospects and then found the equations
Lolagoeslala said:
I looked at all the prospects and then found the equations
I would suggest that you reconsider the net force acting on John and the related rates of the motions of John and Garfield. The pulley arrangement makes a difference.
gneill said:
I would suggest that you reconsider the net force acting on John and the related rates of the motions of John and Garfield. The pulley arrangement makes a difference.
So are you trying to say that for john and the skateboard i would need to include the tenion twice since there is another piece of tension connected to the nail/post.
So like this?
M
Fnet = Fg + Fn + T + T
Fnet = T+ T
Ma = T^2
Lolagoeslala said:
So are you trying to say that for john and the skateboard i would need to include the tenion twice since there is another piece of tension connected to the nail/post.
That's the idea, yes.
So like this?
M
Fnet = Fg + Fn + T + T
Fnet = T+ T
Ma = T^2
Well, you should keep X and Y components separate (Fg and Fn act in a different direction than the T's), but the sum of the forces acting in the X-direction will be T + T. That's 2T (not T squared).
Then consider how the accelerations of the two masses are related; the pulley arrangement will impose a relationship between them --- they are not equal.
gneill said:
That's the idea, yes.
Well, you should keep X and Y components separate (Fg and Fn act in a different direction than the T's), but the sum of the forces acting in the X-direction will be T + T. That's 2T (not T squared).
Then consider how the accelerations of the two masses are related; the pulley arrangement will impose a relationship between them --- they are not equal.
if i do that it does not work, please see my process
m
Fnet = Fg - T
ma = md - T
M
Fnet = Fg + Fn + T + T
Fnet = T + T
Ma = 2T
Combining
a = (mg + T/m+M)
To find the tension
Ma = 2T
M(mg+T/m+M) = 2T
Mmg + MT = 2T (m+M)
Mmg + MT = 2Tm + 2TM
Mmg = 2Tm + TM
Mmg = T(Tm +M)
Like this is confusing me?
Lolagoeslala said:
if i do that it does not work, please see my process
m
Fnet = Fg - T
ma = md - T
What is the 'd' in 'md'? Shouldn't that be mg?
M
Fnet = Fg + Fn + T + T
Fnet = T + T
Ma = 2T
Okay, BUT! The a's for the two masses are NOT THE SAME. They are related, but they are not the same value. The pulley arrangement makes them unequal. Check: If the cat descends by some distance 'y', by how much distance 'x' will the skateboard move?
Write an equation relating the two accelerations (And label them differently. Maybe a1 and a2).
Combining
a = (mg + T/m+M)
I don't see how you can write the above if they don't share the same acceleration. You need to sort out the relationship between the two accelerations before you can meaningfully combine the equations.
Last edited:
gneill said:
Lolagoeslala said:
if i do that it does not work, please see my process
m
Fnet = Fg - T
ma = md - T
[\quote]
What is the 'd' in 'md'? Shouldn't that be mg?
Okay, BUT! The a's for the two masses are NOT THE SAME. They are related, but they are not the same value. The pulley arrangement makes them unequal. Check: If the cat descends by some distance 'y', by how much distance 'x' will the skateboard move?
Write an equation relating the two accelerations (And label them differently. Maybe a1 and a2).
I don't see how you can write the above if they don't share the same acceleration. You need to sort out the relationship between the two accelerations before you can meaningfully combine the equations.
But if the cat moves down then john and its skateboard moves ahead so wouldn't the acceleration be the same.. we did questions similare to this but the acceleration is the same.
But then how would i do it in terms of a1 and a2?
ma1 = mg -T
Ma2 = 2T
ma1 + Ma2 = mg - T + 2T
Lolagoeslala said:
gneill said:
But if the cat moves down then john and its skateboard moves ahead so wouldn't the acceleration be the same.. we did questions similare to this but the acceleration is the same.
Maybe similar, but not the same... At some point you should have done problems involving pulley systems that involve mechanical advantage. The movements of the various parts of the system are related but not equal. For a given distance that the cat moves, the skateboard will move a different (but related) distance. You need to work out what the relationship between the movements is. The same relationship will apply to distance, velocity, and acceleration.
(Hint: Consider that the total length of the rope does not change. This should allow you to work out how far the skateboard moves for a given movement of the cat).
But then how would i do it in terms of a1 and a2?
ma1 = mg -T
Ma2 = 2T
ma1 + Ma2 = mg - T + 2T
You need to determine an equation relating the accelerations. That will give you three equations in three unknowns (a1, a2, T).
gneill said:
Lolagoeslala said:
Maybe similar, but not the same... At some point you should have done problems involving pulley systems that involve mechanical advantage. The movements of the various parts of the system are related but not equal. For a given distance that the cat moves, the skateboard will move a different (but related) distance. You need to work out what the relationship between the movements is. The same relationship will apply to distance, velocity, and acceleration.
(Hint: Consider that the total length of the rope does not change. This should allow you to work out how far the skateboard moves for a given movement of the cat).
You need to determine an equation relating the accelerations. That will give you three equations in three unknowns (a1, a2, T).
so for the cat this is my equation:
fnet = fg - T1
m1a1 = m1g - T1
and for the boy
fnet = T1 + T2
m2a2 = T1 + T2
right?
or..
i could...
make a big huge tension
so itll be like this
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/ddgfffd_zpsbe230706.png.html
Where the equation could become:
cat
Fnet = Fg - T
m1a1 = m1g - T
and for the boy
Fnet = T2
m2a2 = T2
The rope has only one tension for its entire length, since the pulleys are massless and frictionless.
Take a close look at the forces acting on the skateboard pulley. Can you assign a tension to each of the "connections" there? In other words, how does your T2 relate to T?
Also, you still haven't stated the relationship that exists between the motion of the cat and the motion of the skateboard; how far does the skateboard move when the cat moves by some amount?
gneill said:
The rope has only one tension for its entire length, since the pulleys are massless and frictionless.
Take a close look at the forces acting on the skateboard pulley. Can you assign a tension to each of the "connections" there? In other words, how does your T2 relate to T?
Also, you still haven't stated the relationship that exists between the motion of the cat and the motion of the skateboard; how far does the skateboard move when the cat moves by some amount?
Well i have two ways i can do this. Since you said that the "rope has only one tension for its entire length, since the pulleys are massless and frictionless." I am going to assign the T2 to the little rope that connects the skateboard and the pulley. Well if the cat moves down 2m. the skateboard will move ahead 1m right? its a 1:2 ratio
Lolagoeslala said:
Well i have two ways i can do this. Since you said that the "rope has only one tension for its entire length, since the pulleys are massless and frictionless." I am going to assign the T2 to the little rope that connects the skateboard and the pulley.
Yes, but what is the relationship between T2 and T? Isolate that pulley and draw the FBD for it.
Well if the cat moves down 2m. the skateboard will move ahead 1m right? its a 1:2 ratio
Correct. That should tell you the ratio of a2 to a1 also (and v2 to v1 if it's of interest).
gneill said:
Yes, but what is the relationship between T2 and T? Isolate that pulley and draw the FBD for it.
Correct. That should tell you the ratio of a2 to a1 also (and v2 to v1 if it's of interest).
well for the T2 and T1
the ratio i came up with was T2 = 2T1
like this:
http://s1176.beta.photobucket.com/user/LolaGoesLala/media/dfdsdsd_zps2ce0a7cd.png.html
so since T2 = 2T1 and a1 = 2a2
(1) m1a1 = m1g - T
(2) m2a2 = T2
Times the first equation by two
2m1a1 = 2m1g - 2T
m2a2 = 2T
2m1(2a2) = 2m1g
m2a2
4m1a2 + m2a2 = 2m1g
a2 = 2m1g / (4m1 +m2)
when solving
a2 = 2(20kg)(9.8m/s^2) / 4(20kg)+(60kg)
a2 = 2.8 m/s^2
since a1 = 2a2
a1 = 5.6 m/s^2
and using the first equation
m1a1 = m1g - T
T = m1g - m1a1
T = (20kg)(9.8m/s^20 - (20kg)(5.6m/s^2)
T = 84 N
That looks good! Well done.
gneill said:
That looks good! Well done.
THANK YOU :D
it would have not been done without your help. :)
## 1. What caused the tension between Garfield and John at the Olympics parking lot?
The tension between Garfield and John at the Olympics parking lot was caused by a disagreement over a parking spot. Garfield wanted to park in a spot that John was already waiting for, leading to an argument between the two.
## 2. Did the tension between Garfield and John escalate into a physical altercation?
No, the tension between Garfield and John did not escalate into a physical altercation. They were able to resolve their disagreement peacefully.
## 3. How did Garfield and John resolve their tension at the Olympics parking lot?
After their initial argument, Garfield and John were able to calm down and communicate with each other. They were able to come to a compromise and find a different parking spot that worked for both of them.
It is unclear if this was the first time Garfield and John had a disagreement. However, it is likely that they have had disagreements before since they have known each other for some time.
## 5. Did the tension between Garfield and John affect their performance at the Olympics?
It is unknown if the tension between Garfield and John affected their performance at the Olympics. However, it is likely that they were able to put their disagreement behind them and focus on their respective events.
• Introductory Physics Homework Help
Replies
6
Views
8K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
6K
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K | 3,708 | 13,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.903981 |
http://math.stackexchange.com/questions/128018/lie-algebra-using-skew-symmetry | 1,469,440,975,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.33/warc/CC-MAIN-20160723071024-00122-ip-10-185-27-174.ec2.internal.warc.gz | 161,334,528 | 17,885 | # Lie algebra using skew-symmetry
Let g be a Lie algebra such that [[x,y],y]=0 for all $x,y \in g$. Show that 3[[x,y],z]=0 for all $x,y,z \in g$. [Hint: Observe that the mapping (x,y,z) to [[x,y],z] is skew-symmetric in x,y,z and make use of the jacobi identity.]
So I know that [[x,y],z]=[x,[y,z]]+[y,[z,x]] (using the jacobi identity)
So 3[[x,y],z]=3[x,[y,z]]+3[y,[z,x]], I need to use [[x,y],y]=0, but can't see how to do it. Like I would need to elimate x,y or z from 3[x,[y,z]]+3[y,[z,x]], but it just seems impossible.
Was wondering am I using skew symmetric wrong. I assume that skew symmetric is just this [x,y]=-[y,x].
-
Skew symmetry in general means for a function $f$ of $n$ arguments that whenever $(x_1,\ldots, x_n)$ and $(y_1,\ldots, y_n)$ are permutations of each other, then $f(x_1,\ldots,x_n)=\pm f(y_1,\ldots,y_n)$, where $\pm$ means $+$ or $-$ according whether the permutation is even or odd.
To prove that a function is skew symmetric, it is sufficient to show that $$f(x_1,\ldots,x_{i-1},x_i,x_{i+1},x_{i+2},\ldots x_n) = -f(x_1,\ldots,x_{i-1},x_{i+1},x_i,x_{i+2},\ldots x_n)$$ for all $i$ -- that is, that interchanging any two neighbor arguments will flip the sign of the function result. If the function is known to be linear in each argument, it is also enough to prove that $f(x_1,\ldots,x_n)=0$ whenever two of the $x_i$'s are equal.
Here this concept is being applied to the function $f(x,y,z)=[[x,y],z]$. The hint says to prove, based on $\forall xy: [[x,y],y]=0$, that this $f$ is skew symmetric. (You can do this without the Jacobi identity).
Then, write the Jacobi identity in the form $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$, and use the just proved fact about $f$ to see that the left-hand side is equal to $3[[a,b],c]$.
I thought all Lie algebras are skew-symmetric. So in $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$ do you just swap and change signs until you get it as $3[[a,b],c]$. I can see that. However, how would on go about proving f is skew-symmetric. – simplicity Apr 4 '12 at 18:31
The Lie bracket itself is skew-symmetric by definition, that is $[a,b]=-[b,a]$. But it does not follow from this that the three parameter function $(a,b,c)\mapsto [[a,b],c]$ is also skew-symmetric -- that requires not only $[[a,b],c]=-[[b,a],c]$ (which is true for all Lie algebras) but also, say, $[[a,b],c]=-[[a,c],b]$ (which isn't; consider e.g. $(\mathbf e_1 \times \mathbf e_2)\times \mathbf e_1$ in $\mathbb R^3$). The latter relation can be proved if you assume $[[x,y],y]=0$, however. (Hint: let $x=a$, $y=b+c$ and expand the LHS by linearity!) – Henning Makholm Apr 4 '12 at 21:14 | 880 | 2,609 | {"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} | 3.71875 | 4 | CC-MAIN-2016-30 | latest | en | 0.829136 |
http://www.jiskha.com/display.cgi?id=1295840074 | 1,498,226,430,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320063.74/warc/CC-MAIN-20170623133357-20170623153357-00702.warc.gz | 573,635,672 | 3,507 | # math
posted by on .
solve by substitution
(3/7)x + (5/9)y = 27
(1/9)x + (2/7)y = 7 | 41 | 86 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-26 | latest | en | 0.883519 |
https://newsongoogle.com/fast-track-objective-arithmetic-pdf/ | 1,624,515,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00410.warc.gz | 389,709,461 | 64,690 | Thursday, June 24, 2021
### Real Number ( H.C.F): Euclid Division Algorithm #2 Important Topic By Lalit Sir 2021
Euclid Division Algorithm Euclids Division Algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. HCF of two positive...
More
# Fast Track Objective Arithmetic Pdf
Contents
## Fast Track Objective Arithmetic
Today, there is a plethora of books available in the market on Objective Arithmetic which seems to be complete in their way but is still unable to fully satisfy the aspirants.
LET US KNOW SOME OF THE REASONS
Lack of Understanding the Basic Concepts
Mostly, students face a competitive examination on the base of their knowledge about mathematical rules, formulae and concepts. Inspite of having the knowledge, he lacks behind when he faces questions in the examination. Does he realise this inability? Yes, he does but feels confused and blocked when he is unable to solve them and is left with a sense of grudge that he could solve it. The only reason behind this problem is the understanding of basic concepts. If he would have been clear with them, he could solve any of the questions because as a matter of fact, every question is based on a particular concept which is just twisted in the examinations to judge the overall ability of a student.
Inappropriate Use of Short Tricks
This is the second biggest problem in front of the aspirants. The number of questions asked in the competitive examination is much more than the time assigned for diem. This leads die aspirants to use shortcut mediods. Although, diese mediods prove to be beneficial in some cases, but due to time management problems, he gets bound to use these methods irrationally and inappropriately. As a result, he jumbles between all the shortcuts which lead to wrong answers which could have been solved if he knew when and where to apply the shortcut methods.
Inability to Distinguish Between the Applications of Formulae
We all are aware of the amount of stress and pressure a competitive examination creates on the mindset of an aspirant. Succumbed to such pressure, an aspirant is unable to decide the appropriate formula to be applied in a particular step. During the crisis of time, such confusion adds to the problems and squeezes in more time and results to an unsatisfactory score.
Keeping in mind all kinds of problems faced by an aspirant in a competitive examination, we have developed diis book with profound interest in a step-wise method to encounter all your queries and worries. This book named ‘FAST TRACK ARITHMETIC is wordiy to fulfill your expectations and will help you as loyal guide throughout.
OUTSTANDING QUALITIES OF FAST-TRACK OBJECTIVE ARITHMETIC
Use of Fundamental Formulae and Method
In this book, all the fundamental formulae and methods have been presented in such a striking yet friendly and systematic manner that just going through them once will give you an effective grasp. They have been
present in such a manner that they, will never let you get confused between fast track technique and basic method.
Appropriate Short Cut Methods
An important feature of this book is its short cut methods or tricks given in the name of “fast track formulae or techniques”. Each technique is given with its basic or fundamental method. So, that a student can use these tricks according to their desire and save there precious time in exams.
Division of Exercises According to the Difficulty Level
Based on the standard and level of difficulty of various questions, the exercises are divided into two parts i.e., ‘Base level exercise’ for relatively easier questions and ‘Higher skill level exercise’ for difficult questions. ‘Multiconcept questions’ which requires a use of different concepts in a single question have also been incorporated with important chapters.
Special Emphasise on Geometry, Trigonometry and Mensuration
Now-a-days, Questions from geometry, trigonometry and mensuration are asked in large numbers in different exams. So, a large variety and number of questions are provided for tiiese chapters.
Completely Updated with Questions from Recent Exams
This book is incorporated with die questions from all the recent competitive exams, held in year 2013-14.
This book is a brain child of Mr Deepesh Jain, Director, Arihant Publications (India) Limited. Richa Agarwal, Diwakar Sharma and Shivam Mittal have given their best and sincere efforts for die completion and final presentations of die book.
The entire project has been managed and supervised by Mr Mahendra Singh Rawat and Mr Amit Verma. Aas Mohammed and Pradeep are to be complemented for very apt designing to the book cover. Amit Bansal and Mayank Saini have given their expertise in the layout of the book. Everyone’s contribution for tiiis book is very special and is worthy of great applause. Reader’s recommendations will be highly treasured.
With best compliments
Rajesh Verma
### Content of Fast Track Objective Arithmetic
• 1. Number System Numerals • Face Value and Place Value of the Digits in a Number • Types ofNumbers • Operations on Numbers • Divisibility Tests Unit’s Place of an Expression• Basic Number Theory
• 2. Number Series Types of Series • Types of Questions Asked on Number Series
• 3. HCF and LCM Factors and Multiples • Least Common Multiple (LCM)« Highest Common Factor(HCF) • Method to Calculate LCM and HCF of Fractions • Fast Track Techniquesto Solve the Questions • Method to Solve Questions Based on Bells
• 4. Simple and Decimal Fractions Simple Fraction • Decimal Fraction • Operations on Simple Fractions • Operations onDecimal Fractions • Comparison of Simple Fractions • Fast Track Formulae to Solve dieQuestions
• 5. Square Root and Cube Root Square • Square Root • Properties of Squares and Square Roots • Fast Track Formulaeto Solve the Questions • Fast Track Techniques to Solve the Questions •Cube• Cube Root • Properties of Cube and Cube Roots
• 6. Indices and Surds Indices • Surds • Properties of surds • Operations on Surds • Fast Track Techniquesto Solve the Questions
• 7. Simplification VBODMAS Rule • Basic Formulae
• 8. Approximation Basic Rules to Solve the Problems by Approximation
• 9. Word Problems Based on Numbers Types of Word Problems Based on Numbers
• 10. Average Average • Properties of Average • Important Formulae Related to Average of Numbers •Fast Track Techniques to Solve the Questions • Average Speed
• 11. Percentage Percentage • Formulae to Calculate Percentage • Fast Track Techniques to Solve theQuestions
• 12. Profit and Loss Basic Formulae Related to Profit and Loss • Fast Track Techniques to Solve die Questions
• 13. Discount Marked Price • Basic Formulae Related to Discount • Successive Discount • Fast TrackTechniques to Solve die Questions
• 14. Simple Interest Simple Interest (SI) • Instalments • Fast Track Techniques to Solve die Questions
• 15. Compound Interest Basic Formulae Related Compound Interest • Instalments • Fast Track Techniques toSolve die Questions
• 16. True Discount and Banker’s Discount True Discount • Fast Track Formulae to Solve die Questions • Banker’s Discount
• 17. Ratio and Proportion Ratio • Comparison of Ratios • Proportion • Fast Track Techniques to Solve die Questions
• 18. Mixture or Alligation Mixture • Rule of Mixture or Alligation • Fast Track Techniques to Solve die Questions
• 19. Partnership Types of Partnership • Types of Partners • Fast Track Techniques to Solve the Questions
• 20. Unitary Method Direct Proportion • Indirect Proportion
• 21. Problem Based on Ages Important Rules for Problem Based on Ages • Fast Track Techniques to Solve the Questions
• 22. Work and Time Basic Rules Related to Work and Time • Fast Track Techniques to Solve the Questions
• 23. Work and Wages Important Points • Fast Track Formulae to Solve die Questions
• 24. Pipes and Cisterns Important Points • Fast Track Techniques to Solve the Questions
• 25. Speed, Time and Distance Basic Formulae Related to Speed, Time and Distance • Fast Track Techniques to Solvethe Questions
• 26. Problems Based on Trains Basic Rules Related to Problems Based on Trains • Fast Track Techniques to Solve theQuestions
• 27. Boats and Streams Basic Formulae Related to Boats and Streams • Fast Track Techniques to Solve die Questions
• 28. Races and Games of Skill Important Terms • Some Facts about Race • Fast Track Techniques to Solve die Questions
• 29. Clock and Calendar Clock • Important Points Related to Clock • Fast Track Techniques to Solve die Questions• Calender • Ordinary Year • Leap Year • Odd Days
• 30. Linear Equations Linear Equations in One, Two and Three Variables • Metiiods of Solving Linear Equations• Consistency of the System of Linear Equations
• 31. Quadratic Equations Important Points Related to Quadratic Equations • Methods of Solving QuadraticEquations • Fast Track Formulae to Solve die Questions
• 32. Permutations and Combinations Permutation • Cases of Permutation • Combination • Cases of Combination • Factorial •Fundamental Principles of Counting • Fast Track Formulae to Solve die Questions
• 33. Probability Terms Related to Probability • Event • Rules/Theorems Related to Probability. • Types ofQuestions
• 34. Area and Perimeter Area • Perimeter • Triangle • Properties of Triangle • Quadrilateral • Regular Polygon •Circle • Fast Track Techniques to Solve the Questions
• 35. Volume and Surface Area Volume • Surface Area • Cube • Cuboid • Cylinder • Cone • Sphere • Prism • Pyramid •Fast Track Techniques to Solve the Questions
• 36. Geometry Point • Line • Angle • Triangle • Congruency of Triangles • Similarity of Triangles •Quadrilateral • Polygons • Circle
• 37. Coordinate Geometry Rectangular Coordinate Axes • Quadrants • Distance Formula • Basic Points Related toStraight Lines
• 38. Trigonometry Measurement of Angles • Relation between Radian and Degrees • Trigonometric Ratios •Trigonometric Identities • Sign of Trigonometric Functions • Trigonometric Ratios ofCombined Angles
• 39. Height and Distance Line of Sight • Horizontal Line • Angle of Elevation Angle of Depression
• 40. Data Table
• 41. Pie Chart
• 42. Bar Chart Types of Bar Chart
• 43. Line Graph Types of Line Graph
• 44. Mixed Graph
• 45. Data Sufficiency
• Fast Track Practice Sets
Also Check –
### Stay in touch
To be updated with all the latest news, offers and special announcements. | 2,285 | 10,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-25 | latest | en | 0.958098 |
http://intertechtion.com/answers/5-11-2012.html | 1,701,391,271,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00230.warc.gz | 21,353,822 | 3,210 | question
Create a method that checks if an input integer is a palindrome without using an array or converting it to a String.
ex. isPalindrome(1234) returns false
ex 2. isPalindrome(12321) returns true
Try playing with the modulus operator to extract digits.
``` isPalindrome(input) reverse = 0; inTemp = input; while ( inTemp != 0 ) reverse = (reverse * 10) + (inTemp % 10) inTemp = inTemp / 10 if ( reverse == input ) print("true") else print("false") ```
This solution works by first extracting the rightmost digit from the input number (ex. 1234 % 10 = 4) and then adding it to the current reversed number multiplied by ten (ex. (0 * 10) + 4 = 4; in second iteration (4 * 10) + (123 % 10) = 43). inTemp which initially stores the input number is divided by ten with each iteration to move on to the next digit. These steps are repeated until inTemp = 0 at which point all digits have been stored in reverse. | 256 | 962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-50 | latest | en | 0.720209 |
http://scanraid.com/coloursudoku.htm | 1,670,155,443,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00490.warc.gz | 39,225,844 | 9,830 | Solvers
Puzzles
Latest Apps
Str8ts
Other
Colour Sudoku Solver Please report any bugs. Tested in IE, Firefox, Safari and Chrome. Feedback: Form and User Comments/Questions Here
Candidates can be Edited or Highlighted / Shown
Enter numbers here:
1 2 3 4 5 6 7 8 9
Auto Tab Auto Clear
Clues+Solved: 0/81
Pick an example here Easy 1 Easy 2 Easy 3 Moderate 1 Moderate 2 Moderate 3 Tough 1 Tough 2 Tough 3 Quite Hard 1 Quite Hard 2 Quite Hard 3 Extremely Hard 1 Extremely Hard 2 Extremely Hard 3
Puzzle Rules
Enter the numbers 1 to 9 uniquely in every row, column, box and colour. Each colour is in the same position relative to each 3x3 box.
Pick an example from the list and use Take Step to click through the steps of the solver so see how each puzzle solves.
A B C D E F G H J
Version 2.09
See Strategy Overview documentation
Coordinate system: Letter/Number rYcX
1 2 3 4 5 6 7 8 9
Check for Solved Cells Show Possibles 1: Singles in Row/Column 2: Singles in Colour/Box 3: Naked Pairs/Triples 4: Hidden Pairs/Triples 5: Naked Quads 6: Pointing Pairs 7: Box/Line Reduction Tough Strategies 8: X-Wing 9: Simple Colouring 10: Y-Wing 11: Swordfish 12: XYZ Wing Diabolical Strategies 13: X-Cycles 14: XY-Chain 15: 3D Medusa 16: Jellyfish 17: WXYZ Wing 18: Aligned Pair Exclusion Extreme Strategies 19: Grouped X-Cycles 20: Finned X-Wing 21: Finned Swordfish 22: Altern. Inference Chains 23: Sue-de-Coq 24: Digit Forcing Chains 25: Nishio Forcing Chains 26: Cell Forcing Chains 27: Unit Forcing Chains 28: Almost Locked Sets 29: Death Blossom 30: Pattern Overlay Method 31: Quad Forcing Chains "Trial and Error" 32: Bowman's Bingo
Please report any bugs - Thanks.
If you are getting errors please
clear your browser cache
to retrieve the latest script files.
## The Colour Sudoku Solver
Colour Sudoku is the puzzle as normal Sudoku with an extra constraint. In a similar way to the rows, columns an boxes, the colours must also be uniquely 1 to 9. Each colour is in the same position in each 3x3 box, as you will notice with the central white cell, for example. All Sudoku strategies continue to apply - except the ones that rely on the deadly pattern used by Uniqueness strategies. Almost
The extra constraint does not add any extra complexity in the rows and columns - since a uniqueness constraint already exists, but it does effect cells diagonally. For example, B2, C5 and H8 are now connected and a conjugate pair may exist on those cells. So, like Sudoku X, the solver needs to be aware of all the new diagonal opportunities that are now possible.
Interestingly, this constraint forces the Sudoku puzzle to be 'perfect', that is, every number appears in a different position relative to each 3x3 box.
Note also that I have not completed the grading of this type of puzzle so I cannot offer a grading option in this version. The Solution Count program also needs updating to this will be available after the next update.
All feedback, comments, arguments, bug reports and strategy ideas are welcome. There is a new FEEDBACK form with a column displaying comments and questions. Many thanks to all the people who have done so and helped improve this solver.
Original version 1.01 28th May 2005
New in version 1.99 (April 9th 2014)
Unchecked strategies are now saved in a cookie.
Latest version 1.85 (May 12th 2011)
On the small board for number entry I have added an option that automatically clears off candidates as numbers are added. Also, changes to the small board are automatically saved.
Latest version 1.82 (May 10th 2011)
I have redesigned the way cookies are stored and puzzles loaded. You still have a manual save and reload but the solver now automatically saves the board every time it changes. Should you loose the page it will restore the puzzle you were working on. Cookies also retain the difference between clues and solved cells as well.
Latest version 1.74 (June 21st 2011)
I sped up the "Take Step" process by skipping failed strategies and going straight to the first successful one. This applies to the basic strategies which are client-side. If you get "script taking too long" messages from your browser, let me know. It's a bit risky for slower machines and I'd like to find out if this works for everyone.
Full version history here
## Solver help
Use the "Import a Sudoku" button or type in a Sudoku puzzle in the small board.
You can also pick examples from the list above. Click on Take Step to step through the solution. Unknown squares are filled with 'candidates' - possible solutions.
Any cells that are reduced to one possible candidate are solved.
You can use the [<<] button to step back one go. Toggling between Take Step and [<<] helps you see the changes.
Pressing "Enter" on the keyboard after clicking on Take Step is a quick way to activate "Take Step". Details of any solutions will be written out in the text box below the big board. Strategies are ordered by complexity. Any strategy that is successful returns the step-through to the start.
Click on the board to highlight sets of numbers. You can edit the Sudoku at any time - entering solutions in the small board or editing candidates. (Toggle between highlighting and editing using the radio buttons at the top.)
Solver created on 28-May-2005.
This page was last modified on 17-July-2014.
All code and design is copyright and for personal use only and may not be reproduced elsewhere.
Copyright Andrew Stuart @ Syndicated Puzzles Inc, 2005-2014 | 1,341 | 5,474 | {"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-2022-49 | latest | en | 0.728835 |
https://www.dsprelated.com/thread/12557/purity-1 | 1,722,701,768,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00888.warc.gz | 612,649,939 | 16,819 | ## Purity > 1?
Started by 4 years ago26 replieslatest reply 4 years ago258 views
I have an application where I am computing a purity value, because I'm interested in how much noise is present. Since I don't have a full spectrum measurement (I'm using the generalized Goertzel https://doi.org/10.1186/1687-6180-2012-56 to compute a single DTFT), I use Parseval's energy theorem to compute full signal energy in the time domain. This normally works very well, but occasionally I get a purity slightly greater than 1. Is there some good reason for this, or should I start looking for an error in the computation?
I have compared my Goertzel implementation to FFTW so I know there is minimal error there: https://bitbucket.org/hexamer/testgoertzelresonato...
[ - ]
Please give some description and expression for calculation of the purity.
[ - ]
"Total energy" from time domain is the summation of all the squares of samples Σ(x[n])^2
"DTFT energy" is (2/N)*(DTFT magnitude)^2
Purity = "DTFT energy" / "Total energy"
[ - ]
How much is the difference? If it is a factor of two or a factor of sqrt(2) or their inverses, there is an arbitrary scaling factor that can be applied. If it is something very close to 1.0, it is probably just a scale factor error (maybe due to quantization) in the transform coefficients.
[ - ]
Its normally very close to 1 (e.g in a low noise scenario, it's typical for the purity to be > 0.999).
Goertzel is known to have numerical errors with large input sequences, near theta=0 (up to N^2 error growth), but I've applied the Reinsch modifications to mitigate this. With the size of sequences I use and these mitigations, these errors are < 1e-10, as confirmed in this project:https://bitbucket.org/hexamer/testgoertzelresonato...
[ - ]
I think DTFT energy should be (1/N)*(DTFT)^2 .
It all depends as to how DTFT is calculated. Please compare the DTFT with the FFT output. To obtain the Fourier coefficients from FFT output, the DC and Nyquist terms have to be divided by N/2. The other terms by N. Parseval's theorem holds for the Fourier coefficients.
[ - ]
I had been studying this which also notes the special treatment of the 0 and nyquist bins: http://www.dspguide.com/ch10/7.htm
I can definitely see similar effects in my application - as the DTFT is approaching Nyquist, the purity can be near 2.
In the dspguide the factor for the adjusted DFT is 2/N. In my application, for low noise signals the purity is very near (and occasionally more than) 1, so I assume 2/N is right.
[ - ]
If you have access to the input time series, you can apply FFT and get the specific Fourier component you calculate with Goertzel. Do they differ by a factor?
Does most of the energy in your time series reside in the DTFT component? Normally the purity for a single frequency should be well below 1, and as you have noted, not exceed 1.
I suspect you may missing another factor of (1/N),
[ - ]
DFT is defined in math as:
Thus there is no down scaling of accumulated sum (hence output is scaled up by N relative to input). For iDFT the accumulator result is scaled back by 1/N. Apparently the purpose is to make sure if you apply DFT followed by iDFT you get back to your original signal.
This scaling is arbitrary and is not respected by most hardware fft ip vendors (Intel,xilinx...etc). Instead they output(in both cases of fft/ifft) scaling up by sqrt(N).
Either way it is correct. You only need to take scaling effect out of the analysis. accumulation of N samples scales result N times and you might conclude that you need to divide back by N but this approach causes trouble to power results. Matlab fft follows math rule and outputs scaling by N (i.e. just accumulates) and it outputs ifft as 1/N of accumulated result. So if you want to see fft power in matlab you need divide it back by sqrt(N). Tools, books & platforms are not that in sync.
For the case of purity of single tone, that depends on DFT smearing and noise but it can't be > 1 based on the posted definition because you get out what you have put in.
[ - ]
Thank you, Kaz. This is kind of what I suspected. Purity > 1 makes no sense. Wondering if this has anything to do with the use of DTFT vs DFT.
[ - ]
The factor of 2 in the purity expression is not correct. The outside factor should be (1/N)
Please see the simple derivation of Parseval's theorem in
[ - ]
It’s pretty correct. OP is only calculating a single bin value. There would be an identical magnitude value at the negative bin, hence the factor of two
[ - ]
Maybe a dumb question, but are you using the same N for the Goerztel and signal energies? If they were mismatched, it would mostly work on average, but it could for sure cause excursions above 1
[ - ]
Not a dumb question at all, in fact it's been on my mind too. As you probably know, to compute a DFT coefficient with Goertzel, you iterate one final time with a 0 input. So, in my case to compute the signal energy, the factor I use is actually 2/(N+1), not 2/N. That last sample does not affect the time domain energy since it's just a summation of the samples squared. However, if I were to use N instead of N+1, it would only make the purity greater. And, I don't know is skipping the last iteration with 0 input is valid for my applicaiton.
[ - ]
Hi again. My apologies for not providing a more rigorous mathematical analysis, but I have to clock in here in a few minutes. I ran some tests in Python and I think I was able to reproduce what you were seeing. Given an input sine wave, get the RMS, square, sum, divide by length, root. Easy enough. I then did a ‘manual’ DFT bin calculation by modulating the data with a complex sinusoid of the same frequency and summing all the data. The corresponding energy/rms/whatever can be calculated as root of two times the square of the magnitude. For frequencies where an integer number of periods were contained in the input, the values matched exactly. For non integer values, the DFT value was slightly higher, which would correspond to your purity > 1. If I were off the cuff justifying it, I’d say that the window on the Complex modulation results in non-zero side lobes at the corresponding negative frequency, which means something something so it doesn’t work. I’ll try and follow up when I have more time, but at least I can back up that you aren’t crazy
[ - ]
Nice to see that a completely different approach confirms similar results!
[ - ]
Aaron45, are you using 64-bit numerical values in your computations? Also, what is the nature of your input test signal?
[ - ]
Hi Rick,
The inputs are all 16 bit samples, and all subsequent computations are done with doubles. Input signals are very often pure sinusoids with some low background noise, besides quantization.
I'm starting to think that I should start generating some simulated ideal inputs to see if I can characterize when purity goes greater than 1.
I've also started to ponder how using a DTFT is affecting things. With an FFT with discrete bins, and the 0 and fs/2 bins requiring special treatment for for Parseval to hold true, what might that mean for DTFT where there aren't necessarily discrete bins?
Aaron.
[ - ]
Hi Aaron45. I see what's happening here!
The version of Parseval's Theorem that applies to digital signal processing is:
Σ(x[n])^2 = 1/N*Σ(|X[m]|)^2 (1)
where |X[m]| is the N spectral magnitude samples computed using an N-point DFT (or FFT). The above Equation (1) says that the total time-domain energy will be equal to the total spectral magnitude energy divided by N.
Your computation of the value you call "Purity" will only be correct if: (1) the single input sinusoid's frequency is exactly at a DFT bin center (no spectral leakage), and (2) the input sinusoid is noise free.
So, ...for a noise-contaminated signal you must compute an N-point DFT, compute the |X[m]| magnitude sequence, and then use the quantities in my above Equation (1). And then your Purity value will always be unity.
SIDE NOTE: Aaron45, the Wiki page for the Goertzel algorithm states that poles on the z-plane's unit circle make the Goertzel algorithm "marginally stable and vulnerable to numerical-error accumulation". That widely-believed statement is NOT true. The Goertzel algorithm is always guaranteed stable!
[ - ]
Thank you for this insight, Rick. One reason I use the generalized Goertzel (a DTFT and not DFT) is so I don't have to worry about bin centering. But I think you're hinting that the use of the DTFT is a reason purity does not seem to work. I have some ideas for an experiment to see how the DTFT affects purity - stay tuned :)
On Goertzel, I agree with you that the marginally stable part is wrong .. I've read your excellent article on that matter. The part on numerical-error accumulation correct. Not that I really needed to given the amount of literature on the matter, but I independently confirmed it: https://www.dsprelated.com/thread/767/error-growth...
[ - ]
Hi Aaron45. I think of the numerical-error accumulation as being caused by the fact that we cannot place the Goertzel network's poles *exactly* at the desired angles on the z-plane's unit circle due to quantized coefficients. With quantized network coefficients the poles will always be located on the unit circle, but at angles that are very slightly different than what we desire. (By the way, when the desired pole angles are 0, ±pi/2, or pi radians then no coefficient quantization error occurs.)
Aaron45, my previous November 20, 2020 "Reply" was based on my assumption that in your expression:
"DTFT energy" is (2/N)*(DTFT magnitude)^2
the term "DTFT magnitude" was a single number. Was my assumption correct?
[ - ]
Rick, Yes, that is a correct assumption. It's a single frequency point DTFT calculation, most often not at a bin center.
[ - ]
Hi Aaron45. OK. So when a single input sinusoid's frequency is not at an N-point DFT's bin center then all of the DFT's N magnitude samples will be nonzero Then you can now see that using a single DFT magnitude sample will not compute the correct total spectral energy (your "DTFT energy" value).
[ - ]
Purity plot (last 1000).png
Purity plot (first 200).png
Thanks to everyone who took the time to chime in here. Rick's comment got me thinking about how leakage affects the purity calculation in my application, so I tried a couple experiments.
The first experiment I tried was hacking up my application to send in simulated (ideal) data in place of the real data. I noticed some definite trends (see graphs): Purity plot (last 1000), Purity Plot (first 200)
In my second experiment, I made an Octave script (see attached) to visualize where the DTFT is sampled and how the leakage occurs. The results are consistent with my application (see graph): Purity vs Bin Offset N is 16, f is 0.25 fs
1) There is a sinusoidal oscillation in the purity value with a frequency of 2x the DFT bin width.
2) When well away from 0 or Nyquist, regardless of bin center offset, the purity is less than or equal to 1
3) As expected, processing gain minimizes leakage and makes purity higher, all else equal
4) The phase of the purity oscillation within the frequency spectrum depends on whether N is even or odd.
5) When approaching 0 or Nyquist, the purity can become greater than 1. For the near-fs/2 case, It approaches 3/2 then jumps to 2 right at Nyquist. Overall it looks very sinc-like. For the near-0 case, it looks a little more strange.
6) If I force all samples to be at bin center by enforcing an integer number of cycles, the purity at the fs/2 sample is still 2. Consistent with corrections described here?: https://www.dspguide.com/ch10/7.htm
One naive way to address this would be to place extra emphasis on capturing integer cycles to get close to a bin center. I can do this with rational approximation of fs/f and using the denominator as the number of cycles. Unfortunately this is inefficient as it leads to higher samples and loss of consistent bandwidth control.
That has me wondering whether I can construct a correction factor based on inputs like N, bin offset, f/fs.
I initially though about sampling the ideal DTFT (Dirichlet pair) to compute the theoretical purity error and correct for that. But that could be computationally expensive for large N.
Does anyone know of a a way to compute such a "Leakage Correction Factor". I've been searching and found a few articles. The most promising may be this article which attempts to do something similar: http://www.nicholson.com/rhn/dsp.html#4, but it seems that it's finding the sum of the DFT bin magnitudes whereas I need to compute an energy term? Unfortunately there is no derivation there so I can't see how to adapt it.
12/8/2020: Edited Observation 4 above. Being at a bin center always creates purity = 1. Here's what it looks like across half spectrum for even and odd N: Purity Simulation Half Spectrum N is 16.pdf Purity Simulation Half Spectrum N is 17.pdf
[ - ]
Hi, I am sorry to say but I am totally puzzled by your analysis.
If I use Octave FFT (basically floating point) I always get 1 or very close irrespective of frequency. So are you checking your Goertzel algorithm or am I wrong using FFT?
%%%%%%%%%%%%%%%%%%%%%
clear;clc;
N = 100;
ind = 0;
for f = 0:.005:.5
x = cos(2*pi*(0:N-1)*f);
y = fft(x);
ind = ind+1;
pwr_ratio(ind) = mean(abs(y).^2)/sum(x.^2);
end
plot(pwr_ratio,'.-');
%%%%%%%%%%%%%%%%%%%%%%%
[ - ]
Kaz,
Your results make sense to me. Your power ratio is always based on time domain samples or DFT frequency samples. So this script is really demonstrating Parseval's Theorem. What Rick has pointed out is that my application is expecting that my single DTFT sample at the pure frequency is just an ordinary DFT sample. And of course when it's off bin center it's not a DFT sample. For Parseval's to hold, the Frequency domain samples need to be DFT samples. You can visualize what's happening in my case if you execute purity.m. When my pure signal is off bin center, all the DFT samples have some non-0 values (leakage). Since the energy in those non-0 DFT samples is equivalent to the time domain energy (Parseval), it affects the time domain energy (the basis for my total energy denominator), and makes purity != 1.
Aaron
[ - ]
Sorry for dragging this further.
firstly, we agree that we need to force total power of time domain & frequency domain to be same, not just to keep to Parseval's but for this task's common sense.
secondly, we agree that one or few bins of dft should have less power than the sum of all bins.
thirdly we agree that a single tone may end up in one bin or leak.
and finally we agree that if there is additional noise at input then it will go to its bins.
So no matter what resolution you use or what frequency you observe or what leak occurs the power (in one or more bins) as ratio to all bins should be either 1 or less than 1.
If you get >> 1 you are doing something wrong or your dft is going wrong but I checked your gga function (Goertzel) and it is doing well. | 3,643 | 15,121 | {"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} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.900721 |
https://www.statology.org/tips-using-statistical-power-effectively-experiment-design/ | 1,721,781,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00886.warc.gz | 874,599,413 | 22,607 | Tips for Using Statistical Power Effectively in Experiment Design
Designing a successful statistical experiment requires consideration of a variety of factors. Perhaps the most critical is statistical power. Power is the probability that your test will detect an effect, if there is one there to be detected at all. This is a key component of making sure your study will yield reliable and valid results. Without adequate power, even a well-designed experiment will fail to provide conclusive evidence, leading to potentially misleading conclusions. This article will go through five key tips to using statistical power effectively in the design phase of an experiment.
Understanding Statistical Power
Statistical power is the ability to reject the null hypothesis correctly. High statistical power indicates that a study has a greater chance of identifying a true effect while lower power increases the risk of a Type II error.
Power is influenced by a few different factors. First is sample size where increasing the number of observations in a study also increases the power. Second is effect size, or the magnitude of difference between groups being detected, where larger effect sizes are easier to detect and lead to greater power. Third is the predetermined significance level, or alpha, where a lower alpha level is associated with reduced power. Finally is variability in the dataset where decreased variability makes it easier to detect differences, increasing power.
Determining the Appropriate Sample Size
The main goal of a statistical power analysis is to determine the correct sample size needed before a research project is implemented. It is critical to outline the research objectives before calculating the sample size, including knowing the key outcomes and hypotheses to test.
In order to calculate the sample size, the following variables must be known: desired effect size, significance level (alpha), and desired power level. This power level is traditionally set to 80%, which means that there is an 80% probability of correctly rejecting the null hypothesis when it is false. Each type of statistical test being run will then have its own formula for determining the proper sample size using these variables. Various software options are available to conduct these calculations, including R, SPSS, and G*Power.
Choosing the Right Effect Size
Selecting the correct effect size is essential for determining an appropriate sample size and ensuring that the results from the experiment are useful. Effect size quantifies the strength of an effect or relationship in the data. Common effect sizes are Cohen’s d, used for comparing the mean of two groups, the odds ratio, used in logistic regressions, and the correlation coefficient, used to measure the strength and direction of a linear relationship.
Choosing the right effect size can be the most challenging part of power analysis and experiment design. Often, these values are found from reviewing previous research done in the field. If these do not exist, pilot studies with a limited amount of observations can be done. Practical significance can also be considered based on the field and application of the final study results.
Using the Correct Statistical Test
The final element of a power calculation is using all the variables in an equation specific to a statistical test. Ensuring that the correct formula for the proper statistical test is used is critical to a proper power analysis.
