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## Isothermal pressure change
Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$
Samantha Joseph 1F
Posts: 48
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time
### Isothermal pressure change
Why is it that the equation for finding the entropy of a system with an isothermal pressure change uses P1/P2 when temperature and volume are typically T2/T1 or V2/V1?
Ju-Wei Wang 1I
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am
### Re: Isothermal pressure change
According to the ideal gas law, specifically Boyle's Law, pressure is inversely related to volume, so at constant temperature, V2/V1=P1/P2.
Matthew 1C
Posts: 47
Joined: Fri Sep 29, 2017 7:06 am
### Re: Isothermal pressure change
I think the reason P1/P2 is used is because it is inversely proportional to volume, which uses V2/V1. Also since both pressure and volume are directly proportional to temperature, does it matter which one uses 1/2 or 2/1?
Ashley Macabasco 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am
### Re: Isothermal pressure change
The change in pressure equation is P1/P2 opposed to P2/P1 like for the change in temperature and the change in volume because of Boyle's law in which volume and pressure are inversely proportional.
Return to “Entropy Changes Due to Changes in Volume and Temperature”
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## Algebra and Trigonometry 10th Edition
$f^{-1} (x) =-3+\sqrt x$
When we apply the horizontal test, it has been noticed that the function is one-to-one and verifies the horizontal test. Therefore, the function has an inverse function. To compute the inverse, we will have to interchange $y$ and $x$. $x=(y+3)^2 \implies y+3= \pm \sqrt x$ or, $y=-3 \pm \sqrt x$ Since the domain for the function is $x \geq -3$, we neglect $-3-\sqrt x$ So, we have $y=-3+\sqrt x$ Replace $y$ with $f^{-1} (x) =-3+\sqrt x$
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A048401 Primes with consecutive digits that differ exactly by 4. 8
2, 3, 5, 7, 37, 59, 73, 151, 373, 15959, 95959, 515951, 595159, 595951, 9515959, 51515159, 159595151, 159595951, 5151515951, 5159515159, 5159515951, 5951515151, 5951515951, 5959515151, 5959595951, 15151595951, 15951515159 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Sean A. Irvine, Table of n, a(n) for n = 1..1000 MATHEMATICA pd[{a_, b_, c___}]:=Flatten[Table[Select[FromDigits/@Select[Tuples[ {a, b, c}, n], Union[Abs[Differences[#]]]=={4}&], PrimeQ], {n, 11}]]; Union[Join[{2, 3, 5, 7}, pd[{1, 5, 9}], pd[{3, 7}]]] (* Harvey P. Dale, Aug 23 2011 *) CROSSREFS Cf. A048398, A048399, A048400, A048402, A048403, A048404, A048405, A048406. Sequence in context: A090713 A090912 A048416 * A288718 A289861 A160748 Adjacent sequences: A048398 A048399 A048400 * A048402 A048403 A048404 KEYWORD nonn,base AUTHOR Patrick De Geest, Apr 15 1999 EXTENSIONS More terms from Naohiro Nomoto, Jul 28 2001 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified August 10 04:27 EDT 2022. Contains 356029 sequences. (Running on oeis4.)
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# Thin-Film Interference (2 Wavelengths)
• Const@ntine
In summary: I don't know. I have no idea what n is. And m is the order of the fringe, right? So, where to go from there?In summary, the problem involves two glass plaques separated by a wire and illuminated by light with two different wavelengths. The task is to determine the distance from the point of contact where the next dark fringe forms. To solve this, the equations 2nt = mλ and 2nt = (m+1/2)λ for destructive and amplified interference, respectively, are used. The two wavelengths provide an additional constraint, as they must both be dark at the same location for a dark fringe to occur. To find the minimum separation at which it is all dark, the
Const@ntine
## Homework Statement
Two glass plaques of length 10.0 cm osculate, on one end, and are separated by a wire of diameter d = 0.0500 mm on the other. Light with two wavelengths (400 nm & 600 nm) falls on one of them, and we can see it gets reflected. At which distance from the point of contact does the next dark fringe form?
## Homework Equations
2nt = mλ
2nt = (m+1/2)λ
dsinθlight = mλ
dsinθdark = (m+1/2)λ
## The Attempt at a Solution
[/B]
My problem here is that I can't really understand the significance of the two wavelengths, how to use that info and generally how to proceed. My book has a small paragraph on TFI, and only explains the basic stuff (the above formulas for a single instance and that's it). Anyway, I figured:
>d = t : So d, the diameter, plays the role of the thin film's thickness.
>The thin film is the air, and the glass plaques have the same ng, which while bigger than n = 1.00 of the air, is still the same from above and bellow the thin film/air, so I don't have to inverse the two formulas.
>So, 2nt = mλ for destructive confluence, and 2nt = (m+1/2)λ for amplified confluence.
Now, the only thing I came up with was: Take the first formula (for destruction) and try it with both λs, to see which one gives me an integer m. 400 gives me n = 250, whereas 600 gives me 166.6666...7. So, I figured I'd continue with the first wavelength. Then I went back to the chapter about light and dark fringes, and looked at the formulas.
Now it gets murky and I'm just throwing stuff at the wall to see what will stick, but my plan was to put in the data in the "dark fringe" equation, find the angle, then put it in tanθ = y/L and find y. But that didn't work out. So I need some help here.
Any help is appreciated!
Hmmm ... it does seem as if you are throwing stuff against the wall as you say. You only need your second equation, and the point is t varies from one edge of the plate to the other. You can use the information given to parameterize t as a function of position along the plate.
The two wavelengths give an additional constraint. For a dark fringe they both have to be dark at the same location.
Cutter Ketch said:
... You only need your second equation ...
Actually, on second thought, the two reflections being interfered have a 180 degree phase change between them. You only need your first equation.
Cutter Ketch said:
Hmmm ... it does seem as if you are throwing stuff against the wall as you say. You only need your second equation, and the point is t varies from one edge of the plate to the other. You can use the information given to parameterize t as a function of position along the plate.
Ah, so it's t = d at the far let and t ~ 0 at the far right. ...How do I translate that into a function?
Cutter Ketch said:
The two wavelengths give an additional constraint. For a dark fringe they both have to be dark at the same location.
How do I use that though? Do I take the equation two times, once with 400 & once with 600 ? Like I said, I only know the basics (ight fallson thin film, gets reflected) so I'm not that well-versed in the theory behind it all.
Cutter Ketch said:
Actually, on second thought, the two reflections being interfered have a 180 degree phase change between them. You only need your first equation.
You mean 2nt = mλ, right?
For some extra info, there was anothe exercise which went like this:
Same "construction" as in the OP, but here the light has a wavelength of λ = 600 nm, and we are informed that 30 dark fringes are created. Andwe have to find the diameter d. My logic on that one was:
>2nt = mλ, m = 0, 1, 2, 3...
>30 dark fringes, so m = 29
>t = d
>n = 1.00 (air)
Putting that all together I get d = 8.70 μm, which is the book's answer.
I'm writting that here since both the exercise in the OP and some others use the same construction/basic intel, and I figured it'd help if I posted how I handled the previous one (ie didn't use the "t decreases the further we move to the right" bit).
Darthkostis said:
How do I translate that into a function?
You don't need to. First find the minimum separation at which it is all dark, then convert to a distance from the end.
Darthkostis said:
Do I take the equation two times, once with 400 & once with 600 ?
Yes. You can find all the separations at which one will be dark, and all the separations at which the other will be dark. Then figure out where both will be.
haruspex said:
You don't need to. First find the minimum separation at which it is all dark, then convert to a distance from the end.
Yes. You can find all the separations at which one will be dark, and all the separations at which the other will be dark. Then figure out where both will be.
When you say "seperations", what do you mean? I'm asking because due to the translation problem I'm not familiar with the term.
Thanks!
Darthkostis said:
When you say "seperations", what do you mean? I'm asking because due to the translation problem I'm not familiar with the term.
Thanks!
I mean the separation between the two pieces of glass. Gap, if you prefer.
haruspex said:
I mean the separation between the two pieces of glass. Gap, if you prefer.
Oh, okay then. I'll look into it and report back after I've tried my hand at the exercise.
Thanks!
I looked at it again, but I can't come up with something. How do I translate "everything is dark" to the 2nt = mλ formula? λ is the wavelength, t is the gap/seperation, n is a constant (1.00 since the thin film is the air here), which leaves m.
Darthkostis said:
How do I translate "everything is dark" to the 2nt = mλ
If we just stick with one wavelength for the moment, the formula tells you for which values of t it will be dark. Just plug in m=1, 2, ... and solve.
Are you asking why that is true?
Putting in the other wavelength gives a different set of values of t. The question is asking for the smallest value of t (and hence of d) which is dark for both.
haruspex said:
If we just stick with one wavelength for the moment, the formula tells you for which values of t it will be dark. Just plug in m=1, 2, ... and solve.
So I just try different values for m, for both wavelengths, and try to find which ones give me the same t? For example, I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?
haruspex said:
Are you asking why that is true?
Putting in the other wavelength gives a different set of values of t. The question is asking for the smallest value of t (and hence of d) which is dark for both.
To be honest, I'm not really sure what it means. The "question" itself from the book is: "at which distance, from the point of contact, does the next dark fringe form". I took it to mean as "how far from the right side is the next dark fringe", but what does "next dark fringe" mean?
Darthkostis said:
I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?
Looks right to me. What does that correspond to as a distance from where the plates meet?
Darthkostis said:
what does "next dark fringe" mean?
Because of the 180 degree phase difference, the point where the plates meet will be a dark fringe for both wavelengths. You want the next such along.
Const@ntine
Darthkostis said:
So I just try different values for m, for both wavelengths, and try to find which ones give me the same t? For example, I get t = 600 nm for m = 3 for the λ = 400 nm, and t = 600 nm for m = 2 for the λ = 600 nm. Is that it?
To be honest, I'm not really sure what it means. The "question" itself from the book is: "at which distance, from the point of contact, does the next dark fringe form". I took it to mean as "how far from the right side is the next dark fringe", but what does "next dark fringe" mean?
Because of the phase change of the ray reflected from the top surface of the bottom plate dark fringes (lines) are centred at places where the path difference (2t) = 0 or an integral number of wavelengths.
Each wavelength produces its own interference pattern and these two patterns overlap.
The resultant pattern can be rather complicated, for example there will be places where a bright fringe from one wavelength overlaps a dark fringe from the second wavelength. The result is that a fairly bright fringe is seen.
But there will also be places where two bright fringes overlap to make a brighter fringe.
And there will be places where two dark fringes overlap resulting in a dark fringe.
The question is about the dark fringes. Along the line of contact the path difference is zero for both wavelengths and that's where the first dark fringe is observed.
As you move to the left you can reach places where both wavelengths produces other dark fringes, the resultant fringe being dark.
Basically the question is asking you to find the second place at which this happens, the first place being along the line of contact.
But remember m, as in your equation, will have different values for each wavelength. I suggest that you use m for one wavelength, as you have done already, and a different symbol, for example n, for the second wavelength.
Const@ntine
haruspex said:
Looks right to me. What does that correspond to as a distance from where the plates meet?
Because of the 180 degree phase difference, the point where the plates meet will be a dark fringe for both wavelengths. You want the next such along.
Because of the phase change of the ray reflected from the top surface of the bottom plate dark fringes (lines) are centred at places where the path difference (2t) = 0 or an integral number of wavelengths.
Each wavelength produces its own interference pattern and these two patterns overlap.
The resultant pattern can be rather complicated, for example there will be places where a bright fringe from one wavelength overlaps a dark fringe from the second wavelength. The result is that a fairly bright fringe is seen.
But there will also be places where two bright fringes overlap to make a brighter fringe.
And there will be places where two dark fringes overlap resulting in a dark fringe.
The question is about the dark fringes. Along the line of contact the path difference is zero for both wavelengths and that's where the first dark fringe is observed.
As you move to the left you can reach places where both wavelengths produces other dark fringes, the resultant fringe being dark.
Basically the question is asking you to find the second place at which this happens, the first place being along the line of contact.
But remember m, as in your equation, will have different values for each wavelength. I suggest that you use m for one wavelength, as you have done already, and a different symbol, for example n, for the second wavelength.
Thanks a lot for the info and help! It seems the wording of the questiong messed me up a bit. Here's the completed exercise:
>Same kind of materialabove and bellow the air.
>Thus 2nt = mλ is the formula for the dark fringes.
>We're looking for the minimal "t" (height/gap/seperation between the two plaques) for which both wavelengths produce a dark fringe.
λ1: t = m*200 nm
For m = 1: t = 200 nm
For m = 2: t = 400 nm
For m = 3: t = 600 nm
λ2: t = n*300 nm
For m = 1: t = 300 nm
For m = 2: t = 600 nm
>So t = 600 nm is the minimal one.
Now, the two plaques create an tiny angle between them, close to the point of contact, which we can find.
tanθ = d/L <=> ... <=> θ = 5*10-4 rad
The angle remains the same, and using it and the minimal t, we can find how faritis from the point of contact.
tanθ = t/l <=> ... <=> t = 1.20 mm (which is the book's answer)
Thanks a ton for the help and patience everybody!
## 1. What is thin-film interference?
Thin-film interference occurs when light waves reflect off of two different surfaces of a thin film, causing them to interfere with each other. This interference can result in changes in the color or brightness of the reflected light.
## 2. How are two wavelengths involved in thin-film interference?
In thin-film interference, two wavelengths of light are involved because the light waves reflect off of two different surfaces of the thin film. This causes the two waves to interfere with each other, resulting in changes in the color or brightness of the reflected light.
## 3. What is the equation for calculating the path difference in thin-film interference?
The equation for calculating the path difference in thin-film interference is given by 2nt cosθ, where n is the refractive index of the film, t is the thickness of the film, and θ is the angle of incidence.
## 4. How does the thickness of the thin film affect the interference pattern?
The thickness of the thin film affects the interference pattern by changing the path difference between the two light waves. A thicker film will result in a larger path difference, leading to a different interference pattern compared to a thinner film.
## 5. What are some real-life applications of thin-film interference?
Thin-film interference has many practical applications, such as in anti-reflective coatings for glasses and camera lenses, interference filters for optical devices, and in the colorful patterns seen on soap bubbles and oil slicks.
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Since road cars are generally symmetrical (circle-track cars may actually have an asymmetrical chassis), establishing the centerline helps you know how far the suspension is offset from the center of the vehicle.
First, pick two (hopefully) symmetrical points on the front of the car and transfer them to the ground using the plumb bob. Here I used the bolt holes on the front frame rail where the front bumper would attach. Since the frame rails are probably the most symmetrical thing on the car, this is a good bet.
On an old, beat up chassis, are they going to be perfectly symmetrical? Highly unlikely. Is this the best bet you can make given all of the other variables and circumstances? Yes.
Just because the CMM says that the measured point is at X, Y, Z coordinate, the question is really where that coordinate is in relation to the vehicle itself, not where it is with the CMM machine as the origin.
You need to “measure”/calibrate the coordinate system every time you start taking measurements to establish the known vehicle reference point. You noticed that my car was near the ground. The CMM arm is long, and they make longer ones, but it’s not long enough to reach under the car to the center.
Here we have mapped a point 4 feet away from the vehicle centerline. This is at the front of the vehicle. We will then repeat this measure-symmetry-center-point-offset procedure near the back of the car to establish a rear center point and a rear 4-foot point.
Connect the dots on the front and rear 4-foot points, and you now have a line that is a known offset from the actual vehicle center line that is also easy to access with the CMM.
I didn’t want to take off the rear bumper, so I used the inner rear lower control arm bolt holes as my reference points for finding the rear center point. One could argue this isn’t smart because the subframe may be offset from the vehicle due to bushing wear or etc. You’re not going to do much better than this, though.
If the subframe is offset and you measure the chassis center, you’re still ending up measuring the offset of the subframe somehow when you measure the suspension. Seems like six-of-one / half-dozen-of-another to me. There’s lots of slop in all of this process. The goal is to simply minimize the slop as well as is possible within reason.
1. Dan DeRosia says:
There’s places that will do tire testing and it’s… expensive but if you keep the test matrix down, in the 4-figures level. Would be interesting to add that to the mix. I’m figuring that most of the time in sims it’s not that the tire model itself has issues, just that it’s filled with variables that people are guessing at.
1. joe says:
The most important thing is that the driver understands how the tire develops grip, and how to keep it at the proper temp. You can collect as much data as you want, but if the driver doesn’t understand that data, it is worthless. The pneumatic tire has been around for over a hundred years and there’s only one book written about it? Sad.
https://www.sae.org/publications/books/content/r-351/
2. Erik W Lombard says:
Very interesting, great work Erik! I had no idea this type of equipment could be rented and used by near mortals.
The wheels are turning on other ways this type of equipment could be used.
3. @Dan: Where / who does the tire testing? I do have a spare that I could trash…
@Erik L: Oh yeah? What are you thinking of measuring/scanning?
1. Dan DeRosia says:
Calspan does – if you are willing to be flexible on schedule and don’t get too complicated it helps hold the cost down, and the gent I was talking to was willing to deal with it as a smaller project.
1. Dan – if you find me on Facebook or email me at erikmjacobs (gmail) would you tell me who you spoke with? I’d be curious to talk to that person.
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135) During the month of March, Harley's Computer Services made purchases on account totaling \$43,500. Also during the month of March, Harley was paid \$8,000 by a customer for services to be provided in the future and paid \$36,900 of cash on its accounts payable balance. If the balance in the accounts payable account at the beginning of March was \$77,300, what is the balance in accounts payable at the end of March? A) \$83,900. B) \$91,900. C) \$6,600. D) \$75,900. E) \$4,900. Answer: A Explanation: Beginning Accounts Payable Balance + Purchases on Account − Payments on Accounts = Ending Accounts Payable Balance \$77,300 + \$43,500 − \$36,900 = Ending Accounts Payable Balance Ending Accounts Payable = \$83,900
Accounts Payable 77,300 36,900 43,500 83,900 Difficulty: 3 Hard Topic: Analyzing Transactions Learning Objective: 02-A1 Analyze the impact of transactions on accounts and financial statements. Bloom's: Apply AACSB/Accessibility: Analytical Thinking / Keyboard Navigation AICPA: BB Industry; FN Measurement
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# CAT2018-1: 70
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Point $P$ lies between points $A$ and $B$ such that the length of $BP$ is thrice that of $AP$. Car $1$ starts from $A$ and moves towards $B$. Simultaneously, car $2$ starts from $B$ and moves towards $A$. Car $2$ reaches $P$ one hour after car $1$ reaches $P$. If the speed of car $2$ is half that of car $1$, then the time, in minutes, taken by car $1$ in reaching $P$ from $A$ is _________.
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## Related questions
1
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A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in $6$ hours when $6$ filling and $5$ draining pipes are on, ... $6$ draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
2
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While multiplying three real numbers, Ashok took one of the numbers as $73$ instead of $37$. As a result, the product went up by $720$. Then the minimum possible value of the sum of squares of the other two numbers is _________
A right circular cone, of height $12$ ft, stands on its base which has diameter $8$ ft. The tip of the cone is cut off with a plane which is parallel to the base and $9$ ft from the base. With $\pi = 22/7$, the volume, in cubic ft, of the remaining part of the cone is ________
Train $T$ leaves station $X$ for station $Y$ at $3$ pm. Train $S$, traveling at three quarters of the speed of $T$, leaves $Y$ for $X$ at $4$ pm. The two trains pass each other at a station $Z$, where the distance between $X$ and $Z$ is three-fifths of that between $X$ and $Y$. How many hours does train $T$ take for its journey from $X$ to $Y$?
Each of $74$ students in a class studies at least one of the three subjects $H, E$ and $P$. Ten students study all three subjects, while twenty study $H$ and $E$, but not $P$. Every student who studies $P$ also studies $H$ or $E$ or both. If the number of students studying $H$ equals that studying $E$, then the number of students studying $H$ is _________
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# How Many Moles Of Nitrogen, N, Are In 90.0 G Of Nitrous Oxide, N2O?
Contents
=1.93⋅mol .
## How many moles of nitrogen N are in 81.0 g of nitrous oxide n2o?
There are 3.68 mol of N in 81.0 g N2O.
## How many moles of nitrogen N are in 78.0 g of nitrous oxide?
3.54 moles
There are 3.54 moles of N in nitrous oxide. The first step in solving this problem is to convert 78.0 grams of nitrous oxide to moles.
## How many moles of nitrogen N are in 69.0 g of nitrous oxide n2on2o?
And guys that will be our answer. Okay we’re done. We just need to go ahead and calculate this and that will give us 3.14 moles off nitrogen. Okay so in 69 g of nitrous oxide there’s 3.14 g.
## How many moles of N are in 0.201 g of N2O?
First calculate the number of moles of nitrous oxide by its molar mass as follows: 0.201 g / (44.02 g/mol) = 0.004566 moles.
## How many moles of N are in .167 g of N2O?
Explanation: Moles of N2O = 0.167⋅g44.01⋅g⋅mol−1=3.80×10−3⋅mol . And the molar quantity of nitrogen ATOMS is twice this molar quantity: Moles of nitrogen atoms=2×3.80×10−3⋅mol=7.60×10−3⋅mol………
## How many moles of nitrogen nn are in 65.0 GG of nitrous oxide N2O N 2 O?
2.94 mol
Hence 2.94 mol nitrogen is present in 65 grams of nitrous oxide.
## How many moles is 84 g of nitrogen dioxide?
Therefore 6 moles of N atoms contain 84 grams …
## How many molecules are in 9 moles of H2S?
There are 5.42 x 1024 molecules in a 9.00-mole sample of H2 S. Using Avogadro’s number for the number of molecules in a…
## How many molecules are there in 2.509 moles of H2S?
How many molecules are present in 2.509 mol oh H2S? 1.511 x 10^24.
## How many g of sodium chloride are present in 0.236 mol of sodium chloride?
There are 0.236 moles in 13.8 g of sodium chloride.
## How many moles is N2O?
1 mole
Answer and Explanation: Each mole of nitrous oxide N2O N 2 O contains 1 mole of nitrogen atoms. The quantity of moles n of nitrous oxide can be calculated from its mass m by dividing the latter by the molar mass MW of nitrous oxide which is 44.01 g/mol.
15.999 u
## How many moles of N are in 0.197 g of n2o?
0.00896 moles
There are 0.00896 moles of N.
## How many moles of N are in 0.205 g of n2o0 205 g n2o?
00466 mol N2O * ( 2 mol N / 1 mol N2O) = . 00932 mol N!
## How many moles of N are there in 0.251 g of n2o?
Question: How many moles of N are in 0.251 g of N20? Number 00994 mol N There is a hint available!
## What is the mass percent of sodium in the formula naclo4?
This would mean the masses of Na and Cl are 22.99g and 35.45g respectively. Therefore the percent composition of NaCl is 39.3% sodium and 60.7% chloride.
## What is molar mass units?
Kilogram per mole
## How many moles are contained in 14 g of nitrogen dioxide?
grams is the molar mass of nitrogen meaning that one mole of nitrogen atoms is 14 grams. One mole of any substance has 6.02*10^23 atoms/molecules of that subtance therefore there are 6.02*10^23 atoms of nitrogen in 14g of nitrogen.
## How many moles of gas are in 45 g of NO2?
0.978 mol
This is a good example of a dimensional analysis question! The first step is to determine how many moles of NO2 are in 45.0g. For this we need the molar mass of NO2 which is 46.0g. Therefore there are 45.0g NO2 (1 mol of NO2/46.0g) = 0.978 mol of NO2.
## How many moles of Na are in 42g of Na?
Answer: 0.61 moles of Na are in 42 g of Na.
## How many moles are in 6.84 g of CUNH4Cl3?
How many moles are in 6.84g of CuNH C13? – volg Cantilla Imol CuNhuCh 0.0364 mol. 1889 CUNH4Cl3 16.
## How many molecules are in 4 moles of H2S?
And thus for H2S we have 4⋅mol×6.022×1023⋅mol−1 = 24×1023 molecules.
## How many molecules are in 6 moles of H2S?
=3. 61 X 10²⁴ molecules.
## How many molecules of H2O are in 200.0 g of H2O?
Avogadro’s number is approximately 6.022 × 1023. Thus 18.01508 grams of water contain 6.022 × 1023 water molecules. A glass of water or 200 grams of water therefore contains 6.6855 × 1024 water molecules.
## How many H2S molecules are in 10.47 grams of H2S?
How many H2S molecules are in 10.47 grams of H2S? Do the first step: 10.47 grams = moles Now take your answer from the first step and use it in the second step. The final answer is: There are H25 molecules in 10.47 grams of H2S.
## How many molecules of H2S are present in 0.4 mole of H2S?
1mole = 1 gram atom. Therefore 0.40 mole of H2S contains 2 x 0.40 = 0.80 gram atoms of H and 0.40 gram atoms of S.
## What is the molar mass of Cr₃ Po₄ ₂?
The molar mass of chromium is 52.00 g the molar mass of phosphorus is 30.97 g and oxygen has a molar mass of 16.00 g. There are three moles of chromium two moles of phosphorus and eight moles of oxygen per formula unit of chromium phosphate Cr₃(PO₄)₂.
## How many moles of salt are in 13.8 grams of NaCl?
See also a change that occurs when a solid melts or a liquid freezes is an example of a phase change.
Na is 23.0g and Cl is 35.5g which you add together to get the molar mass of the two is 58.5g/mole. Take the 13.8g of NaCl and divide it by the 38.5g will give you the moles: 13.8g NaCl(1 mole/58 5g)=. 236 moles of NaCl.
## How many moles are in 58.5 grams of sodium chloride?
0.855 moles
Na has an atomic mass of 23 and Cl is 35.5. So the Mr(NaCl)= 23 + 35.5 = 58.5. Putting all of this into our moles equation gives us the answer: moles= 50/58.5 = 0.855 moles (3sf).
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### Joule
[crater] This is a worn and eroded crater formation. A pair of smaller craters lies along the northeastern rim, and a crater is intruding into the northwest rim. To the south is an outward projection that has the appearance of a crater partly overlain by Joule. The remainder of the rim and inner wall is somewhat irregular. The interior floo
Found on http://en.wikipedia.org/wiki/Joule_(crater)
### Joule
[programming language] Joule is a concurrent dataflow programming language, designed for building distributed applications. It is so concurrent, that the order of statements within a block is irrelevant to the operation of the block. Statements are executed whenever possible, based on their inputs. Everything in Joule happens by sending mes
Found on http://en.wikipedia.org/wiki/Joule_(programming_language)
### Joule
1) Work done by the force of one neutron when its point of application moves through the distance of one meter in the direction of the force. 2) One watt-second.
Found on http://www.youngco.com/young2.asp?ID=4&Type=3
### Joule
A metric unit of energy or work; 1 joule per second equals 1 watt or 0.737 foot-pounds; 1 Btu equals 1,055 joules.
Found on http://www.encyclo.co.uk/local/21690
### Joule
A unit of energy in the SI system. One joule is 1 kg. m2/s2 which is also 0.2390 calorie.
Found on http://home.nas.net/~dbc/cic_hamilton/dictionary/a.html
### Joule
A basic unit of energy. A 1 Watt transmitter radiates 1 Joule of energy every second.
Found on http://www.coseti.org/glossary.htm
### Joule
The Joule (J) is the SI unit of energy. One Joule is the energy expended when a force of one newton is applied over a displacement of one meter in the direction of the force. The use of the joule is probably limited in Radiation Protection but is used in the definition of Absorbed Dose and the Electron volt .
Found on http://www.ionactive.co.uk/glossary.html
### Joule
Unit of energy in the SI (Système International) system of units. The joule is sometimes used in photography to indicate the output of an electronic flash.
Found on http://www.peterashbyhayter.co.uk/glossaryT-Z.html
### Joule
[n] - a unit of electrical energy equal to the work done when a current of one ampere passes through a resistance of one ohm for one second 2. [n] - English physicist who established the mechanical theory of heat and discovered the first law of thermodynamics (1818-1889)
Found on http://www.webdictionary.co.uk/definition.php?query=Joule
### Joule
[pronounce: jool] The unit for measuring energy (J).
Found on http://www.encyclo.co.uk/local/20442
### Joule
Measurement of energy. Used to define the maximum muzzle energy. The legal limit for Airsoft weapons is 1.35j. See section on the Law for more details.
Found on http://www.tea-and-medals.co.uk/glossary.htm
### Joule
The SI unit of energy is the joule. Defined as:1 joule is the work done by a force of 1 newton moving a distance of 1 metre in the direction of the force.It may also be defined in electrical terms as:the amount of energy needed to sustain 1 amp for 1 sec in a 1-ohm resistance.
Found on http://www.diracdelta.co.uk/science/source/j/o/joule/source.html
### Joule
A unit of energy in the SI system. One joule is 1 kg. m2/s2 which is also 0.2390 calorie.
Found on http://www.allchemicals.info/index/action/detail/keyword/J/id/1059565212.ph
### joule
(J) The SI unit of energy, equal to the work required to move a 1 kg mass against an opposing force of 1 newton. 1 J = 1 kg m2 s-2 = 4.184 calories.
Found on http://antoine.frostburg.edu/chem/senese/101/glossary/j.shtml
### joule
Joule (J) is the SI derived unit of energy, work, and heat. The joule is the work done when the point of application of a force of one newton is displaced a distance of one metre in the direction of the force (J = N m). The unit is named after the British scientist James Prescott Joule (1818-1889).
Found on http://www.ktf-split.hr/periodni/en/abc/j.html
### Joule
Unit of energy. One joule is one watt for one second. It is the measure Of the 'kick' of a pulse. Joules are the most important measure of the power of the energiser.
Found on http://www.electricfence-online.co.uk/ishop/1047/shopscr21.html
### Joule
J A measure of work, energy or cell capacity. For electrical energy, one Joule is one Amp at one Volt for one Second, or one WattSecond. 1 Wh = 3.6kJ. For mechanical energy one Joule is a force of one Newton acting over one metre i.e. One newton metre.
Found on http://www.mpoweruk.com/glossary.htm
### Joule
The SI unit of energy. The release or transfer of one joule per second is one Watt, the SI derived unit of power.
Found on http://www.theiet.org/factfiles/energy/nuclear-terms.cfm?type=pdf
### Joule
Joule (jōl) noun [ From the distinguished English physicist, James P. Joule .] (Physics.) A unit of work which is equal to 10 7 units of work in the C. G. S. system of units (ergs), and is practically equivalent to the energy expended in one second by an electric current of
Found on http://www.encyclo.co.uk/webster/J/13
### joule
<unit> SI unit of energy. ... 1 Joule = 1E7 ergs = 1 Watt of power occurring for one second. 1 Joule is roughly 0.001 BTU and 1 calorie is roughly 4 joules. There are 3.6 million joules in a kilowatt hour. ... (14 Oct 1997) ...
Found on http://www.mondofacto.com/facts/dictionary?joule
### joule
(J) (jldbomacl) the SI unit of energy, being the work done by a force of 1 newton acting over a distance of 1 meter.
Found on http://www.encyclo.co.uk/local/21001
### Joule
• (n.) A unit of work which is equal to 107 units of work in the C. G. S. system of units (ergs), and is practically equivalent to the energy expended in one second by an electric current of one ampere in a resistance of one ohm. One joule is approximately equal to 0.738 foot pounds.
Found on http://thinkexist.com/dictionary/meaning/joule/
### joule
unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. ... [6 related articles]
Found on http://www.britannica.com/eb/a-z/j/24
### Joule
The energy required to push with a force of one Newton for one meter.
Found on http://www.encyclo.co.uk/visitor-contributions.php
### Joule
A unit of energy J such that the heat capacity of water at 15RC is 4.18 J/gRC.
Found on http://www-bdnew.fnal.gov/operations/accgloss/gloss.html
No exact match found
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Cleave Books
Puzzle Trio 27
79 The year 2001 started at the midnight which fell between 31st December 2000 and 1st January 2001.As the millionth second following the start of 2001 came up, what was the day, date and time?
80 The drawing shows a chain of five gear-wheels, identified as A to E, each one meshing properly with its immediate neighbour(s). The number under each one show how many teeth that particular gear-wheel has. When A is turned clockwise ten full turns, in which direction does E turn, and how many times?
81 A solid wooden cuboid is made which measures an exact, and even, number of centimetres along each of its edges.It is then painted black all over its exposed faces and, when the paint is dry, the cuboid is cut up into 1 centimetre cubes.These small cubes are counted and it is then found that the number having NO paint on them at all, is exactly the same as the number which do have some paint on them. What is the LEAST number of small cubes there could be in total?
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Post Go back to editing
# How to ensure zero phase shift for multiple detection channels
Hi,
I need some help from friends in this forum.
I have my system which has 16 channels sensor(E1,E2....,E15,E16 ) made from the conductive plate that put on the outside of the acrylic circle pipe wall. Each sensor is shielded by another conductive plate and this made the sensor function as a cap. Or it is known as electrical capacitance tomography sensor.
Now,what I want to do is to determine the resistive load of the i.e conductive liquid inside the pipe. So, i made one of the sensor as a source,i.e E1 by giving voltage input 10 Vp-p, 2 MHz sine waveform and all the remain sensors,E2...E16 as the detection sensor and measure the current value to the ground.
Thus, in processing the detection current value which is in between 0.1mA to 1mA, i need the I to V converter amplifier and other steps so that I can used it in ADC part. For time being, i used the LT1360 as my op amp to convert current to voltage amplifier with inverting input.. However, I noticed that, if i used the same gain of inverting feedback, i.e Rf=0.9K ohm; the output of course will be the invert of the input. The nearest of detection sensor, E2 showed no phase shift(that is totally inverting of the voltage input). But, the phase shift will keep increasing across the sensor, E3..E9 and for the farthest detection sensor,E9 ; it showed the phase shift is almost 180 deg.Then, it will decreasing again till no phase shift at E16. The output voltage peak to peak showed like a 'smile graph' which is between 0.1V and 1V.
However, the correct signal should be in zero phase shift for all detection sensors with the 'smile' curve of amplitude voltage.
What is the suitable approach/ic that I can use to make it all the signal in zero phase shift with the operating frequency 2Mhz?
I read some of people use phase shift demodulation,PSD technique, but i think it is complicated.
I also read some of them use programmable gain amplifier,PGA but is there any other method that I can use to solve this problem?
So sorry if this post is too long.
Parents
• This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.
Thank you,
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Cambridge University Press The first chapter deals with ancient people and early Greek astronomy. The legs are opened out so that the distance between the feet is 2 m. Despite your best preparation, you might lose the success-train by 2-3 marks. Please help to create a more balanced presentation. Math Introduces Geometry has a light tone, clear layout, and humorous cartoons by Jessica Wolk-Stanley. The Muslim Heritage Site is very interesting. These courses really do allow students to work independently..."
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Simplifying addition in radicals with 2 variables, free algebraic solver, answers to saxon math lesson 85, solving radicals, how to simplify a square root with an exponent. However few, if any, technological innovations seemed to have passed from China to India or Arabia. Children will intuitively discover how to create complex trees from a few slips of virtual construction paper. If the point Q on the terminal side of angle A in standard position has coordinates (x, y), this point will have coordinates (x, −y) when on the terminal side of −A in standard position.
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This book was in excellent condition for being used. The applets are started by clicking the red buttons. He learnt astronomy from Arago and Bouvard and the theory of probability under Joseph Fourier and Pierre Laplace. Before proceeding further we will consider formally by means of the tangent, the relations which exist between the sides and angles of a right-angled triangle. (This is a general method of denoting sides of a right-angled L..) Then, as shown in section 41: Thus anyone of the three quantities a, b, tan B can be determined when the other two are known. (a) As indicated above we sometimes, for brevity, refer to an angle by using only the middle letter of the three which define the angle.
• ### Trigonometry Second Edition
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 13.24 MB
Teachers can capture student performance data, review and understand student progress - thereby saving time, paper, money and enhancing productivity. Center the bubble with the third footscrew. In the logical arguments and constructions strand, students are expected to create formal constructions using a straight edge and compass. In architecture, trigonometry plays a massive role, especially in the compilation of building plans. In short, we may think of MATH 101 as an introductory course in mathematical modeling.
• ### A Treatise of Trigonometry, Plane and Spherical ...: As
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 5.45 MB
If sections are made parallel to the bases, all such sections are identically equal to the bases. If you feel confident about your trig ability and think you can do well enough without MML, and it is enough for you to check the answer to odd-numbered problems then just buy a used text. This will make the user actually do the math in their head every time. The Placement Assessment covers only pre-calculus topics.
• ### Introduction to algebra: Designed for use in our public
Format: Unknown Binding
Language: English
Format: PDF / Kindle / ePub
Size: 5.35 MB
You can play in English, French, or Spanish! 10. If we join A and B to any point D' in the minor segment, then LAD/B is the angle in the minor segment. Students may not take the first two quarters of this sequence for P/F grading. Multiplying and dividing decimals worksheet, McDougal Littell Algebra 2 Answers, does ti-83 solve "first order linear equations", "NJPASS LA 1st Grade Examples", algebra for idiots, sample equation questions 5th grade, a sum of cubes plus a number. 6th grade math test chapters 8,9,10 mcdougal littell, prentice hall algebra 2 with trigonometry answers, hyperbola equation, quadratic equation factors calculator, mixture compoud worksheet ks3.
• ### Contemporary college algebra and trigonometry
Format: Unknown Binding
Language: English
Format: PDF / Kindle / ePub
Size: 10.04 MB
We have the right to terminate your access to the Site if we determine that you have failed to comply with any of the provisions of these Terms and Conditions. Afternotes goes to Graduate school: Lectures on Advanced Numerical Analysis. We have satisfied customers from USA, United Kingdom, Canada, Germany, Australia, New Zealand, etc. Further, you acknowledge that your personal information may, at times, be accessible by individuals may be located worldwide including in countries that may have not been determined to provide the same level of data protection as in your country.
• ### Trigonometry for today
Format: Unknown Binding
Language: English
Format: PDF / Kindle / ePub
Size: 9.28 MB
Find the constant height of the jet. send me 100 questions of trigonometry on my e mail id because i always have problem in solving them. Equal angles are angles with the same measure; i.e., they have the same sign and the same number of degrees. This process is also called long division. A: They will receive an email from The Great Courses notifying them of your eGift. To calculate the distance between London (51.3N 0.5W) and Tokyo (35.7N 139.8E) in kilometers: # Notice the 90 - latitude: phi zero is at the North Pole. @L = (deg2rad(-0.5), deg2rad(90 - 51.3)); @T = (deg2rad(139.8),deg2rad(90 - 35.7)); The answer may be off by few percentages because of the irregular (slightly aspherical) form of the Earth.
• ### Trigonometry for College Students
Format: Hardcover
Language: English
Format: PDF / Kindle / ePub
Size: 6.37 MB
Also when the rotating line has made a half rotation, the angle formed - the straight angle - must contain 180°. You will need to register for a TES account to access this resource, this is free of charge. The student is expected to: (A) determine the slope of a line given a table of values, a graph, two points on the line, and an equation written in various forms, including y = mx + b, Ax + By = C, and y - y1 = m(x - x1); (B) calculate the rate of change of a linear function represented tabularly, graphically, or algebraically in context of mathematical and real-world problems; (C) graph linear functions on the coordinate plane and identify key features, including x-intercept, y-intercept, zeros, and slope, in mathematical and real-world problems; (D) graph the solution set of linear inequalities in two variables on the coordinate plane; (E) determine the effects on the graph of the parent function f(x) = x when f(x) is replaced by af(x), f(x) + d, f(x - c), f(bx) for specific values of a, b, c, and d; (F) graph systems of two linear equations in two variables on the coordinate plane and determine the solutions if they exist; (G) estimate graphically the solutions to systems of two linear equations with two variables in real-world problems; and (H) graph the solution set of systems of two linear inequalities in two variables on the coordinate plane. (4) Linear functions, equations, and inequalities.
• ### Plane Trigonometry and Tables
Format: Paperback
Language: English
Format: PDF / Kindle / ePub
Size: 12.96 MB
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# 2 Times Table Worksheet
Print the two times tables fact target worksheet in pdf. 2 x 4 tables 2 5 10 practice.
Times Tables Worksheets 2 3 4 5 6 7 8 9 10 11 And 12 Eleven Worksheets Fr Times Tables Worksheets 2 Times Table Worksheet First Grade Worksheets
### Table of 2 missing factor.
2 times table worksheet. 2 times table worksheet printable pdf s below are the worksheets we have developed for teachers tutors parents and students to improve multiplication skills. 10 x 100. 2 times table worksheets pdf multiplying by 2 activities download free 2 times table worksheets.
Click on the worksheet to view it in a larger format. 2 x 18. Table of 5 missing factor.
2 times tables worksheet 1 of 5. Completing each of these sheets should only take about a minute. For the 2 times table worksheet you can choose between three different sorts of exercise.
Free and printable 2 times table worksheets are available for you multiplication is one important part in learning math. The target worksheets are made to look like a dartboard. 5 x 2 missing factor questions.
2 x 14 two times multiples of 5. Random order randomly shuffled times table shuffled in random order multiplication worksheets multiply by 1 2 3 4 5 6 7 8 9 10 11. After the 2 times table you will often move on to the 10 times table or 5 times table.
In the first exercise you have to draw a line from the sum to the correct answer. Here is 2 times table worksheets pdf a simple an enjoyable set of x2 times table suitable for your kids or students multiplying by 2 activities is giving to help children to practice their multiplication skills. 5 x 15.
In the second exercise you have to enter the missing number to complete the sum correctly. If you re practising the 2 times table using the number line you will make gradual jumps of 2. The target number is two and it is in the middle of each of the target worksheets.
The trick here is to omit a number each time. For children who want to sharpen their multiplication skills using these time table worksheets is one of the ways for children to practice multiplication. Two times small numbers.
These times tables in pdf format are designed to help students learn how to multiply numbers from 2 to 10. Tables 2 5 10 missing factor x 10 20. You ll also find advanced practice sheets to help reinforce the basics.
How to use these worksheets. These have been specifically designed to assist in learning about multiplication by 2 and range from 2 x 1 10 through to 2 x 1 100 to cover a wide range of abilities or for. 2 4 6 8 10 etc.
The 2 times table is the 2nd table you will learn following the times 1 table. Multiplying by 2. Table of 10 missing factor.
Print 2 times table worksheet.
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# How Far Can A Squirrel Leap
How Far Can a Squirrel Leap?
There are many factors to consider when calculating how far a squirrel can jump. These factors include age, species, and whether the squirrel is hanging upside down or moving. The height at which a squirrel can jump will vary depending on these variables. Read on to find out how far a squirrel can jump. The following video shows some of the heights squirrels can reach. However, these numbers are still indicative and are not to be taken literally.
## The height of a squirrel’s jump depends on several factors
Squirrels have incredible agility. Many species can jump up to 20 feet in the air, but some can leap even higher. Jumping distance depends on age, size and muscle strength. Watch this video to see a flying gray squirrel demonstrate its impressive jumping power. There are also several other factors that can influence how far a squirrel can jump, including its body size and age. Read on to learn more about the factors that determine the length of a squirrel’s jump.
Squirrels’ hind legs are flexible and have sharp claws. Because they can hang upside down from any surface, their paws allow them to absorb the impact of the landing. Their claws allow them to secure themselves on uneven surfaces and withstand the impact of jumping off them. Because their hind legs are reversible, they are also able to bounce back and forth without falling.
Researchers studied how squirrels calculate jump distance by altering several factors, including the stability of the takeoff and landing platforms. In one experiment, researchers changed the stiffness of the branch used for launch. The stiffer the branch, the closer the squirrel jumped. The researchers also studied the effect of backdrops on the distance a squirrel can jump. In some instances, researchers found that squirrels jumped from a branch with a backdrop to slow their momentum.
## It depends on the age, species, position and whether or not a squirrel is moving
The answer depends on many factors including the age, species, position, and whether or not a squirrel is moving. Squirrels often migrate far distances in search of food, but will also move for other reasons. They may need to travel for food during winter, especially if their area is crowded with other squirrels. In some cases, they may have to travel several miles in order to establish a new territory.
The range of a squirrel’s home territory varies from one to 25 acres. Territory boundaries may overlap during mating season, so it depends on the age, species, position, and whether or not a squirrel is moving. In summer, a squirrel may construct a saucer-like nest. During winter, a squirrel will often travel further than a rodent, which may make it an attractive target for predators.
The shape and position of a squirrel’s tail determines the type of movement it makes. Some squirrels use their tail as a rudder and a mini-parachute while they’re airborne. In the wild, a squirrel’s tail can save its life, but it may also break when the predator catches it. The tail also communicates various emotions, including fear, annoyance, anger, and aggression.
## It depends on whether or not a squirrel is hanging upside down
Unlike humans, squirrels are able to hang upside down on trees and climb right side up. Their unique ankle joints allow them to rotate their feet 180 degrees, allowing them to climb in different positions. When they hang upside down, they can also keep their feet pointed in a single direction. But how do these animals climb? Read on to discover the secrets of squirrel grip. We can mimic squirrels’ abilities with robotic systems someday.
## How far can a squirrel leap?
Answer: A squirrel can leap up to 20 feet.
## How does a squirrel jump?
Answer: A squirrel uses its powerful hind legs to jump.
## How long can a squirrel stay airborne?
Answer: A squirrel can stay airborne for up to two seconds.
## How many times can a squirrel jump in a day?
Answer: A squirrel can jump up to 200 times in a day.
## Can a squirrel jump backwards?
Answer: Yes a squirrel can jump backwards.
## What does a squirrel use its tail for?
Answer: A squirrel uses its tail for balance and for changing direction while in midair.
## Do all squirrels jump?
Answer: No not all squirrels jump.
Ground squirrels for example do not jump.
## Where do squirrels jump the most?
Answer: Squirrels jump the most in the spring and fall.
## How high can a squirrel jump?
Answer: A squirrel can jump up to 10 feet high.
## What do baby squirrels do when they want to jump?
Answer: Baby squirrels use their tails and hind legs to propel themselves into the air and onto their mother’s back.
## How many times can a baby squirrel jump in a day?
Answer: A baby squirrel can jump up to 30 times in a day.
## What is the record for the furthest squirrel jump?
Answer: The record for the furthest squirrel jump is 21 feet.
## What is the record for the highest squirrel jump?
Answer: The record for the highest squirrel jump is 10 feet.
## What is the record for the longest time a squirrel has stayed airborne?
Answer: The record for the longest time a squirrel has stayed airborne is 2.
5 seconds.
## What do scientists believe is the reason for a squirrel’s ability to jump?
Answer: Scientists believe that a squirrel’s ability to jump is due to its strong hind legs and tail.
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Search a number
550 = 25211
BaseRepresentation
bin1000100110
3202101
420212
54200
62314
71414
oct1046
9671
10550
11460
1239a
13334
142b4
1526a
hex226
• 550 can be written using four 4's:
550 has 12 divisors (see below), whose sum is σ = 1116. Its totient is φ = 200.
The previous prime is 547. The next prime is 557. The reversal of 550 is 55.
550 = T4 + T5 + ... + T14.
550 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
550 is a Gilda number.
It is a Harshad number since it is a multiple of its sum of digits (10).
It is a super Niven number, because it is divisible the sum of any subset of its (nonzero) digits.
550 is an undulating number in base 7.
It is a plaindrome in base 12, base 13, base 15 and base 16.
It is a nialpdrome in base 5 and base 10.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (557) by changing a digit.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 45 + ... + 55.
It is an arithmetic number, because the mean of its divisors is an integer number (93).
550 is a gapful number since it is divisible by the number (50) formed by its first and last digit.
550 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
It is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (558).
550 is a wasteful number, since it uses less digits than its factorization.
550 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 23 (or 18 counting only the distinct ones).
The product of its (nonzero) digits is 25, while the sum is 10.
The square root of 550 is about 23.4520787991. The cubic root of 550 is about 8.1932127060.
550 divided by its sum of digits (10) gives a palindrome (55).
Adding to 550 its product of nonzero digits (25), we get a palindrome (575).
Subtracting from 550 its product of nonzero digits (25), we obtain a palindrome (525).
550 divided by its product of nonzero digits (25) gives a palindrome (22).
It can be divided in two parts, 5 and 50, that added together give a palindrome (55).
The spelling of 550 in words is "five hundred fifty", and thus it is an aban number and an oban number.
Divisors: 1 2 5 10 11 22 25 50 55 110 275 550
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cancel
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Helper IV
## Taking the average of the sum of values per date
Hi Community! I have a table where I would like to sum up the values per individual date (i.e. move from the table on the left to the table in the middle below) and then take the daily average of those values (value on the right below). The "Date", "Segment 1", and "Segment 2" dimensions are filtered so the solution should respond to any filtering required. TO CLARIFY: I would only like the average value at the end, no need for the intermediate table I just included it for the sake of explanation.
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Super User IV
## Re: Taking the average of the sum of values per date
Try a new measure like
``averagex(summarize(Table[Date],"_Sum",sum(Table[Value])),[_Sum])``
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7 REPLIES 7
Highlighted
Super User IV
## Re: Taking the average of the sum of values per date
Try a new measure like
``averagex(summarize(Table[Date],"_Sum",sum(Table[Value])),[_Sum])``
My Recent Blog -Week is not so Weak Connect on Linkedin
Proud to be a Super User!
Highlighted
Helper IV
## Re: Taking the average of the sum of values per date
`averagex(summarize('Table', [Date],"_Sum",sum(Table[Value])),[_Sum])`
Small change as the format wasn't quite correct before I believe. Huge thank you though!