There are many tools to help determine what statistical test needs to be run. Briefly, it is necessary to understand the types of data available (such as categorical or interval data) as well as what specifically will be compared in the study (means, proportions, counts, etc.).
Planning for Dropouts and Missing Data
Finally, in any real world study using human data, it is also important to account for dropouts and missing data. If a sample size is calculated to meet 80% power, but then a proportion of the study participants recruited drop out or submit missing data, then the study is now underpowered and you may miss identifying an effect.
Typically, this problem is accounted for before the study recruitment process begins. Buffers of 10% or 20% are generally added to any calculated sample size to ensure that even if data is lost in the study process, the final, complete dataset will still yield enough power.
Conclusion
Effectively utilizing statistical power in experiment design is critical for ensuring the reliability and validity of your research findings. Using these tips will help enhance the credibility of results and lead to more meaningful and actionable insights from an experiment. | 858 | 4,699 | {"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} | 2.921875 | 3 | CC-MAIN-2024-30 | latest | en | 0.930358 |
https://metanumbers.com/637327 | 1,643,357,175,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00293.warc.gz | 437,233,823 | 7,362 | # 637327 (number)
637,327 (six hundred thirty-seven thousand three hundred twenty-seven) is an odd six-digits prime number following 637326 and preceding 637328. In scientific notation, it is written as 6.37327 × 105. The sum of its digits is 28. It has a total of 1 prime factor and 2 positive divisors. There are 637,326 positive integers (up to 637327) that are relatively prime to 637327.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 6
• Sum of Digits 28
• Digital Root 1
## Name
Short name 637 thousand 327 six hundred thirty-seven thousand three hundred twenty-seven
## Notation
Scientific notation 6.37327 × 105 637.327 × 103
## Prime Factorization of 637327
Prime Factorization 637327
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 637327 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 13.365 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 637,327 is 637327. Since it has a total of 1 prime factor, 637,327 is a prime number.
## Divisors of 637327
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 637328 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 318664 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 798.328 Returns the nth root of the product of n divisors H(n) 2 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 637,327 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 637,327) is 637,328, the average is 318,664.
## Other Arithmetic Functions (n = 637327)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 637326 Total number of positive integers not greater than n that are coprime to n λ(n) 637326 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 51763 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 637,326 positive integers (less than 637,327) that are coprime with 637,327. And there are approximately 51,763 prime numbers less than or equal to 637,327.
## Divisibility of 637327
m n mod m 2 3 4 5 6 7 8 9 1 1 3 2 1 5 7 1
637,327 is not divisible by any number less than or equal to 9.
## Classification of 637327
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (637327)
Base System Value
2 Binary 10011011100110001111
3 Ternary 1012101020201
4 Quaternary 2123212033
5 Quinary 130343302
6 Senary 21354331
8 Octal 2334617
10 Decimal 637327
12 Duodecimal 2689a7
20 Vigesimal 3jd67
36 Base36 dnrj
## Basic calculations (n = 637327)
### Multiplication
n×y
n×2 1274654 1911981 2549308 3186635
### Division
n÷y
n÷2 318664 212442 159332 127465
### Exponentiation
ny
n2 406185704929 258873116765284783 164986826888668654895041 105150559420474527818291795407
### Nth Root
y√n
2√n 798.328 86.0572 28.2547 14.4835
## 637327 as geometric shapes
### Circle
Diameter 1.27465e+06 4.00444e+06 1.27607e+12
### Sphere
Volume 1.08437e+18 5.10428e+12 4.00444e+06
### Square
Length = n
Perimeter 2.54931e+06 4.06186e+11 901316
### Cube
Length = n
Surface area 2.43711e+12 2.58873e+17 1.10388e+06
### Equilateral Triangle
Length = n
Perimeter 1.91198e+06 1.75884e+11 551941
### Triangular Pyramid
Length = n
Surface area 7.03534e+11 3.05085e+16 520375
## Cryptographic Hash Functions
md5 e44106c07f19b0271ed86e257ac06ae8 dba82be9d6d9ccf303b8268c21806a6e894651c2 d84cceeb10e49f64c995a3423688efc68047f54e715f1935ba2c7f7036c3bda5 779928e05f4c5cdd0dde80c01ae7f97fb1754b8eb8332769d12d9697140a934fd9987af3fd9f8be34b834827f1c4f7d06955176e5a05c73531dd79b96fd3daca 61f36b3aab60111a0f6f1ff3f41ac2200d5377f7 | 1,484 | 4,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-05 | latest | en | 0.819756 |
https://enacademic.com/dic.nsf/enwiki/11537215 | 1,606,152,406,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141163411.0/warc/CC-MAIN-20201123153826-20201123183826-00457.warc.gz | 294,940,536 | 17,466 |
# Christensen Failure Criterion
Failure criteria for isotropic materials has been a long standing problem. Despite many historical attempts, there have not been any successful general forms that span the range from ductile to brittle materials. A new failure criterion that aims to do so has been recently developed. The isotropic material failure theory by Christensen[1] has a two property form calibrated by the uniaxial tensile and compressive strengths T $\left (\sigma_T\right )$ and C $\left (\sigma_C\right )$. It is based upon a sequence of seven related archive journal papers, the first of which was published in 1997[2][3]. The website describing it in considerable detail was started in 2007, and it was first utilized by eFunda[4] in 2008.
The overall failure criterion is composed of two separate subcriteria representing competitive failure mechanisms. when expressed in principal stress components, it is given by :
Polynomial Invariants Failure Criterion
For $0\le\frac{T}{C}\le1$
$\left (\frac{1}{T}-\frac{1}{C} \right )\left (\sigma_1+\sigma_2+\sigma_3\right )+\frac{1}{2TC}\left [\left (\sigma_1-\sigma_2\right )^2+\left (\sigma_2-\sigma_3\right )^2+\left (\sigma_3-\sigma_1\right )^2\right ]\le 1$
(1)
Coordinated Fracture Criterion
For $0\le \frac{T}{C}\le \frac{1}{2}$
$\begin{array}{lcl} \sigma_1 & \le & T \\ \sigma_2 & \le & T \\ \sigma_3 & \le & T \end{array}$
(2)
For plane stresses,σ3 = 0 and T/C=0.3(brittle materials). Blue line is polynomial invariants failure criterion (1). Red line is coordinated fracture criterion(2).
The geometric form of (1) is that of a paraboloid in principal stress space. The fracture criterion (2) (applicable only over the partial range 0 ≤ T/C ≤ 1/2 ) cuts slices off the paraboloid, leaving three flattened elliptical surfaces on it. The fracture cutoff is vanishingly small at T/C=1/2 but it grows progressively larger as T/C diminishes.
The organizing principle underlying the theory is that all isotropic materials admit a distinct classification system based upon their T/C ratio. The comprehensive failure criterion (1) and (2) reduces to the Mises criterion at the ductile limit, T/C = 1. At the brittle limit, T/C = 0, it reduces to a form that cannot sustain any tensile components of stress.
Many cases of verification have been examined over the complete range of materials from extremely ductile to extremely brittle types[1]. Also, examples of applications have been given. Related criteria distinguishing ductile from brittle failure behaviors have been derived and interpreted.
Applications have been given by Ha[5] to the failure of the isotropic, polymeric matrix phase in fiber composite materials.
## References
1. ^ a b Christensen, R. M.,(2010),http://www.failurecriteria.com.
2. ^ Christensen, R.M. (1997).Yield Functions/Failure Criteria for Isotropic Materials, Pro. Royal Soc. London, Vol. 453, No. 1962, pp. 1473-1491
3. ^ Christensen, R.M. (2007), A Comprehensive Theory of Yielding and Failure for Isotropic Materials, J. Engr. Mater. and Technol., 129, 173-181
4. ^ eFunda,(2010),http://www.efunda.com/home.cfm.
5. ^ S. K. Ha, K. K. Jin and Y. C. Huang,(2008), Micro-Mechanics of Failure (MMF) for Continuous Fiber Reinforced Composites. Journal of Composite Materials, vol. 42, no. 18, pp. 1873-1895.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Micro-Mechanics of Failure — Figure 1. Hierarchy of micromechanics based analysis procedure for composite structures. Micro Mechanics of Failure (MMF) is a newly proposed methodology, providing a more logical explanation of failure mechanism of continuous f … Wikipedia
• performing arts — arts or skills that require public performance, as acting, singing, or dancing. [1945 50] * * * ▪ 2009 Introduction Music Classical. The last vestiges of the Cold War seemed to thaw for a moment on Feb. 26, 2008, when the unfamiliar strains … Universalium
• Pedophilia — This article is primarily about the sexual interest in prepubescent children. For the sexual act, see Child sexual abuse. For the primary sexual interest in 11–14 year old pubescents, see Hebephilia. For mid to late adolescents (15 19), see… … Wikipedia
• Denmark — /den mahrk/, n. a kingdom in N Europe, on the Jutland peninsula and adjacent islands. 5,268,775; 16,576 sq. mi. (42,930 sq. km). Cap.: Copenhagen. * * * Denmark Introduction Denmark Background: Once the seat of Viking raiders and later a major… … Universalium
• Darth Vader — Anakin Skywalker / Darth Vader Star Wars character … Wikipedia
• Nichiren — (日蓮) A statue of Nichiren outside Honnoji in the Teramachi District of Kyoto. School Mahayana, Nichiren Personal … Wikipedia
• statistics — /steuh tis tiks/, n. 1. (used with a sing. v.) the science that deals with the collection, classification, analysis, and interpretation of numerical facts or data, and that, by use of mathematical theories of probability, imposes order and… … Universalium
• Idealism (italian) and after — Italian idealism and after Gentile, Croce and others Giacomo Rinaldi INTRODUCTION The history of twentieth century Italian philosophy is strongly influenced both by the peculiar character of its evolution in the preceding century and by… … History of philosophy
• Rheumatoid arthritis — Classification and external resources A hand affected by rheumatoid arthritis ICD 10 M … Wikipedia
• Media and Publishing — ▪ 2007 Introduction The Frankfurt Book Fair enjoyed a record number of exhibitors, and the distribution of free newspapers surged. TV broadcasters experimented with ways of engaging their audience via the Internet; mobile TV grew; magazine… … Universalium | 1,432 | 5,689 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-50 | latest | en | 0.860268 |
https://www.freecodecamp.org/forum/t/return-largest-numbers-in-arrays-please-help/39878 | 1,540,056,611,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513009.81/warc/CC-MAIN-20181020163619-20181020185119-00541.warc.gz | 953,818,757 | 5,369 | 0
#1
function largestOfFour(arr) {
var results = [];
for(var i =0; i<=arr.length; i++){
``````var temp = arr[i].sort().reverse()[0];
results[i] = temp;
``````
}
return results;
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
#2
As a hint: remember that in order to iterate through the actual numbers, you need to use two for loops. You are working with a 2D array.
Also instead of results[i] use results.push(temp), it will be less confusing.
But if you want to continue that way, then use arr[i] not arr[0] because you want to check every array. And you are getting that error because of the condition in your for loop. Set it to i<arr.length instead.
good luck
#3
hi Thanks for reply. Yeah i meant arr[i] not arr[0]. that was a typo.
But with I’m still getting cant read property sort error
#4
Set it to i<arr.length instead. The last iteration is going through an index that doesn’t exist, that’s why.
#5
thanks for helping, you are correct!
#6
function largestOfFour(arr) {
var results = [];
for(var i =0; i < arr.length; i++){
``````var temp = arr[i].sort(function(a,b){return a-b;}).reverse()[0];
results.push(temp);
``````
}
return results;
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]); | 416 | 1,303 | {"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-2018-43 | longest | en | 0.789254 |
https://practice.geeksforgeeks.org/problems/chandler-and-joey/0 | 1,579,407,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594209.12/warc/CC-MAIN-20200119035851-20200119063851-00386.warc.gz | 628,924,530 | 22,938 | Chandler and Joey
##### Submissions: 436 Accuracy: 24.74% Difficulty: Easy Marks: 2
Joey is Chandler's friend and loves to eat pizza. One day on Joey's birthday Chandler planned to take him to a pizza place where Joey wanted to eat all kinds of pizza. Joey started eating pizza from the plates kept nearby. The Plates were numbered from 1 to 10^6. Plate number 1 had 1 pizza, Plate 2 had 1 pizza, Plate 3 had 2 pizzas, Plate 4 had 2 pizzas, Plate 5 had 4 pizzas, Plate 6 had 2 pizzas, Plate 7 had 6 pizzas, Plate 8 had 4 pizzas.... and so on. Joey ate them all. Chandler saw it and found unfair that Joey was getting them so easy. He decided Joey can only eat if any plate has even number of pizzas and he wants to keep a count of pizzas eaten, so he needs to know how much pizzas are eaten from that plate.
Input:
First line of the input file consists of an integer T denoting the number of test cases.Then T test cases follow. Each test case consists of a single line containing an integer N denoting the number of plates.
Output:
Corresponding to each test case, print a string "YES" and an integer denoting the pizzas count if pizzas count is even in that plate, otherwise print only a string "NO".
Output for each test case should be printed in a new line.
Constraints:
1 <= T <= 100
2 <= N <= 10^6
Example:
Input:
2
5
2
Output:
YES 4
NO
Explanation:
Plate 5 had 4 pizzas and 4 is an even number
Plate 2 had 1 pizza and 1 is an odd number
#### ** For More Input/Output Examples Use 'Expected Output' option **
Author: Contribute.Practice
If you have purchased any course from GeeksforGeeks then please ask your doubt on course discussion forum. You will get quick replies from GFG Moderators there. | 452 | 1,724 | {"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} | 3.28125 | 3 | CC-MAIN-2020-05 | longest | en | 0.93547 |
https://research.utu.fi/converis/portal/Publication/2097872?lang=fi_FI | 1,569,119,231,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00507.warc.gz | 632,952,092 | 4,229 | A4 Artikkeli konferenssijulkaisussa
A square root map on Sturmian words (Extended abstract)
Alaotsikko: (Extended abstract)
Julkaisun tekijät: Peltomäki Jarkko,Whiteland Markus
Julkaisuvuosi: 2015
Journal: Lecture Notes in Computer Science
Kirjan nimi *: Combinatorics on Words
Sarjan nimi: Lecture Notes in Computer Science
Volyymi: 9304
Sivujen määrä: 13
ISBN: 978-3-319-23659-9
eISBN: 978-3-319-23660-5
Tiivistelmä
We introduce a square root map on Sturmian words and study its properties. Given a Sturmian word of slope \$alpha\$, there exists exactly six minimal squares in its language (a minimal square does not have a square as a proper prefix). A Sturmian word \$s\$ of slope \$alpha\$ can be written as a product of these six minimal squares: \$s = X_1^2 X_2^2 X_3^2 cdots\$. The square root of \$s\$ is defined to be the word \$sqrt{s} = X_1 X_2 X_3 cdots\$. The main result of this paper is that that \$sqrt{s}\$ is also a Sturmian word of slope \$alpha\$. Moreover, we characterize the Sturmian fixed points of the square root map, and we describe how to find the intercept of \$sqrt{s}\$ and an occurrence of any prefix of \$sqrt{s}\$ in \$s\$. Related to the square root map, we characterize the solutions of the word equation \$X_1^2 X_2^2 cdots X_n^2 = (X_1 X_2 cdots X_n)^2\$ in the language of Sturmian words of slope \$alpha\$ where the words \$X_i^2\$ are minimal squares of slope \$alpha\$.
We also study the square root map in a more general setting. We explicitly construct an infinite set of non-Sturmian fixed points of the square root map. We show that the subshifts \$Omega\$ generated by these words have a curious property: for all \$w in Omega\$ either \$sqrt{w} in Omega\$ or \$sqrt{w}\$ is periodic. In particular, the square root map can map an aperiodic word to a periodic word.
Sisäiset tekijät/toimittajat
Last updated on 2019-21-08 at 22:56 | 562 | 1,884 | {"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-2019-39 | latest | en | 0.592618 |
https://www.physicsforums.com/threads/classical-mechanics-ball-rolling-in-a-hollow-sphere.217135/ | 1,716,918,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00345.warc.gz | 802,707,210 | 16,773 | # Classical mechanics: ball rolling in a hollow sphere
• blalien
In summary, a uniform ball is rolling without slipping inside a fixed hollow sphere. The problem is to find the energy conservation equation and the period of small oscillations of the ball about its equilibrium position. The solution involves variables such as the ball's radius, mass, angle of position, moment of inertia, rotational velocity, kinetic energy, potential energy, total energy, and acceleration due to gravity. After correcting a mistake in the equation for kinetic energy, the correct solution is obtained.
blalien
[SOLVED] Classical mechanics: ball rolling in a hollow sphere
## Homework Statement
This problem is from Gregory:
A uniform ball of radius a and centre G can roll without slipping on the inside surface of a fixed hollow sphere of (inner) radius b and centre O. The ball undergoes planar motion in a vertical plane through O. Find the energy conservation equation for the ball in terms of the variable $$\theta$$, the angle between the line OG and the downward vertical. Deduce the period of small oscillations of the ball about the equilibrium position.
So in summary, we have:
m: mass of ball
$$\theta$$: angle of the ball's position, relative to the vertical line connecting the center and bottom of the hollow sphere
I: moment of inertia of ball
$$\omega$$: rotational velocity of ball
T: kinetic energy of ball
V: potential energy of ball (V=0 at height $$\theta$$=$$\pi$$/2, the center of the sphere)
E: total energy of ball
g: acceleration due to gravity
I = 2/5ma^2
## The Attempt at a Solution
First of all, I'm assuming that $$\omega$$=$$\theta$$'. It sounds intuitive, but I could be wrong there.
I'm given, as a solution, that the period of small oscillation (that is, sin($$\theta$$)=$$\theta$$) is 2$$\pi$$(7(b-a)/5g)^(1/2), which I'm not getting in my results. I have a very strong hunch that my mistake comes from bad energy equations. So, would you mind taking a look of these?
T = 1/2mv^2 + 1/2I$$\omega$$^2
v = $$\omega$$*(b-a)
So T = 1/2m($$\omega$$*(b-a))^2 + 1/2(2/5ma^2)$$\omega$$^2
T = m$$\omega$$^2/10(7a^2-10ab+5b^2)
V = -(b-a)mgcos($$\theta$$)
So E = T + V = that stuff
Am I correct here?
blalien said:
## The Attempt at a Solution
First of all, I'm assuming that $$\omega$$=$$\theta$$'. It sounds intuitive, but I could be wrong there.
That is wrong. Draw a simple diagram to figure it out.
Hah, it's always the little mistakes in the beginning that steal away an hour of my life.
That fixed everything. Thank you so much for catching that.
## 1. What is classical mechanics?
Classical mechanics is a branch of physics that describes the motion of macroscopic objects, such as balls, planets, and cars, using Newton's laws of motion and the principles of energy and momentum.
## 2. How does a ball roll in a hollow sphere?
A ball rolling in a hollow sphere follows the laws of classical mechanics, specifically the principle of conservation of angular momentum. As the ball rolls, it creates a torque that causes it to rotate around the center of the sphere. This rotation causes the ball to follow a circular path inside the sphere.
## 3. What is the role of gravity in a ball rolling in a hollow sphere?
Gravity plays a crucial role in the motion of a ball rolling in a hollow sphere. It provides the force that acts on the ball, causing it to accelerate and start rolling. Additionally, gravity also affects the ball's path and speed as it moves inside the sphere.
## 4. How does the mass of the ball and the size of the sphere affect the motion?
The mass of the ball and the size of the sphere both impact the motion of the ball rolling inside. A heavier ball will have more momentum and will be harder to stop, while a larger sphere will provide a longer path for the ball to follow. These factors can affect the speed and trajectory of the ball as it rolls.
## 5. What are the practical applications of studying ball rolling in a hollow sphere in classical mechanics?
Studying ball rolling in a hollow sphere can have several practical applications, such as understanding the motion of objects in a curved path, designing better sports equipment like bowling balls or golf balls, and predicting the behavior of rolling objects in various scenarios, such as on a sloped surface or in a rotating container.
Replies
1
Views
2K
Replies
3
Views
1K
Replies
4
Views
2K
Replies
26
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
527
• Introductory Physics Homework Help
Replies
60
Views
534
• Introductory Physics Homework Help
Replies
17
Views
2K
• Classical Physics
Replies
5
Views
539 | 1,139 | 4,652 | {"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} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.892729 |
http://freetofindtruth.blogspot.com/2016/04/33-44-74-murder-by-numbers-song-by.html | 1,524,144,885,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936969.10/warc/CC-MAIN-20180419130550-20180419150550-00362.warc.gz | 119,682,481 | 21,624 | ## Friday, April 1, 2016
### 33 44 74 | Murder by Numbers (Song), by the Police, off the album 'Synchronicity'
Synchronicity = 1+7+5+3+8+9+6+5+9+3+9+2+7 = 74/83 (Released in '83) (Masonic = 74)
Synchronicity = 19+25+14+3+8+18+15+14+9+3+9+20+25 = 182
The = 2+8+5 = 15
Police = 7+6+3+9+3+5 = 33 (Sting = 33)
The Police = 48 (Illuminati) (Evil)
The = 20+8+5 = 33
Police = 16+15+12+9+3+5 = 60
The Police = 93 (Saturn)
Notice the released date of the album. Since its release, Rodney King has died on that date, by the numbers, and the Charleston Church shooting has occurred, by the numbers.
6/17/1983 = 6+1+7+1+9+8+3 = 35
6/17/1983 = 6+17+(1+9+8+3) = 44 (Kill)
6/17/1983 = 6+17+19+83 = 125
6/17/83 = 6+17+83 = 106 (Prophecy)
Notice the CD and cassette version runs 44 minutes... with the final song about killing by the numbers....
Murder = 4+3+9+4+5+9 = 34
by = 2+7 = 9
Numbers = 5+3+4+2+5+9+1 = 29/38
Murder by Numbers = 72/81
Murder = 13+21+18+4+5+18 = 79
by = 2+25 = 27
Numbers = 14+21+13+2+5+18+19 = 92
Murder by Numbers = 198 | 477 | 1,038 | {"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-2018-17 | latest | en | 0.864193 |
https://math.stackexchange.com/questions/3356383/is-the-result-textrankab-textrankba-le-textmin-textranka | 1,621,260,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00266.warc.gz | 412,044,431 | 36,793 | # Is the result $|\text{rank(AB)}-\text{rank(BA)}| \le \text{min}(\text{rank}(A),\text{rank}(B))$
Is the result $$|\text{rank(AB)}-\text{rank(BA)}| \le \text{min}(\text{rank}(A),\text{rank}(B))$$ true for all $$n*n$$ matrices $$A,B$$. I only know the result that $$\text{rank(AB)} \le \text{min(rank(A),rank(B))}$$
Yes. Your desired inequality is a consequence of the one you mentioned. In other words, it follows from the fact that the ranks of both $$AB$$ and $$BA$$ are non-negative integers at most equal to that minimum. | 172 | 527 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-21 | latest | en | 0.725222 |
https://www.buicehockey.net/hockey/question-how-many-players-in-a-hockey-team.html | 1,653,397,859,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00257.warc.gz | 787,270,199 | 10,256 | Question: How Many Players In A Hockey Team?
Six players from each team are on the ice at any one time. The line up being; netminder, two defencemen and three forwards. These players can be changed at any time as the game is played at such a speed. A team is usually made up of between 17 and 22 players.
How many players are there in a hockey team?
There are 11 players on a team made up of a goalkeeper, defenders, midfielders and forwards. The only player on the field who is allowed to use their feet and hands as well as their stick is the goalkeeper.
What are the 11 positions in hockey?
Below, we’ll look at just that; each position, their role, and their individual skills.
• Center. The Center in Ice Hockey is generally the main player responsible for dictating the gameplay while the puck is in the offensive zone.
• Left Wing.
• Right Wing.
• Left Defenseman.
• Right Defenseman.
What is a 16 in hockey?
The 16 yard hit is a free hit for the defense 16 yards (for those of us who live in the metric universe, that’s 14.63 meters) from the base line after an opposing player hits the ball over the base line or commits a foul within the shooting circle.
You might be interested: Question: Why Do You Tape Hockey Sticks?
How many players are there in a hockey team including substitutes?
How many players and teams are there on a hockey field? Only two teams of eleven players each are allowed on the field at once, ten field players and one goalie. However, there’s an unlimited amount of times a team can sub in and out at any given time.
What does F mean in hockey?
In ice hockey, a forward is a player position on the ice whose primary responsibility is to score and assist goals. Generally, the forwards try to stay in three different lanes of the ice going from goal to goal.
How many midfielders are there in hockey?
The 3-5-2 formation splits up the 10 field players into two defenders (Fullbacks x 2), five midfielders (Left Half, Right Half, Centre Half, Inside Right and Inside Left), and three strikers (Centre Forward, Right Wing and Left Wing).
What’s the hardest position in hockey?
Goalie: Perhaps the toughest position in all of sports, the goalie is the one player who can control a team’s confidence.
What is bully in hockey?
Bully: Used to restart play when possession is unclear when play was stopped (e.g. injury timeout). Two opposing players start with their sticks on the ground, the ball is placed between them, and they must tap sticks above the ball before they can play the ball.
What are the 5 rules of hockey?
Playing the puck with a high-stick A goal scored by a puck that made contact with a stick that was above the height of the goal crossbar shall be disallowed.
You might be interested: Often asked: Why Do Hockey Players Use Smelling Salts?
What is the ball in hockey called?
The ball. Ice hockey uses a puck. Do not refer to the ball as a puck because it will point you out as a secret ice hockey fan and expose you as a field hockey newbie. The stick. This is what you call that giant stick you carry around.
What is the maximum number of players in hockey team?
How many players are on an NHL team? For the majority of the regular season each NHL team must carry a minimum of 20 players and cannot exceed a maximum of 23 players.
How many people are on a side in hockey?
A field hockey match consists of two halves, usually 35 minutes each, and begins with a pass back (a non-defended pass from one teammate to another at mid-field). There are 11 players to a side, one of whom is a goalkeeper. The object of the game is to score more goals than the opposition. | 816 | 3,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.979757 |
https://graphicmaths.com/gcse/geometry/line-symmetry/ | 1,725,850,148,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00287.warc.gz | 252,666,809 | 7,154 | # Line symmetry
By Martin McBride, 2022-11-28
Tags: symmetry line symmetry reflective symmetry reflection
Categories: gcse geometry
We say that a shape is symmetrical if we can draw a line through the shape that divides it into two parts that are mirror images of each other. We call this type of symmetry line symmetry, or reflective symmetry.
We use the example of triangles to explore this further and then look at some other shapes.
Here is a video on the same topic:
## Line symmetry of triangles
An isosceles triangle has two equal sides and two equal angles. This means that it is symmetrical:
The dotted, vertical line divides the triangle into two halves, and each half is a mirror image of the other. If we were to fold the shape along the line, the two halves would match exactly.
The line goes from the top vertex of the triangle down to the centre of the base.
The dotted line is called a line of symmetry. There is only one way to fold an isosceles triangle, so we say it has one line of symmetry.
Here is a scalene triangle:
In a scalene triangle, all the sides and angles are different, so it is not possible to draw a line that divides the shape into two symmetrical halves. We say that the shape has no lines of symmetry.
Let's look at an equilateral triangle:
In an equilateral triangle, all three sides and angles are equal. We can draw a line of symmetry from the top vertex to the centre base, similar to the isosceles triangle.
We can also draw lines of symmetry from the other two vertices to the opposite side.
An equilateral triangle, therefore, has 3 lines of symmetry. The triangle can be folded along any of the lines of symmetry and the two halves will match.
We normally draw all the lines of symmetry on a single diagram, like this:
## Line symmetry of a square
Here is a square:
A square has a horizontal line of symmetry and a vertical line of symmetry, as shown on the left-hand square. It can be folded along either like to create a rectangle.
The two diagonals of a square are also lines of symmetry as shown on the right-hand square. It can be folded along either like to create a triangle.
This means that a square has four lines of symmetry. Here they are, all shown on the same diagram:
## Line symmetry of a rectangle
A rectangle has two lines of symmetry:
Unlike a square, a rectangle doesn't have diagonal lines of symmetry, as this diagram shows:
The diagonal divides a rectangle into two identical triangles, but they are not mirror images of each other. When the rectangle is folded over the diagonal line, the two parts do not match up, so the diagonal is not a line of symmetry.
## Line symmetry of a trapezium
In general, the two legs of a trapezium do not have the same length or angle. This means that a trapezium has no lines of symmetry:
The exception is an isosceles trapezium. This is a trapezium where the two legs have equal length and have the same angle at the base (but both point inwards). In that case, there is one line of symmetry:
## Line symmetry of a parallelogram
A parallelogram has no lines of symmetry.
## Line symmetry of a rhombus
A rhombus has two lines of symmetry, its diagonals:
## Line symmetry of a kite
A kite has one line of symmetry:
## Line symmetry of a regular polygon
A regular polygon with n sides has n lines of symmetry. However, the way the lines of symmetry are arranged depends on whether the number of sides is odd or even.
We have already seen a regular polygon with an odd number of sides. It is the equilateral triangle, which we looked at earlier. We saw that it has three lines of symmetry. Each line is drawn from a vertex to the middle of the opposite side.
Other regular polygons with odd numbers of sides follow a similar pattern:
For example, the regular pentagon in the middle has 5 vertices and 5 sides. The lines of symmetry are drawn from each vertex to the midpoint of the opposite side.
Polygons with an even number of sides work differently. Again we have already seen an example, the square (a regular quadrilateral).
In the case of a square, there is a line of symmetry between the midpoint of the top and bottom sides. There is a line of symmetry between the midpoint of the left and right sides. The two diagonals are also lines of symmetry, making 4 in total.
Regular polygons with an even number of sides behave in a similar way to squares:
In the case of a hexagon, for example, there are 3 pairs of opposite sides, with a line of symmetry that passes through the centres of each pair. There are also 3 pairs of opposite vertices, with a line of symmetry that passes through each pair.
This means that a hexagon has a total of six lines of symmetry, as you would expect for a six-sided regular polygon. | 1,051 | 4,764 | {"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} | 4.8125 | 5 | CC-MAIN-2024-38 | latest | en | 0.9403 |
http://sunglee.us/mathphysarchive/?cat=93 | 1,566,394,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00534.warc.gz | 176,472,182 | 16,069 | # Why Can’t Speeds Exceed $c$?
This is a guest post by Dr. Lawrence R. Mead. Dr. Mead is a Professor Emiritus of Physics at the University of Southern Mississippi.
It is often stated in elementary books and repeated by many that the reason that an object with mass cannot achieve or exceed the speed of light in vacuum is the “mass becomes infinite”, or “time stops” or even “the object has zero size”. There are correct viewpoints for why matter or energy or any signal of any kind cannot exceed light speed and these reasons have little to do directly with mass changes or time dilations or Lorentz contractions.
Reason Number One
Consider a signal of any kind (mass, energy or just information) which travels at speed $u=\alpha c$ beginning at point $x_A$ at time $t=0$ and arriving at position $x_B$ at later time $t>0$. From elementary kinematics,
$$t=(x_B -x_A)/u = {\Delta x \over \alpha c}.$$
Now suppose the signal travels at a speed exceeding $c$, that is $\alpha > 1$. Let us calculate the elapsed time as measured by a frame going by at speed $V<c$. According to the Lorentz transformation,
$$\label{eqno1}t’ = \gamma (t-{Vx \over c^2}),$$ where $\gamma=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}$.
There are two events: the signal leaves $x=x_A$ at $t=0$, and the signal arrives at $x=x_B$ at time $t=\Delta t$. According to \eqref{eqno1}, these events in the moving frame happen at times,
$$t’_A=\gamma ( 0 -Vx_A/c^2)$$ and
$$t’_B=\gamma (\Delta t – Vx_B/c^2).$$
Thus, the interval between events in the moving frame is, \begin{aligned}\Delta t’ &= t’_B-t’_A\\
&=\gamma \Delta t -\gamma \frac{V}{c^2}(x_B-x_A)\\
&=\gamma \Delta t ( 1-\alpha V/ c).\end{aligned}\label{eqno2}
Now suppose that $\alpha V/c > 1$, which implies that,
$$c > V > c/\alpha .$$ Then for moving frames within that range of speeds it follows from \eqref{eqno2} that,
$$\Delta t’= t’_B-t’_A <0,$$ meaning physically that the signal arrived before it was sent! This is a logical paradox which is impossible on physical grounds; no one will argue that in any frame a person can be shot before the gun is fired, or you obtain the knowledge of the outcome of a horse race before the race has begun.
Well now what if the two events at $x_A$ and $x_B$ are not causally connected but one (say at $x_B$ for definiteness) simply happened after the other? How does the above argument change? How does the above math “know” that there is or is not a causal connection between the events? Everything goes the same up to the second line of equation \eqref{eqno2}:
$$\label{eqno3} \Delta t’ = \gamma \{ \Delta t – V(x_B-x_A)/c^2 \}.$$
Can there be a moving frame of speed $V<c$ for which the event at $x_B$ (the later one in S) happens before the event at $x_A$ (the earlier one in S)? If so, $\Delta t’ < 0$; from \eqref{eqno3} then we find,
$$\Delta t – V(x_B-x_A)/c^2 < 0,$$ or solving for $V$,
$$c > V > c{c\over \Delta x/ \Delta t}.$$ In order for $V$ to be less
than $c$, it must therefore be that, ${c\over \Delta x /\Delta t} < 1$, or
$${\Delta x \over \Delta t} > c.$$ This is possible for sufficiently large $\Delta x$ and/or sufficiently small $\Delta t$ because the ratio ${\Delta x \over \Delta t}$ is not the velocity of any signal, though it has the units of speed.
What is the Speed of “Light” Anyway?
Note that the Lorentz transformation contains the speed $c$ in it. What is this speed? Without referencing Maxwell’s equations of Electromagnetism, one does not know that $c$ is in fact the speed of light itself. But the above analysis shows – without reference to Maxwell – that the speed $c$ cannot be exceeded. And what is the speed talked about in the previous discussion? Well, it is the maximum speed at which one event can influence another with given (fixed) separation – thus, $c$ above isn’t really the speed of light at all; rather it is the speed of causality!
Reason Number Two
Imagine, for example, a constant force $F$ acting on a particle of (rest) mass $m$. Newton’s second Law in its relativistic form gives,
\begin{aligned} F &= {dp \over dt} \\
&= {d \over dt} \, mv\gamma \\
&= m \gamma^3 \, \dot v, \end{aligned}\label{eqno4}
where we have assumed straight line motion. This is an autonomous differential equation whose solution, assuming the object is initially at rest, is,
$$v(t)=at/(1+a^2t^2/c^2)^{1/2},$$
where $a=F/m$. It is clear that as $t \to \infty$, $v(t)$
approaches $c$ and not infinity. Moreover, the differential impulse at arbitrary time $t$ on the particle can be found from taking the derivative of $v(t)$ given in the last equation,
$$\label{eqno5}m\, dv = { F \, dt \over (1+a^2t^2/c^2)^{3/2}}.$$
From equation \eqref{eqno5}, it is clear that the incremental speed increase $dv$ over time $dt$ approaches zero as $t \to \infty$. Thus, from this point of view we see that while the force still does work, the increase in speed for a given interval of time and incremental amount of work, is less and less as time goes on which is why the speed never reaches $c$ over any finite time interval.
Reason Number Three
In the interval of time $dt$ as measured in some inertial frame observing a moving body, the clock attached to the body ticks off proper time
$$\label{eqno6}d\tau = \sqrt{1-v^2/c^2}\, dt.$$
However, for light $v\equiv c$, and therefore $d\tau\equiv 0$. Light takes no proper time to go between two points however distantly separated in space. Thus, no object could travel faster than taking no time. This is the oft-repeated mantra of textbooks, and, while the mathematics verifies it, there are far more fundamental reasons, the best being causality as outlined above.
# Is FTL (Faster-Than-Light) Possible?
Often you hear that Einstein’s relativity prevents FTL (Faster-Than-Light). Is that true? The answer is yes and no. It is not possible for a spaceship to travel faster than the speed of light. But there may be a particle that travels FTL and the existence of such a particle would not violate the principles of relativity if its speed already exceeds the speed of light when it is created. The hypothetical particle that travels FTL is called a tachyon. (tachys means fast in Greek) The name was coined by a Columbia University physicist Gerald Feinberg in 1967. When he was asked why he thought about such a particle Feinberg reportedly quoted a Jewish proverb “Everything which is not forbidden is allowed.” (Author’s note: This is from something I read more than 3 decades ago when I was a high school student so I cannot cite its source. Also I could not find any such Jewish proverb either. It is however a constitutional principle of English law.)
For a Tachyon, the Lorentz transformation is given by the complex coordinates \begin{align*}t’&=-i\frac{t-\frac{v}{c^2}x}{\sqrt{\frac{v^2}{c^2}-1}}\\x’&=-i\frac{x-vt}{\sqrt{\frac{v^2}{c^2}-1}}\end{align*} where $i=\sqrt{-1}$. Although this is a complex transformation, it is still an isometry i.e. it preserves the Minkowski metric. In order for its energy $E$ to be real one has to assume that its rest mass is purely imaginary $im_0$ where $m_0>0$ is real and hence from the relativistic energy $$E=\frac{im_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{im_0c^2}{i\sqrt{\frac{v^2}{c^2}-1}}=\frac{m_0c^2}{\sqrt{\frac{v^2}{c^2}-1}}$$ Imaginary rest mass may sound weird but rest mass is not an observable because a particle is never at rest. What’s important is energy being real as it is an observable. The following figure shows energies of a subluminal particle (in red) and a superluminal particle (in blue) with $m_0=c=1$).
Properties of Tachyons:
1. The speed of light $c$ is the greatest lower bound of Tachyon’s speed. There is no upper limit of Tachyon’s speed.
2. Tachyons have imaginary rest mass (as we discussed above).
3. In order for a tachyon to slow down to the speed of light, it requires infinite amount of energy and momentum.
4. Another peculiar nature of tachyons. If a tachyon looses energy, its speed increases. At $E=0$, $v=\infty$.
How do we detect tachyons if they exist? Since tachyons travel FTL, we can’t see them coming. However if they are charged particles, they will emit electromagnetic Mach shock waves called Tscherenkov (also spelled Cherenkov) radiations. This always happens when charged particles are passing through a medium with a higher speed than the phase speed of light in the medium. By detecting such Tscherenkov radiations we may be able to confirm the existence of tachyons.
An interesting question is “can we use tachyons for FTL communications? ” It was answered by Richard C. Tolman as negative in his book [2] (pp 54-55). In [2]. Tolman considered the following thought experiment. Suppose a signal is being sent from a point $A$ (cause) to another point $B$ (effect) with speed $u$. In an inertial frame $S$ where $A$ and $B$ are at rest, the time of arrival at $B$ is given by $$\Delta t=t_B-t_A=\frac{B-A}{u}$$ In another inertial frame $S’$ moving with speed $v$ relative to $S$ the time of arrival at $B$ is given, according to the Lorentz transformation, by \begin{align*}\Delta t’&=t_{B’}-t_{A’}\\&=\frac{t_B-\frac{v}{c^2}x_B}{\sqrt{1-\frac{v^2}{c^2}}}-\frac{t_A-\frac{v}{c^2}x_A}{\sqrt{1-\frac{v^2}{c^2}}}\\&=\frac{1-u\frac{v}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}\Delta t\end{align*} If $u>c$ then certain values of $v$ can make $\Delta t’$ negative. In other words, the effect occurs before the cause in this frame, the violation of causality!
Food for Thought: Can one possibly use tachyons to send a message (signal) to the past?
References:
1. Walter Greiner, Classical Mechanics, Point Particles and Relativity, Springer, 2004
2. Richard C. Tolman, The Theory of the Relativity of Motion, University of California Press, 1917. A scanned copy is available for viewing online here.
# Time Dilation and Time Travel
In this note, we discuss one of the relativistic effects called Time Dilation namely a clock that is moving relative to an observer will be measured to tick slower than a clock that is at rest in the observer’s reference frame. This is pretty intriguing for those who are familiar with Newtonian notion of time as being a universal parameter for motions. Let us do a thought experiment. Let us consider a frame $K$ at rest and suppose that a light ray is emitted by the light source $Q$ and after reflection by the mirror $S$ is received at $E$. See Figure 1.
Figure 1. Time Dilation
The measured time interval in the frame $K$ is $\Delta t=t_2-t_1=\frac{2l}{c}$. Now consider a frame $K’$ moving at a constant speed $v$ to the right. An observer at rest in $K’$ sees the light ray emerging from $Q$, hitting the mirror (at rest in $K$) at $M$ and reaching the $x’$-axis again at $E$. The observer measures a longer time interval as the light has to travel a longer path to reach the receiver but the speed of light is remained the same according to Einstein’s postulate. How much longer? The time $\Delta t’$ measured by an observer at rest in the frame $K’$ can be easily calculated using the Pythagorean law applied to the isosceles triangle seen in Figure 1. We find
$$\left(\frac{c\Delta t’}{2}\right)^2=l^2+\left(\frac{v\Delta t’}{2}\right)$$
Solving this for $\Delta t’$ we find
\label{eq:timedilation}
\Delta t’=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}
Note that \eqref{eq:timedilation} amounts to the Lorentz transformation into the system $K’$
$$\Delta t’=t_2′-t_1′,$$
where
$$t_i’=\frac{t_i-\frac{v}{c^2}x_i}{\sqrt{1-\frac{v^2}{c^2}}},\ i=1,2$$
Since $x_1=x_2$, we obtain \eqref{eq:timedilation}. In case this whole frame thing is confusing, let us imagine that you are sitting in a train that is running at a constant speed. Since there is no acceleration, you do not feel that you are moving. So inside the train you are at rest (frame $K$). For an observer outside you are moving (frame $K’$) and the observer would measure the time ($\Delta t’$) on your clock ticking slower than what you would measure it ($\Delta t$). In physics $\Delta t$ is called proper time. Simply speaking proper time is the time measured by a clock that is moving along with inertial frame. Mathematically, proper time can be calculated from the arc length $ds^2$ of a worldline, the trajectory of a moving particle or an object in spacetime. Denote by $\tau$ the proper time. Then since the worldline is timelike (meaning leaning more toward time), $ds^2=-c^2d\tau^2$. So the proper time interval is given by
\begin{aligned}
\Delta\tau&=\frac{1}{c}\int\sqrt{-ds^2}\\
&=\frac{1}{c}\int\sqrt{c^2dt^2-dx^2-dy^2-dz^2}\\
&=\int\sqrt{1-\frac{v(t)^2}{c^2}}dt
\end{aligned}\label{eq:propertime}
If $v(t)$ is constant speed $v$, \eqref{eq:propertime} becomes \eqref{eq:timedilation}.
The time dilation effect in \eqref{eq:timedilation} hints us that a time travel to the future may be possible. Here is how. The exoplanet Proxima b is interesting because it is orbiting within the habitable zone of the red dwarf star Proxima Centauri which is a part of triple star system Alpha Centauri in the Constellation of Centaurus, and also because it is relatively close to our world. It is located about 4.2 light-years or 40 trillion km from Earth. In fact, it is the closest known exoplanet to the Solar System.
Artist’s depiction of Proxima b
Let us say we are sending a manned spaceship to Proxima b. Also let us assume that the spaceship can travel at 90% of the speed of light. (It is actually impossible to achieve this due to a physical limitation. I will discuss this in my other note at a later time. In reality, the best we can achieve using nuclear propulsion is about 0.067% of the speed of light.) For people on Earth it would take $\Delta t’=\frac{4\times 10^{13}\mbox{km}}{2.7\times 10^5\mbox{km/sec}}=1.\overline{481}\times 10^8\mbox{sec}$ for the spaceship to get to Proxima b. Since $1\mbox{sec}=3.17\times 10^{-8}\mbox{years}$, it is 4.7 years. Since it would take the same time from Proxima b to Earth, the overall travel time for people on Earth is 9.4 years. In reality, we will have to take some factors into consideration: it takes time for the spaceship to accelerate to reach 90% of the speed of light, once the spaceship is near Proxima b it will have to slow down for stopping or U-turning, etc. But for the sake of simplicity we will disregard those factors. For the crew memebers it took only
\begin{align*}
\Delta t&=\sqrt{1-\frac{v^2}{c^2}}\Delta t’\\
&=\sqrt{1-(0.9)^2}\cdot 1.\overline{481}\times 10^8\mbox{sec}\\
&\approx 0.65\times 10^8\mbox{sec}\\
&\approx 2\mbox{yrs}
\end{align*}
to get to Proxima b. So when they come back home, it’s like they traveled more then 5 years forward in time. I know it is not what you probably think and yes I admit that this is a kind of boring time travel. Can one travel backward in time? This is one of the most intriguing questions. I will come back to this question at another time.
I will finish this note with an example as an application of \eqref{eq:timedilation}. This example was taken from [1].
Example. Muon Decay
The Earth is surrounded by an atmosphere of about 30 km thickness screening us off from cosmic radiation. If a proton from the consmic radiation hits the atmosphere, $\pi$-mesons are produced and several of them decay further into a muon and a neutrino. The muon has a mean lifetime of $\Delta t=2\times 10^{-6}\mbox{sec}$ in its rest system. Classically it would travel even with the speed of light (only massless particles can travel at the speed of light)
\begin{align*}
s&=c\Delta t\\
&=3\times 10^5\mbox{km/sec}\cdot 2\times 10^{-6}\mbox{sec}\\
&=0.6\mbox{km}
\end{align*}
or 600m. If this were true, muon particles would never reach the surface, but they are detected on the surface. In the relativistic approach,
$$s’=v\Delta t’=\frac{v\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$
Muons at rest have a mass of $m_0c^2=10^8$eV (I know it is actually energy but due to mass-energy equivalence physicists customarily call it mass.) The cosmic muons are created at an altitude of about 10km with a total energy of $E=5\times 10^9$eV. In order to apply this information we rewite $S’$ as
\begin{align*}
S’&=\frac{vm_0c^2}{m_0c^2\sqrt{1-\frac{v^2}{c^2}}}\Delta t\\
&=\frac{v}{m_0c^2}E\Delta t\\
&\leq\frac{c}{m_0c^2}E\Delta t\\
&=\frac{3\times 10^5\mbox{km/sec}}{10^8\mbox{eV}}\cdot 5\times 10^9\mbox{eV}\cdot 2\times 10^{-6}\mbox{sec}\\
&=30\mbox{km}
\end{align*}
Here we used $E=mc^2=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$. We will discuss this later in another post. The actual measurement gives a value of 38km.
References:
[1] Walter Greiner, Classical Mechanics, Point Particles and Relativity, Springer, 2004
[2] Paul A. Tipler and Ralph A. Llewellyn, Modern Physics, 5th Edition, W. H. Freeman and Company, 2008
# Lorentz Invariance of Relativistic Equations
Relativistic equations are the equations whose solutions describe certain relativistic motions. Such equations include wave equation, Klein-Gordon equation, Dirac equation etc. A relativistic equation must describe the same physical motion independent of frames i.e. whether an observer is in a frame at rest or in a frame moving at the constant speed $v$. For this reason, all those relativistic equations are required to be invariant under the Lorentz transformation. We show that the wave equation
\label{eq:waveeq}
-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=0
is Lorentz invariant. Here we consider only 1-dimensional wave equation for simplicity. Wave equation has two kinds of solutions. Given boundary conditions its solution describes a vibrating string in which case the boundary conditions are the ends of the string that are held fixed. This is called Fourier’s solution. The other type can be obtained by not imposing any boundary conditions. The resulting solution would describe a propagating wave in vacuum spacetime. Such a propagating wave includes electromagnetic waves. Light is also an electromagnetic wave. This is called a d’Alembert’s solution. The proof is easy. All that’s required is the chain rule.
First let us recall the Lorenz transformation
$$t’=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}},\ x’=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$
Using the chain rule we find
\begin{align*}
\frac{\partial}{\partial x}&=\frac{\partial}{\partial x’}\frac{\partial x’}{\partial x}+\frac{\partial}{\partial t’}\frac{\partial t’}{\partial x}\\
&=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial x’}-\frac{v}{c^2\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial t’}\\
\frac{\partial}{\partial t}&=\frac{\partial}{\partial x’}\frac{\partial x’}{\partial t}+\frac{\partial}{\partial t’}\frac{\partial t’}{\partial t}\\
&=-\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial x’}+\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{\partial}{\partial t’}
\end{align*}
Applying the chain rule again,
\begin{align*}
\frac{\partial^2}{\partial x^2}&=\frac{1}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {x’}^2}+\frac{v^2}{c^4\left(1-\frac{v^2}{c^2}\right)}\frac{\partial^2}{\partial {t’}^2}-2\frac{v}{c^2\left(1-\frac{v^2}{c^2}\right)}\frac{\partial^2}{\partial t’\partial x’}\\
\frac{\partial^2}{\partial t^2}&=\frac{v^2}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {x’}^2}+\frac{1}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial {t’}^2}-2\frac{v}{1-\frac{v^2}{c^2}}\frac{\partial^2}{\partial t’\partial x’}
\end{align*}
It follows that
$$-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=-\frac{1}{c^2}\frac{\partial^2\psi}{\partial {t’}^2}+\frac{\partial^2\psi}{\partial {x’}^2}$$
Therefore, the wave equation is Lorentz invariant.
Would the wave equation be invariant under the Galilean transformation? The answer is no. Recall the Galilean transformation
$$t’=t,\ x’=x-vt$$
We find that under the Galiean transformation
$$-\frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}+\frac{\partial^2\psi}{\partial x^2}=-\frac{1}{c^2}\frac{\partial^2\psi}{\partial {t’}^2}+\left(1-\frac{v^2}{c^2}\right)\frac{\partial^2\psi}{\partial {x’}^2}+\frac{2v}{c^2}\frac{\partial^2\psi}{\partial t’\partial x’}$$
Hence obvisouly the wave equation is not invariant under the Galiean transformation. This implies that there is no light in Euclidean space.
Food for Thought. You can also show that the heat equation (1-dimensional)
$$-\frac{\partial u}{\partial t}+\alpha\frac{\partial^2 u}{\partial x^2}=0$$
is not Lorentz invariant. Is there any relativistic version of the heat equation? There are models of relativistic heat conduction but in my opinion they are more like mathematically augmented equations rather than they are derived in a physically meaningful way. So my question is can we derive a physically meaningful equation of relativistic heat conduction? One may wonder if there is actually any physical phenomenon that exhibits a relativistic heat conduction. As far as I know there isn’t any observed one yet. I speculate though that one may observe a relativistic heat conduction from an extreme physical phenomenon such as a quasar jet.
Update: Of course the Lorentz invariance can be also shown using the Lorentz transformation \begin{align*}t’&=\cosh\phi t-\sinh\phi x\\x’&=-\sinh\phi t+\cosh\phi x\end{align*}
Update: For 3-dimensional case the wave equation is given by $$\Box\psi=0$$
where $\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$ is the d’Alembert’s operator. This case is actually simpler to show its Lorentz invariance. Note $\Box=\nabla\cdot \nabla$ where $\nabla=\frac{1}{c}\frac{\partial}{\partial t}\hat e_0+\frac{\partial}{\partial x}\hat e_x+\frac{\partial}{\partial y}\hat e_2+\frac{\partial}{\partial z}\hat e_3$. Since $\nabla$ is a 4-vector (rigorously it is not a vector but an operator but can be treated as a vector), its squared norm $\Box$ has to be Lorentz invariant.
# Lorentz Transformation 3
Let $\begin{pmatrix} t\\ x \end{pmatrix}$ be a vector in $tx$-plane and $\begin{pmatrix} t’\\ x’ \end{pmatrix}$ denote its rotation by a hyperbolic angle $\phi$. Then as we have seen here we have:
\begin{aligned}
t’&=t\cosh\phi-x\sinh\phi\\
x’&=-t\sinh\phi+x\cosh\phi
\end{aligned}\label{eq:boost1}
$t’$-axis ($x’=0$) is moving at a constant speed
\label{eq:velocity1}
v=\frac{x}{t}=\frac{\sinh\phi}{\cosh\phi}=\tanh\phi
while $x’$-axis ($t’=0$) is moving at a constant speed
\label{eq:velocity2}
v=\frac{x}{t}=\frac{\cosh\phi}{\sinh\phi}=\coth\phi
This means that the time and spacial axes scissor together and hence the speed of light remains the same regardless of the coordinate transformation as illustrated in Figure 1. The picture in Figure 1 is called the spacetime diagram.
Figure 1. Spacetime Diagram
From Euclidean perspective, $t’$ and $x’$ do not appear to be orthogonal. However, from Lorentzian perspective they are orthogonal. To see that let $e_0=\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $e_1=\begin{pmatrix} 0\\ 1 \end{pmatrix}$. Also let $e_0’$ and $e_1’$ be the rotations of $e_0$ and $e_1$ by the hyperbolic angle $\phi$, respectively. Then by \eqref{eq:boost1} we have
$$e_0’=\begin{pmatrix} \cosh\phi\\ -\sinh\phi \end{pmatrix},\ e_1’=\begin{pmatrix} -\sinh\phi\\ \cosh\phi \end{pmatrix}$$
Now \begin{align*}\langle {e_0}’,e_1’\rangle&=(\cosh\phi\ -\sinh\phi)\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
-\sinh\phi\\
\cosh\phi
\end{pmatrix}\\&=\cosh\phi\sinh\phi-\sinh\phi\cosh\phi=0\end{align*} So $e_0’$ and $e_1’$ are orthogonal.
\eqref{eq:boost1} with \eqref{eq:velocity1} can be written as
\begin{align*}
t’&=\cosh\phi(t-vx)\\
x’&=\cosh\phi(x-vt)
\end{align*}
Using the well-known identy $\cosh^2\phi-\sinh^2\phi=1$, we find $\cosh\phi=\frac{1}{\sqrt{1-v^2}}$. Therefore, \eqref{eq:boost1} can be written in terms of $t,x,t’,x’,v$ as
\begin{aligned}
t’&=\frac{t-vx}{\sqrt{1-v^2}}\\
x’&=\frac{x-vt}{\sqrt{1-v^2}}
\end{aligned}\label{eq:boost2}
Remember that we assumed the speed of light to be $c=1$ for simplicity. For the real $c$ value \eqref{eq:boost2} turns into
\begin{aligned}
t’&=\frac{t-\frac{v}{c^2}x}{\sqrt{1-\frac{v^2}{c^2}}}\\
x’&=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}
\end{aligned}\label{eq:boost3}
In physics textbooks, \eqref{eq:boost3} is introduced as the Lorentz transformation. If $v\ll c$ meaning $v$ is significantly slower than the speed of light (such a motion is called a non-relativistic motion), \eqref{eq:boost3} is effectively the Galilean transformation
\begin{aligned}
t’&=t\\
x’&=x-vt
\end{aligned}\label{eq:galilean}
for Newtonian mechanics in Euclidean space. $t’=t$ means that time is independent of the relative motion of different observers and we already know this is the case of Newtonian mechanics. The Galilean transformation \eqref{eq:galilean} can be also obtained by taking the limit $c\to \infty$. This indicates that in Newtonian mechanics the speed of light is presumed to be infinity. | 7,637 | 24,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 4, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.894857 |
https://data-flair.training/blogs/anova-in-r/ | 1,723,562,455,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00691.warc.gz | 150,272,865 | 53,650 | # ANOVA in R – A tutorial that will help you master its Ways of Implementation
Placement-ready Courses: Enroll Now, Thank us Later!
In today’s era, more and more programmers are aspiring to become a Data Scientist. And, you must be aware that R programming is an essential ingredient for mastering Data Science. So, let’s jump to one of the most important topics of R; ANOVA model in R.
In this tutorial, we will understand the complete model of ANOVA in R. Also, we will discuss the One-way and Two-way ANOVA in R along with its syntax. After this, learn about the ANOVA table and Classical ANOVA in R.
Let’s start the tutorial.
### What is ANOVA?
The ANOVA model which stands for Analysis of Variance is used to measure the statistical difference between the means. With the ANOVA model, we assess if the various groups share a common mean. As a result, we have found that it’s used for investigating data by comparing the means of subsets of data.
Anova.glm
Generally, it’s an analysis of Deviance for Generalized Linear Model Fits.
As a result, it’s needed to compute an analysis of deviance table for one or more generalized linear model fits.
Learn everything about the Generalized Linear Models in R
Keywords:
models, regression
Usage:
```# S3 method for glm
anova(object, …, dispersion = NULL, test = NULL)```
Arguments:
1. the object, …
Basically, it’s the result of a call to glm or a list of objects for the “almost” method.
2. dispersion
Dispersion is defined as the parameter for the fitting family.
3. test
It’s a character string. As a result, it should match one of “Chisq”, “LRT”, “Rao”, “F” or “Cp”.
## Implementing ANOVA in R
There are two ways of implementing ANOVA in R:
• One-way ANOVA
• Two-way ANOVA
### One-way ANOVA in R
Let’s take an example of using insect sprays which is a type of data set. We are going to test 6 different insect sprays. As a result, we need to see if there was a difference in the number of insects found in the field after each spraying.
```> attach(InsectSprays)
> data(InsectSprays)
> str(InsectSprays)```
Output:
Do you know about Data Frame Operations in R
#### 1. Descriptive Statistics
1. With the help of descriptive statistics, we calculate the mean, variance and number of elements in each cell.
2. Visualise the data – boxplot; look at the distribution for outliers.
We will be using tapply() here:
tapply() function is a very useful shortcut in processing data. Also, we use it as a function. It should be applied to each subset of the response variable defined by each level of the factor.
#### 2. Run 1-way ANOVA
2.1 Oneway.test ( )
Use, for example:
`> oneway.test(count~spray)`
Output:
Default is equal variances not assumed that is Welch’s correction applied and this explains why the denom df (which is k*{n-1}) is not a whole number in the output O.
In order to alter this, we set the “var.equal =” option to TRUE.
Oneway.test( ) corrects the non-homogeneity but doesn’t give much information.
2.2 Run an ANOVA using aov( )
We use the aov() function to store the output and use extraction functions to obtain what is required.
```> #Author DataFlair
> AOV_Output <- aov(count ~ spray, data=InsectSprays)
> summary(AOV_Output)```
Output:
Have you checked – Survival Analysis in R
### 2. Two-way ANOVA in R
Two-way Analysis of Variance
We use it to compare the means of populations which is classified in two different ways. Besides, we can use lm() to fit two-way ANOVA models in R.
For example, the command:
`> lm(Response ~ FactorA + FactorB)`
fits a two- way ANOVA model without interactions. In contrast, the command:
`> lm(Response ~ FactorA + FactorB + FactorA * FactorB )`
includes an interaction term. Here, both FactorA and FactorB are categorical variables, while Response is quantitative.
Understand the complete concept of Factor Analysis in R
### Classical ANOVA in R
Generally, we start with a simple additive fixed effects model. In this model, we use the built-in function aov:
`aov(Y ~ A + B, data=d)`
Now, to cross these factors or more for interacting with two variables, we use either of:
```aov(Y ~ A * B, data=d)
aov(Y ~ A + B + A:B, data=d)```
So far so good. Furthermore, we make an assumption that B is nested within A:
```aov(Y ~ A/B, data=d)
aov(Y ~ A + B %in% A, data=d)
aov(Y ~ A + A:B, data=d)```
Therefore, in nesting, we add both – the main effect and the interaction.
You must definitely explore the R Graphical Models tutorial
#### Random Effects in Classical ANOVA
aov can also deal with random effects that provides everything which is being balanced. Assume A is alone random effect, e.g. a subject indicator.
`aov(Y ~ Error(A), data=d)`
We make an assumption that A is random, B is fixed as well as nested within A.Â
`aov(Y ~ B + Error(A/B), data=d)`
or B and X are crossed (interacted) within levels of random A.
`aov(Y ~ (B*X) + Error(A/(B*X)), data=d)`
Or B and X within random A are categorized by (non-nested) G and H:
`aov(Y ~ (B*X*G*H) + Error(A/(B*X)) + (G*H), data=d)`
As a result, this Error business can become confusing and the balance requirements, a bit tiresome. Thus, for random effects models, it’s usually easier to move to lme4.
Must Learn – Generalized Linear Models in R
### ANOVA Table in R
Let’s say, we have collected data, and our X values have been entered in R as an array called data.X and Y values as data.Y.
Now, we will find the ANOVA values for the data. Then, follow the below steps:
• First, we will fit our data into a model. > data.lm = lm(data.Y~data.X).
• Next, we will get R to produce an ANOVA table by typing : > anova(data.lm).
• As a result, we will have an ANOVA table!
1. Fitted Values
Type:
`> data.fit = fitted(data.lm)`
to get the fitted values of the model.
As a result, it gives us an array called “data.fit” that contains the fitted values of data.lm.
2. Residuals
We use this to get the residuals of the model.
`> data.res = resid(data.lm)`
Now, as a result, we have an array of residuals.
3. Hypothesis Testing
• In case if we have already found the ANOVA table for our data, then we are able to calculate our test statistic from the numbers given.
• If we want to find the F – quantile given by F(.95;3,24)
Find this by typing:
```> #Author DataFlair
> qf(.97, 5, 23)```
• If we want to find the t – quantile given by t(0.975, 1.2, 20)
Type:
`> qt(0.975, 1.2, 20) #Calculating the t-quantile`
Output:
Don’t forget to check the Creation of Contingency Tables in R
4. P – values
In case if we want to get the p-value for the F – quantile of, say, 2.84, with degrees of freedom 3 and 24, we would type in
`> pf(3.75, 2, 26) Â Â Â Â Â #Calculating p-value for the f-quantile`
5. Normal Q-Q plot
Generally, we use “data.lm to get the normal probability for the standard residuals of our data.
Although, we have already fit our data to a model, but now we will need the studentized residuals:
`> data.stdres = rstandard(data.lm)`
Also, type this to make the plot:
`> qqnorm(data.stdres)`
Then, to see the line, type:
`> qqline(data.stdres)`
## Summary
We have studied ANOVA in R with their different types and properties. It is very much useful in investigating data by comparing the means of subsets of the data. Still, if you have any confusion regarding ANOVA in R, ask in the comment section.
For now, you must have a look at Chi-Square Test in R
Did you know we work 24x7 to provide you best tutorials
### 1 Response
1. Juan says:
Hi,
I wanted to know by the commands (bootstrap) below how could apply Anova to estimate the minimum diference in soybean means.
### Comparación por bootstrap Intacta vs NO en Lit_Sur y Lit_Norte
d19_20 %>% filter(Zona==”Lit_Sur” & Cultivo==”Soja_1″ & Intacta==”SI”) -> SurSI
d19_20 %>% filter(Zona==”Lit_Sur” & Cultivo==”Soja_2″ & Intacta==”SI”) -> SurSI
d19_20 %>% filter(Zona==”Lit_Sur” & Cultivo==”Soja_1″ & Intacta==”NO”) -> SurNO
d19_20 %>% filter(Zona==”Lit_Sur” & Cultivo==”Soja_2″ & Intacta==”NO”) -> SurNO
d19_20 %>% filter(Zona==”Lit_Norte” & Cultivo==”Soja_2″ & Intacta==”SI”) -> NorteSI
d19_20 %>% filter(Zona==”Lit_Norte” & Cultivo==”Soja_1″ & Intacta==”SI”) -> NorteSI
d19_20 %>% filter(Zona==”Lit_Norte” & Cultivo==”Soja_2″ & Intacta==”NO”) -> NorteNO
d19_20 %>% filter(Zona==”Lit_Norte” & Cultivo==”Soja_1″ & Intacta==”NO”) -> NorteNO
rindeProm <- function(dat) return(sum(dat\$Hás.*dat\$Rinde)/sum(dat\$Hás.))
nBoot <- 100000
rindePromSI <- rindePromNO <- rep(0,nBoot)
View(nBoot)
View(rindeProm)
for (i in 1:nBoot){
filasSI <- sample(1:nrow(SurSI), size = nrow(SurSI), replace = TRUE)
filasNO <- sample(1:nrow(SurNO), size = nrow(SurNO), replace = TRUE)
rindePromSI[i] <- rindeProm(SurSI[filasSI,])
rindePromNO[i] <- rindeProm(SurNO[filasNO,])
}
for (i in 1:nBoot){
filasSI <- sample(1:nrow(NorteSI), size = nrow(NorteSI), replace = TRUE)
filasNO <- sample(1:nrow(NorteNO), size = nrow(NorteNO), replace = TRUE)
rindePromSI[i] <- rindeProm(NorteSI[filasSI,])
rindePromNO[i] <- rindeProm(NorteNO[filasNO,])
}
View(rindePromSI)
View(rindePromNO) | 2,667 | 9,109 | {"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} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.895579 |
http://quizlet.com/19493842/cfin-3-chapter-4-flash-cards/ | 1,427,591,758,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131298020.57/warc/CC-MAIN-20150323172138-00137-ip-10-168-14-71.ec2.internal.warc.gz | 218,794,534 | 23,376 | # CFIN 3 Chapter 4
### 22 terms by coreyoconnor91
#### Study only
Flashcards Flashcards
Scatter Scatter
Scatter Scatter
## Create a new folder
Advertisement Upgrade to remove ads
### "If you invest \$500 today in an account that pays 6 percent interest compounded annually, how much will be in your account after two years?"
\n\n
Using a financial calculator, enter N = 2, I/Y= 6, and PV = -500; compute FV = 561.80"
### "What is the present value of an investment that promises to pay you \$1,000 in five years if you can earn 6 percent interest compounded annually?"
\n\n "
Using a financial calculator, enter N = 5, I/Y= 6, and FV = 1,000; compute PV = -747.26"
### Compute the present value of \$1,552.90 due in 10 years at (a) a 12 percent discount rate and (b) a 6 percent rate.
Using a financial calculator, enter N = 10, I/Y= 6, and FV = 1,552.90; compute PV = -867.13
The present value represents the amount that needs to be invested today at the opportunity cost rate to generate the future amount. In essence, we take the interest out of the future value—that is, discount—to determine the current, or present, value. For this problem, then, if \$867.13 is invested today at 6 percent compounded annually, it will grow to \$1,552.90 in 10 years.
### "To the closest year, how long will it take a \$200 investment to double if it earns 7 percent interest? How long will it take if the investment earns 18 percent?"
\n\n "(1)
PV = -200 400
Using a financial calculator, enter I/Y= 7, PV = -200, PMT = 0, and FV = 400; compute N = 10.24 ≈ 10 years
If I/Y= 18%, N = 4.19 ≈ 4 years"
### Which amount is worth more at 14 percent: \$1,000 in hand today or \$2,000 due in six years?
Using a calculator, enter N = 6, I/Y= 14, PMT = 0, and PV = -1,000; compute FV = 2,194.97
PV = 1,000(1.14)6 = 1,000(2.19497) = 2,194.97
\$1,000 today is worth more. The future value of \$1,000 at 14 percent over six years is \$2,194.97, which is greater than the future \$2,000.00.
Alternatively, using a calculator, enter N = 6, I/Y= 14, PMT = 0, and FV = 2,000; compute PV = -911.17
PV = 2,000(1/1.14)6 = 2,000(0.455587) = \$911.17
\$1,000 today is worth more. The present value of \$2,000 at 14 percent over six years is \$911.17, which is less than \$1,000.00."
### "Martell Corporation's sales were \$12 million this year. Sales were \$6 million five years earlier. To the nearest percentage point, at what annual rate have sales grown?"
Using a calculator, enter N = 5, PV = -6, PMT = 0, and FV = 12; compute I/Y= 14.87% ≈ 15%."
### "If you invest \$600 per year for the next 10 years, how much will your investment be worth at the end of 10 years if your opportunity cost is 10 percent? The first \$600 investment will be made at the end of this year."
Using a financial calculator, enter N = 10, I/Y= 10, and PMT = -600; compute FV = 9,562.45"
### "If you invest \$600 per year for the next 10 years, how much will your investment be worth at the end of 10 years if your opportunity cost is 10 percent? The first \$600 investment will be made today."
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute FV = 10,518.70"
### "If you want to pay yourself \$600 per year for the next 10 years, how much must you deposit today in an investment account that will pay 10 percent interest annually? The first \$600 payment will be withdrawn from the account at the end of this year."
Using a financial calculator, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 3,686.74"
### "If you want to pay yourself \$600 per year for the next 10 years, how much must you deposit today in an investment account that will pay 10 percent interest annually? The first \$600 payment will be withdrawn from the account today."
\n\n "The general formula for computing the future value of an annuity due is:
0 1 2 3 4 5 6 7 8 9 10
600 600 600 600 600 600 600 600 600 600
PVA(DUE)10 = ?
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 4,055.41
The general formula for computing the future value of an annuity due is:
0 1 2 3 4 5 6 7 8 9 10
600 600 600 600 600 600 600 600 600 600
PVA(DUE)10 = ?
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 4,055.41
The general formula for computing the future value of an annuity due is:
0 1 2 3 4 5 6 7 8 9 10
600 600 600 600 600 600 600 600 600 600
PVA(DUE)10 = ?
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 4,055.41
The general formula for computing the future value of an annuity due is:
0 1 2 3 4 5 6 7 8 9 10
600 600 600 600 600 600 600 600 600 600
PVA(DUE)10 = ?
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 4,055.41
Using a financial calculator, switch to BEGIN, enter N = 10, I/Y= 10, and PMT = -600; compute PV = 4,055.41"
### "What is the present value of a perpetuity of \$280 per year if the appropriate discount rate is 7 percent? What would happen to the present value of the perpetuity if the appropriate rate rose to 14 percent?"
\n\n "PVP = \$280/0.07 = \$4,000. PVP = \$280/0.14 = \$2,000.
When the interest rate is doubled, the PV of the perpetuity is halved."
### "Find the amount to which \$500 will grow in five years if the investment earns 12 percent compounded (a) annually, (b) semiannually, and (c) monthly."
Using a financial calculator, enter N = 5, I/Y= 12, and PV = -500; compute FV = 881.17
Using a financial calculator, enter N = 10, I/Y= 6, and PV = -500; compute FV = 895.42
Using a financial calculator, enter N = 60, I/Y= 1, and PV = -500; compute FV = 908.35"
### "While Steve Bouchard was a student at the University of Florida, he borrowed \$12,000 in student loans at an annual interest rate of 9 percent. If Steve repays \$1,500 per year, how long, to the nearest year, will it take him to repay the loans?"
\n\n "0 1 2 n-1 n Years
12,000 -1,500 -1,500 -1,500 -1,500
1%
...
...
...
...
...
12%
6%
...
...
...
...
...
r = 9%
...
Solutions
6
−=−=+09.01500,1000,12r1PMTPVAnn)09.1(1)r1(1
Using a calculator, enter I/Y= 9, PV = 12,000, FV = 0, and PMT = -1,500; compute N = 14.77 ≈ 15 years"
### "Hilda invested \$5,000 four years ago. If the investment is now worth \$7,058, what rate of return has Hilda earned on her investment? Assume that interest is compounded annually."
Using a calculator, enter N = 4, PV = -5,000, PMT = 0, and FV = 7,058; compute I/Y= 9%."
### "Jack just discovered that he holds the winning ticket for the \$87 million "mega" lottery in Missouri. Now he must decide which alternative to choose: (a) a \$44 million lump-sum payment today or (b) a payment of \$2.9 million per year for 30 years. With the second option, the first payment will be made today. If Jack's opportunity cost is 5 percent, which alternative should he choose?"
alculator: Switch to begin mode, n = 30, I/Y= 5, PMT = 2,900,000, and FV = 0; compute PV = -46,809,113.
Because PVA(DUE) = \$46,809,113, which is greater than the lump-sum payment of \$44 million, the annuity option should be chosen."
### "Your broker offers to sell you a note for \$13,250 that will pay \$2,345.05 per year for 10 years. If you buy the note, what rate of interest (to the closest percent) will you be earning?"
Using a calculator, enter N = 10, PV = -13,250, PMT = 2,345.05, and FV = 0; compute I/Y= 12%."
### "Brandi just received her credit card bill, which has an outstanding balance equal to \$3,310. The credit card carries an 18 percent simple interest rate, which is compounded monthly. If Brandi pays \$150 each month, how long will it take her to pay off the credit card bill? Assume that the only charge Brandi incurs from month to month is the interest that must be paid on the remaining outstanding balance."
Calculator solution: I/Y= 18/12 = 1.5, PV = 3,310, PMT = -150, and FV = 0; N = ? = 27.0 months, or 2.2 years."
### "Allison wants to pay off her existing automobile loan. Two years ago, Allison borrowed \$35,600 with terms that required her to make monthly payments equal to \$739 for a period of five years. The interest rate on the loan is 9 percent. To the nearest dollar, how much does Allison currently owe on her automobile loan? The most recent payment on the loan was made yesterday."
Calculator solution: N = 36, I/Y= 9/12 = 0.75, PMT = -739, and FV = 0; PV = ? = 23,239."
### "If the appropriate interest rate is 8 percent, what are the present values of the following cash flow streams? Year Cash Stream A Cash Stream B 1 \$100 \$300 2 400 400 3 400 400 4 300 100"
Using a financial calculator, simply enter the cash flows into the cash flow register (be sure to enter CF0 = 0), enter I/Y= 8, and press the NPV key to find NPV = PV = \$973.57 for Cash Stream A. Repeat for Cash Stream B to get NPV = PV = \$1,011.75 when I/Y= 8%."
### "Savannah, who is a recent college graduate, is making plans to pay back the \$46,000 in student loans that she took out during the past four years. The student loans require Savannah to pay interest equal to 5.4 percent. Payments will be made monthly, beginning today, and the loans must be repaid within 20 years. Based on this information, how much must Savannah pay each month?"
\n\n Calculator solution: Switch to begin mode; N = 240, I/Y= 5.4/12 = 0.45, PV = 46,000, and FV = 0; PMT = ? = -312.43.
### "Suppose you have been shopping for mortgages to finance the house that you want to buy. The East Coast Federal Credit Union (ECFCU) has offered a 30-year fixed mortgage that requires you to pay 6 percent interest compounded monthly. The purchase price of the house is \$260,000, and you plan to make a down payment equal to \$28,000. What would your monthly payments be with the ECFCU mortgage?"
\n\n Calculator solution: N = 360, I/Y= 6/12 = 0.5, PV = 232,000, and FV = 0; PMT = ? = -1,390.96.
### "In problem 4-22, you should have found that the monthly payment is \$1,391. Suppose it is now 10 years later, such that you have lived in the house for 10 years, and you are considering paying off your mortgage. How much do you owe on the mortgage if this month's payment was made yesterday?"
Calculator solution: N = 240, I/Y= 6/12 = 0.5, PMT = -1,391, and FV = 0; PV = ? = 194,157."
### Please allow access to your computer’s microphone to use Voice Recording.
Having trouble? Click here for help.
Example:
## Press Ctrl-0 to reset your zoom
### Please upgrade Flash or install Chrometo use Voice Recording.
For more help, see our troubleshooting page.
Create Set | 3,253 | 10,598 | {"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-2015-14 | latest | en | 0.893027 |
https://www.foodandlifelover.com/what-is-175-celsius-in-fahrenheit/ | 1,679,463,328,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943750.71/warc/CC-MAIN-20230322051607-20230322081607-00258.warc.gz | 873,960,517 | 17,778 | # What is 175 Celsius in Fahrenheit?
175 degrees Celsius is equal to 345 degrees Fahrenheit. This conversion can be useful in many different situations, such as cooking or baking. When converting from one temperature scale to another, it is important to remember the difference between the two units of measure.
In this case, 175 degrees Celsius is equivalent to 32 degrees below the boiling point of water on the Fahrenheit scale.
175 Celsius is equal to 347 degrees Fahrenheit. To convert from Celsius to Fahrenheit, multiply the number by 1.8 and add 32.
## Is 180 Degrees Celsius 350 Fahrenheit?
The answer is no. 180 degrees Celsius is 350 degrees Fahrenheit.
## Is 180 Celsius Cold Or Hot?
180 Celsius is hot. At this temperature, water boils and most materials will begin to decompose. Metal will start to melt, and humans will experience severe burns if exposed to this temperature for long periods of time.
180 Celsius is also the highest setting on many ovens, so food can become burnt or charred if left in for too long.
## Is 300 Fahrenheit Cold Or Hot?
300 degrees Fahrenheit is hot. At this temperature, water boils and most metals start to melt. 300 degrees is also the temperature at which many industrial processes take place, such as welding.
## How Hot is 37 in F?
37 in F is about average room temperature. It’s not too hot or too cold, and most people would find it comfortable. However, if you’re used to living in a warmer climate, 37 in F may feel chilly.
## 175 to Fahrenheit
If you need to convert 175 degrees Fahrenheit to Celsius, simply use this formula: take the temperature in Fahrenheit and subtract 32, then multiply by 5/9. In this case, that would give you a result of 79.4 degrees Celsius.
Can You Skateboard on the Sidewalk?
## Celsius to Fahrenheit
For many people, temperature is measured using the Celsius scale. In the United States, however, temperature is often measured in Fahrenheit. So how do you convert between the two?
The formula for converting Celsius to Fahrenheit is: Fahrenheit = (Celsius x 1.8) + 32 To use this formula, simply substitute your desired values for Celsius and Fahrenheit.
For example, let’s say you want to convert 20 degrees Celsius to Fahrenheit. Using the formula above, we get: Fahrenheit = (20 x 1.8) + 32
Fahrenheit = 36 + 32 Fahrenheit = 68 degrees Similarly, you can use the following formula to convert Fahrenheit to Celsius:
Celsius = (Fahrenheit – 32) / 1.8 Let’s say you want to convert 68 degrees Fahrenheit to Celsius. Using the formula above, we get:
Celsius = (68 – 32) / 1.8 Celsius = 36 / 1.8
## 175 Celsius to Gas Mark
If you’re like most people, you probably don’t think much about your oven’s gas mark settings. But if you’re cooking for a large group or trying to bake something delicate, understanding how different temperatures convert between Celsius and Gas Mark can be crucial. Here’s a quick guide to help you make sense of it all:
– 175 degrees Celsius is equivalent to Gas Mark 1 – 250 degrees Celsius is equivalent to Gas Mark 2 – 325 degrees Celsius is equivalent to Gas Mark 3
– 400 degrees Celsius is equivalent to Gas Mark 4
## 17 Celsius to Fahrenheit
The temperature 17 degrees Celsius is equivalent to 62.6 degrees Fahrenheit. To convert a Celsius temperature to Fahrenheit, multiply the Celsius temperature by 1.8 and add 32 degrees. Therefore, 17 multiplied by 1.8 equals 30.6.
Adding 32 to this value gives us the final answer of 62.6 degrees Fahrenheit.
## Conclusion
In order to convert 175 Celsius to Fahrenheit, you must first multiply the Celsius number by 1.8. Once you have done this, you then add 32 to the answer in order to get the final temperature in Fahrenheit. Therefore, 175 Celsius is equal to 351 degrees Fahrenheit.
What Aisle is Rice In? | 870 | 3,799 | {"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-2023-14 | latest | en | 0.872111 |
https://betterlesson.com/lesson/574184/using-number-lines-to-discover-benchmark-fractions?from=consumer_breadcrumb_dropdown_lesson | 1,542,459,759,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743521.59/warc/CC-MAIN-20181117123417-20181117145417-00456.warc.gz | 571,915,317 | 29,094 | # Using Number Lines to Discover Benchmark Fractions
24 teachers like this lesson
Print Lesson
## Objective
SWBAT make generalizations about benchmark fractions using evidence from number lines.
#### Big Idea
Benchmark Fractions help establish quantities, developing fraction number sense and judgment of the reasonableness of estimates.
## Warm-up
7 minutes
The focus of today's lesson is the use of number lines. Number lines are a valuable tool for students to use when building an understanding about fractions. To get started, I want to determine where my students are (in terms of using number lines) so I can meet them there and build on.
The knowledge I am looking for here is the understanding of the layout of a number line, the labeling of a number line, and the placement of values on a number line. The skills students need complete this task is to divide by halves to create halfway points between whole numbers, dividing halves in half to create fourths, etc.
In order to gain this understanding. I provide the students with an open ended prompt for their math journal. "Make a number line and label any fractions that you can. If you want a challenge, extend your number line to the number 5."
Students work for 5 minutes on their own.
Then, we set up at math museum so the students can walk around the room and make observations.
These observations are recorded on the board.
From this activity, I am able to see that students need some direct instruction about making a number line.
## Launch
10 minutes
After the warm -up, I quickly survey the students about their feelings toward number lines. I ask them to put themselves into one of the 3 categories. 1. Number lines are great! 2. Number lines are a tool I can use, but don't always choose too. 3. Number lines are challenging to work with.
The response was pretty balanced with about 1/3 in each of the groups. I explain to the class, that we are going to work with number lines today, because they are an awesome tool. But we have to really understand them in order appreciate them.
I use a structured resource to demonstrate the number lines are not unlike fraction tiles. I draw rectangles over the number line to show that each of the spaces represents a fraction tile. For example, the first number line has 4 spaces. I draw 4 rectangles over these spaces to represent 4ths.
I spend time explaining how a number line is created. "To make a number line that shows 4ths. You first have to identify 0 and 1, then break the whole into 4 equal parts.
I work through the rest of these examples with the whole group. As we move through these number lines, I gradually stop drawing tiles over the number lines. I call students up to label the fraction that is represented in each and ask them to explain how they knew.
## Make a Number Line: Guided Practice
15 minutes
Now that students have a better sense of how a number line can be used as a tool. I help the students work as a class to make their own number lines. I pass out a sheet of paper, cut in 1/2 (horizontally), to each of the students.
Using the following steps, we create a "stacked" number line. That shows equivalent fractions. Be sure to model this activity as you go. Students can get very confused if there is no visual to guide them. Since the goal is to help students appreciate this tool, it is very important they they are not frustrated by this activity.
1. Draw a line horizontally across the middle of your paper. This will be your number line.
2. Label 0 and 1 on this paper. 0 is at the far left and 1 and at the far right.
3. Fold your number line in half. What fraction does this line represent? (1/2). How do you know? (Because there are two equal sections on each side)
4. Label 1/2 on your number line.
5. Now fold your number line in 1/2 again. How many spaces are there between the 0 and 1 mark? (4) What fractional part is this line broken into now? (4ths). Call a student to the board to label 1/4 on the model. Ask students to do the same on theirs. Draw the number tile representation on the model number line to help students understand that the 1/4 label should be placed on the fold, not in a space.
6. Ask students to talk with their group members about where 2/4 should go. Be aware that some students might skip the 1/2 mark and label 2/4 in the 3/4 place. Students with less fraction number sense may need to be reminded that 2/4 and 1/2 are equivalent and therefore are represented by the same place on the number line.
7. Repeat these steps until 8ths are labeled.
8. Some students will need more support than others to get to this point. While helping those who need it, here are some ideas that I use for possible extensions: Give students another strip of paper (already folded into 3rds) ask students to label the fractions. Then add more fractions to the number line. OR allow students to continue to break the 8ths into smaller parts.
Highlight marks on the number line that are represented by more than one fraction. Ask students use some vocabulary to explain why there is more than one fraction for each of these lines. (Equivalent fractions).
## Benchmark Fractions: Guided Practice
10 minutes
The purpose of constructing a number line, with equivalent fractions represented, is to help the students make generalizations about benchmark fractions. The following activities will help students answer these three focus questions:
How do you know if a fraction is 1/2
How do you know if a fraction is 1 whole?
How do you know if a fraction is close to zero, close to 1/2, or close to 1?
Ask students to list the fractions that are equivalent to 1/2 on their number line. Then ask students to share other fractions that are also equivalent to 1/2 (but are not on the number line that was made in class).
Students are able to state fractions that are equivalent to 1/2 with no trouble. After there are about 5 examples on the board, ask students to make generalizations about these fractions. The focus question, "How do you know a fraction is 1/2?" can help start this discussion. The end goal is to get students to recognize that when the numerator is half of the denominator, the fraction always represents one half.
Repeat these same steps with the fractions that make 1 whole. The end goal here is to get students to articulate that fractions that are equal to 1 whole have the same number in the numerator and denominator.
## Practice
15 minutes
Students have now developed a generalization about fractions that are equivalent to 0, ½, 1. These generalizations help students to develop a strong understanding of the basic benchmark fractions. At this point in the lesson, it is important to introduce the term "benchmark fraction" and explain their significant role in estimating the value of fractions.
Note: Benchmark fractions are any fraction that can be easily identified - and at the elementary level this is commonly understood as quarters and thirds as well as halves and wholes. Some interpret it as all the fractional units represented on a ruler (â
, â
, 1/16). Students may have worked with benchmark fractions in the past. It is important to recognize this and determine a working "class definition" of benchmark fractions in your room. Also, if students are able to reason with benchmark fractions aside from halves, it is important to develop this further.
This is a summary of the script I use to explain benchmark fractions to my students. Benchmark fractions are the 3 fractions that everyone in this school knows about. Even those kindergarteners. I said 3 benchmarks fractions. Is 1 a fraction? Is 0 a fraction? (students respond with hesitation, but then prove that 0 is a fraction using their number line as evidence 0/2, 0/4, 0/8) and the same with 1. Benchmark fractions are fractions that we know well, so it is easy to use them to estimate the value of tricker fractions.
Next, to get the students involved and moving, I draw a number line on the SMART Board and label the 3 basic benchmark fractions. I call on one student to say any random fraction and I write that on the board. Then, I call on a different student to come up, and move the fraction to the place on the number line that this fraction belongs.
Do you think it is between 0 and 1/2?
Is is close to 1/2?
Do you think it is between 1/2 and 1?
Is is more than 1?
Then, they are asked to explain why they placed the fraction where they did. This exercise helps students work with the 3 focus questions to estimate fractions using benchmark fractions.
## Ticket Out
5 minutes
Today, I asked to students to look at 7 questions in the textbook about estimating fractions using benchmark fractions. Since time was limited, I had the students meet in pairs to practice without pencils. The students used the 3 focus questions to determine an estimate for these fractions.
1. 9/10 + 5/6
2. 11/12 - 5/6
3. 2/3 -1/8
4. 5/9 +5/6
5. 7/10 +1/3
6. 1/4 + 1/4*
7. 19/20 - 9/6
* Depending on how students estimate 1/4, this will provide interesting discussion. I will connect it with a context to help students determine the most reasonable way to estimate this ( closer to 0 or a 1/2). | 2,207 | 9,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2018-47 | latest | en | 0.923849 |
http://nces.ed.gov/nationsreportcard/tdw/weighting/2000_2001/2001_schoolweights.aspx | 1,475,248,206,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662251.92/warc/CC-MAIN-20160924173742-00168-ip-10-143-35-109.ec2.internal.warc.gz | 190,111,769 | 11,277 | ## Computation of School Weights
The school weight wi (school i) for the 2001 assessment had five basic components:
• a component for PSU selection;
• a component for school selection given PSU selection;
• a component for selection of schools to a field test session only;
• a component to account for school nonresponse (that is, a refusal of the school to participate in the study); and
• a component to trim school weights that contribute too much to overall variance.
For 2001 assessment schools, the overall school weight can be given as
where
• The quantity πi is the overall probability of selection of the school and is the product of the probability of selection of the PSU (see Sampling of Primary Sampling Units) containing the school and the conditional probability of selection of school i given that the PSU was selected on the final PSU-subsetted school frame (see Sampling of Schools). [There were two special cases: one in the eighth-grade sample and one in the twelfth-grade sample. In the eighth-grade sample, a sampled school split into two schools. One part with half probability was selected. A weight factor of 2 is attached. In the twelfth-grade sample, a sampled school was ineligible, but another unsampled school was mistakenly fielded (which was present on the frame but not sampled). The assessments from this school were retained but the conditional probability was reset to 1 to make this school self-representing at this level of sampling.]
• The quantity SUBADJi equals 1 for all originally sampled schools and equals the ratio of estimated grade enrollment of the original to the substitute school for participating substitute schools.
• The quantity FTADJi equals 1 for all schools with two or more sessions and equals the reciprocal of the probability of being selected for the main assessment (rather than the field test assessment) for schools with one session.
• SNk is a weighted response factor in a selected set of nonresponse cells k accounting for school refusals (see School Trimming).
• The trimming factor TRIMi equals 1 for most schools and was set to a value <1 for schools that were contributing too much to overall variance (see School Nonresponse Adjustments).
This table shows the distribution of school base weights πi–1 over the full school sample for each of the three grades. The summations are estimates of the total number of schools in the United States that include the grade.