Highlighted
Helper II
## Re: Taking the average of the sum of values per date
@amitchandak - It worked beautifully. Is there a way, I can add the grouped value in a new table, so that I can maintain a history of the data.
Highlighted
Super User IV
## Re: Taking the average of the sum of values per date
Summarize can be used to create a new table
My Recent Blog -Week is not so Weak Connect on Linkedin
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Highlighted
Helper IV
## Re: Taking the average of the sum of values per date
Hi @amitchandak thank you again hugely for your help. I have one more question related to this. What if I wanted the average to be calculated on only a specific category (e.g. where Segment 2 = "A" or "B"). Where would this filter statement go for the rest of the summarize to work?
Highlighted
Super User IV
## Re: Taking the average of the sum of values per date
@spenot09 , one of the way
averagex(summarize(filter(Table, Table[Segment 2] in {"A","B"}), Table[Date],"_Sum",sum(Table[Value])),[_Sum])
My Recent Blog -Week is not so Weak Connect on Linkedin
Proud to be a Super User!
Highlighted
Helper IV
## Re: Taking the average of the sum of values per date
FANTASTIC as well always @amitchandak , thank you!
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# Write the Room Literacy and Math First Grade Centers – Easter – April
### File Details
File Type: PDF
# of Pages: 62
\$5.00Purchase & earn 5 points!
Categories
## Description
Write the Room math and literacy activities are a great way to get your students up and moving during center time. These first grade centers will keep students engaged while practicing math and ELA skills.
These first grade math centers and first grade literacy centers are seasonal and have fun, themed activities for Easter, Spring and Earth Day.
Pull out your clipboards for students to use while walking around the room. First grade students will love these write the room centers!
APRIL – SPRING – EASTER – EARTH DAY
Literacy Write the Room Center Activities for First Grade
• Rainy Day Sight Words – This is an editable sight word activity. Add your own sight words! There are 2 response sheets for easy differentiation! Students will write the sight words or trace the sight words that are found around the classroom.
• Recycled Diphthongs – Students will say the word and highlight the missing diphthong, then write the whole word.
• Earthy Contractions – Students will say the words on the card and write the contraction they make on the response sheet.
• Gardening with Parts of Speech – Students will say the word and identify it as a verb or noun.
• Hot Air Punctuation- Students will read the sentence and highlight or color in the punctuation that belongs.
• Earthy Tenses – Students will say the verb given and identify it as past or present tense.
Math Write the Room Center Activities for First Grade
• Easter Order – Students will read the numbers and write them in numerical order.
• Eggtastic Number Bonds – Students will solve the number bonds and write an equation to match.
• Earthy Number Sentences – Students will solve for the missing number and write four equations that belong with that fact family.
• Spring Picnic Graphing – Students will count the groups of objects and represent them on the graphs given.
• Spring Picnic Number Sentences – Students will count the groups of objects and write an equation to match.
PREDICTABLE CENTER ACTIVITIES
These write the room activities are purposefully predictable so you don’t need to teach new activities each month. In this way, students can work independently at their center and you can work with a group without interruptions.
Each center activity comes with a response sheet for students to complete. Students walk around the classroom, finding the cards that you placed throughout. This center activity is a fun way to practice those math skills and literacy skills!
EDITABLE CENTER ACTIVITIES
Every month, there is an editable sight word center! Simply add the sight words you are working on with your first grade students, print the cards, and hang them throughout your classroom. There are 2 versions of the response sheet so you can easily differentiate for your first grade students.
❤️ Save more than 30% on the bundle!
Write the Room Math and Literacy Centers Bundle for First Grade
Included in each month of first grade centers:
• Minimum 4 first grade math write the room centers
• Minimum 4 first grade literacy write the room centers
• Recording sheets for the activities
• “I Can” cards to help build student independence
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### Gerwin Sturm
Shared publicly -
I totally love her videos :D
Best sentence in this video:
"e to the i to the e i o is e to the wau to the tau wau wau"
Make sure to check out her other videos, especially the series about spirals.
/via
797
587
I had never even heard of this number before. Some googling ahead of me, I think.
This is wildly amazing and the video deserves more than 306 views.
I think it already got more views, this is a known issue at youtube, that the counter sometimes gets stuck at 306 before being updated to the real number.
Wau
UAU!!!!
I hope so. The video is fantastic.
But I'm now intrigued by why YouTube stops at 306..
Ironically
Ϝ ^ (306 * Ϝ ^ (306 * Ϝ ^ (306 * Ϝ ^ (.... )))) = Ϝ
So that might be the reason...
really amazing! I have never heard of F
can I go to the tau wau wau, too? The letter in Hebrew is ו - Vav in modern pronunciation. The letter functions both as a connector (the word "and" as in "האחד והשני" - the one and the other), and to convert the past tense into the future tense and vice versa as in "ויאמר vs ואמר", or in cases as the word "or". The numerical value of the letter is six, recalling the 6 spacial dimensions and the six days of the week. The form of the letter is a hook (the meaning of the word וו) or a straight line, connecting two points or dividing between them.
wow = wau
does anyone have some kind of proof for that? A link to wikipedia or a online available dissertation?
(But please don't spoil it here...)
yeah, love her stuff too, especially the Moebius strip movie. But this one's singular.
hey i am very missin maths but i love it
Sorry, deleting comments that spoil the fun...
I have to admit I didn't "get it" until she got to the e^(2*pi*i) part..
Not impressed by her British pronunciation of Ptolemy.
The one she did Fibonacci numbers was awesome.
Removed my own comment since I see we don't want to spoil the fun! Everything she says is correct though!
i r no gud at maths
This is THE coolest girl~
Ah. I see. The moderator is deleting comments that say what the answer is. That's a relief. It depressed me that no one seemed to know the answer. It's pretty obvious.
Sort of related: Have you seen her video on Tau? Pi Is (still) Wrong.
She mentions it in the OP video in "e to the i to the e i o equals e to the Wau to the Tau Wau Wau"
thank God I became a doctor !!!
Wau......and this is Important......Why???? :o(
Wau = x^((Real part of y)-1/2) where x is any nonzero number and y is any nontrivial zero of Sum(n^-s,n=1 to infinity). (Assuming the Riemann hypothesis is true!)
its Friday, this just blew my head off..... :)
Scott L
and they say G+ is stuck in a Tech subset of GenPop
Wau!
shit
I'm coming up with nothing. Silly me. Got it now.
Wow, I just felt dumber watching that wau video.... Hahha
If you dial the number Ϝ on your phone, do you get the Most Interesting Man in the World?
Wow, I just stumbled upon a mathematical contraption..
This is AMAZING :D It's so simple once you know the answer :)
This is retarded. Wau is a number? Sounds like another term for infinite or a "number" that ironically fits with everything.....a variable??
lol... what she says is very rythmic and she sounds like a very good singer... just sayin`, she sang a lot
I could just about hear Sesame Street music in the background....lol
no, it's not fake... it's math :) (but don't worry, i didn't get it for the first time either - and i'm studying math :D)
n a wau wau here n a wau wau there..here a wau, there a wau..everywhere a wau wau.. What the F?
e=MC^2 ..... e^Wau/(MC^2)= Wau... WOW Algebra is hard.
Wau ly narrated.
Wau is the loneliest number that you'll ever do.
Wau
The video was moving pretty fast, so couldnt really get what the number, until E^F/MC^2 = F^2, just gave it away. Simple Dimensional Analysis. That one should have been left out.
Eli S.
+
5
6
5
This video explains perfectly why i got an F in maths
Took me a second, but i understand now :P.
In many sentences. In many ways. You use it more than one time a day I guarantee you :)
Thank you for sharing this video and her youtube stream.. Some really amazing stuff she has there..
Had to watch the video a second time, but I got it now.
That's so silly.
I love it! I need a t-shirt of the derivative of e to the wau!
I think she was sniffing those sharpies!
R SS
wau wau wau
dont know the answer at math exam? ...just do some wau at the teacher :D
The best video about Wau than I ever seen!
In fact, you use it here on Google Plus all the time.
just noticed, you cant go to infinity, as ininity doesnt have a definite value
therefore you cannot minus, add, divide or multiply infinite, therefore it cannot be a square number or have a square root
at best, youll get infinite-1, but you cannot have infinite +1, as then the original value of infinite would not be... infinite...
also, it equals 0
wau2 = wau
0 x 2=
no other number works that way xD
OK, mostly true, however, many of her definitions of Wau aren't actually Wau. They just approach Wau.
I agree with Gerwin. That is the best line in the video lol
wow>>> wau,,, wau + - * ^ % wau=wau,,, wau how??
I thought she was making it up until she wrote e^(2*i*pi)... then I thought "Awesome! She's right!". Cool.
Naim Dk
lots of nonsensical bullshit in this video, sadly.
It took me until the last example before I think I figured it out. Good video and yeah, great voice.
They should change the spelling to wow!!!!!
I think my Maths teacher used to smoke Wau.
Really funny once you realize what the value of Wau is :D
wow wau!!
We all love Vi <3
I don't get it, but she sure sounds sexy...
imma check this out in 10 !! dun get low on tha list :P
You need to get out more young lady. Wau doesn't really matter in life :-)
I am a software developer: how could Wau serve me in my job ? hahaha
Wau-to-the-Wau-to-the-Wau-to-the-Wau-to-the-Wau-to-the-Wau-to-the-Wau........and beyond!
Ajith M
+
1
2
1
If I +1 this, would that be a spoiler?
Kind of reminds me of the old Di-hydrogen Monoxide scare lol
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https://math.stackexchange.com/questions/1213763/determining-the-dimensions-of-a-cylindrical-can/1213836
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# Determining the dimensions of a cylindrical can
A cylindrical can (with lid) is required to have a volume of 8000 cm3 . Using calculus, determine the dimensions that minimize the surface area (and hence cost) of the can. Is the can more costly to construct than a closed cubic container of the same volume and made of the same material? Justify your answer.
For a cylinder, $V=\pi r^2h = 8000$
Surface Area $SA = 2 \pi r^2 + 2 \pi r h$
Using the volume formula, solve for h: $$h=\frac{8000}{\pi r^2}$$
Substitute h into the surface area formula, getting $$SA=2\pi r^2 + \frac{16000\pi r}{\pi r^2}$$
Simplify to get $$SA=2\pi r^2 + \frac{16000}{r}$$
Take the derivative and set it equal to zero to find extrema:
$$SA' = 4\pi r - \frac{16000}{r^2} = 0$$
Solve for r to find the radius of the cylinder with the least SA (approximately 10.839). Use this to find the SA.
I leave the rest as an exercise for you.
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2014-04-21T04:13:23-04:00
288=3x(2x+10)
288=6x^2+30x
48=x^2+5x
48+25/4=x^2+5x+25/4
192/4+25/4=(x+5/2)^2
217/4=(x+5/2)^2
14.73/2=x+5/2
9.73/2=x
x=4.87
Length is 14.6 width is19.73
• Brainly User
2014-04-21T05:43:14-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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https://questions.examside.com/past-years/gate/question/fatima-starts-from-point-p-goes-north-for-3-km-and-then-east-gate-ece-2017-set-2-marks-1-5a3987970b449.htm
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1
GATE ECE 2017 Set 2
MCQ (Single Correct Answer)
+1
-0.3
Fatima starts from point P, goes North for 3 km, and then East for 4 km to reach point Q. She then turns to face point P and goes 15 km in that direction. She then goes North for 6 km. How far is she from point P, and in which direction should she go to reach point P?
A
8 km, East
B
12 km, North
C
6 km, East
D
10 km, North
2
GATE ECE 2017 Set 1
MCQ (Single Correct Answer)
+1
-0.3
In the summer, water consumption is known to decrease overall by 25%. A water Board official states that in the summer household consumption decreases by 20%, while other consumption increases by 70%.
Which of the following statement is correct?
A
The ratio of household to other consumption is 8/17
B
The ratio of household to other consumption is 1/17
C
The ratio of household to other consumption is 17/8
D
There are errors in the official's statement
3
GATE ECE 2017 Set 1
MCQ (Single Correct Answer)
+1
-0.3
40% of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart is
A
120
B
144
C
160
D
212
4
GATE ECE 2017 Set 1
MCQ (Single Correct Answer)
+1
-0.3
Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?
(i) At least one bench is a table
(ii) At least one shelf is a bench
(iii) At least one chair is a table
(iv) All benches are chairs
A
Only (i)
B
Only (ii)
C
Only (ii) and (iii)
D
Only (iv)
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
CBSE
Class 12
© ExamGOAL 2024
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https://www.majorfun.com/lonpos-303/
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# Lonpos 303
Lonpos 303. Lonpos, because that’s the name of the inventor. 303 because that’s how many different puzzles there are. Puzzles of two different varieties: the rectangular, 2-dimensional variety, and the 3-D pyramid puzzles. There are 12 pieces, each made of a cluster of small balls, each a different color and shape. The shapes are pentomino-like in their variety (different configurations of clusters of 3, 4 and 5 units), so their mathematical properties are noteworthy – notably to mathematicians. All the pieces fit snugly in the case, which also most neatly serves to house the instruction booklets.
I was concerned, Defender of the Playful that I am, that perhaps the 3-D puzzles would be too, shall we say, challenging. After all, how do you effectively convey a 3-D puzzle in a 2-D booklet? So I tried those first. In fact, I tried the first one first. The illustration very clearly and painstakingly showed me how to place the first 11 pieces. All I had to do was figure out how to place the 12th. I must say that I was experiencing something akin to sensual delight as I built the puzzle – each piece fitting so satisfyingly snugly onto the board or onto other pieces. And, since there was only one piece left to place, and since it so clearly fit in only one possible position, I was able to experience the almost immediate reward of that final click, when everything falls together, and the full glory of pyramid-building manifests itself in multi-colored, opalescence.
Then I tried the next puzzle. Hmmm. A bit more difficult to figure out how to follow the instructions, to envision the proper piece when all you can see is the particular slice of it that appears on each level. And then the next. And another intriguing hmmm. And as I solved each puzzle, I felt I was being taught, carefully, playfully, invitingly, a bit more about the pentomatically puzzling properties of pyramid-building. And it wasn’t really too difficult. I mean it could get difficult. There were many puzzles in the booklet o’ puzzles. And they got progressively more and more, well, challenging. But I could select whatever challenge I was ready for. And I said unto myself, behold, this is fun. And I’m learning things. More than fun, actually. Major fun, even.
Lonpos 303 is very much like Lonpos 101, except Lonpos 101 only has 101 puzzles. And Lonpos 101 is very much like Kanoodle, which is similarly very much like Level Up. But there is only one Lonpos 303. And once you start playing with it, you’ll be grateful for every one of the 202 additional challenges that await. After which you might want to contemplate the significance of knowing that there are actually 360,984 unique rectangle puzzles, and 2,582 similarly unique pyramids puzzles that you could potentially create with your 12 little Lonpos pieces.
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https://fastretrieve.com/kitesurfing/does-a-kite-have-all-equal-sides.html
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# Does a kite have all equal sides?
Contents
## What is a kite with all sides equal?
A kite has two sets of adjacent congruent sides. Rhoumbi are kites where the two sets are also congruent to each other (thus all sides are equal). This means that all Rhombi are kites, but not all kites are rhombi. A square is a rhombus with all right angles.
## Does a kite have all angles equal?
A kite is a polygon with four total sides (quadrilateral). The sum of the interior angles of any quadrilateral must equal: degrees degrees degrees. Additionally, kites must have two sets of equivalent adjacent sides & one set of congruent opposite angles.
## Does a kite have 4 equal angles?
No, because a rhombus does not have to have 4 right angles. Kites have two pairs of adjacent sides that are equal.
## Can a kite have all 4 sides equal?
Kite Angles
∠K = ∠T ∠ K = ∠ T and ∠I = ∠E ∠ I = ∠ E . It is possible to have all four interior angles equal, making a kite that is also a square.
## What are the 5 properties of a kite?
Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Other important polygon properties to be familiar with include trapezoid properties, parallelogram properties, rhombus properties, and rectangle and square properties.
ЭТО ИНТЕРЕСНО: Você perguntou: What is the name of kite paper?
## Are opposite angles equal in a kite?
The two interior angles of a kite that are on opposite sides of the symmetry axis are equal.
## How do you prove a kite?
How to Prove that a Quadrilateral Is a Kite
1. If two disjoint pairs of consecutive sides of a quadrilateral are congruent, then it’s a kite (reverse of the kite definition).
2. If one of the diagonals of a quadrilateral is the perpendicular bisector of the other, then it’s a kite (converse of a property).
## Are opposite sides equal in a kite?
In a kite, two adjoining sides are equal as shown in the figure. … Two pairs of sides known as consecutive sides are equal in length. One pair of diagonally opposite angles is equal in measurement. These angles are said to be congruent with each other.
## Is every kite a rhombus?
For example, kites, parallelograms, rectangles, rhombuses, squares, and trapezoids are all quadrilaterals. Kite: A quadrilateral with two pairs of adjacent sides that are equal in length; a kite is a rhombus if all side lengths are equal.
## Can a kite have exactly two right angles?
Thus the right kite is a convex quadrilateral and has two opposite right angles. If there are exactly two right angles, each must be between sides of different lengths. All right kites are bicentric quadrilaterals (quadrilaterals with both a circumcircle and an incircle), since all kites have an incircle.
## Is a trapezoid a kite?
A trapezoid is a quadrilateral who has two opposite sides which are parallel to each other. In general, a quadrilateral with two pairs of equal adjacent sites (i.e. a kite) mustn’t have a pair of parallel opposite sides (as a trapezoid). … So a kite can be a trapezoid; this is the case when it’s a rhombus.
## Is a rhombus a kite yes or no?
Every rhombus is a kite, and any quadrilateral that is both a kite and parallelogram is a rhombus. A rhombus is a tangential quadrilateral.
## Why is a kite called a kite?
One technical definition is that a kite is “a collection of tether-coupled wing sets“. The name derives from its resemblance to a hovering bird. The lift that sustains the kite in flight is generated when air moves around the kite’s surface, producing low pressure above and high pressure below the wings.
## Why is a rectangle not a kite?
A kite and a rectangle cannot be the same at any time. The reasons are: Two pairs of adjacent sides are equal in a kite, but not so in a rectangle. Two diagonals intersect at right angles in a kite, but not so in a rectangle.
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# Differentiation From First Principles
Differentiating a linear function
A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant.
Example
Consider the straight line y = 3x + 2 shown below
A graph of the straight line y = 3x + 2.
We can calculate the gradient of this line as follows. We take two points and calculate the change in y divided by the change in x.
When x changes from −1 to 0, y changes from −1 to 2, and so
No matter which pair of points we choose the value of the gradient is always 3.
Values of the function y = 3x + 2 are shown below
Look at the table of values and note that for every unit increase in x we always get an increase of 3 units in y. In other words, y increases as a rate of 3 units, for every unit increase in x. We say that “the rate of change of y with respect to x is 3”.
Observe that the gradient of the straight line is the same as the rate of change of y with respect to x.
NOTE: For a straight line: the rate of change of y with respect to x is the same as the gradient of the line.
Differentiation from first principles of some simple curves
For any curve it is clear that if we choose two points and join them, this produces a straight line.
For different pairs of points we will get different lines, with very different gradients. We illustrate below.
Joining different pairs of points on a curve produces lines with different gradients
Example : Suppose we look at y = x2.
Note that as x increases by one unit, from −3 to −2, the value of y decreases from 9 to 4. It has reduced by 5 units. But when x increases from −2 to −1, y decreases from 4 to 1. It has reduced by 3. So even for a simple function like y = x2 we see that y is not changing constantly with x. The rate of change of y with respect to x is not a constant.
Calculating the rate of change at a point
We now explain how to calculate the rate of change at any point on a curve y = f(x). This is defined to be the gradient of the tangent drawn at that point as shown below
The rate of change at a point P is defined to be the gradient of the tangent at P.
NOTE: The gradient of a curve y = f(x) at a given point is defined to be the gradient of the tangent at that point.
We use this definition to calculate the gradient at any particular point.
Consider the graph below which shows a fixed point P on a curve. We also show a sequence of points Q1, Q2, . . . getting closer and closer to P. We see that the lines from P to each of the Q’s get nearer and nearer to becoming a tangent at P as the Q’s get nearer to P.
The lines through P and Q approach the tangent at P when Q is very close to P.
So if we calculate the gradient of one of these lines, and let the point Q approach the point P along the curve, then the gradient of the line should approach the gradient of the tangent at P, and hence the gradient of the curve.
Example : We shall perform the calculation for the curve y = x2 at the point, P, where x = 3.
The graph below shows the graph of y = x2 with the point P marked. We choose a nearby point Q and join P and Q with a straight line. We will choose Q so that it is quite close to P. Point R is vertically below Q, at the same height as point P, so that △PQR is right-angled.
The graph of y = x2. P is the point (3, 9). Q is a nearby point.
Suppose we choose point Q so that PR = 0.1. The x coordinate of Q is then 3.1 and its y coordinate is 3.12. Knowing these values we can calculate the change in y divided by the change in x and hence the gradient of the line PQ.
We can take the gradient of PQ as an approximation to the gradient of the tangent at P, and hence the rate of change of y with respect to x at the point P.
The gradient of PQ will be a better approximation if we take Q closer to P. The table below shows the effect of reducing PR successively, and recalculating the gradient.
The gradient of the line PQ, QR/PR seems to approach 6 as Q approaches P.
Observe that as Q gets closer to P the gradient of PQ seems to be getting nearer and nearer to 6.
We will now repeat the calculation for a general point P which has coordinates (x, y).
The graph of y = x2. P is the point (x, y). Q is a nearby point.
Point Q is chosen to be close to P on the curve. The x coordinate of Q is x + dx where dx is the symbol we use for a small change, or small increment in x. The corresponding change in y is written as dy. So the coordinates of Q are (x + dx, y + dy).
Because we are considering the graph of y = x2, we know that y + dy = (x + dx)2.
As we let dx become zero we are left with just 2x, and this is the formula for the gradient of the tangent at P. We have a concise way of expressing the fact that we are letting dx approach zero. We write
‘lim’ stands for ‘limit ’and we say that the limit, as x tends to zero, of 2x+dx is 2x. Note that when x has the value 3, 2x has the value 6, and so this general result agrees with the earlier result when we calculated the gradient at the point P(3, 9).
We can do this calculation in the same way for lots of curves. We have a special symbol for the phrase
We write this as dy/dx and say this as “dee y by dee x”. This is also referred to as the derivative of y with respect to x.
Use of function notation
We often use function notation y = f(x). Then, the point P has coordinates (x, f(x)). Point Q has coordinates (x + dx, f(x + dx)).
So, the change in y, that is dy is f(x + dx) − f(x). Then,
This is the definition, for any function y = f(x), of the derivative, dy/dx
NOTE: Given y = f(x), its derivative, or rate of change of y with respect to x is defined as
Example
Suppose we want to differentiate the function f(x) = 1/x from first principles.
A sketch of part of this graph shown below. We have marked point P(x, f(x)) and the neighbouring point Q(x + dx, f(x +d x)).
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# Re: st: Questions about matrix operations
From anirban basu To statalist@hsphsun2.harvard.edu Subject Re: st: Questions about matrix operations Date Mon, 17 Jun 2002 14:45:36 -0500 (CDT)
```Hi,
I think you want the -mat score- option...
mat bhat=e(b)
mat score xb=bhat
Remember, it will genenate a scalar for each observation. i.e. xb will be
a new variable containing values of x(i).b for observation (i).
Anirban
______________________________________
ANIRBAN BASU
Doctoral Fellow
Harris School of Public Policy Studies
University of Chicago
(312) 563 0907 (H)
________________________________________________________________
On Mon, 17 Jun 2002, Wolff wrote:
> Hey folks,
> I have a simple problem. I have a regression with three variables. I
> would like to create a new variable equal to the sum of the estimated
> coefficients times the variable values. The coefficients are saved in
> the matrix e(b). I would like to do something like newvar=e(b)'*[var1
> var2 var3]. Since e(b)' is 1X3 and the next matrix is 3X1, I thought it
> would generate a scalar. Is there an easy way to accomplish this?
>
> Thanks,
> Eric
>
>
>
> Eric D. Wolff
> Carnegie-Mellon University
> Tech and Frew
> Pittsburgh, PA 15213
> (412) 268-2939 (office)
> (412) 268-6837 (fax)
> email: wolff@andrew.cmu.edu
>
>
> *
> * For searches and help try:
> * http://www.stata.com/support/faqs/res/findit.html
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
*
* For searches and help try:
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* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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# Solving Proportions Worksheet Doc
Braja Sorensen Team August 9, 2020 Worksheet
The fact is they are very different. Or behind the word is
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# Lebesgue constants of multi-dimensional partial Fourier sums
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Let be an integrable function on , , , -periodic in each variable. Consider its Fourier series , where , , the lattice of points in with integer coordinates, , while
is the th Fourier coefficient of . No natural ordering of Fourier coefficients exists, thus the definition of a multi-dimensional partial Fourier sum presents many problems and points of interest intimately connected to geometry and number theory.
To indicate that the partial sum corresponds to a certain summation domain , one denotes it by
Frequently, sums are considered, where is the th dilatation of a fixed set ; in many cases this is the most natural way of summation. An example of partial Fourier sums that are not of this kind are the rectangular partial sums. By one denotes the partial Fourier sum when the dependence on the parameter , either scalar or vectorial, is of primary importance. As is well-known, if the Fourier series of a continuous function fails to converge at each point, then the sequence of norms of the operators ,
taking into (or, equivalently, into ) is unbounded and measures the rate of divergence of the Fourier series.
This is strongly related to the behaviour of the Fourier transform of the indicator function of the summation domain . For known results on this subject, see, e.g., [a11], [a13], [a18].
For the spherical partial Fourier sums , where , the following order estimate holds: There exist positive constants , and , , depending only on , such that
(a1)
The estimate from below has first appeared in [a14]; the method used there is the main tool for obtaining lower bounds for Lebesgue constants. Nothing is known about existence or non-existence of the limit of as ; this is the main open problem in the subject (as of 2000).
More general summation domains possessing properties of the spherical partial Fourier sums have been considered. E.g., in Yudin's estimate from above [a25], the summation domains are balanced (i.e., along with each point the whole set , , belongs to ), with finite upper Minkowski measure, that is,
where . These natural assumptions provide the bound . It turns out that to satisfy the same estimate from below, only local information is needed ([a17]): Let the boundary of a domain contain a simple (non-intersecting) piece of a surface of smoothness in which there is at least one point with non-vanishing principal curvatures (cf. also Principal curvature). Then there exists a positive constant , depending only on , such that for large.
The estimates in the spherical case and its generalizations are the worst possible if is compact. Once has a point with non-vanishing principal curvatures, the Lebesgue constants are that "bad" . The other side of the scale is called "polyhedral" and is of "logarithmic nature" . Only some natural restrictions have to be put on polyhedra , for example, the hyperplanes that define the sides of the polyhedron do not contain the origin. In that case there exist two positive constants and , , such that for each such polyhedron :
(a2)
Actually, this was proved by E. Belinsky [a6].
There are two important problems concerning the polyhedral case. The first is: Can partial Fourier sums have Lebesgue constants with an intermediate rate of growth (i.e. between (a1) and (a2))? Some trivial solutions were suggested in [a26], where an intermediate growth is achieved by the product of the two mentioned situations. Of course, this is possible only for dimension three and greater. Thus, the problem is to find one for dimension two. It is clear that in this case the boundary can possess no point with non-vanishing curvature. On the other hand, any polyhedron matches (a2). Thus, the solution can only be a (convex) "polyhedron" with infinitely many specially located sides. Such a solution was constructed by A. Podkorytov [a21].
The next question also seems very natural: Is it possible to have a certain asymptotic relation instead of the order estimate (a2)? For rectangular partial sums some special cases were investigated by I. Daugavet [a9] and O. Kuznetsova [a15]. For the sequence of dilated summation domains, an unexpected result was obtained by Podkorytov [a22]. Here also causes the main difficulties. There are two main cases. In the first one, the polygons with sides of rational slopes are dealt with — then the estimates change insignificantly if only one considers instead of sums the corresponding integrals, that is, the Fourier transform of the indicator function of the -dilation of the corresponding set . This allows one to obtain logarithmic asymptotics; namely, the values , and are equivalent. When at least one of the slopes is irrational, the situation changes qualitatively: The upper limit and the lower limit of the ratio , as , may differ. In [a22] and [a20], quantitative estimates of this phenomenon as well as open problems are given.
The paper [a4] started the interest in various questions of approximation theory and Fourier analysis in connected with the study of hyperbolic cross partial Fourier sums (see, e.g., [a24] and Hyperbolic cross). The exact order of growth of their Lebesgue constants, , the same as in the spherical case, was established in the two-dimensional case independently in [a5] and in [a27], and afterwards was generalized to the case of arbitrary dimension in [a16].
Step hyperbolic crosses were introduced by B. Mityagin [a19] and are defined as follows (cf. also Step hyperbolic cross): for such that . These have many important applications too. Belinsky [a8] proved that there exist two positive constants and , , such that .
When is unbounded, it may happen that the operator is unbounded even for fixed . It is proved in [a3] that the Lebesgue constants are either of the usual order of growth or infinite for all values of the parameter , where is a hyperbolic cross, depending on whether the hyperbolic cross is turned at a rational or irrational angle, respectively. For , this was earlier obtained in [a7], which also contains similar results for the strip to be a summation domain.
For other results on Lebesgue constants and related topics, see [a1], [a2], [a10], [a12], [a18], [a23], [a28], [a29].
The ideas used to prove many of the results discussed above have also been applied to estimates of the Lebesgue constants of linear means of multiple Fourier series.
Some results are known for Lebesgue constants in more abstract settings, e.g., for spherical harmonics expansions or Fourier series on compact Lie groups.
#### References
[a1] Sh.A. Alimov, R.R. Ashurov, A.K. Pulatov, "Multiple Fourier series and Fourier integrals" V.P. Khavin (ed.) N.K. Nikolskii (ed.) , Commutative Harmonic Analysis IV , Enc. Math. Sci. , 42 , Springer (1992) pp. 1–95 Itogi Nauki i Tekhn. VINITI Akad. Nauk. SSSR , 42 (1989) pp. 7–104 [a2] Sh.A. Alimov, V.A. Ilyin, E.M. Nikishin, "Convergence problems of multiple Fourier series and spectral decompositions, I, II" Russian Math. Surveys , 31/32 (1976/77) pp. 29–86; 115–139 Uspekhi Mat. Nauk. , 31/32 (1976/77) pp. 28–83; 107–130 [a3] E.S. Belinskii, E.R. Liflyand, "Behavior of the Lebesgue constants of hyperbolic partial sums" Math. Notes , 43 (1988) pp. 107–109 Mat. Zametki , 43 (1988) pp. 192–196 [a4] K.I. Babenko, "Approximation by trigonometric polynomials in a certain class of periodic functions of several variables" Soviet Math. Dokl. , 1 (1960) pp. 672–675 Dokl. Akad. Nauk. SSSR , 132 (1960) pp. 982–985 [a5] E.S. Belinsky, "Behavior of the Lebesgue constants of certain methods of summation of multiple Fourier series" , Metric Questions of the Theory of Functions and Mappings , Nauk. Dumka, Kiev (1977) pp. 19–39 (In Russian) [a6] E.S. Belinsky, "Some properties of hyperbolic partial sums" , Theory of Functions and Mappings , Nauk. Dumka, Kiev (1979) pp. 28–36 (In Russian) [a7] E.S. Belinsky, "On the growth of Lebesgue constants of partial sums generated by certain unbounded sets" , Theory of Mappings and Approximation of Functions , Nauk. Dumka, Kiev (1983) pp. 18–20 (In Russian) [a8] E.S. Belinsky, "Lebesgue constants of step hyperbolic partial sums" , Theory of Mappings and Approximation of Functions , Nauk. Dumka, Kiev (1989) pp. 23–27 (In Russian) [a9] I.K. Daugavet, "On the Lebesgue constants for double Fourier series" Meth. Comput., Leningrad Univ. , 6 (1970) pp. 8–13 (In Russian) [a10] M. Dyachenko, "Some problems in the theory of multiple trigonometric series" Russian Math. Surveys , 47 : 5 (1992) pp. 103–171 Uspekhi Mat. Nauk. , 47 : 5 (1992) pp. 97–162 [a11] I.M. Gelfand, M.I. Graev, N.Ya. Vilenkin, "Generalized functions 5: Integral geometry and problems of representation theory" , Acad. Press (1966) [a12] B.I. Golubov, "Multiple Fourier series and integrals" J. Soviet Math. , 24 (1984) pp. 639–673 Itogi Nauki i Tekhn. VINITI Akad. Nauk. SSSR , 19 (1982) pp. 3–54 [a13] C.S. Herz, "Fourier transforms related to convex sets" Ann. of Math. , 2 : 75 (1962) pp. 81–92 [a14] V.A. Ilyin, "Problems of localization and convergence for Fourier series in fundamental systems of the Laplace operator" Russian Math. Surveys , 23 (1968) pp. 59–116 Uspekhi Mat. Nauk. , 23 (1968) pp. 61–120 [a15] O.I. Kuznetsova, "The asymptotic behavior of the Lebesgue constants for a sequence of triangular partial sums of double fourier series" Sib. Math. J. , 18 (1977) pp. 449–454 Sibirsk. Mat. Zh. , XVIII (1977) pp. 629–636 [a16] E.R. Liflyand, "Exact order of the Lebesgue constants of hyperbolic partial sums of multiple Fourier series" Math. Notes , 39 (1986) pp. 369–374 Mat. Zametki , 39 (1986) pp. 674–683 [a17] E.R. Liflyand, "Sharp estimates of the Lebesgue constants of partial sums of multiple Fourier series" Proc. Steklov Inst. Math. , 180 (1989) pp. 176–177 Trudy Mat. Inst. V.A. Steklov. , 180 (1987) pp. 151–152 [a18] E.R. Liflyand, A.G. Ramm, A.I. Zaslavsky, "Estimates from below for Lebesgue constants" J. Fourier Anal. Appl. , 2 (1996) pp. 287–301 [a19] B.S. Mityagin, "Approximation of functions in and spaces on the torus" Mat. Sb. (N.S.) , 58 (100) (1962) pp. 397–414 (In Russian) [a20] F. Nazarov, A. Podkorytov, "On the behavior of the Lebesgue constants for two-dimensional Fourier sums over polygons" St.-Petersburg Math. J. , 7 (1995) pp. 663–680 Algebra i Anal. , 7 (1995) pp. 214–238 [a21] A.N. Podkorytov, "Intermediate rates of growth of Lebesgue constants in the two–dimensional case" J. Soviet Math. , 36 (1987) pp. 276–282 Numerical Methods and Questions on the Organization of Calculations, Part 7 Notes Sci. Sem. Steklov Inst. Math. Leningrad. Branch Acad. Sci. USSR, Nauka, Leningrad , 139 (1984) pp. 148–155 [a22] A.N. Podkorytov, "Asymptotic behavior of the Dirichlet kernel of Fourier sums with respect to a polygon" J. Soviet Math. , 42 (1988) pp. 1640–1646 Zap. Nauchn. Sem. LOMI , 149 (1986) pp. 142–149 [a23] E.M. Stein, G. Weiss, "Introduction to Fourier analysis on Euclidean spaces" , Princeton Univ. Press (1971) [a24] V.N. Temlyakov, "Approximation of periodic functions" , Nova Sci. (1993) [a25] V.A. Yudin, "Behavior of Lebesgue constants" Math. Notes , 17 (1975) pp. 369–374 Mat. Zametki , 17 (1975) pp. 401–405 [a26] V.A. Yudin, "A lower bound for Lebesgue constants" Math. Notes , 25 (1979) pp. 63–65 Mat. Zametki , 25 (1979) pp. 119–122 [a27] A.A. Yudin, V.A. Yudin, "Discrete imbedding theorems and Lebesgue constants" Math. Notes , 22 (1977) pp. 702–711 Mat. Zametki , 22 (1977) pp. 381–394 [a28] L.V. Zhizhiashvili, "Some problems in the theory of simple and multiple trigonometric and orthogonal series" Russian Math. Surveys , 28 (1973) pp. 65–127 Uspekhi Mat. Nauk. , 28 (1973) pp. 65–119 [a29] L.V. Zhizhiashvili, "Some problems of multidimensional harmonic analysis" , Tbilisi State Univ. (1996) (In Russian) (Edition: Second)
How to Cite This Entry:
Lebesgue constants of multi-dimensional partial Fourier sums. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Lebesgue_constants_of_multi-dimensional_partial_Fourier_sums&oldid=14045
This article was adapted from an original article by E.R. Liflyand (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# How to generate Bayer matrix of arbitrary size?
In ordered dithering Bayer matrix is used.
How is that matrix generated? What algorithm can be used to generate matrix of arbitrary size?
• Have you actually read the wikipedia page you linked? Let me try to summarize it in an answer Sep 29, 2016 at 11:56
(As @Bálint seems not to have gotten around to it… :)
To quote the article as it (likely) was when you initially asked:
Arbitrary size threshold maps can be devised with a simple rule: First fill each slot with a successive integer starting from 1. Then reorder them such that the average distance between two successive numbers in the map is as large as possible, ensuring that the table "wraps" around at edges.
A more detailed explanation has since been added to the article:
For threshold maps whose dimensions are a power of two, the map can be generated recursively via:
The recursive expression can be calculated explicitly using only bit arithmetic:
M(i, j) = bit_reverse(bit_interleave(bitwise_xor(x, y), x)) / n ^ 2
That last bit can be implemented with little trouble in e.g. Python:
def bit_reverse(x, n):
return int(bin(x)[2:].zfill(n)[::-1], 2)
def bit_interleave(x, y, n):
x = bin(x)[2:].zfill(n)
y = bin(y)[2:].zfill(n)
return int(''.join(''.join(i) for i in zip(x, y)), 2)
def bayer_entry(x, y, n):
return bit_reverse(bit_interleave(x ^ y, y, n), 2*n)
def bayer_matrix(n):
r = range(2**n)
return [[bayer_entry(x, y, n) for x in r] for y in r]
print(bayer_matrix(2))
This doesn't produce quite the same matrix as listed in the Wikipedia article, but as the author of the original source writes, "it is good enough in practice".
• For bayer_matrix(1), and the sake of brevity, your code generates [[0, 1], [3, 2]] but for correct results (when doing ordered dithering) I expect it to be [[0, 2], [3, 1]]? Feb 5, 2021 at 14:41
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# Gravitational effect creates the irreversibility
The gravitational force must always act on an object under any circumstances. We have had to supply additional energy against the gravitational force during a closed loop process, leading to an irreversible process.
The individual objects that make up our universe have an inevitable interaction with the environment. The reversible process is an ideal process in which the object can be treated with complete isolation from the environment. In reality, no process is absolutely reversible, and no system can operate in a completely isolated environment. Irreversibility is the most natural characteristic of all phenomena and a natural law. Since every phenomenon is reversible, it’s of great interest to investigate the physical mechanism of the irreversibility and to answer what creates it.
Obviously, we cannot create an environment that can entirely eliminate the influence of gravitational field. An object occupies a position and takes up a certain amount of time to generate the space-time fabric in the universe, and the gravitational force must have always act on it. The position that the object takes depends on the gravitational field, which in turn depends on the position of the object. The object at different positions has different quantized states in a gravitation field and thus, its mass related to its total internal energy is different at different energy states. The situation can become quite complicated because the object always moves from one place to another. To do that some work will be supplied against the gravitational force. In order to tie the gravitational field to the irreversibility, we here make the crucial connections among the theory of relativity, quantum physics, energy conservation and thermodynamics. The energy of the universe is always constant and the law of energy conservation works. We follow these rules to begin our analysis.
We start at position A to move an object to another position B, and then we move it back to position A in a gravitational field. The gravitational force must always have an effect on it under any circumstances, and the object at positions A and B has different quantized states. If the principle of energy conservation works, the work done during the process from position A to position B is not symmetrical for that done during the return trip from position B to position A. We have had to do an additional work against the gravitational force, which may be written mathematically in this way
m is the mass of the object, which is related to its total internal energy E by m=E/c2 , c is the velocity of light, g is the free fall acceleration and H is the height in a gravitational field.
Figure 1 Energy diagram for possible transitions
We first think of the transitions from different conduction bands to the valence band in a quantum well (QW) diode in a gravitational field, for example, from energy state E1 to energy state E0, as show in Fig. 1. This transition happens by emitting light with a frequency of ωc-v. According to Planck’s formula, the energy of a photon related to its frequency will be given by c-v, h is a fundamental physical constant named the Planck constant. During the return trip, a photon with frequency is absorbed to achieve the transition from energy state E0 to energy state E1. The energy of this photon is v-c . Since the transition occurs in a gravitational field, a falling distance H exists from energy state E1 to energy state E0. Therefore, the mass will be given by m1=E1/c2at the energy state E1 and m0=E0/c2 at the energy state E0, respectively. As a result, for a stable free fall acceleration g, the additional work should be done as
Because the height H is not less than zero and energy state E1 is higher than energy state E0, we have
The frequency is higher than the frequency in the gravitational field, suggesting that the QW diode can only detect and modulate higher-energy photons than those emitted by itself. The heights H are normally very small, and consequently, the frequency differences between and are small.
Figure 2 Possible paths between two points in a gravitational field
Proceeding in this way, we consider an object which is moved from position A to position B and brought back to position A along the same path in a gravitational field, as shown in Fig. 2. Both position A and position B are separated by the height H in this gravitational field with the free fall acceleration g. The object has a total internal energy Eb at position B and Ea at position A, respectively, in which Eb is not less than Ea. So, we have done a net amount of work against the gravitational force, which is equal to
Because the height H is not less than zero and Eb is not less than Ea, we have
Figure 3 Change in net work in an irreversible process
In thermodynamics, we use the mean molecular kinetic energy as the definition of the temperature T. Suppose that a system operates from condition A(Ta, Va) to condition B(Tb, Vb), changing energy states A(Ea, Ha) to B(Eb, Hb) in a gravitational field. For simple, Eb is not less than Ea, and Hb is not less than Ha. When the system goes from B(Eb, Hb) to A(Ea, Ha) and returns back from A(Ea, Ha) to B(Eb, Hb), some additional work will be done. Since and are equal to and , respectively, according to the equivalence principle, the net amount of work done is given by
For a stable temperature Ta=Tb, Ea is equal to Eb, the net work done depends on the change of the volume, and will be given by
For a stable volume Va=Vb, Ha is equal to Hb, the net work done depends on the change of the temperature, and will be given by
Our postulations are based on the law of energy conservation and the equivalence principle: (1) The individual objects cannot be completely isolated from the environment in reality, and the gravitational force must always act on them under any circumstances; (2) The objects at different positions have different quantized states in a gravitation field and their masses in turn depends on their energy states. (3) the amount of work done against the gravitational force is different when the object moves from one place to another and goes back to the starting position. The theory of relativity, quantum physics and energy conservation all fit together to connect the gravitational field with the irreversibility. We can conclude that the gravitational effect creates the irreversibility.
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Category: Sports
Stay or Hit, The Risk Remains
Few people realize that the success of the Houston Astros was birthed at the Blackjack tables of a Lake Tahoe casino before Billy Beane employed Moneyball to save the Oakland As.
Working as a dealer at the table, an engineering student majoring in statistical analytics noticed a pattern. Whenever blackjack players “hit” when they already had a score of 17 or more, they almost always busted (exceeded 21, thus losing). However, if they hit on 16 or less, they often got close to 21, sometimes winning, sometimes losing.
This student watched as players lost hundreds, if not thousands, by making emotional decisions, hitting on 18, or staying at 16. To consistently perform well, this student concluded that 17 was the magic number. Stay if more than 17, hit if 16 or less.
From this experience, the student learned the value of rational decisions that are data driven. That same student went on to develop the analytics program of the Houston Astros, which was instrumental in the development of the roster that won the 2017 World Series.
That program analyzed prospective players not only by their stats, but also by their experiences and physical characteristics, using patterns identified by analyzing the prior 30 years of MLB draft picks.
The book, Astroball, by Ben Reiter, outlines how the Astros built a perennial winner out of a team that was built and developed, not bought.
Despite the success of the Astros, their trek was not without failures. The Stros often released players who went on to brilliant careers, and retained players who flamed out.
Despite all the data and direction offered by a sound analytics program, the Astros were unable to eliminate risk from the equation when it came to analyzing prospective players.
Which takes us back to the blackjack table. Even if you religiously follow the data-driven wisdom of when to hit, and when to stay, sometimes you bust, sometimes you win, and sometimes the House wins.
Obviously, as The Chaplain’s Corner, this blog is not about how to win at the card table, and I’ve never been successful as a sports writer. Seeing the intersection of cards and baseball, however, I do see some life lessons.
In either of those situations, there is risk. There’s risk in action, and there’s risk in inaction. There are consequences for taking the leap, and there are consequences for staying put.
The Astros took a risk in spending millions to pioneer a new form of analytics for player evaluation. Had it not worked, the team would have wasted a fortune to stay in last place. But it worked, and the evidence is right there on the field.
Had they stayed put, they may have still built a winning team, but they’d likely have overpaid, the way they did in 2005.
In blackjack, taking the hit can win the hand, or it can bust you. Staying will keep you from busting, but it may or may not win the hand.
And in life, you have choices. To change careers. To relocate to accept a promotion. To start a business, invest in the stock market, or to buy bonds.
Maybe the business succeeds, maybe it fails, or if you don’t start the business, maybe you miss an opportunity. The stocks rise, the stocks fall, or maybe you don’t buy and you miss an opportunity. You buy the bonds with a guaranteed yield, but inflation negates your gain.
Risk is inherent in every decision, every opportunity, and every moment in life. Stay or hit, the risk remains.
Fear and avoidance of risk is futile. The reasonable thing to do is to evaluate risk, choose the risk with the maximum upside and minimum downside, and hedge against losses.
But losses will happen. So will successes. The key is to live life, to move forward (even if that means staying) and to glorify God in the process.
So the choice is yours. Stay, or hit?
When Rome falls
You find the most interesting things on Facebook.
Scrolling through my news feed this morning, I came across this gem, comparing the distraction most Americans enjoy via the NFL with the distractions most Romans enjoyed via chariot races, gladiator “games,” and the Olympics.
And while the Roman government deliberately built elaborate stadiums to distract the masses from the crumbling empire and human rights abuses, in America, we distracted ourselves.
Now, I’m not bashing sports, or the NFL. I enjoy watching football, and even have been able to attend a few Big-12 College Football games, and one Dallas Cowboys’ Thanksgiving Day game. I will probably continue to enjoy watching sports for the foreseeable future.
But for some reason, seeing the above-posted meme on Facebook was kind of an eye-opener for me.
Do you know why the NFL protests were so controversial? And subsequently, why the NFL protests have, at least in part, played a role in the decline of NFL ratings? It’s because, once the players used their platform to advance a socially conscious agenda, they reminded us of the social problems that remain in America.
Whether you agree with Colin Kaepernick or not, seeing he and his followers take a knee during the Star-Spangled Banner reminded you that the reconciliation we thought we had accomplished hadn’t advanced us as far as we had thought. Having that bubble burst, watching football became a reminder of the deep-divides that remain in American society. Once that happened, watching football wasn’t as fun as it was before.
And that’s why the NFL protests were so controversial. People don’t like to be reminded of their problems as they try to escape them. So, we had the controversial debate over the past two years, and we quit watching football.
The good news is that we can use our newly raised awareness to make good things happen. True change will not come through legislation, political action, or by socially-conscious NFL players. It will come through the small, daily decisions made by each individual. So, to borrow a phrase, “be the change.” Extend random acts of kindness to others, and let your light so shine before men that they may see your good works, and glorify your Father in Heaven.
And Go Bulldogs!
Because America enjoys a good train wreck
Let’s be honest. America loves a good train wreck.
You may have heard of Amy Winehouse, but have you ever listened to her music? Most who read this know of Winehouse, fewer can recite her lyrics.
You never heard of Paris Hilton or Kim Kardashian before their sex-films were made public. Tommy Lee’s fame extended beyond his days with Motley Crue as his rocky relationship with Pamela Anderson kept his image on the front of tabloid publications everywhere.
While Lindsey Lohan had a good acting career as a child, most of her press coverage came as a result of her meltdown as she transitioned into adulthood.
These, and other celebrities plagued by personal calamities spawned gossip column articles, magazine covers, reality shows and movies of the week. So, it should come as no surprise that a movie detailing the saga of Tonya Harding and Nancy Kerrigan in the run up to the 1994 Olympics will hit theaters on Dec. 8.
I, Tonya chronicles the life of Tonya Harding leading up to the incident where a hit man hired by her bodyguard struck Nancy Kerrigan above the knee, bruising her thigh and taking her out of the USA National Competition.
The movie chronicles the abuse she endured at the hands of her mother, her dysfunctional relationship with Jeff Gillooly, her struggle to rise to the top of the figure-skating world, the attack on Kerrigan and the fallout thereafter.
Previews of the movie show a jaded Harding character, played by Margot Robbie, struggling through life in the brash fashion that got her labeled as “white trash” back in the 1990s. The depiction of Harding in news reports, TV shows, made-for-TV movies and reality shows in the aftermath of the attack on Kerrigan is one of an unsophisticated white trash girl who somehow stumbled into the talent to make the world figure-skating stage.
The goal of each of these depictions is not necessarily to tell her side of the story, nor is it to tell Nancy’s side, but rather to present another train wreck for America’s entertainment. Judging by the trailers for I, Tonya, this next film promises to be no different.
The saga of Tonya Harding speaks to a blemish on America’s culture at large. The culture is content to thrust a person like Harding into the national spotlight for our amusement, with no regard given for her personal healing and well-being. We laugh at her failure, poke fun at her rural impoverished upbringing, mock her tears, and think of ways we could have done it better.
Such a cultural mentality is not only a shame, but falls into a category of evil described in Romans 1:31-32, “Without understanding, covenantbreakers, without natural affection, implacable, unmerciful: Who knowing the judgment of God, that they which commit such things are worthy of death, not only do the same, but have pleasure in them that do them.”
Tonya Harding was a mess. I’d like to see a revived, redeemed and stronger Tonya emerge. But the fact that we are willing to sit back and find amusement in her demise places us in the same category as those who carried out the attack on Nancy Kerrigan. And folks, that’s not where you want to be on Judgment Day.
Castro, Cruz, and Texas Red vs. Blue
In the debut episode of the “Leland Acker Show” podcast, I examine Joaquin Castro’s decision to stay in the House and not challenge Sen. Ted Cruz in 2018, and what that means for Texas Democrats.
I also discuss what single event could turn Texas blue, the 10 reasons millennials are leaving Christianity, and Jeb Bush’s prospects in owning the Miami Marlins. Check it out, then tell me what you think.
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# Chapter 10. Abstract algebra
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## Transcription
1 Chapter 10. Abstract algebra C.O.S. Sorzano Biomedical Engineering December 17, Abstract algebra December 17, / 62
2 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
3 References J.B. Fraleigh. A first course in Abstract Algebra. Pearson, 7th Ed. (2002) 10. Abstract algebra December 17, / 62
4 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
5 Sets Definition 1.1 (Set) A set is a well-defined collection of elements. We denote the different elements as a S. Definition 1.2 (Empty set) The only set without any element is the empty set ( ). Describing sets We may provide the elements of a set: Intensional definition: by giving a property they all meet (e.g., even numbers from 1 to 10) Extensional definition: by listing all the elements in the set (e.g.,{2, 4, 6, 8, 10}). The order in which the different elements are written has no meaning. 10. Abstract algebra December 17, / 62
6 Sets Definition 1.3 (Subset and proper subset) B is subset of A (denoted B A or A B) if all the elements of B are also elements of A. B is a proper subset of A if B is a subset of A and B is different from A (B A or A B). Properties A is an improper subset of A. is a proper subset of A. Definition 1.4 (Power set (Partes de un conjunto)) The set of all subsets of a set A is called the power set of A. Let A = {1, 2, 3} the power set of A is P(A) = {, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} 10. Abstract algebra December 17, / 62
7 Sets Definition 1.5 (Cartesian product) The cartesian product of the sets A and B is the set of all ordered pairs in which the first element comes from A and the second element comes from B. A B = {(a, b) a A, b B} Note that because of the ordered nature of the pair A B B A. Let A = {1, 2, 3} and B = {4, 5}. A B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} Definition 1.6 (Cardinality) The cardinality of a set is the number of elements it has. 10. Abstract algebra December 17, / 62
8 Sets Definition 1.7 (Disjoint sets) Two sets are disjoint if they do not have any element in common. Some useful sets Integer numbers: Z = {..., 2, 1, 0, 1, 2,...}, Z = ℵ 0 Natural numbers, positive integers: N = Z + = {1, 2, 3,...}, N = ℵ 0 Negative integers: Z = {..., 3, 2, 1}, Z = ℵ 0 Non-null integers: Z = Z {0} = {..., 2, 1, 1, 2,...}, Z = ℵ 0 Rational numbers: Q, Q = ℵ 0 Real numbers: R, R = ℵ 1 Interval: [0, 1], [0, 1] = ℵ 1 Complex numbers: C = {a + bi a, b R}, C = ℵ Abstract algebra December 17, / 62
9 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
10 Relations Definition 2.1 (Relation) A relation arb is a subset of the cartesian product A B. 10. Abstract algebra December 17, / 62
11 Functions Definition 2.2 (Function) A function f : X Y is a relation between X and Y in which each x X appears at most in one of the pairs (x, y). We may write (x, y) f or f (x) = y The domain of f is X, the codomain of f is Y. The support of f is the set of all those values in X for which there exists a pair (x, y). The range of f are all values in Y for which there exists at least one pair (x, y). f : R R f (x) = x 3 (2, 8) f f (2) = 8 + : R R R ((2, 3), 5) + +((2, 3)) = = Abstract algebra December 17, / 62
12 Classification of functions Definition 2.3 Functions can be classified as surjective, injective or bijective: Surjective: A function is surjective if every point of the codomain has at least one point of the domain that maps onto it. They are also called onto functions. Injective: A function is injective if every point of the codomain has at most one point in the domain that maps onto it. They are also called one-to-one functions. Bijective: A function is bijective if it is injective and surjective. 10. Abstract algebra December 17, / 62
13 Inverse function Definition 2.4 (Inverse function) Consider an injective function f : X Y. f 1 : Y X is the inverse of f iff (x, y) f (y, x) f 1 f (x) = x + 3 f 1 (y) = y 3 f (x) = x 3 f 1 (y) = y 1 3 f (x) = x 2 is not invertible because it is not injective (f ( 2) = f (2) = 4) 10. Abstract algebra December 17, / 62
14 Inverse function Theorem 2.1 If f is invertible, its inverse is unique. If f is bijective, so is f 1. X and Y have the same cardinality if there exists a bijective function between the two. Consider the following function f : Z N f = {(0, 0), ( 1, 1), (1, 2), ( 2, 3), (2, 4), ( 3, 5), (3, 6),...} f is bijective. Consequently, Z has the same cardinality as N. 10. Abstract algebra December 17, / 62
15 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
16 Partition Definition 3.1 (Partition) A partition of a set S is a collection of non-empty subsets such that each element of S belongs to one and only one subset (cell) of the partition. We denote as x the subset that contains the element x. All cells in a partition are disjoint to any other cell. s We may partition the set of natural numbers into the subset of even numbers ({2, 4, 6,...}) and the subset of odd numbers ({1, 3, 5,...}). We may partition the set of integer numbers into the subset of all multiples of 3 ({..., 6, 3, 0, 3, 6,...}), the subset of all numbers whose remainder after dividing by 3 is 1 ({..., 5, 2, 1, 4, 7,...}), and the subset of all numbers whose remainder after dividing by 3 is 2 ({..., 4, 1, 2, 5, 8,...}). 10. Abstract algebra December 17, / 62
17 Equivalence relation Definition 3.2 (Equivalence relation) R is an equivalence relation in S if it verifies: 1 R is reflexive: xrx 2 R is symmetric: xry yrx 3 R is transitive: xry, yrz xrz s 1 = is an equivalence relation. 2 Congruence modulo n is an equivalence relation (two numbers are related if they have the same remainder after dividing by n) : 1 and 4 have remainder 1 after dividing by 3. We write 1 4(mod3) 3 n, m Z nrm nm 0 is not an equivalence relationship because it is not transitive (e.g., 3R0, 0R5 but 3 R5). 10. Abstract algebra December 17, / 62
18 Partition and equivalence relation Theorem 3.1 Let S be a non-empty set, and R an equivalence relation defined on S. Then R partitions S with the cells ā = {x S xra} Additionally, we may define another equivalence relation a b ā = b 10. Abstract algebra December 17, / 62
19 Partition and equivalence relation Congruence modulo 3 is an equivalence relation in Z (two numbers are related if they have the same remainder after dividing by 3) Additionally and 0 = {..., 6, 3, 0, 3, 6,...} 1 = {..., 5, 2, 1, 4, 7,...} 2 = {..., 4, 1, 2, 5, 8,...}... = 0 = 3 = 6 = = 1 = 4 = 7 = = 2 = 5 = 8 = Z = Abstract algebra December 17, / 62
20 Partition and equivalence relation Consider the cartesian product Z (Z {0}). Let (m 1, n 1 ) and (m 2, n 2 ) be two ordered sets of this cartesian product. Consider now the equivalence relation (m 1, n 1 ) (m 2, n 2 ) m 1 n 2 m 2 n 1 = 0 The set of rational numbers is formally defined Q as the set of equivalence classes of Z (Z {0}) under the relation. 10. Abstract algebra December 17, / 62
21 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
22 Binary operations Introduction What is addition? Let us assume that we arrive to a classroom in Mars, and that martians are learning to add. The teacher says and the students reply: Then, the teacher says: and the students reply: Gloop, poyt Bimt. Ompt, gaft Poyt. We don t know what they do but it seems that when the teacher gives two elements, students respond with another element. 10. Abstract algebra December 17, / 62
23 Binary operations Introduction (continued) What is addition? This is what we do when we say three plus four, seven. And we may not use any two elements ( three plus apples is not defined). We can only use elements on a given set. This is what we formally call a binary operation. Definition 4.1 (Binary operation) A binary operation on a set S is a function: : S S S (a, b) = a b 10. Abstract algebra December 17, / 62
24 Binary operations s The following binary operations are all different: + : R R R + : Z Z Z + : M m n (R) M m n (R) M m n (R) The following is not a binary operation because it is not well defined + : M(R) M(R) M(R) we don t know how to add a 2 2 matrix with a 3 3 one. 10. Abstract algebra December 17, / 62
25 Closed set Definition 4.2 Let S be a set and H a subset of S. H is said to be closed with respect to the operation defined in S iff a, b H a b H Then we may define the binary operation in H: : H H H (a, b) = a b which is called the binary operation induced in H. 10. Abstract algebra December 17, / 62
26 Closed set Let S = Z and H = {n 2 n Z + } = {1, 4, 9, 16, 25, 36,...}. H is not closed with respect to addition. For example: 1 H 4 H but / H Let S = Z and H = {n 2 n Z + } = {1, 4, 9, 16, 25, 36,...}. H is closed with respect to multiplication. For example: n 2 H m 2 H and n2 m 2 = (nm) 2 H 10. Abstract algebra December 17, / 62
27 Closed set Let S be the set of real-valued functions with a single real argument S = {R R}. Let us define the addition of functions as + : (R R) (R R) R R (f + g)(x) = f (x) + g(x) Similarly for the multiplication and subtraction of functions. Let us define the composition of functions as : (R R) (R R) R R (f g)(x) = f (g(x)) S is closed with respect to addition, subtraction, multiplication and composition. 10. Abstract algebra December 17, / 62
28 Definition of a binary operation To define a binary operation either we give the full table (intensional definition) as in a b b = 0 b = 1 b = 2 a = a = a = or a b b = 0 b = 1 b = 2 a = a = a = or we give a rule to compute it (extensional definition) as in a b = (a + b) mod Abstract algebra December 17, / 62
29 Properties of a binary operation Definition 4.3 (Commutativity) A binary operation is commutative iff a b = b a is commutative because its definition table is symmetric with respect to the main diagonal, but is not commutative. 10. Abstract algebra December 17, / 62
30 Properties of a binary operation Definition 4.4 (Associativity) A binary operation is associative iff (a b) c = a (b c) is not associative because But is associative (0 0) 0 = 1 0 = 1 0 (0 0) = 0 1 = 2 (0 0) 0 = 0 0 = 0 0 (0 0) = 0 0 = 0 We would have to test all possible triples, but after a a little bit of work we could show that is associative. 10. Abstract algebra December 17, / 62
31 Properties of a binary operation Function composition is associative although not commutative. Proof Function composition is not commutative Function composition is associative (f g)(x) = f (g(x)) g(f (x)) = (g f )(x) ((f g) h)(x) = (f g)(h(x)) = f (g(h(x))) = f ((g h)(x)) = (f (g h))(x) 10. Abstract algebra December 17, / 62
32 Properties of a binary operation A function may not be well defined. For instance, is not well defined for b = 0 Q / : Q Q Q a/b = a b A function may not be closed in S. For instance, / : Z Z Z a/b = a b is not closed because a = 1 Z, b = 3 Z but 1 3 / Z. 10. Abstract algebra December 17, / 62
33 Properties of a binary operation Definition 4.5 (Existence of a neutral element) A binary operation has a neutral element, e, iff a S a e = e a = a 0 is the neutral element of addition in R because r R r + 0 = 0 + r = r 1 is the neutral element of multiplication in R because r R r 1 = 1 r = r Addition in N has no neutral element since 0 / N. 10. Abstract algebra December 17, / 62
34 Properties of a binary operation Definition 4.6 (Existence of an inverse element) A binary operation has an inverse element iff a S b S a b = b a = e being e the neutral element of. The inverse element of 2 with respect to addition in R is -2 because 2 + ( 2) = ( 2) + 2 = 0 The inverse element of 2 with respect to multiplication in R is 1 2 because = = 1 Multiplication in N has no inverse element since n N 1 n / N. 10. Abstract algebra December 17, / 62
35 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
36 Groups and subgroups Introduction Groups and subgroups are algebraic structures. They are the ones that allow solving equations like and that the equation x + x = a x = a 2 x x = a does not have a solution in R if a < 0. We ll see that defining a group amounts to define the elements belonging to the group as well as the operations that can be used with them. 10. Abstract algebra December 17, / 62
37 Groups Definition 5.1 (Group) Given a set S and a binary operation defined on S, the pair (S, ) is a group if G is closed under and G1. is associative in S G2. has a neutral element in S G3. has an inverse element in S Definition 5.2 (Abelian group) (S, ) is an abelian group if (S, ) is a group and is commutative. Definition 5.3 (Subgroup) Let (S, ) be a group. Let H be a subset of S, H S, and H be the induced operation in H. The pair (H, H ) is a subgroup of (S, ) if it verifies the conditions to be a group. 10. Abstract algebra December 17, / 62
38 Groups Consider S = {z C z = e iϕ ϕ R}. (U, ) is a group. Proof G1. is associative in S z 1 (z 2 z 3 ) = e iϕ1 (e iϕ2 e iϕ3 ) = e iϕ1 (e i(ϕ2+ϕ3) ) = e i(ϕ1+ϕ2+ϕ3) (z 1 z 2 )z 3 = (e iϕ1 e iϕ2 )e iϕ3 = (e iϕ1+ϕ2 )e iϕ3 = e i(ϕ1+ϕ2+ϕ3) 10. Abstract algebra December 17, / 62
39 Groups (continued) Proof G2. has a neutral element in S 1 = e i0 S z 1 = e iϕ e i0 = e i(ϕ+0) = e iϕ = z 1 z = e i0 e iϕ = e i(0+ϕ) = e iϕ = z G3. has an inverse element in S For each z = e iϕ, its inverse element with respect to is z 1 = e iϕ zz 1 = e iϕ e iϕ = e i(ϕ ϕ) = e i0 = 1 z 1 z = e iϕ e iϕ = e i( ϕ+ϕ) = e i0 = Abstract algebra December 17, / 62
40 Groups (N, +) is not a group because it has no neutral element. (N {0}, +) is not a group because it has no inverse element. (Z, +), (Q, +), (R, +), (C, +) and (R n, +) are abelian groups. (M m n, +) is an abelian group. (R, ) is not a group because 0 has no inverse (M n n (R)), ) is not a group because has no inverse Let S M n n (R) be the set of invertible matrices of size n n. (S, ) is a group (although not abelian). It is called the General Linear Group of degree n (GL(n, R)). 10. Abstract algebra December 17, / 62
41 Groups The existence of groups is what allows us to solve equations. For instance, consider the equation and its solution in the group (Z, +) 5 + x = x = 2 [Addition of the inverse of 5 with respect to + in both 5 + (5 + x) = [Addition is associative ] ( 5 + 5) + x = 3 [Definition of inverse] 0 + x = 3 [Definition of neutral element] x = Abstract algebra December 17, / 62
42 Groups Consider the equation 2x = 3 and its solution in the group (Q, ) 2x = 3 [Multiplication by the inverse of 2 in both sides] 1 2 (2x) = 1 3 [Multiplication is associative ] ( 1 2 2)x = 2 3 [Definition of inverse] 1x = 2 3 [Definition of neutral element] x = Abstract algebra December 17, / 62
43 Groups Theorem 5.1 (Cancellation laws) Given any group (S, ), a, b, c S it is verified Left cancellation: a b = a c b = c Right cancellation: b a = c a b = c Theorem 5.2 (Existance of a unique solution of linear equations) Given any group (S, ), a, b S the linear equations a x = b and y a = b always have a unique solution in S. Theorem 5.3 (Properties of the inverse) Given any group (S, ), a S its inverse is unique and a, b S (a b) 1 = (b 1 ) (a 1 ) 10. Abstract algebra December 17, / 62
44 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
45 Homomorphisms Consider the sets S = {a, b, c} and S = {A, B, C} with the operations : S S S and : S S S x y y = a y = b y = c x = a a b c x = b b c a x = c c a b and x y y = A y = B y = C x = A A B C x = B B C A x = C C A B We may construct a function that translates elements in S into elements in S with the same properties. We note that φ : S S φ(a) = A φ(b) = B φ(c) = C b c = a φ(b) φ(c) = φ(a) B C = A 10. Abstract algebra December 17, / 62
46 Homomorphisms Definition 6.1 (Group homomorphism) Given two groups (S, ) and (S, ), the function φ : S S is a group homomorphism iff a, b S φ(a b) = φ(a) φ(b) Definition 6.2 (Group isomorphism) Given two groups (S, ) and (S, ), the function φ : S S is a group isomorphism iff it is a group homomorphism and it is bijective. 10. Abstract algebra December 17, / 62
47 Homomorphisms Consider the two groups (R n, +) and (R m, +) and a matrix A M m n (R). The application is a group homomorphism because φ : R n R m φ(x) = Ax φ(u + v) = A(u + v) = Au + Av = φ(u) + φ(v) Consider the two groups (GL(n, R), ) and (R, ). The application is a group homomorphism because φ : GL(n, R) R φ(a) = det{a} φ(ab) = det{ab} = det{a} det{b} = φ(a) φ(b) 10. Abstract algebra December 17, / 62
48 Homomorphisms Theorem 6.1 Let φ : S S be a group homomogrphism between two groups. Then, φ(e) = e φ(a 1 ) = (φ(a)) 1 Definition 6.3 (Kernel of a group homomorphism) Let φ : S S be a group homomogrphism between two groups. Then, the kernel of φ is the set Ker{φ} = {x S φ(x) = e } Let φ(x) = Ax. Then, Ker{φ} = {x R n Ax = 0} = Nul{A} 10. Abstract algebra December 17, / 62
49 Isomorphisms Theorem 6.2 (Isomorphisms and cardinality) If two groups (S, ) and (S, ) are isomorph (i.e., there exists an isomorphism between the two groups), then S and S have the same cardinality. 10. Abstract algebra December 17, / 62
50 Isomorphisms Q and R cannot be isomorph because the cardinality of Q is ℵ 0 and the cardinality of R is ℵ 1. There are as many natural numbers as natural even numbers. In other words, the cardinality of N and 2N are the same. The reason is that the function φ(n) = 2n is an isomorphism between N and 2N. Consider the set R c = [0, c) R and the operation x + c y = (x + y) mod c. The pair (R c, + c ) is a group. Consider now the two particular cases (R 2π, + 2π ) and (R 1, + 1 ) and the mapping φ : R 2π R 1 φ(x) = x 2π φ is an isomorphism between (R 2π, + 2π ) and (R 1, + 1 ). In fact, all (R c, + c ) groups are isomorph to any other (R c, + c ) group. 10. Abstract algebra December 17, / 62
51 Isomorphisms Cardinality is a group property. The nice things about isomorphisms is that they preserve group properties. Theorem 6.3 If two groups (S, ) and (S, ) are isomorph, then If is commutative, so is. If there is an order relation in S, it can be translated into an order relation in S. If s S there exists a solution in S of the equation x x = s, then s S there exists a solution in S of the equation x x = s. If a, b S there exists a solution in S of the equation a x = b, then a, b S there exists a solution in S of the equation a x = b. The kernel of any isomorphism φ between (S, ) and (S, ) is Ker{φ} = {e} being e the neutral element of in S. 10. Abstract algebra December 17, / 62
52 Isomorphisms ((Z), +) is not isomorph to ((Q), +) because the equation x + x = s has a solution in Q for any s Q (that is x = s 2 ), but it does not have a solution in Z for any s Z (it only has a solution in Z if s is an even number). ((R), ) is not isomorph to ((C), ) because the equation x x = z has two solution in C for any z C (if z = re iθ, then x = ±re i θ 2 are the two solutions), but it does not have a solution in R for any z R (it only has a solution in R if z is a non-negative number). 10. Abstract algebra December 17, / 62
53 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
54 Algebraic structures Algebraic structures Algebraic structures are tools that help us to define operate on numbers and elements within a set, solve equations, etc. 10. Abstract algebra December 17, / 62
55 Algebraic structures Definition 7.1 (Ring) The tuple (S,, ) is a ring iff R1. (S, ) is an abelian group. R2. is associative. R3. is distributive with respect to, i.e., a, b, c S Left-distributive: a (b c) = (a b) (a c) Right-distributive: (a b) c = (a c) (b c) (Z, +, ), (Q, +, ), (R, +, ), (C, +, ) are rings. (M m n (R), +, ) is a ring. (R R, +, ) is a ring. 10. Abstract algebra December 17, / 62
56 Algebraic structures Theorem 7.1 (Properties of rings) Let (S,, ) be a ring and let e be the neutral element of in S. For any a S, let a be the inverse of a with respect to the operation. Then a, b S a e = e a = e. a b = a b = (a b) a b = a b Consider the ring (R, +, ). We are used to the properties a, b R a 0 = 0 a = 0. a ( b) = ( a) b = (a b) ( a) ( b) = a b But, as stated by the previous theorem, these are properties of all rings. 10. Abstract algebra December 17, / 62
57 Algebraic structures Definition 7.2 (Kinds of rings) A ring (S,, ) is commutative iff is commutative. unitary iff has a neutral element (referred as 1). divisive if it is unitary and a S {e}!a 1 S, a a 1 = a 1 a = 1 That is each element has a multiplicative inverse. (P, +, ) the set of polynomials with coefficients from a ring is a ring. 10. Abstract algebra December 17, / 62
58 Algebraic structures Definition 7.3 (Field (cuerpo)) A divisive, commutative ring is called a field. (Q, +, ), (R, +, ), and (C, +, ) are fields. (Z, +, ) is not a field because multiplication has not an inverse in Z. 10. Abstract algebra December 17, / 62
59 Algebraic structures Definition 7.4 (Vector space over a field) Consider a field (K,, ). A vector space over this field is a tuple (V, +, ) so that V is a set whose elements are called vectors, and + : V V V is a binary operation under which V is closed, : K V V is an operation between scalars in the field (K) and vectors in the vector space (V ) such that a, b K, u, v V V1. (V, +) is an abelian group. V2. (a u) V V3. a (b u) = (a b) u V4. (a b) u = a u + b u V5. a (u + v) = a u + a v V6. 1 u = u 10. Abstract algebra December 17, / 62
60 Algebraic structures s (R n, +, ) and (C n, +, ). (M m n (R), +, ): the set of matrices of a given size with coefficients in a field. (P, +, ): the set of polynomials with coefficients in a field. ({X V }, +, ): the set of all functions from an arbitrary set X onto an arbitrary vector space V. The set of all continuous functions is a vector space. The set of all linear maps between two vector spaces is also a vector space. The set of all infinite sequences of values from a field is also a vector space. 10. Abstract algebra December 17, / 62
61 Algebraic structures Definition 7.5 (Algebra) Consider a vector space (V, +, ) over a field (K,, ) and a binary operation : V V V. (V, +,, ) is an algebra iff a, b K, u, v, w V s A1. Left distributivity: (u + v) w = u w + v w A2. Right distributivity: u (v + w) = u v + u w A3. Compatibility with scalars: (a u) (b v) = (a b) (u v) Real numbers (R) are an algebra ( 1D ). Complex numbers (C) are an algebra ( 2D ). Quaternions are an algebra ( 4D ). 10. Abstract algebra December 17, / 62
62 Outline 10 Abstract algebra Sets Relations and functions Partitions and equivalence relationships Binary operations Groups and subgroups Homomorphisms and isomorphisms Algebraic structures 10. Abstract algebra December 17, / 62
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# Calculator search results
Formula
Factorize the expression
$x ^{ 4 } +3x ^{ 3 } -7x ^{ 2 } -27x-18$
$\left ( x - 3 \right ) \left ( x + 1 \right ) \left ( x + 2 \right ) \left ( x + 3 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 27 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 }$
Do factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$
Do factorization
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
Factorize to use the polynomial formula of sum and difference
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Solution search results
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Updated 2014-02-09 16:14:56 by dkf
MS: moved here the discussion/proposals from Additional math functions
KBK Before cutting-and-pasting these procedures into your code, you probably ought to go have a look at the 'math::combinatorics' module of tcllib. It offers, as of this writing, factorials and their logarithms, binomial coefficients, the Beta and Gamma functions (and the logarithm of the Gamma function), and several others. Incomplete gamma, and normal, chi-square, Poisson, hypergeometric, and F distributions are planned but not yet implemented.
AM (11 january 2007) Many of the statistical distributions mentioned above are available via the statistics module in Tcllib
Binomial coefficient:
``` proc over {n k} {
# binomial coefficient: e.g. [over 5 3] = (5*4*3)/(1*2*3) = 10
set k [expr {((\$n-\$k) < \$k) ? \$n-\$k : \$k}]
if {\$k==0} {return 1}
set num \$n
set den \$k
while {\$k>1} {
incr n -1
incr k -1
set num [expr {\$num*\$n}]
set den [expr {\$den*\$k}]
}
expr {\$num/\$den}
} ;#RS```
Note that combinatoric functions are notorious for rewarding algorithmic "trickiness". In this case, the multiple "expr" invocations will toggle to floating point when n is sufficiently large. This is both slow and inaccurate, in general, although certainly quite adequate for many common cases of interest.
Higher precision is available with algorithms which exploit common factors repeated in the numerator and denominator (and there will be many of them for most calculations). CL will make a point some day of returning here to offer an [over] written in terms of [primefactors] (see below). - RS: Here's a variant which uses mostly lists: collect factors for num and den, remove from num all that are in den, finally computes one integer expression:
``` proc over2 {n k} {
# binomial coefficient: e.g. [over 5 3] = (5*4*3)/(1*2*3) = 10
set k [expr {((\$n-\$k) < \$k) ? \$n-\$k : \$k}]
if {\$k==0} {return 1}
set num \$n
set den \$k
while {\$k>1} {
lappend num [incr n -1]
lappend den [incr k -1]
}
set t 1 ;# neutral element of multiplication
foreach i \$num {
if {[set pos [lsearch \$den \$i]]<0} {
lappend t \$i
} else {
set den [lreplace \$den \$pos \$pos]
}
}
expr [join \$t *]/([join \$den *])
}```
Recursive binomial calculations can be surprisingly well-conditioned (someone compare performance on these):
``` proc binom {m n} {
set diff [expr \$m - \$n]
if {\$diff < \$n} {
return [binom \$m \$diff]
}
switch \$n {
0 {
return 1
}
1 {
return \$m
}
default {
# This could use a bit of common-factor elimination.
return [expr [binom [expr \$m - 1] [expr \$n - 1]] * \$m / \$n]
}
}
} ```
MS: This looks like too much of a good thing, you do not need to recurse over both variables. Recursing over just m gives the following algorithm.
``` proc binom2 {m n} {
set n [expr {((\$m-\$n) > \$n) ? \$m-\$n : \$n}]
set k [expr {\$m - \$n}]
if {\$k < 0} {return 0}
switch \$k {
0 {return 1}
1 {return \$m}
default {
return [expr {(\$m*[binom [expr {\$m - 1}] \$n])/\$k}]
}
}
} ;#MS```
Both recursive algorithms are stable in the sense of not producing unnecessarily large intermediate results. If you are interested in performance, avoid recursion and [switch]. An iterative implementation of binom2 is:
``` proc binom3 {m n} {
set n [expr {((\$m-\$n) > \$n) ? \$m-\$n : \$n}]
if {\$n > \$m} {return 0}
if {\$n == \$m} {return 1}
set res 1
set d 0
while {\$n < \$m} {
set res [expr {(\$res*[incr n])/[incr d]}]
}
set res
} ;#MS```
Remark that this is very much the same as over, with the loop being traversed in reverse order and avoiding the accumulation in den and num which causes an earlier overflow.
---
MS: So, which is best? While trying to compute all coefficients in the expansion of (a+b)^20, both [over] and [over2] fail to produce correct results as some intermediate computations overflow. The timing of the recursive/iterative solutions gives:
``` % time {for {set i 0} {\$i <= 20} {incr i} {set x [binom 20 \$i]}}
51716 microseconds per iteration
% time {for {set i 0} {\$i <= 20} {incr i} {set x [binom2 20 \$i]}}
47071 microseconds per iteration
% time {for {set i 0} {\$i <= 20} {incr i} {set x [binom3 20 \$i]}}
3490 microseconds per iteration```
---
``` % time {for {set i 0} {\$i <= 20} {incr i} {set x [binom4 20 \$i]}}
36949 microseconds per iteration```
The best stability (but low speed) is obtained by using set operations on the lists of prime factors of numerator and denominator:
``` proc binom4 {m n} {
set n [expr {((\$m-\$n) > \$n) ? \$m-\$n : \$n}]
if {\$n > \$m} {return 0}
if {\$n == \$m} {return 1}
# Find the list of prime factors in the numerator and denominator
set num {}
set den {}
set d 0
while {\$n < \$m} {
set num [appendPrimeFactors [incr n] \$num]
set den [appendPrimeFactors [incr d] \$den]
}
# We now need to multiply all factors in \$num that are not in \$den
set num [lsort -integer \$num]
set den [lsort -integer \$den]
set res 1
set i 0 ;# position counter for num
foreach f \$den {
while {[set g [lindex \$num \$i]] != \$f} {
set res [expr {\$res * \$g}]
incr i
} pr {\$res * \$g}]
incr i
}
foreach g [lrange \$num \$i end] {
set res [expr {\$res * \$g}]
}
set res
} ;#MS
proc appendPrimeFactors {n res} {
# a number x is prime if [llength [primefactors \$x]]==1
set f 2
while {\$f<=\$n} {
while {\$n%\$f==0} {
set n [expr {\$n/\$f}]
lappend res \$f
}
set f [expr {\$f+2-(\$f==2)}]
}
set res
} ;#RS```
There should however exist very few cases where binom3 overflows and binom4 doesn't ...
But, if you're worried, read on:
Log gamma -- or, factorials the hard way.
There was a little discussion on the chat about how to do factorials inside [expr]. Larry Virden pointed the questioner to several pages with classical recursive or iterative approaches. Kevin Kenny saw a challenge in the discussion, and decided to do factorials the hard way -- by computing the gamma function:
This is actually useful for avoiding overflows in computing binomial coefficients. It can also be used as an obscure way of computing pi.
``` # Returns log(gamma(x)), where x >= 0
#
# +inf
# _
# | x-1 -t
# gamma(x)= _| t e dt
#
# 0
proc lgamma { xx } {
set x [expr { \$xx - 1.0 }]
set tmp [expr { \$x + 5.5 }]
set tmp [ expr { ( \$x + 0.5 ) * log( \$tmp ) - \$tmp }]
set ser 1.0
foreach cof {
76.18009173 -86.50532033 24.01409822
-1.231739516 .00120858003 -5.36382e-6
} {
set x [expr { \$x + 1.0 }]
set ser [expr { \$ser + \$cof / \$x }]
}
return [expr { \$tmp + log( 2.50662827465 * \$ser ) }]
}
# Return x! given x.
proc fac { x } {
return [expr { exp( [lgamma [expr { \$x + 1 }]] ) }]
}
# Demonstrate by printing a table of factorials
for { set i 0 } { \$i <= 10 } { incr i } {
puts "\$i! == [expr { round( [fac \$i] ) }]"
}
# Compute binomial coefficients.
proc choose { n k } {
set r [expr { [lgamma [expr { \$n + 1 }]]
- [lgamma [expr { \$k + 1 }]]
- [lgamma [expr { \$n - \$k + 1 }]] }]
return [expr { exp( \$r ) }]
}
# Demonstrate binomial coefficients by printing Pascal's triangle.
puts "Pascal's triangle:"
for { set n 0 } { \$n < 15 } { incr n } {
set line {}
for { set k 0 } { \$k <= \$n } { incr k } {
set r [expr { round( [choose \$n \$k] ) }]
append line [format %5d \$r]
}
puts \$line
}
# Compute pi, the hard way!
set sqrtpi [expr { exp( [lgamma 0.5] ) }]
puts "pi is approximately [expr { \$sqrtpi * \$sqrtpi }]"```
For those that don't like ASCII art, the following Tcl script puts the formula on a canvas:
``` grid [canvas .c -width 300 -height 200]
font create big -family times -size 24
font create bigitalic -family times -size 24 -slant italic
font create small -family times -size 18
font create smallitalic -family times -size 18 -slant italic
.c create text 50 100 -anchor w -text "\u0393(" -font big -tags t1
foreach { - - x - } [.c bbox t1] break
.c create text \$x 100 -anchor w -text "x" -font bigitalic -tags t2
foreach { - - x - } [.c bbox t2] break
.c create text \$x 100 -anchor w -text ") = " -font big -tags t3
foreach { - - x - } [.c bbox t3] break
.c create text \$x 100 -anchor w -text "\u222b" -font big -tags t4
foreach { x0 y0 x1 y1 } [.c bbox t4] break
set x2 [expr { ( \$x0 + \$x1 ) / 2 }]
.c create text \$x2 \$y0 -anchor s -text "\u221e" -font small
.c create text \$x2 \$y1 -anchor n -text "0" -font small
.c create text \$x1 100 -anchor w -text " t" -font bigitalic -tags t5
foreach { - y x - } [.c bbox t5] break
.c create text \$x \$y -anchor w -text "x" -font smallitalic -tags t6
foreach { - - x - } [.c bbox t6] break
.c create text \$x \$y -anchor w -text "-1" -font small -tags t7
foreach { - - x - } [.c bbox t7] break
.c create text \$x 100 -anchor w -text "e" -font bigitalic -tags t8
foreach { - y x - } [.c bbox t8] break
.c create text \$x \$y -anchor w -text "-" -font small -tags t9
foreach { - - x - } [.c bbox t9] break
.c create text \$x \$y -anchor w -text "t" -font smallitalic -tags t10
foreach { - - x - } [.c bbox t10] break
.c create text \$x 100 -text " dt" -font bigitalic```
Arjen Markus Of course the gamma function is only one of several hundreds of ways to compute pi, mostly in very mysterious ways. The best (i.e. most unexpected and obscure) one that I know is:
• Get two random integers
• Decide if they have some factor in common or not (i.e. gcd(n,m) > 1)
• Repeat this over and over again: the chance that they have a factor in common is related to pi: 6/pi^2 (if I am not mistaken).
But, more seriously: the above script might lead one to contemplate using the canvas for drawing all manner of mathematical formulae. Has anybody ever done something like that? RS: How about Fun with functions? Arjen Markus I will need to run this one, but I meant more visualising integral signs, partial differential equations and so on, in other words a formula viewer :-)
KBK observes that Arjen answered his own question about rendering mathematical formulae.
DKF: I wanted to get binomial coefficients (for a primality test) but using the math::choose routine described in tcllib's math::combinatorics page didn't work; it had the right right scale, but I needed exact [bigint] math which the tcllib code doesn't use. So I wrote this, using lmap to simplify things:
```proc coeffs {p {signs 1}} {
set clist 1
for {set i 0} {\$i < \$p} {incr i} {
set clist [lmap x [list 0 {*}\$clist] y [list {*}\$clist 0] {
expr {\$x + \$y}
}]
}
if {\$signs} {
set s -1
set clist [lmap c \$clist {expr {[set s [expr {-\$s}]] * \$c}}]
}
return \$clist
}```
(The signs argument is because I was really interested in the polynomial expansion of (x-1)^p.)
I leave it as an exercise to the reader to write a version that caches previous results for lower values so that recalculation is minimised. However, testing in 8.6 indicates that this is only half the speed of binom3 above even without such optimisations.
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# If given a pH of 2.57, how do you find the pKa?
You can find $\left[{H}_{3} {O}^{+}\right]$, but you cannot find $p {K}_{a}$ with the given information.
If $p H$ $=$ $2.57$, then (by definition), $\left[{H}_{3} {O}^{+}\right]$ $=$ ${10}^{- 2.57}$ $m o l \cdot {L}^{-} 1$. We have no handle on the concentration of the acid that gave rise to this concentration of hydronium ion.
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# IMO Level 1- Mathematics Olympiad (SOF) Class 7: Questions 346 - 357 of 1022
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## Question number: 346
Edit
MCQ▾
### Question
What is the value of at ?
### Choices
Choice (4)Response
a.
Discriminant
b.
c
c.
a
d.
0
## Question number: 347
Edit
MCQ▾
### Question
If then a is ________.
### Choices
Choice (4)Response
a.
2
b.
1
c.
0
d.
- 1
## Question number: 348
Edit
MCQ▾
### Question
The value of x, for which is ________.
### Choices
Choice (4)Response
a.
b.
3
c.
d.
1
## Question number: 349
Edit
MCQ▾
### Question
If , then the value of .
### Choices
Choice (4)Response
a.
36
b.
30
c.
33
d.
39
## Question number: 350
Edit
MCQ▾
### Question
A contractor employed 120 men to build a house in 45 days. After 15 days, he was joined by 30 more men. In how many days will the remaining work be finished?
### Choices
Choice (4)Response
a.
20 days
b.
18 days
c.
28 days
d.
24 days
## Question number: 351
Edit
MCQ▾
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 352
Edit
MCQ▾
### Question
Power 5 of is ________.
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 353
Edit
MCQ▾
### Question
The product of is ________.
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 354
Edit
MCQ▾
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 355
Edit
MCQ▾
### Question
The value of is ________.
### Choices
Choice (4)Response
a.
b.
c.
d.
None of the above
## Question number: 356
Edit
MCQ▾
### Choices
Choice (4)Response
a.
b.
c.
d.
## Question number: 357
Edit
MCQ▾
### Question
A man purchased a bag of rice containing 70 kg for Rs. 175. He sold it at the rate of Rs. 2.75 per kg. Then the profit or loss in % is ________.
### Choices
Choice (4)Response
a.
12 % gain
b.
10 % loss
c.
10 % gain
d.
12 % loss
Developed by:
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If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=? (A) -10 (B)
post reply Question banks Downloads My Bookmarks Reviews Important topics
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If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=? (A) -10 (B) [#permalink]
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31 Jul 2008, 10:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
. If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=?
(A) -10
(B) -8
(C) -6
(D) 0
(E) 4
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Joined: 12 Jul 2008
Posts: 514
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Re: DIFF type PROB... [#permalink]
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31 Jul 2008, 10:59
arjtryarjtry wrote:
. If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=?
(A) -10
(B) -8
(C) -6
(D) 0
(E) 4
B
10 = a*5^5 + b*5^3 +1
a*5^5 + b*5^3 = 9
g(-5) = a(-5)^5 + b(-5)^3 + 1
= -a*5^5 - b*5^3 + 1
= -(a*5^5 + b*5^3) + 1
= -9 + 1
= -8
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Current Student
Joined: 11 May 2008
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Re: DIFF type PROB... [#permalink]
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31 Jul 2008, 11:00
...ZOINKK..NICE
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Re: DIFF type PROB... [#permalink]
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31 Jul 2008, 11:05
arjtryarjtry wrote:
. If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=?
(A) -10
(B) -8
(C) -6
(D) 0
(E) 4
clearly -8 is answer :
g(x)+g(-x)= 2
=> g(-5)= 2- g(5)=2-10=-8
_________________
cheers
Its Now Or Never
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Re: DIFF type PROB... [#permalink] 31 Jul 2008, 11:05
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If g(x)=ax^5+bx^3+1, and g(5)=10, then g(-5)=? (A) -10 (B)
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Question
Thu January 24, 2013
# real image of an object is formed on the other side of convex lens on a screen 30 cm from the lens. when the lens is shifted 5cm towards the screen,the screen needs to be shifted by 5cm towards the lens to given another image of the same object.what is the focal length of lens?
Fri February 01, 2013
this problem can be solved using the lens formula:
1/v - 1/u = 1/f
now first case
v = 30cm
thus
1/u -1/30 =1/f -1
second case
v' = 30-5-5 = 20cm
u' = u+5
1/(u+5) -1/20 =1/f -2
from equation 1 and 2 we get that,
1/(u+5) -1/20 = 1/u -1/30
it can be solved to find u and then f
Related Questions
Wed September 20, 2017
# the wall of the room is covered with a perfect plane mirror and two movie films are made one recording the movement of a man and the other of his mirror image while viewing the film later can an outsider tell which is which ?
Fri September 08, 2017
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http://www.mathworks.com/help/matlab/ref/factorial.html?s_tid=gn_loc_drop&requestedDomain=www.mathworks.com&nocookie=true
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# Documentation
### This is machine translation
Translated by
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Note: This page has been translated by MathWorks. Please click here
To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.
# factorial
Factorial of input
## Syntax
``f = factorial(n)``
## Description
example
````f = factorial(n)` returns the product of all positive integers less than or equal to `n`, where `n` is a nonnegative integer value. If `n` is an array, then `f` contains the factorial of each value of `n`. The data type and size of `f` is the same as that of `n`.The factorial of `n` is commonly written in math notation using the exclamation point character as n!. Note that `n!` is not a valid MATLAB® syntax for calculating the factorial of `n`.```
## Examples
collapse all
`f = factorial(10)`
```f = 3628800 ```
```format long f = factorial(22)```
```f = 1.124000727777608e+21 ```
In this case, `f` is accurate up to 15 digits, `1.12400072777760e+21`, because double-precision numbers are only accurate up to 15 digits.
Reset the output format to the default.
`format`
```n = [0 1 2; 3 4 5]; f = factorial(n)```
```f = 1 1 2 6 24 120 ```
```n = uint64([5 10 15 20]); f = factorial(n)```
```f = 1x4 uint64 row vector Columns 1 through 3 120 3628800 1307674368000 Column 4 2432902008176640000 ```
## Input Arguments
collapse all
Input values, specified as a scalar, vector, or array of real, nonnegative integers.
Example: `5`
Example: ```[0 1 2 3 4]```
Example: ```int16([10 15 20])```
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
## Tips
Limitations
• For double-precision inputs, the result is exact when `n` is less than or equal to `21`. Larger values of `n` produce a result that has the correct order of magnitude and is accurate for the first 15 digits. This is because double-precision numbers are only accurate up to 15 digits.
• For single-precision inputs, the result is exact when `n` is less than or equal to `13`. Larger values of `n` produce a result that has the correct order of magnitude and is accurate for the first 8 digits. This is because single-precision numbers are only accurate up to 8 digits.
Saturation
• The table below describes the saturation behavior of each data type when used with the `factorial` function. The values in the last column indicate the saturation point; that is, the first positive integer whose actual factorial is larger than the maximum representable value in the middle column. For `single` and `double`, all values larger than the maximum value are returned as `Inf`. For the integer data types, the saturation value is equal to the maximum value in the middle column.
Data typeMaximum ValueFactorial Saturation Threshold
`double``realmax``factorial(171)`
`single``realmax('single')``factorial(single(35))`
`uint64`264-1`factorial(uint64(21))`
`int64`263-1`factorial(int64(21))`
`uint32`232-1`factorial(uint32(13))`
`int32`231-1`factorial(int32(13))`
`uint16`216-1`factorial(uint16(9))`
`int16`215-1`factorial(int16(8))`
`uint8`28-1`factorial(uint8(6))`
`int8`27-1`factorial(int8(6))`
## See Also
#### Introduced before R2006a
Was this topic helpful?
#### The Manager's Guide to Solving the Big Data Conundrum
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# Raid 5
Tyler Strickland tyler at tylers.org
Fri Dec 2 14:07:34 MST 2005
```On 12/02/2005 01:57 PM, Nicholas Leippe wrote:
> If you consider as I do, that the first drive failure is a given,
> then the odds of failure is magnitudes larger--back up to the same order
> of magnitude as a single drive failing.
>
> This makes the relative differences in failure odds become significant,
> especially when you consider how the odds increase with additional capacity.
What you say is true about the probability increasing with a failed
drive. The probability of two drives failing at once is very small, but
the probability of a second drive failure after the first has occurred
is much larger. RAID 10 can, in most cases, survive multiple failures,
while a RAID 5 cannot. So our lesson for today, kids, is "Always fix
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https://www.brightsurf.com/news/article/081314351062/how-are-queueing-systems-analyzed.html
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# How are queueing systems analyzed?
August 13, 2014
"In the analysis of a physical system, the first step is to derive a mathematical model for the system. However, it is important to note that mathematical models, in general, may assume many different forms," said Emeritus Professor Chan Wah Chun, "Depending on the particular system, one mathematical model may be more suitable than others."
Professor Chan's latest book "An Elementary Introduction to Queueing Systems" aims to highlight the fundamental concepts of queueing systems. He draws from his three decades of experience in teaching and research, to illustrate the fundamental principles and rationale behind complex mathematical concepts of the subject.
Commencing with mathematical modeling of the arrival process (input) of customers to the system, Professor Chan showed that the arrival process can be described mathematically either by the number of arrival customers in a fixed time interval, or by the inter-arrival time between two consecutive arrivals.
In the analysis of queueing systems, he emphasised the importance of exponential service time of customers, where the analysis can be simplified by using the birth and death process as a model, and queueing systems can be analysed by choosing the proper arrival rate and service rate. This approach facilitates the analysis of many queueing systems.
Written in a fairly readable format, students wishing to master the core principles of queueing theory will be able to grasp them quickly.
-end-
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# Bytes to Exbibytes - 256 Byte to EiB Conversion
expand_more
Input Byte (B) - and press Enter.
Byte
Byte --> Exbibyte (binary)
label_important RESULT sentiment_satisfied_alt
256 Byte =0.0000000000000002220446049250313080847263 EiB
( Equal to 2.220446049250313080847263E-16 EiB )
content_copy
Calculated as → 256 ÷ 10246 smart_display Show Stepsexpand_more
## Byte (B) to Exbibyte (EiB) Conversion - Formula & Steps
The Byte to EiB Calculator Tool provides a convenient solution for effortlessly converting data units from Byte (B) to Exbibyte (EiB). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Byte) and target (Exbibyte) data units.
Source Data Unit Target Data Unit
Equal to 8 bits
(Basic Unit)
Equal to 1024^6 bytes
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Byte to Exbibyte in a simplified manner.
÷ 8
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
÷ 1024
x 8
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
x 1024
Based on the provided diagram and steps outlined earlier, the formula for converting the Byte (B) to Exbibyte (EiB) can be expressed as follows:
diamond CONVERSION FORMULA EiB = Byte ÷ 10246
Now, let's apply the aforementioned formula and explore the manual conversion process from Byte (B) to Exbibyte (EiB). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Exbibytes = Bytes ÷ 10246
STEP 1
Exbibytes = Bytes ÷ (1024x1024x1024x1024x1024x1024)
STEP 2
Exbibytes = Bytes ÷ 1152921504606846976
STEP 3
Exbibytes = Bytes x (1 ÷ 1152921504606846976)
STEP 4
Exbibytes = Bytes x 0.0000000000000000008673617379884035472059
By applying the previously mentioned formula and steps, the conversion from 256 Byte (B) to Exbibyte (EiB) can be processed as outlined below.
1. = 256 ÷ 10246
2. = 256 ÷ (1024x1024x1024x1024x1024x1024)
3. = 256 ÷ 1152921504606846976
4. = 256 x (1 ÷ 1152921504606846976)
5. = 256 x 0.0000000000000000008673617379884035472059
6. = 0.0000000000000002220446049250313080847263
7. i.e. 256 Byte is equal to 0.0000000000000002220446049250313080847263 EiB.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Bytes to Exbibytes using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Byte ?
A Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.
arrow_downward
#### What is Exbibyte ?
An Exbibyte (EiB) is a binary unit of digital information that is equal to 1,152,921,504,606,846,976 bytes (or 9,223,372,036,854,775,808 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'exbi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'exabyte' (EB). It is widely used in the field of computing as it more accurately represents the storage size of high end servers and data storage arrays.
## Excel Formula to convert from Byte (B) to Exbibyte (EiB)
Apply the formula as shown below to convert from 256 Byte (B) to Exbibyte (EiB).
A B C
1 Byte (B) Exbibyte (EiB)
2 256 =A2 * 0.0000000000000000008673617379884035472059
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Byte (B) to Exbibyte (EiB) Conversion
You can use below code to convert any value in Byte (B) to Byte (B) in Python.
bytes = int(input("Enter Bytes: "))
exbibytes = bytes / (1024*1024*1024*1024*1024*1024)
print("{} Bytes = {} Exbibytes".format(bytes,exbibytes))
The first line of code will prompt the user to enter the Byte (B) as an input. The value of Exbibyte (EiB) is calculated on the next line, and the code in third line will display the result.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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5 GmatPrep questions
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5 GmatPrep questions [#permalink] 13 Aug 2009, 04:45
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Question Stats:
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Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 05:10
Try uploading them as jpegs instead of docs.....that way members will be able to view the question in their explorer windows itself....
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 05:41
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Q 1 .See the attached figure.
Let the sides of the triangle be a, b and c
Clearly using simple trigo calculations we can see that a=b= 5
and c= 5\sqrt{2}
Now lets draw perpendicular on C with the height h.
This perpendicular will bisect the side C,
So let X= 1/2 x C = 1/2 x 5\sqrt{2}
h^2 + x^2 = a ^2
H^2 = 25- 25/2 = 25/2 = 5/ \sqrt{2}
Area = base x height = 1/2 C x H = 1/2 x 5\sqrt{2} x 5/ \sqrt{2}
Area = 25/2 = 12.5
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 05:47
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
Q 2 See the attached figure.