Distribution of school base weights, by school base weights and grade, national main assessment: 2001
School base weights Grade 4 Grade 8 Grade 12
Number 441 465 522
Minimum 15.48 8.61 1.001
5th percentile 46.67 18.79 8.22
10th percentile 59.14 22.29 9.91
25th percentile 83.83 36.48 14.73
50th percentile 143.91 93.39 41.13
75th percentile 169.31 93.39 60.85
90th percentile 402.95 261.5 174.94
95th percentile 411.17 290.56 184.15
Maximum 1611.79 1129.27 699.76
Sum 80,676 50,881 32,461
Mean 182.94 109.42 62.19
Standard deviation 190.13 145.9 80.38
1 This school was a self-representing school because of a sampling error.
SOURCE: U.S. Department of Education, Institute of Education Sciences, National Center for Education Statistics, National Assessment of Educational Progress (NAEP), 2001.
SUBADJi is added as a weighting factor to allow the student sample from the participating substitute school to fully represent the nonparticipating original school. It equals the ratio of the estimated grade enrollment for the originally sampled school to the estimated grade enrollment for the substitute school. The student sample from the substitute school then "represents" the grade-eligible group from the originally sampled school. As an extreme example, suppose the originally sampled school has one grade-eligible student, with the substitute school having a full sample of 60 students. Then SUBADJi will be 1/60 in this case: the 60 students' assessment are "representing" the one single student from the originally sampled school. The substitute selection process was designed to minimize the differences between the original sampled school enrollment and the substitute's enrollment (see School Substitution for Adjusting School Nonresponse).
This table shows the distribution of SUBADJi values across the substitute schools for each grade.
Distribution of substitute school adjustment factor SUBADJi by substitute school adjustment factor and grade, national main assessment: 2001
Substitute school adjustment factor Grade 4 Grade 8 Grade 12
Number of schools 24 26 29
Minimum 0.613 0.125 0.177
25th percentile 0.934 0.885 0.954
50th percentile 1.028 1.017 1.106
75th percentile 1.258 1.147 1.270
Maximum 5.000 4.714 2.750
SOURCE: U.S. Department of Education, Institute of Education Sciences, National Center for Education Statistics, National Assessment of Educational Progress (NAEP), 2001.
Last updated 26 August 2008 (FW) | 1,138 | 4,894 | {"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-2016-40 | longest | en | 0.971539 |
https://visualfractions.com/calculator/divisible-by/is-64-divisible-by-5/ | 1,686,106,101,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00446.warc.gz | 629,437,731 | 6,810 | # Is 64 Divisible By 5?
In this quick and easy guide, we'll work out whether 64 is divisible by 5. There are some simple rules we can follow to decide whether one number is divisible by another without ever needing to even do the division!
First up, let's clarify what we mean by "64 is divisible by 5". What we want to check is whether 64 can be divided by 5 without any remainder (i.e the answer is a whole number).
Checking whether 64 is divisible by 5 is one of the easiest divisibility checks you can make. Basically, if the number ends with either a 5 or a 0, it is divisible by 5. In this case, it ends with 4.
We can see that 64 DOES NOT end with a 5 or a 0, which means that 64 IS NOT divisible by 5.
Another way you can figure out if 64 is divisible by 5 is by actually doing the calculation and dividing 64 by 5:
64 / 5 = 12.8
As you can see, when we do this division we have a decimal of 0.8. Since the division does not result in a whole number, this shows us that 64 is not divisible by 5.
Hopefully now you know exactly how to work out whether one number is divisible by another. Could I have just told you to divide 64 by 5 and check if it is a whole number? Yes, but aren't you glad you learned the process?
Give this a go for yourself and try to calculate a couple of these without using our calculator. Grab a pencil and a piece of paper and pick a couple of numbers to try it with.
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "Is 64 Divisible By 5?". VisualFractions.com. Accessed on June 7, 2023. http://visualfractions.com/calculator/divisible-by/is-64-divisible-by-5/.
• "Is 64 Divisible By 5?". VisualFractions.com, http://visualfractions.com/calculator/divisible-by/is-64-divisible-by-5/. Accessed 7 June, 2023. | 491 | 1,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.923885 |
http://www.openwetware.org/index.php?title=User:Allison_K._Alix/Notebook/Thesis_Research/2013/07/18&diff=prev&oldid=710357 | 1,448,840,889,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460263.61/warc/CC-MAIN-20151124205420-00032-ip-10-71-132-137.ec2.internal.warc.gz | 598,621,450 | 6,323 | # User:Allison K. Alix/Notebook/Thesis Research/2013/07/18
(Difference between revisions)
Revision as of 10:03, 19 July 2013 (view source) (→Entry title)← Previous diff Revision as of 10:07, 19 July 2013 (view source) (→Calculating %DNA hybridized on AuNP from 7/16)Next diff → Line 6: Line 6: | colspan="2"| | colspan="2"| - ==Calculating %DNA hybridized on AuNP from 7/16== + ==%DNA hybridized on AuNP from 7/16== + + '''Calculating the surface area of 1 AuNP''' - Calculating the surface area of 1 AuNP Surface Area of a sphere = 4πr2, r = 5nm Surface Area of a sphere = 4πr2, r = 5nm + 4π(5x10-9m)2 = 3.14 x 10-16m2 = 314nm2 4π(5x10-9m)2 = 3.14 x 10-16m2 = 314nm2 - Calculating the # of AuNP in 240.7μL 29.67nM solution + '''Calculating the # of AuNP in 144.4uL 29.67nM solution''' - 29.67 x 10-9mol/L x 240 x 10-6L x 6.02x1023 AuNP/mol = 4.299 x 1012 AuNP + - Calculating the Total Surface Area Available for DNA Attachment + 29.67 x 10-9mol/L x 144.4x10-6L x 6.02x1023 AuNP/mol = 2.579x1012 AuNP - 4.299 x 1012 AuNP x 3.14x10-16m2/1 AuNP = 0.001349886 m2 + + '''Calculating the Total Surface Area Available for DNA Attachment''' + + 2.579x1012AuNP x 3.14x10-16m2/1 AuNP = 0.00081 m2 Calculating the Number of Moles of DNA Available to Attach Calculating the Number of Moles of DNA Available to Attach + 2.225x10-6moles/L x 240.7x10-6L = 5.3555575 x 10-10 moles DNA 2.225x10-6moles/L x 240.7x10-6L = 5.3555575 x 10-10 moles DNA
## Revision as of 10:07, 19 July 2013
Project name Main project page
Previous entry Next entry
## %DNA hybridized on AuNP from 7/16
Calculating the surface area of 1 AuNP
Surface Area of a sphere = 4πr2, r = 5nm
4π(5x10-9m)2 = 3.14 x 10-16m2 = 314nm2
Calculating the # of AuNP in 144.4uL 29.67nM solution
29.67 x 10-9mol/L x 144.4x10-6L x 6.02x1023 AuNP/mol = 2.579x1012 AuNP
Calculating the Total Surface Area Available for DNA Attachment
2.579x1012AuNP x 3.14x10-16m2/1 AuNP = 0.00081 m2
Calculating the Number of Moles of DNA Available to Attach
2.225x10-6moles/L x 240.7x10-6L = 5.3555575 x 10-10 moles DNA
Calculating oligonucleotide density on each particle (theoretical) 5.3555575x10-10moles DNA/0.001349886m2 AuNP = 3.967x10-7 moles DNA/m2 AuNP (OR 39.67pmol DNA/cm2 AuNP)
Calculating oligonucleotide density on each particle (actual) Sample A1 moles DNA = (1.827x10-7M x 350 x 10-6L) = 6.40x10-11 moles 6.40x10-11 moles/0.001349886m2 AuNP = 4.74x10-8moles DNA/m2AuNP Sample B moles DNA = 4.44x10-8M x 500 x 10-6L = 2.22x10-11 moles 2.22x10-11 moles/0.001349886m2 AuNP = 1.64x10-8moles DNA/m2AuNP Sample A2 moles DNA = 1.42x10-10 1.42x10-10/0.001349886m2 AuNP = 1.05x10-7moles DNA/m2AuNP
Calculating the percentage of DNA hybridized in each sample Sample A1 4.74x10-8moles DNA/m2AuNP/3.967x10-7 moles DNA/m2 AuNP = 11.95% Sample B 1.64x10-8moles DNA/m2AuNP/3.967x10-7 moles DNA/m2 AuNP = 4.12% Sample A2 1.05x10-7moles DNA/m2AuNP/3.967x10-7 moles DNA/m2 AuNP = 26.5% | 1,245 | 2,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2015-48 | longest | en | 0.724328 |
http://www.reciprocalsystem.org/paper/comments-on-bpom-uom-and-outline | 1,582,790,815,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00337.warc.gz | 217,677,331 | 8,640 | # Comments on BPOM, UOM and Outline
Outline
1. Sec. 2, P. 3, # 65, last lines: “it is an outward motion added to an inward motion.”
Comment: not always. It is inward, in the case of electro-positive elements.
2. #116: “The effect of this restriction ( #57 ) is to bar three-dimensional rotational vibration.”
Comment: How does this follow from #57?
3. #127: “… each unit of rotational vibration combines with a unit of rotation.”
Comment: The electron has one natural unit of rotational displacement. Therefore it can take one unit of electric charge. But the atoms are doubled rotating systems. If Z is the net electric displacement of an atom, its atomic number, then the number of this displacement in natural units is 2Z. If each natural unit of rotation can take on one unit of electric charge, the atom can take 2Z number of charges on complete ionization and not merely Z.
Basic Properties of Matter
1. P. 57, lines 4 & 3 from bottom: “Radiation originates three-dimensionally in the time region, and makes contact one-dimensionally in the outside region. It is thus four-dimensional …”
Comment: gap in the logic.
2. P. 59, lines 3,2 from bottom: “The 3/4 power of 7.20423×1012 is …”
Comment: A dimensional error is involved: °K get altered to °K3/4.
3. P. 60, line 19 from bottom: “1/3 (3.598×109)1/3 degrees K = 510.8 °K.”
Comment: The dimensional balance is thrown to the winds.
4. P. 86, Table 22:
Comment: How are the thermal factors arrived at?
5. P. 87, lines 11, 10 from bottom: “… for rubidium and cesium, there is no effective displacement in the electric dimension…”
Comment: But Rb and Cs have 1 unit of effective displacement.
6. P. 113, 4 para, lines 1-2: “… there are two dimensions of rotation in space… in many of the elements of Division IV…”
Comment: needs explanation.
7. P. 213, para 4, last 4 lines: “On one side of this dividing line the rotation appears CW to observation. The scalar direction of the magnetic charge on this side is therefore outward from a CW direction. A similar charge on the opposite side is a motion outward from a CCW direction.”
Comment: But since the charge is a rotational vibration, in the next half-cycle it reverses.
8. P. 213, para 5, lines 1,2: “The unit of magnetic charge applies to only one of the two rotating systems of the atom. Each atom therefore acquired two charges…”
Comment: What happens in the core of subatomic particles: Don’t these manifest two poles (see line 3)?
9. P. 230, para 3, lines 5,6: “Since this remaining motion is scalar and two-dimensional, it is magnetic…”
Comment: (i) But it is a rotation and not a rotational vibration. Moreover, it is a time displacement and not space displacement. How does it show up as magnetism? It is more like gravitational charge than magnetic charge.
(ii) Should this not also show up as a mass loss?
10. P. 235, Fig. 25
Comment: What does the Theory predict about the force between two currents perpendicular to each other?
11. P. 238, para 2, lines 3,4: “… a two-dimensional magnetic motion… applied in opposition to gravitation will leave one-dimensional residue, an electric current…”
Comment: The scalar direction of this current is inward, what would be its results.
12. P. 241, para 2: “unlike the ferromagnetic charge, (The internal magnetic) charge on the basic rotation of the atom is subject to the electric rotation of the atom in the third scalar dimension…”
Comment: In such case the charge does not display any bipolar effect.
13. P. 241, para 3: “The corresponding factor … is a square root of the product of 1 and 3×1010…”
Comment: What is the rationale?
14. P. 251, line 3,2 from bottom: “each magnetically charged atom exerts a force on its magnetic neighbors, tending to line up these… atoms with its own magnetic direction…”
Comment: But it would be an antiparallel line-up, not a parallel!
15. P. 253, line 2: “… the electron rotation has the inward scalar direction…”
Comment: since it is a space displacement, it has to be an outward scalar direction.
16. P. 253, lines 2-4: “… the electron rotation… the charge. the two motions take place in different scalar dimensions”
Comment: But the charge (motion) modifies the rotation. As such they ought to be in the same scalar dimension. Also see P. 257, lines 20.19: “Addition of an oppositely directed unit of charge… reduces the net displacement to zero, and terminates the existence of the particle.”
17. P. 262, 263: (about the gravitational charge)
Comment: Since the gravitational charge is a two-dimensional rotational vibration like the magnetic charge, there should be a bipolar effect!
18. P. 271, lines 3,4: “Th234 → Pa234; 180-54 →182-52”
Comment: The vibrational mass for Z = 91 is 52.93 (see P. 268) Since 52 < 52.93 why should Pa234 need a beta decay 182-52 → 184-50?
19. P. 283, Table 36: “M 1-1-(1) proton”
Comment: Why *M 1-1-(1) does not take precedence over +M 1-1-(1)? In the case of the neutron the gravitational charge is stated to take precedence, *M ½-½-(1), on the grounds that 2 dimensional motion is more probable.
The Universe of Motion
1. … the inner and the outer gravitational limits…
Comment: One wonders whether similar limits exist for the translational effects in the cases of magnetic and electric charges too.
2. P. 234: The chapter on Pulsars contains many inconsistencies,
Comment: It is not clear what the author wants to say about P and Ṗ.
If P, the period is taken to be proportional to t6 then dP/dt is proportional to t5, that is, P5/6, but not to P5 as depicted. Even if (P)obsd is taken to be (Ṗ)-1cald , we have Ṗobsd proportional to P-5/6.
It is easy to see that the observed relation, Ṗ being proportional to P-5, means that P is proportional to t1/6 (and not to t6 as assumed).
3. P. 235, line 7: …the expression 1/6 (n/P)5
Comment: The above expression does not give the set of radiating lines as shown in Fig. 24. Remembering that Fig. 24 is a log-log plot, the above expression can be seen rather to represent a set of parallel lines with slope = -5 on the diagram. The observed spread of the data on this diagram may be due not only to the different values of n, but also to the differences in the masses of the pulsars.
4. P. 297, para 2, lines 5-2: “… all … sources then Known.”
Comment: That was in 1967. What about in 1984, the year of publication of this Volume? Still no more than 5 conclusive cases ! ?
5. P. 341, bottom para: “… the radiation … travels back to us through time…”
Comment: What does that mean?
6. P. 382, lines -14: “… addition of rotational motion in space to an atom of matter decreases the isotopic mass…”
Comment: decreases or increases?
General
1. Electric charge is a unit of 1-dimensional rotational vibration.
Comment: Two such units make up double ionization. But what would then be a charge with displacement 2?
International Society of Unified Science
Reciprocal System Research Society
Salt Lake City, UT 84106
USA
Theme by Danetsoft and Danang Probo Sayekti inspired by Maksimer | 1,820 | 7,030 | {"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-2020-10 | latest | en | 0.877453 |
https://www.eng-tips.com/viewthread.cfm?qid=27522 | 1,600,751,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00537.warc.gz | 852,697,658 | 12,772 | ×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS
Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.
Posting Guidelines
Promoting, selling, recruiting, coursework and thesis posting is forbidden.
PSD Computing
PSD Computing
(OP)
Hi everybody,
I need to evaluate the PSD from a "time history" signal.
Now I'm able to make the FFT of the signal (the FFT is "normalised", i.e. the spectum of an harmonic signal freq f amplitued A return a single value at freq f and of aplitude A).
Can anybody explain how to go further in order to obtain a PSD?
Thank you very much in advance
Luca
RE: PSD Computing
Multiply the complex value at each spectral line by its conjugate and divide each spectral line by the difference in frequency between two adjacent lines.
M
RE: PSD Computing
The calculation of PSD has many specialities. Simple FFT transformation solves the problem (theoretically), but in practics there are more problems.
E.g.: time signal has finite length, while FFT decompose signal into a number of SIN/COS functions which are infinite, and to make it the time signal is supposed to repeate infinitely which leads to abrupts in begin/end of the signal. Such an abrupt causes additional sinusoidal waves to be added. To resolve this problem time signal is to be multiplied by 'window' and after transformation it is to be scaled to balance the common power.
Another problem: computing PSD has a property that increas of time signal *doesn't increase the precision of PSD*. To overcome it the time signal is diveded into several segments for which PSD is calculated separately and then averaged by segments. This method is known as 'Welch method'.
All these and many other problems are described and resolved in MATLAB program. There is a special matlab's toolbox called 'Signal Processing Toolbox' which involves preprogrammed functions of PSD calculations. I used to take MATLAB and this toolbox in free evaluation; look their website (www.mathworks.com), maybe this possibility still available. There is also free documentation to this toolbox describing PSD computation, including those I have written above: http://www.mathworks.com/access/helpdesk/help/toolbox/signal/signal.shtml
RE: PSD Computing
Just an aside:
The matlab functions psd and csd are the only places I have ever seen this method described as "Welch's method". Almost everyone else calls it "the Fourier transform estimation method". I have sometimes described it to people as Welch's method and they just stare at me blankly!
M
Red Flag This Post
Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.
Red Flag Submitted
Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.
Resources
White Paper - PLM and ERP: Their Respective Roles in Modern Manufacturing
Leading manufacturers are aligning their people, processes, and tools from initial product ideation through to field service. They do so by providing access to product and enterprise data in the context of each person’s domain expertise. However, it can be complicated and costly to unite engineering with the factory and supply chain. Download Now
White Paper - Medical Device Design Control
Medical device product development is a highly integrated and regulated process. Implementation of a requirements tracking solution requires attention to a variety of nuanced topics. When presented with the task of tracking the many concept relationships in a project of this type, the initial software solution of choice tends to be a two-dimensional text systems. Download Now
Close Box
Join Eng-Tips® Today!
Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.
Here's Why Members Love Eng-Tips Forums:
• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...
Register now while it's still free! | 899 | 4,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-40 | latest | en | 0.949245 |
http://mathhelpforum.com/math-software/index25.html | 1,524,527,423,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946256.50/warc/CC-MAIN-20180423223408-20180424003408-00316.warc.gz | 205,179,790 | 14,509 | # Math Software Forum
Math Software Forum: For math software applications
1. ### Diagonal Matrix in Maple
• Replies: 1
• Views: 2,823
Apr 12th 2010, 03:40 AM
2. ### Favourite optimization libraries for c++
• Replies: 0
• Views: 938
Apr 11th 2010, 06:50 PM
3. ### Matlan help (Amos Gilat)
• Replies: 3
• Views: 1,029
Apr 10th 2010, 05:59 AM
4. ### ODE matlab
• Replies: 1
• Views: 720
Apr 9th 2010, 10:27 PM
5. ### Maple 13 - Help with GraphTheory package!
• Replies: 0
• Views: 641
Apr 9th 2010, 05:50 PM
6. ### Differential Eqns in Mathematica help please
• Replies: 0
• Views: 931
Apr 8th 2010, 08:21 PM
7. ### New commands in Mathematica 7
• Replies: 4
• Views: 919
Apr 5th 2010, 09:11 AM
8. ### Matlab function M-Files
• Replies: 1
• Views: 706
Apr 5th 2010, 04:33 AM
9. ### Excel - Uniform Distribution
• Replies: 4
• Views: 5,485
Apr 5th 2010, 02:50 AM
10. ### Noninear system - fixed point method
• Replies: 6
• Views: 1,121
Apr 4th 2010, 04:04 PM
11. ### is there any function in MATLAB?
• Replies: 1
• Views: 514
Apr 4th 2010, 01:57 AM
12. ### Maple making mistake?
• Replies: 3
• Views: 590
Apr 3rd 2010, 02:28 PM
13. ### Matlab - cell array
• Replies: 0
• Views: 956
Apr 3rd 2010, 05:28 AM
14. ### My Matlab Queries Page
• Replies: 1
• Views: 655
Apr 2nd 2010, 03:46 AM
15. ### Matlab - Least squares approximation of f(x) = e^x
• Replies: 5
• Views: 1,903
Apr 2nd 2010, 01:21 AM
16. ### Matlab- Newton's method
• Replies: 4
• Views: 14,551
Mar 31st 2010, 06:50 PM
17. ### Zero power mathematica
• Replies: 1
• Views: 629
Mar 31st 2010, 10:13 AM
18. ### MATLAB help! Functions with integrals
• Replies: 3
• Views: 661
Mar 30th 2010, 11:57 PM
19. ### Matlab Help Tricky pre-defined Matrix
• Replies: 1
• Views: 577
Mar 30th 2010, 06:24 AM
20. ### Matlab - Storing a row for a matrix?
• Replies: 6
• Views: 834
Mar 30th 2010, 02:16 AM
• Replies: 0
• Views: 1,132
Mar 29th 2010, 05:47 AM
22. ### MATLAB - Need some help with errors
• Replies: 1
• Views: 582
Mar 28th 2010, 08:17 PM
23. ### Matlab - Need help on strings
• Replies: 1
• Views: 942
Mar 28th 2010, 10:45 AM
24. ### [SOLVED] Matlab need some help getting on the right track!
• Replies: 2
• Views: 482
Mar 28th 2010, 09:24 AM
25. ### MATLAB - Function M-Files
• Replies: 6
• Views: 910
Mar 27th 2010, 10:20 PM
26. ### Stata
• Replies: 0
• Views: 696
Mar 26th 2010, 03:13 AM
### content
Click on a term to search for related topics.
Use this control to limit the display of threads to those newer than the specified time frame.
Allows you to choose the data by which the thread list will be sorted. | 1,010 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-17 | latest | en | 0.657586 |
https://brainmass.com/economics/perfect-competition/imperfect-competition-optimal-search-model-132354 | 1,579,387,856,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00203.warc.gz | 365,056,424 | 12,407 | Explore BrainMass
Share
# Imperfect Competition
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Consider the consumer's optimal-search model analyzed in section 16.2
Suppose that there are nine types of stores each selling at a different price drawn from a uniform distribution where p is a subset of (1,2,3,4,5,6,7,8,9). Answer the following questions:
1. Construct a table showing the consumer's reservation price, and the expected number of store visits under different values of the search cost parameter. More precisely consider the cases in which s = 0,1,2,3,4,5.
2. What is the value of s that will cause the consumer to purchase the product at his or her first store visit?
3. What is the value of s that will cause the consumer never to buy the product unless the price is p=1.
4. Using the value of s that you found in question 3, calculate the probability that the consumer will search forever. Prove and explain your result.
© BrainMass Inc. brainmass.com October 9, 2019, 7:53 pm ad1c9bdddf
https://brainmass.com/economics/perfect-competition/imperfect-competition-optimal-search-model-132354
#### Solution Preview
Solution-Table 16.1: Reservation price as a function of the search cost parameter
Please see the attached file for the table
2. (a) Using (16.13), we can make the table as follows:
s 0 1 2 3 4 5
p 1 4.77 6.52 7.86 9.00 10
Solution-Table 16.1: Reservation price ...
#### Solution Summary
The solution answers the question below in great detail.
\$2.19 | 401 | 1,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-05 | latest | en | 0.878319 |
http://prewar.mgcc.info/n-type/?rdp_we_resource=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FDiscrete_sine_transform | 1,618,312,197,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00211.warc.gz | 78,393,448 | 35,416 | In mathematics, the discrete sine transform (DST) is a Fourier-related transform similar to the discrete Fourier transform (DFT), but using a purely real matrix. It is equivalent to the imaginary parts of a DFT of roughly twice the length, operating on real data with odd symmetry (since the Fourier transform of a real and odd function is imaginary and odd), where in some variants the input and/or output data are shifted by half a sample.
A family of transforms composed of sine and sine hyperbolic functions exists. These transforms are made based on the natural vibration of thin square plates with different boundary conditions.[1]
The DST is related to the discrete cosine transform (DCT), which is equivalent to a DFT of real and even functions. See the DCT article for a general discussion of how the boundary conditions relate the various DCT and DST types. Generally, the DST is derived from the DCT by replacing the Neumann condition at x=0 with a Dirichlet condition.[2] Both the DCT and the DST were described by Nasir Ahmed T. Natarajan and K.R. Rao in 1974.[3][4] The type-I DST (DST-I) was later described by Anil K. Jain in 1976, and the type-II DST (DST-II) was then described by H.B. Kekra and J.K. Solanka in 1978.[5]
## Applications
DSTs are widely employed in solving partial differential equations by spectral methods, where the different variants of the DST correspond to slightly different odd/even boundary conditions at the two ends of the array.
## Informal overview
Illustration of the implicit even/odd extensions of DST input data, for N=9 data points (red dots), for the four most common types of DST (types I–IV).
Like any Fourier-related transform, discrete sine transforms (DSTs) express a function or a signal in terms of a sum of sinusoids with different frequencies and amplitudes. Like the discrete Fourier transform (DFT), a DST operates on a function at a finite number of discrete data points. The obvious distinction between a DST and a DFT is that the former uses only sine functions, while the latter uses both cosines and sines (in the form of complex exponentials). However, this visible difference is merely a consequence of a deeper distinction: a DST implies different boundary conditions than the DFT or other related transforms.
The Fourier-related transforms that operate on a function over a finite domain, such as the DFT or DST or a Fourier series, can be thought of as implicitly defining an extension of that function outside the domain. That is, once you write a function as a sum of sinusoids, you can evaluate that sum at any , even for where the original was not specified. The DFT, like the Fourier series, implies a periodic extension of the original function. A DST, like a sine transform, implies an odd extension of the original function.
However, because DSTs operate on finite, discrete sequences, two issues arise that do not apply for the continuous sine transform. First, one has to specify whether the function is even or odd at both the left and right boundaries of the domain (i.e. the min-n and max-n boundaries in the definitions below, respectively). Second, one has to specify around what point the function is even or odd. In particular, consider a sequence (a,b,c) of three equally spaced data points, and say that we specify an odd left boundary. There are two sensible possibilities: either the data is odd about the point prior to a, in which case the odd extension is (−c,−b,−a,0,a,b,c), or the data is odd about the point halfway between a and the previous point, in which case the odd extension is (−c,−b,−a,a,b,c)
These choices lead to all the standard variations of DSTs and also discrete cosine transforms (DCTs). Each boundary can be either even or odd (2 choices per boundary) and can be symmetric about a data point or the point halfway between two data points (2 choices per boundary), for a total of possibilities. Half of these possibilities, those where the left boundary is odd, correspond to the 8 types of DST; the other half are the 8 types of DCT.
These different boundary conditions strongly affect the applications of the transform, and lead to uniquely useful properties for the various DCT types. Most directly, when using Fourier-related transforms to solve partial differential equations by spectral methods, the boundary conditions are directly specified as a part of the problem being solved.
## Definition
Formally, the discrete sine transform is a linear, invertible function F : RN -> RN (where R denotes the set of real numbers), or equivalently an N × N square matrix. There are several variants of the DST with slightly modified definitions. The N real numbers x0, xN − 1 are transformed into the N real numbers X0, XN − 1 according to one of the formulas:
### DST-I
The DST-I matrix is orthogonal (up to a scale factor).
A DST-I is exactly equivalent to a DFT of a real sequence that is odd around the zero-th and middle points, scaled by 1/2. For example, a DST-I of N=3 real numbers (a,b,c) is exactly equivalent to a DFT of eight real numbers (0,a,b,c,0,−c,−b,−a) (odd symmetry), scaled by 1/2. (In contrast, DST types II–IV involve a half-sample shift in the equivalent DFT.) This is the reason for the N + 1 in the denominator of the sine function: the equivalent DFT has 2(N+1) points and has 2π/2(N+1) in its sinusoid frequency, so the DST-I has π/(N+1) in its frequency.
Thus, the DST-I corresponds to the boundary conditions: xn is odd around n = −1 and odd around n=N; similarly for Xk.
### DST-II
Some authors further multiply the XN − 1 term by 1/2 (see below for the corresponding change in DST-III). This makes the DST-II matrix orthogonal (up to a scale factor), but breaks the direct correspondence with a real-odd DFT of half-shifted input.
The DST-II implies the boundary conditions: xn is odd around n = −1/2 and odd around n = N − 1/2; Xk is odd around k = −1 and even around k = N − 1.
### DST-III
Some authors further multiply the xN − 1 term by 2 (see above for the corresponding change in DST-II). This makes the DST-III matrix orthogonal (up to a scale factor), but breaks the direct correspondence with a real-odd DFT of half-shifted output.
The DST-III implies the boundary conditions: xn is odd around n = −1 and even around n = N − 1; Xk is odd around k = −1/2 and odd around k = N − 1/2.
### DST-IV
The DST-IV matrix is orthogonal (up to a scale factor).
The DST-IV implies the boundary conditions: xn is odd around n = −1/2 and even around n = N − 1/2; similarly for Xk.
### DST V–VIII
DST types I–IV are equivalent to real-odd DFTs of even order. In principle, there are actually four additional types of discrete sine transform (Martucci, 1994), corresponding to real-odd DFTs of logically odd order, which have factors of N+1/2 in the denominators of the sine arguments. However, these variants seem to be rarely used in practice.
### Inverse transforms
The inverse of DST-I is DST-I multiplied by 2/(N + 1). The inverse of DST-IV is DST-IV multiplied by 2/N. The inverse of DST-II is DST-III multiplied by 2/N (and vice versa).
As for the DFT, the normalization factor in front of these transform definitions is merely a convention and differs between treatments. For example, some authors multiply the transforms by so that the inverse does not require any additional multiplicative factor.
### Computation
Although the direct application of these formulas would require O(N2) operations, it is possible to compute the same thing with only O(N log N) complexity by factorizing the computation similar to the fast Fourier transform (FFT). (One can also compute DSTs via FFTs combined with O(N) pre- and post-processing steps.)
A DST-III or DST-IV can be computed from a DCT-III or DCT-IV (see discrete cosine transform), respectively, by reversing the order of the inputs and flipping the sign of every other output, and vice versa for DST-II from DCT-II. In this way it follows that types II–IV of the DST require exactly the same number of arithmetic operations (additions and multiplications) as the corresponding DCT types.
## References
1. ^ Abedi, M.; Sun, B.; Zheng, Z. (July 2019). "A Sinusoidal-Hyperbolic Family of Transforms With Potential Applications in Compressive Sensing". IEEE Transactions on Image Processing. 28 (7): 3571–3583. Bibcode:2019ITIP...28.3571A. doi:10.1109/TIP.2019.2912355. PMID 31071031.
2. ^ Britanak, Vladimir; Yip, Patrick C.; Rao, K. R. (2010). Discrete Cosine and Sine Transforms: General Properties, Fast Algorithms and Integer Approximations. Elsevier. pp. 35–6. ISBN 9780080464640.
3. ^ Ahmed, Nasir; Natarajan, T.; Rao, K. R. (January 1974), "Discrete Cosine Transform" (PDF), IEEE Transactions on Computers, C-23 (1): 90–93, doi:10.1109/T-C.1974.223784
4. ^ Ahmed, Nasir (January 1991). "How I Came Up With the Discrete Cosine Transform". Digital Signal Processing. 1 (1): 4–5. doi:10.1016/1051-2004(91)90086-Z.
5. ^ Dhamija, Swati; Jain, Priyanka (September 2011). "Comparative Analysis for Discrete Sine Transform as a suitable method for noise estimation". IJCSI International Journal of Computer Science. 8 (Issue 5, No. 3): 162-164 (162). Retrieved 4 November 2019. |issue= has extra text (help) | 2,289 | 9,232 | {"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} | 3.265625 | 3 | CC-MAIN-2021-17 | latest | en | 0.926666 |
https://it.mathworks.com/matlabcentral/answers/343288-auto-fill-zero-matrix-without-row-repetitions | 1,726,338,195,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00040.warc.gz | 294,540,108 | 30,778 | # Auto fill zero matrix without row-repetitions
2 visualizzazioni (ultimi 30 giorni)
JB il 4 Giu 2017
Commentato: Andrei Bobrov il 5 Giu 2017
Please help. I'm new to matlab scripting and need a bit of help. I have a series of numbers: test = [1 1 1 2 2 2 3 3 3 4 4 4 5 5 5] which I wants to randomely fill into a 5x3 matrix without having the same number in the same row. How can I do this??? Potentially I could randomize the test vector and fill it into the 5x3 matrix but I dont know how to do this without getting the same number in the same row. PLEASE help...
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
JB il 5 Giu 2017
MISTAKE, I want the final matrix to be a 3x5 matrix and not a 5x3 matrix...
Accedi per commentare.
### Risposta accettata
JB il 5 Giu 2017
Thanks a lot for your suggestions. However, I made a minor mistake. I want the final matrix to be a 3x5 instead of a 5x3 matrix. Sorry guys, but I still need a bit of help to establish the correct code.
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
JB il 5 Giu 2017
Awesome, thanks KSSV. I have one final question for you which I hope you can help me with. Say test = [1 1 1 2 2 2 3 3 3] and I want to fill it into a 3x5 matix without row repetionen, and leave excess cells blank (or =) how is this done?
KSSV il 5 Giu 2017
test = [1 1 1 2 2 2 3 3 3] ;
A = NaN(3,5) ;
test_unique = unique(test) ;
for i = 1:3
idx = randperm(3) ;
A(i,randperm(5,3)) = test_unique(idx) ;
end
Accedi per commentare.
### Più risposte (2)
KSSV il 5 Giu 2017
You can take the unique matrix of test and pick any three elements out of it and fill in the required 5X3 matrix.
test = [1 1 1 2 2 2 3 3 3 4 4 4 5 5 5] ;
test_unique = unique(test) ;
A = zeros(5,3) ;
for i = 1:size(A,1)
A(i,:) = randsample(test_unique,3) ;
end
randsample needs a statistics toolbox, if you doesn't have it, you may use randperm as shown below.
test = [1 1 1 2 2 2 3 3 3 4 4 4 5 5 5] ;
test_unique = unique(test) ;
A = zeros(5,3) ;
for i = 1:size(A,1)
A(i,:) = test_unique(randperm(length(test_unique),3)) ;
end
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
KSSV il 5 Giu 2017
You mean to say element might repeat in a row?
Walter Roberson il 5 Giu 2017
Elements cannot repeat in any row, but all elements of T must be placed in the 5 x 3 matrix.
Your code prevents elements from repeating in any one row, but the number of copies of any given member of T is not the same as the original. For example your code could randomly create
[1 2 3
1 2 3
1 2 3
1 2 3
1 2 3]
with no 4 or 5 anywhere.
Accedi per commentare.
Andrei Bobrov il 5 Giu 2017
Modificato: Andrei Bobrov il 5 Giu 2017
m = 5;
n = 3;
A = reshape(test,n,m)';
out = A(bsxfun(@plus,hankel(1:m,[m,1:n-1]),m*(0:n-1)));
out = out(randperm(m),:);
out = out(:,randperm(n));
m = 5;
n = 3;
test = repelem(1:3,3);
A = nan(n,m);
A(1:numel(test)) = test;
[~,ii] = sort(rand(n,m),2);
out = A(bsxfun(@plus,n*(ii-1),(1:n)'));
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
JB il 5 Giu 2017
Awesome, thanks Andrei. I have one final question for you which I hope you can help me with. Say test = [1 1 1 2 2 2 3 3 3] and I want to fill it into a 3x5 matix without row repetionen, and leave excess cells blank (or =) how is this done?
Andrei Bobrov il 5 Giu 2017
Accedi per commentare.
### Categorie
Scopri di più su Matrix Indexing in Help Center e File Exchange
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 1,249 | 3,613 | {"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 | 0.722356 |
http://pyevolve.sourceforge.net/wordpress/?p=1319 | 1,419,280,427,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802776563.13/warc/CC-MAIN-20141217075256-00053-ip-10-231-17-201.ec2.internal.warc.gz | 217,993,883 | 11,059 | Any text structure can be represented using Markov Chain states and transition probabilities, for the sake of simplicity I’ll not use probabilities to describe the state transitions, I’ll use only true or false, saying only that this transition is possible or not (100% or 0%).
A Markov Chain in this simple form can be formulated as a set of possible states, lets say $S = \left\{q_1,q_2,q_3,\ldots,q_n\right\}$ as well a dictionary-like data structure in the form of $D(q_x,q_y) = \left\{q_1, q_2, \dots, q_3\right\}$ representing the possible transition states, but let’s put an example text on it to make it clear:
the white rabbit and the white fox
Then the possible states of this text are $S = \left\{ the'', white'', rabbit'', and'', fox''\right\}$, and the dictionary structure should be:
$D(rabbit'', and'') = \left\{the''\right\}$
$D(white'', rabbit'') = \left\{and''\right\}$
$D(and'', the'') = \left\{white''\right\}$
$D(the'', white'') = \left\{rabbit', fox''\right\}$
These transition rules can be incrementally updated. Note that these rules represent the structure of the text and not the text by itself; when updated for instance, with a whole book, it captures the profile of the writer, the characteristics often used by the author, even more enforced when incrementally updated with more than one book of the same author.
I implemented a simple tool using Python and NLTK (just to parse and tokenize the text) that can be used to read an input text and incrementally update a Markov Chain data structure in a way to be reused later in order check similarity with another text.
The most interesting part of these chain is that they also offer a property present in Bloom Filters: false positives are possible but not false negatives. In terms of our implementation and formulation, that property means that if we update a chain with the text of two books from Lewis Carroll, there would be (and of course, is a very low probability) another author (that the chain was not trained with their books) who can have the same chain, creating in this way a false positive, because the same chain from Lewis Carroll books could have the same chain of another different book, but you should note that is almost impossible (actually this fact would be very strange, since the authors style and words used are required to be almost the same) to find books from different authors who share the same chain.
You should note that we can also compare different chains in order to measure similarity between two authors/book profiles. This comparison is done by searching a previous trained chain (with the same author books or not) for the entries of the current chain in order to match the same structure of the chains and then calculate a ratio between the matches found and the total rules of the chain we are checking.
But hey, that’s a lot of blah blah blah and nothing of code, let’s see it in practice. Here is the help of the implemented tool (I created a repository in GitHub if you’re interested in the source-code):
$python -m pymarksim.markov --help pymarksim v.0.1 By Christian S. Perone <christian.perone@gmail.com> http://pyevolve.sourceforge.net/wordpress Type --help parameter for help. Options: -h, --help show this help message and exit -d DUMP_FILE, --df=DUMP_FILE Pickle object file ex. pickle.db -i INPUT_TEXT, --input-text=INPUT_TEXT Input text -m MODE, --mode=MODE Mode of operation: train, check -c COMPRESS_LEVEL, --compress-level=COMPRESS_LEVEL The gzip compression level, default is 9 (max). So let’s train a chain using the “Alice Adventure’s in Wonderland” book (alice_adventures.txt) from Lewis Carroll (took from the Project Gutenberg): $ python -m pymarksim.markov -d lewis.db -i alice_adventures.txt -m train
pymarksim v.0.1
By Christian S. Perone <christian.perone@gmail.com>
http://pyevolve.sourceforge.net/wordpress
Type --help parameter for help.
Updating...
Done !
Here the tool created the file lewis.db which is the chain in a Python dictionary structure (I’ve used the cPickle module to serialize it and gzip compression). Now let’s check how close is this chain from the original text book:
$python -m pymarksim.markov -d lewis.db -i alice_adventures.txt -m check pymarksim v.0.1 By Christian S. Perone <christian.perone@gmail.com> http://pyevolve.sourceforge.net/wordpress Type --help parameter for help. File lewis.db found, loading... Checking... Match 100.00 % As you can see, we have a perfect match between the chain and the text book. Now comes the interesting part, let’s check an unrelated book with this chain trained with the Alice Adventure’s in Wonderland, for this, I took the book “The Adventures of Sherlock Holmes” (doyle.txt) from Sir Arthur Conan Doyle (also from the Project Gutenberg): $ python -m pymarksim.markov -d lewis.db -i doyle.txt -m check
pymarksim v.0.1
By Christian S. Perone <christian.perone@gmail.com>
http://pyevolve.sourceforge.net/wordpress
Type --help parameter for help.
Checking...
Match 7.14 %
See that now the match of the chain from Sherlock Holmes with the Alice Adventure’s is now 7.14%, remember that a perfect match is 100.00 % !