Distance traveled = D = 2 pi R = 2pi = 2 x 3.14
where r = radius of the lake = 1 mile
Distance = speed x time
Speed = 3 miles per hour
time = Distance / speed =(2 x 3.14) / 3
Clearly t >2 and less than 2.5
Hence Option C is the answer.
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 05:49
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
2^x - 2^(x-2) = 3 ( 2^13)
2^(x-2) [ 4-1]= 3 ( 2^13)
Hence x-2= 13
OR X= 15
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 05:54
SOL for Q10)
Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known.
The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by,
1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]
Applying we get,
A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)]
= 1/2 [(-4) - (21)]
= 1/2 [-25] .......... ignore the '-' sign
= 12.5
ANS: A
Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.
Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.
Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles.
A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2)
= 12.5
ANS: A
Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides.
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 06:02
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
The way u can represent the scores = 48, 48 , 70, 70, 70 , 80 , 80, 80, 80, 84 , 84 , 84, 84, 84, 84 , 84, 96, 96, 96,96
So clearly the Median = 84
See the figure for this question .
I don't know how to work on the 5th question may be some body can shed some light on this one.
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 06:04
SOL for Q24
Its given that the student worked for 20 days and earned:
48$- on 2 days 70$ - on 3 days
80$- on 4 days 84$ - on 7 days
96\$ - on 4 days
For Median of even no of amounts, we need to arrange the amounts in ascending order and take the avg of the amounts placed in 10th and 11th positions => (84 + 84)/2 = 84
ANS: B
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 06:05
samrus98 wrote:
SOL for Q10)
Method 1: There is a coordinate geometry formula that uses matrices to give you the area of a triangle whose coordinates are known.
The final formula is: If the coordinates of a triangle are A(x1,y1), B(x2,y2) & C(x3,y3), then the area of the triangle is given by,
1/2* [(x1-x2)(y2-y3) - (y1-y2)(x2-x3)]
Applying we get,
A(PQR) = 1/2* [(4-0)(3-4) - (0-3)(0-7)]
= 1/2 [(-4) - (21)]
= 1/2 [-25] .......... ignore the '-' sign
= 12.5
ANS: A
Note: This method is also useful to decide whether the given 3 points form a triangle or they are collinear. If the area turns out to be zero, then the points are collinear.
Method 2: Drop perpendiculars from point R to the X and Y axes at points S(7,0) and T(0,4) respectively. Thus OSRT is a rectangle.
Now A(PQR) = A(OSRT) - A(OQP) - A(QRT) - A(PRS) where all the three triangles are right triangles.
A(PQR) = 7*4 - (3*4/2) - (7*1/2) - (3*4/2)
= 12.5
ANS: A
Method 3: Find out the lengths of the three sides using distance formula and then use the formula that calculates the area of a triangle from its sides.
Hey Sameer,
I was actually wondering if there is a faster way of solving this one and u just came up with that ..
Cheers
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Re: 5 GmatPrep questions [#permalink] 13 Aug 2009, 06:13
SOL for Q32)
Mean = 8.1
SD = 0.3
Now we need to calculate the acceptable range for the observations measuring 1.5 SDs on either side of the mean => 8.1 + (0.3)*1.5 => 8.1 + 0.45
=> 7.65 to 8.55
Except for 7.51 all the entries are within 1.5 SDs of the mean.
ANS: E
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Re: 5 GmatPrep questions [#permalink] 27 Aug 2009, 15:27
nitishmahajan wrote:
Ibodullo wrote:
Hi everyone. Can anyone help me to solve these questions, please? Sorry, I don't know how to upload questions so everyone can see them. Thanks in advance!
2^x - 2^(x-2) = 3 ( 2^13)
2^(x-2) [ 4-1]= 3 ( 2^13)
Hence x-2= 13
OR X= 15
how did you get what's in red?
thanks
Re: 5 GmatPrep questions [#permalink] 27 Aug 2009, 15:27
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##### Actions
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2020 Mathematics Subject Classification: Primary: 51M04 [MSN][ZBL]
The problem of constructing a square of equal area as the given circle; one of the classical Ancient problems on constructions with a ruler and compass. The side of a square equal in area to a circle of radius $r$ has length $r\sqrt\pi$. Thus the problem of the quadrature of the circle reduces to the following: To construct a line of length $\sqrt\pi$. Such a construction cannot be realized with a ruler and compass since $\pi$ is a transcendental number, as was proved in 1882 by F. Lindemann. However, the problem of the quadrature of a circle is solvable if one extends the means of construction, for example, by using certain transcendental curves, called quadratrices (cf. Quadratrix).
#### References
[1] Yu.I. Manin, "Ueber die Lösbarkeit von Konstruktionsaufgaben mit Zirkel und Lineal" , Enzyklopaedie der Elementarmathematik , 4. Geometrie , Deutsch. Verlag Wissenschaft. (1969) pp. 205–230 (Translated from Russian)
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# What Is Child Support Based On In Washington State?
The child support schedule is based on Washington’s state child support guidelines. You can determine your child support obligation based on the parents’ combined income, the number of children, and the children’s ages. In any event, a judge must approve the final child support amount.
How do you calculate child support in Washington State?
• The Basic Concepts. Good news: Washington has simplified child support calculations by adopting a standardized child support formula.
• Extending the Economic Table. Washington’s child support formula only extends to a combined monthly income of\$12,000 for the parties.
• Deviations.
• Extraordinary Expenses.
• Post-Secondary Support.
## What factors determine child support amount?
How is Child Support Determined?
• Parent’s gross income.
• Amount of time the child spends with each parent.
• Cost of child care.
• Any available tax deductions, such as mortgage interest.
• Each parent’s mandatory dues, such as pension or health insurance.
## How much is average child support in Washington state?
Washington State uses a child support formula to determine the base monthly child support amount. The factors include the number of children, their ages, and the incomes of the parents. Depending on the variables plugged into the formula, the base child support payment will be anywhere from \$200 up to \$3,500 per month.
## What are child support payments based on?
Child support – how much is paid by each parent for the needs of the child – is calculated by looking at each parent’s income, each parent’s expenses and the amount of care each parent provides. First each parent’s “child support income” is considered.
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## What is the average child support payment?
According to the Census Bureau Reports, the average monthly child support payment is \$430.
## Does child support factor in bills?
Generally, child support payments are for the ordinary expenses of food, shelter, clothing, education and medication needs for the children only. Generally your ability to pay does not include calculations of bills and debts such as car payments, credit cards, etc.
## How do they calculate child support in Washington state?
To calculate child support, use the combined net income of both parents (in other words, add your net income to the other parent’s net income) to determine the total amount of child support due.
## Is child support based on gross or net income?
Child support is determined by a formula that is based on an individual’s net income rather than an individual’s gross income. The amount of child support changes as the circumstances evolve for the children and both parents. Individuals must petition the court to change the amount of child support.
## Does child support go up if you make more money?
If you are requesting increased payments because your ex-spouse is earning more, the court will recalculate the child support amount using their new income. If the new amount is at least 10 percent higher than the previous one, the court will update the support order accordingly so you get higher monthly payments.
## What percentage is child support?
Texas child support laws provide the following Guideline calculations: one child= 20% of Net Monthly Income (discussed further below); two children = 25% of Net Monthly Income; three children = 30% of Net Monthly Income; four children = 35% of Net Monthly Income; five children = 40% of Net Monthly Income; and six
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## How does the child support system work?
To calculate child support, the final step is to multiply the Costs of the Children value by the difference between your Cost % and Income %. You receive child support if your Care % is > 35% and your Cost % > your Income %. You pay child support when your Care % < 65% and your Income % > your Cost %.
## Does father have to pay child support?
Regardless of marital status, if you are the father of a child, then you are obligated to pay child support. If you are questioning if the child is yours, then a paternity test may be in order.
## What’s the most child support can take?
Because child support is so important, the law sets a very high limit on the amount that can be withheld from your paycheck for this purpose. If you are not currently supporting another child or spouse who are not the subject of the order, up to 60% of your wages can be garnished.
## Does child support go down if the father has another baby?
When another child is born to that parent, they have now become responsible for the support of two children. Thus, the court is likely to divide the amount of overall support so that each of the children receives an equal percentage for their care.
## What state has the highest child support rate?
Why child support varies so much Massachusetts is first, and Nevada second. According to the study, the Northeast region ranks higher, while Rocky Mountain states rate the lowest.
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CC-MAIN-2022-27
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latest
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en
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https://myaptitude.in/jee/physics/a-mass-m-is-supported-by-a-massless-string-wound-around-a-uniform
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A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
1. g
2. 5g/6
3. g/2
4. 2g/3
For the mass m,
mg - T = ma
For the cylinder,
TR = mR2 a/R
T = ma
mg = 2ma
a = g/2
The correct option is C.
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CC-MAIN-2020-10
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latest
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en
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https://ai-scholar.tech/en/articles/data-augmentation/ensemble
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crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00777.warc.gz
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Are There Pitfalls To Using Ensembles And Data Augmentation Together?
3 main points
✔️ Discovered that calibration may be degraded when combining Ensemble Data Augmentation
✔️ Investigated the above issues and identified the reasons for poor calibration performance
✔️ Proposed a new data augmentation method "CAMixup" to avoid the above problems
written by Yeming WenGhassen JerfelRafael MullerMichael W. DusenberryJasper SnoekBalaji LakshminarayananDustin Tran
(Submitted on 19 Oct 2020)
Comments: Accepted to ICLR 2021.
Subjects: Machine Learning (cs.LG); Machine Learning (stat.ML)
code：
First of all
Ensemble methods, which use the predictive average of multiple models, and Data Augmentation, which increases the amount of data used for training, are often used to improve the calibration and robustness of models.
However, in the paper presented in this article, it is shown that the combination of these two methods can have a negative impact on the calibration of the model. In addition, the phenomenon of such calibration degradation was investigated and a method to avoid it, CAMixup, was proposed.
Preliminary preparation (calibration, ensemble, data enhancement)
Before explaining the negative effects of combining ensemble data augmentation, some preliminary knowledge is introduced.
Calibration
The calibration error is useful for evaluating the reliability of the model predictions. In the paper presented in this article, we use ECE, which is explained below, as a measure for evaluating the reliability of the model.
ECE(Expected Calibration Error)
The class prediction of a classifier and the confidence (indicating the predicted probability of the model) are denoted by $(\hat{Y},\hat{P})$. The ECE is then an approximation of the difference between the confidence and the expectation of accuracy $E_{\hat{P}}[|P(\hat{Y}=Y|]\hat{P}=p)-p|]$).
This is done by binning the [0,1] predictions into $M$ equal intervals (a quantization process that replaces the values within a certain interval (bin) with a specific value, such as the center value, as is done in a histogram) and then finding a weighted average of the accuracy/confidence difference for each bin.
Let $B_m$ be the set of $m$-th bins whose predicted confidence falls into the interval $(\frac{m-1}{M},\frac{m}{M}]$, then the accuracy and confidence of bin $B_m$ can be expressed by the following equation
$Acc(B_m)=\frac{1}{|B_m|}\sum_{x_i \in B_m} I(\hat{y_i}=y_i)$
$Conf(B_m)=\frac{1}{|B_m|}\sum_{x_i \in B_m} \hat{p_i}$
$hat{y_i},y_i$ denote the predicted and true labels, respectively, and $\hat{p_i}$ denotes the confidence level of $x_i$. Given $n$ examples, the ECE is $\sum^M_{m=1}\frac{|B_m|}{n}|Acc(B_m)-Conf(B_m)|$.
Ensemble method
Ensemble methods are methods that aggregate predictions from multiple models. In our experiments, we focus on investigating the interaction of three ensemble methods, BatchEnsemble, MC-Dropout, and Deep Ensembles, with data augmentation methods.
Data Augmentation Methodology
Data Augmentation is a method to improve the generalization performance by augmenting the input dataset with various transformations (e.g., image clipping). In our experiments, we examine the following two methods.
Mixup
Given an example $(x_i,y_i)$, Mixup is represented by the following equation.
$\tilde{x}_i=\lambda x_i+(1-\lambda)x_j$
$\tilde{y}_i=\lambda y_i+(1-\lambda)y_j$
where $x_j$ is a sample from the TRAIN set (obtained from a mini-batch) and $\lambda \in [0,1]$ is sampled from the beta distribution $\beta(a,a)$ ($a$ is a hyperparameter).
AugMix
Let $O$ be a set of data augmentation operations and $k$ be the number of AugMix iterations. In this case, the augmentation operations $op_1,... ,op_k$ and their weights $w_1,... ,w_k$(Dirichlet distribution (a,... ,a)), the augmentation by augmix is expressed by the following equation.
$\tilde{x}_{augmix}=mx_{orig}+(1-m)x_{aug}$
$x_aug=\sum^k_{i=1}w_iop_i(x_{orig})$
Experiment
In the following experiments, we investigate the calibration of ensembles combined with data augmentation. To begin, the results of applying Mixup to an ensemble on CIFAR-10/100 are shown below.
These results show the average of the results run for the five random seeds. Red is for ensemble only, blue is for Mixup + ensemble, and orange is for neither. In figures (a) and (b), we can see that the combination of Mixup and Ensemble improves the test performance (decreases Error).
On the other hand, Figs. (c) and (d) show that the calibration is worse (ECE is increased) when Mixup is combined with Ensemble.
Why do Mixup ensembles worsen calibration?
We investigate in more detail the phenomenon of calibration degradation when ensembles are combined with data augmentation. The following figure shows the difference between the average precision and the average confidence calculated for different confidence intervals when BatchEnsemble and Mixup are combined.
If the value of the difference between accuracy and confidence (vertical axis) is positive, it means that the confidence level is low for accuracy (confidence is underestimated), and if it is negative, it means that the confidence level is high (confidence is overestimated).
This figure shows that the accuracy-reliability difference becomes larger and approaches zero for the BatchEnsemble-only and Mixup-only cases compared to the Single network case.
In the case of Misup+BatchEnsemble, the overall accuracy-reliability difference is biased in the positive direction, indicating that the reliability is underestimated relative to the accuracy. In other words, although the data augmentation and ensemble methods have the effect of preventing the overestimation of the confidence level, the simultaneous use of both methods rather underestimates the confidence level, which seems to be the cause of the calibration deterioration.
As a further visualization example, the confidence (softmax probability) of training a three-layer MLP on a simple dataset consisting of five clusters is as follows
In the Mixup/no ensemble case (c), the overall probability is predicted to be high (yellow). This is mitigated by the introduction of Mixup, and by using the ensemble at the same time, we can see that the overall confidence is predicted to be much lower (green).
In addition, label smoothing is one of the most effective methods to suppress overestimation of the confidence level, and the same phenomenon occurs when this label smoothing is used together with an ensemble. The same phenomenon occurs when label smoothing is used together with the ensemble. This is illustrated in the following figure, where the stronger the label smoothing is applied, the larger the increase in ECE.
Confidence Adjusted Mixup Ensembles (CAMIXUP)
In the paper, we propose CAMixup as a method to prevent such calibration degradation due to the underestimation of the confidence level. the idea underlying CAMixup is that in a classification task, the difficulty of prediction can vary from class to class. In this case, it is desirable to increase the confidence level in classes where prediction is easy and to prevent the confidence level from increasing too much in classes where prediction is difficult.
CAMixup is based on this idea, and varies the degree of Mixup application for each class, especially for classes where model confidence is likely to be overestimated (difficult to predict). This is illustrated in the following figure.
As shown in the left figure, if the Accuracy-Confidence difference is positive, Mixup is not applied, and if it is negative, Mixup is applied. In the right figure, the number of times Mixup is applied for each class in 250 epochs is shown. In this case, we can see that Mixup is applied very often to the classes which are difficult to predict (dog and cat). The results of using CAMixup are as follows.
Red shows the results for ensemble only, blue for Mixup+ensemble, and green for CAMixup+ensemble. Figures (a) and (b) show that the test accuracy is slightly reduced compared to the regular Mixup, but the ECE can be reduced significantly. In the following table, we also show the results on ImageNet.
It was shown that ECE can be significantly improved while the loss of accuracy is negligible.
Performance during distribution shift
The results of the evaluation with CIFAR-10-C/CIFAR-100-C (C indicates corruption) are as follows.
As shown in the figure, CAMixup is shown to be effective even in task settings where distributional shifts occur. CAMixup is also shown to work well on AugMix, a state-of-the-art data augmentation method. The results are as follows.
AugMixup is a method that combines AugMix and Mixup (details are omitted). The modified version of AugMix combined with CAMixup (AugCAMisup) was shown to improve ECE significantly, as well as the normal CAMixup case.
Summary
In the paper below, it was shown that calibration can be degraded when ensembles are combined with data augmentation. This is likely due to the fact that ensembles and data augmentation underestimate the confidence level. To avoid this, we proposed CAMixup, which varies the application of Mixup depending on the difficulty of predicting the class.
While both data augmentation and ensembling are effective methods for improving performance, this is an important study in which we discovered a phenomenon that can be harmful by combining them, and showed a solution to it.
If you have any suggestions for improvement of the content of the article,
please contact the AI-SCHOLAR editorial team through the contact form.
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CC-MAIN-2023-06
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https://www.coursehero.com/file/5837945/Chapter-18/
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# Chapter 18 - Chapter 18 Two-port Networks the four...
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Chapter 18 Two-port Networks
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four-terminal network the four terminals have been paired into ports two-port network At all times, the instantaneous current flowing into one terminal is equal to the instantaneous current flowing out the other. KCL i1 i2 i3 i4 i1 i4 i1+i2+i3+i4=0(KCL) i1=-i2 ; i3=-i4
2 22 1 21 2 2 12 1 11 1 2 22 1 21 2 2 12 1 11 1 V y V y I V y V y I I z I z V I z I z V + = + = + = + = 2 22 1 21 2 2 12 1 11 1 2 22 1 21 2 2 12 1 11 1 I g V g V I g V g I V h I h I V h I h V + = + = + = + = 1 1 2 1 1 2 2 2 1 2 2 1 dI cV I bI aV V DI CV I BI AV V - = - = - = - = The network is linear(without independent sources).
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Impedance or z parameters are defined by impedance matrix Z 2 22 1 21 2 2 12 1 11 1 I z I z V I z I z V + = + = 0 22 0 21 0 12 0 11 1 2 2 2 1 2 1 2 1 2 1 1 = = = = = = = = I I V I I V I I V I I V z z z z Open-circuit input impedance. Open-circuit transfer impedance from port 1 to port 2 Open-circuit transfer impedance from port 2 to port 1 Open-circuit output impedance. 2
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https://en.wikipedia.org/wiki/Splay_tree
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# Splay tree
Splay tree
TypeTree
Invented1985
Invented byDaniel Dominic Sleator and Robert Endre Tarjan
Complexities in big O notation
Space complexity
Space O(n)
Time complexity
Function Amortized Worst Case
Search O(log n)[1]: 659 O(n)[2]: 1
Insert O(log n)[1]: 659 O(n)
Delete O(log n)[1]: 659 O(n)
A splay tree is a binary search tree with the additional property that recently accessed elements are quick to access again. Like self-balancing binary search trees, a splay tree performs basic operations such as insertion, look-up and removal in O(log n) amortized time. For random access patterns drawn from a non-uniform random distribution, their amortized time can be faster than logarithmic, proportional to the entropy of the access pattern. For many patterns of non-random operations, also, splay trees can take better than logarithmic time, without requiring advance knowledge of the pattern. According to the unproven dynamic optimality conjecture, their performance on all access patterns is within a constant factor of the best possible performance that could be achieved by any other self-adjusting binary search tree, even one selected to fit that pattern. The splay tree was invented by Daniel Sleator and Robert Tarjan in 1985.[1]
All normal operations on a binary search tree are combined with one basic operation, called splaying. Splaying the tree for a certain element rearranges the tree so that the element is placed at the root of the tree. One way to do this with the basic search operation is to first perform a standard binary tree search for the element in question, and then use tree rotations in a specific fashion to bring the element to the top. Alternatively, a top-down algorithm can combine the search and the tree reorganization into a single phase.
Good performance for a splay tree depends on the fact that it is self-optimizing, in that frequently accessed nodes will move nearer to the root where they can be accessed more quickly. The worst-case height—though unlikely—is O(n), with the average being O(log n). Having frequently-used nodes near the root is an advantage for many practical applications (also see locality of reference), and is particularly useful for implementing caches and garbage collection algorithms.
• Comparable performance: Average-case performance is as efficient as other trees.[3]
• Small memory footprint: Splay trees do not need to store any bookkeeping data.
The most significant disadvantage of splay trees is that the height of a splay tree can be linear.[2]: 1 For example, this will be the case after accessing all n elements in non-decreasing order. Since the height of a tree corresponds to the worst-case access time, this means that the actual cost of a single operation can be high. However the amortized access cost of this worst case is logarithmic, O(log n). Also, the expected access cost can be reduced to O(log n) by using a randomized variant.[4]
The representation of splay trees can change even when they are accessed in a 'read-only' manner (i.e. by find operations). This complicates the use of such splay trees in a multi-threaded environment. Specifically, extra management is needed if multiple threads are allowed to perform find operations concurrently. This also makes them unsuitable for general use in purely functional programming, although even there they can be used in limited ways to implement priority queues.
Finally, when the access pattern is random, the additional splaying overhead adds a significant constant factor to the cost compared to less-dynamic alternatives.
## Operations
### Splaying
When a node x is accessed, a splay operation is performed on x to move it to the root. A splay operation is a sequence of splay steps, each of which moves x closer to the root. By performing a splay operation on the node of interest after every access, the recently accessed nodes are kept near the root and the tree remains roughly balanced, so it provides the desired amortized time bounds.
Each particular step depends on three factors:
• Whether x is the left or right child of its parent node, p,
• whether p is the root or not, and if not
• whether p is the left or right child of its parent, g (the grandparent of x).
There are three types of splay steps, each of which has two symmetric variants: left- and right-handed. For the sake of brevity, only one of these two is shown for each type. (In the following diagrams, circles indicate nodes of interest and triangles indicate sub-trees of arbitrary size.) The three types of splay steps are:
Zig step: this step is done when p is the root. The tree is rotated on the edge between x and p. Zig steps exist to deal with the parity issue, will be done only as the last step in a splay operation, and only when x has odd depth at the beginning of the operation.
Zig-zig step: this step is done when p is not the root and x and p are either both right children or are both left children. The picture below shows the case where x and p are both left children. The tree is rotated on the edge joining p with its parent g, then rotated on the edge joining x with p. Zig-zig steps are the only thing that differentiate splay trees from the rotate to root method introduced by Allen and Munro[5] prior to the introduction of splay trees.
Zig-zag step: this step is done when p is not the root and x is a right child and p is a left child or vice versa (x is left, p is right). The tree is rotated on the edge between p and x, and then rotated on the resulting edge between x and g.
### Join
Given two trees S and T such that all elements of S are smaller than the elements of T, the following steps can be used to join them to a single tree:
• Splay the largest item in S. Now this item is in the root of S and has a null right child.
• Set the right child of the new root to T.
### Split
Given a tree and an element x, return two new trees: one containing all elements less than or equal to x and the other containing all elements greater than x. This can be done in the following way:
• Splay x. Now it is in the root so the tree to its left contains all elements smaller than x and the tree to its right contains all element larger than x.
• Split the right subtree from the rest of the tree.
### Insertion
To insert a value x into a splay tree:
As a result, the newly inserted node x becomes the root of the tree.
Alternatively:
• Use the split operation to split the tree at the value of x to two sub-trees: S and T.
• Create a new tree in which x is the root, S is its left sub-tree and T its right sub-tree.
### Deletion
To delete a node x, use the same method as with a binary search tree:
• If x has two children:
• Swap its value with that of either the rightmost node of its left sub tree (its in-order predecessor) or the leftmost node of its right subtree (its in-order successor).
In this way, deletion is reduced to the problem of removing a node with 0 or 1 children. Unlike a binary search tree, in a splay tree after deletion, we splay the parent of the removed node to the top of the tree.
Alternatively:
• The node to be deleted is first splayed, i.e. brought to the root of the tree and then deleted. leaves the tree with two sub trees.
• The two sub-trees are then joined using a "join" operation.
## Implementation and variants
Splaying, as mentioned above, is performed during a second, bottom-up pass over the access path of a node. It is possible to record the access path during the first pass for use during the second, but that requires extra space during the access operation. Another alternative is to keep a parent pointer in every node, which avoids the need for extra space during access operations but may reduce overall time efficiency because of the need to update those pointers.[1]
Another method which can be used is based on the argument that the tree can be restructured during the way down the access path instead of making a second pass. This top-down splaying routine uses three sets of nodes – left tree, right tree and middle tree. The first two contain all items of original tree known to be less than or greater than current item respectively. The middle tree consists of the sub-tree rooted at the current node. These three sets are updated down the access path while keeping the splay operations in check. Another method, semisplaying, modifies the zig-zig case to reduce the amount of restructuring done in all operations.[1][6]
Below there is an implementation of splay trees in C++, which uses pointers to represent each node on the tree. This implementation is based on bottom-up splaying version and uses the second method of deletion on a splay tree. Also, unlike the above definition, this C++ version does not splay the tree on finds – it only splays on insertions and deletions, and the find operation, therefore, has linear time complexity.
#include <functional>
#ifndef SPLAY_TREE
#define SPLAY_TREE
template<typename T, typename Comp = std::less<T>>
class splay_tree {
private:
Comp comp;
unsigned long p_size;
struct node {
node *left, *right;
node *parent;
T key;
node(const T& init = T()) : left(nullptr), right(nullptr), parent(nullptr), key(init) { }
~node() {
}
} *root;
void left_rotate(node *x) {
node *y = x->right;
if (y) {
x->right = y->left;
if (y->left) y->left->parent = x;
y->parent = x->parent;
}
if (!x->parent) root = y;
else if (x == x->parent->left) x->parent->left = y;
else x->parent->right = y;
if (y) y->left = x;
x->parent = y;
}
void right_rotate(node *x) {
node *y = x->left;
if (y) {
x->left = y->right;
if (y->right) y->right->parent = x;
y->parent = x->parent;
}
if (!x->parent) root = y;
else if (x == x->parent->left) x->parent->left = y;
else x->parent->right = y;
if (y) y->right = x;
x->parent = y;
}
void splay(node *x) {
while (x->parent) {
if (!x->parent->parent) {
if (x->parent->left == x) right_rotate(x->parent);
else left_rotate(x->parent);
} else if (x->parent->left == x && x->parent->parent->left == x->parent) {
right_rotate(x->parent->parent);
right_rotate(x->parent);
} else if (x->parent->right == x && x->parent->parent->right == x->parent) {
left_rotate(x->parent->parent);
left_rotate(x->parent);
} else if (x->parent->left == x && x->parent->parent->right == x->parent) {
right_rotate(x->parent);
left_rotate(x->parent);
} else {
left_rotate(x->parent);
right_rotate(x->parent);
}
}
}
void replace(node *u, node *v) {
if (!u->parent) root = v;
else if (u == u->parent->left) u->parent->left = v;
else u->parent->right = v;
if (v) v->parent = u->parent;
}
node* subtree_minimum(node *u) {
while (u->left) u = u->left;
return u;
}
node* subtree_maximum(node *u) {
while (u->right) u = u->right;
return u;
}
public:
splay_tree() : root(nullptr), p_size(0) { }
void insert(const T &key) {
node *z = root;
node *p = nullptr;
while (z) {
p = z;
if (comp(z->key, key)) z = z->right;
else z = z->left;
}
z = new node(key);
z->parent = p;
if (!p) root = z;
else if (comp(p->key, z->key)) p->right = z;
else p->left = z;
splay(z);
p_size++;
}
node* find(const T &key) {
node *z = root;
while (z) {
if (comp(z->key, key)) z = z->right;
else if (comp(key, z->key)) z = z->left;
else return z;
}
return nullptr;
}
void erase(const T &key) {
node *z = find(key);
if (!z) return;
splay(z);
if (!z->left) replace(z, z->right);
else if (!z->right) replace(z, z->left);
else {
node *y = subtree_minimum(z->right);
if (y->parent != z) {
replace(y, y->right);
y->right = z->right;
y->right->parent = y;
}
replace(z, y);
y->left = z->left;
y->left->parent = y;
}
delete z;
p_size--;
}
/* //the alternative implementation
void erase(const T &key) {
node *z = find(key);
if (!z) return;
splay(z);
node *s = z->left;
node *t = z->right;
delete z;
node *sMax = NULL;
if (s) {
s->parent = NULL;
sMax = subtree_maximum(s);
splay(sMax);
root = sMax;
}
if (t) {
if (s)
sMax->right = t;
else
root = t;
t->parent = sMax;
}
p_size--;
}
*/
const T& minimum() { return subtree_minimum(root)->key; }
const T& maximum() { return subtree_maximum(root)->key; }
bool empty() const { return root == nullptr; }
unsigned long size() const { return p_size; }
};
#endif // SPLAY_TREE
## Analysis
A simple amortized analysis of static splay trees can be carried out using the potential method. Define:
• size(r) = the number of nodes in the sub-tree rooted at node r (including r).
• rank(r) = log2(size(r)).
• Φ = the sum of the ranks of all the nodes in the tree.
Φ will tend to be high for poorly balanced trees and low for well-balanced trees.
To apply the potential method, we first calculate ΔΦ: the change in the potential caused by a splay operation. We check each case separately. Denote by rank' the rank function after the operation. x, p and g are the nodes affected by the rotation operation (see figures above).
### Zig step
ΔΦ = rank'(p) − rank(p) + rank'(x) − rank(x) [since only p and x change ranks] = rank'(p) − rank(x) [since rank'(x)=rank(p)] ≤ rank'(x) − rank(x) [since rank'(p)
### Zig-zig step
ΔΦ = rank'(g) − rank(g) + rank'(p) − rank(p) + rank'(x) − rank(x) = rank'(g) + rank'(p) − rank(p) − rank(x) [since rank'(x)=rank(g)] ≤ rank'(g) + rank'(x) − 2 rank(x) [since rank(x)rank'(p)] ≤ 3(rank'(x)−rank(x)) − 2 [due to the concavity of the log function]
### Zig-zag step
ΔΦ = rank'(g) − rank(g) + rank'(p) − rank(p) + rank'(x) − rank(x) ≤ rank'(g) + rank'(p) − 2 rank(x) [since rank'(x)=rank(g) and rank(x)
The amortized cost of any operation is ΔΦ plus the actual cost. The actual cost of any zig-zig or zig-zag operation is 2 since there are two rotations to make. Hence:
amortized-cost = cost + ΔΦ ≤ 3(rank'(x)−rank(x))
When summed over the entire splay operation, this telescopes to 1 + 3(rank(root)−rank(x)) which is O(log n), since we use The Zig operation at most once and the amortized cost of zig is at most 1+3(rank'(x)−rank(x)).
So now we know that the total amortized time for a sequence of m operations is:
${\displaystyle T_{\mathrm {amortized} }(m)=O(m\log n)}$
To go from the amortized time to the actual time, we must add the decrease in potential from the initial state before any operation is done (Φi) to the final state after all operations are completed (Φf).
${\displaystyle \Phi _{i}-\Phi _{f}=\sum _{x}{\mathrm {rank} _{i}(x)-\mathrm {rank} _{f}(x)}=O(n\log n)}$
where the big O notation can be justified by the fact that for every node x, the minimum rank is 0 and the maximum rank is log(n).
Now we can finally bound the actual time:
${\displaystyle T_{\mathrm {actual} }(m)=O(m\log n+n\log n)}$
### Weighted analysis
The above analysis can be generalized in the following way.
• Assign to each node r a weight w(r).
• Define size(r) = the sum of weights of nodes in the sub-tree rooted at node r (including r).
• Define rank(r) and Φ exactly as above.
The same analysis applies and the amortized cost of a splaying operation is again:
${\displaystyle 1+3(\mathrm {rank} (root)-\mathrm {rank} (x))}$
where W is the sum of all weights.
The decrease from the initial to the final potential is bounded by:
${\displaystyle \Phi _{i}-\Phi _{f}\leq \sum _{x\in tree}{\log {\frac {W}{w(x)}}}}$
since the maximum size of any single node is W and the minimum is w(x).
Hence the actual time is bounded by:
${\displaystyle O\left(\sum _{x\in sequence}{\left(1+3\log {\frac {W}{w(x)}}\right)}+\sum _{x\in tree}{\log {\frac {W}{w(x)}}}\right)=O\left(m+\sum _{x\in sequence}{\log {\frac {W}{w(x)}}}+\sum _{x\in tree}{\log {\frac {W}{w(x)}}}\right)}$
## Performance theorems
There are several theorems and conjectures regarding the worst-case runtime for performing a sequence S of m accesses in a splay tree containing n elements.
Balance Theorem — The cost of performing the sequence S is ${\displaystyle O\left[m\log n+n\log n\right]}$.
Proof
Take a constant weight, e.g. ${\displaystyle w(x)=1}$ for every node x. Then ${\displaystyle W=n}$.
This theorem implies that splay trees perform as well as static balanced binary search trees on sequences of at least n accesses.[1]
Static Optimality Theorem — Let ${\displaystyle q_{x}}$ be the number of times element x is accessed in S. If every element is accessed at least once, then the cost of performing S is ${\displaystyle O\left[m+\sum _{x\in tree}q_{x}\log {\frac {m}{q_{x}}}\right]}$
Proof
Let ${\displaystyle w(x)=q_{x}}$. Then ${\displaystyle W=m}$.
This theorem implies that splay trees perform as well as an optimum static binary search tree on sequences of at least n accesses.[7] They spend less time on the more frequent items.[1] Another way of stating the same result is that, on input sequences where the items are drawn independently at random from a non-uniform probability distribution on n items, the amortized expected (average case) cost of each access is proportional to the entropy of the distribution.[8]
Static Finger Theorem — Assume that the items are numbered from 1 through n in ascending order. Let f be any fixed element (the 'finger'). Then the cost of performing S is ${\displaystyle O\left[m+n\log n+\sum _{x\in sequence}\log(|x-f|+1)\right]}$.
Proof
Let ${\displaystyle w(x)=1/(|x-f|+1)^{2}}$. Then ${\displaystyle W=O(1)}$. The net potential drop is O (n log n) since the weight of any item is at least ${\displaystyle 1/n^{2}}$.[1]
Dynamic Finger Theorem — Assume that the 'finger' for each step accessing an element y is the element accessed in the previous step, x. The cost of performing S is ${\displaystyle O\left[m+n+\sum _{x,y\in sequence}^{m}\log(|y-x|+1)\right]}$.[9][10]
Working Set Theorem — At any time during the sequence, let ${\displaystyle t(x)}$ be the number of distinct elements accessed before the previous time element x was accessed. The cost of performing S is ${\displaystyle O\left[m+n\log n+\sum _{x\in sequence}\log(t(x)+1)\right]}$
Proof
Let ${\displaystyle w(x)=1/(t(x)+1)^{2}}$. Note that here the weights change during the sequence. However, the sequence of weights is still a permutation of ${\displaystyle 1,{\tfrac {1}{4}},{\tfrac {1}{9}},\cdots ,{\tfrac {1}{n^{2}}}}$. So as before ${\displaystyle W=O(1)}$. The net potential drop is O (n log n).
This theorem is equivalent to splay trees having key-independent optimality.[1]
Scanning Theorem — Also known as the Sequential Access Theorem or the Queue theorem. Accessing the n elements of a splay tree in symmetric order takes O(n) time, regardless of the initial structure of the splay tree.[11] The tightest upper bound proven so far is ${\displaystyle 4.5n}$.[12]
## Dynamic optimality conjecture
Unsolved problem in computer science:
Do splay trees perform as well as any other binary search tree algorithm?
In addition to the proven performance guarantees for splay trees there is an unproven conjecture of great interest from the original Sleator and Tarjan paper. This conjecture is known as the dynamic optimality conjecture and it basically claims that splay trees perform as well as any other binary search tree algorithm up to a constant factor.
Dynamic Optimality Conjecture:[1] Let ${\displaystyle A}$ be any binary search tree algorithm that accesses an element ${\displaystyle x}$ by traversing the path from the root to ${\displaystyle x}$ at a cost of ${\displaystyle d(x)+1}$, and that between accesses can make any rotations in the tree at a cost of 1 per rotation. Let ${\displaystyle A(S)}$ be the cost for ${\displaystyle A}$ to perform the sequence ${\displaystyle S}$ of accesses. Then the cost for a splay tree to perform the same accesses is ${\displaystyle O[n+A(S)]}$.
There are several corollaries of the dynamic optimality conjecture that remain unproven:
Traversal Conjecture:[1] Let ${\displaystyle T_{1}}$ and ${\displaystyle T_{2}}$ be two splay trees containing the same elements. Let ${\displaystyle S}$ be the sequence obtained by visiting the elements in ${\displaystyle T_{2}}$ in preorder (i.e., depth first search order). The total cost of performing the sequence ${\displaystyle S}$ of accesses on ${\displaystyle T_{1}}$ is ${\displaystyle O(n)}$.
Deque Conjecture:[11][13][14] Let ${\displaystyle S}$ be a sequence of ${\displaystyle m}$ double-ended queue operations (push, pop, inject, eject). Then the cost of performing ${\displaystyle S}$ on a splay tree is ${\displaystyle O(m+n)}$.
Split Conjecture:[6] Let ${\displaystyle S}$ be any permutation of the elements of the splay tree. Then the cost of deleting the elements in the order ${\displaystyle S}$ is ${\displaystyle O(n)}$.
## Variants
In order to reduce the number of restructuring operations, it is possible to replace the splaying with semi-splaying, in which an element is splayed only halfway towards the root.[1][2]
Another way to reduce restructuring is to do full splaying, but only in some of the access operations – only when the access path is longer than a threshold, or only in the first m access operations.[1]
The CBTree augments the splay tree with access counts at each node and uses them to restructure infrequently. A variant of the CBTree called the LazyCBTree does at most one rotation on each lookup. This is used along with an optimistic hand-over-hand validation scheme to make a concurrent self-adjusting tree.[15]
Using pointer-compression techniques,[16] it is possible to construct a succinct splay tree.
## References
• Afek, Yehuda; Kaplan, Haim; Korenfeld, Boris; Morrison, Adam; Tarjan, Robert E. (2014). "The CB tree: a practical concurrent self-adjusting search tree". Distributed Computing. 27 (6): 393–417. doi:10.1007/s00446-014-0229-0.
• Albers, Susanne; Karpinski, Marek (28 February 2002). "Randomized Splay Trees: Theoretical and Experimental Results" (PDF). Information Processing Letters. 81 (4): 213–221. doi:10.1016/s0020-0190(01)00230-7.
• Allen, Brian; Munro, Ian (October 1978). "Self-organizing binary search trees". Journal of the ACM. 25 (4): 526–535. doi:10.1145/322092.322094. S2CID 15967344.
• Brinkmann, Gunnar; Degraer, Jan; De Loof, Karel (January 2009). "Rehabilitation of an unloved child: semi-splaying" (PDF). Software: Practice and Experience. 39 (1): 33–45. CiteSeerX 10.1.1.84.790. doi:10.1002/spe.v39:1. hdl:11382/102133. The results show that semi-splaying, which was introduced in the same paper as splaying, performs better than splaying under almost all possible conditions. This makes semi-splaying a good alternative for all applications where normally splaying would be applied. The reason why splaying became so prominent while semi-splaying is relatively unknown and much less studied is hard to understand.
• Cole, Richard; Mishra, Bud; Schmidt, Jeanette; Siegel, Alan (January 2000). "On the Dynamic Finger Conjecture for Splay Trees. Part I: Splay Sorting log n-Block Sequences". SIAM Journal on Computing. 30 (1): 1–43. CiteSeerX 10.1.1.36.4558. doi:10.1137/s0097539797326988.
• Cole, Richard (January 2000). "On the Dynamic Finger Conjecture for Splay Trees. Part II: The Proof". SIAM Journal on Computing. 30 (1): 44–85. CiteSeerX 10.1.1.36.2713. doi:10.1137/S009753979732699X.
• Elmasry, Amr (April 2004). "On the sequential access theorem and Deque conjecture for splay trees". Theoretical Computer Science. 314 (3): 459–466. doi:10.1016/j.tcs.2004.01.019.
• Goodrich, Michael; Tamassia, Roberto; Goldwasser, Michael (2014). Data Structures and Algorithms in Java (6 ed.). Wiley. p. 506. ISBN 978-1-118-77133-4.
• Grinberg, Dennis; Rajagopalan, Sivaramakrishnan; Venkatesan, Ramarathnam; Wei, Victor K. (1995). "Splay trees for data compression". In Clarkson, Kenneth L. (ed.). Proceedings of the Sixth Annual ACM-SIAM Symposium on Discrete Algorithms, 22–24 January 1995. San Francisco, California, USA. ACM/SIAM. pp. 522–530. Average depth of access in a splay tree is proportional to the entropy.
• Knuth, Donald (1997). The Art of Computer Programming. Vol. 3: Sorting and Searching (3rd ed.). Addison-Wesley. p. 478. ISBN 0-201-89685-0. The time needed to access data in a splay tree is known to be at most a small constant multiple of the access time of a statically optimum tree, when amortized over any series of operations.
• Lucas, Joan M. (1991). "On the Competitiveness of Splay Trees: Relations to the Union-Find Problem". On-line Algorithms: Proceedings of a DIMACS Workshop, February 11–13, 1991. Series in Discrete Mathematics and Theoretical Computer Science. Vol. 7. Center for Discrete Mathematics and Theoretical Computer Science. pp. 95–124. ISBN 0-8218-7111-0.
• Pettie, Seth (2008). Splay Trees, Davenport-Schinzel Sequences, and the Deque Conjecture (PDF). Proc. 19th ACM-SIAM Symposium on Discrete Algorithms. Vol. 0707. pp. 1115–1124. arXiv:0707.2160. Bibcode:2007arXiv0707.2160P.
• Sleator, Daniel D.; Tarjan, Robert E. (1985). "Self-Adjusting Binary Search Trees" (PDF). Journal of the ACM. 32 (3): 652–686. doi:10.1145/3828.3835. S2CID 1165848.
• Sundar, Rajamani (1992). "On the Deque conjecture for the splay algorithm". Combinatorica. 12 (1): 95–124. doi:10.1007/BF01191208. S2CID 27422556.
• Tarjan, Robert E. (1985). "Sequential access in splay trees takes linear time". Combinatorica. 5 (4): 367–378. doi:10.1007/BF02579253. S2CID 34757821.
• Bender, Michael A.; Conway, Alex; Farach-Colton, Martin; Kuszmaul, William; Tagliavini, Guido (2023). "Tiny Pointers". Proceedings of the 2023 Annual ACM-SIAM Symposium on Discrete Algorithms (SODA): 477–508. doi:10.1137/1.9781611977554.ch21. ISBN 978-1-61197-755-4. S2CID 244709005.
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# 99028 in Roman Numerals | How to write 99028 in roman numerals
Do you want to know what 99028 is in Roman numerals or how to write 99028 in Roman numerals? If so, you have come to the right corner of the internet. In this article, we'll show you the correct way to write 99028 in Roman numerals, including explanations and rules with pictures and text. Along with this, you'll also get a Roman numeral converter tool to convert different numbers.
## 99028 in Roman numerals
The Hindu-Arabic (Indo-Arabic) number 99028 in Roman numerals is X̅C̅MX̅XXVIII. Here the X̅C̅ stands for 90000, MX̅ stands for 9000, XX stands for 20 and VIII stands for 8.
Roman Numerals Tool
## How to write 99028 to Roman numerals
To write 99028 in Roman numerals, you need to know the basics of the Roman number system and some basic math, such as which Roman symbol represents what value and how to do addition and subtraction.
The Roman number system is based on seven letters or symbols: I, V, X, L, C, D, and M. These symbols represent the values 1, 5, 10, 50, 100, 500, and 1000, respectively. For easier and faster calculations, you can memorize 4 as IV, 9 as IX, 40 as XL, 90 as XC, 400 as CD, and 900 as CM. You can convert numbers like 2, 3, 20, and 30 by adding two or three symbols that are the same (e.g., X + X + X = 30). And for numbers like 6, 7, 8, 17, and 46, you can convert them by adding smaller symbols followed by larger symbols (e.g., V + III = 5 + 3 = 8, X + V + II = 10 + 5 + 2 = 17, and XL + V + I = 40 + 5 + 1 = 46).
Note - The value of these symbols is multiplied by 1000 by adding a bar above them (e.g., V̅ = 5000).
To convert any number to Roman numerals, including 99028, follow these two steps:
• Step 1:- Split the number into ones, tens, hundreds, thousands and so on like this.
90000 + 9000 + 20 + 8 = 99028
• Step 2:- Then replace those numbers with equivalent Roman symbols like this.
X̅C̅ + MX̅ + XX + VIII = X̅C̅MX̅XXVIII
Here’s a table and some calculations for better understanding.
X̅C̅ = C̅ - X̅
or, X̅C̅ = 100000 - 10000
or, X̅C̅ = 90000
MX̅ = X̅ - M
or, MX̅ = 10000 - 1000
or, MX̅ = 9000
XX = X + X
or ,XX = 10 + 10
or , XX = 20
VIII = V + III
or, VIII = 5 + 1 + 1 + 1
or, VIII = 8
### Final Word
Thanks for visiting our website. We hope this article has helped you. If so, feel free to leave a comment or feedback about this article.
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# chem
posted by .
this is my question
Calculate the hydronium ion consentration and the hydroxide concentration in blood in which the PH is 7.3.
I don't have invlog on my calculator so I hit log(7.3)= .863 for hydronium. Is this dumb for me to do?
That is dumb for you to do.
Get a calc that has an INV key. Mine costs eight dollars.
concentration= invlog (-7.3)= 5.01?E-08
Do you have a 2nd key, and a log key?
normally, 2nd Log is same as INVLOG
yes when I press 2nd then log I get 10^, is this correct?
In order to get antilog on mine I punch the 10x key. And a pH of 7.3 is hydroniuim of 5.01E-8.
ok I got that part. now how can I find Hydroxide concentration from this?
Two ways.
#1. (H^+)(OH^-) = 1 x 10^-14.
You know (H^+).
#2. pH + pOH = pKw = 14
You know pH, so subtract from 14 to get pOH. Then pOH = -log(OH^-).
PH + pOH = pKw =14
7.3 + pOH =pKw =14
7.3 - 14 = 6.7
pOH = -log(OH^-)
6.7 = -log(1.00 x 10^-14)
pOH = 1.92 x 10^6? I'm stuck
PH + pOH = pKw =14
7.3 + pOH =pKw =14
7.3 - 14 = 6.7
pOH = -log(OH^-) OK to here
6.7 = -log(1.00 x 10^-14)
pOH = 1.92 x 10^6? I'm stuck
pOH = -log(OH^-)
6.70 = -log(OH^-)
-6.70 = log(OH^-)
1.996 x 10^-7 = round to
2 x 10^-7 = (OH^-)
## Similar Questions
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Calculate the hydronium ion consentration and the hydroxide concentration in blood in which the PH is 7.3. Hydronium = [H3O+] = 10-pH or [H3O+] = antilog (- pH) and I got 1.00 x 10^7.3 for both concentrations is this correct?
2. ### Chemistry
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1. ## Inequalities
Solve the inequality. State the solution set using interval notation and graph.
x-3/x+4>7
x^2-5x-7<0
2. Originally Posted by pashah
Solve the inequality. State the solution set using interval notation and graph.
x-3/x+4>7
x^2-5x-7<0
Hello, pashah,
I assume that your first problem reads: $\frac{x-3}{x+4}>7$
Multiply by denominator. The next steps depends on the sign of the denominator!
$x-3>7x+28 \ \wedge\ x>-4$
$-31>6x \ \wedge\ x>-4$
$-\frac{31}{6}>x \ \wedge\ x>-4$. that means $x\notin \mathbb{R}$
or
$x-3<7x+28 \ \wedge\ x<-4$
$-31<6x \ \wedge\ x<-4$
$-\frac{31}{6}. So the solution is:
$-\frac{31}{6}
2nd problem:
Solve this equation: $x^2-5x-7=0$. You'll get:
$x=\frac{5}{2}-\frac{\sqrt{53}}{2}\ \vee\ x=\frac{5}{2}+\frac{\sqrt{53}}{2}$
Because this equation belongs to a quadratic funtion which open upward, the part of the function between the zeros is below the x-axis, that means it is smaller than zero. Therefore your solution is:
$\frac{5}{2}-\frac{\sqrt{53}}{2}
Greetings
EB
3. Originally Posted by pashah
Solve the inequality. State the ... graph....
Hello, pashah,
I've attached the graph of the 1rst problem.
Greetings
EB
4. Originally Posted by pashah
Solve the inequality. State the ...graph.
x^2-5x-7<0
Hello, pashah,
I've attached the graph of your 2nd problem.
Greetings
EB
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# Sudoku
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Size 4 x 4 9 x 9 16 x 16 25 x 25 Difficulty Easy Medium Hard MENSA Genius Lex Luthor Characters Numbers Letters Theme Puzzles Theme (choose) Birthday Christmas Earth Day Easter Halloween New Years St. Patricks Super Bowl Valentines
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Name-based puzzles like the one for Farwa below are a fun twist on traditional puzzles but they only work for names with a useable amount of unique letters.
If you like our Farwa Sudoku Puzzles, remember to add us to your online bookmarks, mention us on Facebook, or give us a Tweet by clicking one of the buttons to the left.
Farwa
(F a r w )
### Sudoku Instructions
Rules for solving sudoku puzzles are simple. Complete each blank square in a puzzle with the correct number. There are three very simple constraints to follow. In a 9x9 sudoku puzzle using numbers, the following must contain all digits 1-9 in any order. Each:
• 9-block row
• 9-block column
• 3x3 subsection
The same constraints can be applied to puzzles of different sizes. For example, 4x4 puzzles must have the numerals 1-4 in each row, column and 2x2 subsection. Larger 16x16 sudoku puzzles must have numerals 1-16 in each row, column and subsection. The principles are the same whatever the size of the game and also when letters are used. With letters, each letter must be used but only one time in each row, column, and subsection. 4x4, 9x9, 16x16, 25x25.
Every sudoku puzzle begins with some blocks already filled in. Puzzle difficulty is largely a function of how many squares are filled in. The more squares that are known, the easier it is to figure out which numbers or letters go in the empty blocks. As you fill in blocks correctly, options for the remaining blocks reduces and it becomes easier to fill them in. This site offers the following difficulty levels: Easy, Medium, Hard, MENSA, Genius, Lex Luthor.
Solutions can be generated online and printed if you need help or to check your progress.
### Sudoku Strategy
Need help solving sudoku puzzles? Try these tips. Begin by scanning for a weakness. It will be obvious in easy sudoku puzzles with many blocks already completed. Some may have entire rows or columns filled in which make it easy to figure out remaining empty blocks in a subsection.
### History of Sudoku
Our Japanese friends tell us "sudoku" means something like "single number" referring to the solitary position of each number in the puzzle grid. Although the sound "doku" may also mean, "poison" which would be interesting; "poison number". Maybe it seems that way when a wrong number is selected. As in most languages, one sound can be associated with several meanings. And when translating into English, it has also been spelled either "su doku", "soduko", "sudoko", or "suduko". Sometimes puzzle is mispelled as "puzzel". Anyway, although sudoku puzzles were popularized in Japan, they're actually of European origin being similar to a mathematic concept called "Latin Squares". Each square is a table or grid in which a number or symbol appears only once in each row and column. In 1979, a puzzle maker (Howard Garnes) expanded the Latin Squares concept to create what was then called "Number Place". Then, Maki Kaji, President of a Japanese puzzle company, began publishing Number Place puzzles there but called them "Suji wa dokushin ni kagiru" which translates roughly as "the numbers must be single". Later he limited the number of clues, imposed rules about the symmetry of their appearance, and shortened the puzzle's name to just sudoku. Since these changes, soduko puzzles became increasingly popular in Japan. Later, with the help of Wayne Gould, sudoku's spread back to America and now globally. So, although the puzzles have European and American roots, many regard Maki Kaji as the Father of Sudoku.
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# Fractions Pack Adding Subtracting Math Worksheets Classcrown And Color By Number Grade4 Pack1 Page21 900w Stunning
Published at Friday, 10 May 2019. Color by Number. By .
Since tomorrow is National Coloring Book Day, today seems like a good day to talk about color by number exercises, and some of the principal benefits of color by number activities, and why they are good exercises to give to your kids, either at home or in a classroom setting. So here we go: Color by number exercises encourage creativity, But wait, you might say… color by number exercises give children defined colors and limits… why would this support creativity and imagination? Well, if you have children who don’t naturally want to draw or color, or feel timid doing so, color by number exercises offer a “safe zone” that kids can use to practice working with color and design. This can lead to future drawing, painting, or coloring activities.
Children learn the meaning of symbols. Color by number worksheets are fantastic in helping children understand that symbols have meaning. Not only do children get better at color recognition when using color by number pages, children also learn that numbers can be used to represent other things and aren’t only just for counting. Later on, children will be able to grasp that different objects in their surroundings can actually be used as a symbol for other things and have other meanings. It’s just undeniable that symbolic understanding is important to function well in the society we live in.
First off, even though it may seem that children are quite restrained when coloring these coloring pages in terms of shapes, choice of colors and what not, color by number images can really be encouraging for some kids. Namely, there are children who feel frustrated when given too many choices and opportunities to make decisions. For some of them, having to pick colors and plan the way to color a picture might come as torture, making them feel anxious and nervous. That is why these color by number coloring pages are more than welcome: you are already told what to do, therefore you can relax and perform the actions. One more major benefit of color by number coloring pages is that not only young kids learn how to discern colors (in terms of primary and secondary ones), but they also learn how to put numbers in contexts other than counting. Here, numbers are presented as mere symbols that can and will occur anywhere else in real life.
File name:
### Fractions Pack Adding Subtracting Math Worksheets Classcrown And Color By Number Grade4 Pack1 Page21 900w Stunning
Image Size: 900 x 1165 Pixels
File Type: Image/jpg
Total Gallery: 45 Pictures
File Size: 201 kb
#### Adding And Subtracting Fractions Color By Number Gallery
63 of 100 by 343 users
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Online JudgeProblem SetAuthorsOnline ContestsUser
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## 怎么可能是TLE呢！！！！！怎么可能。。。才10^6！！！
Posted by severous at 2011-08-15 14:44:25 on Problem 3304
```#include<cstdio>
using namespace std;
struct point {
double x;
double y;
};
bool judge(point p1,point p2,point p3)
{
if((p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x)==0)
return true;
else
return false;
}
point inter(point u1,point u2,point v1,point v2)
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
main(){
int t,cont,flag,n,i,j,k;
struct point a[220],ret;
double x1,x2,y1,y2;
scanf("%d",&t);
while(t--){
flag=1;
scanf("%d",&n);
for(i=0;i<=2*n-1;i++){
scanf("%lf%lf",&a[i].x,&a[i].y);
}
for(i=0;i<=2*n-1 && flag;i++){
for(j=i+1;j<=2*n-1 && flag;j++){
cont=0;
for(k=0;k<=2*n-1 && flag;k=k+2){
if(judge(a[i],a[j],a[k])&&judge(a[i],a[j],a[k+1]))
cont++;
else{
ret=inter(a[i],a[j],a[k],a[k+1]);
if(a[k].x>a[k+1].x){
x1=a[k+1].x;
x2=a[k].x;
}
else{
x1=a[k].x;
x2=a[k+1].x;
}
if(a[k].y>a[k+1].y){
y1=a[k+1].y;
y2=a[k].y;
}
else{
y1=a[k].y;
y2=a[k+1].y;
}
if(ret.x<=x2 && ret.x>=x1 && ret.y<=y2 && ret.y>=y1)
cont++;
}
}
if(cont==n)
flag=0;
}
}
if(flag==0)
printf("Yes!\n");
else
printf("No!\n");
}
}
```
Followed by:
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https://www.etf.bg.ac.rs/en/fis/karton_predmeta/OF2SIS-2006
| 1,723,515,136,000,000,000
|
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# OF2SIS - Signals and Systems
Course specification
Course title Signals and Systems
Acronym OF2SIS
Study programme Electrical Engineering and Computing
Module Physical Electronics
Type of study bachelor academic studies
Lecturer (for classes)
Lecturer/Associate (for practice)
Lecturer/Associate (for OTC)
ESPB 6.0 Status mandatory
Condition none
The goal Objective of the course is to inform the students about the modelling of continuous and discrete systems as well as about the characterization of continuous and discrete signals. Also, the objective is for the students to understand basic tools for signals and systems analysis in time, frequency and complex domain.
The outcome After the passing this exam, student will have skills to classify and analyze the experimentally measured signals, to make basic experiments on systems, and to apply different techniques for signals and systems analysis (Fourier series, Fourier transformation, Laplace transformation, zed transformation).
Contents
Contents of lectures Classification of signals and systems; Systems properties; Fourier series of periodic signals; Fourier transformation; Bode diagrams; Laplace transformation; Transfer function; Stability and causality of continuous and discrete LTI systems; Zed transformation of discrete signals; Shannon theorem about signals sampling; Discrete Fourier transform.
Contents of exercises Within this course, students have obligation to solve three practical problems: 1. Calculation of signals convolution; 2. approximation of periodic signals using MATLAB; 3. calculation of system frequency response.
Literature
1. Signali i sistemi, B. Kovačević, Ž. Đurović, S. Stanković, Akademska misao, 2007. (Original title)
2. Signals, Systems, and Transforms, Charles Phillips, John Paar, Eve Riskih, Prentice Hall, 2003. (Original title)
Number of hours per week during the semester/trimester/year
Lectures Exercises OTC Study and Research Other classes
3 1 1
Methods of teaching 45 hours of lectures + 15 hours of auditory exercising + 15 hours of practical exercising with computers
Knowledge score (maximum points 100)
Pre obligations Points Final exam Points
Activites during lectures 0 Test paper 44
Practical lessons 12 Oral examination 0
Projects 0
Colloquia 44
Seminars 0
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| 2.71875
| 3
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CC-MAIN-2024-33
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latest
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en
| 0.868616
|
https://www.physicsforums.com/threads/why-do-we-need-the-s-2-when-describing-force-energyetc.430251/
| 1,544,424,875,000,000,000
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crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00038.warc.gz
| 1,019,409,391
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|
# Why do we need the s-2 when describing force/energyetc?
1. Sep 19, 2010
### Femme_physics
For force, for instance, why can't we just use m x kg without the s-2?
2. Sep 19, 2010
### bp_psy
This question is very weird and my attempt of an answer will fail miserably. Force as it is defined and understood is something that is at the base of the concept of motion (the main subject of pretty much all physics). Motion is only meaningful if you have the notion of time. So there is no quantity that is equal to distance*mass because it is not really useful for describing anything about motion. If that quantity would be very useful than it would get some name.
3. Sep 19, 2010
### Pythagorean
Because force and energy are associated with motion (especially if we want to do anything interesting with them) which requires a change in the system. Seconds are a unit of change.
Of course, realize that no motion is a special case of motion in this context (we don't exclude v = 0 for instance, on a graph that we'd plot motion on). And in the case of energy, there is potential energy which describes a potential for motion.
4. Sep 19, 2010
### Femme_physics
Right, so there has to be a defined scale of time in order for it to work, I see. Does it mean if the same m x kg is applied at 5 seconds as opposed to 1 second, the force would be lesser.
Correct?
5. Sep 19, 2010
### Pythagorean
Well, it's good thinking, but there's other things to consider. Consider the momentum formulation of force:
F = dp/dt = m*(dv/dt) = m*(dx^2/dt^2)
which requires calculus to understand in full. If you haven't seen any calculus yet, we'll look at
F = m*(dv/dt)
the mass (m) multiplied by the change in velocity (dv) over the change in time (dt). To treat dv/dt like division is incorrect in general, but I'm trying to demonstrate it algebraicly.
we can change this to
F*dt = dv
So if you replace dt with 1 sec or 5 sec, you could still have the same F if you also changed dv to make it match.
But for a fixed dv, yes your statement would be true.
The way to imagine this is a chunk of clay splatting against the wall. It has some velocity, v when it hits the wall. Over the next couple milliseconds (or whatever dt is), it slows down to 0. So it's change in velocity, dv = v-0 = v. F tells you the force that the clay and the wall imparted on each other throughout the event.
6. Sep 19, 2010
### Pythagorean
oh, and if you're wondering where the other s went, it's implicit in the v:
v = dx/dt (velocity is the change in position with respect to time)
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| 3.703125
| 4
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CC-MAIN-2018-51
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latest
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en
| 0.958376
|
http://www.halfbakery.com/idea/Aeroelectric_20Calculator
| 1,556,137,104,000,000,000
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text/html
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| 239,597,767
| 5,295
|
h a l f b a k e r y
Where life irritates science.
meta:
account: browse anonymously, or get an account and write.
user: pass:
register,
# Aeroelectric Calculator
Pioneering in renewable energy, I present the latest advancement that will save the planet.
(+2, -1) [vote for, against]
Oil prices are rocketing, and in desperate times we need to source our energy from renewable, "green" ideas. Solar powered calculators are available, but where's the dignity in that? I present to you: the aeroelectric calculator. To fuel it, blow air into the balloon at the top via the one-way valve. In the central mechanism, a miniature Francis turbine (the most efficient turbine in the world!) takes advantage of the air flow, and generates a tiny current - giving the calculator enough life to perform even the most intense calculations. For about 3 seconds. Then you need to blow more air in.
— phoenix, Oct 04 2008
Hydroelectric Calculator Hydroelectric_20Calculator
The inspiration. [phoenix, Oct 04 2008]
A Francis turbine would be virtually useless for air.
— mitxela, Oct 07 2008
/Francis turbine (the most efficient turbine in the world!)/
Whoa there, big fella. Show me a Francis turbine that outperforms a Kaplan in low head, high flow situations. For that matter, show me a Francis turbine that out performs a well-designed Pelton at very high heads. Horses for courses, my exclamatory friend.
— Texticle, Oct 07 2008
"A Francis turbine would be virtually useless for air."
As long as you enjoy the premise of an air-powered calculator.
— phoenix, Oct 08 2008
Man, this new calculator I bought really blows...
— Canuck, Oct 08 2008
I'm well aware of different turbines for different situations, my original choice of it was almost sarcastic. With so little power needed, efficiency is barely noticeable.
The point of the hydroelectric calculator is to demonstrate, on a miniature scale, a green fuel source. If you wanted this to be anything more than a silly way of powering a calculator, it should be designed like a tiny but realistically shaped wind turbine. Having that on your desk would be entertaining, or at least aesthetically pleasing compared to a balloon.
— mitxela, Oct 08 2008
[annotate]
back: main index
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en
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https://docs.relational.ai/preview/graph-analytics/overview
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Skip to content
# Overview of Graph Algorithms
Rel’s Graph Library implements a wide range of algorithms for common graph analytics tasks.
## Centrality
Centrality algorithms assign ranks to nodes in a graph, often for the purpose of measuring a node’s influence or importance in the graph. The Graph Library implements three centrality algorithms:
RelationDescription
`eigenvector_centrality`Measures a node’s importance in such a way that connections to more important nodes contribute more to a node’s score than connections to less important nodes.
`pagerank`Measures a node’s importance in a graph. `pagerank` is similar to `eigenvector_centrality`, but with an additional scaling factor.
`degree_centrality`Measures a node’s importance based on its degree. Unlike `pagerank` and `eigenvector_centrality`, `degree_centrality` does not consider the importance of a node’s neighbors when ranking a node.
## Similarity
Similarity algorithms are used to cluster nodes and predict links between nodes. The Graph Library implements a number of algorithms related to similarity:
RelationDescription
`jaccard_similarity`Measures the similarity of two nodes based on the number of neighbors common to both nodes. Values range from 0 to 1, inclusive.
`cosine_similarity`Measures the similarity of two nodes as a function of the angle between vector representations of their neighborhoods. Values range from -1 to 1, inclusive.
`preferential_attachment`Computes the “closeness” of two nodes `u` and `v` as the number of neighbors of `u` times the number of neighbors of `v`. Preferential attachment is used to predict the likelihood of two nodes receiving a new link when modeling network growth. Higher scores indicate that two nodes are “closer” than lower scores.
`adamic_adar`Computes the “closeness” of two nodes by computing the inverse logarithmic sum of the degrees of neighbors common to both nodes. Like `preferential_attachment`, `adamic_adar` is used to predict the likelihood that two nodes receive a link as a network grows.
`common_neighbor`Finds common neighbors of nodes in a graph.
## Community
These algorithms are used to determine how nodes are clustered in a graph:
RelationDescription
`is_connected`Computes whether or not a graph is connected.
`weakly_connected_component`Computes the weakly connected components of a graph.
`triangle`Computes triples of nodes that form a triangle in a graph. Use `triangle` to check whether three nodes in a graph form a triangle.
`unique_triangle`Computes triples of nodes, unique up to order, that form a triangle in the graph. Use `unique_triangle` to find unique triangles containing a given node or pair of nodes.
`num_triangles`Computes the number of unique triangles contained in a graph.
`triangle_count`Computes the number of unique triangles containing each node in a graph.
`triangle_distribution`Computes the distribution of unique triangles among nodes in a graph.
`diameter_range`Estimates the diameter of a graph by giving a minimum and maximum bound.
`local_clustering_coefficient`Computes the clustering coefficient for each node in a graph.
`average_clustering_coefficient`Computes the average clustering coefficient over all nodes in a graph.
`triangle_community`Assigns community labels to nodes using the `K`-clique algorithm with `K=3`.
## Paths
The Graph Library implements the following algorithms related to paths in graphs:
RelationDescription
`shortest_path_length`Computes the length of the shortest path between nodes in a graph.
`transitive_closure`Computes the transitive closure of the edges in a graph and may be used to determine which nodes are reachable from each node in the graph.
Was this doc helpful?
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# 算法题：如何计算模数除以2的幂？
2021年4月1日15:40:56 发表评论 507 次浏览
## C ++
#include<stdio.h>
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, …
unsigned int getModulo(unsigned int n, unsigned int d)
{
return ( n & (d - 1) );
}
// Driver Code
int main()
{
unsigned int n = 6;
// d must be a power of 2
unsigned int d = 4;
printf ( "%u moduo %u is %u" , n, d, getModulo(n, d));
getchar ();
return 0;
}
## Java
// Java code for Compute modulus division by
// a power-of-2-number
class GFG {
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, static int getModulo( int n, int d)
{
return ( n & (d- 1 ) );
}
// Driver Code
public static void main(String[] args)
{
int n = 6 ;
/*d must be a power of 2*/
int d = 4 ;
System.out.println(n+ " moduo " + d +
" is " + getModulo(n, d));
}
}
// This code is contributed
// by Smitha Dinesh Semwal.
## Python3
# Python code to demonstrate
# modulus division by power of 2
# This function will
# return n % d.
# d must be one of:
# 1, 2, 4, 8, 16, 32, …
def getModulo(n, d):
return ( n & (d - 1 ) )
# Driver program to
# test above function
n = 6
#d must be a power of 2
d = 4
print (n, "moduo" , d, "is" , getModulo(n, d))
# This code is contributed by
# Smitha Dinesh Semwal
## C#
// C# code for Compute modulus
// division by a power-of-2-number
using System;
class GFG {
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, …
static uint getModulo( uint n, uint d)
{
return ( n & (d-1) );
}
// Driver code
static public void Main ()
{
uint n = 6;
uint d = 4; /*d must be a power of 2*/
Console.WriteLine( n + " moduo " + d +
" is " + getModulo(n, d));
}
}
// This code is contributed by vt_m.
## 的PHP
<?php
// This function will return n % d.
// d must be one of: 1, 2, 4, 8, 16, 32, …
function getModulo( \$n , \$d )
{
return ( \$n & ( \$d - 1) );
}
// Driver Code
\$n = 6;
// d must be a power of 2
\$d = 4;
echo \$n , " moduo" , " " , \$d , " is " , " " , getModulo( \$n , \$d );
// This code is contributed by vt_m.
?>
http://graphics.stanford.edu/~seander/bithacks.html#ModulusDivisionEasy
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# Volume of a parallelepiped with three adjacent vectors
Given a parallelepiped in $$\mathbb{R}^3$$ with the three adjacent vectors corresponding to three adjacent edges of the parallelepiped,
to find the volume, we just take any two vectors $$\vec{u},\vec{v}$$ from the three adjacent vectors and take the crossproduct $$\vec{u}\times\vec{v}$$ and then take the dot product with the other vector $$\vec{w}$$, that is, $$(\vec{u}\times\vec{v})\cdot\vec{w}$$.
And then taking the absolute value, we have the volume $$|(\vec{u}\times\vec{v})\cdot\vec{w}|$$.
The reason for the above argument is because
1. if we take any two adjacent vectors from the given three adjacent vectors, they form a base of a parallelepiped
2. and by taking the dot product with the other one, we have the volume or the negative the volume of the parallelepiped. So we take the absolute value.
Is this the correct argument?
• Parallelogram (note spelling) is two-dimensional. Do you mean parallelipiped? – Gerry Myerson Feb 10 at 4:00
• @GerryMyerson Thanks! I just corrected the question and body. – Junpyo Choi Feb 10 at 4:12
• parallelepiped You're missing the "r" (and we both had the wrong vowel between the "l" and the "p"). – Gerry Myerson Feb 10 at 4:29
• As for your question, the whole problem is, why is your statement 2 true? That's what needs to be established, not just asserted. – Gerry Myerson Feb 10 at 4:31
Enclosed volume is given by the scalar triple product and is unaffected when their order is unchanged.
$$|(\vec{u}\times\vec{v})\cdot\vec{w}|=|\vec{u}\cdot(\vec{v}\times\vec{w})|$$
When order is interchanged, the volume becomes negative.
Btw, triclinic mineral crystals form in nature with such a 3d edge geometry.
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http://crypto.stackexchange.com/questions/5725/why-is-hmk-insecure
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# Why is h(m||k) insecure?
Here is the post that explains the failure for doing h(k||m) and I understand it.
But I don't understand how h(m||k) is subjected to collison attack, or birthday attack. Please explain?
-
– CodesInChaos Dec 17 '12 at 7:37
The birthday attack can be used with every hash function. It's a simple matter of probability (see: birthday problem). However, that only means that a hash function has to generate $2n$ of output to achieve $n$ bits of security.
It's fairly obvious that $H(m||k)$ is collision-resistant provided that $H$ itself is collision-resistant, since $H(m_1||k)=H(m_2||k)$ would imply that $H(x)=H(y)$ for $x=m_1||k$ and $y=m_2||k$.
However, the big advantage of HMAC over $H(m||k)$ is that collision-resistance of the underlying hashing function is not needed. Bellare proves in New Proofs for NMAC and HMAC: Security without Collision-Resistance that HMAC is secure as long as $H$ is a pseudorandom function, which is considerably weaker than collision-resistance.
Thus, even though the once popular hashing functions $MD5$ and $SHA$ are already broken, $HMAC_{MD5}$ and $HMAC_{SHA}$ are stil considered secure.
-
OKay. I read it again. Basically, if the assumption that the hash function is not crypto-safe anymore, then we will see collision at some point. But what about H(K||X). If we use MD5, we not only experience length extension attack, but also collision attack, right? – CppLearner Dec 17 '12 at 6:17
@CppLearner A collision attack for $H(K||X)$ with secret (short) $K$ would also lead to a collision attack on $HMAC_H(K', X)$ with a similar $K'$ (just $K$ XORed with the ipad constant). For now, there is no known way to do this. – Paŭlo Ebermann Dec 17 '12 at 8:29
The MAC algorithm you describe is called "the secret suffix method" in that paper. See the following paper, which shows how to attack the secret suffix method:
The paper describes how to use internal collisions to attack the MAC algorithm you describe. They use a birthday attack to find internal collisions, and then show how this allows attacks on the secret-suffix MAC. Crucially, almost all of the workload of the attack can be done entirely offline.
In contrast, HMAC and other good MAC algorithms are designed to make those kinds of attacks much more expensive. Yes, you can do birthday attacks on HMAC, but it has to be an online attack, and the attack requires a lot of chosen plaintext -- which makes the corresponding attacks on HMAC much harder to mount in practice. This is one reason why HMAC and other modern MAC algorithms are considered strongly preferable over "the secret suffix method", and why cryptographers recommend that no one should use the secret suffix method.
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# irrational number
(redirected from Irrational numbers)
Also found in: Thesaurus, Encyclopedia.
Related to Irrational numbers: Imaginary numbers
## irrational number
n.
A real number that cannot be expressed as a ratio between two integers.
## irrational number
n
(Mathematics) any real number that cannot be expressed as the ratio of two integers, such as π
## irra′tional num′ber
n.
a number that cannot be exactly expressed as a ratio of two integers.
[1545–55]
## ir·ra·tion·al number
(ĭ-răsh′ə-nəl)
A real number that cannot be expressed as a ratio between two integers. If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without repetition. Pi and the square root of 2 (√2) are irrational numbers.
ThesaurusAntonymsRelated WordsSynonymsLegend:
Noun 1 irrational number - a real number that cannot be expressed as a rational numberirrationalreal, real number - any rational or irrational numbertranscendental number - an irrational number that is not algebraicalgebraic number - root of an algebraic equation with rational coefficients
Translations
irrationaaliluku
iracionalni broj
óræð tala
References in periodicals archive ?
Drawing on a variety of genres for examples, each chapter explains musical topics first, such as rhythm, music theory, sound, tuning and temperament, musical group theory, change ringing, 12-tone music, and modern mathematical music by Steve Reich, Peter Maxwell Davies, and Iannis Xenakis, then related math concepts like geometric series and sequences, fractions, rational and irrational numbers, and multiplication tables.
It would be appropriate to use irrational numbers in figuring this out.
These three chapters provide a quick introduction to algebra, sufficient to exhibit irrational numbers or to gain a taste of cryptography.
The Year 8 achievement standard (ACARA, 2014) includes the following: "They [students] describe rational and irrational numbers .
Since ancient times, the concept of irrational numbers (any real number that cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer) has fascinated mathematicians and students alike.
The ancient Pythagoreans tried to prevent irrational numbers from coming out of "concealment.
r [member of] Q*, where Q- is set of irrational numbers, resonance is impossible.
369): "This rule also works for irrational numbers of similar type like 2[square root to 2]/3, 3[square root to 2]/3 etc.
Niven: Irrational Numbers, MAA, John Wiley & Sons, Inc.
Nine studies consider representing and defining irrational numbers, student use of Derive software in comprehending and making sense of definite integral and area concepts, perspectives by mathematicians on teaching and learning proof, case studies from a transition-to-proof course on referential and syntactic approaches to proving, infinite iterative processes and actual infinity, teaching assistants learning how students think, the knowledge base about teaching among teachers of calculus in higher education, modeling students' conceptions, and strategies for controlling the work in mathematics textbooks for introductory calculus.
As one pupil explained irrational numbers at the front of the class, the prince was seen to grimace as he battled with the concept.
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## Posts
### Simplifying Fractions
Please watch short video on simplifying fractions: Here is another example: So, the steps are: 1. What are the factors for the numerator and denominator? 2. Are there any numbers that can be cancelled out? 3. Are there any composite numbers, if so choose only prime numbers if possible. Your answer should be simplified, meaning you can't make the fraction smaller. Do the following assignment: Loading…
### Factors January 29,2021
Please watch video on factors: Factors: What are the factors of 12? To solve this , you ask yourself what numbers can I use to multiply to equal 12? 2x6= 12, so 2, 6 are factors of 12. 3x4=12, 3,4 are factors of 12. 1x12=12, so 1, 12 are factors of 12. So, all the numbers 1,2,3,4,6,12 are all factors of 12. The key is, when you are asked to find the factors of 12, you ask yourself what numbers can I multiply with to get 12. Loading…
### Math: Equivalent Fractions: Using Equivalency Blocks- Jan.26,2010
Here is another way to show equivalency using cubes: If you look at 1/2, is 1/2 equivalent to 1/3? No, it isn't. If you look at 1/2 and 1/3, the way to check if they are equivalent using the cubes is to see if there's a line that goes through 1/2 and 1/4: How many 1/4's do you need to make 1/2? 2/4 is the answer. You can see that there is a line that goes straight through 1/2 and 2/4. You can also multiply the numerator and denominator of 1/2 =2/4. So, to get 2/4, you multiply 2 with the numerator and then denominator of 1/2. Or you can do this: Now it's time to do some equivalent fractions. Loading…
### Math: Equivalent Fractions: January 25,2021
January 25,2021 Watch the video on Equivalent Fractions, then follow by doing worksheet. Rule to find equivalent fractions: Equivalent fractions can only be made by multiplying or dividing and Whatever you do the numerator (top #), you do to the denominator ( bottom #). You can use any number you want as long as you multiply the numerator and denominator with the same number. Loading…
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# Proof of the sine angle addition identity
Sal proves the identity sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y). Created by Sal Khan.
## Want to join the conversation?
• I understand how this video proves the angle addition for sine, but not where this formula comes from to begin with, I feel like somewhere I missed a step. It seems like a very complex proof for such a simple concept, why can't we just add sine a + sine b directly? What is it about trig functions that makes angle additions so complicated? I felt like I was grasping all the trig identities/unit circle definitions, etc up to this point but just crashed & burned here... The video is very clear but it seems like there should be some sort of introductory video to the concept of adding angles & why we can't do it more easily... Maybe it's just me?
• The proof is where the formula comes from. We can't do sin(a + b) = sin(a) + sin(b) because sine does not distribute. It's similar to x^2: (a + b)^2 isn't a^2 + b^2, it's a^2 + 2ab + b^2. The same thing applies to sin(a + b): sin(a + b) = sin(a)cos(b) + cos(a)sin(b).
• It seems that this (very nice) proof only covers the case where x, y, and (x+y) are all acute angles. How would one generalize this to cover all x and y?
• This is a good question. Here's a proof I just came up with that the angle addition formula for sin() applies to angles in the second quadrant:
Given: pi/2 < a < pi and pi/2 < b < pi // a and b are obtuse angles less than 180°.
Define: c = a - pi/2 and d = b - pi/2 // c and d are acute angles.
Theorem: sin(c + d) = sin(c)*cos(d) + cos(c)*sin(d) // angle addition formula for sin().
Substitute: sin((a - pi/2) + (b - pi/2)) = sin(a - pi/2)*cos(b - pi/2) + cos(a - pi/2)*sin(b - pi/2)
Simplify: sin((a - pi/2) + (b - pi/2)) = sin(a + b - pi)
sin(a + b - pi) = -sin(a + b) // from unit circle
sin(a - pi/2) = -cos(a) and sin(b - pi/2) = -cos(b) // from unit circle
cos(a - pi/2) = sin(a) and cos(b - pi/2) = sin(b) // from unit circle
Substitute: -sin(a + b) = -cos(a)*sin(b) + sin(a)*(-cos(b))
Simplify and rearrange: sin(a + b) = sin(a)*cos(b) + cos(a)*sin(b) // Done
You can also use this method to generalize to angles greater than 180°, and it also works for the cos() addition formula.
• I understood all the parts of the proof, but is it okay to generalize this after proving it with only a right triangle that has a side of unit measure? Isn't there a way to prove it holds true for all angles, without taking the measure of a side as 1?
• The trigonometric functions are all based on the ratios of the sides of right triangles and for a similar triangle (which is the case for triangles with the same angles) this ratio will remain constant as you scale up or down the length of the hypotenuse.
• I tried to post this as a "mistake", but when I hit send, it just hung with the little status indicator in constant motion:
Current convention is that the end point names with a horizontal bar over them names the segment that is the geometric figure defined as the two points plus all points between them on the line that passes through the two points. The end point names with no horizontal bar over them names the segment's length, which is a number. Horizontal line names something spatial; no horizontal line names something quantitative. In the video, horizontal bars are used counter to this convention to name segment lengths.
• at why is that angle labeled 90 - y?
• We don't know what y is, but we know that those two angles add up to 90. That means the other one has to be y less than 90.
y + (that angle) = 90
(that angle) = 90 - y
• Very clear and detailed proof, but where does this diagram even come from? I feel like I need to know where there diagram is derived from as well.
• Hello Omar,
If you have a protractor I encourage you to construct the triangles to show the sum of SIN(30 degrees) plus SIN(45 degrees).
Tips: Let the length of the upper triangle's hypotenuse equal one. Observe that this is the only length that equals one. Take your time to do the procedure even if you need start over multiple times like I did. It's worth the effort and you will see how the triangles came to be. You will also have a nice drawing you can hang on you wall because you will "own" the identity.
Regards,
APD
• sin(∠ABC−60∘) , im confused whether I should be subtracting or adding the two because I'm subtracting -60∘ instead of adding 60∘ : sin(theta)*cos(60) + cos(theta)*sin(60)
or: sin(theta)*cos(60) - cos(theta)*sin(60)
Do i always add sinacosb and cosasinb regardless of positive or negative value of 'b' in sin(a+b)??
Also what about for cos(a+b)? ( Where the value of b is negative)
Sorry if this doesn't make any sense...
Sin(a+b)=sin(a)cos(b)+cos(a)sin(b).
Cos(a+b)=cos(a)cos(b) - sin(a)sin(b).
Tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b)).
Subtraction Formulas:
Sin(a-b)=sin(a)cos(b) - cos(a)sin(b).
Cos(a-b)=cos(a)cos(b)+sin(a)sin(b).
Tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b)).
• How to prove sin0/cos0+cos0/sin0=sec0 cos ec0?
I mean i'm putting 0 because i couldnt type theater sign..
(1 vote)
• Ok, first the variable is called theta (θ).
https://en.wikipedia.org/wiki/Theta
``sin(θ)/cos(θ) + cos(θ)/sin(θ) = sec(θ)•csc(θ)sin(θ)/cos(θ) + cos(θ)/sin(θ) = 1/cos(θ)•1/sin(θ)sin^2(θ)/cos(θ) + cos(θ) = 1/cos(θ)•sin(θ)/sin(θ)sin^2(θ) + cos^2(θ) = cos(θ)/cos(θ)•sin(θ)/sin(θ)sin^2(θ) + cos^2(θ) = 1•1sin^2(θ) + cos^2(θ) = 1 QED``
• Not sure where to ask, but where can I find video/article(s) about these formulas?
sin(x) + sin(y) = 2sin((x+y)/2)*cos((x-y))/2)
sin(x) - sin(y) = 2cos((x+y)/2)*sin((x-y))/2)
cos(x) + cos(y) = 2cos((x+y)/2)*cos((x-y))/2)
cos(x) - cos(y) = -2sin((x+y)/2)*sin((x-y))/2)
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Topic: X/O Grid
Replies: 2 Last Post: Dec 23, 2012 3:16 AM
Messages: [ Previous | Next ]
William Elliot Posts: 2,637 Registered: 1/8/12
Re: X/O Grid
Posted: Dec 23, 2012 3:16 AM
On Fri, 21 Dec 2012, James Waldby wrote:
> On Tue, 18 Dec 2012 22:32:43 -0800, William Elliot wrote:
>
> > Consider an n by m grid of x's and o's.
n is width, m is height.
> > An x-path from the bottom squares of the grid to the top squares of
> > the grid constitutes a sequence of horizontally or vertically adjacent
> > x-squares from some bottom x-square to some top x-square. Similar
> > with o-paths.
> >
> > An exit is a top square that is connected to a bottom square by
> > either an x-path or an o-path. For example, in the 8 by 3, x/o grid
> >
> > x x o o x x o o x
> > x o o x o x x x x
> > o o x x x o x o x
> >
> > has two o-exits and three x-exits for a total of five exits.
> >
> > In an n by m, x/o grid, what is the expected number of exits?
>
> <http://pat7.com/jw/pathcounts/sample-results> for more lines).
>
> m,n: 3 2 Total: 54 Avg.: 0.84375
> m,n: 3 3 Total: 1194 Avg.: 2.33203125
> m,n: 3 4 Total: 18306 Avg.: 4.46923828125
> m,n: 3 5 Total: 231634 Avg.: 7.06890869141
> m,n: 3 6 Total: 2614194 Avg.: 9.97235870361
>
Yes, as the width n increases, the expected number of exits will incease.
> The other program, expCount.py, computes data like the following
> via a few milliseconds of work.
>
> p= 0.5
> m n= 2 3 4 5 6 7 8
> 3 0.8438 2.3320 4.4692 7.0689 9.9724 13.0664 16.2766
> 4 0.6328 2.1489 4.6596 7.9910 11.9236 16.2681 20.8833
> 5 0.4746 1.9803 4.8587 9.0363 14.2647 20.2720 26.8264
> 6 0.3560 1.8249 5.0666 10.2193 17.0690 25.2702 34.4792
> 7 0.2670 1.6817 5.2834 11.5575 20.4259 31.5051 44.3256
> 8 0.2002 1.5498 5.5094 13.0711 24.4436 39.2805 56.9899
>
I don't see why as the height m increases, that the expected number of
exists would increase. In fact, I'd expect just the opposite.
> Note that the paths considered by the two programs only continue
> (from a given square) by going left, right, or upward to a
> not-yet-traversed square. That is, they count the following grid
> as having no paths of either kind:
>
> 0 0 0 0 1
> 1 1 1 0 1
> 1 0 1 1 1
> 1 0 0 0 0
Oh oh. That grid has one exit. Note that the grid
1 1 1 0 1
1 0 1 1 1
1 0 0 0 0
has four exits.
Date Subject Author
12/21/12 James Waldby
12/23/12 William Elliot
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1. ## Division By Zero
How would you explain to a student in grades 6 to 8 that division by zero is not possible? Middle school students struggle with this math fact. If x is a number, then x/0 is undefined. What is the best way to make this clear to a child?
2. ## Re: Division By Zero
I'd start by looking at patterns
$\displaystyle 1 \div \frac{1}{10} = 10$
$\displaystyle 1 \div \frac{1}{100} = 100$
$\displaystyle 1 \div \frac{1}{1000} = 1000$
$\displaystyle 1 \div \frac{1}{10 000} = 10 000$
As the divisor gets closer and closer to zero the quotient just keeps increasing.
3. ## Re: Division By Zero
When I was in middle school (grade 7-9), my home tutor explained to me with the example 5 ÷ 0. He said, so... Because there's no number which if multiplied by 0 becomes 5, the result is undefined.
4. ## Re: Division By Zero
The point Monoxdifly makes is the one I would use: x= a/b means bx= a. If b= 0, then bx= 0x= 0 which is NOT equal to non-zero a.
Many texts make a distinction between "a/0" where a is non-zero and "0/0", calling the first "undefined", for the reason above, and the second "undetermined" or "indeterminate", since "0/0= x" would mean 0x= 0 which is true but for all x so that "0/0" does not determine a specific value for x.
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# Metamath Proof Explorer
## Theorem sbexi
Description: Discard class substitution in an existential quantification when substituting the quantified variable, in inference form. (Contributed by Giovanni Mascellani, 27-May-2019)
Ref Expression
Hypothesis sbexi.1 ${⊢}{A}\in \mathrm{V}$
Assertion sbexi
### Proof
Step Hyp Ref Expression
1 sbexi.1 ${⊢}{A}\in \mathrm{V}$
2 nfe1 ${⊢}Ⅎ{x}\phantom{\rule{.4em}{0ex}}\exists {x}\phantom{\rule{.4em}{0ex}}{\phi }$
3 1 2 sbcgfi
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https://getallcalculator.com/kelvin-to-rankine/
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# Kelvin to Rankine
First, we know about the conversion formula of Kelvin to Rankine. Here, the method is written below, with the help of method you will convert them. In this article let’s understand what Kevin and ranking are? And try to learn how to turn them
## Kelvin to Rankine
Convert 100 degrees Kelvin to degrees Rankine
After putting the value in the above formula we get the answer is 180 R. this way you convert Kelvin to Rankine and vice versa without any issue, but remember to pay attention to the conversion which you are doing it.
• The temperature T in degrees Rankine (°R) is equal to the temperature T in Kelvin (K) times 5/9. In many application, you have asked how to convert Kelvin into Rankine. So, we have to know about conversion. First, we explain the formula which we wrote below T(K) is the temperature in Kelvin and T(R) is the temperature in Rankine.
• The Rankine scale was proposed by Scottish Engineer and physicist William John Macquorn Rankine in 1859. It is used for scientific calculation. The Kelvin is a base unit of the International System and it is used for measuring thermodynamic temperature. Both the temperature are related by the scale factor 1R (5/9)K. The Temperature on the Kelvin scale is not harmful because William Kelvin decided to place zero on his level as the lowest temperature possible. The zero on the Kelvin scale represents the Absolute Zero.
Formula
T(k) * 5/9= T(R)
In this formula, the T(K) represents the temperature in Kelvin and T(R) represent the temperature in Rankine. To convert the given value into Rankine first one has to know about this symbol that is written in the above formula.
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# Chem Help!
posted by .
I really need help for my homework on balancing equations please show all the work including the coeffients if needed.
I need help with 3 questions,
1) NaN3 -> Na + N2
2) NO2 + O2 -> NO + 03
3) LiOH + CO2 -> Li2CO3 + H2O
• Chem Help! -
1)
Well, to make those nitrogen atoms come out, we will need the least common multiple of 2 and 3 which is 6, so try 5 nitrogen atoms on each side. That means 2 of NaN3 and 3 of N2, then work on the sodium
2 NaN3 --> ? Na + 3 N2
I guess we can all see that our question mark is really 2
2)
I have 1 Nitrogen on the left and one on the right
I have 4 Oxygens on the left and 4 on the right
That will do as is
3)
I have 1 Li left and 2 on the right so try 2 LiOh
2LiOH + CO2 -> Li2CO3 + H2O
I now have 2 H atoms left and 2 right -- ok
I now have 4 O atoms left and 4 right-- ok
I now have 1 C atom left and one right -- ok
so I am finished
• typo repair 5-->6 -
1)
Well, to make those nitrogen atoms come out, we will need the least common multiple of 2 and 3 which is 6, so try 66 nitrogen atoms on each side.
• I mean 6 -
Or maybe just 6
## Similar Questions
1. ### algebra 2
A. I need to write an equation to the linear equation..(-2,-2), (3,3). B. I need to graph the function of y=|x| on a coordinate plane. C. I need to give one similarity and one differnce for the equations and graphs in parts A and B. …
2. ### Help needed for Chem!
How can I balance the equation and place coefficients where they are needed for this question: Na + Cl2 -> NaCl please show all the correct steps for this answer because I really need help in understanding it.
4. ### Urgent for Chem!
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Can someone help me please? I have 4 equations that need to be matched up to different graphs. I need to know how to work the equations so I can match them to the graphs. 6y>15-3x 5x-3<15 3x>=2y -x y+x<=-3 If you can show
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supose 1.25 mol of N2 and 50.0g of O2 are moxied together. a)which one is the limiting reactant?
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10. ### chem
Use the balanced equation, N2 + O2 --> 2NO, to answer the following questions... A) How many grams of NO are formed from 10.0g of N2?
More Similar Questions
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Question 50
# A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as shown in the given two matrices. The columns and rows of Matrix-I we numbered from 0 to 4 and that of Matrix-II are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, for example, '0' can be represented by 30, 23, etc., and 'D' can be represented by 76, 88, etc. Similarly, you have to identify the set for the word "POND".Â
Solution
(A) :Â 00, 04, 67, 57 =Â POND
(B) :Â 23, 12, 86, 69 = OPND
(C) :Â 43, 24, 98, 95 = PPND
(D) :Â 30, 42, 55, 87 = OOND
=> Ans - (A)
• Free SSC Study Material - 18000 Questions
• 230+ SSC previous papers with solutions PDF
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# Distance Accumulation (Map Viewer Classic)
The Distance Accumulation tool calculates accumulated distance from each cell to input sources.
## Examples
Example applications include answering questions such as the following:
• What is the distance to the closest town?
• What is the cost distance to the closest road?
• What is the distance around barriers to the closest water source?
## Usage notes
The input source data can be a feature class or a raster. The feature class can be point, line or polygon.
When the input source data is a raster, the set of source cells consists of all cells in the source raster that have valid values. Cells that have NoData values are not included in the source set. The value 0 is considered a legitimate source.
When the input source data is a feature, the source locations are converted internally to a raster before performing the analysis. The resolution of the raster can be controlled with the Cell Size environment. By default, if no other rasters are specified in the tool, the resolution will be determined by the shorter of the width or height of the extent of the input feature in the input spatial reference, divided by 250.
To avoid this situation, as an intermediate step, you could rasterize the input features directly with the Convert Feature to Raster tool and set the Field parameter. Then use the resulting output as input to the particular distance tool you want to use.
If a source falls on NoData in any of the corresponding input rasters, it is ignored in the analysis, and therefore no distance from that source will be calculated.
If the input surface raster has a vertical coordinate system (VCS), the values of the surface raster are considered to be in the units of the VCS. If the input surface raster does not have a VCS and the data is projected, the surface values are considered to be in the linear units of the spatial reference. If the input surface raster does not have a VCS and the data is not projected, the surface values are considered to be in meters. The final distance accumulation result is in cost per linear unit, or in linear units if no cost is introduced.
Barriers are obstacles that must be routed around. They can be defined in two ways.
For the Choose barrier raster or feature parameter, barriers can be represented either by cells that have a valid value or by feature data that is converted to a raster. Where barriers are connected only by diagonal cells, the barriers will be thickened to make them impermeable.
Barriers are also defined by locations where NoData cells exist in the following inputs:Choose cost raster, Choose surface raster, Choose vertical raster, and Choose horizontal raster. Where NoData is connected only by diagonal cells, it will be thickened with additional NoData cells to make it an impermeable barrier.
The cost raster cannot contain values of zero since the algorithm is a multiplicative process. If your cost raster does contain values of zero, and these values represent areas of lowest cost, change those cells to a small positive value (such as 0.01) before running this tool.
The default values for the Vertical factor modifiers are the following:
```Keyword Zero Low High Slope Power Cos Sec
factor cut cut power power
angle angle
------------------------ ------ ----- ----- ----- ----- ----- -----
Binary 1.0 -30 30 ~ ~ ~ ~
Linear 1.0 -90 90 1/90 ~ ~ ~
Symmetric linear 1.0 -90 90 1/90 ~ ~ ~
Inverse linear 1.0 -45 45 -1/45 ~ ~ ~
Symmetric inverse linear 1.0 -45 45 -1/45 ~ ~ ~
Cos ~ -90 90 ~ 1.0 ~ ~
Sec ~ -90 90 ~ 1.0 ~ ~
Cos_sec ~ -90 90 ~ ~ 1.0 1.0
Sec_cos ~ -90 90 ~ ~ 1.0 1.0```
The output of the Derive Aspect tool can be used as input for the Choose horizontal raster parameter.
The default values for the Horizontal factor modifiers are the following:
```Keywords Zero factor Cut angle Slope Side value
-------------- ----------- ----------- ----- ---------
Binary 1.0 45 ~ ~
Forward 0.5 45 (fixed) ~ 1.0
Linear 0.5 181 1/90 ~
Inverse linear 2.0 180 -1/90 ~```
The characteristics of the source, or the movers from or to a source, can be controlled by specific parameters.
• Initial accumulation sets the initial cost before the movement begins.
• Maximum accumulation specifies how much cost a source can accumulate before reaching its limit.
• Multiplier to apply to costs specifies the mode of travel or magnitude at the source.
• Travel direction identifies whether the mover is starting at a source and moving to nonsource locations or starting at nonsource locations and moving back to a source.
If any of the source characteristics parameters are specified using a field, the source characteristic will be applied on a source-by-source basis, according to the information in the given field for the source data. When a keyword or a constant value is given, it will be applied to all sources.
If Initial accumulation is specified, the source locations on the output cost distance surface will be set to the Initial accumulation value; otherwise, the source locations on the output cost distance surface will be set to zero.
When no Extent environment setting is specified, the processing extent is determined in the following way:
If only the source and the barrier data are specified, the union of the inputs, expanded by two cell widths on each side, will used as the processing extent. The reason the output raster is expanded by two rows and columns is so that when the outputs can be used in Optimal Path As Raster and Optimal Path As Line the generated paths can move around the barriers. To use the extent as an implicit barrier, you must explicitly set the Extent value in the environment settings.
The processing extent will be the intersection of surface raster, cost raster, vertical raster, or horizontal raster, if specified.
The analysis Mask environment can be set to a feature class or a raster dataset. If the mask is a feature, it will be converted to a raster. The cells that have a value define the locations that are within the mask area. NoData cells define the locations that are outside the mask area and will be treated as a Barriers.
When the Cell Size or Snap Raster environment settings are not specified, and there are multiple rasters specified as inputs, Cell Size and Snap Raster environments are set based on an order of precedence: cost raster, surface raster, vertical raster, horizontal raster, source data, and barrier data.
The parameters for this tool are listed in the following table:
ParameterExplanation
Choose source raster or features
A raster or feature layer that identifies the sources to which the distance will be calculated.
If the input is a raster, it must consist of cells that have valid values (zero is a valid value) for the sources, and the remaining cells must be assigned NoData.
If the input is a feature layer, it can be point, line, or polygon.
Choose barrier raster or feature (optional)
The dataset that defines the barriers.
For a raster, the input type can be integer or float. Any cells that have a value (including zero) will be treated as a barrier. Any cells that are NoData will not be treated as a barrier.
For a feature service, the input can be point, line, or polygon.
Choose surface raster (optional)
Specifies whether the distance will be calculated using a planar (flat earth) or a geodesic (ellipsoid) method.
The values are used to calculate the actual surface distance covered when passing between cells.
Choose cost raster (optional)
A raster defining the impedance, or cost, to move planimetrically through each cell.
The value at each cell location represents the cost-per-unit distance for moving through the cell. Each cell location value is multiplied by the cell resolution while also compensating for diagonal movement to obtain the total cost of passing through the cell.
The values of the cost raster can be integer or floating point, but they cannot be negative or zero (you cannot have a negative or zero cost).
Choose vertical raster (optional)
A raster defining the z-values for each cell location.
The values are used for calculating the slope used to identify the vertical factor incurred when moving from one cell to another.
Vertical factor (optional)
The Vertical factor defines the relationship between the vertical cost factor (VF) and the vertical relative moving angle (VRMA).
This option is only available if Choose vertical raster (optional) is specified.
There are several factors with modifiers that identify a defined vertical factor graph. The graphs are used to identify the vertical factor used in calculating the total cost of moving into a neighboring cell.
In the descriptions below, the VF defines the vertical difficulty encountered in moving from one cell to the next, and the VRMA modifier identifies the slope angle between the From source cell and the To source cell.
• Binary—If the VRMA is greater than the low-cut angle and less than the high-cut angle, the VF is set to the value associated with the zero factor; otherwise, it is infinity.
• Linear—The VF is a linear function of the VRMA.