Let’s check now this chain trained only with the Alice Adventure’s in Wonderland of Lewis Carroll with another book from the same author, called “Through the looking-glass” (looking-glass.txt):
$python -m pymarksim.markov -d lewis.db -i looking-glass.txt -m check pymarksim v.0.1 By Christian S. Perone <christian.perone@gmail.com> http://pyevolve.sourceforge.net/wordpress Type --help parameter for help. File lewis.db found, loading... Checking... Match 25.93 % See that now we have a match of 25.93% ! Let’s check with a more related book also from Lewis Carroll, called “Alice Adventure’s Under Ground” (underground.txt): $ python -m pymarksim.markov -d lewis.db -i underground.txt -m check
pymarksim v.0.1
By Christian S. Perone <christian.perone@gmail.com>
http://pyevolve.sourceforge.net/wordpress
Type --help parameter for help. | 1,459 | 5,992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 7, "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-2014-52 | latest | en | 0.887685 |
http://www.algebra.com/algebra/homework/Trigonometry-basics/Addition_and_subtraction_of_trigonometric_functions_Examples.lesson?content_action=show_source | 1,368,965,212,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697442043/warc/CC-MAIN-20130516094402-00031-ip-10-60-113-184.ec2.internal.warc.gz | 301,875,136 | 5,688 | # Lesson Addition and subtraction of trigonometric functions - Examples
Algebra -> Algebra -> Trigonometry-basics -> Lesson Addition and subtraction of trigonometric functions - Examples Log On
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Algebra: Trigonometry Solvers Lessons Answers archive Quiz In Depth
### Source code of 'Addition and subtraction of trigonometric functions - Examples'
This Lesson (Addition and subtraction of trigonometric functions - Examples) was created by by ikleyn(4) : View Source, Show
Addition and subtraction of trigonometric functions - Examples
The addition and subtraction trigonometric functions formulas are: {{{sin(alpha) + sin(beta) = 2*sin((alpha+beta)/2)*cos((alpha-beta)/2)}}} {{{sin(alpha) - sin(beta) = 2*sin((alpha-beta)/2)*cos((alpha+beta)/2)}}} {{{cos(alpha) + cos(beta) = 2*cos((alpha+beta)/2)*cos((alpha-beta)/2)}}} {{{cos(alpha) - cos(beta) = -2*sin((alpha+beta)/2)*sin((alpha-beta)/2)}}} {{{tan(alpha) +- tan(beta) = sin(alpha +- beta)/(cos(alpha)*cos(beta))}}} {{{cot(alpha) +- cot(beta) = sin(alpha +- beta)/(sin(alpha)*sin(beta))}}} The proofs of these formulas are presented in the lesson Addition and subtraction of trigonometric functions in this module. Below are examples of application of these formulas. Example 1 Find sin(75°) + sin(15°). Solution Use the addition formula for sines: {{{sin(alpha) + sin(beta) = 2*sin((alpha+beta)/2)*cos((alpha-beta)/2)}}}. You have sin(75°) + sin(15°) = 2*sin((75°+15°)/2)*cos((75°-15°)/2) = 2*sin(45°)*cos(30°) = {{{2*sqrt(2)/2*sqrt(3)/2 = sqrt(6)/2}}}. Example 2 Prove that {{{(sin(alpha)+sin(3alpha))/(cos(alpha)+cos(3alpha)) = tan(2alpha)}}}. Solution Use the addition formula for sines and cosines: {{{sin(alpha) + sin(beta) = 2*sin((alpha+beta)/2)*cos((alpha-beta)/2)}}}, {{{cos(alpha) + cos(beta) = 2*cos((alpha+beta)/2)*cos((alpha-beta)/2)}}}. You have {{{(sin(alpha)+sin(3alpha))/(cos(alpha)+cos(3alpha)) = 2*sin((alpha+3alpha)/2)*cos((alpha-3alpha)/2)/(2*cos((alpha+3alpha)/2)*cos((alpha-3alpha)/2)))}}} = {{{2*sin(2alpha)*cos(-2alpha)/(2*cos(2alpha)*cos(-2alpha)) = sin(2alpha)/cos(2alpha) = tan(2alpha)}}}. The proof is completed. Example 3 Prove that {{{1 + cos(2alpha) + cos(4alpha) + cos(6alpha) = 4*cos(alpha)*cos(2alpha)*cos(3alpha)}}}. Solution Using the addition formula for cosines you have {{{1 + cos(2alpha) = cos(0) + cos(2alpha) = 2*cos((0+2alpha)/2)*cos((0-2alpha)/2) = 2*cos^2(alpha)}}}, {{{cos(4alpha) + cos(6alpha) = 2*cos((4alpha+6alpha)/2)*cos((4alpha-6alpha)/2) = 2*cos(5alpha)*cos(-alpha) = 2*cos(alpha)*cos(5alpha)}}}. By summing the left and the right sides of these two equalities, you get {{{1 + cos(2alpha) + cos(4alpha) + cos(6alpha) = 2*cos^2(alpha) + 2*cos(alpha)*cos(5alpha)}}}. Furthermore, you can transform the right side as follows: {{{2*cos^2(alpha) + 2*cos(alpha)*cos(5alpha) = 2*cos(alpha)*(cos(alpha) + cos(5alpha))}}} = {{{2*cos(alpha)*(cos(alpha+5alpha)/2)*(cos(alpha-5alpha)/2)}}} = {{{2*cos(alpha)*cos(3alpha)*cos(-2alpha) = 2*cos(alpha)*cos(2alpha)*cos(3alpha)}}}. The proof is completed. Example 4 Prove yourself that {{{1 - cos(2alpha) + cos(4alpha) - cos(6alpha) = 4*sin(alpha)*cos(2alpha)*sin(3alpha)}}}. Solution The proof is similar to that of the Example 3. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ For your convenience, below is the list of my lessons on Trigonometry in this site in the logical order. They all are under the current topic Trigonometry in the section Algebra II. Addition and subtraction formulas
{{{cos(alpha + beta) = cos(alpha)*cos(beta) - sin(alpha)*sin(beta)}}}, {{{cos(alpha - beta) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta)}}}, {{{sin(alpha + beta) = sin(alpha)*cos(beta) + cos(alpha)*sin(beta)}}}, {{{sin(alpha - beta) = sin(alpha)*cos(beta) - cos(alpha)*sin(beta)}}}, {{{tan(alpha + beta) = (tan(alpha) + tan(beta))/(1 - tan(alpha)*tan(beta))}}}, {{{tan(alpha - beta) = (tan(alpha) - tan(beta))/(1 + tan(alpha)*tan(beta))}}}. The lessons Addition and subtraction formulas and Addition and subtraction formulas - Examples
Addition and subtraction of trigonometric functions
{{{sin(alpha) + sin(beta) = 2*sin((alpha+beta)/2)*cos((alpha-beta)/2)}}}, {{{sin(alpha) - sin(beta) = 2*sin((alpha-beta)/2)*cos((alpha+beta)/2)}}}, {{{cos(alpha) + cos(beta) = 2*cos((alpha+beta)/2)*cos((alpha-beta)/2)}}}, {{{cos(alpha) - cos(beta) = -2*sin((alpha+beta)/2)*sin((alpha-beta)/2)}}}, {{{tan(alpha) +- tan(beta) = sin(alpha +- beta)/(cos(alpha)*cos(beta))}}}, {{{cot(alpha) +- cot(beta) = sin(alpha +- beta)/(sin(alpha)*sin(beta))}}}. The lessons Addition and subtraction of trigonometric functions and Addition and subtraction of trigonometric functions - Examples (this lesson)
Product of trigonometric functions
{{{sin(alpha)*sin(beta) = (1/2)*(cos(alpha-beta) - cos(alpha+beta))}}}, {{{cos(alpha)*cos(beta) = (1/2)*(cos(alpha-beta) + cos(alpha+beta))}}}, {{{sin(alpha)*cos(beta) = (1/2)*(sin(alpha-beta) + sin(alpha+beta))}}}. The lessons Product of trigonometric functions and Product of trigonometric functions - Examples
Powers of trigonometric functions
{{{cos^2(alpha) = (1/2)*cos(2alpha) + 1/2}}}, {{{sin^2(alpha) = -(1/2)*cos(2alpha) + 1/2}}}, {{{cos^3(alpha) = (1/4)*cos(3alpha) + (3/4)*cos(alpha)}}}, {{{sin^3(alpha) = -(1/4)*sin(3alpha) + (3/4)*sin(alpha)}}}. The lessons Powers of Trigonometric functions and Powers of Trigonometric functions - Examples
Trigonometric functions of multiply argument
{{{cos(2alpha) = 2*cos^2(alpha) - 1}}}, {{{sin(2alpha) = 2*sin(alpha)*cos(alpha)}}}, {{{cos(3alpha) = 4*cos^3(alpha) - 3*cos(alpha)}}}, {{{sin(3alpha) = -4*sin^3(alpha) + 3*sin(alpha)}}}. The lessons Trigonometric functions of multiply argument and Trigonometric functions of multiply argument - Examples
Trigonometric functions of half argument
{{{sin^2(alpha/2) = (1-cos(alpha))/2}}}, {{{cos^2(alpha/2) = (1+cos(alpha))/2}}}, {{{tan(alpha/2) = sin(alpha)/(1+cos(alpha)) = (1-cos(alpha))/sin(alpha)}}}, {{{sin(alpha) = 2*tan(alpha/2)/(1+tan^2(alpha/2))}}}, {{{cos(alpha) = (1-tan^2(alpha/2))/(1+tan^2(alpha/2))}}}, {{{tan(alpha) = 2*tan(alpha/2)/(1-tan^2(alpha/2))}}}. The lessons Trigonometric functions of half argument and Trigonometric functions of half argument - Examples
Miscellaneous Trigonometry problems The lesson Miscellaneous Trigonometry problems | 2,001 | 6,781 | {"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} | 4.21875 | 4 | CC-MAIN-2013-20 | latest | en | 0.576299 |
https://amazonartstudio.com/qa/is-the-square-root-of-81-a-rational-number.html | 1,620,589,223,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00397.warc.gz | 116,103,069 | 7,942 | # Is The Square Root Of 81 A Rational Number?
## Is 64 a perfect square?
The perfect squares are the squares of the whole numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 ….
## What is the real square root of 81?
Explanation: 81=9⋅9 then the square root of √81=9 .
## Is 3.456 a rational number?
a number that can be written as a fraction Any number that is not an irrational number Examples: 2.34, 3.456, 6.323 232 32… Examples: 400, +8, 0, 29, 49578 • Whole numbers ﴾W﴿: all of the positive integers and zero Examples: 0, 1, 2, 3, 4, etc.
## What is the root square of 324?
18Answer and Explanation: The square root of 324 is 18.
## Is 2.718 a rational number?
The sequence is bounded below by the sum of the first two terms, which will be a positive number. Therefore, the right-hand side is strictly between 0 and 1, so it is a fractional number. This is a contradiction! Therefore, e cannot be a rational number and it is an irrational number.
## What are the factors of 82?
82 is a composite number. 82 = 1 x 82 or 2 x 41. Factors of 82: 1, 2, 41, 82. Prime factorization: 82 = 2 x 41.
## Is the square root of 89 a rational number?
The square root of 89 is a rational number if 89 is a perfect square. It is an irrational number if it is not a perfect square. Since 89 is not a perfect square, it is an irrational number.
## Is 82 rational or irrational?
The number 82 is a rational number if 82 can be expressed as a ratio, as in RATIOnal. A quotient is the result you get when you divide one number by another number. For 82 to be a rational number, the quotient of two integers must equal 82.
## What is the square of 82?
However, note that 82=81+1=92+1 . Since this is of the form n2+1 , the square root has a very regular form as a continued fraction: √82=[9;¯¯¯¯18]=9+118+118+118+118+…
## What is the square root of 72 simplified?
This result is equivalent to the square root we are trying to simplify! In this example, √72=2⋅3√2.
## Is 0 a rational number?
Zero Is a Rational Number As such, if the numerator is zero (0), and the denominator is any non-zero integer, the resulting quotient is itself zero.
## Is the square root of 82 a rational number?
The square root of 82 is a rational number if 82 is a perfect square. It is an irrational number if it is not a perfect square. Since 82 is not a perfect square, it is an irrational number.
## What is the square of 84?
The two square roots of 84 are denoted √84 and −√84 . Since 84 is 319 ‘s of the way between 81 and 100 , we can approximate √84 as 319≈16 th of the way between 9 and 10 , about 916=556 .
## Is 3.142 a irrational number?
The number pi, denoted π, is a well-known irrational number that is generally known to have value 3.14. However, 3.14 is actually the number π rounded to two decimal places, and is not the true value of π. | 819 | 2,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-21 | latest | en | 0.890348 |
http://www.convertunits.com/dates/from/Nov+25,+1911/to/May+23,+1934 | 1,501,187,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549429485.0/warc/CC-MAIN-20170727202516-20170727222516-00231.warc.gz | 409,318,684 | 11,051 | # ››Date difference from Nov 25, 1911 to May 23, 1934
The total number of days between Saturday, November 25th, 1911 and Wednesday, May 23rd, 1934 is 8,215 days.
This is equal to 22 years, 5 months, and 28 days.
This does not include the end date, so it's accurate if you're measuring your age in days, or the total days between the start and end date. But if you want the duration of an event that includes both the starting date and the ending date, then it would actually be 8,216 days.
If you're counting workdays or weekends, there are 5,867 weekdays and 2,348 weekend days.
If you include the end date of May 23, 1934 which is a Wednesday, then there would be 5,868 weekdays and 2,348 weekend days including both the starting Saturday and the ending Wednesday.
8,215 days is equal to 1,173 weeks and 4 days.
The total time span from 1911-11-25 to 1934-05-23 is 197,160 hours.
This is equivalent to 11,829,600 minutes.
You can also convert 8,215 days to 709,776,000 seconds.
# ››November, 1911 calendar
Su M Tu W Th F Sa 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
November 25th, 1911 is a Saturday. It is the 329th day of the year, and in the 47th week of the year (assuming each week starts on a Sunday), or the 4th quarter of the year. There are 30 days in this month. 1911 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 11/25/1911, and almost everywhere else in the world it's 25/11/1911.
# ››May, 1934 calendar
Su M Tu W Th F Sa 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
May 23rd, 1934 is a Wednesday. It is the 143rd day of the year, and in the 21st week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 31 days in this month. 1934 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 5/23/1934, and almost everywhere else in the world it's 23/5/1934.
# ››Enter dates
Enter two dates below to find the number of days between them. For best results, avoid entering years before 1753. Examples include 1933-04-12 or Mar 8, 2032. You can also type words like today or yesterday, or use the American format, 7/27/2017.
Number of days between:
and
I'm feeling lucky, show me a random date difference.
# ››Date calculator
This site provides an online date calculator to help you find the difference in the number of days between any two calendar dates. Simply enter the start and end date to calculate the duration of any event. You can also use this tool to determine how many days have passed since your birthday, or measure the amount of time until your baby's due date. The calculations use the Gregorian calendar, which was created in 1582 and later adopted in 1752 by Britain and the eastern part of what is now the United States. For best results, use dates after 1752 or verify any data if you are doing genealogy research. Historical calendars have many variations, including the ancient Roman calendar and the Julian calendar. Leap years are used to match the calendar year with the astronomical year. If you're trying to figure out the date that occurs in X days from today, switch to the Days From Now calculator instead. | 953 | 3,314 | {"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} | 3.203125 | 3 | CC-MAIN-2017-30 | latest | en | 0.95457 |
https://sciencedocbox.com/Chemistry/78696908-Why-all-the-repeating-why-all-the-repeating-why-all-the-repeating-why-all-the-repeating.html | 1,621,137,165,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00534.warc.gz | 529,151,185 | 25,946 | # Why all the repeating Why all the repeating Why all the repeating Why all the repeating
Size: px
Start display at page:
Download "Why all the repeating Why all the repeating Why all the repeating Why all the repeating"
Transcription
1 Why all the repeating Why all the repeating Why all the repeating Why all the repeating
2 Patterns What Patterns have you observed in your life?
3 Where to Get Help If you don t understand concepts in chapter 6, what can you do? Read your book Youtube or the internet A friend in class or your parents Last resort, come talk to me.
4 History of Elements First 10 elements to be discovered- Copper-9000 BC Lead-7000 BC Gold 6000 BC Silver BC Iron BC Carbon-3750 BC Tin-3500 BC Sulfur-2000 BC Mercury-2000 BC Zinc-1000 BC
5 What patterns do you see in the periodic table.
6 History of the Periodic 1700 s Lavoisier complied a list of all known elements. It contained 33 elements 1870 There were 60 elements. With all the elements, scientists needed to organize them. Organization is the key to success -Bailey Table
7 Develop of the Periodic John Newlands- He proposed a scheme to organize the known elements. Arranged elements by increasing atomic mass. He noticed the properties repeated every 8 th element. Law of Octaves Table
8 Newton s Law of Octaves
9 Development of the Dmiti Mendeleev- Russian Chemist Demonstrated a connection between atomic mass and the properties of elements. Arranged elements in order of increasing atomic mass into columns with similar properties. Periodic Table
10 Mendeleev s P-Table
11 Development of the Henry Mosely- Periodic Table Noticed that Medeleev s table wasn t totally correct. Discovered that each element had a unique number of protons. # of protons is equal to the atomic number.
12 Development of the Mosley arranged the elements based on increasing atomic number and grouped elements together that had the same properties. Periodic Law- There is a periodic repetition of chemical and physical properties of elements when they are arranged by increasing atomic number. Periodic Table
13 Modern Periodic Table Arrangement- Groups- 18 groups Columns of elements Periods- 7 Periods Families or rows Representative elements- Groups 1, 2, Transition Elements- Groups 3-12
14 Example- Element Symbol
15 Modern Periodic Table Broken into Four groups- Metals Transition Metals Inner Transition Non-metals Halogens Noble gases Metalloids
16 Metals Al Ga In Sn Bi Cn
17 Modern Periodic Table Metals- Solid at room temp. Good conductors of heat and electricity. Malleable and ductile. Groups Alkali Metals Group 1 Metal, except Hydrogen Alkaline Earth Metals Group 2 metals Found in the crust of the earth.
18 Alkali Metals Li Na K Rb Cs Fr
19 Alkaline Earth Metals Li Na K Rb Cs Fr Be Mg Ca Sr Ba Ra
20 Modern Periodic Table Transition/Inner transition metals- Transition Metals- d-block metals Less reactive than Group I and II Inner transition metals- f-block metals Lanthanide series (4f)-rare earth metals. Actinide series (5f)-Actinium All radioactive
21 Transition Metals Li Be Na Mg Al K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn
22 Inner Transitional Metals-Lanthanide Li Be Na Mg Al K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
23 Inner Transitional-Actinide Li Be Na Mg Al K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
24 Modern Periodic Table Nonmetals- Gases or brittle Poor conductors of heat and electricity Group 17 Halogens Always found in compounds React vigorously with metals Group 18- Noble Gases Nonreactive Full s and p orbitals 8 valence electrons
25 Nonmetals Al Ga In Sn Bi Cn
26 Halogens Li Be F Na Mg Al Cl K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Br Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn I Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
27 Nonmetals H Li Be C N O F Na Mg Al P S Cl K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Se Br Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn I Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
28 Noble Gas H He Li Be C N O F Ne Na Mg Al P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
29 Modern Periodic Table Metalloids- Have physical properties of both metals and nonmetals. Properties lie between metals and nonmetals B, Si, Ge, As, Sb, Te, Po, and At
30 Metalloids H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
31 Temporary elements-yet to be named H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Uut UUq UUp Uuh UUo Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
32 I would classify elements as very elegant!
33 Valence Electrons Electrons in the highest principle energy level. Group I Elements: Each has one valence electron. ns 1 All share the same chemical properties. This is true for Group II and Groups Try it-
34 Hydrogen and Helium Hydrogen has an electron configuration of 1s 1, but despite the ns 1 configuration, it does not share the same properties of the elements in of Group 1 Hydrogen is a unique element. Like Group 2 elements, helium has an ns 2 group configuration. Yet it is part of Group 18. Because its highest occupied energy level is filled by two electrons, helium possesses special chemical stability.
35 Valence Electrons and Period The energy level of an elements valence electrons indicates the period in which it is found. Mg valence electrons are in the 3 rd energy level, Mg is in the 3 rd period. Draw the electron configuration of Mg. Compare this to that of Gold. Compare this to Oxygen. Is this correct?
36 Electron Configuration and the Periodic Table Relationship Between Periodicity and Electron Configuration
37 What elements is this? Practice First, write down the principle energy level. [Ne]3s 2 3p 2 [Rn]7s 1 [Kr]5s 2 4d 10 5p 4 [He]2s 2
38 Group I and II S-Block Elements Very reactive metals Always found in compounds. Group II metals are denser and have a higher melting point.
39 P-Block Elements Properties vary greatly in the P-block: Halogen-very reactive Metalloids/Noble Gases-nonreactive Metals are harder and denser than s-block, but softer and less dense than d-block metals
40 d-block Elements Less reactive than other metals. Do not usually from compounds. Do not always follow the Aufbau principle- Ni= [Ar]4s 2 3d 8 [Kr]5s 0 4d 10 Pd= Pt= [Xe]6s 1 4f 14 5d 9
41 f-block Elements Wedged between the 6 th and 7 th period. Lanthanide are shiny and have similar reactivity as Group II metal. The first 4 actinides are natural, the rest man made.
42 The Trends are looking UP!
43 Periodic Trends There are six trends that we will be looking at: Atomic Radii Ionization Energy Electron Affinity Ionic Radii Valence Electrons Electronegativity
44 Atomic Radii Periodic Trends Measurement between two nuclei of two atoms that are chemically bonded together. ½ of the above distance See the picture.
45 Periodic Trends Atomic nuclei- The trend is a increase from right to left, and from top to bottom. Smallest is He Biggest is Fr The trend is caused by an increase in the positive charge of the nucleus. Draw the trend-
46 Periodic Trends What is up with Al and Ga? What comes before Ga? 3d 10
47 Periodic Trends
48 Periodic Trends Ionization Energy- Removing an electron from an atom. A + energy = A + + e - ( Requires energy) Ion Negative or positived charged atom Energy to remove the 1 st electron-ie 1 Increase from left to right, and from bottom to top Easiest is Fr Hardest is He Draw the trend-
49 Periodic Trends Note: IE 1 compared to IE 2 What Trend do you see?
50 Periodic Trends
51 Periodic Trends Electron Affinity- An electron acquired by a neutral atom. A + e - = A - + energy (Release energy) Cl + e - = Cl - + energy Trend is- Increase from left to right, increases from bottom to top. Group 17 loves electrons Group 1 hates them Group 18 has an electron affinity of 0 Draw the trend-
52 Periodic Trends
53 Periodic Trends Ionic Radii- Atoms that have gained or lost electrons; Cation-lost an electron Na + Ca ++ Anion-gained an electron O - - F - Trend: Hold on! Draw the trend-
54 Periodic Trends
55 Electron Configuration and Periodic Properties Summary i n c r e a s e s increases Ionic Radii Increase Ionic Radii increase as you move diagonally from Fluorine to Francium
56 Periodic Trends Valence electrons- Compounds are formed when: Electrons are lost Electrons are gained Electrons are shared These electrons are called valence electrons.
57 Valence Electrons Group Number Group Configuration # of Valence electrons 1 ns ns ns 2 p ns 2 p ns 2 p ns 2 p ns 2 p ns 2 p 6 8
58 Periodic Trends Electronegativity- Measure of the ability of an atom in a chemical compound to attract electrons. Most electronegative element (Fluorine) is given a 4 Trend-increases from bottom to top, increases from left to right.
59 Periodic Trends
60 Periodic Trends
61 Periodic Trends
62 Periodic Trends i n c r e a s e s increases Electronegativity Electronegativity increases as you move diagonally from Francium to Fluorine Fluorine-Most electronegative element Francium-Least electronegative element
63 Electron Configuration and Periodic Properties Summary Hints: Metals Nonmetals IE LOW HIGH EN LOW HIGH Metals lose electrons and don t want them Nonmetals gain electrons and want them very much.
### Atoms and the Periodic Table
Atoms and the Periodic Table Parts of the Atom Proton Found in the nucleus Number of protons defines the element Charge +1, mass 1 Parts of the Atom Neutron Found in the nucleus Stabilizes the nucleus
More information
### The Periodic Table of Elements
The Periodic Table of Elements 8 Uuo Uus Uuh (9) Uup (88) Uuq (89) Uut (8) Uub (8) Rg () 0 Ds (9) 09 Mt (8) 08 Hs (9) 0 h () 0 Sg () 0 Db () 0 Rf () 0 Lr () 88 Ra () 8 Fr () 8 Rn () 8 At (0) 8 Po (09)
More information
### DO NOW: Retrieve your projects. We will be reviewing them again today. Textbook pg 23, answer questions 1-3. Use the section 1.2 to help you.
DO NOW: Retrieve your projects. We will be reviewing them again today. Textbook pg, answer questions. Use the section. to help you. Chapter test is FRIDAY. The Periodic Table of Elements 8 Uuo Uus Uuh
More information
### Made the FIRST periodic table
Made the FIRST periodic table 1869 Mendeleev organized the periodic table based on the similar properties and relativities of certain elements Later, Henri Moseley organized the elements by increasing
More information
### 9/20/2017. Elements are Pure Substances that cannot be broken down into simpler substances by chemical change (contain Only One Type of Atom)
CAPTER 6: TE PERIODIC TABLE Elements are Pure Substances that cannot be broken down into simpler substances by chemical change (contain Only One Type of Atom) The Periodic Table (Mendeleev) In 1872, Dmitri
More information
### 6.3 Classifying Elements with the Periodic Table
6.3 Classifying Elements with the Periodic Table The Periodic Table was developed by scientists to organize elements in such a way as to make sense of the growing information about their properties. The
More information
### Using the Periodic Table
MATH SKILLS TRANSPARENCY WORKSHEET Using the Periodic Table 6 Use with Chapter 6, Section 6.2 1. Identify the number of valence electrons in each of the following elements. a. Ne e. O b. K f. Cl c. B g.
More information
### Advanced Chemistry. Mrs. Klingaman. Chapter 5: Name:
Advanced Chemistry Mrs. Klingaman Chapter 5: The Periodic Law Name: _ Mods: Chapter 5: The Periodic Law Reading Guide 5.1 History of the Periodic Table (pgs. 125-129) 1) What did Dimitri Mendeleev notice
More information
### The Periodic Table. Periodic Properties. Can you explain this graph? Valence Electrons. Valence Electrons. Paramagnetism
Periodic Properties Atomic & Ionic Radius Energy Electron Affinity We want to understand the variations in these properties in terms of electron configurations. The Periodic Table Elements in a column
More information
### A little history. When and How? Sir William Ramsey. ü 12/5/13. ü 1. Who put together the first useable Periodic Table?
ü // A little history Johahann Dobereiner (80-89) o Triads John Newlands (8-898) o Law of Octaves Who put together the first useable ic Table? Mendeleev you remember him right? When and How? You know it
More information
### What is the periodic table?
The periodic table of the elements represents one of the greatest discoveries in the history of science that certain elements, the basic chemical substances from which all matter is made, resemble each
More information
### SCIENCE 1206 UNIT 2 CHEMISTRY. September 2017 November 2017
SCIENCE 1206 UNIT 2 CHEMISTRY September 2017 November 2017 UNIT OUTLINE 1. Review of Grade 9 Terms & the Periodic Table Bohr diagrams Evidence for chemical reactions Chemical Tests 2. Naming & writing
More information
### Name Unit 4: Periodic Table Period. Unit 4 Vocabulary.Due Test Day
Name Unit 4: Periodic Table Period 1. History and Language of the Periodic Table 2. Identifying PROPERTIES OF METALS, METALLOIDS, & NONMETALS 3. Identifying GROUP PROPERTIES 4. Classifying elements 5.
More information
### The Periodic Table of the Elements
The Periodic Table of the Elements All matter is composed of elements. All of the elements are composed of atoms. An atom is the smallest part of an element which still retains the properties of that element.
More information
### MANY ELECTRON ATOMS Chapter 15
MANY ELECTRON ATOMS Chapter 15 Electron-Electron Repulsions (15.5-15.9) The hydrogen atom Schrödinger equation is exactly solvable yielding the wavefunctions and orbitals of chemistry. Howev er, the Schrödinger
More information
### Radiometric Dating (tap anywhere)
Radiometric Dating (tap anywhere) Protons Neutrons Electrons Elements on the periodic table are STABLE Elements can have radioactive versions of itself called ISOTOPES!! Page 1 in your ESRT has your list!
More information
### The Periodic Law Notes (Chapter 5)
The Periodic Law Notes (Chapter 5) I. History of the Periodic Table About 70 elements were known by 1850 (no noble gases) but there didn t appear to be a good way of arranging or relating them to study.
More information
### Chapter 12 The Atom & Periodic Table- part 2
Chapter 12 The Atom & Periodic Table- part 2 Electrons found outside the nucleus; negatively charged Protons found in the nucleus; positive charge equal in magnitude to the electron s negative charge Neutrons
More information
### Nucleus. Electron Cloud
Atomic Structure I. Picture of an Atom Nucleus Electron Cloud II. Subatomic particles Particle Symbol Charge Relative Mass (amu) protons p + +1 1.0073 neutrons n 0 1.0087 electrons e - -1 0.00054858 Compare
More information
### CHEM 10113, Quiz 5 October 26, 2011
CHEM 10113, Quiz 5 October 26, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
More information
### Modified from: Larry Scheffler Lincoln High School IB Chemistry 1-2.1
Modified from: Larry Scheffler Lincoln High School IB Chemistry 1-2.1 The development of the periodic table brought a system of order to what was otherwise an collection of thousands of pieces of information.
More information
### 1 Genesis 1:1. Chapter 10 Matter. Lesson. Genesis 1:1 In the beginning God created the heavens and the earth. (NKJV)
1 Genesis 1:1 Genesis 1:1 In the beginning God created the heavens and the earth. (NKJV) 1 Vocabulary Saturated having all the solute that can be dissolved at that temperature Neutron a particle with no
More information
### Solutions and Ions. Pure Substances
Class #4 Solutions and Ions CHEM 107 L.S. Brown Texas A&M University Pure Substances Pure substance: described completely by a single chemical formula Fixed composition 1 Mixtures Combination of 2 or more
More information
### Instructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More information
### Atomic Structure & Interatomic Bonding
Atomic Structure & Interatomic Bonding Chapter Outline Review of Atomic Structure Atomic Bonding Atomic Structure Atoms are the smallest structural units of all solids, liquids & gases. Atom: The smallest
More information
### Essential Chemistry for Biology
1 Chapter 2 Essential Chemistry for Biology Biology and Society: More Precious than Gold A drought is a period of abnormally dry weather that changes the environment and one of the most devastating disasters.
More information
### Chemistry 431 Practice Final Exam Fall Hours
Chemistry 431 Practice Final Exam Fall 2018 3 Hours R =8.3144 J mol 1 K 1 R=.0821 L atm mol 1 K 1 R=.08314 L bar mol 1 K 1 k=1.381 10 23 J molecule 1 K 1 h=6.626 10 34 Js N A = 6.022 10 23 molecules mol
More information
### Topic 3: Periodicity OBJECTIVES FOR TODAY: Fall in love with the Periodic Table, Interpret trends in atomic radii, ionic radii, ionization energies &
Topic 3: Periodicity OBJECTIVES FOR TODAY: Fall in love with the Periodic Table, Interpret trends in atomic radii, ionic radii, ionization energies & electronegativity The Periodic Table What is the periodic
More information
### K. 27 Co. 28 Ni. 29 Cu Rb. 46 Pd. 45 Rh. 47 Ag Cs Ir. 78 Pt.
1 IA 1 H Hydrogen 1.01 Atomic number Element symbol Element name Atomic mass VIIIA 1 H 1.01 IIA IIIA IVA VA VIA VIIA 2 He 4.00 Metalloids 3 Li 6.94 4 Be 9.01 5 B 10.81 6 C 12.01 7 N 14.01 8 O 16.00 9 F
More information
### Ch. 9 NOTES ~ Chemical Bonding NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics.
Ch. 9 NOTES ~ Chemical Bonding NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. I. Review: Comparison of ionic and molecular compounds Molecular compounds Ionic
More information
### Secondary Support Pack. be introduced to some of the different elements within the periodic table;
Secondary Support Pack INTRODUCTION The periodic table of the elements is central to chemistry as we know it today and the study of it is a key part of every student s chemical education. By playing the
More information
### Discovery of Elements. Dmitri Mendeleev Stanislao Canizzaro (1860) Modern Periodic Table. Henry Moseley. PT Background Information
Discovery of Elements Development of the Periodic Table Chapter 5 Honors Chemistry 412 At the end of the 1700 s, only 30 elements had been isolated Included most currency metals and some nonmetals New
More information
### Element Cube Project (x2)
Element Cube Project (x2) Background: As a class, we will construct a three dimensional periodic table by each student selecting two elements in which you will need to create an element cube. Helpful Links
More information
### Lab Day and Time: Instructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More information
### If anything confuses you or is not clear, raise your hand and ask!
CHM 1045 Dr. Light s Section December 10, 2002 FINAL EXAM Name (please print) Recitation Section Meeting Time This exam consists of six pages. Make sure you have one of each. Print your name at the top
More information
### CHM 101 PRACTICE TEST 1 Page 1 of 4
CHM 101 PRACTICE TEST 1 Page 1 of 4 Please show calculations (stuffed equations) on all mathematical problems!! On the actual test, "naked answers, with no work shown, will receive no credit even if correct.
More information
### CHEM 130 Exp. 8: Molecular Models
CHEM 130 Exp. 8: Molecular Models In this lab, we will learn and practice predicting molecular structures from molecular formulas. The Periodic Table of the Elements IA 1 H IIA IIIA IVA VA VIA VIIA 3 5
More information
### Lab Day and Time: Instructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More information
### Chapter 3: Elements and Compounds. 3.1 Elements
Chapter 3: Elements and Compounds 3.1 Elements An element is a fundamental substance that cannot be broken down by chemical or physical methods to simpler substances. The 118 known elements are nature
More information
### Periodicity & Many-Electron Atoms
Chap. 8 ELECTRON CONFIGURAT N & CEMICAL PERIODICITY 8.1-8.2 Periodicity & Many-Electron Atoms Understand the correlation of electron configuration and the periodic character of atomic properties such as
More information
### PERIODIC TABLE OF THE ELEMENTS
Useful Constants and equations: K = o C + 273 Avogadro's number = 6.022 x 10 23 d = density = mass/volume R H = 2.178 x 10-18 J c = E = h = hc/ h = 6.626 x 10-34 J s c = 2.998 x 10 8 m/s E n = -R H Z 2
More information
### Periodic Table. Modern periodic table
41 Periodic Table - Mendeleev (1869): --- When atoms are arranged in order of their atomic weight, some of their chemical and physical properties repeat at regular intervals (periods) --- Some of the physical
More information
### (C) Pavel Sedach and Prep101 1
(C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 1 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 2 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach and Prep101 3 (C) Pavel Sedach
More information
### Lab Day and Time: Instructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More information
### CHEMICAL COMPOUNDS MOLECULAR COMPOUNDS
48 CHEMICAL COMPOUNDS - Dalton's theory does not mention this, but there is more than one way for atoms to come together to make chemical compounds! - There are TWO common kinds of chemical compound, classified
More information
### Atomic terms. Example: Helium has an atomic number of 2. Every helium atom has two protons in its nucleus.
Atomic terms - ATOMIC NUMBER: The number of protons in the atomic nucleus. Each ELEMENT has the SAME NUMBER OF PROTONS in every nucleus. In neutral atoms, the number of ELECTRONS is also equal to the atomic
More information
### Last 4 Digits of USC ID:
Chemistry 05 B Practice Exam Dr. Jessica Parr First Letter of last Name PLEASE PRINT YOUR NAME IN BLOCK LETTERS Name: Last 4 Digits of USC ID: Lab TA s Name: Question Points Score Grader 8 2 4 3 9 4 0
More information
### 1 Arranging the Elements
CHAPTER 12 1 Arranging the Elements SECTION The Periodic Table BEFORE YOU READ After you read this section, you should be able to answer these questions: How are elements arranged on the periodic table?
More information
### Periodic Table. Modern periodic table
41 Periodic Table - Mendeleev (1869): --- When atoms are arranged in order of their atomic weight, some of their chemical and physical properties repeat at regular intervals (periods) --- Some of the physical
More information
### Periodic Table. - Mendeleev was able to predict the properties of previously unknown elements using his "periodic law" Modern periodic table
74 Periodic Table - Mendeleev (1869): --- When atoms are arranged in order of their atomic weight, some of their chemical and physical properties repeat at regular intervals (periods) --- Some of the physical
More information
### NAME: FIRST EXAMINATION
1 Chemistry 64 Winter 1994 NAME: FIRST EXAMINATION THIS EXAMINATION IS WORTH 100 POINTS AND CONTAINS 4 (FOUR) QUESTIONS THEY ARE NOT EQUALLY WEIGHTED! YOU SHOULD ATTEMPT ALL QUESTIONS AND ALLOCATE YOUR
More information
### 8. Relax and do well.
CHEM 1314 3;30 pm Theory Exam III John III. Gelder November 13, 2002 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 8 different pages. The last page include a periodic
More information
### Reporting Category 1: Matter and Energy
Name: Science Teacher: Reporting Category 1: Matter and Energy Atoms Fill in the missing information to summarize what you know about atomic structure. Name of Subatomic Particle Location within the Atom
More information
### 1 Electrons and Chemical Bonding
CHAPTER 13 1 Electrons and Chemical Bonding SECTION Chemical Bonding BEFORE YOU READ After you read this section, you should be able to answer these questions: What is chemical bonding? What are valence
More information
### Example: Helium has an atomic number of 2. Every helium atom has two protons in its nucleus.
59 Atomic terms - ATOMIC NUMBER: The number of protons in the atomic nucleus. Each ELEMENT has the SAME NUMBER OF PROTONS in every nucleus. In neutral atoms, the number of ELECTRONS is also equal to the
More information
### Chem Exam 1. September 26, Dr. Susan E. Bates. Name 9:00 OR 10:00
Chem 1711 Exam 1 September 26, 2013 Dr. Susan E. Bates Name 9:00 OR 10:00 N A = 6.022 x 10 23 mol 1 I A II A III B IV B V B VI B VII B VIII I B II B III A IV A V A VI A VII A inert gases 1 H 1.008 3 Li
More information
### Part 2. Multiple choice (use answer card). 90 pts. total. 3 pts. each.
1 Exam I CHEM 1303.001 Name (print legibly) Seat no. On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Part 1. Nomenclature. 10 pts. total. 2 pts. each. Fill in
More information
### CHEMICAL COMPOUNDS MOLECULAR COMPOUNDS
48 CHEMICAL COMPOUNDS - Dalton's theory does not mention this, but there is more than one way for atoms to come together to make chemical compounds! - There are TWO common kinds of chemical compound, classified
More information
### Guide to the Extended Step-Pyramid Periodic Table
Guide to the Extended Step-Pyramid Periodic Table William B. Jensen Department of Chemistry University of Cincinnati Cincinnati, OH 452201-0172 The extended step-pyramid table recognizes that elements
More information
### CHEM 172 EXAMINATION 1. January 15, 2009
CHEM 17 EXAMINATION 1 January 15, 009 Dr. Kimberly M. Broekemeier NAME: Circle lecture time: 9:00 11:00 Constants: c = 3.00 X 10 8 m/s h = 6.63 X 10-34 J x s J = kg x m /s Rydberg Constant = 1.096776 x
More information
### NUCLEAR MODEL. Electron cloud. Electron cloud. Nucleus. Nucleus
37 NUCLEAR MODEL - Atoms are mostly empty space - NUCLEUS, at the center of the atom, contains protons and neutrons. This accounts for almost all the mass of an atom - Electrons are located in a diffuse
More information
### Atomic structure. The subatomic particles. - a small, but relatively massive particle that carres an overall unit POSITIVE CHARGE
35 Atomic structure - Until the early 20th century, chemists considered atoms to be indivisible particles. - The discovery of SUBATOMIC PARTICLES changed the way we view atoms! PROTON NEUTRON ELECTRON
More information
### 8. Relax and do well.
CHEM 1014 Exam I John I. Gelder September 16, 1999 Name TA's Name Lab Section Please sign your name below to give permission to post your course scores on homework, laboratories and exams. If you do not
More information
### Fall 2011 CHEM Test 4, Form A
Fall 2011 CHEM 1110.40413 Test 4, Form A Part I. Multiple Choice: Clearly circle the best answer. (60 pts) Name: 1. The common constituent in all acid solutions is A) H 2 SO 4 B) H 2 C) H + D) OH 2. Which
More information
### 7. Relax and do well.
CHEM 1014 Exam III John III. Gelder November 18, 1999 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 7 different pages. The last page includes a periodic table and
More information
lectures accompanying the book: Solid State Physics: An Introduction, by Philip ofmann (2nd edition 2015, ISBN-10: 3527412824, ISBN-13: 978-3527412822, Wiley-VC Berlin. www.philiphofmann.net 1 Bonds between
More information
### INSTRUCTIONS: Exam III. November 10, 1999 Lab Section
CHEM 1215 Exam III John III. Gelder November 10, 1999 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 7 different pages. The last page includes a periodic table and
More information
### Honors Chemistry Unit 4 ( )
Honors Chemistry Unit 4 (2017-2018) Families (research and present) Metals/nonmetals Trends o Atomic radius o Electronegativity o Ionization energy o Metallic and nonmetallic character Review Ions Oxidation
More information
### 1. Following Dalton s Atomic Theory, 2. In 1869 Russian chemist published a method. of organizing the elements. Mendeleev showed that
20 CHEMISTRY 11 D. Organizing the Elements The Periodic Table 1. Following Dalton s Atomic Theory, By 1817, chemists had discovered 52 elements and by 1863 that number had risen to 62. 2. In 1869 Russian
More information
### VIIIA He IIA IIIA IVA VA VIA VIIA. Li Be B C N O F Ne. Na Mg VIB VIIB VIIIB IB IIB S. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
188 THE FIRST TWO PERIODIC TRENDS IN A NUTSHELL LARGER IONIZATION ENERGY SMALLER RADIUS IA H IIA IIIA IVA VA VIA VIIA VIIIA He Li Be B C N O F Ne Na Mg IIIB IVB VB Al Si P VIB VIIB VIIIB IB IIB S Cl Ar
More information
### 4.1 Atomic structure and the periodic table. GCSE Chemistry
4.1 Atomic structure and the periodic table GCSE Chemistry All substances are made of atoms this is cannot be chemically broken down it is the smallest part of an element. Elements are made of only one
More information
### POLYTECHNIC OF NAMIBIA
POLYTECHNIC OF NAMIBIA DEPARTMENT OF HEALTH SCIENCES BACHELOR OF ENVIRONMENTAL HEALTH SCIENCES HEALTH SCIENCE CHEMISTRY (HSC 511S) NQF level 5 SECOND OPPORTUNITY EXAMINATION November 2014 TIME: MARKS:
More information
### HANDOUT SET GENERAL CHEMISTRY I
HANDOUT SET GENERAL CHEMISTRY I Periodic Table of the Elements 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 IA VIIIA 1 2 H He 1.00794 IIA IIIA IVA VA VIA VIIA 4.00262 3 Li 6.941 11 Na 22.9898
More information
### 7. Relax and do well.
CHEM 1215 Exam II John II. Gelder October 7, 1998 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 5 different pages. The last page includes a periodic table and a solubility
More information
### 8. Relax and do well.
CHEM 1225 Exam I John I. Gelder February 4, 1999 Name KEY TA's Name Lab Section Please sign your name below to give permission to post your course scores on homework, laboratories and exams. If you do
More information
### 8. Relax and do well.
CHEM 1215 Exam III John III. Gelder November 11, 1998 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 7 different pages. The last page includes a periodic table and
More information
### K. 27 Co. 28 Ni. 29 Cu Rb. 46 Pd. 45 Rh. 47 Ag Cs Ir. 78 Pt.
1 IA 1 ydrogen 1.01 Atomic number Element symbol Element name Atomic mass VIIIA 1 1.01 IIA IIIA IVA VA VIA VIIA 2 e 4.00 Metalloids 3 Li 6.94 4 Be 9.01 5 B 10.81 6 C 12.01 7 N 14.01 8 O 16.00 9 F 19.00
More information
### Putting it together... - In the early 20th century, there was a debate on the structure of the atom. Thin gold foil
36 Putting it together... - In the early 20th century, there was a debate on the structure of the atom. RUTHERFORD EXPERIMENT Where do the particles go? Radioactive material A few bounce back A few particles
More information
### Atomic weight: This is a decimal number, but for radioactive elements it is replaced with a number in parenthesis.
47 Blocks on the periodic table 11 Sodium 22.99 Atomic number: This is always a whole number. The periodic table is arranged by atomic number! Element symbol: A one or two letter abbreviation for the name
More information
### (please print) (1) (18) H IIA IIIA IVA VA VIA VIIA He (2) (13) (14) (15) (16) (17)
CHEM 10113, Quiz 3 September 28, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
More information
### 1 Arranging the Elements
CHAPTER 11 1 Arranging the Elements SECTION The Periodic Table BEFORE YOU READ After you read this section, you should be able to answer these questions: How are elements arranged on the periodic table?
More information
### (FIRST) IONIZATION ENERGY
181 (FIRST) IONIZATION ENERGY - The amount of energy required to remove a single electron from the outer shell of an atom. - Relates to reactivity for metals. The easier it is to remove an electron, the
More information
### Lab Day and Time: Instructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More information
### Chemistry Standard level Paper 1
Chemistry Standard level Paper 1 Thursday 12 May 2016 (morning) 45 minutes Instructions to candidates Do not open this examination paper until instructed to do so. Answer all the questions. For each question,
More information
### CHEMICAL COMPOUNDS MOLECULAR COMPOUNDS
48 CHEMICAL COMPOUNDS - Dalton's theory does not mention this, but there is more than one way for atoms to come together to make chemical compounds! - There are TWO common kinds of chemical compound, classified
More information
### - Dalton's theory sets LIMITS on what can be done with chemistry. For example:
34 - Dalton's theory sets LIMITS on what can be done with chemistry. For example: Chemistry can't convert lead (an element) into gold (another element). Sorry, alchemists! You can't have a compound form
More information
### Atomic weight: This is a decimal number, but for radioactive elements it is replaced with a number in parenthesis.
47 Blocks on the periodic table 11 Sodium 22.99 Atomic number: This is always a whole number. The periodic table is arranged by atomic number! Element symbol: A one or two letter abbreviation for the name
More information
### נושא מס' 8: המבנה האלקטרוני של אטומים. Electronic Structure of Atoms. 1 Prof. Zvi C. Koren
נושא מס' 8: המבנה האלקטרוני של אטומים Electronic Structure of Atoms 1 Prof. Zvi C. Koren 19.07.10 The Electron Spin From further experiments, it was evident that the e had additional magnetic properties
More information
### 02/05/09 Last 4 Digits of USC ID: Dr. Jessica Parr
Chemistry 05 B First Letter of PLEASE PRINT YOUR NAME IN BLOCK LETTERS Exam last Name Name: 02/05/09 Last 4 Digits of USC ID: Dr. Jessica Parr Lab TA s Name: Question Points Score Grader 2 2 9 3 9 4 2
More information
### Speed of light c = m/s. x n e a x d x = 1. 2 n+1 a n π a. He Li Ne Na Ar K Ni 58.
Physical Chemistry II Test Name: KEY CHEM 464 Spring 18 Chapters 7-11 Average = 1. / 16 6 questions worth a total of 16 points Planck's constant h = 6.63 1-34 J s Speed of light c = 3. 1 8 m/s ħ = h π
More information
### - Some properties of elements can be related to their positions on the periodic table.
179 PERIODIC TRENDS - Some properties of elements can be related to their positions on the periodic table. ATOMIC RADIUS - The distance between the nucleus of the atoms and the outermost shell of the electron
More information
### Chapter 3: Stoichiometry
Chapter 3: Stoichiometry Chem 6A Michael J. Sailor, UC San Diego 1 Announcements: Thursday (Sep 29) quiz: Bring student ID or we cannot accept your quiz! No notes, no calculators Covers chapters 1 and
More information
### ORBITAL DIAGRAM - A graphical representation of the quantum number "map" of electrons around an atom.
178 (MAGNETIC) SPIN QUANTUM NUMBER: "spin down" or "spin up" - An ORBITAL (region with fixed "n", "l" and "ml" values) can hold TWO electrons. ORBITAL DIAGRAM - A graphical representation of the quantum
More information
### Unit 1 Part 2 Atomic Structure and The Periodic Table Introduction to the Periodic Table UNIT 1 ATOMIC STRUCTURE AND THE PERIODIC TABLE
UNIT 1 ATOMIC STRUCTURE AND THE PERIODIC TABLE PART 2 INTRODUCTION TO THE PERIODIC TABLE Contents 1. The Structure of the Periodic Table 2. Trends in the Periodic Table Key words: group, period, block,
More information
### ORBITAL DIAGRAM - A graphical representation of the quantum number "map" of electrons around an atom.
160 ORBITAL DIAGRAM - A graphical representation of the quantum number "map" of electrons around an atom. 4p 3d 4s 3p 3s 2p 2s 1s Each blank represents an ORBITAL, and can hold two electrons. The 4s subshell
More information
### 5 questions, 3 points each, 15 points total possible. 26 Fe Cu Ni Co Pd Ag Ru 101.
Physical Chemistry II Lab CHEM 4644 spring 2017 final exam KEY 5 questions, 3 points each, 15 points total possible h = 6.626 10-34 J s c = 3.00 10 8 m/s 1 GHz = 10 9 s -1. B= h 8π 2 I ν= 1 2 π k μ 6 P
More information
### FINAL EXAM April 26, 2004
CM 1045 (11:15 am Lecture) Dr. Light FINAL EXAM April 26, 2004 Name (please print) Check your recitation section: Sec. 21 5:30-6:20 pm (Popovic) Sec. 24 3:30-4:20 pm (Giunta) Sec. 22 6:30-7:20 pm (Popovic)
More information
### Chemistry 1 First Lecture Exam Fall Abbasi Khajo Levine Mathias Mathias/Ortiz Metlitsky Rahi Sanchez-Delgado Vasserman
Chemistry 1 First Lecture Exam Fall 2011 Page 1 of 9 NAME Circle the name of your recitation/lab instructor(s) Abbasi Khajo Levine Mathias Mathias/Ortiz Metlitsky Rahi Sanchez-Delgado Vasserman Before
More information
### - Some properties of elements can be related to their positions on the periodic table.
186 PERIODIC TRENDS - Some properties of elements can be related to their positions on the periodic table. ATOMIC RADIUS - The distance between the nucleus of the atoms and the outermost shell of the electron
More information
### Periods: horizontal rows (# 1-7) 2. Periodicity the of the elements in the same group is explained by the arrangement of the around the nucleus.