• Inverse linear—The VF is an inverse linear function of the VRMA.
• Symmetric linear—The VF is a linear function of the VRMA in either the negative or positive side of the VRMA, and the two linear functions are symmetrical with respect to the VF (y) axis.
• Symmetric inverse linear—The VF is an inverse linear function of the VRMA in either the negative or positive side of the VRMA, and the two linear functions are symmetrical with respect to the VF (y) axis.
• Cos—The VF is the cosine-based function of the VRMA.
• Sec—The VF is the secant-based function of the VRMA.
• Cos-Sec—The VF is the cosine-based function of the VRMA when the VRMA is negative and is the secant-based function of the VRMA when the VRMA is not negative.
• Sec-Cos—The VF is the secant-based function of the VRMA when the VRMA is negative and is the cosine-based function of the VRMA when the VRMA is not negative.
The modifiers to the vertical parameters are as follows:
• Zero factor—The vertical factor used when the VRMA is zero. This factor positions the y-intercept of the specified function. By definition, the zero factor is not applicable to any of the trigonometric vertical functions (Cos, Sec, Cos-Sec, or Sec-Cos). The y-intercept is defined by these functions.
• Low cut angle—The VRMA angle below which the VF will be set to infinity.
• High cut angle—The VRMA angle above which the VF will be set to infinity.
• Slope—The slope of the straight line used with the Linear and Inverse linear parameters. The slope is specified as a fraction of rise over run (for example, 45 percent slope is 1/45, which is input as 0.02222).
Choose horizontal raster (optional)
A raster defining the horizontal direction at each cell.
The values on the raster must be integers ranging from 0 to 360, with 0 degrees being north, or toward the top of the screen, and increasing clockwise. Flat areas should be given a value of -1. The values at each location will be used in conjunction with the Horizontal factor parameter to determine the horizontal cost incurred when moving from a cell to its neighbors.
Horizontal factor (optional)
The Horizontal factor defines the relationship between the horizontal cost factor and the horizontal relative moving angle.
This option is only available if Choose horizontal raster (optional) is specified.
There are several factors with modifiers that identify a defined horizontal factor graph. The graphs are used to identify the horizontal factor used in calculating the total cost of moving into a neighboring cell.
In the descriptions below, horizontal factor (HF) defines the horizontal difficulty encountered when moving from one cell to the next, and horizontal relative moving angle (HRMA) identifies the angle between the horizontal direction from a cell and the moving direction.
The definitions and parameters of these HF are as follows:
• Binary—If the HRMA is less than the cut angle, the HF is set to the value associated with the zero factor; otherwise, it is infinity.
• Forward—Only forward movement is allowed. The HRMA must be greater than or equal to 0 and less than 90 (0 <= HRMA < 90). If the HRMA is greater than 0 and less than 45 degrees, the HF for the cell is set to the value associated with the zero factor. If the HRMA is greater than or equal to 45 degrees, the side value modifier value is used. The HF for any HRMA equal to or greater than 90 degrees is set to infinity.
• Linear—The HF is a linear function of the HRMA.
• Inverse linear—The HF is an inverse linear function of the HRMA.
The modifiers to the horizontal keywords are as follows:
• Zero factor—The horizontal factor to be used when the HRMA is 0. This factor positions the y-intercept for any of the horizontal factor functions.
• Cut angle—The HRMA angle beyond which the HF will be set to infinity.
• Slope—The slope of the straight line used with the Linear and Inverse linear horizontal factor keywords. The slope is specified as a fraction of rise over run (for example, 45 percent slope is 1/45, which is input as 0.02222).
• Side Value—The HF when the HRMA is greater than or equal to 45 degrees and less than 90 degrees when the Forward horizontal factor keyword is specified.
Initial accumulation (optional)
The initial accumulative cost that will be used to begin the cost calculation.
This allows for the specification of the fixed cost associated with a source. Instead of starting at a cost of zero, the cost algorithm will begin with the value set by Initial accumulation.
The values must be zero or greater. The default is 0.
Maximum accumulation (optional)
The maximum accumulation for the traveler for a source.
The cost calculations continue for each source until the specified accumulation is reached.
The values must be greater than zero. The default accumulation is to the edge of the output raster.
Cost multiplier (optional)
The multiplier that will be applied to the cost values.
This allows for control of the mode of travel or the magnitude at a source. The greater the multiplier, the greater the cost to move through each cell.
The values must be greater than zero. The default is 1.
Travel direction (optional)
Specifies the direction of the traveler when applying horizontal and vertical factors.
From source—The horizontal factor and vertical factor will be applied beginning at the input source and travel out to the nonsource cells. This is the default.
To source—The horizontal factor and vertical factor will be applied beginning at each nonsource cell and travel back to the input source.
Specify the From source or To source keyword, which will be applied to all sources, or specify a field in the source data that contains the keywords to identify the direction of travel for each source. That field must contain the FROM_SOURCE or TO_SOURCE string.
Distance method (optional)
Specifies whether the distance will be calculated using a planar (flat earth) or a geodesic (ellipsoid) method.
• Planar—The distance calculation will be performed on a projected flat plane using a 2D Cartesian coordinate system. This is the default method.
• Geodesic—The distance calculation will be performed on the ellipsoid. Regardless of input or output projection, the results will not change.
Result distance accumulation raster name
The distance accumulation raster contains the accumulative distance for each cell from, or to, the least-cost source.
The output raster is of type float.
The name of the layer that will be created in My Content and added to the map. The default name is based on the tool name and the input layer name. If the layer already exists, you will be prompted to provide another name.
You can specify the name of a folder in My Content where the result will be saved using the Save result in drop-down box.
Result back direction raster name (optional)
The back direction raster contains the calculated direction in degrees. The direction identifies the next cell along the shortest path back to the closest source while avoiding barriers.
The range of values is from 0 degrees to 360 degrees, with 0 reserved for the source cells. Due east (right) is 90, and the values increase clockwise (180 is south, 270 is west, and 360 is north).
The output raster is of type float.
The name of the layer that will be created in My Content and added to the map. The default name is based on the tool name and the input layer name. If the layer already exists, you will be prompted to provide another name.
You can specify the name of a folder in My Content where the result will be saved using the Save result in drop-down box.
Result source direction raster name (optional)
The source direction raster identifies the direction of the least accumulated cost source cell as an azimuth in degrees.
The range of values is from 0 degrees to 360 degrees, with 0 reserved for the source cells. Due east (right) is 90, and the values increase clockwise (180 is south, 270 is west, and 360 is north).
The output raster is of type float.
The name of the layer that will be created in My Content and added to the map. The default name is based on the tool name and the input layer name. If the layer already exists, you will be prompted to provide another name.
You can specify the name of a folder in My Content where the result will be saved using the Save result in drop-down box.
Result source location raster name (optional)
The source location raster is a multiband output. The first band contains a row index, and the second band contains a column index. These indexes identify the location of the source cell that is the least accumulated cost distance away.
The name of the layer that will be created in My Content and added to the map. The default name is based on the tool name and the input layer name. If the layer already exists, you will be prompted to provide another name.
You can specify the name of a folder in My Content where the result will be saved using the Save result in drop-down box.
## Environments
Analysis environment settings are additional parameters that affect a tool's results. You can access the tool's analysis environment settings by clicking the gear icon at the top of the tool pane.
This tool honors the following Analysis Environments:
• Output coordinate system—Specifies the coordinate system of the output layer.
• Extent—Specifies the area to be used for analysis.
• Snap Raster—Adjusts the extent of the output so it matches the cell alignment of the specified snap raster layer.
• Cell size—The cell size to use in the output layer.
• Mask—Specifies a mask layer, where only the cells that fall within the mask area will be used for analysis.
• Parallel processing factor—Controls the raster processing CPU or GPU instances.
## Similar tools and raster functions
The Distance Accumulation tool calculates accumulated distance from each cell to input sources. Other tools may be useful in solving similar problems.
### Map Viewer Classic analysis tools and raster functions
Use the Distance Allocation tool or the Distance Allocation raster function if a distance allocation image service output is required.
Distance Accumulation is also available as a raster function.
### ArcGIS Pro analysis tools and raster functions
The Distance Accumulation and Distance Allocation geoprocessing tools are available in the Spatial Analyst toolbox.
The Distance Accumulation and Distance Allocation geoprocessing tools are available in the Raster Analysis toolbox.
Distance Accumulation and Distance Allocation are available as raster functions.
### ArcGIS Enterprise developers resources
If you are working in the ArcGIS REST API, use the Distance Accumulation and the Distance Allocation tasks.
If you are working in ArcGIS API for Python, use distance_accumulation and distance_allocation from the arcgis.raster.functions.gbl module.
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# ParametricPlot3D coloring based on whether function located inside/outside of another parametric function
I have two parametric shapes: an inclined ring and an elliptic cone whose vertex lies along the x axis.
ring[u_,v_,ψ_,α_]:={v Cos[u] Cos[α] Cos[ψ]-v Sin[u] Sin[α],v Cos[α] Sin[u]+v Cos[u] Cos[ψ] Sin[α],-v Cos[u] Sin[ψ]}
cone[u_,v_,Req_,obl_,r_,ψ_,α_]:={u,(Req (r-u) Cos[v] Cos[ψ Sin[α]])/r+(Req (r-u) Sin[v] Sin[ψ Sin[α]])/(r Sqrt[Cos[ψ Cos[α]]^2/(1-obl)^2+Sin[ψ Cos[α]]^2]),(Req (r-u) Cos[ψ Sin[α]] Sin[v])/(r Sqrt[Cos[ψ Cos[α]]^2/(1-obl)^2+Sin[ψ Cos[α]]^2])-(Req (r-u) Cos[v] Sin[ψ Sin[α]])/r};
when plotted, these two shapes intersect each other something like this (depending on input parameters):
Hopefully it's clear from the image that the cone intersects the ring, leaving part of the ring inside of the cone and part of the ring outside of it.
What I actually want is, instead of displaying both ring and cone, I want to color parts of the ring differently depending on whether it is inside or outside of the cone.
Your answer does not have to be specific to the messy equations I have listed above. What I want boils down to, "how do I set the colors of a parametric plot as a conditional of a different parametric equation?" I have tried all sorts of MeshShading options, but I can't seem to get it to work.
In case it matters: the arguments of ring are: the azimuthal angle u, the radial extent of the ring v, the magnitude of the tilt out of the XY plane psi, and the XY orientation of the ring's normal vector alpha. The unique arguments of cone are: the semimajor axis of the cone at x=0 Req, the distance of the vertex away from the origin r, and the oblateness of a YZ cross-section of the cone obl.
EDIT: as requested, here is the exact code to make the above figure.
Req = 60300000.;
obl = .25;
au = 1.496*10^11;
r = 10*au;
\[Psi] = 27*Pi/180.;
\[Alpha] = 0*Pi/180.;
ringstart = Req + 7000000.;
ringend = Req + 80000000;
lightcone[u_,v_,Req_,obl_,r_,\[Psi]_,\[Alpha]_]:={u,(Req (r-u) Cos[v] Cos[\[Psi] Sin[\[Alpha]]])/r+(Req (r-u) Sin[v] Sin[\[Psi] Sin[\[Alpha]]])/(r Sqrt[Cos[\[Psi] Cos[\[Alpha]]]^2/(1-obl)^2+Sin[\[Psi] Cos[\[Alpha]]]^2]),(Req (r-u) Cos[\[Psi] Sin[\[Alpha]]] Sin[v])/(r Sqrt[Cos[\[Psi] Cos[\[Alpha]]]^2/(1-obl)^2+Sin[\[Psi] Cos[\[Alpha]]]^2])-(Req (r-u) Cos[v] Sin[\[Psi] Sin[\[Alpha]]])/r};
rings[u_,v_,\[Psi]_,\[Alpha]_]:={v Cos[u] Cos[\[Alpha]] Cos[\[Psi]]-v Sin[u] Sin[\[Alpha]],v Cos[\[Alpha]] Sin[u]+v Cos[u] Cos[\[Psi]] Sin[\[Alpha]],-v Cos[u] Sin[\[Psi]]};
lightconeplot[Req_,obl_,r_,\[Psi]_,\[Alpha]_,start_,stop_]:=ParametricPlot3D[lightcone[u,v,Req,obl,r,\[Psi],\[Alpha]],{v,0,2Pi},{u,start,stop},
Mesh->None,
PlotStyle->{Blue,Opacity[0.3]},
PlotPoints->30
];
Show[{
ParametricPlot3D[rings[u, v, \[Psi], \[Alpha]], {u, 0, 2 Pi}, {v,ringstart,ringend},PlotStyle -> White, PlotPoints -> 30],
lightconeplot[Req, obl, 8*Req, \[Psi], \[Alpha], -4*Req, 4*Req]
},
ViewPoint -> 10*{0, 1, 1},
Axes -> False,
Boxed -> False
]
I am working in Mathematica 10. This code should be directly pastable.
• Can you please share all the code you are using to make that plot? May 31, 2018 at 21:38
A simpler example is that of two spheres intersecting, so I'll use that:
sphere1[u_, v_] := {Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]}
sphere2[u_, v_] := {0.8, 0, 0} + {Sin[u] Cos[v], Sin[u] Sin[v], Cos[u]}
ParametricPlot3D[{
sphere1[u, v],
sphere2[u, v]
}, {u, 0, Pi}, {v, 0, 2 Pi}]
A trick that can be used for this case and for your case is to plot the region that intersects separately from the region that doesn't intersect. Like this:
Show[
ParametricPlot3D[
sphere1[u, v], {u, 0, Pi}, {v, 0, 2 Pi},
RegionFunction -> (Norm[{0.8, 0, 0} - {#, #2, #3}] > 1 &),
Mesh -> {Subdivide[Pi, 15], Subdivide[2 Pi, 30]}
],
ParametricPlot3D[
sphere1[u, v], {u, 0, Pi}, {v, 0, 2 Pi},
RegionFunction -> (Norm[{0.8, 0, 0} - {#, #2, #3}] < 1 &),
PlotStyle -> Red,
Mesh -> {Subdivide[Pi, 15], Subdivide[2 Pi, 30]}
],
PlotRange -> All
]
All that is needed is to be able to write a function that says whether a point should be considered part of the region being plotted or not.
The purpose of using the Mesh option is to make sure that the meshes in the two different regions align.
If you can't find a region function for the region you're interested in then you may resort to creating the corresponding regions (e.g. with ImplicitRegion) and then using the RegionMember function.
• Thanks for the reply, this is a huge step in the right direction. I was unaware that those functions exist. I don't think a single inequality can describe the boundary set by my intersection. How would I actually use ImplicitRegion and RegionMember in a ParametricPlot3D environment? I am only seeing them be used as setup functions for RegionPlot. May 31, 2018 at 22:21
• @ahle6481 In this particular case it would work to replace the given RegionFunction with (RegionMember[Ball[{0.8, 0, 0}, 1], {#, #2, #3}] &). This is because Ball can be used as a region. I mentioned ImplicitRegion, which can take the place of Ball, just to point out that the region functionality offers a very general solution. As long as you can write down inequalities to describe the region, it will work. May 31, 2018 at 22:48
• Apologies for still not getting it. This approach makes sense to me, but I am struggling with the syntax. With ImplicitRegion, I tried to make one of the above spheres only partially display with the option RegionFunction -> (RegionMember[ ImplicitRegion[#1^2 + #2^2 + #3^2 == 1 && #1 < 0. && #2 > 0.0, {#1, #2, #3}]] &) but I a get an error stating that it is not a boolean. What is the correct syntax? May 31, 2018 at 23:32
• First define reg = Region@ImplicitRegion[x^2 + y^2 + z^2 <= 1 && x < 0. && y > 0, {x, y, z}]; and then use (RegionMember[reg, {#, #2, #3}] &). May 31, 2018 at 23:52
• Great that worked! Note for other users: I am using Mathematica 10, which does not include Region. However, simply removing the Region function (reg=ImplicitRegion...) it worked exactly how I expected it to. Jun 1, 2018 at 0:09
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15,905,616 members
See more:
I have 2 dictionaries and one of them is nested. I would like to create another nested dictionary by checking keys and values of these two dictionaries. 'dict1' gives the connections between the key and the elements of the list:
Python
`dict1={'A': ['K', 'J'], 'C': ['A'], 'D': ['B', 'C']}`
dict2 has same keys (less or more) compared to dict1 as:
Python
`dict2={'D': {'D': '0.20','B': '0.20','C': '0.00','A': '0.06','K': '0.00','J': '0.02'},'A':{'K':'3','J':'3'}}`
For a given key in the first level and all the other keys in the second level of dict2, I would like to create a nested dictionary which checks dict1 and its list to see the connections. If an element from the list of dict1 (which matches with second level of the nested dict2) has zero value in dict2, it should checks its list connection in dict1, and do the connection with dict1 elements until it reaches to non-zero value. For instance, 'D' in dict1 has list of ['B','C']. 'B':0.20 which is non-zero in dict2 so we can do connection as ('D','B',[]). Then, we check for 'D' and 'C'. Given 'D' and 'C':'0.00', we check dict1 for 'C' and see that it has list of ['A'] which is non-zero in dict2 ('A': '0.06') so we get ('D','A',[C]). If 'A' was zero in dict2 then we should have checked dict1 to see that 'A': [['K', 'J'] then check whether 'K' and 'J' are non-zero. If they were non-zero then we should have gotten (D,K,[]) and (D,J,[]). Given these, I would like to obtain a nested dictionaries like below so that I can have the connections and list of keys which are like intermediary within my connections:
Python
`result={'D': {('D','B',[]),('D','A',['C']),('A','J',[])}'A':{('A','K',[]),('A','J',[])}}`
'result' should give me the connections and also the list that shows if I have had any element between the connected ones while creating the connections. This is how far I could get but I cant get my intended output whereas I have bigger data unlike the example here:
What I have tried:
```output = []
for target in dict2:
for key in dict1:
for i in dict1[key]:
if dict2[target][i]!='0.00':
output.append((key, i))```
thanks
Posted
Updated 31-Aug-22 4:19am
v4
Richard MacCutchan 31-Aug-22 6:46am
You can only use recursion by writing a function that calls itself. But you really need to explain what the recursion will actually do.
Member 15753358 31-Aug-22 6:52am
Actually, using recursion was my guess in order to solve this problem since I dont know how to achieve the result1 or result2.
## Solution 1
Recursion is a poor idea, particularly if you have a large set of data to process: each recursive call uses more stack space to store the return address and the local variables, and stacks are generally pretty small - only around 1MB.
So if your data set gets large, you will very quickly exhaust all the available stack space, and your app with crash.
Only use recursive methods when processing genuinely recursive data and when the sample size is guaranteed to be pretty small.
I'd start by doing the job manually on paper to get a good idea what you need to do to generate your outputs from sample inputs. When you have that sorted out, computerizing it should be relatively simple.
This may help: How to Write Code to Solve a Problem, A Beginner's Guide[^]
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#### Welcome to the Smartsheet Forum Archives
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# Lookup function in Countifs statement?
Options
✭✭✭✭
edited 10/09/17
In my sheet I have a list of projects with various statuses and I need to do summary numbers by month and reduce the number of statuses to 2 (scheduled and unscheduled). I know how to do this if I use the Lookup function to populate a column with the converted statuses, then use Countifs to get the numbers for each month.
My question is, is there a way to do this in a single formula?
Here are samples of the formulas I'm using now:
Converting a row in the Proj State column (with lots of statuses) to one of 2 statuses using a Lookup (this formula is in the ConvertedStatus column on every project row):
=LOOKUP([Proj State]12, Primary\$1:Result\$5, 2, false)
Calculating a monthly number in a range of cells using the converted status and month # - the sample is for October:
=COUNTIFS(\$Month12:\$Month62, 10, \$ConvertedStatus12:\$ConvertedStatus62, ="Unscheduled")
I'm trying to calculate the values in the yellow shaded columns in the screen shot.
Thanks!
Tags:
• ✭✭✭
Options
Try these - keeping with the October example:
Top yellow cell:
=COUNTIFS(\$Month11:\$Month17, 10, \$ConvertedStatus11:\$ConvertedStatus17, "Unscheduled")
Bottom yellow cell:
=COUNTIFS(\$Month11:\$Month17, 10, \$ConvertedStatus11:\$ConvertedStatus17, "Scheduled")
I think all you needed to do was remove the last "=" that is before "Unscheduled".
Obviously you'll need to adjust the row numbers in the above equations. See this screenshot for verification.
• ✭✭✭✭
edited 10/10/17
Options
Hmm...I looked at my post and realized it wasn't clear...
What I'm trying to do is avoid having to use the ConvertedStatus column...I'd like to do that determination on the fly by using the Lookup function in the Countifs formula.
Here's what I tried, but it gives me this error: #INCORRECT ARGUMENT SET
=COUNTIFS(\$Month12:\$Month18, 10, LOOKUP([Proj State]12:[Proj State]18, Primary1:Result5, 2, false), "Unscheduled")
• ✭✭✭
Options
Are you trying to create an IF statement that counts the number of "Scheduled" and "Unscheduled" values, which by the way would be derived from the number of times other particular values appear? If so, this nested IF statement would rely heavily on two different OR functions.
• Employee
Options
Hello,
It sounds like you're looking to have these counts happen without having to use the "ConvertedStatus" column. In order to make this possible, you would also not be using the lookup table anymore. It also requires a longer formula.
Since we know what values mean both "Scheduled" and "Unscheduled" in the "Proj State" column, you can write your COUNTIFS formulas to look for these values instead. Since there are multiple values, we would want to write a formula that says "Count rows where the month column is 10, and the Proj State column is EITHER Holding Pen OR Overdue to Start" to get the equivalent of "Unscheduled". Since a COUNTIFS formula is inherintly an AND statement, we can't just use the OR function. Instead we would created one COUNTIFS statement for each Proj State value, and add them together. This is what the formula would look like for October Unscheduled:
=COUNTIFS(CHILDREN(Month11), 10, CHILDREN([Proj State]11), "Holding Pen") + COUNTIFS(CHILDREN(Month11), 10, CHILDREN([Proj State]11), "Overdue to Start")
You'll also notice that I replaced the cell ranges with the CHILDREN function. This is possible because the rows that contain the data are indented under row 11. This range will include all new rows that are added and indented under row 11. If you'd prefer to just use a normal cell range reference like you currently are, you can replace CHILDREN(Month11) with your Month column cell range, and CHILDREN([Proj State]11) with your Proj State column cell range.
This discussion has been closed.
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What Is The Maximum Amount Transfer Through Google Pay In India?
Can I transfer 50000 through Google pay? For example in SBI UPI transaction limit per day is Rs 1,00,000, while in Bank of Baroda UPI transaction limit is Rs 50,000....Use Cashfree Payouts -to send money to any UPI ID, bank account, Paytm wallet or AmazonPay.No UPI transaction day frequency limit.. How many rupees transfer by Google pay in one day? Daily limits You try to send more than ₹1,00,000 in one day across all UPI apps. What is the per day limit of UPI? ₹1 LakhThe transaction limit per day for UPI transaction is ₹1 Lakh. The maximum limit for BHIM UPI is ₹10,000 per transaction and ₹20,000 in a 24 hour window. What is Paytm transfer limit? Every Paytm users can transfer up to Rs. 5,000 at a time, with 25,000 per month the limit. If you are a merchant, you can transfer up to Rs. 50,000; with the…
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Does A 401k Loan Reduce Your Balance?
Should I pay off 401k loan early? If you want to invest for retirement, pay back the loan and invest that money inside your 401(k).If you leave your job, the 401(k) loan needs to be paid back in full, or else taxes and penalties will apply.If you have put the funds in an IRA, they won't be available to you should you need to pay back the loan early.. Do you pay yourself back the interest on a 401k loan? Borrowing against your 401K means, you are borrowing from yourself. Unlike borrowing from a bank, the interest you pay, you pay to yourself. The amount you borrowed is no longer invested so rather than getting investment gains; your “gain” is the interest you payback. How long after paying off 401k Loan Can I borrow again? Borrowing limitations are placed on a 12-month period, even if you've paid the amount back…
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Question: Which Home Inspection Certification Is Best?
What is the best home inspection training? 6 Best Home Inspector Training Schools 2020Home Inspector TrainingBest ForAmerican Home Inspectors Training (AHIT)(Best Overall) Inspectors who want top-notch training with lifetime supportATI TrainingIndependent home inspectors who want an all-in-one platform for inspection training and professional resources4 more rows•Aug 25, 2020. What certifications should a home inspector have? An approved home inspection designation includes: CMI (Certified Master Inspector) granted by the Master Inspector Certification Boards, Inc. RHI (Registered Home Inspector) granted by CAHPI (Alberta) (Canadian Association of Home & Property Inspectors (Alberta)) dated prior to October 1, 2017. Is home inspector a good career? Home inspection is a great career choice for so many reasons including high-income potential, a stable industry, flexible schedules, and the ability to work for yourself. With over 50,000 AHIT alumni, we've seen this career improve the lives of so many of our students. Can you perform your own…
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Question: Is HDFC Home Loan Mandatory?
What is the eligibility for home loan in HDFC? Home Loan Eligibility Criteria Age Limit for Salaried Individuals: 21 to 65 years .Age Limit for Self-Employed Individuals: 21 to 65 years.Minimum Salary: ₹10,000 p.m.Minimum business income: ₹2 lac p.a.. Which bank is best for home loan? These 10 banks are offering the lowest home loan interest rates for salaried individuals.BANK NAMERLLRMinimum Interest Rate (%)Union Bank of India6.806.85Bank of India6.856.85Central Bank of India6.856.85Canara Bank6.906.906 more rows•Oct 23, 2020 Is HDFC home loan processing fee refundable? Fee refundable if loan not sanctioned/disbursed Processing Fee paid by the Customer for availing the loan is non-refundable. ... submit all relevant documents as mentioned in the Sanction Letter/Loan Agreement. b. intimate HDFC of any change in his employment/contact details. What is HDFC home loan processing fee? For salaried individuals and self-employed professionals, HDFC charges 0.50% of the loan amount or Rs. 3,000 whichever is higher…
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Question: How Much Is A Walk In Tub Shower Combo?
How much does a Kohler walk in bathtub cost? Tub Cost Comparison by BrandBrandPrice RangeKohler\$4,000 - \$9,000American Standard\$1,500 - \$10,000Universal\$2,000 - \$7,000Premier Care\$3,000 - \$8,0001 more row. Are walk in showers a good idea? A walk-in shower increases your home's value, especially if you are conscious of the accessibility issue. It also is easier to clean and can be used in every decor style, from the soft traditional to rough industrial. Are walk in tubs safe for seniors? In general, walk in tubs are safe because they have low thresholds and grab bars to help prevent falls. Built-in seats, anti-slip flooring and hand-held shower wands add to their safety. Some walk in tubs even have anti-scald technology. Are walk in tubs covered by Medicare? Unfortunately for most seniors, walk in bathtubs are not considered to be durable medical equipment by original Medicare. Thus, original Medicare will not pay for the…
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Is Gold Loan Better Than Personal Loan?
Does personal loan affect home loan eligibility? In most cases, having a personal loan won't make or break your chances of getting approved for a mortgage....And if you have time, consider working on paying down some loans and credit cards to potentially decrease your DTI.Finally, consider taking some time to increase your down payment amount.. What credit score do you need for a personal loan? 660FICO credit scores range from 300 to 850. The higher the number, the lower the perceived risk. Typically, the credit score for a personal loan that you'll want to aim for is 660 or higher. Which bank is best for gold loan? SynopsisBank / NBFCGold Loan Interest RateProcessing FeeKotak Mahindra Bank10.5% to 17%Upto 2%HDFC Bank9.50% to 17.55%1.50% + GSTBandhan Bank10.99% to 18.00%1% + GSTICICI Bank10% to 19.76%1% of loan amount25 more rows•2 days ago Which bank is best for home loans? These 10 banks are…
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Question: What Does A Typical Homeowners Policy Cover?
Is a refrigerator leak covered by homeowners insurance? Unfortunately, slow leaks (like ones caused by a toilet, ice maker, freezer, or fridge) are NOT covered by a typical homeowner's insurance policy.. What does State Farm homeowners policy cover? Unless the cause of loss is excluded in the policy, a homeowners policy provides coverage for personal liability, medical payments to others, and accidental direct physical loss to your dwelling. In addition, the policy provides coverage for your personal property for specific perils including, but not limited to: Fire. When should you use homeowners insurance? Homeowners insurance provides financial relief if a covered event damages your home, property or personal belongings. It can also pay out when you're held responsible for an accident or injury. It has three main functions: Repair your house, yard and other structures. What is excluded in a homeowners policy? The standard HO-3 policy contains these exclusions: Ordinance…
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What Is The Difference Between PTSD And Moral Injury?
What might injure or harm or damage your conscience? Moral injury is the damage done to one's conscience or moral compass when that person perpetrates, witnesses, or fails to prevent acts that transgress one's own moral beliefs, values, or ethical codes of conduct.. What is moral distress? Moral distress is the emotional state that arises from a situation when a nurse feels that the ethically correct action to take is different from what he or she is tasked with doing. When policies or procedures prevent a nurse from doing what he or she thinks is right, that presents a moral dilemma. What does moral injury mean? Moral injury is the distressing psychological, behavioral, social, and sometimes spiritual aftermath of exposure to such events (3). A moral injury can occur in response to acting or witnessing behaviors that go against an individual's values and moral beliefs. Did PTSD exist in ancient…
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Which Type Of Home Loan Is Best?
What is a good mortgage rate right now? Current Mortgage and Refinance RatesProductInterest RateAPRConforming and Government Loans30-Year Fixed Rate2.625%2.726%30-Year Fixed-Rate VA2.25%2.455%20-Year Fixed Rate2.5%2.671%6 more rows. What is a 10 over 30 mortgage? It provides you the security of an interest rate and a monthly payment that is fixed for the first 10 years; then, makes available the option of paying the outstanding balance in full or elect to amortize the remaining balance over the final 20 years at our current 30-year fixed rate, but no more than 3% above your ... Which home loan is best in SBI? SBI Home Loan Interest Rates 2020SBI Home Loan SchemesInterest Rates for SalariedInterest Rates for Self EmployedSBI Smart Home Top Up Loan (Term Loan)8.50%8.55%SBI Smart Home Top Up Loan (Overdraft)8.55%9.05%Insta Home Top Up Loan8.20%8.20%SBI Earnest Money Deposit (EMD)10.45% onwards–7 more rows•Oct 20, 2020 Which loan is best for home? SynopsisBANK NAMERLLRMinimum Interest Rate…
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Quick Answer: Which Bank Is Better For Home Loan?
What is current rate of interest on home loan? Current Home Loan Interest Rates in IndiaLendersMinimum Interest RateEMI/Lakh**HDFC Bank6.90%Rs.659ICICI Bank6.90%Rs.659LIC Housing Finance6.90%Rs.659Punjab & Sind Bank6.90%Rs.65925 more rows•4 days ago. Which home loan is best in SBI? SBI Home Loan Interest Rates 2020SBI Home Loan SchemesInterest Rates for SalariedInterest Rates for Self EmployedSBI Smart Home Top Up Loan (Term Loan)8.50%8.55%SBI Smart Home Top Up Loan (Overdraft)8.55%9.05%Insta Home Top Up Loan8.20%8.20%SBI Earnest Money Deposit (EMD)10.45% onwards–7 more rows•Oct 20, 2020 How much house loan can I get on 50000 salary? How much home loan can I get on my salary?Net Monthly incomeHome Loan AmountRs.30,000Rs.22,37,206Rs.40,000Rs.29,82,941Rs.50,000Rs.37,28,676Rs.70,000Rs.52,20,1461 more row What are the 4 types of loans? There are 4 main types of personal loans available, each of which has their own pros and cons.Unsecured Personal Loans. Unsecured personal loans are offered without any collateral. ... Secured Personal Loans. Secured personal loans are backed by collateral.…
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Question: How Do You Know If A Home Is FHA Approved?
Are all homes FHA approved? It's a common misconception that all properties need to be FHA approved.While that may be true with condos, all single-family homes and townhouses are eligible for FHA financing.As you may know, FHA Loans allow you to purchase a home with as little as 3.5% down and eased credit requirements.. What type of homes qualify for FHA loans? An FHA home loan can be used to buy or refinance single-family houses, two- to four-unit multifamily homes, condominiums and certain manufactured and mobile homes. Specific types of FHA loans can also be used for new construction or for renovating an existing home. What does it mean when a house is not FHA approved? The house appraises below the purchase price. A property appraisal is usually required when a person uses an FHA loan to buy a house. ... If the appraisal “comes in low” (meaning the house…
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What Are Ways To Help Veterans?
Why should we help our veterans? American Veterans are brave and heroic – and these soldiers put their lives on the line to defend the lives of their fellow citizens....They protect the freedoms we often take for granted, and uphold the principles and laws that govern us and make our nation unique.. Why are Vietnam veterans homeless? In addition to the complex set of factors influencing all homelessness – extreme shortage of affordable housing, livable income and access to health care – a large number of displaced and at-risk veterans live with lingering effects of post-traumatic stress disorder (PTSD) and substance abuse, which are compounded by a ... What are the best cities to be homeless in? Best Cities to be Homeless in AmericaKey West, Florida. The first city on our list of the best cities for homeless people in Key West, Florida. ... Austin, Texas. Weather in Austin, Texas…
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How Does A Home Equity Loan Affect Your Taxes?
Who has the lowest home equity loan rates? Best home equity loan ratesLenderLoan amountAPR RangeNavy Federal Credit Union\$10,000–\$500,000Starting at 4.99%Frost\$2,000 and up4.49%–5.64%Connexus Credit Union\$5,000 and upStarting at 4.482%Regions Bank\$10,000–\$250,0003.25%–11.625% (with autopay)6 more rows. Can you get a fixed rate on a home equity loan? With a home equity loan, you receive one lump sum and repay the loan with regular payments for the loan repayment term, usually five to 30 years. Most home equity loans offer fixed interest rates, which means your interest rate never changes, and you'll have a fixed monthly payment. Is it better to refinance or take out a home equity loan? A home equity loan may be a better option since you won't have to pay hefty refinance closing costs but you'll still receive the funds as a lump sum. ... A cash-out refinance might have a lower interest rate, but it'll take several years to…
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Question: What Is The Current Interest Rate On A Home Equity Line Of Credit?
What are the disadvantages of a home equity line of credit? Below are three disadvantages you'll want to seriously consider before you commit to a HELOC.Possible Foreclosure: When a lender grants a home equity line of credit, the borrower's home is secured as collateral....Risk of More Debt: Among the biggest problems associated with HELOCs is the potential to rack up more debt.More items.... Which bank is offering lowest interest rate on home loan? These 10 banks are offering the lowest home loan interest rates for salaried individuals.BANK NAMERLLRMinimum Interest Rate (%)ICICI Bank6.957.05Indian Bank6.807.05Kotak Mahindra Bank7.407.05Indian Overseas Bank6.857.056 more rows•Oct 23, 2020 Are mortgage rates trending up or down? However, the possibility of rates falling to 2.5 percent or lower has faded as the U.S. economy has rebounded. The average 30-year fixed mortgage rate reached an all-time low of 3.09 percent in September 2020, according to Bankrate's weekly survey of large…
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How Do I Get Out Of Credit Card Debt Without Ruining My Credit?
How much credit card debt is OK? But ideally you should never spend more than 10% of your take-home pay towards credit card debt.So, for example, if you take home \$2,500 a month, you should never pay more than \$250 a month towards your credit card bills.. How long does debt consolidation stay on your credit report? seven yearsIf the settled debt has no history of late payments—called delinquencies—the account will remain on the credit report for seven years from the date it was reported settled. How do I get out of credit card debt without hurting my credit? 3 alternatives to debt consolidation loansDebt settlement. Debt settlement could be an option if a low credit score has prevented you from securing a debt consolidation loan. ... Balance transfer credit card. A balance transfer credit card essentially puts your debt on hold. ... Rework your budget. Will credit card companies…
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Quick Answer: Will The VA Help Me Move?
How much does the VA pay for assisted living? As of December 2018, a single veteran who qualifies for A&A can receive up to \$1,881 per month, a married vet can receive up to \$2,230 per month and a surviving spouse can receive up to \$1,209 per month to pay for needed care at home, in an assisted living community, memory care or in a nursing home.. Can I lose my VA disability? VA can stop a veteran's disability benefits if it severs service connection for the veteran's disability. ... However, if VA does find that severance of service connection is warranted, it will discontinue the veteran's disability payments as the veteran will no longer be service connected for that condition. Does Va help with assisted living? Although the VA does not directly pay for assisted living nor offer its own assisted living residences, there are several ways veterans can…
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# Just what Lottery Techniques Are usually the Best? Whenever Finding Lottery Quantities Your own Method will be Almost everything
At any time ask by yourself, ‘Which lottery approaches are the greatest for my lottery?’
I know I have. In https://qq998slot.com/togel , I request myself that question all the time due to the fact, it is the most crucial question a serious lottery player can request. But, incredibly, most lottery gamers depart that concern unanswered.
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We can answer this query ourselves and it is quite simple.
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The magic formula to knowing this complete procedure is uncovered when we solution this issue, ‘Better than what?’ In other words and phrases, we have to have a reference level. After we have it, then all of our techniques are calculated in opposition to that reference. And, when playing the lottery, the reference stage is often found the identical way.
How properly would we do if we randomly selected the figures?
This is an instance. I’ll use the Mega Millions lottery, a 5/fifty six recreation, to show. If we randomly select 5 figures to perform, that signifies eight.9% of the 56 number pool. Consequently, above several drawings of the lotteries heritage, we would anticipate to get 8.nine% of the successful numbers correct on the regular. This signifies we would common .forty five proper quantities for each lottery drawing by guessing. If you randomly picked ten numbers to perform, you would typical .ninety appropriate quantities and so on.
So, right here is the great news you’ve been waiting for. Any strategy that we select that averages much better than .forty five has outperformed random variety assortment and vice versa. The methods with the maximum averages are the very best and we should give them severe thing to consider. For illustration, if you find a lottery technique that averages .fifty six profitable quantities per drawing, it is carrying out a whopping 24% far better than random number selection! I do not know of any gambler that wouldn’t like a 24% edge. I contact this lottery technique the Best Lottery Predictions technique.
The Ideal Lottery Predictions Method
1. Determine the reference position.
2. Appraise how nicely your strategies did at selecting lottery figures.
three. Are your lottery techniques far better or worse than the reference.
Warning: Cease employing any lottery method if you don’t know what its overall performance report is!
Do you know how properly your preferred lottery approach has performed?
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# Pure Prolog vs. Extralogical Prolog Pure Prolog programs can be interpreted as logical statements about what they compute. Pure Prolog programs can be.
## Presentation on theme: "Pure Prolog vs. Extralogical Prolog Pure Prolog programs can be interpreted as logical statements about what they compute. Pure Prolog programs can be."— Presentation transcript:
Pure Prolog vs. Extralogical Prolog Pure Prolog programs can be interpreted as logical statements about what they compute. Pure Prolog programs can be interpreted as logical statements about what they compute. they are self-describing! You can read them at a high-level. they are self-describing! You can read them at a high-level. term for such programs: declarative term for such programs: declarative (Well, this is an ideal... not always true!) (Well, this is an ideal... not always true!) However, real-world requirements mean we have to use non-logical operations in Prolog programs in order for them to be practical. However, real-world requirements mean we have to use non-logical operations in Prolog programs in order for them to be practical. (a) input/output: reading from user, writing to screens, file I/O,... (b) control: making execution more efficient (c) low-level operations: inspecting the construction of constants, structures,... (d) database operations: creating, deleting, altering program clauses (e) plus other advanced tools Unfortunately, these practical needs usually have no direct logical meaning like pure Prolog programs. Unfortunately, these practical needs usually have no direct logical meaning like pure Prolog programs. Their inclusion therefore ruins our logical interpretation Their inclusion therefore ruins our logical interpretation 1COSC 2P93 : Extralogical Prolog
*** Rule of Thumb *** Use declarative predicates when they are practical. Use declarative predicates when they are practical. Practical can mean... Practical can mean... It’s possible to do what is required. It’s possible to do what is required. Efficient (speed, resources). Efficient (speed, resources). Small programs. Small programs. 2COSC 2P93 : Extralogical Prolog
listing, clause After you consult your program file(s), the clauses are asserted into the program database. After you consult your program file(s), the clauses are asserted into the program database. To look at the clauses (*** after consult, not compile ***). To look at the clauses (*** after consult, not compile ***). (a) listing. - lists the whole program database (b) listing(functor/arity): lists one particular predicate functor: predicate name functor: predicate name arity (optional): number of arguments arity (optional): number of arguments eg. listing(append/3). eg. listing(append/3). listing(append). <-- acceptable if only one ‘append’ exists listing(append). <-- acceptable if only one ‘append’ exists (c) clause(H, B): unifies H with a clause head, and B with its body backtracking will let it unify with the next clause unifying with H and B backtracking will let it unify with the next clause unifying with H and B useful for grabbing clauses one after another useful for grabbing clauses one after another 3COSC 2P93 : Extralogical Prolog
clause (c) clause (cont) (c) clause (cont) eg. if our database has: member(X, [X|_]). eg. if our database has: member(X, [X|_]). member(X, [_|Y]) :- member(X, Y). member(X, [_|Y]) :- member(X, Y). ?- clause(X, Y). X = member(X,[X|_]) Y = true ; X = member(X,[_|Y]) Y= member(X,Y) ; no ?- clause(member(a,[a,b,c,d]), Y). Y = true ; Y = member(a, [b,c,d]). facts are stored as: member(X,[X|_]) :- true. facts are stored as: member(X,[X|_]) :- true. true is a builtin that always succeeds (its definition: true. ) true is a builtin that always succeeds (its definition: true. ) 4COSC 2P93 : Extralogical Prolog
Assert, retract (d) assert(X) : asserts the clause X into the program database (d) assert(X) : asserts the clause X into the program database asserta(X): asserts clause X as the first clause in its predicate asserta(X): asserts clause X as the first clause in its predicate assertz(X): asserts clause X as the lasst clause in its predicate assertz(X): asserts clause X as the lasst clause in its predicate eg. if program is: parent(tom, bob). eg. if program is: parent(tom, bob). parent(mary, bob). parent(mary, bob). ?- asserta(parent(kim, nixon)). yes. ?- assertz(parent(X,Y) :- father(X.Y)). yes. ?- listing(parent). parent(kim, nixon). parent(tom, bob). parent(tom, bob). parent(mary, bob). parent(mary, bob). parent(X,Y) :- father(X,Y). 5COSC 2P93 : Extralogical Prolog
Assert, retract (e) retract(X): finds first clause that unifies with X, and removes it (e) retract(X): finds first clause that unifies with X, and removes it ?- retract(parent(mary,X)), X = bob <-- note that it shows result of unification ?- listing(parent). parent(kim, nixon). parent(tom, bob). parent(tom, bob). parent(X,Y) :- father(X,Y). (f) abolish(functor/arity): removes entire predicate (f) abolish(functor/arity): removes entire predicate ?- abolish(parent/2). yes ?- listing(parent). yes using assert and retract, you can create & assert new facts and rules in your program, which can be executed as normal Prolog clauses using assert and retract, you can create & assert new facts and rules in your program, which can be executed as normal Prolog clauses 6COSC 2P93 : Extralogical Prolog
Assert, Retract eg. create a list of numbers from 1 to N, and assert that list for use later make_num_list(N) :- make_list(N, List), make_list(N, List), asserta(my_list(List)). asserta(my_list(List)). make_list(N, [ ]) :- N =< 0. make_list(N, [N | Rest]) :- N > 0, N > 0, N2 is N - 1, N2 is N - 1, make_list(N2, Rest). make_list(N2, Rest). Thereafter, there exists a fact: my_list([1,2,3,... ] ): Thereafter, there exists a fact: my_list([1,2,3,... ] ): ?- make_num_list(10), my_list(L). L=[1,2,3,4,5,6,7,8,9,10] 7COSC 2P93 : Extralogical Prolog
Assert/Retract Consider the following: Consider the following: ?- make_num_list(12). yes ?- listing(my_list). my_list([1,2,3,4,5,6,7,8,9,10,11,12]).yes ?- make_num_list(8). yes ?- listing(my_list). my_list([1,2,3,4,5,6,7,8]).my_list([1,2,3,4,5,6,7,8,9,10,11,12]). If we only want one my_list, we need to remove the previous one: If we only want one my_list, we need to remove the previous one: make_num_list(N) :- abolish(my_list/1), % always succeeds, even if my_list doesn’t exist abolish(my_list/1), % always succeeds, even if my_list doesn’t exist make_list(N, List), make_list(N, List), asserta(my_list(List)). asserta(my_list(List)). 8COSC 2P93 : Extralogical Prolog
Warning about assert & retract “Danger! Danger Will Robinson!” You need to be disciplined when you use assert and retract You need to be disciplined when you use assert and retract you should never use it as a regular means to pass data among predicates you should never use it as a regular means to pass data among predicates eg. in the last example, we could just as easily create the list and pass it to other predicates... eg. in the last example, we could just as easily create the list and pass it to other predicates... ?- make_list(N,L), do_something(L), do_something_else(L),... ?- make_list(N,L), do_something(L), do_something_else(L),... often assert & retract are used for very large data items which are ‘global’ in nature, and in which passing the data is too inconvenient often assert & retract are used for very large data items which are ‘global’ in nature, and in which passing the data is too inconvenient assert & retract are side effects! assert & retract are side effects! side effect: with respect to Prolog, any activity other than unifying logical variables side effect: with respect to Prolog, any activity other than unifying logical variables when you call make_num_list, a side effect is the creation of a new fact. when you call make_num_list, a side effect is the creation of a new fact. There is no way to know make_num_list will do this from looking at its arguments; you must look at its code in detail to know this There is no way to know make_num_list will do this from looking at its arguments; you must look at its code in detail to know this Very bad effect on program clarity, declarativity, etc Very bad effect on program clarity, declarativity, etc 9COSC 2P93 : Extralogical Prolog
Prolog Input & Output Writing terms: Writing terms: write(X) - writes the term X to screen write(X) - writes the term X to screen upon backtracking, write will fail upon backtracking, write will fail nl - write a newline nl - write a newline eg. simple debugging: make_list(N, [ ]) :- N =< 0, write(first),write(N),nl. make_list(N, [N | Rest]) :- N > 0, N > 0, write(second), write(N), write(second), write(N), N2 is N - 1, N2 is N - 1, make_list(N2, Rest), make_list(N2, Rest), write(second),write(‘finished make_list!’),nl. write(second),write(‘finished make_list!’),nl. Note: ‘finish make_list!’ is one constant; need single quotes! Note: ‘finish make_list!’ is one constant; need single quotes! 10COSC 2P93 : Extralogical Prolog
I/O eg. write out all the parent clauses to the screen eg. write out all the parent clauses to the screen write_parent :- clause(parent(X,Y), B), clause(parent(X,Y), B), write(parent(X,Y)), write(‘:-’), write(B), write(‘.’), nl, write(parent(X,Y)), write(‘:-’), write(B), write(‘.’), nl, fail. fail. write_parent. % when clause above finally finishes, this will let write_parent % gracefully succeed (rather than fail) % gracefully succeed (rather than fail) This is a failure-driven loop. This is a failure-driven loop. fail: always fails! (a builtin clause that isn’t defined anywhere) fail: always fails! (a builtin clause that isn’t defined anywhere) Improvements we can make: Improvements we can make: 1. facts are printed strangely: parent(bob, bill) :- true. 2. tedious to write things term by term 3. would be nice to make this more general, and not specific to parent 11COSC 2P93 : Extralogical Prolog
Improved write_clauses ?- dynamic write_list/1. % to use as example clause write_clauses(P) :- clause(P, B), clause(P, B), write_one_clause(P,B), write_one_clause(P,B), fail. fail.write_clauses(_). write_one_clause(P, true) :- write_list([P, '.', nl]). write_list([P, '.', nl]). write_one_clause(P, B) :- \+ (B == true), \+ (B == true), write_list([P, ':-', nl, tab(5), B, '.', nl]). write_list([P, ':-', nl, tab(5), B, '.', nl]).write_list([]). write_list([nl|R]) :- nl, write_list(R). write_list([tab(N)|R]) :- tab(N), write_list(R). write_list([T|R]) :- \+ (T=nl; T=tab(_)), write(T), write_list(R). 12COSC 2P93 : Extralogical Prolog
tab tab(0). tab(K) :- K > 0, K > 0, write(' '), write(' '), K2 is K-1, K2 is K-1, tab(K2). tab(K2). 13COSC 2P93 : Extralogical Prolog
Other I/O read(X) - reads next term from input stream read(X) - reads next term from input stream input terminated by ‘.’ input terminated by ‘.’ fails upon backtracking fails upon backtracking Prolog will try to unify user input with read argument; Prolog will try to unify user input with read argument; eg. p(X) :- write(“What is your name?”), read(X). get(X) - read a single character get(X) - read a single character put(X) - write a single character put(X) - write a single character...plus many more 14COSC 2P93 : Extralogical Prolog
File I/O see (X) - opens input stream to come from file ‘X’ error if X uninstantiated, or file X doesn’t exist error if X uninstantiated, or file X doesn’t exist default: ‘user’ (terminal keyboard, standard input) default: ‘user’ (terminal keyboard, standard input) seeing(X) - indicates where you are currently reading input from seen - closes input stream, resets input as ‘user’ tell(X) - opens file X as target for output stream default: ‘user’ default: ‘user’ telling(X) - where you are writing to told - closes output stream, resets to ‘user’ 15COSC 2P93 : Extralogical Prolog
Example file I/O % write_to_file(File, List): writes entire List to File, one line per entry write_to_file(File, List) :- telling(Old), % save old output stream name telling(Old), % save old output stream name tell(File),% open new output stream tell(File),% open new output stream dump_list(List), dump_list(List), told, told, tell(Old). % this resets output to previous destination tell(Old). % this resets output to previous destination write_to_file(File, _) :- write(‘unable to write to file ‘), write(File). dump_list([ ]). dump_list([X| T]) :- write(X), nl, dump_list(T). 16COSC 2P93 : Extralogical Prolog
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## air slide design calculation pdf
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# Thread: If x=e^t, what is d^2 y/dx^2 ?
1. ## If x=e^t, what is d^2 y/dx^2 ?
Given $\displaystyle x=e^t$, what is the expression for $\displaystyle \frac{d^2 y}{dx^2}$?
I need it to solve a second order differential equation.
I get $\displaystyle \frac{dy}{dx}=e^{-t} \frac{dy}{dt}$, but when i tried to differentiate the expression $\displaystyle e^{-t} \frac{dy}{dt}$ i am stuck at d/dx (dy/dt), help me out please.
Given $\displaystyle x=e^t$, what is the expression for $\displaystyle \frac{d^2 y}{dx^2}$?
I need it to solve a second order differential equation.
I get $\displaystyle \frac{dy}{dx}=e^{-t} \frac{dy}{dt}$, but when i tried to differentiate the expression $\displaystyle e^{-t} \frac{dy}{dt}$ i am stuck at d/dx (dy/dt), help me out please.
$\displaystyle \frac{d^2 y}{dx^2} = \frac{d}{dx} \left[\frac{dy}{dx} \right] = \frac{d}{dt} \left[\frac{dy}{dx} \right] \cdot \frac{dt}{dx}$.
Now note that $\displaystyle \frac{d}{dt} \left[\frac{dy}{dx} \right] = \frac{d}{dt} \left[e^{-t} \cdot \frac{dy}{dt}\right] = - e^{-t} \frac{dy}{dt} + \frac{d^2 y}{dt^2} e^{-t}$.
3. Originally Posted by mr fantastic
$\displaystyle \frac{d^2 y}{dx^2} = \frac{d}{dx} \left[\frac{dy}{dx} \right] = \frac{d}{dt} \left[\frac{dy}{dx} \right] \cdot \frac{dt}{dx}$.
Now note that $\displaystyle \frac{d}{dt} \left[\frac{dy}{dx} \right] = \frac{d}{dt} \left[e^{-t} \cdot \frac{dy}{dt}\right] = - e^{-t} \frac{dy}{dt} + \frac{d^2 y}{dt^2} e^{-t}$.
Ok, got the required form. Thanks for the helping hand
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Explore BrainMass
# SML Fund: required rate of return; should new stock be purchased?
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
An investment fund has total capital of \$500 million invested in 5 stocks:
STOCK INVESTMENT STOCK'S BETA COEFFICIENT
A \$60 million 0.5
B \$120 million 2.0
C \$80 million 4.0
D \$80 million 1.0
E \$60 million 3.0
The beta coefficient for the fund can be found as a weighted average of the fund's investments. The current risk-free rate is 6%, whereas market returns have the following estimated probability distribution for the next period:
PROBABILITY MARKET RETURN
0.1 ---------- 7%
0.2 ---------- 9%
0.4 ---------- 11%
0.2 ---------- 13%
0.1 ---------- 15%
a. What is the estimated equation for the Security Market Line (SML)?
b. Compute the fund's required rate of return for the next period.
c. Assume that Jose Croesact, the fund's president, receives a proposal for a new stock. The
investment needed to take a position in the stock is \$50 million, it would have an expected
return of 15%, and its estimated beta coefficient is 2.0. Should the new stock be purchased?
d. At what expected rate of return should the fund be indifferent to purchasing the stock?
Show all work and all steps.
#### Solution Preview
a. What is the estimated equation for the Security Market Line (SML)?
The expected market return is
E(Rm)=SUM(Pr*Rm)=.1*7+.2*9+.4*11+.2*13+.1*15= 11
And the Security Market Line is
E(Rstock)=Rf + (Rm-Rf)*Beta = 6+(11-6)*Beta = 6 + 5 Beta (%)
b. Compute the fund's required rate of return for the next period.
The fund's beta coefficient is the weighted average of the fund's investments
*However, there might be some type error in your question: the sum of ...
#### Solution Summary
The solution presents a quality narrative to understand the concepts and then shows the formulas used to compute the answers.
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Reading comprehension Arrays Worksheets for Ages 4-8
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• 4-8
• Interactive
• Arrays
Arrays of George Washington Carver’s Creations Worksheet
Remind your child of what George Washington Carver created. Can they tell you? This worksheet has arrays with his creations. Match the array to the multiplication fact and help your child to circle the product.
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Ben Franklin’s Invention Arrays Worksheet
Have your child guess some of the inventors of the popular items we use today. For instance, the light bulb was invented by Thomas Edison. Look at Ben Franklin's inventions and help your kids match the arrays to the correct multiplication fact. Check the box and circle the product.
Ben Franklin’s Invention Arrays Worksheet
Worksheet
Thomas Edison’s Invention Arrays Worksheet
Test your students' knowledge on history by asking them what Thomas Edison invented. If they get the right answer, provide more information on other inventors. Check out this printout of Edison's light bulb inventions laid out in arrays. Get them to match each array to the correct multiplication fact.
Thomas Edison’s Invention Arrays Worksheet
Worksheet
Adding with Arrays: Chocolate Bars Worksheet
Kids love chocolates! Gauge how enthused your kids get when you mention them. This worksheet is a fun exercise about chocolates - get your kids to check the correct number sentences and totals for each chocolate bar. Reward their hard work with some chocolates - extra incentive for a job well done!
Adding with Arrays: Chocolate Bars Worksheet
Worksheet
Independence Day Arrays: Rows and Columns Worksheet
Does your child know when Independence Day is? They might love the family picnics and fireworks, or they may not understand the significance of the holiday. After this worksheet, they'll comprehend it better. Help them look at the images and count the number of rows and columns in each array.
Independence Day Arrays: Rows and Columns Worksheet
Worksheet
Learning Skills
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IBPS PO/Upcoming Exams 2016-Aptitude Questions (Data Interpretation)
IBPS PO/Upcoming Exams 2016-Aptitude Questions (Data Interpretation) Set-6:
Dear Readers, Important Practice Aptitude Questions for IBPS PO and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.
Directions (Q. 1-5): Study the following graph carefully to answer the given questions.
Number of students playing Chess, Football and Tennis from different schools
1).The total number of students playing Chess and Tennis together from school A is what per cent of the total no. of students playing these two games together from school C?
a) 68 3/16
b) 62 3/13
c) 69 3/13
d) 63 3/13
e) 62 3/16
2).If the no. of students playing each game in school D is increased by 15% and the no. of students playing each game in school B is decreased by 5%, what will be the difference between the no. of students in school D and B?
a) 54
b) 218
c) 356
d) 224
e) 205
3).If out of the students playing Football from schools B, D and E, 40%, 35% and 45% respectively got selected for State level competition, what is the total no. of students who got selected for State level competition from these three schools together?
a) 346
b) 241
c) 292
d) 284
e) 268
4).The total number of students playing Tennis from all schools together is approximately what per cent of the total number of students playing Football from all schools together?
a) 84
b) 74
c) 72
d) 79
e) 70
5).From school A, out of the students playing Chess, 40% got selected for State level competition, out of which 25% further got selected for National level competition. From school E, out of the students playing Chess, 45% got selected for State level competition, out of which two-thirds further got selected for National level competition. What is the total no. of students playing Chess from these two schools who got selected for National level competition?
a) 106
b) 98
c) 112
d) 108
e) 96
Directions (Q. 6โ10): Following line-graph shows the number of boys and the number of girls admitted in a college in different years. Answer the questions given below based on this graph.
Number of girls / boys (in hundred)
6).What is the difference between the total number of boys and that of girls admitted in all eight years together?
a) 228
b) 230
c) 232
d) 234
e) 236
7).The number of girls admitted in the year 2009 and 2010 together is what percentage of the number of boys admitted in the year 2013 and 2016 together? (Answer in approximate value)
a) 52.4%
b) 54.3%
c) 56.8%
d) 58%
e) 62.4%
8).What is the approximate percentage increase in the number of girls admitted in the year 2012 and 2013?
a) 42.8%
b) 38.6%
c) 36.48%
d) 35%
e) 32%
9).In which of the following years is the percentage rise in the number of boys the maximum compared to its previous year?
a) 2010
b) 2012
c) 2013
d) 2014
e) None of these
10).The number of girls admitted in the year 2016 is what percentage more than the average number of girls admitted during the entire period of eight years?
a) 8.26%
b) 10.34%
c) 12.24%
d) 16%
e) 17.5%
1).c 2).e 3).c 4).d 5).a 6).b 7).c 8).a 9).b 10).b
Solution:
1).Required %
= Total number of students playing Chess and Tennis from School A / Total number of students playing Chess and Tennis from School C ร 100
= 140 + 220 / 200 + 320 ร 100
= 360 / 520 ร 100 = 900 / 13 = 69 3/13%
2).Required difference
= (260 ร 115 / 100 + 320 ร 115 / 100 + 160 ร 115 / 100) โ (260 ร 95 / 100 + 180 ร 95 / 100 + 240 ร 95 / 100)
= (2.60 ร 115 + 3.20 ร 115 + 1.60 ร 115) โ (2.40 ร 95 + 1.80 ร 95 + 2.60 ร 95)
= (299 + 368 + 184) โ (228 + 171 + 247)
= 851 โ 646 = 205
3).Total number of students selected for State level competition from B, D and E
= 180 ร 40 / 100 + 320 ร 35 / 100 + 240 ร 45 / 100
= 1.80 ร 40 + 3.20 ร 35 + 2.40 + 45
= 72 + 112 + 108 = 292
4).Total number of students playing Tennis from all schools together
= 140 + 260 + 320 + 160 + 180 = 1060
Total number of students playing Football from all schools together
= 360 + 180 + 240 + 320 + 240 = 1340
Required % = 1060 / 1340 ร 100
= 79.10 = 79%
5).From School A the students playing Chess selected for National level competition
= 220 ร 40 / 100 ร 25 / 100 = 22
From School E the students playing Chess selected for National level competition
= 280 ร 45 / 100 ร 2 / 3 = 84
Total number of students selected for National level competition from School A and E together
= 22 + 84 = 106
6).Difference = 4870 – 4640 = 230
7).Number of girls = 300+ 450 = 750
Number of boys = 720 + 600 = 1320
Required % = 750 / 1320 ร 100 = 56.8%
8).Girls 2012 = 560
Girls 2013 = 800
Required % = 800 – 560 / 560 ร 100 = 24000 / 560 = 42.8%
9).% rise = 600 โ 400 / 400 ร 100 = 50%
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## An explanation for the Emdrive
Beyond the boundaries of established science an avalanche of exotic ideas compete for our attention. Experts tell us that these ideas should not be permitted to take up the time of working scientists, and for the most part they are surely correct. But what about the gems in the rubble pile? By what ground-rules might we bring extraordinary new possibilities to light?
Moderators: bboyer, MGmirkin
Keith Ness
Posts: 41
Joined: Fri Jun 06, 2014 6:53 am
### An explanation for the Emdrive
This thread is to continue where I left off posting on my explanation for the Emdrive in the Electric Universe sub-forum thread below, since my explanation suggests a significant limitation to Newton’s third law, and doesn’t explicitly refer to electric universe theory:
http://www.thunderbolts.info/forum/phpB ... 42#p116242
...Shawyer also says the two end plates are spherical sections (concentric, from what I can see) a microwave resonant distance apart, and it's only the interaction between the end plates which generates the thrust, at least ideally. I still say the fact that group velocity decreases with aperture size represents a momentum transfer, but if you're going to consider just the end plates, then there's still the following to consider.
When you strike a convex surface, the impact drives the convex surface together, reinforcing and hardening the surface against the impact, and distributing the impact over the convex surface more effectively than in less convex surfaces. When you strike a concave surface, just the opposite happens. The impact pushes the impact site away from the rest of the concave surface, weakening and softening the surface against the impact, and localizing the impact to the impact site more effectively than in less concave surfaces. So again, the smaller, convex end spreads less easily and receives more momentum, and the larger, concave end spreads more easily and receives more heat.
Here's Shawyer's explanation video:
His explanation of how the large end gets hit harder than the small end but the chamber moves in the direction of the small end thoroughly does not satisfy me.
Here's a link containing a diagram of the heat distribution:
http://www.nextbigfuture.com/2015/02/mo ... ation.html
Also, it occurs to me that if you throw a racquet ball directly at a wall in a ship so that the racquet ball bounces directly back at you, then the racquet ball, by virtue of it still being in motion relative to the ship, did not dump all the momentum of the throw into the ship when it bounced off the wall, and did instead provide its own internal angular deflectors by virtue of its own bounciness/stretchiness, and will not have dumped all its momentum into the ship until it stops moving relative to the ship (and even then, its bounciness on impacts will have laterally dissipated some of the momentum given to it by the throw).
So, in order to provide an example of my explanation which hopefully makes the allocation of deflectiveness and momentum clearer, consider having your back up against the inside front wall of a spaceship in zero-g, placing a bowling ball under your feet with you directly between the ball and the wall behind you, and extending your whole body as hard as you can directly away from the wall to push the bowling ball directly towards the back wall of the ship, giving the ship a good amount of forward momentum, and the ball a good amount of rearward momentum. The bowling ball travels straight for a while down a tube, then comes to a gentle curve in the tube, gentle enough to change the direction of the bowling ball 90 degrees, heading towards the side wall of the ship, while reducing the momentum of the bowling ball by only half, and converting some of that loss of momentum into thrust into the ship’s rear wall, thus reducing but not eliminating the ship’s forward momentum. The bowling ball is then heading directly towards the side wall at about half its former speed, and it passes through a gently springing gate which takes a little more momentum out of the ball to get out of the ball’s way and then shut behind the ball again. Then the bowling ball hits a big spring on the side wall, and bounces gently back towards the gate, which has a big spring on the side the ball is heading towards, and the gate does not swing in the direction the ball is now heading. The ball then dissipates its remaining momentum harmlessly in successive lateral bounces off the gate and side wall. When the ball comes to rest, an operator picks it up and throws it to you at a speed which exactly cancels out the forward momentum the ship gained. Oh no, is the process ruined? Nope, you’ve got a bowling ball coming at you, and it is easy to see that, the more exactly you catch the ball in the reverse process of how it was thrown to you, then the more exactly the forward momentum will be restored when you catch the ball. You are then free to repeat the process, this time with the ship moving forward faster relative to the last time you did it. Of course, as before, in order that your ship does move straight forward in this example, you have another team doing the same thing at the same time in a bilaterally symmetrical chamber next to yours.
And again, as usual, if I am missing something in one or more of my examples or elsewhere, please let me know.
Keith Ness
Posts: 41
Joined: Fri Jun 06, 2014 6:53 am
### Re: An explanation for the Emdrive
I posted the above example of the bowling ball, which is essentially the same as the racquetball and fans-in-rings examples, in NASASpaceflight forums, and it got shot down. Although no clear reason was given, in retrospect I suspect it's because all I was doing in such examples was just converting ALL forward/rearward momentum of the system to lateral momentum. Oh well.
I've more recently posted my original billiard ball example (modified so the billiard balls remain fixed to the ship's movement until the cue ball stops accelerating, although I'm not sure that's necessary), in the NASASpaceflight forums, and am awaiting feedback on that...
mamuso
Posts: 25
Joined: Wed Mar 21, 2012 5:13 pm
### Re: An explanation for the Emdrive
Think about the act of throwing the ball.
Keith Ness
Posts: 41
Joined: Fri Jun 06, 2014 6:53 am
### Re: An explanation for the Emdrive
mamuso wrote:Think about the act of throwing the ball.
You mean in step 4? Okay, so he throws the ball forward, giving the ball forward momentum, and giving the ship equal and opposite rearward momentum. Then the front operator catches the ball, eliminating the forward momentum of the ball, as well as the rearward momentum the throw gave the ship. Thus the inertial frame of reference from before the throw is restored. Anything else?
And, as for this statement I made:
Although no clear reason was given, in retrospect I suspect it's because all I was doing in such examples was just converting ALL forward/rearward momentum of the system to lateral momentum.
...Well, in further retrospect, maybe I don't suspect that that is all I was doing so much as I suspect that the example too strongly left open the possibility that that is all I was doing.
Keith Ness
Posts: 41
Joined: Fri Jun 06, 2014 6:53 am
### Re: An explanation for the Emdrive
Yeah, I'm definitely on shaky ground here, with some weak considerations of momentum exchange during collision. It may well be nothing more than a mad idea.
You can see the thread here (I start on page 14):
https://forum.nasaspaceflight.com/index ... ic=41732.0
Apparently my expectation for the billiard balls is totally wrong according to their descriptions of basic classical mechanics, but I can't figure out how it is empirically wrong. Oh well, I leave it a mystery for me to solve some other time.
Keith Ness
Posts: 41
Joined: Fri Jun 06, 2014 6:53 am
### Re: An explanation for the Emdrive
After looking into basic mechanics, I've seen that the false assumption that energy and momentum must covary is where my explanation went wrong.
Well this demonstrates points I've made elsewhere about what happens when you leap before you look. But more than that, even though I had decided to take this leap without looking just for fun, once I thought I'd found something within this leap, I still managed to get really carried away with it...
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## Playdoh Math
Everyone knows that kids LOVE when they get to create something, mold something, touch something, etc. I decided to spend more time “diving” into the concept of fractions. What is a fraction? Can the students draw it? Can they represent the value/meaning in a different way, other then simply writing the numbers (which is essentially a symbol to communicate a certain amount/value)?
I keep 3 miniature cans of playdoh in each tub of supplies at every group. When they work with the playdoh, they always work with their strategically assigned shoulder parters. (I haven’t spent the money yet to have each kid have their own playdoh.) In this first lesson, I simply wanted them to show me 1/4. I explained that inside each playdoh can was one whole “unit” of playdoh. I wanted them to show me 1/4 of that whole. Some of the kids got it right away and took that whole amount of playdoh, flattened it into a workable shape i.e. square, rectangle, circle, and divided it equally into four parts. Awesome!
Some other students did not do it correctly at first, which I LOVE because it creates a perfect teaching opportunity for me and lends itself to an authentic discussion. One strategy I witnessed was students that ripped some playdoh off the “whole” and then divided that chunk into 4 parts. What I had to explain to them was that they just divided a part of a whole into fourths (basically ripped off 1/2 the playdoh, and then divided that 1/2 into 1/4 which was not accurate). Another strategy I saw was students did not divide the whole into equal parts, which is a crucial concept to master about fractions.
The second playdoh math lesson was a little different. I wanted the students to be able to show me in pictures (playdoh molds) what an equation represented. I thought this would be super easy because I began with an addition sentence: 7 + 3 = 10. The directions were to show me this equation, this value using your playdoh. I didn’t want them to use any numbers (symbols). Only a few students laid out 7 units of playdoh and 3 other units of playdoh on their desk to show 10 units in all.
The other students did some things that I found so interesting! As you can see from the picture, this group laid out the units of playdoh, but then molded an addition and equal sign and added the 10 units also. When I asked the students to count the units or pieces of playdoh on their desks, they counted the 7 + 3 + 10 for a total of 20. They realized their error.
This last picture shows a struggle that MANY of my students had! I walked over and couldn’t help, but laugh. I told these students that they weren’t showing me what this addition sentence meant in pictures. They simply just molded the symbols, or numbers! The students asked me repeatedly if they could just mold the numbers because they couldn’t see it any other way. I had to prompt multiple times to finally get them to orally explain the sentence using a story as an example, and then they figured out how to “show” what the addition sentence meant.
It was very eye opening for me and, I can’t lie, really fun!
## Teaching Narrative with Crazy Pictures
Prior to beginning our writing unit on narrative text, I like to get the creative juices flowing!
My students use a composition book for their year long journal. Mini-writing prompts, debates of the week, and any other informal writing is done in their journal. The purpose of the journal and the activities I choose are to get the students comfortable with writing. I’m not grading for spelling and/or grammar. I have found that when the students are worried about spelling, they won’t make the attempt to spell a challenging word. They’ll opt to write “sad” instead of “depressed”.
The idea is the more they write (more meaning frequently, not the amount required for a writing task), the less they dread it. BUT, as a teacher, it is our job to make it relevant, new, and exciting for them. They cannot just respond to “google-like-journal-prompt-lists” all year long. Gotta keep it interesting.
The “Funny Pic Prompt” (that’s what I call it in my lesson plans) is an activity that the kids LOVE! When I tell them I have a picture for them to glue in their journals, I immediately hear the hushed “Yes!” from the group. I find interesting pictures on the computer by typing things like “funny animals”, “awkward family photos”, and “scary animal encounters”. The chosen image is then copied and pasted onto a word doc to be printed in color. Tip: Don’t make pictures too large. I try to fit 6-10 on a page so I’m not wasting ink.
After pictures are glued in, I set my timer for 10 minutes. Students are NOT allowed to stop writing for 10 minutes! I like to give them timed task to practice time management skills. Once the timer beeps, the students do a round robin share of their stories. During this time, students are NOT reading what they wrote verbatim. That would take forever. They must sum up their story in one summary statements (another skill we practice in class).
I cannot explain how this one weekly activity has greatly improved my students’ attitudes towards writing. They don’t even realize that I’m prepping their minds for the upcoming narrative unit.
It’s the perfect segue to writing a short story.
Click below for the pictures to use in class. Have fun!
## Matter Flipbooks
We just wrapped up a unit of study on the three states of matter. As their final assessment, my students had to create these matter flipbooks. I gave the students the headings (Solid, Liquid, and Gas), a white sheet of paper (we make flipbooks alot), and some fruit loops. *I must note that I am not a fan of having the students use any food item as art, but these fruit loops were stale and three years old. I was going to throw these away.
Anyway, the instructions were as follows:
Assemble the fruit loops as particles of that particular state of matter. The idea was to see if they would glue the solid particles closely packed together, the liquid particles should be a little more spread out and so on. Under the front flap, they were to write the properties of each state of matter–all of them. And finally, the students had to find pictures of a solid, liquid, and a gas in magazines to cut out and glue in the appropriate spot.
All in all, I’d say the kids had a good time and for the most part seemed to understand the basic concepts of the three states of matter.
Success.
Warning: I’ve done this project for the past couple of years and never encountered a problem with ants, except for this year. This is why some students ended up simply drawing the particles with markers..the ants were on the attack! Be careful.
## We love MLK!
Happy Black History Month Everyone!
This post is to share the wonderful and easy project my students do for Dr. Martin Luther King Jr. Day. I found this awesome post while searching Pinterest (Follow me!) one day.
My only tip is to make a sample prior to the lesson. Sometimes I think I can model an art project for the first time in front of the class, but not this one. This one has a couple of tricky areas so be sure to complete it first so you’ll better be able to help your students. They need to follow these directions exactly. I think the end products are so cool, even the ones that were done “wrong”.
The other project is the “I Have a Dream” picture project. I basically run off copies of three thought bubbles. One bubble says…”For my school”, the other says…”For my community”, and the last bubble says…”For my world”. After we have a discussion and brainstorm some quality ideas, the students write their dreams on a piece of paper. Once I look that over and proofread it, they get permission to write on the thought bubbles (final draft). A couple days before we complete and assemble this project, I take a picture of each student holding a MLK picture looking up to the sky. I know! It’s so cheesy, but they turn out hilarious. I think the kids secretly like when I do these cheesy photo projects, at least that’s what I tell myself.
Once the pictures are developed, the students assemble them quickly and “Voila!”, an awesome project is complete.
Finally, my students all memorized a portion of Dr. Martin Luther King’s famous speech. I was a little nervous with the complexity of the speech, but my students killed it! Truly a job well done for all of them–so impressed. After all the students finished, we watched the actual footage from The March on Washington. The kids were so into it, and saying the words right along with Dr. King!
Awesome.
## My New and Improved Library
Thanks to a wonderful teacher friend, my library arrangement has been significantly enhanced!
Anyone that knows me, knows that I am a little “crazy” about my class library. I think “crazy” in a good way meaning I’m always trying to get more books for the kids, add little displays, rugs, chairs, etc. I just want to lure the students in there and, let’s be honest, it has to look nice. Books need to be updated constantly which is why I’m a major pusher of the Scholastic Book Orders. I love this company’s deals! For every student that orders online, the class gets \$3 to spend on books! What?! One order I got over \$60 in free books! If you are not already using Scholastic, I suggest you get on it.
Back to the original point, my lovely friend gave me some new shelves and displays that actually spin (ooooooh!). The library is looking sweet!
How do you build your class library?
Upcoming….just had a project funded on Donorschoose.org–8 kindles are being shipped to add to our library. Details to come.
## Stained Glass Art!
When I was teaching first grade, a very amazing teacher introduced me to this craft and I’ve done it every year since. The results are always the same. The artwork looks great and the kids LOVE doing it, regardless of their age. We usually do this during the holiday season, so that is why the pictures you will see are Christmas-y. You could run copies of any stained glass design to fit your needs. I just like my Christmas pics.
Ideally, I try to pair this assignment with an informational text article explaining stained glass windows, i.e. the history of it, the technique, most renowned pieces, etc. I also have a short poem titled “Stained Glass” that the students memorize for an oral presentation. Now, notice I said ideally….this year, I simply ran out of time. I ended up completing this activity during our holiday party. It actually worked out well, but it was purely for the fun of it–no text needed.
## Our Holiday Party!
Even though January is coming to an end, I still feel the need to share our awesome holiday party. The kids had such a blast and I’m always looking for new activities/crafts/games for the students.
Some teachers at my school do an ornament exchange or cookie exchange, but gor the past couple of years, I have conducted a Holiday Book Exchange. A letter goes home asking parents for permission to bring one, wrapped book to participate. Of course, I always have extra wrapped books for the students who are unable to bring one. It’s important that no on misses out. And, I must admit, I have some wonderful parents each year. It never fails that students that are able to bring a wrapped book bring extras for those who cannot. It’s a beautiful thing.
As far as the actual exchange activity, I put all the books in the center of my circle of students. My students (34) all pick a number out of a hat to determine what order they will select books. This is the fun part! We run this with the same rules as the always-entertaining-White Elephant. For me, this is a classic Christmas. *Helpful Hint: The first year I did this, I allowed a book to be stolen three times. This made the game last forever, and for me, it was too long. So my rule is a book can only be stolen twice and it’s “dead”–meaning belongs to the person that stole it for a second time.
What kinds of activities do you have for your class parties??
## Writing Rubrics, finally!
I don’t know about anybody else, but it took me FOREVER to create a writing rubric that worked for me. My district uses a particular writing program called “Step Up to Writing” which is really helpful, but I was not satisfied with the rubrics. I was more drawn to rubrics that assessed each of the six traits of writing separately. I feel that this is a more effective way to assess writing, so that the students can become aware of their strengths and weaknesses. In the past, when their writing was given just one letter grade, the students simply based that one grade on whether they were a “good” writer or not. But we, as teachers, know that students can have wonderful ideas, but poor conventions. Or a student might demonstrate perfect spelling, grammar, organization, etc., but lack substantive ideas. You get the picture.
This rubric has helped me give better and more effective feedback to the students. Now, they are able to choose one area to grow and feel proud of the areas in which they excel. I am loving the results!
The rubric packet is on sale right now! Hope you like it!
## Thank you, Rafe Esquith.
Code of Ethics Levels
I’ve been meaning to write about this man, his book, and the things I learned from it for a while now. There is so much in this book that it would be impossible to cover in one post. I will focus now on his interpretation of Kohlberg’s Levels of Moral Development. As I read the following explanation of these levels, I was dying to start implementing this in my classroom, and who am I kidding, my life with my own son. The following is a copy of his article from http://www.superconscious.com:
Rafe Esquith on Moral Development
Reprinted by arrangement with Viking Penguin, a member of Penguin Group (USA) Inc., from TEACH LIKE YOUR HAIR’S ON FIRE: THE METHODS AND MADNESS INSIDE ROOM 56 by Rafe Esquith.
With experience, patience, and lessons learned from failure, you can create a classroom based on trust. The students know you to be fair. You’re dependable. The kids know that with you around, they’re safe and they’re going to learn something. A classroom based on trust and devoid of fear is a fantastic place for kids to learn.
But a foundation of trust is not an end result. It is not even a middle ground; it is only a good first step. We’ve all seen this time and time again: Students do a terrific job with a fine teacher, but one day the teacher calls in sick or has to attend a meeting. A substitute takes over, and the classroom that had previously functioned so well turns into a scene from Animal House.
A classroom based on trust and devoid of fear is a fantastic place for kids to learn. But a foundation of trust is not an end result. It is not even a middle ground; it is only a good first step.
Over the years, I have tried many different ways to develop a classroom culture in which students behaved well for all the right reasons. Most teaching victories come as a result of years of difficult and painful labor – there are very few “educational eurekas,” where the light bulb blazes over your head and you know where to go. But one glorious evening it happened to me.
I had been planning lessons around my favorite book, To Kill a Mockingbird, and was reading a study guide that analyzed the novel’s characters in relation to Lawrence Kohlberg’s Six Levels of Moral Development. The Six Levels were simple, easy to understand, and, most important, perfectly applicable to teaching young people exactly what I wanted them to learn. I quickly incorporated the Six Levels into my class, and today they are the glue that holds it together. Trust is always the foundation, but the Six Levels are the building blocks that help my kids grow as both students and people.
I teach my students the Six Levels on the first day of class. I do not expect the kids to actually apply them to their own behavior immediately. Unlike simplistic approaches that tell us, “If you follow these twenty-seven rules, you too can have a successful child,” the Six Levels take a lifetime of effort. They are a beautiful road map, and I am constantly amazed at how well my students respond to them.
Level I: I Don’t Want to Get in Trouble
Most students are trained from the minute they enter school to be Level I thinkers. Practically all of their behavior is based on the fact that they want to avoid trouble. “Quiet down!” they frantically tell one another. “The teacher’s coming!” They do homework to stay out of trouble. They walk in a line to keep the teacher happy. They listen in class to stay in the good graces of the instructor. And we teachers and parents reinforce this constantly by promising them trouble if they don’t toe the line.
But is this good teaching? Level I thinking is based on fear. Eventually, we want our children to behave well not because they fear punishment, but because they believe it is right.
Level I thinking is based on fear. Eventually, we want our children to behave well not because they fear punishment, but because they believe it is right.
Level II: I Want a Reward
Eventually children begin to make decisions for reasons other than avoiding trouble. But teachers are especially guilty of reinforcing what in our class is identified as Level II thinking. We learned that if children are rewarded for good behavior, they are more likely to repeat behavior we deem acceptable. There is, of course, truth in this. Whether the reward is candy, toys, or more time for sports, a dangling carrot can be a powerful inducement for good behavior
I have visited middle school classrooms in which teachers use Level II thinking to encourage their students to finish homework. One history teacher I met pits his classes against each other in a competition to see which of them can complete the most homework. The winning class gets a prize at the end of the year. Apparently this teacher had forgotten that a knowledge of history is supposed to be the prize. When I spoke to the class that did the most homework, I learned that they were terrific at completing assignments and turning them in, but their understanding of history was shockingly limited.
Level III: I Want to Please Somebody
As they grow up, kids learn to do things to please people: “Look, Mommy, is this good?” They do the same thing with teachers, chiefly with the charismatic or popular ones. They sit up straight and behave the way we hope they’ll behave. But they do it for the wrong reasons.
When kids want to please you, it gives your ego a jolt. It’s nice to have students show you what you think of as respect, to have them jump when you say jump. But we can still do better. This is a point on which I simultaneously tease and challenge my students. Do you brush your teeth for me? Do you tie your shoes for me? Do you see how silly that sounds? And yet many children still spend their days trying to please their teachers.
I still think we can do better.
When kids want to please you, it gives your ego a jolt. It’s nice to have students show you what you think of as respect, to have them jump when you say jump. But we can still do better.
Level IV thinking is very popular these days. With so many young people behaving badly, most teachers are trained to lay down the law on the first day of class. After all, it is essential that kids know the rules. The better teachers take the time to explain the “why” of certain rules, and many creative teachers get their students involved in the creation of class standards. The theory is that kids who are involved in generating classroom rules will be more invested in following them. There is truth in this.
I have no problem with rules. Obviously, children need to learn about boundaries and behavioral expectations. But if we want our children to receive a meaningful education, do we really want them to do things because Rule 27 says they should?
I met a teacher who had an interesting way of teaching his kids to say “thank you.” One of his rules was that if the teacher gave you something – a calculator or a baseball or a candy bar – you had three seconds to acknowledge his kindness by saying “Thank you.” If you didn’t do this, the gift was immediately taken back.
And it worked. The kids said it constantly. The only problem was that they had no real appreciation for the gifts they received. They were merely following a rule. Also, the “lesson” did not carry over into other areas of the kids’ lives. One night I took those same children to see a play, and they were no more or less gracious than the other children in the theater.
Level V: I Am Considerate of Other People
Level V is rarefied air for both children and adults. If we can help kids achieve a state of empathy for the people around them, we’ve accomplished a lot.
After many years of trying to get this idea across to my students, I finally found success by introducing them to Atticus Finch and To Kill a Mockingbird. At one point in the novel, Atticus gives his daughter, Scout, a piece of advice that perfectly illustrates Level V thinking: “You never really understand a person until you consider things from his point of view . . . until you climb inside his skin and walk around in it.” Many of my students took this advice to heart and before long the idea began to snowball. During these years, I received extraordinary thank-you notes from my substitute teachers. They were amazed that my students were able to modulate their voices throughout the day. When one sub asked the class why they spoke in whispers, the kids told him they did not want to disturb the kids in the next room. Announcements were made by grateful pilots on airplanes that the Hobart Shakespeareans were on board, and planeloads of people applauded their quiet demeanor and extraordinary manners. I was happy and very proud to be their teacher.
But . . . you guessed it. I still think we can do better. I know we can do better because I’ve seen it happen.
Level VI: I Have a Personal Code of Behavior and I Follow It
The Level VI thinker is someone who knows himself. He does not base his actions on fear, or desire to please someone, or even on rules. He has his own rules.
Level VI behavior is the most difficult to attain and just as difficult to reach. This is because a personal code of behavior resides within the soul of an individual. It also includes a healthy dose of humility. This combination makes it almost impossible to model; by definition, Level VI behavior cannot be taught by saying, “Look at what I’m doing. This is how you should behave.”
A personal code of behavior resides within the soul of an individual. It also includes a healthy dose of humility. This combination makes it almost impossible to model; by definition, Level VI behavior cannot be taught by saying, “Look at what I’m doing. This is how you should behave.”
I teach my students about level VI in several ways. Since I cannot discuss my own personal codes, I try to help the kids identify them in others. There are any number of outstanding books and films in which the Level VI individual exists. It’s fun for parents and teachers to find this type of thinker – they’re all over the place once you begin looking.
If you are skeptical about trying to get kids to this level of thinking, I don’t blame you. Any teacher who is sincere and ambitious about what he does opens himself up to colossal failures and heartbreaking disappointments. But that’s what I do. It’s what all good teachers and parents do. We ask a lot of our kids and do the best we can.
A few years ago, I missed a day of school in order to speak to a group of teachers in another state. I told my class in advance and did not discuss consequences if they behaved poorly for the substitute. I did not promise any rewards if they behaved well. I told them I’d miss them and would see them the day after my talk.
When I returned, I found a note from the substitute to the effect that my students were wonderful. About an hour later, there was a knock at the door of my classroom, and a short woman came in, holding hands with her six-year-old son. Something had happened to her little boy, a first grader, the day before. Walking home from school, he had been beaten up and robbed of his backpack. While this was happening, other students, as is so often the case, only watched or continued on their way home. But a little girl who was walking by had picked him off the sidewalk, taken him to a fountain, cleaned him up and walked him home to make sure he arrived safely. The boy’s mother was going around trying to find the girl who had helped her son, to thank her.
I asked my class if anyone knew about this. Nobody knew anything. They left and continued their search. Most of the kids were speculating on which school bully had perpetrated the crime, but Brenda kept working on her math. I noticed this because Brenda hated math.
I stared at her as she hunched over her math problems in the back of the room. And for one oh-so-brief moment she looked up, unaware that I was watching her. She looked up because she had a secret and wanted to know if anyone knew it. I didn’t until our eyes met for a split second. Her eyes narrowed and she gave me a serious shake of her head that told me to mind my own business.
It was Brenda. She had helped the little boy, but her plan for anonymity had been foiled by the mother and my brief glance. The rest of the day was a blur. Brenda had reached Level VI and no one would ever know. She and I have remained very close over the years, but we have never discussed that day.
I don’t think we can do any better than this.
My only regret is that I did not read this book sooner and that I was not able to teach these “levels” at the beginning of the year. Will that really matter? Who knows, but there’s no better time to start then now.
*I made the signs shown in the picture below to serve as a reminder for me to constantly evaluate our decisions, the students and my own. I’m even going to use them to analyze characters in the literature we read. I can’t wait!
## Kid President, I Love You!
Okay, okay… I know I’m supposed to be planning right now, but I just watched another one of this kid’s videos and I cannot help, but share. If you haven’t seen his videos, do yourself a favor and find him on youtube.com. My personal favorite was one shared by our principal at a staff meeting. It’s a “Pep Talk from Me to You”. I immediately shared it with my students, and they journaled and discussed it. He leaves you with such a positive, uplifting feeling. It’s totally contagious!
Anyway, as I was “planning” I came across this: Pep Talk for Teachers and Students. You must watch. Even though he is so entertaining and cute, his message is strong and beyond his years.
Let us all reflect more often: What are we teaching the world? I, too, hope it’s to be awesome.
Thanks again, Kid President.
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https://nl.mathworks.com/matlabcentral/cody/problems/43016-find-the-next-fibonacci-number/solutions/2059219
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Cody
# Problem 43016. Find the next Fibonacci number
Solution 2059219
Submitted on 16 Dec 2019 by Nando Käslin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 6; out = 8; assert(isequal(next_fibonacci(n),out));
2 Pass
n = [12 40 50]; out = [13 55 55]; assert(isequal(next_fibonacci(n),out));
3 Pass
n = 10.^(1:5); out = [13 144 1597 10946 121393]; assert(isequal(next_fibonacci(n),out));
4 Pass
n = round(7.^(3:.5:7)); out = [377 987 2584 6765 17711 46368 121393 317811 832040]; assert(isequal(next_fibonacci(n),out));
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https://www.studypool.com/questions/292160/Applications-of-Discrete-Mathematics-and-Statistics-in-IT-computer-science-homework-help
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# Applications of Discrete Mathematics and Statistics in IT
Anonymous
Question description
High-level computer languages are created to be understood by humans. As a result, the keywords and the commands of these languages are easy to understand. Machine languages are harder to understand and operate.
For this assignment, you should assume that the memory cells at addresses F0 to F9 are in the machine described , and that it contains the hexadecimal bit patterns described in the following table:
F0 20 F1 C0 F2 30 F3 F8 F4 20 F5 00 F6 30 F7 F9 F8 FF F9 FF
1. Explain (in detail) each step of the machine cycle. Show the contents of each of the registers and each of the memory cells after the execution of the code.
Machine Described here...
Op-Code Operand Description
1 RXY LOAD the register R with the bit pattern found in the memory cell whose address is XY
2 RXY LOAD the register R with the bit XY
3 RXY STORE the bit pattern found in register R in the memory cell whose address is XY
4 0RS MOVE the bit pattern found in register R to register S
5 RST ADD the bit patterns in registers S and T as though they were two’s complement representations and leave the result in register R
6 RST ADD the bit patterns in registers S and T as though they represented values in floating-point notation and leave the result in register R
7 RST OR the bit pattern in registers S and T and place the result in register R
8 RST AND the bit patterns in register S and T and place the result in register R
9 RST Exclusive OR the bit patterns in registers S and T and place the result in register R
A R0X ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that started at the low-order end at the high-order end.
B RXY JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of execution.
C 000 HALT execution
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https://www.ukessays.com/essays/environmental-sciences/the-conservation-of-momentum-environmental-sciences-essay.php
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# The Conservation Of Momentum Environmental Sciences Essay
2900 words (12 pages) Essay
1st Jan 1970 Environmental Sciences Reference this
Tags:
Disclaimer: This work has been submitted by a university student. This is not an example of the work produced by our Essay Writing Service. You can view samples of our professional work here.
Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UKEssays.com.
The conservation of momentum was shown in three types of collisions, elastic, inelastic and explosive. By getting mass and velocities for two carts during the collision the change in momentum and kinetic energy was found. In an elastic collision of equal massess ΔP = Pf-Pi =-8.595 and ΔKE = KEf-Kei = -4.762. In an inelastic collision of equal massess ΔP = -12.989 and ΔKE = -43.14. In an explosive collision of equal massess ΔP = -448.038 and ΔKE = -118.211.
This shows that conservation of momentum is conserved in elastic and inelastic equations due to their very low change in momentum; however kinetic energy is conserved in the elastic collision but not in the inelastic collision. In an explosive collision momentum is not conserved since the two objects start at rest with no momentum and gain momentum once moving opposite.
## Introduction
Just like Newton’s laws, the conservation of momentum is a fundamental principal in physics that is integral in daily life. However unlike Newton’s laws, the conservation of momentum does not seem to be entirely intuitive. If a ball is thrown in the air some momentum seems to be loss to the air. This makes proving the conservation of momentum tricky and difficult to do in a real life setting.
To measure the conservation of momentum in the lab, two carts will be used along a frictionless track. This allows calculation to be easier since the vectors will be moving along only one axis. This way positive direction can be movement to the right while negative direction can be movement to the left. One cart will have a plunger which is ejected by a spring that will convert its potential energy to kinetic energy of the cart. This will knock the other cart and its momentum will be transferred either partially or entirely. These velocities of the two carts will be measured by a graphing device. This is shown in diagram 1.
Diagram 1.
Momentum is produced by mass and velocity, in other words:
p = mv.
It is important to point out that momentum is not conserved on an object by object basis, however it is conserved for the isolated system. This is shown in the equation:
Psystem = P1 + P2.
Therefore if momentum is conserved then the initial momentum of the entire system should equal the final momentum of the entire system. Thus this can be shown in the equation where:
Psystem, initial = Psystem, final
M1 X V1i + M2 X V2i = M1 X V1f + M2 X V2f
In the lab collisions will be shown to illustrate the conservation of momentum. In elastic collisions energy is always conserved. Unfortunately for this lab kinetic energy can be converted into heat so that energy is lost to viable measurements. If the energy is conserved, the collision is considered to be elastic, but if the energy is not conserved, then the collision is considered inelastic.
Kinetic energy is energy associated with motion where an object with mass and moving with a certain velocity the equation is:
KE = ½ m |v|2
This allows to find the loss or gain in energy of a system much like for momentum where the change in kinetic energy of a system is determined by the equation:
ΔKESYS = KEsys,final – KEsys,intial
For the two collisions stated earlier if ΔKESYS is equal to zero the collision is considered elastic, however if ΔKESYS does not equal zero then the collision is considered inelastic. There is also another type of collision that will be determined in this lab called an explosive collision. This can be considered the opposite of an inelastic collision since the energy is not conserved because the kinetic energy is transformed for potential energy to kinetic energy. These three types of collisions will be measured in the lab under differing conditions and the change in momentum and kinetic energy of the system will be calculated.
## Procedure
In the lab the momentum and kinetic energy will be calculated by measuring different velocities for the two carts at different masses. Two carts will be set along a frictionless track. As stated earlier this allows for easier calculations since it allows working only in one dimension. One of the carts used has a plunger while the other car is just a regular car. Both carts have different sides which will allow the emulation of the different collision types.
For and elastic collision the plunger cart will be placed against the side of the ramp and then set off by a small piece of wood. It will the knock the other cart and emulate a elastic collision because the carts have magnets facing each other that will help conserve energy and momentum by having the opposite sides face each other. Having magnets of opposite charge face each other help keep the collision elastic since major contact between the two carts can convert kinetic energy into heat and will be lost. This will be done in three different ways, first having equal mass carts, second having the plunger cart heavier than the regular cart, and lastly by having the plunger cart lighter than the regular cart. The velocities for these carts will be measured for the different variable for six different trails and averaged.
For the inelastic the set up will be identical except to emulate this collision the carts will have Velcro sides that will be facing each other and cause the carts to stick together once they hit each other. This will be done in three different ways similar to the elastic collision, first having equal mass carts, second having the plunger cart heavier than the regular cart, and lastly by having the plunger cart lighter than the regular cart. The velocities for these carts will be measured for the different variable for six different trails and averaged also.
For the explosive collision the two carts will be sitting next to each other. The plunger car will have its plunger faced toward the adjacent regular car so when the button is pressed the will move away from each other in opposite directions. This will only be done in two different ways, one way having the carts equal in mass and one ways have one cart heavier than the other cart. The velocities for these carts will be measured for the different variable for six different trails and averaged as well.
## Results
Table 1. Elastic Collision Data
Elastic –
Equal Mass
regular car (g) –
506.2
plunger car (g) –
503.3
## Kef= .5m1vf1 + .v5m2vf2
0.5
0
0.483
251.65
244.4946
62.9125
59.04545
0.494
0
0.482
248.6302
243.9884
61.41166
58.8012
0.574
0
0.505
288.8942
255.631
82.91264
64.54683
0.422
0
0.405
212.3926
205.011
44.81484
41.51473
ΔP = Pf-Pi
0.482
0
0.496
242.5906
251.0752
58.46433
62.26665
-8.595433333
0.516
0
0.498
259.7028
252.0876
67.00332
62.76981
ΔKE = KEf-KEi
average
250.6434
242.048
62.91988
58.15744
-4.762437183
Elastic –
Heavy Int.
regular car (g) –
506.2
plunger car (g) –
1000.9
## Kef= .5m1vf1 + .v5m2vf2
0.412
0
0.501
294.3059
237.5554
84.94838
63.52835
0.502
0
0.59
310.6885
245.6916
126.1154
88.10411
0.321
0
0.466
324.3081
244.3456
51.56687
54.96218
0.462
0
0.544
337.2292
242.4102
106.818
74.9014
ΔP = Pf-Pi
0.51
0
0.602
354.5463
242.5007
130.167
91.72445
-81.71491849
0.486
0
0.52
324.2156
242.5007
118.2043
68.43824
ΔKE = KEf-KEi
average
324.2156
242.5007
102.97
73.60979
-29.36021623
Elastic –
Light Int.
regular car (g) –
1003.8
plunger car (g) –
503.3
## Kef= .5m1vf1 + .v5m2vf2
0.563
0
0.309
468.8014
310.1742
79.76525
47.92191
0.396
0
0.243
495.1158
243.9234
39.46275
29.63669
0.697
0
0.351
523.2297
352.3338
122.2538
61.83458
0.554
0
0.296
563.0325
297.1248
77.23541
43.97447
ΔP = Pf-Pi
0.596
0
0.343
610.7959
344.3034
89.39011
59.04803
-227.7090311
0.493
0
0.278
532.195
279.0564
61.16328
38.78884
ΔKE = KEf-KEi
average
532.195
304.486
78.21177
46.86742
-31.34434946
For the elastic collision with equal masses the change in momentum and kinetic energy is every small. Where as in the other two methods the change in momentum is much larger since the masses where different then the change in kinetic energy.
Table 2. Inelastic Collision Data
Inelastic –
Equal Mass
regular car (g) –
506.2
plunger car (g) –
503.3
## Kef= .5m1vf1 + .v5m2vf2
0.622
0.292
0.297
313.0526
297.305
97.35936
43.78238
0.481
0.242
0.243
242.0873
244.8052
58.222
29.68293
0.619
0.289
0.289
311.5427
291.7455
96.42247
42.15722
0.602
0.276
0.274
302.9866
277.6096
91.19897
38.17143
ΔP = Pf-Pi
0.51
0.236
0.237
256.683
238.7482
65.45417
28.23227
-12.98885
0.502
0.248
0.249
252.6566
250.8622
63.41681
31.16993
ΔKE = KEf-KEi
average
279.8348
266.846
78.67896
35.5327
-43.14626406
Inelastic –
Heavy Int.
regular car (g) –
506.2
plunger car (g) –
1000.9
## Kei = .5m1vi1 + .v5m2vi2
0.495
0.322
0.321
319.6722
484.78
122.6228
77.96833
0.506
0.343
0.342
323.0093
516.4291
128.1332
88.48103
0.497
0.317
0.318
336.2746
478.2569
123.6157
75.8842
0.499
0.312
0.312
352.9982
470.2152
124.6126
73.35357
ΔP = Pf-Pi
0.323
0.211
0.208
367.6309
316.4795
52.21145
33.23065
115.4745216
0.486
0.31
0.308
339.917
466.1886
118.2043
72.10332
ΔKE = KEf-KEi
average
339.917
455.3916
111.5667
70.17019
-41.39646683
Inelastic –
Light Int.
regular car (g) –
1003.8
plunger car (g) –
503.3
## Kei = .5m1vi1 + .v5m2vi2
0.575
0.181
0.181
480.8526
272.7851
83.20178
24.68705
0.589
0.172
0.163
506.4235
250.187
87.30267
20.77979
0.555
0.179
0.183
534.182
273.7861
77.51449
24.87125
0.563
0.186
0.186
573.035
280.3206
79.76525
26.06982
ΔP = Pf-Pi
0.367
0.115
0.113
619.6586
171.3089
33.89449
9.736832
-289.887818
0.574
0.178
0.179
542.8304
269.2676
82.91264
24.05466
ΔKE = KEf-KEi
average
542.8304
252.9426
74.09855
21.6999
-52.3986526
For the inelastic collision the change in kinetic energy is much larger then it was in elastic collision. This holds true for the other all three methods used.