The Modern Periodic Table 1. An arrangement of the elements in order of their numbers so that elements with properties fall in the same column (or group). Groups: vertical columns (#1-18) Periods: horizontal
More information | 10,590 | 39,645 | {"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-2021-21 | longest | en | 0.882298 |
https://gmatclub.com/forum/geometry-gmat-questions-master-list-79792.html?sort_by_oldest=true | 1,498,376,463,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320443.31/warc/CC-MAIN-20170625064745-20170625084745-00121.warc.gz | 732,458,883 | 56,063 | It is currently 25 Jun 2017, 00:41
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Geometry GMAT Questions - Master List
Author Message
TAGS:
### Hide Tags
GMAT Club team member
Joined: 16 Mar 2009
Posts: 115
Location: Bologna, Italy
Geometry GMAT Questions - Master List [#permalink]
### Show Tags
17 Jun 2009, 08:12
8
KUDOS
17
This post was
BOOKMARKED
Geometry GMAT Questions
This thread is a part of the GMAT Diagnostic Test project. Feel free to suggest other GMAT questions that you feel are helpful
Theory and Formulas:
- Geometry section from Math formulas and shortcuts thread by Bandit
- Understanding Coordinate Geometry thread by Sumithra
Easy DS - 77245 (**)
Easy PS - 77697 (**)
Easy PS - 78840 (**)
Medium PS - 78117 (***)
Medium PS - 78429 (***)
Medium PS - 78413 (***)
Medium DS - 74168 (***)
Medium DS - 78878 (***)
Medium DS - 79105 (***)
Medium DS - 77047 (***)
Medium DS - 69309 (***)
Medium PS - 42105 (***)
HardPS - 44496 (****)
_____________________________________________________________________
Many other useful Math resources can be found through following thread:
_________________
My Recent Posts:
All GMAT CAT Practice Tests - links, prices, reviews
Review: GMATPrep (GMAT Prep) & PowerPrep (Power Prep) Tests
Manager
Joined: 18 May 2010
Posts: 80
Re: Geometry GMAT Questions - Master List [#permalink]
### Show Tags
10 Dec 2011, 04:38
Re: Geometry GMAT Questions - Master List [#permalink] 10 Dec 2011, 04:38
Similar topics Replies Last post
Similar
Topics:
3 The E-GMAT Question Series on Geometry 1 15 Dec 2016, 22:36
6 Useful Articles and Practice Questions to ace GMAT Geometry 1 10 Dec 2016, 00:55
2 Fractions GMAT Questions - Master List 3 31 Jan 2017, 19:21
9 Work and Rate GMAT Questions - Word Problems Master List 6 10 Nov 2016, 19:56
43 Number Properties GMAT Questions - Master List 7 31 Jan 2017, 03:31
Display posts from previous: Sort by | 655 | 2,421 | {"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-2017-26 | longest | en | 0.766483 |
https://manishvermaofficial.com/nios-class-10th-mathematics-211-solved-tma-in-english-medium-pdf-session-2022-23/ | 1,701,392,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00150.warc.gz | 439,397,798 | 20,651 | ## TMA Questions
(a) Puneet goes for morning walk to correct his BMI. He uses a fitness app to note down
Monday Tuesday Wednesday Thursday Friday Saturday Sunday
3500 3000 2700 3200 2300 2800 4000
his average steps per day whose results for a week are given in the table below.
i. Calculate the total steps taken by Puneet in a week?
ii. On which day of the week did he walk the most?
iii. How many steps did he walk on an average in a week?
) Refer to above table. A teacher told ten students to write polynomials on black board.
i. How many students wrote cubic polynomials?
ii. Divide the polynomial x2 −2x+1 by polynomial.
(a)On the number line take points O, A, B, C, D, E and F representing integer 0,1,2, −1, −2 and −2 −3 respectively. At B draw BB’ perpendicular to OB such that BB’ = 1 unit. Taking O as a center and radius OB’ draw an arc which intersect number line at P. Measure the length of OP.
(b)The arrow shows the path of a billiard ball from X to H. XY and ZH are parallel path. Find the value of angles a, b, c, d and e. Show your working process and give reasons.
Ques3:-Answer any one of the following questions in about 40- 60 words.
In a quadrilateral ABCD,AOand BO are the bisector of ∠A and ∠B respectively. Prove that∠ AOB =〖1/2〗(∠𝐶 + ∠𝐷)
(b)If x + 1/x =5, find the value of x3+1/x3
In figure AB ∥ CD, ∠A = 1550, ∠E = 1350 and ∠FCD = x0 , then find thevalue of x.
(b) In figure ABC is a triangle in which it is given that ∠ A = 800 , ∠ C = 600 , ∠B = 2×0 , and ∠ BDC = y0 . BD and CD bisects ∠ B and ∠C respectively. Calculate the value of x and y.
Ques :- 5Answer any one of the following questions in about 100-150 words:
In the figure, DE ∥BC and DF∥BE then prove thatEF/FR = CE/EB.
(b) AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB ∥ CD. If the distance between AB and CD is 3 cm, find the radius of circle.
Ques 6:- Prepare any one of the project out of two given below.
(a) Conduct the survey of 50 households from your locality/ village, regarding number of person in family and family income.
i.Present the data related to family members in tabular form mentioning frequencies.
ii.Calculate the average family size. How many families are above the average family size?
iii.Draw the Bar graph for top 10 earning families.
(b)Construct a rectangle ABCD whose side AB = DC = 6 cm and AD = BC = 2 cm.
(c)Construct a trapezium EFGH at a distance of 1 cm from both the vertices of the side DC whoseother parallel side is 2 cm and is at a height of 3 cm from the DC.
(d)Construct a square FIJG with the same dimension as the side FG.
(e) Join I and J to the point k situated at a height of 3 cm from the mid point of the side IJ. | 790 | 2,702 | {"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} | 3.609375 | 4 | CC-MAIN-2023-50 | latest | en | 0.926425 |
https://www.wyzant.com/resources/answers/5559/how_to_get_smallest_amount_from_this_math_problem_3_3_1_1_27_1_4_9_1_6_81_2_3_9_2_3_3_2_1_1_3_2_243_1_4 | 1,524,670,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947822.98/warc/CC-MAIN-20180425135246-20180425155246-00449.warc.gz | 919,323,710 | 12,878 | 0
# How to get smallest amount from this math problem? (.3)^3\1 (1\27)^1\4 (9)^1\6 (.81)^2\3 \ (.9)^2\3 (3)^-2\1(1\3) ^-2 (243) ^1\4
thank you very much.
Nish,
Hmmm...I don't see any variables, just numbers so this appears to be a single valued expression.
Also, the notation is ambiguous with operators and factor not being given any priorities relative to each other.
Try expressing this question more specifically and clearly.
Good Luck,
BruceS | 134 | 455 | {"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-2018-17 | longest | en | 0.862221 |
http://reference.iucr.org/mediawiki/index.php?title=Laue_equations&diff=next&oldid=1355 | 1,576,213,854,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00452.warc.gz | 120,446,791 | 6,272 | # Laue equations
(Difference between revisions)
Revision as of 12:10, 25 January 2006 (view source)← Older edit Revision as of 08:58, 27 February 2006 (view source)Newer edit → Line 10: Line 10: The three Laue equations give the conditions to be satisfied by an incident wave to be diffracted by a crystal. Consider the three basis vectors, '''OA''' = '''a''', '''OB''' = '''b''' , '''OC''' = '''c''' of the crystal and let '''so''' and '''sh''' be unit vectors along the incident and reflected directions, respectively. The conditions that the The three Laue equations give the conditions to be satisfied by an incident wave to be diffracted by a crystal. Consider the three basis vectors, '''OA''' = '''a''', '''OB''' = '''b''' , '''OC''' = '''c''' of the crystal and let '''so''' and '''sh''' be unit vectors along the incident and reflected directions, respectively. The conditions that the waves scattered by ''O'' and ''A'', ''O'' and ''B'', ''O'' and ''C'', respectively, be in phase are that waves scattered by ''O'' and ''A'', ''O'' and ''B'', ''O'' and ''C'', respectively, be in phase are that +
'''a''' . ('''sh''' - '''so''') = ''h'' λ '''a''' . ('''sh''' - '''so''') = ''h'' λ Line 16: Line 17: '''c''' . ('''sh''' - '''so''') = ''l'' λ '''c''' . ('''sh''' - '''so''') = ''l'' λ +
If these three conditions are simultaneously satisfied, the incoming wave is reflected on the set of lattice planes of Miller indices ''h/n'', ''k/n'', ''l/n''. ''h'', ''k'', ''l'' are the indices of the reflection. If these three conditions are simultaneously satisfied, the incoming wave is reflected on the set of lattice planes of Miller indices ''h/n'', ''k/n'', ''l/n''. ''h'', ''k'', ''l'' are the indices of the reflection. Line 21: Line 23: The three Laue equations can be generalized by saying that the diffraction condition is satisfied if the scalar product '''r''' . ('''sh'''/λ - '''so'''/λ) is an integer for any vector '''r''' = ''u'' '''a''' + ''v'' '''b''' + ''w'' '''c''' (''u'', ''v'', ''w'' integers) of the direct lattice. This is the case if The three Laue equations can be generalized by saying that the diffraction condition is satisfied if the scalar product '''r''' . ('''sh'''/λ - '''so'''/λ) is an integer for any vector '''r''' = ''u'' '''a''' + ''v'' '''b''' + ''w'' '''c''' (''u'', ''v'', ''w'' integers) of the direct lattice. This is the case if +
('''sh'''/λ - '''so'''/λ) = ''h'' '''a*''' + ''k'' '''b*''' + ''l'' '''c*''', ('''sh'''/λ - '''so'''/λ) = ''h'' '''a*''' + ''k'' '''b*''' + ''l'' '''c*''', - +
where ''h'', ''k'', ''l'' are integers, namely if the diffraction vector '''OH''' = '''sh,'''/λ - '''so'''/λ is a vector of the [[reciprocal lattice]]. where ''h'', ''k'', ''l'' are integers, namely if the diffraction vector '''OH''' = '''sh,'''/λ - '''so'''/λ is a vector of the [[reciprocal lattice]]. Line 30: Line 33: - === See also === + == See also == [[reciprocal lattice]]
[[reciprocal lattice]]
Line 38: Line 41: ---- ---- - [[Category:Fundamental crystallography]]
+ [[Category:X-rays]]
# Laue equations
### Other languages
Equations de Laue (Fr). Ecuaciones de Laue (Sp).
## Definition
The three Laue equations give the conditions to be satisfied by an incident wave to be diffracted by a crystal. Consider the three basis vectors, OA = a, OB = b , OC = c of the crystal and let so and sh be unit vectors along the incident and reflected directions, respectively. The conditions that the waves scattered by O and A, O and B, O and C, respectively, be in phase are that
a . (sh - so) = h λ
b . (sh - so) = k λ
c . (sh - so) = l λ
If these three conditions are simultaneously satisfied, the incoming wave is reflected on the set of lattice planes of Miller indices h/n, k/n, l/n. h, k, l are the indices of the reflection.
The three Laue equations can be generalized by saying that the diffraction condition is satisfied if the scalar product r . (sh/λ - so/λ) is an integer for any vector r = u a + v b + w c (u, v, w integers) of the direct lattice. This is the case if
(sh/λ - so/λ) = h a* + k b* + l c*,
where h, k, l are integers, namely if the diffraction vector OH = sh,/λ - so/λ is a vector of the reciprocal lattice.
## History
The three Laue conditions for diffraction were first given in Laue, M. von (1912). Eine quantitative Prüfung der Theorie für die Interferenz-Erscheinungen bei Röntgenstrahlen. Sitzungsberichte der Kgl. Bayer. Akad. der Wiss 363--373, reprinted in Ann. Phys. (1913), 41, 989 where he interpreted and indexed the first diffraction diagram (Friedrich, W., Knipping, P., and Laue, M. von (1912). Interferenz-Erscheinungen bei Röntgenstrahlen, Sitzungsberichte der Kgl. Bayer. Akad. der Wiss, 303--322, reprinted in Ann. Phys., (1913), 41, 971, taken with zinc-blende, ZnS. For details, see P. P. Ewald, 1962, IUCr, 50 Years of X-ray Diffraction, Section 4, page 52. | 1,419 | 4,864 | {"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} | 2.9375 | 3 | CC-MAIN-2019-51 | latest | en | 0.868433 |
https://www.coursehero.com/file/6641249/911-Mechanics-SolutionInstructors-SolManual-Mechanics-Materials-7ebook-Gere-light1/ | 1,526,923,746,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00558.warc.gz | 724,577,961 | 75,998 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
911_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1
# 911_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1
This preview shows page 1. Sign up to view the full content.
Problem 11.9-25 A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P 60 k. The bar has pinned supports and is made of alloy 2014-T6. (a) If the diameter d 2.0 in., what is the maximum allowable length L max of the bar? (b) If the length L 30 in., what is the minimum required diameter d min ? SECTION 11.9 Aluminum Columns 905 Solution 11.9-25 Aluminum bar Alloy 2014-T6 Pinned ends (a) F IND L max IF d 2.0 in. Assume L/r is less than 55: Eq. (11-84 a ): L max (50.43) r 25.2 in. ; Solve for L / r : L r 50.43 L r 6 55 ok or 19.10 30.7 0.23( L / r ) s allow 30.7 0.23( L / r ) ksi s allow P A 60 k 3.142 in. 2 19.10 ksi r A I A d 4 0.5 in. A p d 2 4 3.142 in. 2 I p d 4 64 ( K 1). P 60 k (b) F IND d min IF L 30 in. Assume L/r is greater than 55: L / r 120/ d 120/2.12 56.6 7 55 ok d 4 20.37 in. 4 d min 2.12 in. ; or 76.39 d 2 54,000 (120/ d ) 2 Eq.(11-84 b ): s allow 54,000 ksi ( L / r ) 2 s allow P A 60 k p d 2 /4 76.39 d 2 (ksi) A p d 2 4 r d r L r 30 in. d /4 120 in. d Problem 11.9-26 A solid round bar of aluminum having diameter
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: d (see figure) is compressed by an axial force P ± 175 kN. The bar has pinned supports and is made of alloy 2014-T6. (a) If the diameter d ± 40 mm, what is the maximum allowable length L max of the bar? (b) If the length L ± 0.6 m, what is the minimum required diameter d min ? ( Hint: Convert the given data to USCS units, determine the required quantities, and then convert back to SI units.) d Probs. 11.9-25 through 11.9-28 Solution 11.9-26 Aluminum bar Alloy 2014-T6 Pinned supports (a) F IND L max IF d ± 40 mm ± 1.575 in. A ± p d 2 4 ± 1.948 in. 2 I ± p d 4 64 ( K ± 1). P ± 175 kN ± 39.34 k s allow ± P A ± 39.34 k 1.948 in. 2 ± 20.20 ksi r ± A I A ± d 4 ± 0.3938 in. 11Ch11.qxd 9/27/08 2:27 PM Page 905...
View Full Document
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 1,034 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.702352 |
https://www.airmilescalculator.com/distance/kus-to-naq/ | 1,716,412,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00016.warc.gz | 551,174,350 | 10,787 | # How far is Qaanaaq from Kulusuk?
The distance between Kulusuk (Kulusuk Airport) and Qaanaaq (Qaanaaq Airport) is 1060 miles / 1706 kilometers / 921 nautical miles.
1060
Miles
1706
Kilometers
921
Nautical miles
2 h 30 min
155 kg
## Distance from Kulusuk to Qaanaaq
There are several ways to calculate the distance from Kulusuk to Qaanaaq. Here are two standard methods:
Vincenty's formula (applied above)
• 1060.071 miles
• 1706.018 kilometers
• 921.176 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 1056.158 miles
• 1699.721 kilometers
• 917.776 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Kulusuk to Qaanaaq?
The estimated flight time from Kulusuk Airport to Qaanaaq Airport is 2 hours and 30 minutes.
## Flight carbon footprint between Kulusuk Airport (KUS) and Qaanaaq Airport (NAQ)
On average, flying from Kulusuk to Qaanaaq generates about 155 kg of CO2 per passenger, and 155 kilograms equals 341 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path from Kulusuk to Qaanaaq
See the map of the shortest flight path between Kulusuk Airport (KUS) and Qaanaaq Airport (NAQ).
## Airport information
Origin Kulusuk Airport
City: Kulusuk
Country: Greenland
IATA Code: KUS
ICAO Code: BGKK
Coordinates: 65°34′24″N, 37°7′24″W
Destination Qaanaaq Airport
City: Qaanaaq
Country: Greenland
IATA Code: NAQ
ICAO Code: BGQQ
Coordinates: 77°29′18″N, 69°23′19″W | 502 | 1,746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.829215 |
https://artscolumbia.org/evolution-of-computers-essay-2-70529/ | 1,627,618,273,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00329.warc.gz | 124,758,659 | 15,466 | We use cookies to give you the best experience possible. By continuing we’ll assume you’re on board with our cookie policy
# Evolution of Computers Essay
How many inventions in your lifetime can you think of that have changed everything in our society today? Computers have taken over today’s society. From everyday tasks to moving satellites in space, PCs have revolutionized almost everything in our society.
Computers weren’t always this complicated though, and were around a long time before anyone even knew what the word “computer” meant. The Abacus was the first known machine developed to help perform mathematical equations. From what researchers have discovered it was invented around 500 to 600 BC in an area around China or Egypt. This early tool was used to perform addition and subtraction and can still be found used in some of today’s Middle Eastern cultures.
We will write a custom essay on Evolution of Computers specifically for you
for only \$16.38 \$13.9/page
Order now
In 650 AD the Hindus invented a written symbol for zero. Before this no true written calculations could be made, making this one of the most essential inventions to help computers. In 830 AD the first mathematics textbook was invented by a man named Mohammed Ibn Musa Abu Djefar. The subject of this textbook he wrote was “Al Gebr We’l Mukabala” which in today’s society is known as “Algebra” (History of Computers). So what does all of this have to do with computers? Well without numbers computers wouldn’t exist or have any reason to exist.
The whole point of a computer is to perform mathematical computations. Computers weren’t the first to do these mathematical calculations though. In 1623 AD Wilhelm Schickard invented “The Calculating Clock” which would perform operations like addition, subtraction, multiplication, and division. In the year 1801 Jacquard Loom devised a punch card system with a power loom and an automatic card reader. Later that century in 1890 Herman Hollerith invented a census calculator that put each person’s information on a punch card and sent it through an electrical/mechanical tabulating machine. This sped up the process from about 7 years to 2 years making this a very efficient method of performing a census, which in turn helped spread it around the world (History of Computers).
Jump to the year 1937 when John V. Atanasoff invented the first electronic computer. This computer and others below, unless otherwise stated, were made using vacuum tubes, “an electronic device in which conduction by electrons takes place through a vacuum within a sealed glass or metal container and which has various uses based on the controlled flow of electrons” (Dictionary. com). From 1941 to 1954 the governments of various countries started developing different computers for different purposes (Sandiego 1). The year of 1941 was a very important year for computers.
It marks the year the first fully functional program controlled computer was invented. This pc was developed in secret by Konrad Zuse and was called the Z3. It was the first to introduce the general architecture for today’s microprocessor. In the picture below the items seem simple, but at the time this was very advanced. This was the start of the true evolution of computers. After this, from 1943 to 1954, governments and research teams continued to pump out different computers.
The last of the vacuum tube computers was created in 1954, and was called the SAGE aircraft warning system. This was the largest vacuum tube computer system every built (Sandiego 1). These were all first generation computers. Second generation computers utilized transistors instead of vacuum tubes and were invented from around 1954-1959.
In 1950 NBS (The National Bureau of Standards) created SEAC (Standards Eastern Automatic Computer). This system used over 10,000 germanium diodes, germanium is a semiconductor that is more expensive than silicon but better suited and more efficient than silicon, and was used to solve over 50 unrelated scientific problems per day. In 1959 GE, General Electric Corporation, made an ERMA (Electronic Recording Machine Accounting) computer system for the Bank of America in California. This system introduced automation in banking, which later helped with the creation of ATMs (Sandiego 2).
READ: Computers and Ethics in the Workplace
The era of third generation computers was from 1959-1971 and they utilized ICs (Integrated Circuits) for these computers. In 1959 Jack Kilby, of Texas instruments, patented the first IC. The first commercial IC product was a hearing aid made and produced in 1963. IBM produced SABRE in 1964 for American Airlines.
It’s a tracking system for ticket reservations, which helped speed up the reservation process considerably. DEC was the creator of the first “mini-computer” called the PDP-8. It was one of the first mini-computers made in mass production that pretty much anyone could afford at the time. In 1969 the DOD, Department of Defense, developed the precursor to the internet which was called ARPANet (Advanced Research Projects Agency Network). This was an experimental WAN (Wide Area Network) that would survive a nuclear war. (Sandiego 3)Fourth generation PCs were the first to use microprocessors and were in the era of 1971 to 1987.
Gilbert Hyatt patented the microprocessor in 1971. Later that year Ted Hoff, of Intel, introduced new microprocessors to use in calculators. IBM arrived with the first 8 in. floppy disk. They also started using these microprocessors in unison with LCD screens in calculators and watches.
November of that year Intel introduced the first microcomputer to the public called the MCS-4. In 1972 Nolan Bushnell introduced the “arcade game” to the public with “Computer Space. ” Later that year he also created and introduced Atari and the game “Pong” to the public which became the beginnings of today’s “Video Gamer. ” The next year in 73′ IBM developed the first hard disk drive utilizing two 30MB platters.
Two years later they started selling Altair personal computers. This is when Steve Jobs and Steve Wozniak became influenced and developed the Apple personal computer. Also within the same year the 5. 25 in. floppy disk was introduced (Sandiego 4). It seems that computers began to progress exponentially.
In 1977 Nintendo started creating video games on plastic cartridges with chips inside. Throughout the 80’s typewriters were slowly being replaced with PCs and word processing software was created to help this process along. In 1980 Microsoft signed a contract with IBM to supply their PCs with an OS (Operating System). In 1984 Steve Jobs and his company Apple created the Macintosh personal computer, which is still to this day a popular alternative to PCs (Sandiego 4). Fifth generations PCs are present day and beyond computers.
The WWW (World Wide Web) was created in 1991 by Tim Berners-Lee. In 1993 the first web browser called Mosaic was created. Later on two more browsers took over the internet browsing industry. These were called Netscape Navigator (which was free) and Microsoft’s Internet explorer (Sandiego 5). These allow people to communicate their views around the world and express themselves without fear of being outcast.
After this, computers still progress with newer standards and updates to standards being released every year. This will continue until performance increases level off and consumers start to leave the computer market. Throughout history computers have affected everything. From international commerce to international security computers make everything simpler.
They make it easy to communicate with others and even help when researching projects for school. Overall the good points for computers and the internet outweigh most downsides. The computer has changed most things in the world today whether bad or good. Computers are not done evolving yet and may never beWorks Cited”The History of Computers. ” Florida State University.
October 10, 2004http://www. scri. fsu. edu/odyssey/cyberkids/computers/history/”Evolution of the Computer.
” University of Sandiego. October 10, 2004http://history. sandiego. edu/gen/recording/computer1. html”The Z3.
Choose Type of service
Choose writer quality
Page count
1 page 275 words
Order Essay Writing
\$13.9 Order Now
Sara from Artscolumbia
Hi there, would you like to get such an essay? How about receiving a customized one?
Check it out goo.gl/Crty7Tt
Evolution of Computers Essay
Artscolumbia
Artscolumbia
How many inventions in your lifetime can you think of that have changed everything in our society today? Computers have taken over today's society. From everyday tasks to moving satellites in space, PCs have revolutionized almost everything in our society. Computers weren't always this complicated though, and were around a long time before anyone even knew what the word "computer" meant. The Abacus was the first known machine developed to help perform mathematical equations. From wha
2021-02-09 09:46:31
\$ 13.900 2018-12-31
artscolumbia.org
In stock
Rated 5/5 based on 1 customer reviews | 1,930 | 9,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.953023 |
https://aviation.stackexchange.com/questions/42205/are-airliners-cruising-altitudes-we-know-indicated-or-true-altitudes/42206 | 1,576,006,125,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540528490.48/warc/CC-MAIN-20191210180555-20191210204555-00264.warc.gz | 281,979,731 | 32,027 | # Are airliners' cruising altitudes we know indicated or true altitudes?
Are airliners' crusing altitudes, which commonly range from 25000 to 45000 feet depending on various conditions, indicated altitudes or true altitudes? If they are indicated altitudes, what are the range of cruising true altitudes?
• What do you mean by "indicated altitude"? We have "indicated airspeed" and "true airspeed". I believe the term you're looking for is "pressure altitude". – kevin Jul 22 '17 at 3:03
• @kevin Although you're right in saying that the indicated altitude is the pressure altitude at crusing levels, there IS indicated altitude. It's a known concept. – lemonincider Jul 22 '17 at 4:26
• Note that there are 3 kinds of altitude: geometric altitude (measured by GPS), indicated altitude (measured by barometric altimeter set to local reference datum) and pressure altitude (measured by barometric altimeter set to 1013 hPa (29.92 inHg)). The first two are different and the third is different unless local reference datum happens to be standard pressure. – Jan Hudec Jul 23 '17 at 19:50
The delta between "indicated altitude" and "true altitude" depends on how much the actual pressure differs from the set 29.92. If the actual pressure is 29.92, then indicated = true altitude.
For each 0.01 of altimeter setting, you change the altimeter reading by about 10 feet; for each 0.10 of altimeter setting, 100 feet, and for each 1.00, 1000 feet. So if your actual pressure is 30.92, then with 29.92 set in your altimeter, you'll be 1,000' higher than indicated, so at FL410 (read on your altimeter with 29.92 set), your "true altitude" would be about 42,000' MSL.
Anything approaching 31.00 as an altimeter setting is pretty rare, so I'd say that 42,000' is about the top end of the range for most commercial operations, although if you find a jet that can cruise above FL410 (at least some models of the 747 can, IIRC), then add 1000' to their max ceiling for your answer.
As far as the low end, anything below about 29.00 is a pretty rare altimeter setting, so subtract 1000' from whatever you want to use as your range for "indicated" altitude, and that will be the bottom of your corresponding range of "true" altitudes.
That all said, I've used altitudes below FL 180 for a cruise altitude in airline operations plenty of times, so there my indicated would essentially match my true altitude. Most often, this is on short flights where you can get more direct routing by staying low & out of more congested airspace up higher (U.S. east coast); sometimes the headwinds/tailwinds or the rides (turbulence) is enough better down low to make it worthwhile to fly below the flght levels.
So, if you want to consider what is the range of commercial operations, I'd say that FL410 is mostly a very firm upper limit, with very few airline flights above that, while the lower limit would be much less clearly demarcated. And converting to true altitude, add/subtract about 1,000' from your indicated (flight level) altitude to cover almost all of the likely variation in atmospheric pressure.
• Thank you so much. May I ask you a follow-up question? When a jet is certified for cruising altitudes up to 45100 feet (B747), is it pressure altitude? – lemonincider Jul 22 '17 at 5:56
• Always a Flight Level, as far as I've seen. For example, the 737 is good to FL410, which isn't changed no matter how high or low the.local pressure may be. – Ralph J Jul 22 '17 at 15:38
• The difference between “indicated altitude” and “true altitude” actually depends on the actual temperature, humidity and bunch of other atmospheric properties that are never corrected for. The difference you describe is between “indicated altitude” (corrected for local pressure (but not density)) and “pressure altitude” (measured with standard setting of 1013 hPa (29.92 inHg)). “True altitude” is measured with GPS and only used for terrain proximity warning. – Jan Hudec Jul 23 '17 at 19:59
The altitudes are "indicated." That is the altitude shown on the aircraft's altimeter.
In the U.S. at and above 18,000 feet (known as Flight Level [FL] 180) all aircraft altimeters are set to 29.92 in. Therefore the "indicated" altitude on the aircraft's altimeter (at and above FL180) is also "Pressure Altitude." In other parts of the world "Flight Levels" (where all altimeters are set at 29.92) varies.
"True Altitude" (actual altitude above mean sea level) is not, in an operational sense, used other than a reference for weather reports, terrain, obstacle clearance, etc. It is important, but when aircraft are flying (below FL 180) their altimeters are set to a "local" altimeter setting which adjusts the "indicated" altitude to correspond closely to "true altitude." This eliminates "true altitude" as being an operational consideration.
It sounds complicated, but it really is not. Just a bit difficult to explain.
The range of cruising altitudes varies considerably for air carrier aircraft. It would be fair to say that the range is generally between FL290 (about 29,000 ft. msl) to FL 400 (about 40,000 ft. msl). This varies due to ATC operational considerations, wind speed and direction aloft, and aircraft weight and performance capabilities. There are times when some aircraft will fly a bit lower or, if the capability exists, fly higher, say FL430.
• Thank you for your detailed reply. If the cruising altitudes (FL 290~400) are indicated altitudes, do you know what their true altitude range is? The numbers might be varied depending on the atmospheric conditions, but can you give me approximate numbers? – lemonincider Jul 22 '17 at 4:32
• I can't give you an approximate range because the altimeters (above Fl180) are set to 29.92 (so all aircraft are being separated by ATC based on the same altimeter setting) and the atmospheric conditions (e.g. temp) could vary considerably. – 757toga Jul 22 '17 at 4:42 | 1,448 | 5,904 | {"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} | 3.078125 | 3 | CC-MAIN-2019-51 | latest | en | 0.904232 |
https://www.casinocitytimes.com/donald-catlin/article/cardano-revisited-1237 | 1,713,183,512,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00601.warc.gz | 626,239,509 | 10,650 | Stay informed with the
Recent Articles
Best of Donald Catlin
# Cardano Revisited
1 December 1999
In my June 1999 column, Cardano's Gaff Lives On, I looked at a letter sent in to Marilyn vos Savant's popular column, Ask Marilyn, which appears every Sunday in Parade Magazine. In it the writer, who I called Chuck, wrote to inquire about the old game of Chuck-a-Luck and the probability of winning it. I pointed out that Chuck made an error that was first made over 400 years ago by an Italian named Gerlamo Cardano (1501-1576). Although I straightened out Chuck's inquiry in my June article, I noted at the time that we didn't have all of the tools necessary to address Cardano's problem. With the introduction of independence in last month's column, we now have the tools. Well, almost.
Let me review Cardano's problem with you. Cardano was interested in determining how many rolls of a pair of dice are needed in order to have an even chance of rolling at least one total of 2. Cardano didn't have all of the tools we have. There was no Theory of Probability. It is understandable, therefore, that Cardano answered his question incorrectly. I want to formulate his question in modern terminology so that it will be easy for us to see what the error was. Then we are going to fix it.
Let me define a collection of events { Ei | i = 1, 2, 3, 4, ... } as follows. Ei is defined to be the event that a total of 2 is rolled on the ith roll of the dice. This said, it follows that the event defined by E1 or E2 or E3 or ... or En represents the event that either a 2 is rolled on the first roll or a 2 is rolled on the second roll or a two is rolled on the third roll or ... or a two is rolled on the nth roll. Now, it is possible for a total of 2 to be rolled on more than one roll, so another way of describing the event E1 or E2 or E3 or ... or En is to say that it is the event that in n rolls of a pair of dice that at least one total of 2 occurs. So Cardano's question is equivalent to determining n so that
P(E1 or E2 or E3 or ... or En) = 1/2 (1)
So far so good. Now here is where Cardano made his error. He calculated the left hand side of (1) by adding the probabilities of the individual events Ei. In other words he (mentally) changed equation (1) into
P(E1) + P(E2 ) + P(E3 ) + ... + P(En) = 1/2 (2)
This is wrong! As I pointed out in the June 1999 article, the only time you can calculate the probability of a series of disjunctions (events connected by 'or's) by adding the individual probabilities is when the collection of events involved is pairwise disjoint. The collection in (1) is definitely not a pairwise disjoint collection since a 2 can occur on more than one roll. For example, if a 2 occurred on the third and on the eighth roll, then the event E3 and E8 would be non-empty (see my September 1999 article An Ear Full of Cider for the technical definition of 'and'). Since the probability of each Ei is just 1/36, Cardano set the left hand side of (2) equal to n/36 and obtained
n/36 = 1/2 (3)
which produces the result n = 18. This is the wrong answer for the reason given above. It is also intuitively clear that this reasoning is wrong. If the probability of rolling at least one total of 2 in n rolls is n/36, then for 36 rolls the probability would be 1. This would imply that in 36 rolls a total of 2 will show up with certainty. If n = 37 we would be more than certain, whatever that means.
Very well, (2) and (3) are just plain wrong. We have to figure out a correct way to calculate the left-hand side of (1). To do so we will use a couple of set identities and a generalization of the notion of independence that was introduced last month.
Recall that the complement of an event A is just the set of all outcomes that are not in A. The complement of A is denoted by A'. For example, if A is the event of rolling a 2 with a pair of dice, then A' is the event of rolling any total other than 2. A and A' have no outcomes in common, so {A, A'} is a pairwise disjoint collection. This means that we can carry out a computation such as the one that Cardano wanted to do. The collection is also exhaustive, that is, every outcome in the sample space S is in either A or A'. In other words, S = A or A'. Putting all of this together we have
P(A) + P(A') = P(A or A') = P(S) = 1 (4)
and from (4) we can conclude that
P(A) = 1 - P(A') (5)
The other set identity that we will need is called DeMorgan's Law and is really very simple. I'll state it for you and then we'll discuss it. Here it is:
(E1 or E2 or E3 or ... or En)' = E1' and E2' and E3' and ... and En' (6)
This looks complicated, but it is really an easy consequence of language usage. Let me say it in words and you'll see what I mean. The left hand side of (6) is just the set of all outcomes that are not in E1 or E2 or E3 or ... or En. This is the same as the set of outcomes that are not in E1 and are not in E2 and are not in E3 and ... and are not in En. That's all there is to it.
Finally, I have to generalize the idea of independence from last month's column. I am not going to do a formal treatment since that is too messy, that is, lots of subscripts dripping from other subscripts. The useful ideas are easily understood without all of the notation. First, a collection of events {E1, E2, E3, ... , En} is said to be causally independent if no sub-collection of events has any causal effect on any remaining sub-collection. Just remember, this has to work for any and all sub-collections and any and all choices of the remaining sub-collections. The generalization of the Independence Metaprinciple from my last column is simply that if the collection is causally independent, then the probability of any conjunction of events (meaning they are connected by 'and's) in the collection can be calculated by multiplying the individual probabilities.
Before we get to Cardano's problem, let me emphasize the importance of the words 'any and all' in the above paragraph with a simple example. Consider the experiment of flipping a coin twice. Let A be the event that the first flip is heads, let B be the event that the second flip is heads, and let C be the event that the first and second flips are different. The sets are:
S = {hh, ht, th, tt}
A = {hh, ht}
B = {hh, th}
C = {ht, th}
so
P(A) = 1/2 P(B) = 1/2 P(C) = 1/2
Are A and B independent? Yes, the outcome of heads on the first flip in no way influences whether or not heads apears on the second flip. Although lack of causality between A and C or B and C is, I believe, not compelling, each pair is mathematically independent. In fact
P(A and B) = P({hh}) = 1/4 = 1/2 x 1/2 = P(A)P(B)
P(A and C) = P({ht}) = 1/4 = 1/2 x 1/2 = P(A)P(C)
P(B and C) = P({th}) = 1/4 = 1/2 x 1/2 = P(B)P(C)
Does this mean {A, B, C} is an independent collection? No indeed. Look at the event A and B. If this occurs we can say with certainty that C does not occur. In fact, note the failure of multiplying probabilities in this case:
P(A and B and C) = P(empty set) = 0
whereas
P(A)P(B)P(C) = 1/2 x 1/2 x 1/2 = 1/8
I think you see the point.
Well, what about Cardano? We're all set to solve his problem with some really slick mathematics; just watch how nicely this goes. Remember that in equation (1) we wanted to find an expression for P(E1 or E2 or E3 or ... or En). Using (5) with A = E1or E2 or E3 or ... or En we have
P(E1 or E2 or E3 or ... or En) = 1 - P((E1 or E2 or E3or ... or En)' ) (7)
Next, using DeMorgan's Law (6) applied to the right-hand term in (7), it follows that
P(E1 or E2 or E3 or ... or En) = 1 - P(E1' and E2' and E3' and ... and En' ) (8)
Each of the events Ei' represents the event of not rolling a total of 2 on the ith roll. Does not rolling a 2 on any particular set of rolls influence whether or not a 2 does or doesn't occur on any of the remaining rolls? No, indeed. The collection {E1', E2', E3', ... , En' } is causally independent. From the Independence Metaprinciple we can rewrite the last term in (8) to obtain
P(E1or E2 or E3 or ... or En) = 1 - P(E1' ) P(E2' ) P(E3' ) ... P(En' ) (9)
The probability of not rolling a total of 2 on a single roll of a pair of dice is 35/36, so each of the terms on the right-hand side of (9) is 35/36. If follows that (9) can be rewritten as
P(E1 or E2 or E3 or ... or En) = 1 - (35/36)n (10)
Substituting (10) into (1) we obtain
1 - (35/36)n = 1/2 (11)
which is equivalent to
(35/36)n = 1/2 (12)
Using our trusty calculator we discover that (35/36)24 = 0.50859 whereas (35/36)25 = 0.49447 so that (12), hence (10), cannot be solved exactly. In other words Cardano's problem was ill posed. We can rephrase it as follows. How many rolls of a pair of dice do we need to have at least an even chance of rolling at least one total of 2? The answer is 25 rolls and the probability of at least one total of 2 in those 25 rolls is 0.50553 (which is 1 - 0.49447).
The calculation you have just seen in (7), (8), and (9) is an extremely useful one and we will use it again; you might want to save it somewhere. As for Cardano, I think he would have been thrilled to see these simple ideas of independence, complementation, and DeMorgan's Law being brought to bear on his problem. He was a very bright man, certainly brighter than yours truly. But, unlike Cardano who never even heard the words 'probability theory', I have the the legacy of knowledge from many, many clever people who contributed to the subject. How lucky I am! How lucky we all are. See you next month.
Recent Articles
Best of Donald Catlin
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.
#### Books by Donald Catlin:
Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.
#### Books by Donald Catlin:
Lottery Book: The Truth Behind the Numbers | 2,881 | 10,597 | {"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-2024-18 | latest | en | 0.953274 |
http://www22.pair.com/csdc/car/carfre70.htm | 1,544,535,001,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00082.warc.gz | 511,486,837 | 2,547 | Stimulated Absorption and Self-Organization In his (1926) book ''The Internal Constitution of the Stars'', Sir Arthur Eddington not only stated that the BKG radiation of the Universe is about 3 degrees Kelvin, but also deduced the Planck distribution formula, but from a direction opposite to that used by Einstein. Eddington presumed the principles that: (1.) Thermodynamic equilibrium requires reversible exchanges, and (2.) Electromagnetic energy exchanges occur in a ''quantized'' fashion. He then deduced Boltzman's formula and Planck's formula from ''pure quantum theory'' (Einstein presumed Boltzman's formula to deduce the quantization concept.). See pages 44 -53. Eddington's argument was based upon the absorption and emission coefficients, a12 and a21 between quantum states 1 and 2, and the spontaneous coefficient (of decay from 2 to 1) given by the coefficient, +b21. Balancing the transitions between states 1 and 2 (the equilibrium assumption) he formulates the famous Einstein balance equation. Eddington then considers the detailed balance between 3 states, and ultimately deduces as the only possible solution, the Bose-Einstein distribution formula. Eddington then demonstrates that at a temperature T >0 for the atom in the higher energy state, the ''emission is stimulated by the presence of radiation in the field. This stimulated emission is called by Einstein negative absorption''. Quotes by Eddington. With further characteristic insight, Eddington examined equilibrium between four states, and obtained a slightly modified distribution law which was supposedly useful to include the dissociation and combination of atoms which might take place in a radiation field. All of this was done before 1926! and before the Fermi-Dirac distribution was known. What is remarkable is that if the coefficient +b12 is taken to be negative, then all of Eddington's arguments are still valid, and the result for the minus sign, -b12, yields the Fermi-Dirac distribution law!!! By comparison, the negative coefficient, -b21, could be interpreted as "stimulated absorption" (in the same sense as when the positive coefficient is interpreted as stimulated emission). Stimulated absorption of what? Could it be the stimulated absorption of components to form topologically coherent Self-Organized complex systems? Is Stimulated Absorption the basis of molecular self-organization into 1-100 nanometer sized aggregates from proteins to viruses? | 509 | 2,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.936172 |
https://www.esaral.com/q/mark-against-the-correct-answer-in-each-of-the-following-66754 | 1,716,129,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00621.warc.gz | 674,216,679 | 11,408 | # Mark against the correct answer in each of the following:
Question:
Mark $(\sqrt{)}$ against the correct answer in each of the following:
$\int \frac{\sin ^{6} x}{\cos ^{8}} d x=?$
A. $\frac{1}{7} \tan ^{7} \mathrm{x}+\mathrm{C}$
B. $\frac{1}{7} \sec ^{7} \mathrm{x}+\mathrm{C}$
C. $\log \left|\cos ^{6} x\right|+C$
D. none of these
Solution: | 126 | 352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.676821 |
http://spssx-discussion.1045642.n5.nabble.com/unable-to-test-linearity-assumption-of-logistic-regression-td1069295.html | 1,579,864,125,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00465.warc.gz | 155,740,761 | 14,977 | # unable to test linearity assumption of logistic regression
5 messages
Open this post in threaded view
|
## unable to test linearity assumption of logistic regression
Hello, I am doing a simple binary logistic regression with the following structure: logit(Y) = a + bX X is a categorical variable with 5 categories, which is entered into the model as indicator with the last category as reference. To test for linearity, I performed the box-tidwell transformation on X (= X*ln(x) ) and added this variable as a covariate (I actually use the multinomial logistic regression procedure of SPSS as a binary logistic regression to achieve more information). First Question: Is it ok to enter this new variable as covariate (= non categorical)? Or should it be entered as a Factor (= categorical). However, the results are strange. When taking a look at the Likelihood Ratio Test, SPSS tells us that removing X ( or X*ln(X) ) does not increase the degrees of freedom, hence no significance is calculated and I cannot check for linearity. Second Question: How comes that removing this parameter from the model, does not increase the degrees of freedom? Third Question: How can i test for linearity ? I'm really puzzled here. Also looking at the parameter estimates it seems that no parameters are calculated for two of the five categories of X (they are set to 0). I understand that the parameters for the fifth category of X is redundant because this is the reference category. But somehow, also the fourth category becomes redundant? When I enter X as a covariate (= non-categorical variable), none of these problems occur. Thanks in advance
Open this post in threaded view
|
## Re: unable to test linearity assumption of logistic regression
Hi Benoît, I might be wrong, but the assumption of linearity of the logit is important for quantitative/ordinal variables, not for categorical ones... HTH Marta BD> I am doing a simple binary logistic regression with the following structure: BD> logit(Y) = a + bX BD> X is a categorical variable with 5 categories, which is entered into the BD> model as indicator with the last category as reference. BD> To test for linearity, I performed the box-tidwell transformation on X (= BD> X*ln(x) ) and added this variable as a covariate (I actually use the BD> multinomial logistic regression procedure of SPSS as a binary logistic BD> regression to achieve more information). BD> First Question: Is it ok to enter this new variable as covariate (= non BD> categorical)? Or should it be entered as a Factor (= categorical). BD> However, the results are strange. When taking a look at the Likelihood Ratio BD> Test, SPSS tells us that removing X ( or X*ln(X) ) does not increase the BD> degrees of freedom, hence no significance is calculated and I cannot check BD> for linearity. BD> Second Question: How comes that removing this parameter from the model, does BD> not increase the degrees of freedom? BD> Third Question: How can i test for linearity ? BD> I'm really puzzled here. Also looking at the parameter estimates it seems BD> that no parameters are calculated for two of the five categories of X (they BD> are set to 0). I understand that the parameters for the fifth category of X BD> is redundant because this is the reference category. But somehow, also the BD> fourth category becomes redundant? BD> When I enter X as a covariate (= non-categorical variable), none of these BD> problems occur.