Table 3. Explosive Collision Data
Explosive –
Equal
regular car (g) –
506.2
plunger car (g) –
503.3
## Kef= .5m1vf1 + .v5m2vf2
0
0.482
0.503
0
497.2092
0
122.4709
0
0.448
0.471
0
463.8986
0
106.6245
0
0.489
0.512
0
505.2881
0
126.4901
0
0.438
0.469
0
457.8532
0
103.9089
ΔP = Pf-Pi
0
0.478
0.492
0
489.6278
0
118.7447
488.0378833
0
0.506
0.513
0
514.3504
0
131.0292
ΔKE = KEf-KEi
average
0
488.0379
0
118.2114
118.2113751
Explosive-
Unequal
regular car (g) –
506.2
plunger car (g) –
1000.9
## Kef= .5m1vf1 + .v5m2vf2
0
0.297
0.615
0
608.5803
0
139.8729
0
0.34
0.618
0
653.1376
0
154.517
0
0.292
0.619
0
605.6006
0
139.6484
0
0.307
0.633
0
627.7009
0
148.5813
ΔP = Pf-Pi
0
0.276
0.574
0
566.8072
0
121.5127
599.3574667
0
0.24
0.581
0
534.3182
0
114.2626
ΔKE = KEf-KEi
average
0
599.3575
0
136.3992
136.399151
For the explosive collision the change in momentum is much larger than in the other two collisions. There is no initial momentum for this collision since the two carts started together at rest.
## Conclusion
From momentum and the kinetic energies calculated from the formulas the different trails were averaged to find the initial and final momentum and kinetic energy for each of the eight conditions. They the change in momentum of the system was calculated for the system by subtracting the final momentum minus the initial momentum. This was then done for kinetic energy to find the change in kinetic energy by subtracting final minus initial as well. This produced different values for the different conditions.
For the elastic collision the momentum and kinetic energy are supposed to be conserved. As table 1 shows, the momentum and kinetic energy for the equal mass carts is very close to zero, much closer than for the other conditions. For the heavier plunger cart, the initial force had much more inertia and caused the lighter second car to move much further. This is opposite in the other conditions where the plunger cart was much light. It had a harder time moving the second heavier cart. The main difference for the change in momentum and kinetic energy for the two unequal mass cart conditions was due to the fact the final velocity for cart one was never measured properly. It was assumed that the velocity was zero when in fact the plunger cart moved slightly after the collision. The assumption was due to careless human error.
For the inelastic collision kinetic energy is not conserved. This is evident very much in the results for the change in kinetic energy. There is a much larger value or this change then in the elastic counterpart since the carts stick together and move as one unit. This close interaction allows for the loss of energy as heat. As for the explosive collision, the change in momentum is by far the largest. Since the system start at rest it is entirely potential energy. When the collision happened the carts move apart and become kinetic energy. Since the final momentum is subtracted by an initial momentum of zero, it is obvious why the change is so large.
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Vous êtes sur la page 1sur 6
# Chapter # 34 Magnetic Field [1]
manishkumarphysics.in
Objective - I
1. Apositively charged particle projected towards east os deflected towards north by a magnetic field. The field
may be -
(A) towards west (B) towards south (C) upward (D*) downward
( i i iln )ii(lin i +i ii ii -i i i l(ilnri ini r| ii ri ni r -
(A) l- i i (B) lii i i (C) + i i (D*) i i i
Sol. A
( )
F q V B =
W E
N
S
j
i
( )
j q i B B =
## The magnetic field may be down ward direction.
2. A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one
point. If a magnetic field is switched on in the vertical direction, the tension in the string.
(A) will increase (B) will decrease (C) will remain the same
(D*) may increase or decrease
)i(lini i i +i i ;i )ii i lrn- + i ln( -i - i -ii ini r , i i i )li )l+ l-i rni
r| l (ii lii - +iii iilnli i, ii - ni( -
(A) +^i (B) - ri^i (C) -ir^i (D*) lii - ri ni r|
Sol. D
B=B
0
j
The tension is the strong may increases or decreases.
3. Which of the following particles will describe the smallest circle when projected with the same velocity
perpendicular to a magnetic field ?
(A) electron (B) proton (C) He
+
(D*) Li
+
+iii +(n -i (^ iln l - ii i lin- +i+ (l-ii -) i( ^i -
(A) ;-i (B) i-i (C) He
+
(D*) Li
+
Sol. D
F qVB =
charge of Li
++
> charge of (He
+
, proton, electron)
4. Which of the following particles will have minimum frequency of revolution when projected with the same
velocity perpendicular to a magnetic field ?
(A*) electron (B) proton (C) He
+
(D) Li
+
+iii +(n -i (^ iln ii i + i- (-i ^ln ^i -
(A*) ;-i (B) i-i (C) He
+
(D) Li
+
Chapter # 34 Magnetic Field [2]
manishkumarphysics.in
Sol. A
2
mv
F qVB
r
= =
mV
r
qB
=
charge electron = charge of proton = charge of He
+
= charge of Li
+
But mass of electron is
Lowest.
(the electron so smallest + circle made by)
5. Which of the following particles will have minimum frequency of revolution when projected with the same
velocity perpendicular to a magnetic field ?
(A) electron (B) proton (C) He
+
(D*) Li
+
+iii +(n -i (^ iln l - i i i l-i i(l-i n- ri^i -
(A) ;-i (B) i-i (C) He
+
(D*) Li
+
Sol. D
2 r
T ....(i)
v
t
=
1
mv
r
qB
=
r m
....(ii)
v qB
=
## from eq. (i) & (ii) we get
2 m
T
qB
t
=
1 qB
f
T 2 m
= =
t
Charge of all these particles are same but mass of Li
+
is Highest.
mass , f | +
6. Acircular loop of area 1 cm
2
, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicular
to the plane of the loop. The torque on the loop due to the magnetic field is
1 -i
2
ii(ii (-iii , l- 10Aiii (ilrnri ri r, 0.1 --i ni(ni +iii - ii ri r, +iii
ii + iii r -
(A*) zero i (B) 10
-4
N-m (C) 10
2
N-m (D) 1 N-m
Sol. A
B 0.1T =
Area = 1 cm
2
Net torque on the loop due to the uniform magnetic
field is always zero.
I=10A
7. A beam consisting of protons and electrons moving at the same speed goes through a thin region in which
there is a magnetic field perpandicular to the beam. The protons and the electrons
(A) will go undeviated
(B) will be deviated by the same angle will not separate
(C*) will be deviated by different angles and hence separate
(D) will be deviated by the same angle but will separate.
-ii ^lnii; - ii nii i -i i i ) , )ii ) ii ^ni r, ri +ii i i lii +(n
r| i-i nii ;-i -
(A) l+i l(iln r) i^|
(B) -iii l(ilnri^ nii iri ri^|
(C*) lili iii l(lnri^, n iri i^|
(D) -iii l(iln ri^, ln i ri i^|
Chapter # 34 Magnetic Field [3]
manishkumarphysics.in
Sol. C
( )
F q V B =
## Charge proton is poritive = e
F
P
= evB
Charge of electron is negative = -e
F
e
= -evB
They will be deviated by different angles and Hence separate.
8. A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an
acute angle with the magnetic field. The path of the particle will be.
(A) a straight line (B) a circle
(C*) a helix with uniform pitch (D) a helix with nonuniform pitch.
) i(lin i - +iii - ^ln ni r| li ii i (^ i lii, +iii ii +ini r| i
i i ri^i -
(A) ii (B) (-i
(C*) )-i i ni (ii rl (D) -ii ni (ii rl
Sol. C
B
V
( )
F q V B qvBsin = = u
Megnetic force doesnt change the speed of the particle. It change the direction of the
velocity of the particle.
Vcosu provide the displacement of the particle in Horizontal direction & force is provide the
centripetal acceleration of the particle.
So the path of the particle will be ahelix with uniform pitch.
9. A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some
instant, the velocity of the particle is perpendiculars to the field direction. The path of the particle will be
(A) a straight line (B) a circle
(C) a helix with uniform pitch (D*) a helix with nonuniform pitch.
)i ) ii - ^lnni r, ri - +i ii r nii -iin l(nn ii r| li ii i (^i lii, ii i
lii +(n r| i i i ri^i -
(A) ii (B) (-i
(C) )-i i ni(ii rl (D*) -ii ni (ii rl
Sol. D
( )
F qE q V B = + +
F qE =
provides the acceleration in x direction.
E
B
V
( ) 2
F q V B =
provides the centripetal Force.
The path of the particle will be ahelix with nonuniform pitch.
10. An electric current i enters and leaves a uniform circular wire of radius a through diametrically opposite
points. Acharged particle q moving along the axis of the circular wire passes through its centre at speed u.
The magnetic force acting on the particle when it passes through the centre has a magnitude
)- ( -iii ni i lii a r , ;- in li- i l+i i iii ( i ni r nii +ir lni r | )i( lini
q (-iii ni i i li u i ^lnni ri ; ^ ni r| + i ^ni r ni ; ^ (i
+i+i l-ii ri^i -
Chapter # 34 Magnetic Field [4]
manishkumarphysics.in
(A) q
a
v
0
2
i
(B) q
a
v
t
0
2
i
(C) q
a
v
0
i
(D*) zero
Sol. D
Zero
Objective - II
1. If a charged particle at rest experieces no electromagnetic force,
(A*) the electric field must be zero (B) the magnetic field must be zero
(C) the electric field may or may not be zero (D*) the magnetic field may or may not be zero
ll(i-i(-ii - l-in i i; l(nn+i+ i( ri ni r -
(A*) l(nnii iri^i| (B) +iii iri^i|
(C) l(nnii iri ii ni r i ri ii| (D*) +iii iri ii ni r i ri ii|
The electric field must be zero.
The Magnetic field may or may not be zero.
2. If a charged particle kept at list expreiences an electromagnetic force,
(A*) the electric field must not be zero (B) the magnetic field must not be zero
(C) the electric field may or may not be zero (D*) the magnetic field may no may not be zero
l l(i-i(-ii - l-in)i l(nn+i+i( ni r -
(A*) l(nn ii i ri ri^i (B) +iii i ri ri^i|
(C) l(nnii iri ii ni r i ri ii| (D*) +iii iri ii ni r i ri ii|
The electric field must not be zero.
The Magnetic field may or may not be zero.
3. If a charged particle projected in a gravity-free room deflects,
(A) there must be an electric field (B) there must be a magnetic field
(C*) both field cannot be zero (D*) both fields can be nonzero
) ^-(ri- - iln li ^i i(lini l(iln rini r, ni -
(A) (ri l(nn ii ri^i| (B) (ri +iii ri^i|
(C*) ii ii iri ri n r| (D*) ii ii iri n r|
Sol. CD
4. A charged particle moves in a gravity-free space without change in veloicty. Which of the following is/are
possible ?
)i(lin i ^-(-n ii - (^ - l(n l+i ^ln ni r| l - iii i( r -
(A*) E = 0, B = 0 (B*) E = 0, B = 0 (C) E = 0, B = 0 (D*) E = 0, B = 0
Sol. ABD
Particle move with constant velocity in ay direction.
So B = 0, E = 0
Particle move in a circle with constant speed.
Magnetic force is provide the centripetal force that causes particle is move in a circle.
If qE = qvB and Magnetic & Electric force in opposite direction in this case particle
also move with uniform speed.
5. A charged particle moves alonh a circle under the action of possible constant electric and magnetic fields.
Which of the following are possible ?
)i(lin i i( )( ln l(nnnii +iiii - (-iii i ^lnni r| l - ii i( r-
(A) E = 0, B = 0 (B*) E = 0, B = 0 (C) E = 0, B = 0 (D) E = 0, B = 0
Sol. B
A charged particle moves along a circle that mean Magnetic force is provides centripetal
force that causes particle is move in a circle.
So, E = 0, B 0 =
Chapter # 34 Magnetic Field [5]
manishkumarphysics.in
6. A charged particle goes undelflected in a region containing electric and magnetic field. It is possible that
(A*)
E ||
B,
u ||
E (B*)
E is not parallel to
B
(C)
u ||
B but
E is not parallel to
B (D)
E ||
B but
u is not parallel to
E
li ii - ri l(nn)( +iii l-in r| )i(lin i l(iln r) l+i ^lnni r| r i( r l-
(A*)
E ||
B,
u ||
E (B*)
E
B -iin ri r|
(C)
u ||
B ln
E ,
B -iin ri r| (D)
E ||
B ln
u ,
E -iin ri r|
Sol. AB
V E , B E
In this case Magnetic force on the particle is zero &
V
is paralle to
E
. So charged particle
goes undeflected in a region.
E
is not parallel to
B
, But
V
is parallel to
E
.
7. If a charged particle goes unacceleration in a region containing electric and magnetic fields,
(A*) E
must be perpeandicular to B
(B*)
u
must be perpendicular to E
(C)
u
## (D) E must be equla to uB.
l )i(lin i ) ii - -(iri^ln ni r, ri l(nn )( +iii r -
(A*) E
, B
u , E
+(n ri^i|
(C)
u , B
## +(n ri^i| (D) E, uB +i+ ri^i|
Sol. AB
E B & V E
8. Two ions have equla masses but one is singly-ionized and other is douly-ionized. The are project from the
same place in a uniform magnetic field with the same veloicty perpendicular to the field.
(A) Both ions will go along circles of equal radii.
(B*) The circle described by the single-ionized charge will have a radius double that of the other
circle
(C) The two circles do not touch each other
(D*) The two circles touch each other
i ii -i )-i r, ln ))ii iln nii i l,iln r| ;i )ri -ii )-i +i
ii - )-i (^ +iii +(n iln li ini r -
(A) ii i -iliii (-iii ii ^ln^|
(B*) )ii iln i ,ii +i ^ (-iii
(C) ii (-i ) -i ri ^|
(D*) ii (-i ) i -i ^|
Sol. BD
mv
r
qB
=
If charge of singly ionized = e
Then charge of doubly ionized = ze
The circle described by the singly - ionized charge will have a radius double that of the other
circle.
The two circle touch each other because brojected from the same place.
9. An electron is moving along the positive X-axis. You want to apply a magnetic field for a short time so that the
electron may reverse its direction and move parallel to the negative X-axis. This can be done by applying the
magentic field along.
(A*) Y-axis (B*) Z-axis (C) Y-axis only (D) Z-axis only
) ;-i ii--X-i li ^ln ri r| i - l )+iii ^ii irn r, nil ;-i
i lii l(in ~ii--X-i i i ^ln | rli i ni r l +i i l li ^ii i -
(A*) Y-i (B*) Z-i (C) (Y-i (D) (Z-i
Chapter # 34 Magnetic Field [6]
manishkumarphysics.in
Sol. AB
( )
F q V B =
This can be done by applying the Magnetic field along y axis or z axis.
10. Let E
and B
and B
## in another frame S moving
with respect to S at a velocity
u . Two of the following equations are wrong. Identify them.
-ii l E
nii B
## ) li ni S- l(nn nii +i iii i n n r nii S ii
u (^ ^lnii li
ni S - E
nii B
| l - i -ii ^n r| ;i llrn il -
(A) B
y
1
+ B
y
+
uE
c
2
2
(B*) E
y
1
+ E
y
+
uB
c
2
2
(C*) B
y
1
+ B
y
+ uE
2
(D) B
y
1
+ B
y
+ uB
2
Sol. BC
qE =qvB
e vB =
By dimensionally b & c are wrong.
2
vE v B =
2
vE
B
v
=
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| 2.84375
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http://oeis.org/A000305
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A000305 Number of certain rooted planar maps. (Formerly M3543 N1435) 2
1, 4, 18, 89, 466, 2537, 14209, 81316, 473338, 2793454, 16674417, 100487896, 610549829, 3735850007, 23000055178, 142370597601, 885521350882, 5531501612071, 34686798239678, 218273864005214, 1377897874711437 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 REFERENCES N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..200 W. G. Brown, Enumeration of non-separable planar maps, Canad. J. Math., 15 (1963), 526-545. W. G. Brown, Enumeration of non-separable planar maps [Annotated scanned copy] MAPLE with(linalg): T := proc(n, k) if k<=n then k*sum((2*j-k+1)*(j-1)!*(3*n-k-j)!/(j-k+1)!/(j-k)!/(2*k-j-1)!/(n-j)!, j=k..min(n, 2*k-1))/(2*n-k+1)! else 0 fi end:A := matrix(30, 30, T): seq(sum(A[i, j], j=1..i), i=1..30); R := RootOf(x-t*(t-1)^2, t); ogf := series((R+1)/((1-R-R^2)*(R-1)^2), x=0, 20); # Mark van Hoeij, Nov 08 2011 MATHEMATICA t[n_, k_] := If[k <= n, k*Sum[(2*j-k+1)*(j-1)!*(3*n-k-j)!/(j-k+1)!/(j-k)!/ (2*k-j-1)!/(n-j)!, {j, k, Min[n, 2*k-1]}]/(2*n-k+1)!, 0]; a[n_] := Sum[ t[n, k], {k, 1, n}]; Array[a, 21] (* Jean-François Alcover, Feb 07 2016 after Herman Jamke in A046652 *) CROSSREFS Row sums of A046652. Sequence in context: A127394 A046984 A129323 * A200029 A020070 A197650 Adjacent sequences: A000302 A000303 A000304 * A000306 A000307 A000308 KEYWORD nonn AUTHOR EXTENSIONS More terms from Emeric Deutsch, Mar 03 2004 STATUS approved
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Last modified April 22 22:19 EDT 2021. Contains 343197 sequences. (Running on oeis4.)
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# infinite loop for mario.c
I'm almost there in solving the pset1 of cs50 mario.c the problem in my code is when i enter a number more than 1 and less than 23 , it goes in infinite loop and prints spaces all the way without stop. I'm quite sure the problem in the condition of my "spaces for loop" but i don't know where exactly the problem is if anyone can provide me tips to overcome this problem , it will be great ! thanks !
``````#include <cs50.h>
#include <stdio.h>
int main(void)
{
int m ;
int row , hash , space ;
// we will need to prompt the user at least once so , do-while loop
do
{
printf("Height: ");
m = GetInt() ;
}
while ((m < 0) || (m > 23)) ;
//Drawing the pyramid
for (row = 0 ; row < m ; row ++)
{
for (space = 0 ; space < (m-1) ; space--)
{
printf(" ");
}
for (hash = 0 ; hash < ((m+1)-(m-1)) ; hash++)
{
printf("#");
}
printf("\n");
}
}
``````
if you followed the logic of your code, you'd find that if you entered a number that is > 1 for the height, `m - 1` would become at least 1. since `m - 1` will always produce the same value and `spaces` is initialized with a `0` and gets decreased after each iteration of the for loop, the condition `spaces < (m - 1)` will be always `true` and therefore, an infinite loop.
as a side note, you probably wanna give your variables more descriptive names. it's generally a bad practice to give your variables names like `m`. you also probably don't wanna make your code difficult to understand by placing all sorts of magic arithmetic expressions (e.g., `m - 1` and `(m + 1) - (m - 1)`.
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https://www.convert-measurement-units.com/convert+Pico+to+Zepto.php
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Convert Pico to Zepto (SI-prefixes)
## Pico into Zepto
Measurement Categorie:
Original value: Original unit: AttoCentiDecaDeciExaFemtoGigaHectoKiloMegaMicroMilliNanoPetaPicoTeraYoctoYottaZeptoZetta Target unit: AttoCentiDecaDeciExaFemtoGigaHectoKiloMegaMicroMilliNanoPetaPicoTeraYoctoYottaZeptoZetta numbers in scientific notation
https://www.convert-measurement-units.com/convert+Pico+to+Zepto.php
# Convert Pico to Zepto:
1. Choose the right category from the selection list, in this case 'SI-prefixes'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Pico'.
4. Finally choose the unit you want the value to be converted to, in this case 'Zepto'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '777 Pico'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'SI-prefixes'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '83 Pico to Zepto' or '75 Pico into Zepto' or '56 Pico -> Zepto' or '50 Pico = Zepto'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(61 * 2) Pico'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '777 Pico + 2331 Zepto' or '46mm x 42cm x 18dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 5.056 790 077 44×1022. For this form of presentation, the number will be segmented into an exponent, here 22, and the actual number, here 5.056 790 077 44. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 5.056 790 077 44E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 50 567 900 774 400 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
## How many Zepto make 1 Pico?
1 Pico = 1 000 000 000 Zepto - Measurement calculator that can be used to convert Pico to Zepto, among others.
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# Ultimate Power: Lithium-Ion Batteries In Series
At some point, the 3.6 V of a single lithium ion battery just won’t do, and you’ll absolutely want to stack LiIon cells in series. When you need high power, you’ve either got to increase voltage or current, and currents above say 10 A require significantly beefed up components. This is how you’re able to charge your laptop from your USB-C powerbank, for instance.
Or maybe you just need higher voltages, and don’t feel like using a step-up converter, which brings along with it some level of inefficiency. Whatever your reasons, it’s time to put some cells into series. Continue reading “Ultimate Power: Lithium-Ion Batteries In Series”
# Arduino Measures Remaining Battery Power With Zero Components, No I/O Pin
[Trent M. Wyatt]’s CPUVolt library provides a fast way to measure voltage using no external components, and no I/O pin. It only applies to certain microcontrollers, but he provides example Arduino code showing how handy this can be for battery-powered projects.
The classical way to measure a system’s voltage is to connect one of your MCU’s ADC pins to a voltage divider made from a couple resistors. A simple calculation yields a reading of the system’s voltage, but this approach has two disadvantages: one is that it constantly consumes power, and the other is that it ties up a pin that you might want to use for something else.
There are ways to mitigate these issues, but it would be best to avoid them entirely. Microchip application note 2447 describes a method of doing exactly that, and that’s precisely what [Trent]’s Arduino library implements.
What happens in this method is one selects Vbg (a fixed internal voltage reference that is temperature-independent) as Vin, and selects Vcc as the ADC’s voltage reference. This is essentially backwards from how the ADC is normally used, but it requires no external hookup and is only a bit of calculation away from determining Vcc in millivolts. There is some non-linearity in the results, but for the purposes of measuring battery power in a system or deciding when to send a “low battery” signal, it’s an attractive solution.
Being an Arduino library, CPUVolt makes this idea very easy to use, but the concept and method is actually something we have seen before. If you’re interested in the low-level details, then check out our earlier coverage which goes into some detail on exactly what is going on, using an ATtiny84.
# Power Supplies Without Transformers
For one-off projects or prototyping, it’s not too hard to find a wall wart or power supply to send a few joules of energy from the wall outlet to your circuit. Most of these power supplies use a transformer to step down the voltage to a more usable level and also to provide some galvanic isolation to the low voltage circuit. But for circuits where weight, volume, or cost are a major concern, a transformer may be omitted in the circuit design in favor of some sort of transformerless power supply.
While power supplies with this design do have many advantages, some care needs to be taken with regard to safety. The guide outlines four designs of increasing complexity which first puts out a basic transformerless power supply, using a series capacitor to limit current. To bring the voltage to an acceptable level, a recognizable bridge rectifier is paired with a capacitor as well as a zener diode. The second circuit presented adds voltage stabilization using a transistor and 78XX regulator. From there, zero-crossing detection is added to limit inrush surge currents, and the final design uses the venerable 555 timer to build a switching power supply.
Although it is noted several times throughout the guide, we’ll still point out here that transformerless designs like these introduce several safety issues since a mistake or fault can lead to the circuit being exposed to the mains voltage. However, with proper care and design it’s possible to make use of these designs to build more effective power supplies that can be safe to use for powering whatever circuit might energy but might not require the cost or weight of a transformer. For more on the theory of these interesting circuits and a few examples of where they are often found, check out the shocking truth about transformerless power supplies.
Thanks to [Stephen] for the tip!
# Car Security System Monitors Tiny Voltage Fluctuations
As the old saying goes, there’s no such thing as a lock that can’t be picked. However, it seems like there are plenty of examples of car manufacturers that refuse to add these metaphorical locks to their cars at all — especially when it comes to securing the electronic systems of vehicles. Plenty of modern cars are essentially begging to be attacked as a result of such poor practices as unencrypted CAN busses and easily spoofed wireless keyfobs. But even if your car comes from a manufacturer that takes basic security precautions, you still might want to check out this project from the University of Michigan that is attempting to add another layer of security to cars.
The security system works like many others, by waiting for the user to input a code. The main innovation here is that the code is actually a series of voltage fluctuations that are caused by doing things like turning on the headlights or activating the windshield wipers. This is actually the secondary input method, though; there is also a control pad that can mimic these voltage fluctuations as well without having to perform obvious inputs to the vehicle’s electrical system. But, if the control pad isn’t available then turning on switches and lights to input the code is still available for the driver. The control unit for this device is hidden away, and disables things like the starter motor until it sees these voltage fluctuations.
One of the major selling points for a system like this is the fact that it doesn’t require anything more complicated than access to the vehicle’s 12 volt electrical system to function. While there are some flaws with the design, it’s an innovative approach to car security that, when paired with a common-sense approach to securing modern car technology, could add some valuable peace-of-mind to vehicle ownership in areas prone to car theft. It could even alleviate the problem of cars being stolen via their headlights.
# Rapid Charging Supercapacitors
Battery technology is the talk of the town right now, as it’s the main bottleneck holding up progress on many facets of renewable energy. There are other technologies available for energy storage, though, and while they might seem like drop-in replacements for batteries they can have some peculiar behaviors. Supercapacitors, for example, have a completely different set of requirements for charging compared to batteries, and behave in peculiar ways compared to batteries.
This project from [sciencedude1990] shows off some of the quirks of supercapacitors by showing one method of rapidly charging one. One of the most critical differences between batteries and supercapacitors is that supercapacitors’ charge state can be easily related to voltage, and they will discharge effectively all the way to zero volts without damage. This behavior has to be accounted for in the charging circuit. The charging circuit here uses an ATtiny13A and a MP18021 half-bridge gate driver to charge the capacitor, and also is programmed in a way that allows for three steps for charging the capacitor. This helps mitigate the its peculiar behavior compared to a battery, and also allows the 450 farad capacitor to charge from 0.7V to 2.8V in about three minutes.
If you haven’t used a supercapacitor like this in place of a lithium battery, it’s definitely worth trying out in some situations. Capacitors tolerate temperature extremes better than batteries, and provided you have good DC regulation can often provide power more reliably than batteries in some situations. You can also combine supercapacitors with batteries to get the benefits of both types of energy storage devices.
# The Best Voltage And Current Reference This Side Of A Test Lab
When you measure a voltage, how do you know that your measurement is correct? Because your multimeter says so, of course! But how can you trust your multimeter to give the right reading? Calibration of instruments is something we often trust blindly without really thinking about, but it’s not always an impossible task only for a high-end test lab. [Petteri Aimonen] had enough need for a calibrated current source to have designed his own, and he’s shared the resulting project for all to see.
The cost of a reference source goes up with the degree of accuracy required, and can stretch into the many millions of dollars if you are seeking the standards of a national metrology institute, but fortunately [Petteri]’s requirements were considerably more modest. 0.02% accuracy would suffice. An Analog Devices precision voltage reference driving a low-offset op-amp with a driver transistor supplies current to a 0.01% precision resistor, resulting in a reference current source fit for his needs. The reference is available in a range of voltages, his chosen 2.048 volts gave a 2.048 mA current sink with a 100 ohm resistor.
In a way it is a miracle of technology that the cheapest digital multimeter on the market can still have a surprisingly good level of calibration thanks to its on-chip bandgap voltage reference, but it never hurts to have a means to check your instruments. Some of us still rather like analogue multimeters, but beware — calibration at the cheaper end of that market can sometimes be lacking.
# A Division In Voltage Standards
During my recent trip to Europe, I found out that converters were not as commonly sold as adapters, and for a good reason. The majority of the world receives 220-240 V single phase voltage at 50-60 Hz with the surprisingly small number of exceptions being Canada, Colombia, Japan, Taiwan, the United States, Venezuela, and several other nations in the Caribbean and Central America.
While the majority of countries have one defined plug type, several countries in Latin America, Africa, and Asia use a collection of incompatible plugs for different wall outlets, which requires a number of adapters depending on the region traveled.
Although there is a fair degree of standardization among most countries with regards to the voltage used for domestic appliances, what has caused the rift between the 220-240 V standard and the 100-127 V standards used in the remaining nations?
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Let f and g be functions satisfying f(x) = ex g(x), f (x + y) = f (x) + f(y),g(0) = 0, g’(0) = 4 g and g’ are continuous at 0. Then : Kaysons Education
# Let f and g be Functions Satisfying f(x) = ex g(x), f (x + y) = f (x) + f(y),g(0) = 0, g’(0) = 4 g and g’ are Continuous At 0. Then
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## Question
easy
### Solution
Correct option is
So f(x) = 4x.
#### SIMILAR QUESTIONS
Q1
Let f be a twice differentiable function such that f’’(x) = –f(x) and f’(x) = g(x). If h’(x) = [f(x)]2 + [g(x)]2h(1) = 8 and
h(0) = 2, then h(2) is equal to
Q2
If y2 = P(x) is a polynomial of degree 3, then
is equal to
Q3
If then the set of all points where the derivative exist is
Q4
The value of y’’ (1) if x3 – 2x2y2 + 5x + y – 5 = 0 when y(1) = 1, is equal to
Q5
If f(x, then f’(1) equals
Q6
If f (x) = (1 + x)n, then the value of
Q7
The solution set of f’(x) > g’(x) where f(x) = (1/2)52x + 1 and g(x) = 5x + 4x log 5 is
Q8
Let f(x) = sin xg(x) = x2 and h(x) = log x.
Q9
up to nterms, then y’(0) is equal to
Q10
Let f : R R is a function which is defined by (x) = max {xx3}. The set of all points on which (x) is not differentiable is
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The Rookie’s Guide to Options: Intrdoduction
This is an excerpt from the Introduction to The Rookie’s Guide to Options.
You are about to enter the exciting world of stock options. These versatile investment tools possess properties not found elsewhere in the investment universe: limited lifetimes with explicit expiration dates. Options were invented as hedging, or risk-reducing tools, allowing specific risks (described by the Greeks) associated with owning any position to be identified. Thus, each risk factor can be controlled to suit your needs.
Options allow investors to use leverage to take control of far more valuable stock positions with less cash at risk.
Options are versatile and can be used in a variety of strategies, ranging from ultra conservative to outright gambles. I encourage readers to adopt strategies between the extremes.
It is possible to use advanced mathematics when discussing options, but in keeping with the goal of making this an enjoyable learning experience, this book uses nothing more complicated than elementary algebra. Let’s leave the advanced math to the academics.
Equity options are related to stocks. The term used to describe that relationship is derivative. The value of an option is derived from the value of an individual stock or group of stocks (an index). If this sounds complicated, it is not. Computers and calculators do the math for us, and our job is to understand how to use the numbers — just as we learn to use any tool.
This book delivers the background information needed to understand why options do what they do. Note that key word: ‘understand.’ I’m not going to define a term without explaining how it relates to trading options. I’m not going to tell you how to open a trade and then leave you stranded. You will learn to open, manage and exit positions. I do not provide rules to follow. Instead, you get detailed explanations and suggestions that enable you to make your own decisions.
The book contains a great deal of background information (Part I), lessons on three basic strategies (Part II), as well as explanations of how to adopt more advanced strategies (Part III).
These lessons are designed to help you use options effectively. That means trading with less risk, increasing the frequency of winning trades, and earning more money (when compared with trading without options). There is one important point: both the basic concepts and basic strategies are easy to understand. As with any other endeavor, the more sophisticated you become, the more you can do. Consider this book to be your college level course—perhaps even an elective course. However, it is not graduate school. Option trading can get very sophisticated and today’s top experts are quants with a PhD in math or physics. The good news is that you do not have to compete directly with them. Option trading is so widespread that there is ample opportunity for everyone.
If you want to become an expert trader, this book will not get you there. However, it is an excellent starting point. And if your objective is to enhance your income by generating earnings with less chance of suffering large losses, then you have come to the right place. You do not have to compete with the professionals. Most of us can succeed by adopting the most basic strategies —if we have the discipline to manage risk. While I appreciate advanced strategies, I use only the methods discussed in this book when trading my personal money.
This guide takes you from the novice stage through the intermediate trader stage. Although intended for option rookies, there is enough meat in The Rookie’s Guide to Options for the investor who already trades options. Re-reading these pages as you gain experience will provide insights you may have missed the first time.
My objective is for you, the reader, to gain a solid understanding of options and learn to use them to improve your investment results. You will not learn everything there is to know about options, but, you will be prepared to trade profitably. If we each do our jobs well, you will come away with a clear understanding of options—how they work and how you can make money by incorporating option strategies into your investing methods.
Be prepared for discussions on risk, including definitions (how much money can be lost vs. the probability of losing), setting risk limits (position size), using calculators to discover the odds that something specific will go wrong (stock doesn’t move your way), etc. Included are ideas on how to handle risky situations. Is it better to get out of the trade or use an ‘adjustment’ trade that reduces risk to an acceptable level (compared with the potential reward)? These are all part of risk management, and included are my thoughts on why survival should be any trader’s top priority. Earning money is important—in fact it is our reason for trading—but it ranks behind risk management, unless you plan to have a very short trading career.
Rookies
In the sports world, a rookie is someone in his/her first year of professional play. The term also refers to someone who is new to a profession. This book was written for newcomers to the world of options—not necessarily investment rookies, but option rookies. The strategies detailed are not the only ones available, but they were chosen because they can be understood and put into practice by traders who have patience and discipline. I stress discipline throughout the book because without it you have almost no chance of becoming a successful trader. Most investors who enter this realm are familiar with stock investing from the standpoint of owning individual stocks (mutual fund ownership does not count, but ETF trading does). If that is your experience, it should be a smooth transition when you add options to your arsenal of investment tools.
If you are brand new to investing, then you have more to learn. However, the good news is that you can get started without having formed any difficult-to-break bad habits.
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# Relief Valve Tailpipe Design
1. Oct 5, 2009
### 0507476h
Hi,
I'm a final year master's student doing a study on relief valve tailpipe design for my final year project. The system which the relief valve is fitted to is a high pressure breathing air gas cylinder (300 bar) on board a large ship and it vents to atmosphere via piping which can be between 30 and 100 metres long and include bends and changes in section. I am looking for any advice on how to go about this study... From what I have read etc so far I think that when overpressure occurs and the relief valve opens the gas will flow through the tailpipe and increase in velocity until the flow is choked. This choked flow will create large resultant forces against the inner walls of the piping and it is the magnitude of this force that I should be trying to work out in order to determine pipe diameters/mountings requirements etc. Does this sound like I am along the correct tracks? Also I am working under the assumption that the gas involved is compressible and unsteady and therefore I will have to use isothermal gas laws rather than bernoullis? Any suggestions on literature/resources etc would be gratefully appreciated since I am a little lost with this project so far!
Cheers.
2. Oct 5, 2009
### Q_Goest
Hi. Very briefly, ASME code (and virtually all RV manufacturers) prohibit pressure in excess of 10% of set pressure at the outlet of a relief device. So if the relief valve is set at 100 psi, outlet pressure during flow should not exceed 10 psi.
To accomplish this, the flow through the pipe should be calculated using standard methods. Generally, Crane Technical Paper #410 is used in industry as a guide to flow. They use the Darcy-Weisbach equation with Bernoulli's to determine pressure drop through a piping network. So basically, you start with the known flow rate which is determined by calculating the flow through the relief device. Use ASME or API code equations for this.
Next is to determine the flow restriction and pressure drop. For air, the pressure drop is generally sufficient that density changes must be taken into consideration, so you will probably need to break your line up into smaller sections and calculate each section individually, taking into account the change in density and temperature. To determine fluid states, you will also need to apply the first law of thermo and possibly heat transfer with the walls of the pipe. Or you can assume adiabatic or isothermal conditions, but that's often a judgment call and for a student that's doing a master's program, probably isn't advisable. I'd expect a grad student to be able to create a numerical analysis by creating a computer program that can handle changes in state as the air travels along the pipe. Start at the outlet of your RV header, you will need to determine if a shock wave exists. Sometimes, the flow through a vent header is too high to exit at atmospheric pressure. Regardless, you can work backwards from the outlet, determining the pressure drop from each individual flow restriction.
That's a very short description of what's done. Start with Crane TP #410. Also attached is a copy of a general discussion. Also go to your library and pull out ASME Section VIII, Div 1 and go through paragraphs UG-125 through UG-136. I also have a lot of old papers written on the topic which provide a cook-book methodology if you'd like.
File size:
1.2 MB
Views:
1,869
3. Oct 5, 2009
### 0507476h
Q_Goest, thank you very much for your reply. I will start by trying to obtain a copy of Crane's Technical Paper 410 through our library at University and also the ASME document you mentioned (I already have ASME b31.3). Any old papers on the topic would be very much appreciated, thank-you very much.
4. Oct 6, 2009
### Q_Goest
The attached "Calculation of Flow Losses in Inlet and Discharge Headers Associated with Safety Relief Valves" has been used quite a bit throughout industry.
(Note: Pg 7 is blank and has no information on it.)
File size:
1.6 MB
Views:
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5. Oct 6, 2009
### FredGarvin
What a great reference Q. Thanks for posting that.
6. Oct 6, 2009
### 0507476h
Again thanks, much appreciated...
I am now busying myself with pressure transient surges of compressible gas caused by the choked flow, much like a water hammer effect apparently
7. Oct 6, 2009
### CFDFEAGURU
Q_Goest,
Excellent paper on relief valves. I have a waste heat boiler project at work right now and I put that paper to good use. The head of my department (by default) let our subscription to the journals run out years ago and now we aren't allowed to renew them because of the cost.
Anyways,
Thanks
Matt
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Cody
# Problem 2914. Matlab Basics II - Count rows in a matrix
Solution 1981603
Submitted on 18 Oct 2019 by Le Cong Hieu
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(count_rows(x),y_correct))
2 Pass
x = [1 2 2; 15 3 4]; y_correct = 2; assert(isequal(count_rows(x),y_correct))
3 Pass
x = [3.1 7.2;5.4 2.2;7 2.1;5 8.6]; y_correct = 4; assert(isequal(count_rows(x),y_correct))
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https://forums.solar2d.com/t/proper-traveling-positions-of-bullet/354416
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# Proper traveling Positions of Bullet
Hi. I got this code from the documentation, then i tried to make it rectangular.
`````` local physics = require("physics")
physics.setDrawMode("hybrid")
physics.start()
local bullet = display.newRect( 0, display.contentCenterY, 100, 20)
bullet.gravityScale = 0
-- Make the object a "bullet" type object
bullet.isBullet = true
bullet:setLinearVelocity( 800, 0 )
``````
I understand that I may repeatedly produce this through timer.performWithDelay, or “enterframe” listener, or through transition.to instead of Linear Velocity, particle system, ray casting and etc.
As a training and practice for me as a rookie, I will make a shooting object. But I don’t know how to start because of thinking-when the gun aims to shoot vertically, the bullet may travel in horizontal position, and more bad is when the gun randomly shoot slantingly.
How can I make the rectangular objects travels correspondingly to it’s rotation position, or travel accordingly where it’s rotation is pointing?
Please, I really need help from you. And a sample code or project will be indeed an elevation for me, its really a BIG help. thank you so much.
This comes down to trigonometry. You use the sine and cosine functions to get the vertical and horizontal components of the angle. For example, a measurement of one unit at a 30 degree angle corresponds to moving .866 units on the x-axis and .5 units on the y-axis. These are the figures that the cosine and sine functions will give you. If you multiply them by the distance you want to move, it will give you horizontal and vertical distances required to make that move in the given direction. Hope I’m not explaining it too badly. Anyway, maybe these examples will help:
``````-- with an enterFrame listener
local bullet = display.newRect(display.contentCenterX, display.contentCenterY, 20, 10)
bullet.rotation = math.random(0, 360)
local movementPerFrame = 1
local function update()
bullet.x = bullet.x + xMovement
bullet.y = bullet.y + yMovement
end
-- with transition.to
local bullet = display.newRect(display.contentCenterX, display.contentCenterY, 20, 10)
bullet.rotation = math.random(0, 360)
local travelDist = 200
local xDestination, yDestination = bullet.x + xDist, bullet.y + yDist
transition.to(bullet, {x = xDestination, y = yDestination, time = 2000})
``````
You could do it with timer.performWithDelay also, but… I wouldn’t.
1 Like
This is very helpful. Thank you very much, hasty.
Hello!
I hope this code helps you in some way! Maybe you can take something Good from it. It is a gun which shoots in the direction of a tap on the screen! Both the gun and bullet rotate in the direction of the tap/target. And bullet is sent in the direction of the tap.
``````local math = math
local physics = require("physics")
local background, gun
local bullets = {}
local function shootGun(event)
local bullet = display.newRect( 0, display.contentCenterY, 100, 20)
bullet.width, bullet.height = gun.width*0.2, gun.height*0.2
bullet.x, bullet.y = gun.x, gun.y
bullet:setFillColor(1, 0, 0)
gun:toFront()
bullet.gravityScale = 0
-- Make the object a "bullet" type object
bullet.isBullet = true
local direction = {}
direction.x = event.x - gun.x
direction.y = event.y - gun.y
local axis = {}
axis.x = 0
axis.y = direction.y
local directionLength = math.sqrt(direction.x*direction.x + direction.y*direction.y)
local axisLength = math.sqrt(axis.y*axis.y)
local cosA = axisLength/directionLength
local finalAngle = 0
if direction.y <= 0 then
finalAngle = angleInDegrees
else
finalAngle = 90 + (90 - angleInDegrees)
end
if direction.x < 0 then finalAngle = -finalAngle end
gun.rotation, bullet.rotation = finalAngle, finalAngle
bullet:setLinearVelocity( direction.x, direction.y )
end
physics.start()
background = display.newRect( display.contentCenterX, display.contentCenterY, display.actualContentWidth, display.actualContentHeight)
gun = display.newRect( 0, display.contentCenterY, 20, 100)
gun:setFillColor(0, 0, 0)
``````
1 Like
Hi again guys, thank you very much for your help.
I have innovated hasty’s code a very little bit.
`````` local function Berserk ()
local bullet = display.newRect(display.contentCenterX, display.contentCenterY, 40, 20)
bullet.rotation = math.random(0, 360)
local travelDist = display.screenOriginY
local xDestination, yDestination = bullet.x + xDist, bullet.y + yDist
transition.to(bullet, {x = xDestination, y = yDestination, time = 200, onComplete = function() display.remove(bullet) end})
end
timer.performWithDelay(250, Berserk, 0)
``````
I have notice a problem that when the app crashes, the timer.performWithDelay() will still execute it’s function, which result in many many more bullets.
this can be intentionally done when you hold down the minimize button or the close button.
I’m now thinking that maybe firing repeated bullets is much better using the enterFrame listener.
can I have an example of shooting bullets in enterFrame listener? please. Or how can the timer.performWithDelay() also solve this problem if possible? I really need help from you, thank you very much.
I don’t know why it behaves this way when you hold the minimize or close buttons. I did a windows build and saw what you were talking about. I have no idea what the cause of this is. I would suggest starting a new thread about it if no-one chimes in here in the next few days.
By the way, travelDist is the distance for the bullet to travel. If you set it to display.screenOriginY you will get varying results, depending on device. On some devices it’s 0, so the bullets won’t move at all!
2 Likes
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jon has to pay 7.00 admission for the skating rink and 1.50 per hour to rent rollerblades.how many hours can he skate for \$19?
level of service of any roadway is a factor of volume/capacity
I need help with this, I have to turn this in tomorrow help would be very appreciated. This question is designed to be a garage sale type of question.
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https://discourse.mc-stan.org/t/gam-contrainted-to-pass-through-fixed-value-add-argument-parapen/34592
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# Gam contrainted to pass through fixed value - add argument paraPen?
Hi, I am trying to use the strategy outlined here [R] Use pcls in "mgcv" package to achieve constrained cubic spline to fit a gam constrained to pass through a fixed value. I can get this to work using mgcv, but am unable to get this to work using brms because the paraPen argument has not been implemented in brms. Does anyone know a way around this, if there is a different strategy that can be used to achieve the same thing or could this be added to brms?
Example constraining spline to pass through a particular point (0,.6)…
## Fake some data…
library(mgcv)
library(brms)
set.seed(0)
n ← 100
x ← runif(n)4-1;x ← sort(x);
f ← exp(4
x)/(1+exp(4*x));y ← f+rnorm(100)*0.1;plot(x,y)
dat ← data.frame(x=x,y=y)
## Create a spline basis and penalty, making sure there is a knot at the constraint point, (0 here, but could be anywhere)
knots ← data.frame(x=seq(-1,3,length=9)) ## create knots
## set up smoother…
sm ← smoothCon(s(x, k=9, bs=“cr”), dat, knots=knots)[[1]]
## set it to 0 by dropping…
X ← sm\$X[,-3] ## spline basis
S ← sm\$S[[1]][-3,-3] ## spline penalty
off ← y*0 + .6 ## offset term to force curve through (0, .6)
## fit spline constrained through (0, .6)
b ← gam(y ~ X - 1 + offset(off), paraPen=list(X=list(S)))
lines(x,predict(b))
dat\$X ← sm\$X[,-3]
dat\$off ← off
b2 ← brm(y ~ X - 1 + offset(off), paraPen=list(X=list(S)), data = dat)
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CC-MAIN-2024-18
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https://en.wikipedia.org/wiki/Madryga
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In cryptography, Madryga is a block cipher published in 1984 by W. E. Madryga. It was designed to be easy and efficient for implementation in software.[1] Serious weaknesses have since been found in the algorithm, but it was one of the first encryption algorithms to make use of data-dependent rotations,[citation needed] later used in other ciphers, such as RC5 and RC6.
In his proposal, Madryga set forth twelve design objectives that are generally considered to be good goals in the design of a block cipher. DES had already fulfilled nine of them. The three that DES did not fulfill were:
1. Any possible key should produce a strong cipher. (Meaning no weak keys, which DES has.)
2. The length of the key and the text should be adjustable to meet varying security requirements.
3. The algorithm should be efficiently implementable in software on large mainframes, minicomputers, and microcomputers, and in discrete logic. (DES has a large amount of bitwise permutations, which are inefficient in software implementations.)
## The algorithm
Madryga met the objective of being efficient in software: the only operations it uses are XOR and rotations, both operating only on whole bytes. Madryga has a variable-length key, with no upper limit on its length.
Madryga is specified with eight rounds,[1] but this can be increased to provide more security if need be. In each round, the algorithm passes over the entire plaintext n times, where n is the length of the plaintext in bytes. The algorithm looks at three bytes at a time, so Madryga is a 24-bit block cipher. It XORs a key byte with the rightmost byte, and rotates the other two as one block. The rotation varies with the output of the XOR. Then, the algorithm moves to the right by one byte. So if it were working on bytes 2, 3 and 4, after it finished rotating and XORing them, it would repeat the process on bytes 3, 4 and 5.
The key schedule is very simple. To start with, the entire key is XORed with a random constant of the same length as the key, then rotated to the left by 3 bits. It is rotated again after each iteration of rotation and XOR. The rightmost byte of it is used in each iteration to XOR with the rightmost byte of the data block.
The decryption algorithm is simply the reverse of the encryption algorithm. Due to the nature of the XOR operation, it is reversible.
## Cryptanalysis
At a glance, Madryga seems less secure than, for example, DES. All of Madryga's operations are linear. DES's S-boxes are its only non-linear component, and flaws in them are what both differential cryptanalysis and linear cryptanalysis seek to exploit. While Madryga's rotations are data-dependent to a small degree, they are still linear.
Perhaps Madryga's fatal flaw is that it does not exhibit the avalanche effect. Its small data block is to blame for this. One byte can only influence the two bytes to its left and the one byte to its right.
Eli Biham has reviewed the algorithm without making a formal analysis. He noticed that "the parity of all the bits of the plaintext and the ciphertext is a constant, depending only on the key. So, if you have one plaintext and its corresponding ciphertext, you can predict the parity of the ciphertext for any plaintext." Here, parity refers to the XOR sum of all the bits.
In 1995, Ken Shirriff found a differential attack on Madryga that requires 5,000 chosen plaintexts.[2] Biryukov and Kushilevitz (1998) published an improved differential attack requiring only 16 chosen-plaintext pairs, and then demonstrated that it could be converted to a ciphertext-only attack using 212 ciphertexts, under reasonable assumptions about the redundancy of the plaintext (for example, ASCII-encoded English language). A ciphertext-only attack is devastating for a modern block cipher; as such, it is probably more prudent to use another algorithm for encrypting sensitive data.[1]
## References
1. ^ a b c Alex Biryukov; Eyal Kushilevitz (1998). From Differential Cryptanalysis to Ciphertext-Only Attacks. CRYPTO. pp. 72–88. CiteSeerX 10.1.1.128.3697.
2. ^ Ken Shirriff (October 1995). "Differential Cryptanalysis of Madryga". `{{cite journal}}`: Cite journal requires `|journal=` (help) Unpublished manuscript.
• W. E. Madryga, "A High Performance Encryption Algorithm", Computer Security: A Global Challenge, Elsevier Science Publishers, 1984, pp. 557–570.
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