Open this post in threaded view
|
## Re: unable to test linearity assumption of logistic regression
In reply to this post by Beno=?ISO-8859-1?Q?=EEt?= Depaire If X is an ordered, numeric categorical variable, then it might make sense to test for deviations from linearity. In OLS regression, this means comparing the model where X is categorical with one where X is assumed to be linearly relayed. PAul Paul R. Swank, Ph.D. Professor, Developmental Pediatrics Director of Research, Center for Improving the Readiness of Children for Learning and Education (C.I.R.C.L.E.) Medical School UT Health Science Center at Houston -----Original Message----- From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of Marta García-Granero Sent: Friday, June 23, 2006 11:36 AM To: [hidden email] Subject: Re: unable to test linearity assumption of logistic regression Hi Benoît, I might be wrong, but the assumption of linearity of the logit is important for quantitative/ordinal variables, not for categorical ones... HTH Marta BD> I am doing a simple binary logistic regression with the following structure: BD> logit(Y) = a + bX BD> X is a categorical variable with 5 categories, which is entered into BD> the model as indicator with the last category as reference. BD> To test for linearity, I performed the box-tidwell transformation on BD> X (= BD> X*ln(x) ) and added this variable as a covariate (I actually use the BD> multinomial logistic regression procedure of SPSS as a binary BD> logistic regression to achieve more information). BD> First Question: Is it ok to enter this new variable as covariate (= BD> non categorical)? Or should it be entered as a Factor (= categorical). BD> However, the results are strange. When taking a look at the BD> Likelihood Ratio Test, SPSS tells us that removing X ( or X*ln(X) ) BD> does not increase the degrees of freedom, hence no significance is BD> calculated and I cannot check for linearity. BD> Second Question: How comes that removing this parameter from the BD> model, does not increase the degrees of freedom? BD> Third Question: How can i test for linearity ? BD> I'm really puzzled here. Also looking at the parameter estimates it BD> seems that no parameters are calculated for two of the five BD> categories of X (they are set to 0). I understand that the BD> parameters for the fifth category of X is redundant because this is BD> the reference category. But somehow, also the fourth category becomes redundant? BD> When I enter X as a covariate (= non-categorical variable), none of BD> these problems occur. | 1,342 | 6,016 | {"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-2020-05 | latest | en | 0.91111 |
https://oneclass.com/textbook-notes/ca/utsc/psyc/psyb51h3/7961-chapter-9.en.html | 1,524,389,660,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945552.45/warc/CC-MAIN-20180422080558-20180422100558-00575.warc.gz | 670,430,481 | 42,299 | Textbook Notes (368,317)
Psychology (9,695)
PSYB51H3 (306)
Chapter 9
# Chapter 9
5 Pages
54 Views
Department
Psychology
Course
PSYB51H3
Professor
Matthias Niemeier
Semester
Winter
Description
Chapter 9: Hearing: Physiology and Psychoacoustics What Is Sound? Sounds created when objects vibrate Vibrations of object cause molecules in objects surrounding medium to vibrate as well o Vibration causes pressure changes in medium o Pressure changes are waves Sound waves travel at particular speed depending on medium, moving faster through denser substances When jet plane travels faster than speed of sound, plane catches up to and passes fronts of sound waves it is creating o Sound waves combine into shock wave, or huge pressure fluctuation o When shock wave reaches ground, we hear sonic boom Basic Qualities of Sound Waves: Frequency and Amplitude Amplitude (intensity): magnitude of pressure change in sound wave- difference between highest pressure area and lowest pressure area Pressure fluctuations can be very close together or spread apart over longer periods Frequency: how quickly pressure fluctuates Hertz (Hz): measure for frequency. 1 Hz equals one cycle per second Amplitude associated with loudness: the more intense a sound wave is, the louder it will sound Frequency associated with pitch: low-frequency sounds correspond to low pitches and high-frequency sound correspond to high pitches If young and careful about exposure can hear from 20 to 20 000Hz Decibels (dB): measure for physical intensity of sound o Define difference between 2 sounds as ratio between 2 sound pressures o Each 10:1 sound pressure ratio equals 20dB and 100:1 ratio equals 40dB Sine Waves, Complex Tones, Fourier Analysis Sine wave (pure tone): simplest kinds of sounds o Air pressure in sine wave changes continuously at same frequency Period: time taken for one complete cycle of sine wave Phases: relative position of 2 or more sine waves Complex tones: sound wave consisting of more than one sinusoidal component of different frequencies o Human voices, birds Fourier analysis: individual sine wave components of complex sound o Spectrum: shows intensity of each sine wave frequency found in complex tone Summarize results of Fourier Harmonic spectra: typically caused by simple vibrating source o i.e. string of guitar or reed of saxophone o each frequency component in such sound called harmonic Fundamental frequency: lowest frequency component of sound Timbre: describe the quality of sound that depends on relative energy levels of harmonic components Basic Structure of Mammalian Auditory System Outer Ear Sounds collected from environment by pinna: curly structure on side of head that we typically call ear Only mammals have pinnae Ear canal: sound waves funnelled by pinna into and through canal o Length and shape of ear canal enhance sound frequencies between 2000 and 6000 Hz o Main purpose to insulate structure at end tympanic membrane (eardrum) from damage Damaged eardrum will heal itself, like skin Pinna and ear canal makes outer ear www.notesolution.com
More Less
Related notes for PSYB51H3
Me
OR
Join OneClass
Access over 10 million pages of study
documents for 1.3 million courses.
Join to view
OR
By registering, I agree to the Terms and Privacy Policies
Just a few more details
So we can recommend you notes for your school. | 708 | 3,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-17 | latest | en | 0.851209 |
https://intelligencemission.com/free-electricity-invention-free-electricity-houston.html | 1,611,434,268,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00235.warc.gz | 381,114,064 | 17,004 | I am currently designing my own magnet motor. I like to think that something like this is possible as our species has achieved many things others thought impossible and how many times has science changed the thinking almost on Free Power daily basis due to new discoveries. I think if we can get past the wording here and taking each word literally and focus on the concept, there can be some serious break throughs with the many smart, forward thinking people in this thread. Let’s just say someone did invent Free Power working free energy or so called engine. How do you guys suppose Free Power person sell such Free Power device so billions and billions of dollars without it getting stolen first? Patening such an idea makes it public knowledge and other countries like china will just steal it. Such Free Power device effects the whole world. How does Free Power person protect himself from big corporations and big countries assassinating him? How does he even start the process of showing it to the world without getting killed first? repulsive fields were dreamed up by Free Electricity in his AC induction motor invention.
Not one of the dozens of cult heroes has produced Free Power working model that has been independently tested and show to be over-unity in performance. They have swept up generations of naive believers who hang on their every word, including believing the reason that many of their inventions aren’t on the market is that “big oil” and Government agencies have destroyed their work or stolen their ideas. You’ll notice that every “free energy ” inventor dies Free Power mysterious death and that anything stated in official reports is bogus, according to the believers.
The magnitude of G tells us that we don’t have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to Free Power. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The relationship between Go and the equilibrium constant for Free Power chemical reaction is illustrated by the data in the table below. As the tube is cooled, and the entropy term becomes less important, the net effect is Free Power shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when Free Power sealed tube containing NO2 gas is immersed in liquid nitrogen. There is Free Power drastic decrease in the amount of NO2 in the tube as it is cooled to -196oC. Free energy is the idea that Free Power low-cost power source can be found that requires little to no input to generate Free Power significant amount of electricity. Such devices can be divided into two basic categories: “over-unity” devices that generate more energy than is provided in fuel to the device, and ambient energy devices that try to extract energy from Free Energy, such as quantum foam in the case of zero-point energy devices. Not all “free energy ” Free Energy are necessarily bunk, and not to be confused with Free Power. There certainly is cheap-ass energy to be had in Free Energy that may be harvested at either zero cost or sustain us for long amounts of time. Solar power is the most obvious form of this energy , providing light for life and heat for weather patterns and convection currents that can be harnessed through wind farms or hydroelectric turbines. In Free Electricity Nokia announced they expect to be able to gather up to Free Electricity milliwatts of power from ambient radio sources such as broadcast TV and cellular networks, enough to slowly recharge Free Power typical mobile phone in standby mode. [Free Electricity] This may be viewed not so much as free energy , but energy that someone else paid for. Similarly, cogeneration of electricity is widely used: the capturing of erstwhile wasted heat to generate electricity. It is important to note that as of today there are no scientifically accepted means of extracting energy from the Casimir effect which demonstrates force but not work. Most such devices are generally found to be unworkable. Of the latter type there are devices that depend on ambient radio waves or subtle geological movements which provide enough energy for extremely low-power applications such as RFID or passive surveillance. [Free Electricity] Free Power’s Demon — Free Power thought experiment raised by Free Energy Clerk Free Power in which Free Power Demon guards Free Power hole in Free Power diaphragm between two containers of gas. Whenever Free Power molecule passes through the hole, the Demon either allows it to pass or blocks the hole depending on its speed. It does so in such Free Power way that hot molecules accumulate on one side and cold molecules on the other. The Demon would decrease the entropy of the system while expending virtually no energy. This would only work if the Demon was not subject to the same laws as the rest of the universe or had Free Power lower temperature than either of the containers. Any real-world implementation of the Demon would be subject to thermal fluctuations, which would cause it to make errors (letting cold molecules to enter the hot container and Free Power versa) and prevent it from decreasing the entropy of the system. In chemistry, Free Power spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of Free Power diamond turning into graphite, which can be written as the following reaction: Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be Free Power chemical reaction in Free Power beaker. Free Power we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?) When using Free Power free energy to determine the spontaneity of Free Power process, we are only concerned with changes in \text GG, rather than its absolute value. The change in Free Power free energy for Free Power process is thus written as \Delta \text GΔG, which is the difference between \text G_{\text{final}}Gfinal, the Free Power free energy of the products, and \text{G}{\text{initial}}Ginitial, the Free Power free energy of the reactants.
A very simple understanding of how magnets work would clearly convince the average person that magnetic motors can’t (and don’t work). Pray tell where does the energy come from? The classic response is magnetic energy from when they were made. Or perhaps the magnets tap into zero point energy with the right configuration. What about they harness the earth’s gravitational field. Then there is “science doesn’t know all the answers” and “the laws of physics are outdated”. The list goes on with equally implausible rubbish. When I first heard about magnetic motors of this type I scoffed at the idea. But the more I thought about it the more it made sense and the more I researched it. Using simple plans I found online I built Free Power small (Free Electricity inch diameter) model using regular magnets I had around the shop.
Any ideas on my magnet problem? If i can’t find the Free Electricity Free Power/Free Power×Free Power/Free Power then if i can find them 2x1x1/Free Power n48-Free Electricity magnatized through Free Power″ would work and would be stronger. I have looked at magnet stores and ebay but so far nothing. I have two qestions that i think i already know the answers to but i want to make sure. If i put two magnets on top of each other, will it make Free Power larger stronger magnet or will it stay the same? Im guessing the same. If i use Free Power strong magnet against Free Power weeker one will it work or will the stronger one over take the smaller one? Im guessing it will over take it. Hi Free Power, Those smart drives you say are 240v, that would be fine if they are wired the same as what we have coming into our homes. Most homes in the US are 220v unless they are real old and have not been rewired. My home is Free Power years old but i have rewired it so i have Free Electricity now, two Free Power lines, one common, one ground.
Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
But I will send you the plan for it whenever you are ready. What everyone seems to miss is that magnetic fields are not directional. Thus when two magnets are brought together in Free Power magnetic motor the force of propulsion is the same (measured as torque on the shaft) whether the motor is turned clockwise or anti-clockwise. Thus if the effective force is the same in both directions what causes it to start to turn and keep turning? (Hint – nothing!) Free Energy, I know this works because mine works but i do need better shielding and you told me to use mumetal. What is this and where do you get it from? Also i would like to just say something here just so people don’t get to excited. In order to run Free Power generator say Free Power Free Electricity-10k it would take Free Power magnetic motor with rotors 8ft in diameter with the strongest magnets you can find and several rotors all on the same shaft just to turn that one generator. Thats alot of money in magnets. One example of the power it takes is this.
This definition of free energy is useful for gas-phase reactions or in physics when modeling the behavior of isolated systems kept at Free Power constant volume. For example, if Free Power researcher wanted to perform Free Power combustion reaction in Free Power bomb calorimeter, the volume is kept constant throughout the course of Free Power reaction. Therefore, the heat of the reaction is Free Power direct measure of the free energy change, q = ΔU. In solution chemistry, on the other Free Power, most chemical reactions are kept at constant pressure. Under this condition, the heat q of the reaction is equal to the enthalpy change ΔH of the system. Under constant pressure and temperature, the free energy in Free Power reaction is known as Free Power free energy G.
If power flows from the output shaft where does it flow in? Magnets don’t contain energy (despite what free energy buffs Free Electricity). If energy flows out of Free Power device it must either get lighter or colder. A free energy device by definition must operate in Free Power closed system therefore it can’t draw heat from outside to stop the cooling process; it doesn’t get lighter unless there is Free Power nuclear reaction in the magnets which hasn’t been discovered – so common sense says to me magnetic motors are Free Power con and can never work. Science is not wrong. It is not Free Power single entity. Free Electricity or findings can be wrong. Errors or corrections occur at the individual level. Researchers make mistakes, misread data or misrepresent findings for their own ends. Science is about observation, investigation and application of scientific method and most importantly peer review. Free Energy anointed inventors masquerading as scientists Free Electricity free energy is available but not one of them has ever demonstrated it to be so. Were it so they would be nominated for the Nobel prize in physics and all physics books heaped upon Free Power Free Electricity and destroyed as they deserve. But this isn’t going to happen. Always try to remember.
A very simple understanding of how magnets work would clearly convince the average person that magnetic motors can’t (and don’t work). Pray tell where does the energy come from? The classic response is magnetic energy from when they were made. Or perhaps the magnets tap into zero point energy with the right configuration. What about they harness the earth’s gravitational field. Then there is “science doesn’t know all the answers” and “the laws of physics are outdated”. The list goes on with equally implausible rubbish. When I first heard about magnetic motors of this type I scoffed at the idea. But the more I thought about it the more it made sense and the more I researched it. Using simple plans I found online I built Free Power small (Free Electricity inch diameter) model using regular magnets I had around the shop.
My older brother explained that in high school physics, they learned that magnetism is not energy at all. Never was, never will be. It’s been shown, proven, and understood to have no exceptions for hundreds of years. Something that O. U. should learn but refuses to. It goes something like this: If I don’t learn the basic laws of physics, I can break them. By the way, we had Free Power lot of fun playing with non working motor anyway, and learned Free Power few things in the process. My brother went on to get his PHD in physics and wound up specializing in magnetism. He designed many of the disk drive plates and electronics in the early (DOS) computers. bnjroo Harvey1 Thanks for the reply! I’m afraid there is an endless list of swindlers and suckers out there. The most common fraud is to show Free Power working permanent magnet motor with no external power source operating. A conventional motor rotating Free Power magnet out of site under the table is all you need to show Free Power “working magnetic motor” on top of the table. How could I know this? Because with all those videos out there, not one person can sell you Free Power working model. Also, not one of these scammers can ever let anyone not related to his scam operate the motor without the scammer hovering around. The believers are victims of something called “Confirmation Bias”. Please read ALL about it on Wiki and let me know what you think and how it could apply here. This trap has ensnared some very smart people. Harvey1 bnjroo Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been created! NOTHING IS IMPOSSIBLE! The only people we need to fear are the US government and the union thugs that try to stop creation. Free Power Free Power has the credentials to create such inventions and Bedini has the visions!
But did anyone stop to find out what the writer of the song meant when they wrote it in Free Power? Yes, actually, some did, thankfully. But many didn’t and jumped on the hate bandwagon because nowadays many of us seem to have become headline and meme readers and take all we see as fact without ever questioning what we’re being told. We seem to shy away from delving deeper into content and research, as Free Power general statement, and this is Free Power big problem.
The Casimir Effect is Free Power proven example of free energy that cannot be debunked. The Casimir Effect illustrates zero point or vacuum state energy , which predicts that two metal plates close together attract each other due to an imbalance in the quantum fluctuations. You can see Free Power visual demonstration of this concept here. The implications of this are far reaching and have been written about extensively within theoretical physics by researchers all over the world. Today, we are beginning to see that these concepts are not just theoretical but instead very practical and simply, very suppressed.
### I am doing more research for increasing power output so that it can be used in future in cars. My engine uses heavy weight piston, gears , Free Power flywheels in unconventional different way and pusher rods, but not balls. It was necessary for me to take example of ball to explain my basic idea I used in my concept. (the ball system is very much analogous to the piston-gear system I am using in my engine). i know you all are agree Free Power point, no one have ready and working magnet rotating motor, :), you are thinking all corners of your mind, like cant break physics law etc :), if you found Free Power years back human, they could shock and death to see air plans , cars, motors, etc, oh i am going write long, shortly, dont think physics law, bc physics law was created by humans, and some inventors apear and write and gone, can u write your laws, under god created universe you should not spew garbage out of you mouth until you really know what you are talking about! Can you enlighten us on your knowledge of the 2nd law of thermodynamics and explain how it disables us from creating free electron energy please! if you cant then you have no right to say that it cant work! people like you have kept the world form advancements. No “free energy magnetic motor” has ever worked. Never. Not Once. Not Ever. Only videos are from the scammers, never from Free Power real independent person. That’s why only the plans are available. When it won’t work, they blame it on you, and keep your money.
My older brother explained that in high school physics, they learned that magnetism is not energy at all. Never was, never will be. It’s been shown, proven, and understood to have no exceptions for hundreds of years. Something that O. U. should learn but refuses to. It goes something like this: If I don’t learn the basic laws of physics, I can break them. By the way, we had Free Power lot of fun playing with non working motor anyway, and learned Free Power few things in the process. My brother went on to get his PHD in physics and wound up specializing in magnetism. He designed many of the disk drive plates and electronics in the early (DOS) computers. bnjroo Harvey1 Thanks for the reply! I’m afraid there is an endless list of swindlers and suckers out there. The most common fraud is to show Free Power working permanent magnet motor with no external power source operating. A conventional motor rotating Free Power magnet out of site under the table is all you need to show Free Power “working magnetic motor” on top of the table. How could I know this? Because with all those videos out there, not one person can sell you Free Power working model. Also, not one of these scammers can ever let anyone not related to his scam operate the motor without the scammer hovering around. The believers are victims of something called “Confirmation Bias”. Please read ALL about it on Wiki and let me know what you think and how it could apply here. This trap has ensnared some very smart people. Harvey1 bnjroo Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been created! NOTHING IS IMPOSSIBLE! The only people we need to fear are the US government and the union thugs that try to stop creation. Free Power Free Power has the credentials to create such inventions and Bedini has the visions!
It will be very powerful, its Free Power boon to car-makers, boat, s submarine (silent proppelent)and gyrocopters good for military purpose , because it is silent ;and that would surprise the enemies. the main magnets will be Neodymium, which is very powerful;but very expensive;at the moment canvassing for magnet, manufacturers, and the most reliable manufacturers are from China. Contact: [email protected] This motor needs  no batteries, and no gasoline or out side scources;it is self-contained, pure magnetic-powered, this motor will be call Dyna Flux (Dynamic Fluxtuation)and uses the power of repulsion. Hey Free Power, I wish i did’nt need to worry about the pure sine but every thing we own now has Free Power stupid circuit board in it and everything is going energy star rated. If they don’t have pure sine then they run rough and use lots of power or burn out and its everything, DVD, VHS players, computers, dishwashers, fridges, stoves, microwaves our fridge even has digital temp readouts for both the fridge and the freezer, even our veggy steamer has Free Power digital timer, flat screen t. v’s, you can’t get away from it anymore, the world has gone teck crazzy. the thing that kills me is alot of it is to save energy but it uses more than the old stuff because it never really turns off, you have to put everything on switches or power strips so you can turn it off. I don’t know if i can get away from using batteries for my project. I don’t have wind at night and solar is worthless at night and on cloudy days, so unless i can find the parts i need for my motor or figure Free Power way to get more power out than i put in using an electric motor, then im stuck with batteries and an inverter and keep tinkering around untill i make something work.
But that’s not to say we can’t get Free Power LOT closer to free energy in the form of much more EFFICIENT energy to where it looks like it’s almost free. Take LED technology as Free Power prime example. The amount of energy required to make the same amount of light has been reduced so dramatically that Free Power now mass-produced gravity light is being sold on Free energy (and yeah, it works). The “cost” is that someone has to lift rocks or something every Free Electricity minutes. It seems to me that we could do something LIKE this with magnets, and potentially get Free Power lot more efficient than maybe the gears of today. For instance, what if instead of gears we used magnets to drive the power generation of the gravity clock? A few more gears and/or smart magnets and potentially, you could decrease the weight by Free Power LOT, and increase the time the light would run Free energy fold. Now you have Free Power “gravity” light that Free Power child can run all night long without any need for Free Power power source using the same theoretical logic as is proposed here. Free energy ? Ridiculous. “Conservation of energy ” is one of the most fundamental laws of physics. Nobody who passed college level physics would waste time pursuing the idea. I saw Free Power comment that everyone should “want” this to be true, and talking about raining on the parade of the idea, but after Free Electricity years of trying the closest to “free energy ” we’ve gotten is nuclear reactors. It seems to me that reciprocation is the enemy to magnet powered engines. Remember the old Mazda Wankel advertisements?
##### Vacuums generally are thought to be voids, but Hendrik Casimir believed these pockets of nothing do indeed contain fluctuations of electromagnetic waves. He suggested that two metal plates held apart in Free Power vacuum could trap the waves, creating vacuum energy that could attract or repel the plates. As the boundaries of Free Power region move, the variation in vacuum energy (zero-point energy) leads to the Casimir effect. Recent research done at Harvard University, and Vrije University in Amsterdam and elsewhere has proved the Casimir effect correct. (source)
These were Free Power/Free Power″ disk magnets, not the larger ones I’ve seen in some videos. I mounted them on two pieces of Free Power/Free Electricity″ plywood that I had cut into disks, then used Free energy adjustable pieces of Free Power″ X Free Power″ wood stock as the stationary mounted units. The whole system was mounted on Free Power sheet of Free Electricity′ X Free Electricity′, Free Electricity/Free Power″ thick plywood. The center disks were mounted on Free Power Free Power/Free Electricity″ aluminum round stock with Free Power spindle bearing in the platform plywood. Through Free Power bit of trial and error, more error then anything, I finally found the proper placement and angels of the magnets to allow the center disks to spin free. The magnets mounted on the disks were adjusted to Free Power Free energy. Free Electricity degree angel with the stationary units set to match. The disks were offset by Free Electricity. Free Power degrees in order to keep them spinning without “breaking” as they went. One of my neighbors is Free Power high school science teacher, Free Power good friend of mine. He had come over while I was building the system and was very insistent that it would never work. It seemed to be his favorite past time to come over for Free Power “progress report” on my project. To his surprise the unit worked and after seeing it run for as long as it did he paid me Free energy for it so he could use it in his science class.
The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
What is the name he gave it for research reasons? Thanks for the discussion. I appreciate the input. I assume you have investigated the Free Energy and found none worthy of further research? What element of the idea is failing? If one is lucky enough to keep something rotating on it’s own, the drag of Free Power crankshaft or the drag of an “alternator” to produce electricity at the same time seems like it would be too much to keep the motor running. Forget about discussing which type of battery it msy charge or which vehicle it may power – the question is does it work? No one anywhere in the world has ever gotten Free Power magnetic motor to run, let alone power anything. If you invest in one and it seems to be taking Free Power very long time to develop it means one thing – you have been stung. Free Energy’t say you haven’t been warned. As an optimist myself, I want to see it work and think it can. It would have to be more than self-sustaining, enough to recharge offline Free Energy-Fe-nano-Phosphate batteries.
The torque readings will give the same results. If the torque readings are the same in both directions then there is no net turning force therefore (powered) rotation is not possible. Of course it is fun to build the models and observe and test all of this. Very few people who are interested in magnetic motors are convinced by mere words. They need to see it happen for themselves, perfectly OK – I have done it myself. Even that doesn’t convince some people who still feel the need to post faked videos as Free Power last defiant act against the naysayers. Sorry Free Power, i should have asked this in my last post. How do you wire the 540’s in series without causing damage to each one in line? And no i have not seen the big pma kits. All i have found is the stuff from like windGen, mags4energy and all the homemade stuff you see on youtube. I have built three pma’s on the order of those but they don’t work very good. Where can i find the big ones? Free Power you know what the 540 max watts is? Hey Free Power, learn new things all the time. Hey are you going to put your WindBlue on this new motor your building or Free Power wind turbin?
But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
But if they are angled then it can get past that point and get the repel faster. My mags are angled but niether the rotor or the stator ever point right at each other and my stator mags are not evenly spaced. Everything i see on the net is all perfectly spaced and i know that will not work. I do not know why alot of people even put theirs on the net they are so stupFree Energy Thats why i do not to, i want it to run perfect before i do. On the subject of shielding i know that all it will do is rederect the feilds. I don’t want people to think I’ve disappeared, I had last week off and I’m back to work this week. I’m stealing Free Power little time during my break to post this. Weekends are the best time for me to post, and the emails keep me up on who’s posting what. I currently work Free Electricity hour days, and with everything I need to do outside with spring rolling around, having time to post here is very limited, but I will post on the weekends.
We need to stop listening to articles that say what we can’t have. Life is to powerful and abundant and running without our help. We have the resources and creative thinking to match life with our thoughts. Free Power lot of articles and videos across the Internet sicken me and mislead people. The inventors need to stand out more in the corners of earth. The intelligent thinking is here and freely given power is here. We are just connecting the dots. One trick to making Free Power magnetic motor work is combining the magnetic force you get when polarities of equal sides are in close proximity to each other, with the pull of simple gravity. Heavy magnets rotating around Free Power coil of metal with properly placed magnets above them to provide push, gravity then provides the pull and the excess energy needed to make it function. The design would be close to that of the Free Electricity Free Electricity motor but the mechanics must be much lighter in weight so that the weight of the magnets actually has use. A lot of people could do well to ignore all the rules of physics sometimes. Rules are there to be broken and all the rules have done is stunt technology advances. Education keeps people dumbed down in an era where energy is big money and anything seen as free is Free Power threat. Open your eyes to the real possibilities. Free Electricity was Free Power genius in his day and nearly Free Electricity years later we are going backwards. One thing is for sure, magnets are fantastic objects. It’s not free energy as eventually even the best will demagnetise but it’s close enough for me.
# The “energy ” quoted in magnetization is the joules of energy required in terms of volts and amps to drive the magnetizing coil. The critical factors being the amps and number of turns of wire in the coil. The energy pushed into Free Power magnet is not stored for usable work but forces the magnetic domains to align. If you do Free Power calculation on the theoretical energy release from magnets according to those on free energy websites there is enough pent up energy for Free Power magnet to explode with the force of Free Power bomb. And that is never going to happen. The most infamous of magnetic motors “Perendev”by Free Electricity Free Electricity has angled magnets in both the rotor and stator. It doesn’t work. Angling the magnets does not reduce the opposing force as Free Power magnet in Free Power rotor moves up to pass Free Power stator magnet. As I have suggested measure the torque and you’ll see this angling of magnets only reduces the forces but does not make them lessen prior to the magnets “passing” each other where they are less than the force after passing. Free Energy’t take my word for it, measure it. Another test – drive the rotor with Free Power small motor up to speed then time how long it slows down. Then do the same test in reverse. It will take the same time to slow down. Any differences will be due to experimental error. Free Electricity, i forgot about the mags loseing their power.
A device I worked on many years ago went on television in operation. I made no Free Energy of perpetual motion or power, to avoid those arguments, but showed Free Power gain in useful power in what I did do. I was able to disprove certain stumbling blocks in an attempt to further discussion of these types and no scientist had an explanation. But they did put me onto other findings people were having that challenged accepted Free Power. Dr. Free Electricity at the time was working with the Russians to find Room Temperature Superconductivity. And another Scientist from CU developed Free Power cryogenic battery. “Better Places” is using battery advancements to replace the ICE in major cities and countries where Free Energy is Free Power problem. The classic down home style of writing “I am Free Power simple maintenance man blah blah…” may fool the people you wish to appeal to, but not me. Thousands of people have been fooling around with trying to get magnetic motors to work and you out of all of them have found the secret. | 7,285 | 35,327 | {"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} | 3 | 3 | CC-MAIN-2021-04 | latest | en | 0.966904 |
https://www.techylib.com/el/view/monkeybeetle/chapter_4_data-level_parallelism_in_vector_simd_and_gpu_architect | 1,529,820,218,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00050.warc.gz | 906,222,946 | 16,861 | # Chapter 4 - Data-Level Parallelism in Vector, SIMD, and GPU Architectures
Λογισμικό & κατασκευή λογ/κού
2 Δεκ 2013 (πριν από 4 χρόνια και 7 μήνες)
109 εμφανίσεις
Chapter
4
-
Data
-
Level Parallelism in
Vector, SIMD, and GPU
Architectures
CSCI/ EENG
641
-
W01
Computer Architecture 1
Dr.
Babak Beheshti
Slides based on the PowerPoint Presentations created by David Patterson as
part of the Instructor Resources for the textbook by Hennessy &
Patterson
Taxonomy
Flynn’s Taxonomy
Classify by
Instruction Stream
and
Data Stream
SISD
Single Instruction Single Data
Conventional processor
SIMD
Single Instruction Multiple Data
One instruction stream
Multiple data items
Several Examples Produced
MISD
Multiple Instruction Single Data
Systolic Arrays
MIMD Multiple Instruction Multiple Data
General Parallel Processors
SIMD
-
Single Instruction Multiple Data
Originally thought to be the ultimate
massively parallel
machine!
Some machines built
Illiac IV
Thinking Machines CM2
MasPar
Vector processors
(special category!)
SIMD
-
Single Instruction Multiple Data
Each PE is a
simple ALU
(1 bit in CM
-
1,
small processor in some)
Control Proc
issues same
instruction to
each PE in each
cycle
Each PE has
different data
SIMD
SIMD performance depends on
Mapping problem
processor architecture
Image processing
Maps naturally to 2D processor array
Calculations on individual pixels trivial
Combining data is the problem!
Some matrix operations also
SIMD
Matrix
multiplication
Each PE
*
then
+
PE
ij
C
ij
Note the B matrix
is transposed!
Parallel Processing
Communication patterns
If the system provides the “correct” data paths,
then good performance is obtained
even with slow PEs
Without effective communication
bandwidth,
even fast PEs are starved of data!
In a multiple PE system, we have
Raw communication bandwidth
Equivalent processor
memory bandwidth
Communications patterns
Imagine the Matrix Multiplication problem if the matrices are not already
transposed!
Network topology
Vector Processors
-
The Supercomputers
Optimized for vector & matrix operations
Conventional”
scalar processor
section not shown
Example
Dot product
or
in terms of the elements
Fetch each element of each vector in turn
Stride
“Distance” between successive elements of a vector
1 in dot
-
product case
Vector Processors
-
Vector operations
y = A
l
B
y =
S
a
k
* b
k
Vector Processors
-
Vector operations
Example
Matrix multiply
or
in terms of the elements
C = A B
c
ij
=
S
a
ik
* b
kj
Vector Operations
Fetch data into vector register
Very high effective
bandwidth
to memory
Long “burst”
accesses with
AGU managing
SIMD
SIMD architectures can exploit significant data
-
level parallelism for:
matrix
-
oriented scientific computing
media
-
oriented image and sound processors
SIMD is more energy efficient than MIMD
Only needs to fetch one instruction per data operation
Makes SIMD attractive for personal mobile devices
SIMD allows programmer to continue to think
sequentially
Introduction
SIMD Parallelism
Vector architectures
SIMD extensions
Graphics Processor Units (GPUs)
For x86 processors:
Expect two additional cores per chip per year
SIMD width to double every four years
Potential speedup from SIMD to be twice that
from MIMD!
Introduction
Vector Architectures
Basic idea:
Read sets of data elements into “vector registers”
Operate on those registers
Disperse the results back into memory
Registers are controlled by compiler
Used to hide memory latency
Leverage memory bandwidth
Vector Architectures
VMIPS
Example architecture: VMIPS
Loosely based on Cray
-
1
Vector registers
Each register holds a 64
-
element, 64 bits/element vector
Register file has 16 read ports and 8 write ports
Vector functional units
Fully pipelined
Data and control hazards are detected
-
store unit
Fully pipelined
One word per clock cycle after initial latency
Scalar registers
32 general
-
purpose registers
32 floating
-
point registers
Vector Architectures
VMIPS Instructions
Example: DAXPY
L.D
F0,a
LV
V1,Rx
MULVS.D
V2,V1,F0
; vector
-
scalar multiply
LV
V3,Ry
V4,V2,V3
SV
Ry,V4
; store the result
Requires 6 instructions vs. almost 600 for MIPS
Vector Architectures
Vector Execution Time
Execution time depends on three factors:
Length of operand vectors
Structural hazards
Data dependencies
VMIPS functional units consume one element per
clock cycle
Execution time is approximately the vector length
Convoy
Set of vector instructions that could potentially
execute together
Vector Architectures
Chimes
-
after
-
write dependency
hazards can be in the same convey via
chaining
Chaining
Allows a vector operation to start as soon as the
individual elements of its vector source operand
become available
Chime
Unit of time to execute one convoy
m
convoys executes in
m
chimes
For vector length of
n
, requires
m
x
n
clock cycles
Vector Architectures
Example
LV
V1,Rx
MULVS.D
V2,V1,F0
;vector
-
scalar multiply
LV
V3,Ry
V4,V2,V3
SV
Ry,V4
;store the sum
Convoys:
1
LV
MULVS.D
2
LV
3
SV
3 chimes, 2 FP ops per result, cycles per FLOP = 1.5
For 64 element vectors, requires 64 x 3 = 192 clock cycles
Vector Architectures
Challenges
Start up time
Latency of vector functional unit
Assume the same as Cray
-
1
Floating
-
point add => 6 clock cycles
Floating
-
point multiply => 7 clock cycles
Floating
-
point divide => 20 clock cycles
Vector load => 12 clock cycles
Improvements
:
> 1 element per clock cycle
Non
-
64 wide vectors
IF statements in vector code
Memory system optimizations to support vector processors
Multiple dimensional matrices
Sparse matrices
Programming a vector computer
Vector Architectures
Multiple Lanes
Element
n
of vector register
A
is “hardwired” to element
n
of
vector register
B
Allows for multiple hardware lanes
Vector Architectures
Vector Length Register
Vector length not known at compile time?
Use Vector Length Register (VLR)
Use strip mining for vectors over the maximum length:
low = 0;
VL = (n % MVL);
/*find odd
-
size piece using modulo op % */
for (j = 0; j <= (n/MVL); j=j+1)
{
/*outer loop*/
for (
i
= low;
i
< (
low+VL
);
i
=i+1)
/*runs for length VL*/
Y[
i
] = a * X[
i
] + Y[
i
] ;
/*main operation*/
low = low + VL;
/*start of next vector*/
VL = MVL;
/*reset the length to maximum
vector length*/
}
Vector Architectures
Consider:
for (i = 0; i < 64; i=i+1)
if (X[
i
] != 0)
X[
i
] = X[
i
]
Y[
i
];
Use vector mask register to “disable” elements:
LV
V1,Rx
LV
V2,Ry
L.D
F0,#0
SNEVS.D
V1,F0
;sets VM(
i
) to 1 if V1(
i
)!=F0
SUBVV.D
V1,V1,V2
SV
Rx,V1
;store the result in X
GFLOPS rate decreases!
Vector Architectures
Programming
Vec
. Architectures
Compilers can provide feedback to programmers
Programmers can provide hints to compiler
Vector Architectures
SIMD Implementations
Implementations:
Intel MMX (1996)
Eight 8
-
bit integer ops or four 16
-
bit integer ops
Streaming SIMD Extensions (SSE) (1999)
Eight 16
-
bit integer ops
Four 32
-
bit integer/
fp
ops or two 64
-
bit integer/
fp
ops
Four 64
-
bit integer/
fp
ops
Operands must be consecutive and aligned memory
locations
SIMD Instruction Set Extensions for Multimedia
Example SIMD Code
Example DXPY:
L.D
F0,a
MOV
F1, F0
;copy a into F1 for SIMD MUL
MOV
F2, F0
;copy a into F2 for SIMD MUL
MOV
F3, F0
;copy a into F3 for SIMD MUL
R4,Rx,#512
Loop:
L.4D F4,0[Rx]
i
], X[i+1], X[i+2], X[i+3]
MUL.4D
F4,F4,F0
;
a
×
X
[
i
],
a
×
X
[i+1],
a
×
X
[i+2],
a
×
X
[i+3]
L.4D
F8,0[
Ry
]
F8,F8,F4
;a
×
X[i]+Y[i], ..., a
×
X[i+3]+Y[i+3]
S.4D
0[
Ry
],F8
;store into Y[
i
], Y[i+1], Y[i+2], Y[i+3]
Rx,Rx,#32
;increment index to X
Ry,Ry,#32
;increment index to Y
DSUBU
R20,R4,Rx
;compute bound
BNEZ
R20,Loop
;check if done
SIMD Instruction Set Extensions for Multimedia
Roofline Performance Model
Basic idea:
Plot peak floating
-
point throughput as a function
of arithmetic intensity
Ties together floating
-
point performance and
memory performance for a target machine
Arithmetic intensity
Floating
-
SIMD Instruction Set Extensions for Multimedia
Examples
Attainable GFLOPs/sec Min = (Peak Memory BW
×
Arithmetic
Intensity, Peak Floating Point
Perf
.)
SIMD Instruction Set Extensions for Multimedia
Loop
-
Level Parallelism
Focuses on determining whether data accesses in later
iterations are dependent on data values produced in earlier
iterations
Loop
-
carried dependence
Example 1:
for (i=999; i>=0; i=i
-
1)
x[
i
] = x[
i
] + s;
No loop
-
carried dependence
Detecting and Enhancing Loop
-
Level Parallelism
Loop
-
Level Parallelism
Example 2:
for (
i
=0;
i
<100;
i
=i+1)
{
A[i+1] = A[
i
] + C[
i
]; /* S1 */
B[i+1] = B[
i
] + A[i+1]; /* S2 */
}
S1 and S2 use values computed by S1 in previous
iteration
S2 uses value computed by S1 in same iteration
Detecting and Enhancing Loop
-
Level Parallelism
Loop
-
Level Parallelism
Example 3:
for (
i
=0;
i
<100;
i
=i+1) {
A[
i
] = A[
i
] + B[
i
]; /* S1 */
B[i+1] = C[
i
] + D[
i
]; /* S2 */
}
S1 uses value computed by S2 in previous iteration but dependence is not
circular so loop is parallel
Transform to:
A[0] = A[0] + B[0];
for (
i
=0;
i
<99;
i
=i+1) {
B[i+1] = C[
i
] + D[
i
];
A[i+1] = A[i+1] + B[i+1];
}
B[100] = C[99] + D[99];
Detecting and Enhancing Loop
-
Level Parallelism
Loop
-
Level Parallelism
Example 4:
for (
i
=0;i<100;i=i+1) {
A[
i
] = B[
i
] + C[
i
];
D[
i
] = A[
i
] * E[
i
];
}
Example 5:
for (
i
=1;i<100;i=i+1) {
Y[
i
] = Y[i
-
1] + Y[
i
];
}
Detecting and Enhancing Loop
-
Level Parallelism
VECTOR PROCESSING
-
EXAMPLE
Consider the following vector
-
multiplication problem:
X * Y = Z, where X, Y, and Z are 100
-
value
vectors (arrays of size 100).
In
C (to
help visualize the connection to the Vector and
MIPS Pseudo
-
Code) this would be written as:
for (
i
=0;
i
<100;
i
++)
Z(
i
)
=
X(
i
)
*
Y(
i
)
Example (Cont’d)
Were this to be implemented in a MIPS machine,
each addition would take 4 clock
-
cycles. The entire
loop would be in excess of 400 cycles.
Were this to be implemented in a Vector Processing
machine, first, a number of elements from X and a
number from Y would be loaded into separate vector
registers (can be done simultaneously).
Example (Cont’d)
Next, the multiply pipeline would begin taking in elements
from X and Y. After a single clock
-
cycle, another set of
elements would be fed into this pipeline. After 4 clock
-
cycles
the first result would be completed and stored in vector
register Z. The second result would be completed in clock
-
cycle 5, and so on.
Finally, once all this is complete, the values are taken from
vector register Z and stored in main memory.
The time it takes for the multiplication by itself is a mere 103
clock
-
cycles.
PSEUDO CODE
-
VECTOR PROCESSING
//
register
//
register
VMULT VR1 VR2 VR3 //vector multiplying VR1
by
VR2, storing results
in
VR3
VSTORE VR3 Z //store vector register VR3 into
main memory as Z
PSEUDO CODE
MIPS
LW X[
i
], \$a0 //load first element of X into a
register
LW Y[
i
], \$a1 //load first element of Y into a
register
“MULT” \$a2, \$a0, \$a1 //multiply \$a0 and
\$a1 and store
result in \$a2
SW \$a2, Z[
i
] //store \$a2 into memory
//Repeat 100 times
SUMMARY
The Vector machine is faster at performing
mathematical operations on larger vectors
than is the MIPS machine.
The Vector processing computer’s vector
register architecture makes it better able to
compute vast amounts of data quickly. | 3,352 | 11,657 | {"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-2018-26 | latest | en | 0.730775 |
https://number.academy/2552598 | 1,685,800,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00444.warc.gz | 468,720,325 | 12,511 | # Number 2552598
Number 2,552,598 spell 🔊, write in words: two million, five hundred and fifty-two thousand, five hundred and ninety-eight, approximately 2.6 million. Ordinal number 2552598th is said 🔊 and write: two million, five hundred and fifty-two thousand, five hundred and ninety-eighth. The meaning of the number 2552598 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 2552598. What is 2552598 in computer science, numerology, codes and images, writing and naming in other languages
## What is 2,552,598 in other units
The decimal (Arabic) number 2552598 converted to a Roman number is (M)(M)(D)(L)MMDXCVIII. Roman and decimal number conversions.
#### Weight conversion
2552598 kilograms (kg) = 5627457.6 pounds (lbs)
2552598 pounds (lbs) = 1157850.9 kilograms (kg)
#### Length conversion
2552598 kilometers (km) equals to 1586111 miles (mi).
2552598 miles (mi) equals to 4108010 kilometers (km).
2552598 meters (m) equals to 8374564 feet (ft).
2552598 feet (ft) equals 778042 meters (m).
2552598 centimeters (cm) equals to 1004959.8 inches (in).
2552598 inches (in) equals to 6483598.9 centimeters (cm).
#### Temperature conversion
2552598° Fahrenheit (°F) equals to 1418092.2° Celsius (°C)
2552598° Celsius (°C) equals to 4594708.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
2552598 seconds equals to 1 month, 1 day, 13 hours, 3 minutes, 18 seconds
2552598 minutes equals to 5 years, 3 months, 1 week, 1 day, 15 hours, 18 minutes
### Codes and images of the number 2552598
Number 2552598 morse code: ..--- ..... ..... ..--- ..... ----. ---..
Sign language for number 2552598:
Number 2552598 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 2552598
### Multiplications
#### Multiplication table of 2552598
2552598 multiplied by two equals 5105196 (2552598 x 2 = 5105196).
2552598 multiplied by three equals 7657794 (2552598 x 3 = 7657794).
2552598 multiplied by four equals 10210392 (2552598 x 4 = 10210392).
2552598 multiplied by five equals 12762990 (2552598 x 5 = 12762990).
2552598 multiplied by six equals 15315588 (2552598 x 6 = 15315588).
2552598 multiplied by seven equals 17868186 (2552598 x 7 = 17868186).
2552598 multiplied by eight equals 20420784 (2552598 x 8 = 20420784).
2552598 multiplied by nine equals 22973382 (2552598 x 9 = 22973382).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 2552598
Half of 2552598 is 1276299 (2552598 / 2 = 1276299).
One third of 2552598 is 850866 (2552598 / 3 = 850866).
One quarter of 2552598 is 638149,5 (2552598 / 4 = 638149,5 = 638149 1/2).
One fifth of 2552598 is 510519,6 (2552598 / 5 = 510519,6 = 510519 3/5).
One sixth of 2552598 is 425433 (2552598 / 6 = 425433).
One seventh of 2552598 is 364656,8571 (2552598 / 7 = 364656,8571 = 364656 6/7).
One eighth of 2552598 is 319074,75 (2552598 / 8 = 319074,75 = 319074 3/4).
One ninth of 2552598 is 283622 (2552598 / 9 = 283622).
show fractions by 6, 7, 8, 9 ...
### Calculator
2552598
#### Is Prime?
The number 2552598 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 2552598 are 2 * 3 * 3 * 141811
The factors of 2552598 are 1, 2, 3, 6, 9, 18, 141811, 283622, 425433, 850866, 1276299, 2552598.
Total factors 12.
Sum of factors 5530668 (2978070).
#### Powers
The second power of 25525982 is 6.515.756.549.604.
The third power of 25525983 is 16.632.107.137.006.071.808.
#### Roots
The square root √2552598 is 1597,685201.
The cube root of 32552598 is 136,666102.
#### Logarithms
The natural logarithm of No. ln 2552598 = loge 2552598 = 14,752622.
The logarithm to base 10 of No. log10 2552598 = 6,406982.
The Napierian logarithm of No. log1/e 2552598 = -14,752622.
### Trigonometric functions
The cosine of 2552598 is -0,846253.
The sine of 2552598 is -0,532781.
The tangent of 2552598 is 0,629576.
### Properties of the number 2552598
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 2552598 in Computer Science
Code typeCode value
2552598 Number of bytes2.4MB
Unix timeUnix time 2552598 is equal to Friday Jan. 30, 1970, 1:03:18 p.m. GMT
IPv4, IPv6Number 2552598 internet address in dotted format v4 0.38.243.22, v6 ::26:f316
2552598 Decimal = 1001101111001100010110 Binary
2552598 Decimal = 11210200111200 Ternary
2552598 Decimal = 11571426 Octal
2552598 Decimal = 26F316 Hexadecimal (0x26f316 hex)
2552598 BASE64MjU1MjU5OA==
2552598 SHA1b5d3f2dc6b8a0a49799f5104459afb2d3b0dbfbc
2552598 SHA2247330684ff9b162c3db544596297917f822a6d1aba5f3f335af293cbd
2552598 SHA38425debf6b07bd2121da5302c8c4b7789066783f4481e7f7e658e0cfe415b284404d9d13997fdc41b6177942681f25f362
More SHA codes related to the number 2552598 ...
If you know something interesting about the 2552598 number that you did not find on this page, do not hesitate to write us here.
## Numerology 2552598
### Character frequency in the number 2552598
Character (importance) frequency for numerology.
Character: Frequency: 2 2 5 3 9 1 8 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 2552598, the numbers 2+5+5+2+5+9+8 = 3+6 = 9 are added and the meaning of the number 9 is sought.
## № 2,552,598 in other languages
How to say or write the number two million, five hundred and fifty-two thousand, five hundred and ninety-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 2.552.598) dos millones quinientos cincuenta y dos mil quinientos noventa y ocho German: 🔊 (Nummer 2.552.598) zwei Millionen fünfhundertzweiundfünfzigtausendfünfhundertachtundneunzig French: 🔊 (nombre 2 552 598) deux millions cinq cent cinquante-deux mille cinq cent quatre-vingt-dix-huit Portuguese: 🔊 (número 2 552 598) dois milhões, quinhentos e cinquenta e dois mil, quinhentos e noventa e oito Hindi: 🔊 (संख्या 2 552 598) पच्चीस लाख, बावन हज़ार, पाँच सौ, अट्ठानवे Chinese: 🔊 (数 2 552 598) 二百五十五万二千五百九十八 Arabian: 🔊 (عدد 2,552,598) مليونان و خمسمائة و اثنان و خمسون ألفاً و خمسمائةثمانية و تسعون Czech: 🔊 (číslo 2 552 598) dva miliony pětset padesát dva tisíce pětset devadesát osm Korean: 🔊 (번호 2,552,598) 이백오십오만 이천오백구십팔 Danish: 🔊 (nummer 2 552 598) to millioner femhundrede og tooghalvtredstusindfemhundrede og otteoghalvfems Dutch: 🔊 (nummer 2 552 598) twee miljoen vijfhonderdtweeënvijftigduizendvijfhonderdachtennegentig Japanese: 🔊 (数 2,552,598) 二百五十五万二千五百九十八 Indonesian: 🔊 (jumlah 2.552.598) dua juta lima ratus lima puluh dua ribu lima ratus sembilan puluh delapan Italian: 🔊 (numero 2 552 598) due milioni e cinquecentocinquantaduemilacinquecentonovantotto Norwegian: 🔊 (nummer 2 552 598) to million, fem hundre og femti-to tusen, fem hundre og nitti-åtte Polish: 🔊 (liczba 2 552 598) dwa miliony pięćset pięćdziesiąt dwa tysiące pięćset dziewięćdziesiąt osiem Russian: 🔊 (номер 2 552 598) два миллиона пятьсот пятьдесят две тысячи пятьсот девяносто восемь Turkish: 🔊 (numara 2,552,598) ikimilyonbeşyüzelliikibinbeşyüzdoksansekiz Thai: 🔊 (จำนวน 2 552 598) สองล้านห้าแสนห้าหมื่นสองพันห้าร้อยเก้าสิบแปด Ukrainian: 🔊 (номер 2 552 598) два мільйони п'ятсот п'ятдесят дві тисячі п'ятсот дев'яносто вісім Vietnamese: 🔊 (con số 2.552.598) hai triệu năm trăm năm mươi hai nghìn năm trăm chín mươi tám Other languages ...
## News to email
I have read the privacy policy
## Comment
If you know something interesting about the number 2552598 or any other natural number (positive integer), please write to us here or on Facebook.
#### Comment (Maximum 2000 characters) *
The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,848 | 8,194 | {"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-2023-23 | latest | en | 0.678155 |
http://gmatclub.com/forum/640-q48-v31-end-of-a-long-story-23244.html?fl=similar | 1,484,586,970,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00524-ip-10-171-10-70.ec2.internal.warc.gz | 112,211,645 | 61,637 | 640 (Q48 V31)...end of a long story : Share GMAT Experience
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 16 Jan 2017, 09:16
# Jan 16th:
All GMAT Club CATs and Quizzes are Open Free for 24 hrs. See our Holiday Policy to learn more
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# 640 (Q48 V31)...end of a long story
Author Message
TAGS:
### Hide Tags
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7
Kudos [?]: 45 [0], given: 0
640 (Q48 V31)...end of a long story [#permalink]
### Show Tags
18 Nov 2005, 02:56
I am an old GMAt taker
I've experienced the Gmat test 4 times and I got the following final scores in the order :
550 (Q39 V27) in Sept 04
500 (Q38 V23) in Nov 04
620 (Q48 V27) in Sept 05
640 (Q48 V31) in Nov 05
I think that it's all for me folks
I don't think I am able to reach something like 670, so I have more chances to get a low score and thereforce damage my profile than to get something that will really help me
I am just disappointed because I really wanted to reach 650 but you know, I did what I could
Quant was ok but no C/P, not even one
Many DS on algebra and long PS
In Verbal it was ok but I was too slow, really, I had the last 12 mintues to answer 10 questions (including a long RC) so it was tough, had to guess on 4 questions I assumed. I got a strange RC called "Sand Dunes", long and technical, very tought, I think it is experimental, hope so for you guys I got it just after another RC ! it was like 2 RC back to back
I'll see what I can do with my profile and my scores and I'll let you know guys...Thanks everybody for your support and I'll definitely stick around to promote the forum
Many thanks to Paul, Praetorian for this incredible website
Special thanks to Ywilfred, HongHu, Hjort, Bhai, Banerjeea_98, Carsen, Mandy, Himalaya, Gmatt73 (Japan rules), Nakib77, MA, Venksune, laxieqv (go Hanoi go), praveen, gsr, newkid, Macedon (ruben forza italia)
I really hope we could meet for real around a nice french diner
VP
Joined: 14 Feb 2005
Posts: 1001
Location: New York
Followers: 5
Kudos [?]: 20 [0], given: 0
### Show Tags
18 Nov 2005, 04:24
Congrats mate...
640 is a good score ...
All the best for your Applications
Cheers
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 40
Kudos [?]: 429 [0], given: 0
### Show Tags
18 Nov 2005, 04:40
Congratulation Antmavel! What a journey it must have been and I'm glad that you were on a consistently increasing trend. I am planning to go to HK/Guan Zhou next march. I could perhaps drop by Shen Zhen and pay you a visit I'll keep you posted. Send me your email via PM if you are interested. Take care.
_________________
Best Regards,
Paul
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16
Kudos [?]: 324 [0], given: 0
### Show Tags
18 Nov 2005, 06:13
Good job Antmavel! Congratulations! Your persistence and the resulted success is examplary for all of us. Good luck for your next step!
ps say hello to my Chinese fellows at ShenZhen.
_________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5
Kudos [?]: 74 [0], given: 0
Re: 640 (Q48 V31)...end of a long story [#permalink]
### Show Tags
18 Nov 2005, 06:24
Antonio,
Congratulations. 640 is definitly a respectable score. it can take you any school if you have strength in other aspects such as GPA, essays, strong recommendations etc..
hope to see you arround here.
himalaya
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 25
Kudos [?]: 376 [0], given: 0
### Show Tags
18 Nov 2005, 06:48
Your persistance is contagious Antoine. Remember that 640 is better than 80% of testtakers worldwide, that in itself is commendable, especially for someone who has to work full-time. We all appreciate your profound math strategies so please stick around and help the rest of us if you can. About the apps, I sent you a PM.
Congratulations!
SVP
Joined: 16 Oct 2003
Posts: 1810
Followers: 4
Kudos [?]: 136 [0], given: 0
### Show Tags
18 Nov 2005, 08:11
Antmavel,
Congraulations to you for getting 640. This is a respactable score and with some power push in the application you can be competitive. I can understand the dissappointment that you may be experiencing because you could not reach 650 mark. There are many good schools in top 20 that accept student in that GMAT range. I am sure you will find out the right school for you. Put your energy in the applications. Good luck
--Bhai
Director
Joined: 21 Aug 2005
Posts: 793
Followers: 2
Kudos [?]: 24 [0], given: 0
### Show Tags
18 Nov 2005, 08:46
Hey Antmavel, Congrats!! From 500 to 640 is truly a terrific effort! Hats-off to your persistence and hard-work. This kind of determination will see you through with success in every walk of life. 640 is a good score. Hope your profile and the application will get you in to a top School!
See you around and keep us all posted on you application and admits!
Good Luck with the next steps!!
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6
Kudos [?]: 327 [0], given: 0
### Show Tags
18 Nov 2005, 09:32
good luck !!!
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1
Kudos [?]: 102 [0], given: 0
### Show Tags
18 Nov 2005, 09:41
Congratulations, friend!!!
I am inspired by your persistence for continuous improvement. This is what matters at the end.
_________________
Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)
Last edited by duttsit on 18 Nov 2005, 15:53, edited 1 time in total.
Senior Manager
Joined: 30 Oct 2004
Posts: 284
Followers: 1
Kudos [?]: 62 [0], given: 0
### Show Tags
18 Nov 2005, 11:15
All the best with your apps!
_________________
-Vikram
VP
Joined: 18 Nov 2004
Posts: 1440
Followers: 2
Kudos [?]: 37 [0], given: 0
### Show Tags
18 Nov 2005, 11:16
Antmavel,
Sorry I have not been around much in the forum lately. This app season is killing me - will be back in full swing when I am done....but I was wondering abt your progress the other day.....u did gr8 frnd....nearly a 100 point increase is not a small feat. 640 is a good score, so if u r apping in Round 2, write those gr8 essays and u will be all set to get into a gr8 school. Remember GMAT is just a mean to an end (MBA) and not an end itself.....I am in the app process now, trust me MBA app is much much more than just GMAT/GPA....clear goals, passion, essays are the key.
Good luck and party for now!
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19
Kudos [?]: 291 [0], given: 0
### Show Tags
18 Nov 2005, 17:59
Antoine, 640 is a good score, you raised your score in such a short time . All the best to your applications~!!! ^_^ ...if I have time, I'd love to drop by Shenzhen soon ..or you can come to Guangzhou, I'll make you feel homely ^_~
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7
Kudos [?]: 45 [0], given: 0
### Show Tags
18 Nov 2005, 20:13
Thank you guys
thanks for your support and your help during all the time I studied the GMAT.
I need now to select several schools and apply.
I admit that my target score was 660 but with 640 it's just not enough for the best universities and at the same time I would like to get soemthing quite high too is top 10 too tight with 640 ...
I am french, caucasian
27 years old when i will enter the MBA (2006)
Born and lived in Africa for 15 years (Cameroon 7 + Ivory Coast 8)
International Busines School in Paris for 4 years with 1.5 year internship (Spain + France + USA)
3 years of experience in China (Sales Manager for a manufactruing company, my sales account are Philips, Nokia, Sony Ericsson, Valeo, ...)
Gmat results :
550 (Q39 V27) in Sept 04
500 (Q38 V23) in Nov 04
620 (Q48 V27) in Sept 05
640 (Q48 V31) in Nov 05
Toefl : 270
GPA : 3
I want to target as high as i can
I am not interested by Marketing but except that any other general MBA or Finance MBA would be interesting for me. Which schools could I target in Round 2 ?
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 40
Kudos [?]: 429 [0], given: 0
### Show Tags
18 Nov 2005, 20:44
LBS has asked us if we would like to send out an information package to any acquaintance for the program. If you would like, I could have them send you a package for the masters in finance which is a 1 year program with a focus, obviously, in finance. My roommate got in with a 640 GMAT so do not worry too much if the rest of your application is fine. Furthermore, we can also meet when you'll be back to Paris for your xmas holidays. We can then talk about the program at length.
_________________
Best Regards,
Paul
SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5
Kudos [?]: 74 [0], given: 0
### Show Tags
18 Nov 2005, 21:23
Paul wrote:
LBS has asked us if we would like to send out an information package to any acquaintance for the program. If you would like, I could have them send you a package for the masters in finance which is a 1 year program with a focus, obviously, in finance. My roommate got in with a 640 GMAT so do not worry too much if the rest of your application is fine. Furthermore, we can also meet when you'll be back to Paris for your xmas holidays. We can then talk about the program at length.
paul is a real ambassdor (LBS). more than that he is always willing to support/help people arround him.
you are really nice and gr8, paul. you will sucess in every aspects of life.
wish you all the best to you to paul at LBS.
himalaya..
Director
Joined: 31 Aug 2004
Posts: 610
Followers: 3
Kudos [?]: 120 [0], given: 0
### Show Tags
19 Nov 2005, 08:11
Congrats Antoine for this steep increase !
Just work hard your essays : in retrospect I spent 9 months of my life on the gmat, whereas it took me "only" 2 months to complete two application forms.
The end of the story is Ok as I am in at my 1st pick but if I were not, I probably would have been sorry that I spent a so unbalanced part of my time on the most important point of the apps.
About the choice of schools which fit your ambition I do think it is something really personal and it does not have to be altered by gmat in any way (unless the score is out of range which is not your case).
Keep posting!
Take care.
F
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2
Kudos [?]: 57 [0], given: 0
### Show Tags
19 Nov 2005, 10:17
Good luck Antmavel. I wish you the better in your application time. I have a question for you. How long and how difficult did you find RC in the actual test??
Regards,
automan
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7
Kudos [?]: 45 [0], given: 0
### Show Tags
19 Nov 2005, 12:03
automan wrote:
Good luck Antmavel. I wish you the better in your application time. I have a question for you. How long and how difficult did you find RC in the actual test??
Regards,
automan
PM me if you want more details but, for the record, RC was definitely doable.
4 total Rc's
2 very long + 2 very short
I would say that 1 was really hard but the others were definitely easy to get
by the way, I am not sure if all of you are aware of this, but some people got 5 RC' during their test
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7
Kudos [?]: 98 [0], given: 0
### Show Tags
20 Nov 2005, 11:16
Hi Antmavel,
Congratulations!!!
i am happy to learn about your gmat score. 640 is a good score. i wish your application with top business School will be a sucess.
Good Luck to u..
MA
20 Nov 2005, 11:16
Go to page 1 2 Next [ 23 posts ]
Similar topics Replies Last post
Similar
Topics:
feeling ok about 640 - my somewhat unique and long story 1 20 Mar 2015, 22:45
10 640 to 720 in a month (q48, v41) 22 21 Dec 2009, 15:56
640(Q: 48, V 31) - Need advise 2 14 Aug 2009, 01:19
1 640-Q 48, Verbal 31 5 05 Dec 2008, 03:48
The story of my 700 (Q48, V37) 6 20 Dec 2007, 06:42
Display posts from previous: Sort by | 3,736 | 12,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-04 | latest | en | 0.938919 |
https://www.physicsforums.com/threads/it-is-a-electromagnetics-problem.101890/ | 1,606,499,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00246.warc.gz | 799,142,689 | 16,209 | # It is a Electromagnetics problem ~
It is a Electromagnetics problem.. plz help~
In certain experiments it is desirable to have a region of constant magnetic
flux density. This can be created in an off-center cylindrical cavity that is
cut in a very long cylindrical conductor carrying a uniform current density.
The uniform axial current density is J=az j
Find the magnitude and direction of B in the cylindrical cavity whose axis
is dispalced that of the conducting part by a distance d
I am depressed because of this problem. I have no know idea...... ㅜㅜ
so... I submit this forum. plz help~
reference : David K. Cheng, Field and Wave Electromagnetics, Second Edition, Addison-Wesley(1989)
Problem 6-15
#### Attachments
• 25.5 KB Views: 236
## Answers and Replies
Related Engineering and Comp Sci Homework Help News on Phys.org
berkeman
Mentor
Hi fufu, Sorry it took me a while to get back to your question -- probably you already have the answer from your prof by now. But in case you're still working on this problem, I'll say a couple of things:
-- My initial reaction was to use Ampere's Law to figure out the B field distribution in the enclosed hollow cylinder that is caused by the uniform J current density in the conducting part of the cylinder. But Ampere's Law works better for symmetric geometries (like coaxial cable), so that's probably not the way to go on this problem.
-- Instead, I'd suggest solving the Biot-Savart Law integral for each point in the hollow cylinder slice, and using the following trick. Think of the geometry of this question as two conducting cylinders, superimposed on one another. The outer cylinder is a uniform conductor with radius b, and the current density is uniformly J across it. Then superimpose a conductor of radius a, displaced from the 1st conductor's center by the distance d as shown in your figure, but its current density is uniformly -J. So inside the outer cylinder's wall the current density is J, but inside the inner cylinder wall, the total current density adds up J-J=0. This is a way to set up your Biot-Savart integrals so that they are easier to do. Just figure out the field at each point inside the inner cylinder as the sum of all the current source elements throughout a uniform outer cylinder with +J, plus the current source elements throughout a uniform inner cylinder with -J. Using this trick to set up the integrations makes them a lot easier to do, with reasonable limits in cylindrical coordinates.
-- Even easier is probably to do the initial simplification from Biot-Savart for calculating the B field next to an infinitely long wire (uI/2PI*R), and use that with the opposing -J trick above to calculate the B field in the hollow cylinder area as the integral of all the little current elements in the overall cylinder(s). Hope that makes sense.
Good luck! Let us know if you get the right answer. -Mike-
Last edited: | 653 | 2,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-50 | latest | en | 0.934501 |
https://prod-eks-app-alb-1037681640.ap-south-1.elb.amazonaws.com/blog/machine-learning-and-its-breakthrough-applications/ | 1,725,913,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00165.warc.gz | 454,534,218 | 69,495 | Programs
# 5 Breakthrough Applications of Machine Learning
Machine Learning is the latest buzzword floating around, and quite rightly so. It’s one of the most interesting and fastest growing subfields of Computer Science. To put it simply, Machine Learning is what makes your Artificial Intelligence intelligent. Most people find the inner-workings of Machine Learning mysterious – but that’s far from the truth. If you’re just beginning to understand Machine Learning, let us make it easier by using an analogy:
## Top Machine Learning and AI Courses Online
Master of Science in Machine Learning & AI from LJMU Executive Post Graduate Programme in Machine Learning & AI from IIITB Advanced Certificate Programme in Machine Learning & NLP from IIITB Advanced Certificate Programme in Machine Learning & Deep Learning from IIITB Executive Post Graduate Program in Data Science & Machine Learning from University of Maryland To Explore all our certification courses on AI & ML, kindly visit our page below. Machine Learning Certification
You’re trying to throw a paper-ball into a dustbin.
After one attempt, you’ll get a fair idea of the amount of force you need to put. You put the required force in your second attempt, but the angle seems to be wrong. What is essentially happening here is that with each throw you’re learning something and bringing your outcome closer to the desired result. That is because we, humans, are inherently programmed to learn and grow from our experiences.
## Trending Machine Learning Skills
AI Courses Tableau Certification Natural Language Processing Deep Learning AI
Join the Artificial Intelligence Course online from the World’s top Universities – Masters, Executive Post Graduate Programs, and Advanced Certificate Program in ML & AI to fast-track your career.
Suppose you replace yourself with a machine. Now, we have two ways of going forward:
### Non-Machine Learning Approach
A generic, non-machine learning approach would be to measure the angle and distance and then use a formula to calculate the optimal force required. Now, suppose we add another variable – a fan that adds some wind force. Our non-ML program will fail almost certainly owing to the added variable. If we’re to get it work, we need to reprogram it keeping the wind factor in mind and the formula.
### Machine Learning Approach
Now, if we were to device a Machine Learning based approach for the same problem, it’d also begin with a standard formula – but, after every experience, it’d update/refractor the formula. The formula will get improved continuously using more experiences (known as ‘data points’ in the world of Machine Learning) – this will lead to improvements in the outcome as well. You experience these things on a daily basis in the form of your Facebook newsfeed, or custom curated YouTube suggestions or other things of this sort – you get the gist.
## What is Machine Learning?
The above analogy should make it clear that Machine Learning is simply using algorithms and processes to train your system to get better with experience. However, for the sake of a technical definition, a system is said to learn from the experiences with respect to a set of tasks, if its performance at the said tasks improves with time and experience.
What this essentially means is that in Machine Learning, the system improves its performance with experience. This is precisely what we noticed in our analogy as well.
### Types of Machine Learning
Depending on your problem statement, you can use either of the three techniques to train your system:
#### Supervised Learning
Supervised Machine Learning should be applied to datasets where the label/class of each data is known. Let us imagine we want to teach our system how to distinguish between the images of a dog and a human. Suppose we have a collection of pictures that are labeled as either human or dog (labeling is done by human annotators to ensure a better quality of data). Now, we can use this data set and data classes to train our algorithm to learn the right way. Once our algorithm learns how to classify images, we can use it on different data sets- to predict the label of any new data point.
#### Unsupervised Learning
As you can guess from the name, unsupervised Machine Learning is devoid of any supervising classes or labels. We just provide our system with a large amount of data and characteristics of each data piece. For example, suppose in our earlier example we just fed a number of images (of humans and dog) to our system giving each image a characteristic. Clearly, the characteristics of humans will be similar and different from dogs. Using these characteristics, we can train our system to group data into two categories. An unsupervised version of “classification” is called as “clustering”. In clustering, we don’t have any labels. We group the datasets on the basis of common characteristics.
#### Reinforcement Learning
In reinforcement learning, there are no classes or characteristics, there’s just an end-point – pass or fail. To understand this better, consider the example of learning to play chess. After every game, the system is informed of the win/loss status. In such a case, our system does not have every move labeled as “right” or “wrong”, but only has the end-result. As our algorithm plays more games during the training, it’ll keep giving bigger “weights” (importance) to the combination of those moves that resulted in a win.
## Breakthrough Applications in the field of Machine Learning
From our above discussion, it’s clear that Machine Learning can indeed solve a lot of problems that traditional computers just can not. Let’s look at some of the applications of Machine Learning that have changed the world as we know it:
### 1. Fighting Webspam
Google is using “deep learning” – it’s neural network, to fight spam both online and offline. Deep Learning uses data from the users and applies natural-language processing to conclude about the emails it encountered. Not only does it help the web-users, but also the SEO companies trying to help legitimate websites rank higher using white-hat techniques.
### 2. Imitation Learning
Imitation learning is very similar to observational learning – something we do as infants. This is extensively used in field robotics and in industries like agriculture, search, construction, rescue, military, and others. In all such situations, it’s extremely difficult to manually program the robots. To help with that, programming by demonstration – also known as collaborative methods is used coupled with Machine Learning. Take a look at this video published by Arizona state, which shows a humanoid robot learning to grasp different objects.
### 3. Assistive and Medical Tech
Assistive robots are robots that are capable of processing sensory information, and performing actions in times of need. The Smart Tissue Autonomous Robot (STAR) was created using this type of machine learning and real-world collaborations. STAR uses ML and 3D sensing and can stitch together pig intestines (used for testing) better than any surgeon. While STAR wasn’t developed to replace the surgeons, it does offer a collaborative solution for delicate steps in medical procedures.
Machine Learning also finds applications in the form of predictive measures. Like a colleague can look at a doctor’s prescription and find out what they might have missed, an artificially intelligent system too can find out the missing links in a prescription if trained well. Not only this, but AI can also look for patterns that point to possible heart failures. This can prove to be extremely helpful to doctors as they can collaborate with the virtual robot A.I to better diagnose a fatal heart condition before it strikes. The extra pair of eyes (and intelligence) can do more good than harm. Studies thus far also promise for the future application of this technology.
### 4. Automatic Translation/Recognition
Although it looks like a simple concept, ML can also be used to translate text (even from images) into any language. Using neural networks will help in the extraction of text from an image which can then be translated into the required language before putting it back into the picture. Other than this, ML is also used in every application that deals with any kind of recognition – voice, images, text, you name it!
### 5. Playing Video Games Automatically
This is one of the cooler applications of Machine Learning although it might not have that much of social utility like the others mentioned in the list. Machine Learning can be used to train Neural Networks to analyse the pixels on a screen and play a video game accordingly. One of the initial attempts at this was Google’s Deepmind.
## Popular AI and ML Blogs & Free Courses
IoT: History, Present & Future Machine Learning Tutorial: Learn ML What is Algorithm? Simple & Easy Robotics Engineer Salary in India : All Roles A Day in the Life of a Machine Learning Engineer: What do they do? What is IoT (Internet of Things) Permutation vs Combination: Difference between Permutation and Combination Top 7 Trends in Artificial Intelligence & Machine Learning Machine Learning with R: Everything You Need to Know AI & ML Free Courses Introduction to NLP Fundamentals of Deep Learning of Neural Networks Linear Regression: Step by Step Guide Artificial Intelligence in the Real World Introduction to Tableau Case Study using Python, SQL and Tableau
In Conclusion…
Having said that, Machine Learning isn’t the solution to all your problems. You don’t need machine learning to figure out a person’s age from his DOB, but you certainly need ML to figure out a person’s age from his music preferences. For example, you’ll find that fans of Johnny Cash and the Doors are mostly 35+ in age, whereas most of the Selena Gomez fans are under 20. Machine Learning *can* be used for any problem around you, but should it? Not really. Never use machine learning as a solution to your problems without being sure that you really need your machine to learn. Otherwise, it’d be like killing mosquitoes using machine guns – they might get killed, they might not, but at the end of the day, was it worth it?
## Is it beneficial to have fine knowledge of machine learning?
Not everything that is popular and expanding is suitable for everyone. Your growth may be hampered even if you enter an emerging field with no interest in or passion for it. As a result, you should make an informed decision about whether machine learning is something that actually interests you. If you enjoy coding and learning new programming languages, you should consider giving machine learning a go. A job as a machine learning engineer might be a good fit if you like problem-solving, are fascinated by data, and are a good communicator.
## Is it important to have a fine knowledge of mathematics to do well in machine learning?
Linear algebra, statistics, calculus, and probability are some of the areas of mathematics that are required in machine learning. If you want to grasp the ML concepts well and comprehend all the machine learning algorithms, you should know at least the basics of these areas. You do not have to be a mathematics wizard, but just knowing the fundamentals would make the work easier for you.
## What are the limitations of using machine learning in the education sector?
Machine learning takes away the scope of interaction from students, thus exaggerating their ability to engage socially. In grading papers, using a computer is not enough, as a teacher’s manual grade is still required to give a justified comment and result. It is also expensive to deploy machine learning to make education more personalized. | 2,301 | 11,743 | {"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-2024-38 | latest | en | 0.883861 |
http://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/2004/1/ | 1,600,684,423,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00725.warc.gz | 204,450,902 | 109,730 | # Properties
Label 2004.1 Level 2004 Weight 1 Dimension 26 Nonzero newspaces 1 Newform subspaces 6 Sturm bound 223104 Trace bound 0
## Defining parameters
Level: $$N$$ = $$2004\( 2004 = 2^{2} \cdot 3 \cdot 167$$ \) Weight: $$k$$ = $$1$$ Nonzero newspaces: $$1$$ Newform subspaces: $$6$$ Sturm bound: $$223104$$ Trace bound: $$0$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{1}(\Gamma_1(2004))$$.
Total New Old
Modular forms 1716 358 1358
Cusp forms 56 26 30
Eisenstein series 1660 332 1328
The following table gives the dimensions of subspaces with specified projective image type.
$$D_n$$ $$A_4$$ $$S_4$$ $$A_5$$
Dimension 26 0 0 0
## Trace form
$$26q + 4q^{4} - 4q^{6} + 4q^{9} + O(q^{10})$$ $$26q + 4q^{4} - 4q^{6} + 4q^{9} + 4q^{16} - 4q^{24} - 18q^{25} + 4q^{36} - 11q^{42} - 11q^{48} - 18q^{49} + 7q^{54} + 4q^{64} + 11q^{72} + 4q^{81} + 11q^{84} - 8q^{85} - 4q^{96} - 8q^{97} + O(q^{100})$$
## Decomposition of $$S_{1}^{\mathrm{new}}(\Gamma_1(2004))$$
We only show spaces with odd parity, since no modular forms exist when this condition is not satisfied. Within each space $$S_k^{\mathrm{new}}(N, \chi)$$ we list the newforms together with their dimension.
Label $$\chi$$ Newforms Dimension $$\chi$$ degree
2004.1.d $$\chi_{2004}(1337, \cdot)$$ None 0 1
2004.1.e $$\chi_{2004}(1669, \cdot)$$ None 0 1
2004.1.f $$\chi_{2004}(1003, \cdot)$$ None 0 1
2004.1.g $$\chi_{2004}(2003, \cdot)$$ 2004.1.g.a 1 1
2004.1.g.b 1
2004.1.g.c 2
2004.1.g.d 2
2004.1.g.e 10
2004.1.g.f 10
2004.1.k $$\chi_{2004}(23, \cdot)$$ None 0 82
2004.1.l $$\chi_{2004}(7, \cdot)$$ None 0 82
2004.1.m $$\chi_{2004}(13, \cdot)$$ None 0 82
2004.1.n $$\chi_{2004}(29, \cdot)$$ None 0 82
## Decomposition of $$S_{1}^{\mathrm{old}}(\Gamma_1(2004))$$ into lower level spaces
$$S_{1}^{\mathrm{old}}(\Gamma_1(2004)) \cong$$ $$S_{1}^{\mathrm{new}}(\Gamma_1(167))$$$$^{\oplus 6}$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ ($$1 + T$$)($$1 - T$$)($$( 1 + T )^{2}$$)($$( 1 - T )^{2}$$)($$1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10}$$)($$1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10}$$)
$3$ ($$1 + T$$)($$1 - T$$)($$( 1 - T )^{2}$$)($$( 1 + T )^{2}$$)($$1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10}$$)($$1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10}$$)
$5$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$7$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)
$11$ ($$( 1 - T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )^{2}$$)($$( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )^{2}$$)
$13$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$17$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$19$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)
$23$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$29$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)
$31$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)
$37$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$41$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$43$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$47$ ($$( 1 - T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )^{2}$$)($$( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )^{2}$$)
$53$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$59$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$61$ ($$( 1 + T )^{2}$$)($$( 1 + T )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 + T^{2} )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )^{2}$$)
$67$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$71$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$73$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$79$ ($$1 + T^{2}$$)($$1 + T^{2}$$)($$1 + T^{4}$$)($$1 + T^{4}$$)($$( 1 + T^{2} )^{10}$$)($$( 1 + T^{2} )^{10}$$)
$83$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)($$( 1 - T )^{10}( 1 + T )^{10}$$)
$89$ ($$( 1 - T )( 1 + T )$$)($$( 1 - T )( 1 + T )$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T )^{2}( 1 + T )^{2}$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)($$( 1 - T + T^{2} - T^{3} + T^{4} - T^{5} + T^{6} - T^{7} + T^{8} - T^{9} + T^{10} )( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )$$)
$97$ ($$( 1 - T )^{2}$$)($$( 1 - T )^{2}$$)($$( 1 + T )^{4}$$)($$( 1 + T )^{4}$$)($$( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )^{2}$$)($$( 1 + T + T^{2} + T^{3} + T^{4} + T^{5} + T^{6} + T^{7} + T^{8} + T^{9} + T^{10} )^{2}$$) | 4,485 | 7,896 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-40 | latest | en | 0.411918 |
https://www.coursehero.com/file/221909/lecture24/ | 1,527,268,516,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867140.87/warc/CC-MAIN-20180525160652-20180525180652-00346.warc.gz | 738,662,814 | 70,293 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
lecture24
# lecture24 - Taking Stock IE170 Algorithms in Systems...
This preview shows pages 1–3. Sign up to view the full content.
IE170: Algorithms in Systems Engineering: Lecture 24 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University March 28, 2007 Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 1 / 24 Taking Stock Last Time Transitive Closure (Fast) Flows in Networks This Time Flows, Flows, Flows Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 2 / 24 Flows Flows in Networks G = ( V, E ) directed. Each edge ( u, v ) E has a capacity c ( u, v ) 0 If ( u, b ) E c ( u, v ) = 0 We will typically have a special source vertex s V , a sink vertex t V , and we will assume there exists paths from s v t v V The combination of all of these things ( G, s, t, c ) is known as a flow network . Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 3 / 24 Flows Net Flows A net flow is a function f : V × V R | V |×| V | that satisfies three conditions: 1 Capacity Constraints: f ( u, v ) c ( u, v ) 2 Skew Symmetry: f ( u, v ) = - f ( v, u ) u V, v V 3 Flow Conservation: v V f ( u, v ) = 0 u V \ { s, t } Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 4 / 24
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Flows More Flow An important value we will be worried about is the value of flow f = | f | = v V f ( s, v ) : The total flow out of the source. The Maximum Flow Problem Given G = ( V, E ) . source node s V , sink node t V , edge capacities c . Find a flow whose value is maximum. Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 5 / 24 Flows Lemma, Lemma, Lemma Recall Shorthand f ( X, Y ) = x X y Y f ( x, y ) . 1 f ( X, X ) = 0 X V 2 f ( X, Y ) = - f ( Y, X ) X, Y V 3 Let X, Y, Z V be such that X Y = , then f ( X Y, Z ) = f ( X, Z ) + f ( Y, Z ) f ( Z, X Y ) = f ( Z, X ) + f ( Z, Y ) 4 | f | = f ( V, t ) Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 6 / 24 Flows Cuts A cut of a (flow) network G = ( V, E ) is a partition of V into S and T = V \ S such that s S and t T For flow f , net flow across a cut is f ( S, T ) and the cuts capacity is c ( S, T ) = u S v T c ( u, v ) A minimum cut of G is a cut whose capacity is minimum Jeff Linderoth (Lehigh University) IE170:Lecture 24 Lecture Notes 7 / 24 Flows A Simple Upper Bound Flow Across Cuts Lemma For any cut ( S, T ) , f ( S, T ) = | f | Coronary :-) The value of any flow is no more than the capacity of any cut | f | = f ( S, T ) = u S v T f ( u, v ) u
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 1,063 | 3,709 | {"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} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.748902 |
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.20/share/doc/Macaulay2/InvariantRing/html/_relations_lp__Finite__Group__Action_rp.html | 1,674,904,260,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2023-06/segments/1674764499541.63/warc/CC-MAIN-20230128090359-20230128120359-00356.warc.gz | 261,728,637 | 2,600 | # relations(FiniteGroupAction) -- relations of a finite group
## Synopsis
• Function: relations
• Usage:
relations G
• Inputs:
• Outputs:
• a list, a list of relations of the group
## Description
This function is provided by the package InvariantRing.
Use this function to get the relations among elements of a group. Each element is represented by a word of minimal length in the Coxter generators. And each relation is represented by a list of two words that equates the group element represented by those two words.
The following example defines the permutation action of a symmetric group on three elements using three transpositions.
i1 : R = QQ[x_1..x_3] o1 = R o1 : PolynomialRing i2 : L = {matrix {{0,1,0},{1,0,0},{0,0,1}}, matrix {{0,0,1},{0,1,0},{1,0,0}}, matrix {{1,0,0},{0,0,1},{0,1,0}} } o2 = {| 0 1 0 |, | 0 0 1 |, | 1 0 0 |} | 1 0 0 | | 0 1 0 | | 0 0 1 | | 0 0 1 | | 1 0 0 | | 0 1 0 | o2 : List i3 : G = finiteAction(L, R) o3 = R <- {| 0 1 0 |, | 0 0 1 |, | 1 0 0 |} | 1 0 0 | | 0 1 0 | | 0 0 1 | | 0 0 1 | | 1 0 0 | | 0 1 0 | o3 : FiniteGroupAction i4 : relations G o4 = {{{}, {1, 1}}, {{}, {2, 2}}, {{1}, {0, 1, 2}}, {{1}, {0, 2, 0}}, {{}, ------------------------------------------------------------------------ {0, 0}}, {{0, 2}, {1, 0}}, {{0, 2}, {2, 1}}, {{0, 1}, {1, 2}}, {{0, 1}, ------------------------------------------------------------------------ {2, 0}}, {{2}, {0, 1, 0}}, {{2}, {0, 2, 1}}} o4 : List | 537 | 1,437 | {"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} | 2.9375 | 3 | CC-MAIN-2023-06 | latest | en | 0.674864 |
https://howmanyis.com/area/192-ha-in-ac/46041-10-hectares-in-acres | 1,656,660,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00017.warc.gz | 344,040,685 | 6,267 | How many is
Conversion between units of measurement
Rating 4.00 (3 Votes)
You can easily convert 10 hectares into acres using each unit definition:
Hectares
hectoare = 1e2 are = 10000 m²
Acres
usacre = 10 surveychain² = 4046.8726 m²
With this information, you can calculate the quantity of acres 10 hectares is equal to.
## ¿How many ac are there in 10 ha?
In 10 ha there are 24.710439 ac.
Which is the same to say that 10 hectares is 24.710439 acres.
Ten hectares equals to twenty-four acres. *Approximation
### ¿What is the inverse calculation between 1 acre and 10 hectares?
Performing the inverse calculation of the relationship between units, we obtain that 1 acre is 0.040468726 times 10 hectares.
A acre is zero times ten hectares. *Approximation | 205 | 763 | {"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-2022-27 | latest | en | 0.848449 |
http://www.brightstorm.com/tag/degree/page/4 | 1,432,612,115,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928757.11/warc/CC-MAIN-20150521113208-00044-ip-10-180-206-219.ec2.internal.warc.gz | 349,837,555 | 13,521 | • #### Graphing Polynomial Functions - Concept
##### Math›Precalculus›Polynomial and Rational Functions
How we identify the end behavior of a polynomial functions.
• #### Limits at a Glance - Problem 1
##### Math›Precalculus›Polynomial and Rational Functions
How to quickly find the limit of a rational function as x goes to infinity.
• #### Polynomial Vocabulary - Problem 2
##### Math›Algebra›Polynomials
How to give a polynomial a two word name, including monomial, binomial, trinomial, polynomial, constant, linear, quadratic, cubic, quartic, or n-th degree.
• #### Factoring: Special Cases Part II - Concept
##### Math›Algebra›Factoring
How to factor a third degree polynomial by grouping.
• #### Limits at a Glance - Problem 2
##### Math›Precalculus›Polynomial and Rational Functions
How to quickly find the limit of a rational function (cubic polynomials) as x goes to infinity.
• #### Calculating Coordinates in the Unit Circle - Problem 2
##### Math›Trigonometry›Pythagorean Theorem
How to find the coordinates on the unit circle, given an angle of 60 degrees.
• #### Graphing Polynomial Functions - Problem 1
##### Math›Precalculus›Polynomial and Rational Functions
How we graph a factored cubic (3rd degree polynomial) function.
• #### Graphing Polynomial Functions - Problem 3
##### Math›Precalculus›Polynomial and Rational Functions
How we graph a factored quintic (5th degree polynomial) function.
• #### Calculating Coordinates in the Unit Circle - Problem 1
##### Math›Trigonometry›Pythagorean Theorem
How to find the coordinates on the unit circle, given an angle of 225 degrees.
• #### Calculating Coordinates in the Unit Circle - Problem 2
##### Math›Geometry›Pythagorean Theorem
How to find the coordinates on the unit circle, given an angle of 60 degrees.
• #### Calculating Coordinates in the Unit Circle - Problem 1
##### Math›Geometry›Pythagorean Theorem
How to find the coordinates on the unit circle, given an angle of 225 degrees.
• #### Graphing Polynomial Functions - Problem 2
##### Math›Precalculus›Polynomial and Rational Functions
How we graph a factored quartic (4th degree polynomial) function.
• #### Factoring: Special Cases Part II - Problem 2
##### Math›Algebra›Factoring
How to factor by grouping if given a third degree polynomial.
• #### Graphing Rational Functions, n=m - Concept
##### Math›Precalculus›Polynomial and Rational Functions
How to recognize when a rational function has a horizontal asymptote, and how to find its equation.
• #### The Second Derivative Test for Relative Maximum and Minimum - Problem 3
##### Math›Calculus›Applications of the Derivative
How to use the second derivative test to find the relative maxima and minima of a fourth degree polynomial function with five terms.
Tags:
• #### The Second Derivative Test for Relative Maximum and Minimum - Problem 1
##### Math›Calculus›Applications of the Derivative
How to find the relative maxima and minima of a fourth degree polynomial function with three terms, using the second derivative test.
Tags:
• #### Simplifying Rational Functions with Factoring and GCFs - Problem 10
##### Math›Algebra›Rational Expressions and Functions
Factoring higher degree polynomials to reduce rational expressions and find excluded values.
Tags:
• #### The Resultant of Two Forces - Problem 2
##### Math›Precalculus›Vectors and Parametric Equations
How to find the resultant of two forces at an acute angle (45 degrees).
• #### Graphing Rational Functions, n>m - Concept
##### Math›Precalculus›Polynomial and Rational Functions
How to find the oblique asymptote of a rational function, if it has one.
• #### Simplifying Radicals using Rational Exponents - Problem 1
##### Math›Algebra 2›Roots and Radicals
How to simplify a higher degree radical using rational exponents. | 874 | 3,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2015-22 | longest | en | 0.710844 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.